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Research Article Hermitian Operators and Isometries on Banach Algebras of Continuous Maps with Values in Unital Commutative -Algebras Osamu Hatori Department of Mathematics, Faculty of Science, Niigata University, Niigata 950-2181, Japan Correspondence should be addressed to Osamu Hatori; [email protected] Received 21 April 2018; Accepted 26 July 2018; Published 2 September 2018 Academic Editor: Miguel Martin Copyright © 2018 Osamu Hatori. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We study isometries on algebras of the Lipschitz maps and the continuously differentiable maps with the values in a commutative unital -algebra. A precise proof of a theorem of Jarosz concerning isometries on spaces of continuous functions is exhibited. 1. Introduction In this paper an isometry means a complex-linear isometry. de Leeuw [1] probably initiated the study of isometries on the algebra of Lipschitz functions on the real line. Roy [2] studied isometries on the Banach space Lip() of Lipschitz functions on a compact metric space , equipped with the max norm ‖‖ = max{‖‖ , ()}, where () denotes the Lipschitz constant of . Cambern [3] has considered isometries on spaces of scalar-valued continuously differentiable functions 1 ([0, 1]) with norm given by ‖‖ = max ∈[0,1] {|()| + | ()|} for 1 ([0, 1]) and determined a representation for the surjective isometries supported by such spaces. Rao and Roy [4] proved that surjective isometries on Lip([0, 1]) and 1 ([0, 1]) with respect to the norm ‖‖ = ‖‖ + ‖ are of canonical forms in the sense that they are weighted composition operators. ey asked whether a surjective isometry on Lip() with respect to the sum norm ‖‖ = ‖‖ + () for Lip() is induced by an isometry on (note that ‖‖ ∞([0,1]) + () = ‖‖ ∞([0,1]) + ‖ ∞([0,1]) for every Lip([0, 1]). e reason is as follows. Let Lip([0, 1]). en is absolutely continuous. Hence the derivative exists almost everywhere on [0, 1], and it is integrable by the theory of the absolutely continuous functions. Furthermore the equality () − () = ∫ () , 0≤≤≤1 (1) holds. As () < ∞ we see that is essentially bounded. In fact, ( 0 ) = lim 0 ( 0 ) − () 0 ≤ () (2) assures that ∞([0,1]) ≤ (). By (1) we have () − () = 1 () 1 ∞([0,1]) = ∞([0,1]) . (3) It follows that () ∞([0,1]) . We conclude that () = ‖ ∞([0,1]) . us ‖‖ ∞([0,1]) + () = ‖‖ ∞([0,1]) + ∞([0,1]) .) Jarosz [5] and Jarosz and Pathak [6] studied a problem when an isometry on a space of continuous functions is a weighted composition operator. ey provided a unified approach for certain function spaces including 1 (), Lip(), lip (), and [0, 1]. In particular, Jarosz [5, eorem] proved that a unital isometry between unital semisimple commutative Banach algebras with natural norms is canonical. By a theorem of Jarosz [5] a surjective unital isometry on Lip() is an algebra isomorphism when the Hindawi Journal of Function Spaces Volume 2018, Article ID 8085304, 14 pages https://doi.org/10.1155/2018/8085304
Transcript
Page 1: Hermitian Operators and Isometries on Banach Algebras of ...downloads.hindawi.com/journals/jfs/2018/8085304.pdfJournalofFunctionSpaces normiseitherthemaxnormorthesumnorm.esituation

Research ArticleHermitian Operators and Isometries onBanach Algebras of Continuous Maps with Values inUnital Commutative 119862lowast-Algebras

Osamu Hatori

Department of Mathematics Faculty of Science Niigata University Niigata 950-2181 Japan

Correspondence should be addressed to Osamu Hatori hatorimathscniigata-uacjp

Received 21 April 2018 Accepted 26 July 2018 Published 2 September 2018

Academic Editor Miguel Martin

Copyright copy 2018 OsamuHatoriThis is an open access article distributed under theCreative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We study isometries on algebras of the Lipschitz maps and the continuously differentiable maps with the values in a commutativeunital 119862lowast-algebra A precise proof of a theorem of Jarosz concerning isometries on spaces of continuous functions is exhibited

1 Introduction

In this paper an isometry means a complex-linear isometryde Leeuw [1] probably initiated the study of isometries on thealgebra of Lipschitz functions on the real line Roy [2] studiedisometries on the Banach space Lip(119870) of Lipschitz functionson a compact metric space 119870 equipped with the max norm119891119872 = max119891infin 119871(119891) where 119871(119891) denotes the Lipschitzconstant of 119891 Cambern [3] has considered isometries onspaces of scalar-valued continuously differentiable functions1198621([0 1]) with norm given by 119891 = max119909isin[01]|119891(119909)| +|1198911015840(119909)| for 119891 isin 1198621([0 1]) and determined a representationfor the surjective isometries supported by such spaces Raoand Roy [4] proved that surjective isometries on Lip([0 1])and 1198621([0 1]) with respect to the norm 119891 = 119891infin + 1198911015840infinare of canonical forms in the sense that they are weightedcomposition operators They asked whether a surjectiveisometry on Lip(119870) with respect to the sum norm 119891119871 =119891infin + 119871(119891) for 119891 isin Lip(119870) is induced by an isometry on119870 (note that 119891infin([01]) + 119871(119891) = 119891infin([01]) + 1198911015840infin([01])

for every 119891 isin Lip([0 1]) The reason is as follows Let119891 isin Lip([0 1]) Then 119891 is absolutely continuous Hencethe derivative 1198911015840 exists almost everywhere on [0 1] andit is integrable by the theory of the absolutely continuousfunctions Furthermore the equality

119891 (119887) minus 119891 (119886) = int119887

1198861198911015840 (119905) 119889119905 0 le 119886 le 119887 le 1 (1)

holds As 119871(119891) lt infin we see that 1198911015840 is essentially bounded Infact

100381610038161003816100381610038161198911015840 (1199050)10038161003816100381610038161003816 = lim119905997888rarr1199050

100381610038161003816100381610038161003816100381610038161003816119891 (1199050) minus 119891 (119905)1199050 minus 119905

100381610038161003816100381610038161003816100381610038161003816 le 119871 (119891) (2)

assures that 1198911015840infin([01]) le 119871(119891) By (1) we have10038161003816100381610038161003816100381610038161003816119891 (119887) minus 119891 (119886)119887 minus 119886

10038161003816100381610038161003816100381610038161003816 =1003816100381610038161003816100381610038161003816100381610038161119887 minus 119886 int

119887

1198861198911015840 (119905) 119889119905100381610038161003816100381610038161003816100381610038161003816

le 1119887 minus 119886 int119887

119886

10038171003817100381710038171003817119891101584010038171003817100381710038171003817infin([01])119889119905

= 10038171003817100381710038171003817119891101584010038171003817100381710038171003817infin([01])

(3)

It follows that 119871(119891) le 1198911015840infin([01]) We conclude that119871(119891) = 1198911015840infin([01]) Thus 119891infin([01]) + 119871(119891) = 119891infin([01]) +1198911015840infin([01])) Jarosz [5] and Jarosz and Pathak [6] studieda problem when an isometry on a space of continuousfunctions is a weighted composition operator They provideda unified approach for certain function spaces including1198621(119870) Lip(119870) lip120572(119870) and 119860119862[0 1] In particular Jarosz[5 Theorem] proved that a unital isometry between unitalsemisimple commutative Banach algebras with natural normsis canonical By a theorem of Jarosz [5] a surjective unitalisometry on Lip(119870) is an algebra isomorphism when the

HindawiJournal of Function SpacesVolume 2018 Article ID 8085304 14 pageshttpsdoiorg10115520188085304

2 Journal of Function Spaces

norm is either the max norm or the sum normThe situationis very different without assuming the unitality for theisometry with respect to the max norm There is a simpleexample of a surjective isometry which is not canonical [7p242] On the other hand Jarosz and Pathak exhibited in[6 Example 8] that a surjective isometry on Lip(119870) withrespect to the sum norm is canonical After the publicationof [6] some authors expressed their suspicion about theargument there and the validity of the statement there hadnot been confirmed until quite recently Hence the problemon isometries with respect to the sum norm has not been wellstudied

Jimenez-Vargas and Villegas-Vallecillos in [8] have con-sidered isometries of spaces of Lipschitz maps on a compactmetric space taking values in a strictly convex Banach spaceequippedwith the norm 119891 = max119891infin 119871(119891) see also [9]Botelho and Jamison [10] studied isometries on 1198621([0 1] 119864)with max119909isin[01]119891(119909)119864 + 1198911015840(119909)119864 See also [11ndash27] Referalso to a book of Weaver [28]

We propose a unified approach to the study of isome-tries with respect to the sum norm on Banach algebrasLip(119870119862(119884)) lip120572(119870119862(119884)) and 1198621(119870119862(119884)) where 119870 isa compact metric space [0 1] or T (T denotes the unitcircle on the complex plane) and 119884 is a compact Hausdorffspace We study isometries without assuming that theypreserve unit As corollaries of a general result we describeisometries on Lip(119870119862(119884)) lip120572(119870119862(119884)) 1198621([0 1] 119862(119884))and 1198621(T 119862(119884)) respectively

The main result in this paper is Theorem 14 which givesthe form of a surjective isometry 119880 with respect to the sumnorm between certain Banach algebras with the values in acommutative unital 119862lowast-algebra The proof of the necessityof the isometry in Theorem 14 comprises several steps Thecrucial part of the proof ofTheorem 14 is to prove that119880(1) =1 otimes ℎ for an ℎ isin 119862(1198842) with |ℎ| = 1 on 1198842 (Proposition 15)To prove Proposition 15 we apply Choquetrsquos theory (cf [29])with measure theoretic arguments A proof of Proposition 15is completely the same as that of [30 Proposition 9] Pleaserefer to it By Proposition 15 we have that 1198800 = (1 otimes ℎ)119880is a surjective isometry fixing the unit Then by applying atheorem of Jarosz [5] (Theorem 1 in this paper) we see that1198800 is also an isometry with respect to the supremum normBy the Banach-Stone theorem 1198800 is an algebra isomorphismThen by applying Lumerrsquos method (cf [30]) we see that 1198800 isa composition operator of type BJ (cf [31])

Our proofs in this paper make substantial use of thetheorem of Jarosz [5 Theorem] The author believes that itis convenient for the readers to show a precise proof becausethere need to be some ambitious changes in the original proofby Jarosz

2 Preliminaries

Let 119884 be a compact Hausdorff space Let 119864 be a real orcomplex Banach space The space of all 119864-valued continuousmaps on 119884 is denoted by 119862(119884 119864) When 119864 = C (resp R)119862(119884 119864) is abbreviated by119862(119884) (resp119862R(119884)) For a subset 119878 of119884 the supremum norm of 119865 on 119878 is 119891infin(119878) = sup119909isin119878119865(119909)119864

for 119865 isin 119862(119884 119864) When no confusion will result we omit thesubscript 119878 and write only sdot infin Let 119870 be a compact metricspace and 0 lt 120572 le 1 For 119865 isin 119862(119870 119864) put

119871120572 (119865) = sup119909 =119910

1003817100381710038171003817119865 (119909) minus 119865 (119910)1003817100381710038171003817119864119889 (119909 119910)120572 (4)

Then 119871120572 is called an 120572-Lipschitz number of 119865 or just aLipschitz number of 119865 When 120572 = 1 we omit the subscript120572 and write only 119871(119865) The space of all 119865 isin 119862(119870 119864) suchthat 119871120572(119865) lt infin is denoted by Lip120572(119870 119864) When 120572 = 1 thesubscript is omitted and it is written as Lip(119870 119864)

When 0 lt 120572 lt 1 the closed subspace

lip120572 (119870 119864) = 119865

isin Lip120572 (119870 119864) lim119909997888rarr1199090

1003817100381710038171003817119891 (1199090) minus 119891 (119909)1003817100381710038171003817119864119889 (1199090 119909)120572= 0 for every 1199090 isin 119870

(5)

of Lip120572(119870 119864) is called a little Lipschitz space There are avariety of complete norms on Lip120572(119870 119864) and lip120572(119870 119864) Inthis paper we are mainly concerned with the norm sdot 119871 ofLip120572(119870 119864) (resp lip120572(119870 119864)) which is defined by

10038171003817100381710038171198911003817100381710038171003817119871 = 10038171003817100381710038171198911003817100381710038171003817infin(119870) + 119871120572 (119865) 119865 isin Lip120572 (119870 119864) (resp lip120572 (119870 119864)) (6)

The norm sdot 119872 of Lip120572(119870 119864) (resp lip120572(119870 119864)) is defined by

10038171003817100381710038171198911003817100381710038171003817119872 = max 10038171003817100381710038171198911003817100381710038171003817infin(119870) 119871120572 (119865) 119865 isin Lip120572 (119870 119864) (resp lip120572 (119870 119864)) (7)

Note that Lip(119870 119864) (resp lip120572(119870 119864)) is a Banach space withrespect to sdot 119871 and sdot 119872 respectively If 119864 is a Banachalgebra the norm sdot 119871 is multiplicative Hence Lip(119870 119864)(resp lip120572(119870 119864)) is a (unital) Banach algebra with respect tothe norm sdot 119871 if 119864 is a (unital) Banach algebra The normsdot119872 fails to be submultiplicative even if119864 is a Banach algebraFor a metric 119889(sdot sdot) on 119870 the Holder metric is defined by 119889120572for 0 lt 120572 lt 1 Lip120572((119870 119889) 119864) is isometrically isomorphic toLip((119870 119889120572) 119864)

We are mainly concerned with 119864 = 119862(119884) in this paperThen Lip120572(119870119862(119884)) and lip120572(119870119862(119884)) are unital semisimplecommutative Banach algebras with sdot 119871 when 119864 =C119871ip(119870C) (resp lip120572(119870C)) is abbreviated to Lip(119870) (resplip120572(119870))

Let 119865 isin 119862(119870 119862(119884)) for 119870 = [0 1] or T We say that 119865is continuously differentiable if there exists 119866 isin 119862(119870 119862(119884))such that

lim119905997888rarr1199050

100381710038171003817100381710038171003817100381710038171003817119865 (1199050) minus 119865 (119905)1199050 minus 119905 minus 119866 (1199050)

100381710038171003817100381710038171003817100381710038171003817infin(119884)

= 0 (8)

Journal of Function Spaces 3

for every 1199050 isin 119870 We denote 1198651015840 = 119866 Put1198621 (119870 119862 (119884)) = 119865isin 119862 (119870119862 (119884)) 119865 is continuously differentiable (9)

Then1198621(119870119862(119884)) with norm 119865 = 119865infin + 1198651015840infin is a unitalsemisimple commutative Banach algebra If 119884 is singleton wemay suppose that 119862(119884) is isometrically isomorphic to C andwe abbreviate 1198621(119870119862(119884)) by 1198621(119870)

By identifying 119862(119870 119862(119884)) with 119862(119870times119884) we may assumethat Lip(119870119862(119884)) (resp lip120572(119870119862(119884))) is a subalgebra of119862(119870 times 119884) by the correspondence

119865 isin Lip (119870 119862 (119884)) larrrarr ((119909 119910) 997891997888rarr (119865 (119909)) (119910)) isin 119862 (119870 times 119884) (10)

Throughout the paper we may suppose that

Lip (119883119862 (119884)) sub 119862 (119883 times 119884) lip120572 (119883119862 (119884)) sub 119862 (119883 times 119884) 1198621 (119870 119862 (119884)) sub 119862 (119870 times 119884)

(11)

We say that a subset 119876 of 119862(119884) is point separating if 119876separates the points of 119884 The unit of commutative Banachalgebra 119861 is denoted by 1 The maximal ideal space of 119861 isdenoted by 119872119861 Suppose that 119861 is a unital point separatingsubalgebra of 119862(119884) equipped with a Banach algebra normThen119861 is semisimple because 119891 isin 119861 119891(119909) = 0 is amaximalideal of 119861 for every 119909 isin 119883 and the Jacobson radical of 119861vanishes

3 A Theorem of Jarosz RevisitedIsometries Preserving Unit

Whether an isometry between unital semisimple commu-tative Banach algebras is of the canonical form dependsnot only on the algebraic structures of these algebras butalso on the norms in these algebra in most cases A simpleexample is a surjective isometry on theWiener algebra whichneed not be canonical Jarosz [5] defined natural norms andprovided a theorem that isometries between a variety ofalgebras equipped with natural norms are of canonical formsFor the sake of completeness we outline the notations and theterminologies which are due to [5]The set of all norms onR2

with 119901(1 0) = 1 is denoted byP For 119901 isin P we put

119863(119901) = lim119905997888rarr+0

119901 (1 119905) minus 1119905 (12)

Recently Tanabe pointed out by a private communication that119863(119901) exists and it is finite for every 119901 isin P (In fact it iseasy to see that (119901(1 119905) minus 1)119905 is increasing since 119901(1 119905) isconvex We also see that inf 119905gt0((119901(1 119905) minus 1)119905) gt minusinfin) Let119883be a compact Hausforff space and 119860 a liner subspace of 119862(119883)which contains constant functions A seminorm lsaquo sdot lsaquo on 119860 iscalled one-invariant (in the sense of Jarosz) if lsaquo119891+ 1lsaquo = lsaquo119891lsaquofor all 119891 isin 119860 Let 119901 isin P A norm sdot on 119860 is called a 119901-norm if there is a one-invariant seminorm lsaquo sdot lsaquo on 119860 suchthat sdot = 119901( sdot infin lsaquo sdot lsaquo) A natural norm is a 119901-norm forsome 119901 isin P

Theorem 1 (Jarosz [5]) Let 119883 and 119884 be compact Hausdorffspaces let 119860 and 119861 be complex-linear subspaces of 119862(119883) and119862(119884) respectively and let 119901 119902 isin P Assume 119860 and 119861 containconstant functions and let sdot119860 sdot119861 be a 119901-norm and 119902-normon 119860 and 119861 respectively Assume next that there is a linearisometry 119879 from (119860 sdot 119860) onto (119861 sdot 119861) with 1198791 = 1 Then if119863(119901) = 119863(119902) = 0 or if 119860 and 119861 are regular subspaces of 119862(119883)and 119862(119884) respectively then 119879 is an isometry from (119860 sdot infin)onto (119861 sdot infin)

In the sequel a unital semisimple commutative Banachalgebra 119860 is identified via the Gelfand transforms with a sub-algebra of 119862(119872119860) A unital semisimple commutative Banachalgebra is regular (in the sense of Jarosz [5]) Hence we haveby a theorem of Nagasawa [32] (cf [33]) that the followingholds

Corollary 2 Let 119860 and 119861 be unital semisimple commutativeBanach algebras Assume they have natural norms respectivelySuppose that 119879 119860 997888rarr 119861 is a surjective complex-linearisometry with 1198791 = 1 Then there exists a homeomorphism120593 119872119861 997888rarr 119872119860 such that

119879 (119891) (119909) = 119891 ∘ 120593 (119909) 119891 isin 119860 119909 isin 119872119861 (13)

Proof A unital semisimple commutative Banach algebra isregular by Proposition 2 in [5]ThenTheorem 1 ensures that119879is a surjective linear isometry from (119860 sdotinfin) onto (119861 sdotinfin) Itis easy to see that 119879 is extended to a surjective linear isometry from the uniform closure 119860 of 119860 onto the uniform closure119861 of 119861 Then a theorem of Nagasawa asserts that there existsa homeomorphism 120593 119872119861 997888rarr 119872119860 such that (119891)(119909) =119891 ∘ 120593(119909) (119891 isin 119860 119909 isin 119872119861) As |119860 = 119879 we have theconclusion

Corollary 3 Let 119870119895 be a compact metric space for 119895 = 1 2Suppose that 119879 Lip(1198701) 997888rarr Lip(1198702) is a surjective complex-linear isometry with respect to the norm sdot 119871 Assume 1198791 = 1Then there exists a surjective isometry 120593 1198832 997888rarr 1198831 such that

119879119891 (119909) = 119891 ∘ 120593 (119909) 119891 isin Lip (1198701) 119909 isin 1198702 (14)

Conversely if 119879 Lip(1198701) 997888rarr Lip(1198702) is of the form as (14)then 119879 is a surjective isometry with respect to both of sdot 119872 and sdot 119871 such that 1198791 = 1

Proof As (Lip(119870119895) sdot 119871) is a unital semisimple commutativeBanach algebra with maximal ideal space 119870119895 Corollary 2asserts that there is a homeomorphism 120593 1198702 997888rarr 1198701 suchthat

119879119891 (119909) = 119891 ∘ 120593 (119909) 119891 isin Lip (1198701) 119909 isin 1198702 (15)

Then by a routine argument we see that 120593 is an isometryConverse statement is trivial

Without assuming 1198791 = 1 we have that 119879 is aweighted composition operator We exhibit a general resultas Theorem 14 (see also [30])(Lip(119870) sdot 119872) need not be a Banach algebra since sdot 119872need not be submultiplicative On the other hand sdot 119872

4 Journal of Function Spaces

is a natural norm in the sense of Jarosz (see [5]) such thatlim119905997888rarr+0((max1 119905 minus 1)119905) = 0 Then by Theorem 1 we havethe following

Corollary 4 Let 119870119895 be a compact metric space for 119895 = 1 2Suppose that 119879 Lip(1198701) 997888rarr Lip(1198702) is a surjective complex-linear isometry with respect to the norm sdot119872 Assume 1198791 = 1Then there exists a surjective isometry 120593 1198832 997888rarr 1198831 such that

119879119891 = 119891 ∘ 120593 119891 isin Lip (1198701) (16)

Conversely if 119879 Lip(1198701) 997888rarr Lip(1198702) is of a similar form as(16) then 119879 is a surjective isometry with respect to both of sdot119872and sdot 119871 such that 1198791 = 1

Proof As sdot 119872 is a natural norm we have by Corollary 2 thatthere is a homeomorphism 120593 1198702 997888rarr 1198701 such that

119879119891 (119909) = 119891 ∘ 120593 (119909) 119891 isin Lip (1198701) 119909 isin 1198702 (17)

Then by a routine argument we see that 120593 is an isometryConverse statement is trivial

Without the assumption that 1198791 = 1 in Corollary 4 onemay expect that 119879 is a weighted composition operator But itis not the case A simple counterexample is given by Weaver[7 p242] (see also [28])

As is pointed out in [34] the original proof of Theorem 1needs a revision in some part and a proof when 119860 and 119861are algebras of Lipschitz functions is revised [34 Proposition7] Although a revised proof for a general case is similar tothat of Proposition 7 in [34] we exhibit it here for the sakeof completeness of this paper To prove Theorem 1 we needLemma 2 in [5] in the sameway as the original proof of JaroszThe following is Lemma 2 in [5]

Lemma5 (Jarosz [5]) Assume119860 is a regular subspace of119862(119883)with 1 isin 119860 and let 1199090 isin 119862ℎ(119860) Then for any 120576 gt 0 and anyopen neighborhood 119880 of 1199090 there is an 119891 isin 119860 such that

10038171003817100381710038171198911003817100381710038171003817infin le 1 + 120576119891 (1199090) = 11003816100381610038161003816119891 (119909) + 11003816100381610038161003816 le 120576

119909 isin 119883 119880(18)

and |Im119891(119909)| le 120576 for all 119909 isin 119883Proof The proof is essentially due to the original proof ofLemma 2 in [5] Several minor changes are needed Weitemize them as follows

(i) Five 1205762rsquos between 11 lines and 5 lines from the bottomof page 69 read as 1205763

(ii) Next 119909 isin 1198831198801 reads as 119909 isin 1198801 on the bottom of page69

(iii) We point out that the term sum1198960minus1119895=1 (119891119895(119909) minus 1) which

appears on the first line of the first displayed inequal-ities on page 70 reads 0 if 1198960 = 1

(iv) The term 1 + 120576 on the right hand side of the secondline of the same inequalities reads as 1 + 1205763

(v) Two 1205762rsquos on the same line read as 1205763(vi) On the next line ((119899 + 1)119899)(1205762) reads as 1205763(vii) For any 1 le 1198960 le 119899 we infer that

1 ge 1 minus 21198960 minus 1119899 ge 1 minus 2119899 minus 1119899 gt minus1 (19)

Hence we have |119891(119909)| le 1 + 120576 if 119909 isin 1198801 by the firstdisplayed inequalities of page 70

(viii) The inequality 119891infin le 120576 on the fifth line on page 70reads as 119891infin le 1 + 120576

Let119870 be a nonempty convex subset of the complex planeand 120593 isin [0 2120587) Put

119888 (119870 120593) = sup 119886 isin R there is a 119887isin R with (119886 + 119894119887) 119890119894120593 isin 119870 (20)

Note that we may write

119888 (119870 120593) = sup Re120582 120582 isin 119890minus119894120593119870 (21)

Let 119860 be a subspace of 119862(119883) for a compact Hausdorff spaceFor 119891 isin 119860 we put 120590(119891) = 119891(119883) and 120590(119891) = co(119891(119883)) whereco(sdot) denotes the closed convex-hull We define the functions

119888119860 119860 times [0 2120587) 997888rarr R119888119860 (119891 120593) = 119888 ( (119891) 120593)

119903119860 119860 times R+ times [0 2120587) 997888rarr R

+119903119860 (119891 119905 120593) = 10038171003817100381710038171003817119891 + 119890119894120593119905110038171003817100381710038171003817infin

(22)

Proof of Theorem 1 Let 119891 isin 119860 First we note that10038171003817100381710038171198911003817100381710038171003817infin = sup 119911 isin 120590 (119891) = sup 119911 isin (119891) (23)

since (119891) is the closed convex-hull of a compact set 120590(119891) =119891(119883) We prove the inequalities

119905 + 119888119860 (119891 120593) le 119903119860 (119891 119905 120593)le radic(119905 + 119888119860 (119891 120593))2 + 100381710038171003817100381711989110038171003817100381710038172infin

119905 ge 0 120593 isin [0 2120587) (24)

which appear on p 68 in [5] Put 119904 = 119888119860(119891 120593) As 120590(119891) iscompact there exists 119887 isin R such that (119904 + 119894119887)119890119894120593 isin 120590(119891)Hence

119890119894120593 (119905 + 119904 + 119894119887) isin 119890119894120593119905 + 120590 (119891) = 120590 (1198901198941205931199051 + 119891) (25)

Journal of Function Spaces 5

As

119905 + 119904 le |119905 + 119904| le |119905 + 119904 + 119894119887| = 10038161003816100381610038161003816119890119894120593 (119905 + 119904 + 119894119887)10038161003816100381610038161003816le 100381710038171003817100381710038171198901198941205931199051 + 11989110038171003817100381710038171003817infin = 119903119860 (119891 119905 120593) (26)

we have

119905 + 119888119860 (119891 120593) le 119903119860 (119891 119905 120593) (27)

Let 119909 isin 119883 By the definition of 119904 = 119888119860(119891 120593) we infer thatRe(119890minus119894120593119891(119909)) le 119904 hence we have Re(119905 + 119890minus119894120593119891(119909)) le 119905 + 119904 forevery 119905 ge 0 Then

Re 119890minus119894120593119891 (119909) le Re (119905 + 119890minus119894120593119891 (119909)) le 119905 + 119904 119905 ge 0 (28)

Letting119872 = max|119905 + 119904| |Re(119890minus119894120593119891(119909))| we have10038161003816100381610038161003816119890119894120593119905 + 119891 (119909)100381610038161003816100381610038162 = 10038161003816100381610038161003816119905 + 119890minus119894120593119891 (119909)100381610038161003816100381610038162

= (Re (119905 + 119890minus119894120593119891 (119909)))2+ (Im (119905 + 119890minus119894120593119891 (119909)))2

le 1198722 + (Im (119905 + 119890minus119894120593119891 (119909)))2le |119905 + 119904|2 + (Re (119890minus119894120593119891 (119909)))2+ (Im (119890minus119894120593119891 (119909)))2

= |119905 + 119904|2 + 10038161003816100381610038161003816119890minus119894120593119891 (119909)100381610038161003816100381610038162

(29)

As 119909 isin 119883 is arbitrary we have

119903119860 (119891 119905 120593) = 10038171003817100381710038171003817119905 + 119890minus11989412059311989110038171003817100381710038171003817infin le radic|119905 + 119904|2 + 100381710038171003817100381711989110038171003817100381710038172infin= radic(119905 + 119888119860 (119891 120593))2 + 100381710038171003817100381711989110038171003817100381710038172infin

(30)

It follows that (24) holds In the same way we have

119905 + 119888119861 (119892 120593) le 119903119861 (119892 119905 120593) le radic(119905 + 119888119861 (119892 120593))2 + 100381710038171003817100381711989210038171003817100381710038172infin119905 ge 0 120593 isin [0 2120587) (31)

for every 119892 isin 119861 By (24) and (31) we infer that

lim119905997888rarrinfin

(119903119860 (119891 119905 120593) minus 119905) = 119888119860 (119891 120593) lim119905997888rarrinfin

(119903119861 (119879119891 119905 120593) minus 119905) = 119888119861 (119879119891 120593) (32)

As lsaquo sdot lsaquo119860 is 1-invariant we have119901 (119903119860 (119891 119905 120593) lsaquo119891lsaquo119860)= 119901 (10038171003817100381710038171003817119891 + 119890119894120593119905110038171003817100381710038171003817infin lsaquo119891 + 1198901198941205931199051lsaquo119860) (33)

As 119879 is an isometry 119879(1) = 1 and lsaquo sdot lsaquo119861 is 1-invariant wehave

119901 (10038171003817100381710038171003817119891 + 119890119894120593119905110038171003817100381710038171003817infin lsaquo119891 + 1198901198941205931199051lsaquo119860)= 119902 (10038171003817100381710038171003817119879 (119891 + 1198901198941205931199051)10038171003817100381710038171003817infin lsaquo119879 (119891 + 1198901198941205931199051)lsaquo119861)= 119902 (10038171003817100381710038171003817119879 (119891) + 119890119894120593119905110038171003817100381710038171003817infin lsaquo119879 (119891) + 1198901198941205931199051lsaquo119861)= 119902 (119903119861 (119879119891 119905 120593) lsaquo119879119891lsaquo119861)

(34)

Thus

119901 (119903119860 (119891 119905 120593) lsaquo119891lsaquo119860) = 119902 (119903119861 (119879119891 119905 120593) lsaquo119879119891lsaquo119861) (35)

It follows that0 = lim

119905997888rarrinfin(119901 (119903119860 (119891 119905 120593) lsaquo119891lsaquo119860)

minus 119902 (119903119861 (119879119891 119905 120593) lsaquo119879119891lsaquo119861))= lim

119905997888rarrinfin(119903119860 (119891 119905 120593) 119901(1 lsaquo119891lsaquo119860119903119860 (119891 119905 120593))

minus 119903119860 (119891 119905 120593)) + lim119905997888rarrinfin

(119903119861 (119879119891 119905 120593)minus 119903119861 (119879119891 119905 120593) 119902(1 lsaquo119879119891lsaquo119861119903119861 (119879119891 119905 120593)))+ lim

119905997888rarrinfin(119903119860 (119891 119905 120593) minus 119905 minus (119903119861 (119879119891 119905 120593) minus 119905))

= lsaquo119891lsaquo119860119863(119901) minus lsaquo119879119891lsaquo119861119863(119902) + 119888119860 (119891 120593)minus 119888119861 (119879119891 120593)

(36)

Recall that 119863(119901) = lim119905997888rarr+0(119901(1 119905) minus 1)119905 and 119863(119902) =lim119905997888rarr+0(119902(1 119905) minus 1)119905 It follows thatlsaquo119891lsaquo119860119863 (119901) minus lsaquo119879119891lsaquo119861119863(119902) = 119888119861 (119879119891 120593) minus 119888119860 (119891 120593)

119891 isin 119860 120593 isin [0 2120587) (37)

Suppose that 119863(119901) = 119863(119902) = 0 Then we have by (37)that 119888119861(119879119891 120593) = 119888119860(119891 120593) for every 119891 isin 119860 and 120593 isin [0 2120587)By Lemma 1 in [5] we infer that 120590(119879119891) = 120590(119891) Thus we have119879119891infin = 119891infin We have proved that 119879 is an isometry from(119860 sdot infin) onto (119861 sdot infin) if119863(119901) = 119863(119902) = 0

Suppose that 119860 and 119861 are regular subspaces of 119862(119883) and119862(119884) respectively Let 119891 isin 119860 PutΔ119891 = lsaquo119891lsaquo119860119863(119901) minus lsaquo119879119891lsaquo119861119863(119902) (38)

Suppose that Δ119891 ge 0 For any 119903 ge 0 and any nonemptycompact convex subset 119870 sub C we have that

119888 (119870 + 119870 (119903) 120593) = 119888 (119870 120593) + 119903 (39)

for all 120593 isin [0 2120587) where 119870(119903) = 119911 isin C |119911| le 119903 Then by(37) we have

119888 (120590 (119879119891) 120593) = 119888119861 (119879119891 120593) = 119888119860 (119891 120593) + Δ119891= 119888 (120590 (119891) 120593) + Δ119891= 119888 (120590 (119891) + 119870 (Δ119891) 120593)

(40)

6 Journal of Function Spaces

for all 120593 isin [0 2120587) It follows by Lemma 1 in [5] that

(119879119891) = 120590 (119891) + 119870 (Δ119891) (41)

and therefore10038171003817100381710038171198791198911003817100381710038171003817infin = 10038171003817100381710038171198911003817100381710038171003817infin + Δ119891 (42)

If Δ119891 le 0 then a similar calculation shows that

(119891) = (119879119891) + 119870 (minusΔ119891) (43)

and10038171003817100381710038171198911003817100381710038171003817infin = 10038171003817100381710038171198791198911003817100381710038171003817infin + (minusΔ119891) (44)

It follows that in any case (Δ119891 ge 0 Δ119891 le 0) we obtainΔ119891 = 10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin 119891 isin 119860 (45)

We will prove that1003817100381710038171003817119879 (119891)1003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin = Δ119891 ge 0 (46)

for all 119891 isin 119860 Once it is proved applying the same argumentfor119879minus1 instead of119879 we see that 119879minus1119892infinminus119892infin ge 0 for every119892 isin 119861 As 119879 is a bijection it follows that 119891infin minus 119879119891infin ge 0for every 119891 isin 119860 It will follow that 119879119891infin = 119891infin for every119891 isin 119860 A proof of (46) is the following For every 120576 gt 0denote

A120576 = 119891 isin 119860 120588 (120590 (119891)) le 120576 (47)

The inequality in (46) is deduced by the following assertionswhich appear in the proof of [5 Theorem]

(1) 119879 is a continuous mapping from (119860 sdot infin) onto (119861 sdotinfin)(2) For each 120576 gt 0 the setA120576 is dense in (119860 sdot infin)(3) For each 120576 gt 0 and each 119891 isin A120576 it holds that119879(119891)infin ge 119891infin minus 120576

Suppose that these assertions are proved Let 119891 isin 119860 By (2)for any 120576 gt 0 there is a sequence 119891119899 of functions inA120576 suchthat 119891119899 minus 119891infin 997888rarr 0 as 119899 997888rarr infin By (3) we have

10038171003817100381710038171198791198911198991003817100381710038171003817infin minus 10038171003817100381710038171198911198991003817100381710038171003817infin ge minus120576 (48)

for every 119899 Letting 119899 997888rarr infin we have10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin ge minus120576 (49)

by (1) As 120576 gt 0 is arbitrary we have that10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin ge 0 (50)

We show proofs of three assertions (1) (2) and (3) aboveprecisely The proof of (1) is slightly different from the corre-sponding one in [5 p 70] This change is rather ambitiousWe also point out that the terms minus1205872 and 1205872 which appearin the formulae (7) and (8) in [5] seem inappropriate theyread for example as 31205874 and 1205874 respectively

We now proceed to prove the first statement Aiming fora contradiction suppose that 119879 is not continuous from (119860 sdotinfin) to (119861 sdot infin) Let 120576 be a positive real number less than1100 Then there is a function 1198910 isin 119860 such that 1198910infin le 120576and 119879(1198910)infin = 1 Then there exist 1199100 isin Ch(119861) such that|119879(1198910)(1199100)| = 1 by [29 Proposition 63] Since 119879 is complex-linear we may suppose that 119879(1198910)(1199100) = 1

By (41) and (45) we deduce that Δ1198910 = 119879(1198910)infin minus1198910infin ge 1 minus 120576 and 120590(119879(1198910)) = 120590(1198910) + 119870(Δ1198910) As 1198910infin le 120576and 119879(1198910)infin = 1 we infer that 120590(1198910) sub 119870(120576) and (119879(1198910) sub119870(1) Let 119911 isin 119870(1 minus 2120576) Let 119909 isin 119883 Then |1198910(119909)| le 120576 assertsthat 119908 = 119911 minus 1198910(119909) isin 119870(1 minus 120576) sub 119870(Δ1198910) Thus

119911 = 1198910 (119909) + 119908 isin 120590 (1198910) + 119870 (Δ1198910) = 120590 (119879 (1198910)) (51)

Hence119870(1 minus 2120576) sub 120590(1198791198910) As 119879(1198910)infin = 1 we have119870(1 minus 2120576) sub 120590 (119879 (1198910)) sub 119870 (1) (52)

Consider the open neighborhood 1198800 of 1199100 in1198832 given by

1198800 = 119910 isin 1198832 1003816100381610038161003816119879 (1198910) (119910) minus 11003816100381610038161003816 lt 120576 (53)

We infer that 1198800 is a proper subset of1198832 by (52) Then by [5Lemma 2] there exists 119892 isin 119861 such that 119892infin le 1 + 120576 119892(1199100) =1 |119892(119910) + 1| lt 120576 for every 119910 isin 1198832 1198800 and |Im119892(119910)| lt 120576 forall 119910 isin 1198832 If 119867 denotes the closed rectangle whose verticesare the four points plusmn(1 + 120576) plusmn 120576119894 we have

120590 (119892) sub 119867 (54)

Consider now the set

119871 = 11989031205871198944119911 |119911| le 1 Re 119911 ge 1 minus 2120576 (55)

We claim that119879(1198910)(1198832)cap119871 = 0 Suppose that119879(1198910)(1198832)cap119871 =0 As 119879(1198910)(1198832) is compact there exists a positive integer119899 such that 119879(1198910)(1198832) cap 119871119899 = 0 where 119871119899 = 11989031205871198944119911 |119911| le 1Re 119911 gt 1 minus 2120576 minus 1119899 Then (52) gives 119879(1198910)(1198832) sub119870(1)119871119899 As (119879(1198910)) is the closed convex-hull of 119879(1198910)(1198832)it is contained in the closed convex set 119870(1) 119871119899 On theother hand (1 minus 2120576)11989031205871198944 isin 119870(1 minus 2120576) sub 120590(119879(1198910)) by (52) As(1minus2120576)11989031205871198944 isin 119871 sub 119871119899 this contradicts 120590(119879(1198910)) sub 119870(1) 119871119899and this proves our claim Hence there is 1199101 isin 1198832 with119879(1198910)(1199101) isin 119871 As 120576 le 1100 it follows that |119879(1198910)(1199101) minus1| ge 120576 and so 1199101 isin 1198832 1198800 Hence |119892(1199101) + 1| lt 120576 Thus119892(1199101) + 119879(1198910)(1199101) is in 119871 minus 1 + K(120576) Thus we have

1 + radic22 minus 3120576 le 119888119861 (119892 + 119879 (1198910) 31205874 ) (56)

We claim that

120590 (119892 + 119879 (1198910)) sub co (119870 (minus1 1) cup 2) + 119870 (3120576) (57)

where119870(1199110 119903) = 119911 isin C |119911 minus 1199110| le 119903 Let 119910 isin 1198832 Supposefirst that |119879(1198910)(119910)minus1| lt 120576 Since 119892(1198832) sub 119867 by (54) we have

119879 (1198910) (119910) + 119892 (119910) isin 119870 (1 120576) + 119867 = (119867 + 1) + 119870 (120576) (58)

Journal of Function Spaces 7

Suppose next that |119879(1198910)(119910)minus1| ge 120576Then 119910 isin 1198832 1198800 and so|119892(119910) + 1| lt 120576 Moreover |119879(1198910)(119910)| le 1 Therefore we have

119892 (119910) + 119879 (1198910) (119910) isin 119870 (1) minus 1 + 119870 (120576)= 119870 (minus1 1) + 119870 (120576) (59)

It follows from (58) and (59) that

(119892 + 119879 (1198910)) (1198832) sub 119870 (minus1 1) cup ((119867 + 1)) + 119870 (120576) (60)

and hence

120590 (119892 + 119879 (1198910)) sub co (119870 (minus1 1) cup 2) + 119870 (3120576) (61)

as is claimed Therefore we have

119888119861 (119892 + 119879 (1198910) 1205874 ) le radic2 + 3120576 (62)

Put 1198911 = 119879minus1(119892) We claim that Δ1198911 le 120576 If Δ1198911 lt 0 thereis nothing to prove Suppose that Δ1198911 ge 0 Then by (41) wehave

120590 (119892) = 120590 (1198911) + 119870 (Δ1198911) (63)

Since 120590(119892) sub 119867 by (54) we have

119867 sup 120590 (1198911) + 119870 (Δ1198911) (64)

As 119867 does not include a closed disk with the radius greaterthan 120576 we conclude that Δ1198911 le 120576

In the following we will consider two cases 0 le Δ1198911 le 120576and Δ1198911 le 0 Suppose first that 0 le Δ1198911 le 120576 Then (64) yields(1198911) sub 119867 From 1198910infin le 120576 we deduce that 120590(1198910) sub 119870(120576)Hence we have

120590 (1198911 + 1198910) sub 120590 (1198911) + 120590 (1198910) sub (1198910) + 120590 (1198910)sub 119867 + 119870 (120576) (65)

Since119867 +119870(120576) is convex we have (1198911 + 1198910) sub 119867 + 119870 (120576) (66)

From (39) we infer that

119888119860 (1198911 + 1198910 31205874 ) = 119888 (120590 (1198911 + 1198910) 31205874 )le 119888 (119867 + 119870 (120576) 31205874 )= 119888 (119867 31205874 ) + 120576= radic22 + (1 + radic2) 120576

(67)

Since 119879(1198911 + 1198910) = 119892 + 119879(1198910) from (56) and (67) we obtainthat

1 minus (4 + radic2) 120576 le 119888119861 (119879 (1198911 + 1198910) 31205874 )minus 119888119860 (1198911 + 1198910 31205874 )

(68)

By (63) and 1 = 119892(1199100) we deduce that 1 isin 120590(1198911) + 119870(Δ1198911)Thus there is 119911 isin 120590(1198911) such that |119911 minus 1| le Δ1198911 It follows thatradic22 minus Δ1198911 le 119888119860(1198911 1205874) hence we have

radic22 minus 2120576 le 119888119860 (1198911 + 1198910 1205874 ) (69)

as 1198910infin le 120576 and 0 le Δ1198911 le 120576 We get by (62) and (69) that

119888119861 (119879 (1198911 + 1198910) 1205874 ) minus 119888119860 (1198911 + 1198910 1205874 ) leradic22 + 5120576 (70)

On the other hand 119888119861(119879(119891) 120593)minus119888119860(119891 120593) is invariant for any 120593by (37) From (68) and (70)we deduce that 120576 ge (2minusradic2)2(9+radic2) and this contradicts that 120576 le 1100

For the second case suppose next that Δ1198911 le 0 Then by(43) we have

120590 (1198911) = 120590 (119892) + 119870 (minusΔ1198911) (71)

and by (54) it follows that (1198911) sub 119867 + 119870(minusΔ1198911) Moreover(1198910) sub 119870(120576) since 1198910infin le 120576 Then

120590 (1198911 + 1198910) sub 120590 (1198911) + 120590 (1198910) sub 120590 (1198911) + 120590 (1198910)sub 119867 + 119870(minusΔ1198911) + 119870 (120576) (72)

Hence (1198911+1198910) sub 119867+119870(minusΔ1198911)+119870(120576) Using (39) we inferthat

119888119860 (1198911 + 1198910 31205874 ) le 119888 (119867 + 119870(minusΔ1198911) + 119870 (120576) 31205874 )= 119888 (119867 31205874 ) + (minusΔ1198911) + 120576= radic22 + (1 + radic2) 120576 + (minusΔ1198911)

(73)

By (71) we obtain that 120590(1198911) sup 119892(1198832) + 119870(minusΔ1198911) and as119892(1199100) = 1 we infer that 120590(1198911) sup 1 + 119870(minusΔ1198911) Hence radic22 +(minusΔ1198911) le 119888119860(1198911 1205874) so thatradic22 + (minusΔ1198911) minus 120576 le 119888119860 (1198911 + 1198910 1205874 ) (74)

as 1198910infin le 120576 Since 119879(1198911 +1198910) = 119892 +119879(1198910) we obtain by (56)and (73) that

1 minus (4 + radic2) 120576 minus (minusΔ1198911)le 119888119861 (119879 (1198911 + 1198910) 31205874 ) minus 119888119860 (1198911 + 1198910 31205874 )

(75)

We also obtain by (62) and (74) that

119888119861 (119879 (1198911 + 1198910) 1205874 ) minus 119888119860 (1198911 + 1198910 1205874 )le radic22 + 4120576 minus (minusΔ1198911)

(76)

Since 119888119861(119879(119891) 120593)minus119888119860(119891 120593) is invariant for any 120593 by (37) from(75) and (76) we deduce that 120576 ge (2 minusradic2)2(8 +radic2) and thisis impossible since 120576 le 1100

8 Journal of Function Spaces

Next we show a proof of the second assertion (2) Let119891 isin 119860 We prove that there exists a sequence 119891119899 sub 119860 whichuniformly converges to 119891 such that 120588(120590(119891119899)) 997888rarr 0 as 119899 997888rarrinfin Without loss of generality we may assume that 119891infin = 1Then there exists 1199100 isin Ch(119860) such that |119891(1199100)| = 119891infinby [29 Proposition 63] We may assume that 119891(1199100) = 1Suppose that 119899 ge 4 Put

119880119899 = 119909 isin 119883 1003816100381610038161003816119891 (119909) minus 11003816100381610038161003816 le 11198992 (77)

and

Ω = co(119870 (1) cup 1 + 1119899) + 119870( 11198992) (78)

(In the following we identify R2 and C that is we identify(119909 119910) and 119909 + 119894119910 for every 119909 119910 isin R) Since we assume that119899 ge 4 we infer by a simple calculation that

[1 minus 11198992 minus 120575119899119899 1 + 1119899 + 120575119899119899 ] times [minus 11198992 minus 120575119899119899 11198992 + 120575119899119899 ]sub Ω

(79)

for 120575119899 with 0 lt 120575119899 le 11198992 We assume that 0 lt 120575119899 le 11198992 By[5 Lemma 2] there exists ℎ119899 isin 119860 such that ℎ119899infin le 1 + 2120575119899ℎ119899(1199100) = 1 |ℎ119899 + 1| le 2120575119899 on119883 119880119899 and |Im ℎ119899| le 2120575119899 on119883Put 119892119899 = (ℎ119899 + 1)2 Then 119892119899 isin 119860 and

10038171003817100381710038171198921198991003817100381710038171003817infin le 1 + 120575119899119892119899 (1199100) = 110038161003816100381610038161198921198991003816100381610038161003816 le 120575119899 on 119883 1198801198991003816100381610038161003816Im1198921198991003816100381610038161003816 le 120575119899 on 119883

minus120575119899 le Re119892119899 le 1 + 120575119899 on 119883

(80)

Let 119909 isin 119880119899 Then we have

1 minus 11198992 le Re119891 (119909) le 1minus 11198992 le Im119891 (119909) le 11198992

(81)

and

minus120575119899119899 le Re119892119899 (119909)119899 le 1119899 + 120575119899119899

minus120575119899119899 le Im119892119899 (119909)119899 le 120575119899119899

(82)

Hence

1 minus 11198992 minus 120575119899119899 le Re(119891 (119909) + 119892119899 (119909)119899 ) le 1 + 1119899 + 120575119899119899 minus 11198992 minus 120575119899119899 le Im(119891 (119909) + 119892119899 (119909)119899 ) le 11198992 + 120575119899119899

(83)

It follows that we have

(119891 + 119892119899119899 ) (119909) isin [1 minus 11198992 minus 120575119899119899 1 + 1119899 + 120575119899119899 ]times [minus 11198992 minus 120575119899119899 11198992 + 120575119899119899 ] sub Ω

(84)

for 119909 isin 119880119899 Suppose that 119909 isin 119883 119880119899 Then10038161003816100381610038161003816100381610038161003816119891 (119909) +

119892119899 (119909)11989910038161003816100381610038161003816100381610038161003816 le 1 +

120575119899119899 le 1 + 11198993 lt 1 + 11198992 (85)

and hence

119891 (119909) + 119892119899 (119909)119899 isin 119870 (1) + 119870( 11198992) sub Ω (86)

for 119909 isin 119883 119880119899 Since 1 + 1119899 = 119891(1199100) + 119892119899(1199100)119899 we have bycombining (84) and (86) that

1 + 1119899 isin (119891 + 119892119899119899 ) (119883) sub Ω (87)

AsΩ is convex we obtain

1 + 1119899 isin 120590 (119891 + 119892119899119899 ) sub Ω (88)

Recall that for 119903 ge 0 and a complex number 1199110119870 (1199110 119903) = 119908 isin C 1003816100381610038161003816119908 minus 11991101003816100381610038161003816 le 119903 (89)

denotes the closed disk with center 1199110 and radius 119903 Weobserve that 120588(Ω 1 + 1119899) = 1119899 + 11198992 Recall that

120588(Ω 1 + 1119899)= sup119903 119911 isin Ω 1 + 1119899 isin 119870 (119911 119903) sub Ω

(90)

Let ℓ1 be the line defined by the equation

119910 = minus119899radic2119899 + 1 (119909 minus (1 +1119899 + 11198992 + 11198993)) (91)

part of which is a part of the boundary ofΩ Let ℓ2 be the linedefined by the equation

119910 = minus119899radic2119899 + 1 (119909 minus (1 minus11198992)) (92)

By some calculation we have that the distance between 1 minus11198992 and 1+1119899 is 1119899+11198992 and it coincides with the distancebetween the point 1 minus 11198992 and the line ℓ1 Hence we see that

1 + 1119899 isin 119870(1 minus 11198992 1119899 + 11198992 ) sub Ω (93)

Thus 1119899 + 11198992 le 120588(Ω 1 + 1119899)Next we prove that 119870(119901 119903) cap Ω119888 = 0 for every 119901 isin Ω

with 119903 = |119901 minus (1 + 1119899)| gt 1119899 + 11198992 It will follow that1119899 + 11198992 ge 120588(Ω 1 + 1119899) and the equality will hold Let

Journal of Function Spaces 9

119901 = (119909119901 119910119901) = 119909119901 + 119894119910119901 isin Ω with 119903 = |119901 minus (1 + 1119899)| gt1119899 + 11198992 We prove the case where 119910119901 ge 0 A proof for thecase where 119910119901 le 0 is the same and we omit it We divideΩ+ =119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902 isin Ω 119910119902 ge 0 into two parts

Ω1 = 119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902isin Ω+ minus119899radic2119899 + 1 (119909119902 minus (1 minus

11198992)) le 119910119902le minus119899radic2119899 + 1 (119909 minus (1 +

1119899 + 11198992 + 11198993))(94)

and

Ω2 = 119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902 isin Ω+ 119910119902le minus119899radic2119899 + 1 (119909119902 minus (1 minus

11198992)) (95)

Suppose that 119901 isin Ω1 and 119903 = |119901minus (1+1119899)| gt 1119899+11198992Thedistance between ℓ1 and ℓ2 is 1119899+11198992 Hence119870(119901 119903)capΩ119888 =0 Suppose that 119901 isin Ω2 and 119903 = |119901 minus (1 + 1119899)| gt 1119899 + 11198992that is119901 = 1minus11198992 Let ℓ1015840 be the line passing through119901whichis parallel to ℓ2 Let 1199011015840 be the unique point in the intersectionof ℓ1015840 and the 119909-axis Then 1199011015840 = 1 minus 11198992 minus 119906 for some 119906 ge 0Then the distance between ℓ1 and ℓ1015840 is 1119899+11198992+119906(1+1119899)which is equal to the distance between the point119901 and the lineℓ1 On the other hand1003816100381610038161003816100381610038161003816119901 minus (1 + 1119899)

1003816100381610038161003816100381610038161003816 ge10038161003816100381610038161003816100381610038161199011015840 minus (1 + 1119899)

1003816100381610038161003816100381610038161003816 = 1119899 + 11198992 + 119906 (96)

It follows that 119870(119901 119903) cap Ω119888 = 0 We conclude that if 119901 isin Ωsatisfies 119903 = |119901minus(1+1119899)| gt 1119899+11198992 then119870(119901 119903)capΩ119888 = 0Thus we have

120588 (Ω 1 + 1119899) = 1119899 + 11198992 (97)

Since 1 + 1119899 isin 120590(119891 + 119892119899119899) sub Ω we have120588 (120590 (119891 + 119892119899119899 )) le 120588 (120590 (119891 + 119892119899119899 ) 1 + 1119899)

le 120588 (Ω 1 + 1119899) = 1119899 + 11198992 (98)

On the other hand 119892119899119899infin 997888rarr 0 as 119899 997888rarr infin ensures that119891 + 119892119899119899 minus 119891infin 997888rarr 0 as 119899 997888rarr infin It follows that for every120576 gt 0A120576 is dense in (119860 sdot infin)Finally we show a proof of the third assertion (3) As is

pointed out in the proof of [5Theorem] 120588(co(119870)+119870(119888)) ge 119888for any 119870 sub C and any 119888 ge 0 Let 120576 gt 0 and 119891 isin A120576 Supposethat Δ119891 le 0 Then by (43) we have (119891) = (119879119891) + 119870(minusΔ119891)Hence we have

120588 ( (119891)) = 120588 (120590 (119879119891) + 119870 (minusΔ119891)) ge minusΔ119891 (99)

As 120576 ge 120588(120590(119891)) we conclude by (45) that10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin = Δ119891 ge minus120576 (100)

This completes the proof of the theorem

4 Hermitian Operators on a BanachAlgebras of Continuous Maps WhoseValues Are in a Uniform Algebras

Let 119883 and 119884 be compact Hausdorff spaces Let 119861 be aunital subalgebra of 119862(119883) which separates the points of 119883Throughout this sectionwe assume119861 is a Banach algebrawiththe norm sdot 119861 and 119860 is a uniform algebra on 119884 Recall thata uniform algebra on 119884 is a uniformly closed subalgebra of119862(119884) which contains constants and separates the points of 119884For functions 119891 isin 119862(119883) and 119892 isin 119862(119884) let 119891 otimes 119892 isin 119862(119883 times 119884)be the function defined by119891otimes119892(119909 119910) = 119891(119909)119892(119910) for (119909 119910) isin119883 times 119884 and for a subspace 119864119883 of 119862(119883) and a subspace 119864119884 of119862(119884) put

119864119883 otimes 119864119884

= 119899sum

119895=1

119891119895 otimes 119892119895 119899 isin N 119891119895 isin 119864119883 119892119895 isin 119864119884

(101)

and

1 otimes 119864119884 = 1 otimes 119892 119892 isin 119864119884 (102)

Throughout the section 119861 is a unital subalgebra of 119862(119883 times 119884)with a Banach algebra norm sdot 119861 We assume that 119861otimes119860 sub 119861Note that 119861 separates the points of119883times119884 since119860 separates thepoints of 119884 and 119861 separates the points of 119883 We assume thatthere exists a compact Hausdorff space M and a complex-linear map 119863 119861 997888rarr 119862(M) such that ker119863 = 1 otimes 119860 Weassume that 119865119861 = 119865infin(119883times119884)+119863(119865)infin(M) for every 119865 isin 119861Hence119863 is continuous Defining

lsaquo119865lsaquo = 119863 (119865)infin(M) 119865 isin 119861 (103)

lsaquo sdot lsaquo is a one-invariant seminorm in the sense of Jarosz lsaquo sdot lsaquois a seminorm on 119861 such that lsaquo119865+ 1lsaquo = lsaquo119865lsaquo for every 119865 isin 119861Hence the norm sdot 119861 is a natural norm (see [5 p67]) Notethat 119861 is a regular subspace of 119862(119883times119884) in the sense of Jarosz[5 Proposition 2]

Lumerrsquos seminal paper [35] opened up a useful methodof finding isometries which is often referred to as Lumerrsquosmethod It involves the notion ofHermitian operators and thefact that 119880119867119880minus1 must be Hermitian if119867 is Hermitian and 119880is a surjective isometry

Definition 6 Let A be a unital Banach algebra We say that119890 isin A is a Hermitian element if1003817100381710038171003817exp (119894119905119890)1003817100381710038171003817A = 1 (104)

for every 119905 isin R The set of all Hermitian elements of A isdenoted by119867(A)

If A is a unital 119862lowast-algebra then 119867(A) is the set of allself-adjoint elements of A Hence 119867(119872119899(C)) is the set of allHermitian matrices and119867(119862(119884)) = 119862R(119884)Definition 7 Let 119864 be a complex Banach space The Banachalgebra of all bounded operators on 119864 is denoted by 119861(119864) Wesay that 119879 isin 119861(119864) is a Hermitian operator if 119879 isin 119867(119861(119864))

10 Journal of Function Spaces

Note that a Hermitian element of a unital Banach algebraand a Hermitian operator are usually defined in terms ofnumerical range or semi-inner product Here we define themby an equivalent form (see [36]) By the definition of aHermitian operator we have the following

Proposition 8 Let 119864119895 be a complex Banach space for 119895 = 1 2Suppose that 119881 1198641 997888rarr 1198642 is a surjective isometry and 119867 1198641 997888rarr 1198641 is a Hermitian operator Then 119881119867119881minus1 1198642 997888rarr 1198642

is a Hermitian operator

Proposition 9 An element 119865 isin 119861 is Hermitian if and only ifthere exists 119891 isin 119860 cap 119862R(119884) such that 119865 = 1 otimes 119891Proof Suppose that 119865 isin 119861 is a Hermitian element Then1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884) + 1003817100381710038171003817119863 (exp (119894119905119865))1003817100381710038171003817infin(M)

= 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817 = 1 (105)

for every 119905 isin R Suppose that there exists a point (119909 119910) isin119883 times 119884 with Im119865(119909 119910) = 0 where Im denotes the imaginarypart of a complex number Suppose that Im119865(119909 119910) gt 0Then1003817100381710038171003817exp (minus119894119865)1003817100381710038171003817infin(119883times119884) ge 1003816100381610038161003816exp (minus119894119865 (119909 119910))1003816100381610038161003816

= exp (Im119865 (119909 119910)) gt 1 (106)

Suppose that Im119865(119909 119910) lt 0 Then1003817100381710038171003817exp (119894119865)1003817100381710038171003817infin(119883times119884) ge 1003816100381610038161003816exp (119894119865 (119909 119910))1003816100381610038161003816= exp (minusIm119865 (119909 119910)) gt 1 (107)

In any case we have there exists 119905 isin R such that1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817119861 ge 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884) gt 1 (108)

which contradicts our assumption We have that

119865 isin 119862R (119883 times 119884) (109)

Thus for every (119904 119905) isin 119883 times 119884 and 119905 isin R |exp(119894119905119865(119904 119905))| = 1Hence exp(119894119905119865)infin(119883times119884) = 1 for every 119905 isin R By (105)we have119863(exp(119894119905119865)infin(M) = 0 which ensures that 119863(exp(119894119905119865)) = 0for every 119905 isin R Thus exp(119894119905119865) isin 1 otimes 119860 for every 119905 isin R Wehave

exp (119894119905119865) minus 1119905 minus 119894119865 = infinsum119899=2

((119894119865)119899 119905119899minus2119899 ) 119905 (110)

and hence for every 119905 isin R with |119905| le 1 we have10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817 le (

infinsum119899=2

119865119899119861|119905|119899minus2119899 ) |119905|

le (infinsum119899=2

119865119899119861119899 ) |119905| le (exp 119865119861) |119905|

(111)

It follows that10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817infin(119883times119884)

le 10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817119861

997888rarr 0(112)

as 119905 997888rarr 0 Since (exp(119894119905119865)minus1)119905 isin 1otimes119860 for each 119905 isin R thereexists 119892119905 isin 119860 such that

exp (119894119905119865) minus 1119905 = 1 otimes 119892119905 (113)

By (112) we have10038171003817100381710038171 otimes 1198921119899 minus 1198941198651003817100381710038171003817infin(119883times119884)

997888rarr 0 (114)

as 119899 997888rarr infin We have that 1 otimes 1198921119899 is a Cauchy sequencein 119862(119883 times 119884) thus we infer that1198921119899 is a Cauchy sequence in119862(119884) Since 119860 is uniformly closed as it is a uniform algebrathere exists 119892 isin 119860 such that

10038171003817100381710038171198921119899 minus 1198921003817100381710038171003817infin(119884)997888rarr 0 (115)

and hence10038171003817100381710038171 otimes 1198921119899 minus 1 otimes 1198921003817100381710038171003817infin(119883times119884)

997888rarr 0 (116)

as 119899 997888rarr infin It follows by (114) that 119894119865 = 1 otimes 119892 thus119865 = 1 otimes (minus119894119892) isin 1 otimes 119860 (117)

By (109) we see that minus119894119892 isin 119862R(119884) thus we have 119891 = minus119894119892 isin119860 cap 119862R(119884) and 119865 = 1 otimes 119891Suppose conversely that 119861 ni 119865 = 1otimes119891 for119891 isin 119860cap119862R(119884)

We infer that 119865 isin 119862R(119883 times 119884) and |exp(119894119905(119865(119909 119910))| = 1 forevery 119905 isin R and (119909 119910) isin 119883 times 119884 Hence exp(119894119905119865)infin(119883times119884) = 1for every 119905 isin R Since

exp (119894119905119865) = exp (119894119905 (1 otimes 119891)) = 1 otimes exp (119894119905119891) isin 1 otimes 119860 (118)

we have 119863(exp(119894119905119865)) = 0 It follows that1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817 = 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884)

+ 1003817100381710038171003817119863 (exp (119894119905119865))1003817100381710038171003817infin(M) = 1 (119)

for every 119905 isin R We conclude that 119865 is a Hermitian elementin 119861

Note that 119891 isin 119860 is Hermitian if and only if119891 isin 119860cap119862R(119884)by [37 Proposition 5] Hence Proposition 9 asserts that 119865 is aHermitian element in119861 if and only if119865 = 1otimes119891 for aHermitianelement 119891 in 119860Proposition 10 Suppose that 119880 119861 997888rarr 119861 is a surjectiveunital isometry Then 119880 is an algebra isomorphism

Proof As we have already mentioned 119861 is a regular subspace(in the sense of Jarosz) with a natural norm Then byTheorem 1 119880 is also an isometry with respect to the supre-mum norm on 119883 times 119884 Then 119880 is uniquely extended to asurjective isometry with respect to the supremum norm from the uniform closure 119861 onto itself Since 119861 is a uniformalgebra a theorem of Nagasawa [32] asserts that is analgebra isomorphism since (1) = 1 Thus 119880 is an algebraisomorphism from 119861 onto itself

Journal of Function Spaces 11

Theorem 11 A bounded operator 119879 119861 997888rarr 119861 is a Hermitianoperator if and only if 119879(1) is a Hermitian element in 119861 and119879 = 119872119879(1) the multiplication operator by 119879(1)Proof By Proposition 10 every surjective unital isometry on119861 is multiplicative Then by [37 Theorem 4] we have theconclusion

5 Banach Algebras of 119862(119884)-Valued Maps

Suppose that119883 is a compact Hausdorff space Suppose that 119861is a unital point separating subalgebra of119862(119884) equipped witha Banach algebra norm Then 119861 is semisimple because 119891 isin119861 119891(119909) = 0 is a maximal ideal of 119861 for every 119909 isin 119883 and theJacobson radical of 119861 vanishes The inequality 119891infin le 119891119861for every 119891 isin 119861 is well known We say that 119861 is natural if themap 119890 119884 997888rarr 119872119861 defined by 119910 997891997888rarr 120601119910 where 120601119910(119891) = 119891(119910)for every 119891 isin 119861 is bijective We say that 119861 is self-adjoint if 119861 isnatural and conjugate-closed in the sense that 119891 isin 119861 impliesthat 119891 isin 119861 for every 119891 isin 119861 where sdot denotes the complexconjugation on 119884Definition 12 Let 119883 and 119884 be compact Hausdorff spacesSuppose that 119861 is a unital point separating subalgebra of119862(119883)equipped with a Banach algebra norm sdot 119861 Suppose that119861 is self-adjoint Suppose that 119861 is a unital point separatingsubalgebra of 119862(119883 times 119884) such that 119861 otimes 119862(119884) sub 119861 equippedwith a Banach algebra norm sdot 119861 Suppose that 119861 is self-adjoint We say that 119861 is a natural 119862(119884)-valuezation of 119861 ifthere exists a compact Hausdorff space M and a complex-linear map 119863 119861 997888rarr 119862(M) such that ker119863 = 1 otimes 119862(119884) and119863(119862R(119883 times 119884) cap 119861) sub 119862R(M) which satisfies

119865 = 119865infin(119883times119884) + 119863 (119865)infin(M) 119865 isin 119861 (120)

The term ldquoa natural 119862(119884)-valuezation of 119861rdquo comes fromthe natural norm defined by Jarosz [5] In fact the norm sdot 119861is a natural norm in the sense of Jarosz [5]

Note that (119883119862(119884) 119861 119861) need not be an admissiblequadruple defined by Nikou and OrsquoFarrell [38] (cf [31]) sincewe do not assume that 119865(sdot 119910) 119865 isin 119861 119910 isin 119884 sub 119861 whichis a requirement for the admissible quadruple On the otherhand if (119883119862(119884) 119861 119861) is an admissible quadruple of type Ldefined in [30] then 119861 is a natural 119862(119884)-valuezation of 119861 dueto Definition 12

Example 13 Let 119861 = 1198621([0 1]) and 119861 = Lip([0 1] 119862(119884))for 119884 = 119901 a singleton Then Lip([0 1]) is algebraically iso-morphic to Lip([0 1] 119862(119884)) Suppose thatM is the maximalideal space of 119871infin([0 1]) and 119863 119861 997888rarr 119862(M) is definedby 119891 997891997888rarr Γ(1198911015840) where Γ denotes the Gelfand transformin 119871infin([0 1]) Then 119861 is a natural 119862(119884)-valuezation of 119861The Banach algebra Lip([0 1]) with the norm 119891infin([01]) +1198911015840infin([01]) is isometrically isomorphic to 119861

Let 119884 be a compact Hausdorff space Note that a closedsubalgebra 119861 of Lip((119870 119889120572) 119862(119884))which appears in Example12 in [30] is an example of a natural 119862(119884)-valuezation of 119861

The Banach algebras1198621([0 1] 119862(119884)) and 1198621(T 119862(119884)))whichappear in Examples 16 and 17 in [30] respectively are alsoexamples of natural 119862(119884)-valuezations of 1198621([0 1])6 Isometries on Natural 119862(119884)-ValuezationsThemain theorem in this paper is the following

Theorem 14 Suppose that 119861119895 is a natural 119862(119884119895)-valuezationof 119861119895 sub 119862(119883119895) for 119895 = 1 2 We assume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (121)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for119895 = 1 2 Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linear isometry Then there exists ℎ isin 119862(1198842) such that |ℎ| = 1on 1198842 a continuous map 120593 1198832 times 1198842 997888rarr 1198831 such that120593(sdot 119910) 1198832 997888rarr 1198831 is a homeomorphism for each 119910 isin 1198842and a homeomorphism 120591 1198842 997888rarr 1198841 which satisfies119880 (119865) (119909 119910) = ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842 (122)

for every 119865 isin 1198611

In short a surjective isometry between 119862(119884)-valuezationsis a weighted composition operator of a specific form thehomeomorphism 1198832 times 1198842 997888rarr 1198831 times 1198841 (119909 119910) 997891997888rarr(120593(119909 119910) 120591(119910)) has the second coordinate that depends onlyon the second variable 119910 isin 1198842 A composition operatorinduced by such a homeomorphism is said to be of type BJin [31 37] after the study of Botelho and Jamison [39]

Quite recently the author of this paper and Oi [30 Theo-rem 8] proved a similar result of Theorem 14 for admissiblequadruples of type L To prove it we apply Proposition 32and the following comments in [31] Instead of this we proveTheorem 14 by Lumerrsquosmethod with which a proof is simplerthan that in [30]

In the following in this section we assume that 119861119895 is anatural 119862(119884119895)-valuezation of 119861 sub 119862(119883119895) for 119895 = 1 2 Weassume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (123)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for 119895 = 1 2Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linearisometry A crucial part of a proof of Theorem 14 is to proveProposition 15

Proposition 15 Suppose that 1198832 is not a singleton Thereexists ℎ isin 119862(1198842) with |ℎ| = 1 on 1198842 such that 119880(11198611) = 11198612 otimesℎ

A similar result for admissible quadruples of type L isproved in [30 Proposition 9] If we assumed that

119865 (sdot 119910) 119865 isin 119861119895 119910 isin 119884119895 sub 119861119895 (124)

then 119861119895 were an admissible quadruple of type L Although119861119895 in this paper need not be an admissible quadruple of type

12 Journal of Function Spaces

L a proof of Proposition 15 is completely the same as that in[30 Proposition 9] since we do not make use of the condition(124) in the proof of [30 Proposition 9] The condition (124)is needed in [30] when we apply Proposition 32 and thefollowing comments in [31]

7 Proof of Theorem 14 An Application ofLumerrsquos Method

Proof of Theorem 14 A proof for the case where 1198831 = 1199091and 1198832 = 1199092 are singletons is the same as the proof ofTheorem 8 in [30]

Suppose that1198832 is not a singleton By Proposition 15 thereexists ℎ isin 119862R(1198842) with |ℎ| = 1 on 1198842 such that 119880(1) =1 otimes ℎ Letting 1198800 1198611 997888rarr 1198612 by 1198800(119865) = (1 otimes ℎ)119880(119865)119865 isin 1198611 we see by the hypothesis (1 otimes ℎ)1198651198612 = 1198651198612 forevery 119865 isin 1198612 that 1198800 is a surjective unital isometry from1198611 onto 1198612 Then Corollary 2 asserts that 1198800 is an algebraisomorphism Let 119891 isin 119862R(1198841) By Proposition 9 1 otimes 119891 isa Hermitian element in 1198611 Then by Theorem 11 1198721otimes119891 is aHermitian operator on 1198611 By Proposition 8 11988001198721otimes119891119880minus1

0 is aHermitian operator on 1198612 Then by Theorem 11 there exists119878(119891) isin 119862R(1198842) such that 11988001198721otimes119891119880minus1

0 = 1198721otimes119878(119891) Hence anoperator 119878 119862R(1198841) 997888rarr 119862R(1198842) is defined Since 1198800 is analgebra isomorphism it is easy to see that 119878 is a real algebraisomorphism from 119862R(1198841) onto 119862R(1198842) Then 119878 119862(1198841) 997888rarr119862(1198842) defined by 119878(119891) = 119878(Re119891) + 119894119878(Im119891) for 119891 isin 119862(1198841)gives a complex algebra isomorphism Gelfand theory assertsthat there is a homeomorphism 120591 1198842 997888rarr 1198841 such that119878(119891) = 119891 ∘ 120591 119891 isin 119862(1198841) It follows that

11988001198721otimes119891119880minus10 = 1198721otimes119891∘120591 119891 isin 119862 (1198841) (125)

Since 119880minus10 (1) = 1 we have

1198800 (1 otimes 119891) = 1 otimes 119891 ∘ 120591 119891 isin 119862 (1198841) (126)

Define Φ 1198611 997888rarr 1198612 by Φ(119886) = 1198800(119886 otimes 1) 119886 isin 1198611Since 1198800 is an algebra isomorphism the map Φ is a unitalhomomorphism Since the maximal ideal space of 1198611 is 1198831

and the maximal ideal space of 1198612 is 1198832 times 1198842 there is acontinuous map 120593 1198832 times 1198842 997888rarr 1198831 such that

Φ(119886) (119909 119910) = 119886 (120593 (119909 119910)) 119886 isin 1198611 (119909 119910) isin 1198832 times 1198842 (127)

It follows by (126) and (127) that

1198800 (119886 otimes 119891) (119909 119910) = 1198800 ((119886 otimes 1) (1 otimes 119891)) (119909 119910)= 1198800 (119886 otimes 1) (119909 119910)1198800 (1 otimes 119891) (119909 119910)= 119886 (120593 (119909 119910))119891 (120591 (119910))= (119886 otimes 119891) (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198832 times 1198842

(128)

for every 119886 isin 1198611 and 119891 isin 119862(1198841) Thus

1198800 (sum(119886119895 otimes 119891119895)) (119909 119910)= (sum(119886119895 otimes 119891119895)) (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842(129)

for every sum(119886119895 otimes 119891119895) isin 1198611 otimes 119862(1198841) By the Stone-Weierstrasstheorem 1198611 otimes 119862(1198841) is uniformly dense in 119862(1198831 times 1198841) henceany element in 1198611 is uniformly approximated by 1198611 otimes 119862(1198841)As 1198800 is also an isometry with respect to the uniform normwe see that

1198800 (119865) (119909 119910) = 119865 (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198831 times 1198841 (130)

for every 119865 isin 1198611 and

119880 (119865) (119909 119910) = (1 otimes ℎ) (119909 119910) 119865 (120593 (119909 119910) 120591 (119910))= ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198831 times 1198841(131)

As 1198800 is an algebra isomorphism the map 1198832 times 1198842 997888rarr1198831 times 1198841 defined by (119909 119910) 997891997888rarr (120593(119909 119910) 120591(119910)) gives ahomeomorphism Therefore for every 119910 isin 1198842 the map

120593 (sdot 119910) 1198832 997888rarr 1198831 (132)

is a homeomorphismSuppose that 1198831 is not a singleton By the same way as

in the last part of the proof of Theorem 8 in [30] we havethat 1198832 is not a singleton Then we have the conclusion bythe previous argument

8 Application of Theorem 14

We exhibit applications of Theorem 14

Corollary 16 ([4 Theorem 33]) Suppose that119880 Lip([0 1]) 997888rarr Lip([0 1]) is a surjective isometrywith respect to the norm defined by 119891infin([01]) + 1198911015840infin([01])

for 119891 isin Lip([0 1]) Then 119880(1) is a constant function of unitmodulus such that

119880(119891) (119909) = 119880 (1) 119891 (119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (133)

or

119880(119891) (119909) = 119880 (1) 119891 (1 minus 119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (134)

The converse statement also holds

Proof By Example 13 we may suppose that Lip([0 1]) is aBanach algebra of 119862(119884)-valuezation Applying Theorem 14we have that119880(1) = 1otimesℎ for ℎ isin 119862(119884)with |ℎ| = 1 Since our

Journal of Function Spaces 13

119884 is a singleton 119880(1) is a constant function of unit modulusWe also see that the corresponding continuous map 120593 [0 1] times 119884 997888rarr [0 1] can be considered as a homeomorphismfrom [0 1] onto [0 1] therefore we have that119880(119891) (119909) = 119880 (1) 119891 (120593 (119909))

119891 isin Lip ([0 1]) 119909 isin [0 1] (135)

The rest is a routine argument to prove that 120593 is an isometryhence 120593(119909) = 119909 119909 isin [0 1] or 120593(119909) = 1 minus 119909 119909 isin [0 1]

The converse statement is trivial

Corollaries 14 15 18 and 19 in [30 Section 6] follow herewith a similar proof

Data Availability

The data used to support the findings of this study areavailable from the corresponding author upon request

Conflicts of Interest

The author declares that they have no conflicts of interest

Acknowledgments

This work was supported by JSPS KAKENHI Grants Num-bers JP16K05172 and JP15K04921

References

[1] K de Leeuw ldquoBanach spaces of Lipschitz functionsrdquo StudiaMathematica vol 21 pp 55ndash66 19611962

[2] A K Roy ldquoExtreme points and linear isometries of the Banachspace of Lipschitz functionsrdquoCanadian Journal of MathematicsJournal Canadien de Mathematiques vol 20 pp 1150ndash11641968

[3] M Cambern ldquoIsometries of certain Banach algebrasrdquo StudiaMathematica vol 25 pp 217ndash225 19641965

[4] N V Rao and A K Roy ldquoLinear isometries of some functionspacesrdquo Pacific Journal of Mathematics vol 38 pp 177ndash192 1971

[5] K Jarosz ldquoIsometries in semisimple commutative BanachalgebrasrdquoProceedings of the AmericanMathematical Society vol94 no 1 pp 65ndash71 1985

[6] K Jarosz and V D Pathak ldquoIsometries between functionspacesrdquo Transactions of the AmericanMathematical Society vol305 no 1 pp 193ndash206 1988

[7] NWeaver ldquoIsometries of noncompact Lipschitz spacesrdquoCana-dian Mathematical Bulletin Bulletin Canadien de Mathemat-iques vol 38 no 2 pp 242ndash249 1995

[8] A Jimenez-Vargas and M Villegas-Vallecillos ldquoLinear isome-tries between spaces of vector-valued Lipschitz functionsrdquoProceedings of the American Mathematical Society vol 137 no4 pp 1381ndash1388 2009

[9] A Jimenez-Vargas and M Villegas-Vallecillos ldquoInto linearisometries between spaces of Lipschitz functionsrdquo HoustonJournal of Mathematics vol 34 no 4 pp 1165ndash1184 2008

[10] F Botelho and J Jamison ldquoSurjective isometries on spacesof differentiable vector-valued functionsrdquo Studia Mathematicavol 192 no 1 pp 39ndash50 2009

[11] E Mayer-Wolf ldquoIsometries between Banach spaces of Lipschitzfunctionsrdquo Israel Journal of Mathematics vol 38 no 1-2 pp 58ndash74 1981

[12] A Jimenez-Vargas M Villegas-Vallecillos and Y-S WangldquoBanach-Stone theorems for vector-valued little Lipschitz func-tionsrdquo Publicationes Mathematicae vol 74 no 1-2 pp 81ndash1002009

[13] J Araujo andLDubarbie ldquoNoncompactness andnoncomplete-ness in isometries of Lipschitz spacesrdquo Journal of MathematicalAnalysis and Applications vol 377 no 1 pp 15ndash29 2011

[14] F Botelho R J Fleming and J Jamison ldquoExtreme points andisometries on vector-valued Lipschitz spacesrdquo Journal of Math-ematical Analysis and Applications vol 381 no 2 pp 821ndash8322011

[15] H Koshimizu ldquoLinear isometries on spaces of continuouslydifferentiable and Lipschitz continuous functionsrdquo NihonkaiMathematical Journal vol 22 no 2 pp 73ndash90 2011

[16] F Botelho J Jamison and B Zheng ldquoIsometries on spaces ofvector valued Lipschitz functionsrdquo Positivity An InternationalMathematics Journal Devoted to Theory and Applications ofPositivity vol 17 no 1 pp 47ndash65 2013

[17] A Ranjbar-Motlagh ldquoA note on isometries of Lipschitz spacesrdquoJournal of Mathematical Analysis and Applications vol 411 no2 pp 555ndash558 2014

[18] F Botelho and J Jamison ldquoSurjective isometries on spaces ofvector valued continuous and Lipschitz functionsrdquo PositivityAn International Mathematics Journal Devoted to Theory andApplications of Positivity vol 17 no 3 pp 395ndash405 2013Erratum to Surjective isometries on spaces of vector valuedcontinuous and Lipschitz functions by F Botelho 20 (2016)757ndash759

[19] T Miura and H Takagi ldquoSurjective isometries on the Banachspace of continuously differentiable functionsrdquo ContemporaryMathematics vol 687 pp 181ndash192 2017

[20] K Kawamura ldquoBanach-Stone type theorems for 1198621-functionspaces over Riemannian manifoldsrdquo Acta Universitatis Szegedi-ensis Acta Scientiarum Mathematicarum vol 83 no 3-4 pp551ndash591 2017

[21] K Kawamura ldquoPerturbations of norms on 1198621-function spacesand associated isometry groupsrdquo Topology Proceedings vol 51pp 169ndash196 2018

[22] K Kawamura ldquoA Banach-Stone type theorem for 1198621-functionspaces over the circlerdquo Topology Proceedings vol 53 pp 15ndash262019

[23] L Li D Chen Q Meng and Y-S Wang ldquoSurjective isometrieson vector-valued differentiable function spacesrdquo Annals ofFunctional Analysis vol 9 no 3 pp 334ndash343 2018

[24] K Kawamura H Koshimizu and T Miura ldquoNorms onC1([01]) and there isometriesrdquo Acta Scientiarum Mathemati-carum vol 84 no 12 pp 239ndash261 2018

[25] L Li A M Peralta L Wang and Y-S Wang ldquoWeak-2-local isometries on uniform algebras and Lipschitz algebrasrdquohttpsarxivorgabs170503619

[26] A Jimenez-Vargas L LiAMPeralta LWang andY-SWangldquo2-local standard isometries on vector-valued Lipschitz func-tion spacesrdquo Journal of Mathematical Analysis and Applicationsvol 461 no 2 pp 1287ndash1298 2018

[27] A Ranjbar-Motlagh ldquoIsometries of Lipschitz type functionspacesrdquoMathematischeNachrichten vol 291 no 11-12 pp 1899ndash1907 2018

14 Journal of Function Spaces

[28] N Weaver Lipschitz Algebras World Scientific Publishing CoInc River Edge NJ USA 1999

[29] R R Phelps Lectures on Choquetrsquos theorem vol 1757 of LectureNotes in Mathematics Springer-Verlag Berlin Germany 2ndedition 2001

[30] O Hatori and S Oi ldquoIsometries on Banach algebras of vector-valued mapsrdquo Acta Scientiarum Mathematicarum vol 84 no12 pp 151ndash183 2018

[31] O Hatori S Oi and H Takagi ldquoPeculiar homomorphismsbetween algebras of vector-valued mapsrdquo Studia Mathematicavol 242 no 2 pp 141ndash163 2018

[32] M Nagasawa ldquoIsomorphisms between commutative Banachalgebras with an application to rings of analytic functionsrdquoKodai Mathematical Seminar Reports vol 11 pp 182ndash188 1959

[33] K de LeeuwW Rudin and J Wermer ldquoThe isometries of somefunction spacesrdquo Proceedings of the American MathematicalSociety vol 11 pp 694ndash698 1960

[34] OHatori A Jimenez-Vargas andMVillegas-Vallecillos ldquoMapswhich preserve norms of non-symmetrical quotients betweengroups of exponentials of Lipschitz functionsrdquo Journal of Math-ematical Analysis and Applications vol 415 no 2 pp 825ndash8452014

[35] G Lumer ldquoOn the isometries of reflexiveOrlicz spacesrdquoAnnalesde lrsquoInstitut Fourier vol 68 pp 99ndash109 1963

[36] R J Fleming and J E Jamison Isometries on Banach SpacesFunction Spaces vol 129 ofMonographs and Surveys in Pure andApplied Mathematics Chapman amp Hall CRC Boca Raton FlaUSA 2003

[37] OHatori and SOi ldquoHermitian operators onBanach algebras ofvector-valued Lipschitzmapsrdquo Journal ofMathematical Analysisand Applications vol 452 no 1 pp 378ndash387 2017 Corrigendumto rdquoHermitian operators on Banach algebras of vector-valuedLipschitz mapsrdquo Journal of Mathematical Analysis and Appli-cations 452 (2017) 378ndash387 MR3628025

[38] A Nikou andA G OrsquoFarrell ldquoBanach algebras of vector-valuedfunctionsrdquo Glasgow Mathematical Journal vol 56 no 2 pp419ndash426 2014

[39] F Botelho and J Jamison ldquoHomomorphisms on a class ofcommutative Banach algebrasrdquo Rocky Mountain Journal ofMathematics vol 43 no 2 pp 395ndash416 2013

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Page 2: Hermitian Operators and Isometries on Banach Algebras of ...downloads.hindawi.com/journals/jfs/2018/8085304.pdfJournalofFunctionSpaces normiseitherthemaxnormorthesumnorm.esituation

2 Journal of Function Spaces

norm is either the max norm or the sum normThe situationis very different without assuming the unitality for theisometry with respect to the max norm There is a simpleexample of a surjective isometry which is not canonical [7p242] On the other hand Jarosz and Pathak exhibited in[6 Example 8] that a surjective isometry on Lip(119870) withrespect to the sum norm is canonical After the publicationof [6] some authors expressed their suspicion about theargument there and the validity of the statement there hadnot been confirmed until quite recently Hence the problemon isometries with respect to the sum norm has not been wellstudied

Jimenez-Vargas and Villegas-Vallecillos in [8] have con-sidered isometries of spaces of Lipschitz maps on a compactmetric space taking values in a strictly convex Banach spaceequippedwith the norm 119891 = max119891infin 119871(119891) see also [9]Botelho and Jamison [10] studied isometries on 1198621([0 1] 119864)with max119909isin[01]119891(119909)119864 + 1198911015840(119909)119864 See also [11ndash27] Referalso to a book of Weaver [28]

We propose a unified approach to the study of isome-tries with respect to the sum norm on Banach algebrasLip(119870119862(119884)) lip120572(119870119862(119884)) and 1198621(119870119862(119884)) where 119870 isa compact metric space [0 1] or T (T denotes the unitcircle on the complex plane) and 119884 is a compact Hausdorffspace We study isometries without assuming that theypreserve unit As corollaries of a general result we describeisometries on Lip(119870119862(119884)) lip120572(119870119862(119884)) 1198621([0 1] 119862(119884))and 1198621(T 119862(119884)) respectively

The main result in this paper is Theorem 14 which givesthe form of a surjective isometry 119880 with respect to the sumnorm between certain Banach algebras with the values in acommutative unital 119862lowast-algebra The proof of the necessityof the isometry in Theorem 14 comprises several steps Thecrucial part of the proof ofTheorem 14 is to prove that119880(1) =1 otimes ℎ for an ℎ isin 119862(1198842) with |ℎ| = 1 on 1198842 (Proposition 15)To prove Proposition 15 we apply Choquetrsquos theory (cf [29])with measure theoretic arguments A proof of Proposition 15is completely the same as that of [30 Proposition 9] Pleaserefer to it By Proposition 15 we have that 1198800 = (1 otimes ℎ)119880is a surjective isometry fixing the unit Then by applying atheorem of Jarosz [5] (Theorem 1 in this paper) we see that1198800 is also an isometry with respect to the supremum normBy the Banach-Stone theorem 1198800 is an algebra isomorphismThen by applying Lumerrsquos method (cf [30]) we see that 1198800 isa composition operator of type BJ (cf [31])

Our proofs in this paper make substantial use of thetheorem of Jarosz [5 Theorem] The author believes that itis convenient for the readers to show a precise proof becausethere need to be some ambitious changes in the original proofby Jarosz

2 Preliminaries

Let 119884 be a compact Hausdorff space Let 119864 be a real orcomplex Banach space The space of all 119864-valued continuousmaps on 119884 is denoted by 119862(119884 119864) When 119864 = C (resp R)119862(119884 119864) is abbreviated by119862(119884) (resp119862R(119884)) For a subset 119878 of119884 the supremum norm of 119865 on 119878 is 119891infin(119878) = sup119909isin119878119865(119909)119864

for 119865 isin 119862(119884 119864) When no confusion will result we omit thesubscript 119878 and write only sdot infin Let 119870 be a compact metricspace and 0 lt 120572 le 1 For 119865 isin 119862(119870 119864) put

119871120572 (119865) = sup119909 =119910

1003817100381710038171003817119865 (119909) minus 119865 (119910)1003817100381710038171003817119864119889 (119909 119910)120572 (4)

Then 119871120572 is called an 120572-Lipschitz number of 119865 or just aLipschitz number of 119865 When 120572 = 1 we omit the subscript120572 and write only 119871(119865) The space of all 119865 isin 119862(119870 119864) suchthat 119871120572(119865) lt infin is denoted by Lip120572(119870 119864) When 120572 = 1 thesubscript is omitted and it is written as Lip(119870 119864)

When 0 lt 120572 lt 1 the closed subspace

lip120572 (119870 119864) = 119865

isin Lip120572 (119870 119864) lim119909997888rarr1199090

1003817100381710038171003817119891 (1199090) minus 119891 (119909)1003817100381710038171003817119864119889 (1199090 119909)120572= 0 for every 1199090 isin 119870

(5)

of Lip120572(119870 119864) is called a little Lipschitz space There are avariety of complete norms on Lip120572(119870 119864) and lip120572(119870 119864) Inthis paper we are mainly concerned with the norm sdot 119871 ofLip120572(119870 119864) (resp lip120572(119870 119864)) which is defined by

10038171003817100381710038171198911003817100381710038171003817119871 = 10038171003817100381710038171198911003817100381710038171003817infin(119870) + 119871120572 (119865) 119865 isin Lip120572 (119870 119864) (resp lip120572 (119870 119864)) (6)

The norm sdot 119872 of Lip120572(119870 119864) (resp lip120572(119870 119864)) is defined by

10038171003817100381710038171198911003817100381710038171003817119872 = max 10038171003817100381710038171198911003817100381710038171003817infin(119870) 119871120572 (119865) 119865 isin Lip120572 (119870 119864) (resp lip120572 (119870 119864)) (7)

Note that Lip(119870 119864) (resp lip120572(119870 119864)) is a Banach space withrespect to sdot 119871 and sdot 119872 respectively If 119864 is a Banachalgebra the norm sdot 119871 is multiplicative Hence Lip(119870 119864)(resp lip120572(119870 119864)) is a (unital) Banach algebra with respect tothe norm sdot 119871 if 119864 is a (unital) Banach algebra The normsdot119872 fails to be submultiplicative even if119864 is a Banach algebraFor a metric 119889(sdot sdot) on 119870 the Holder metric is defined by 119889120572for 0 lt 120572 lt 1 Lip120572((119870 119889) 119864) is isometrically isomorphic toLip((119870 119889120572) 119864)

We are mainly concerned with 119864 = 119862(119884) in this paperThen Lip120572(119870119862(119884)) and lip120572(119870119862(119884)) are unital semisimplecommutative Banach algebras with sdot 119871 when 119864 =C119871ip(119870C) (resp lip120572(119870C)) is abbreviated to Lip(119870) (resplip120572(119870))

Let 119865 isin 119862(119870 119862(119884)) for 119870 = [0 1] or T We say that 119865is continuously differentiable if there exists 119866 isin 119862(119870 119862(119884))such that

lim119905997888rarr1199050

100381710038171003817100381710038171003817100381710038171003817119865 (1199050) minus 119865 (119905)1199050 minus 119905 minus 119866 (1199050)

100381710038171003817100381710038171003817100381710038171003817infin(119884)

= 0 (8)

Journal of Function Spaces 3

for every 1199050 isin 119870 We denote 1198651015840 = 119866 Put1198621 (119870 119862 (119884)) = 119865isin 119862 (119870119862 (119884)) 119865 is continuously differentiable (9)

Then1198621(119870119862(119884)) with norm 119865 = 119865infin + 1198651015840infin is a unitalsemisimple commutative Banach algebra If 119884 is singleton wemay suppose that 119862(119884) is isometrically isomorphic to C andwe abbreviate 1198621(119870119862(119884)) by 1198621(119870)

By identifying 119862(119870 119862(119884)) with 119862(119870times119884) we may assumethat Lip(119870119862(119884)) (resp lip120572(119870119862(119884))) is a subalgebra of119862(119870 times 119884) by the correspondence

119865 isin Lip (119870 119862 (119884)) larrrarr ((119909 119910) 997891997888rarr (119865 (119909)) (119910)) isin 119862 (119870 times 119884) (10)

Throughout the paper we may suppose that

Lip (119883119862 (119884)) sub 119862 (119883 times 119884) lip120572 (119883119862 (119884)) sub 119862 (119883 times 119884) 1198621 (119870 119862 (119884)) sub 119862 (119870 times 119884)

(11)

We say that a subset 119876 of 119862(119884) is point separating if 119876separates the points of 119884 The unit of commutative Banachalgebra 119861 is denoted by 1 The maximal ideal space of 119861 isdenoted by 119872119861 Suppose that 119861 is a unital point separatingsubalgebra of 119862(119884) equipped with a Banach algebra normThen119861 is semisimple because 119891 isin 119861 119891(119909) = 0 is amaximalideal of 119861 for every 119909 isin 119883 and the Jacobson radical of 119861vanishes

3 A Theorem of Jarosz RevisitedIsometries Preserving Unit

Whether an isometry between unital semisimple commu-tative Banach algebras is of the canonical form dependsnot only on the algebraic structures of these algebras butalso on the norms in these algebra in most cases A simpleexample is a surjective isometry on theWiener algebra whichneed not be canonical Jarosz [5] defined natural norms andprovided a theorem that isometries between a variety ofalgebras equipped with natural norms are of canonical formsFor the sake of completeness we outline the notations and theterminologies which are due to [5]The set of all norms onR2

with 119901(1 0) = 1 is denoted byP For 119901 isin P we put

119863(119901) = lim119905997888rarr+0

119901 (1 119905) minus 1119905 (12)

Recently Tanabe pointed out by a private communication that119863(119901) exists and it is finite for every 119901 isin P (In fact it iseasy to see that (119901(1 119905) minus 1)119905 is increasing since 119901(1 119905) isconvex We also see that inf 119905gt0((119901(1 119905) minus 1)119905) gt minusinfin) Let119883be a compact Hausforff space and 119860 a liner subspace of 119862(119883)which contains constant functions A seminorm lsaquo sdot lsaquo on 119860 iscalled one-invariant (in the sense of Jarosz) if lsaquo119891+ 1lsaquo = lsaquo119891lsaquofor all 119891 isin 119860 Let 119901 isin P A norm sdot on 119860 is called a 119901-norm if there is a one-invariant seminorm lsaquo sdot lsaquo on 119860 suchthat sdot = 119901( sdot infin lsaquo sdot lsaquo) A natural norm is a 119901-norm forsome 119901 isin P

Theorem 1 (Jarosz [5]) Let 119883 and 119884 be compact Hausdorffspaces let 119860 and 119861 be complex-linear subspaces of 119862(119883) and119862(119884) respectively and let 119901 119902 isin P Assume 119860 and 119861 containconstant functions and let sdot119860 sdot119861 be a 119901-norm and 119902-normon 119860 and 119861 respectively Assume next that there is a linearisometry 119879 from (119860 sdot 119860) onto (119861 sdot 119861) with 1198791 = 1 Then if119863(119901) = 119863(119902) = 0 or if 119860 and 119861 are regular subspaces of 119862(119883)and 119862(119884) respectively then 119879 is an isometry from (119860 sdot infin)onto (119861 sdot infin)

In the sequel a unital semisimple commutative Banachalgebra 119860 is identified via the Gelfand transforms with a sub-algebra of 119862(119872119860) A unital semisimple commutative Banachalgebra is regular (in the sense of Jarosz [5]) Hence we haveby a theorem of Nagasawa [32] (cf [33]) that the followingholds

Corollary 2 Let 119860 and 119861 be unital semisimple commutativeBanach algebras Assume they have natural norms respectivelySuppose that 119879 119860 997888rarr 119861 is a surjective complex-linearisometry with 1198791 = 1 Then there exists a homeomorphism120593 119872119861 997888rarr 119872119860 such that

119879 (119891) (119909) = 119891 ∘ 120593 (119909) 119891 isin 119860 119909 isin 119872119861 (13)

Proof A unital semisimple commutative Banach algebra isregular by Proposition 2 in [5]ThenTheorem 1 ensures that119879is a surjective linear isometry from (119860 sdotinfin) onto (119861 sdotinfin) Itis easy to see that 119879 is extended to a surjective linear isometry from the uniform closure 119860 of 119860 onto the uniform closure119861 of 119861 Then a theorem of Nagasawa asserts that there existsa homeomorphism 120593 119872119861 997888rarr 119872119860 such that (119891)(119909) =119891 ∘ 120593(119909) (119891 isin 119860 119909 isin 119872119861) As |119860 = 119879 we have theconclusion

Corollary 3 Let 119870119895 be a compact metric space for 119895 = 1 2Suppose that 119879 Lip(1198701) 997888rarr Lip(1198702) is a surjective complex-linear isometry with respect to the norm sdot 119871 Assume 1198791 = 1Then there exists a surjective isometry 120593 1198832 997888rarr 1198831 such that

119879119891 (119909) = 119891 ∘ 120593 (119909) 119891 isin Lip (1198701) 119909 isin 1198702 (14)

Conversely if 119879 Lip(1198701) 997888rarr Lip(1198702) is of the form as (14)then 119879 is a surjective isometry with respect to both of sdot 119872 and sdot 119871 such that 1198791 = 1

Proof As (Lip(119870119895) sdot 119871) is a unital semisimple commutativeBanach algebra with maximal ideal space 119870119895 Corollary 2asserts that there is a homeomorphism 120593 1198702 997888rarr 1198701 suchthat

119879119891 (119909) = 119891 ∘ 120593 (119909) 119891 isin Lip (1198701) 119909 isin 1198702 (15)

Then by a routine argument we see that 120593 is an isometryConverse statement is trivial

Without assuming 1198791 = 1 we have that 119879 is aweighted composition operator We exhibit a general resultas Theorem 14 (see also [30])(Lip(119870) sdot 119872) need not be a Banach algebra since sdot 119872need not be submultiplicative On the other hand sdot 119872

4 Journal of Function Spaces

is a natural norm in the sense of Jarosz (see [5]) such thatlim119905997888rarr+0((max1 119905 minus 1)119905) = 0 Then by Theorem 1 we havethe following

Corollary 4 Let 119870119895 be a compact metric space for 119895 = 1 2Suppose that 119879 Lip(1198701) 997888rarr Lip(1198702) is a surjective complex-linear isometry with respect to the norm sdot119872 Assume 1198791 = 1Then there exists a surjective isometry 120593 1198832 997888rarr 1198831 such that

119879119891 = 119891 ∘ 120593 119891 isin Lip (1198701) (16)

Conversely if 119879 Lip(1198701) 997888rarr Lip(1198702) is of a similar form as(16) then 119879 is a surjective isometry with respect to both of sdot119872and sdot 119871 such that 1198791 = 1

Proof As sdot 119872 is a natural norm we have by Corollary 2 thatthere is a homeomorphism 120593 1198702 997888rarr 1198701 such that

119879119891 (119909) = 119891 ∘ 120593 (119909) 119891 isin Lip (1198701) 119909 isin 1198702 (17)

Then by a routine argument we see that 120593 is an isometryConverse statement is trivial

Without the assumption that 1198791 = 1 in Corollary 4 onemay expect that 119879 is a weighted composition operator But itis not the case A simple counterexample is given by Weaver[7 p242] (see also [28])

As is pointed out in [34] the original proof of Theorem 1needs a revision in some part and a proof when 119860 and 119861are algebras of Lipschitz functions is revised [34 Proposition7] Although a revised proof for a general case is similar tothat of Proposition 7 in [34] we exhibit it here for the sakeof completeness of this paper To prove Theorem 1 we needLemma 2 in [5] in the sameway as the original proof of JaroszThe following is Lemma 2 in [5]

Lemma5 (Jarosz [5]) Assume119860 is a regular subspace of119862(119883)with 1 isin 119860 and let 1199090 isin 119862ℎ(119860) Then for any 120576 gt 0 and anyopen neighborhood 119880 of 1199090 there is an 119891 isin 119860 such that

10038171003817100381710038171198911003817100381710038171003817infin le 1 + 120576119891 (1199090) = 11003816100381610038161003816119891 (119909) + 11003816100381610038161003816 le 120576

119909 isin 119883 119880(18)

and |Im119891(119909)| le 120576 for all 119909 isin 119883Proof The proof is essentially due to the original proof ofLemma 2 in [5] Several minor changes are needed Weitemize them as follows

(i) Five 1205762rsquos between 11 lines and 5 lines from the bottomof page 69 read as 1205763

(ii) Next 119909 isin 1198831198801 reads as 119909 isin 1198801 on the bottom of page69

(iii) We point out that the term sum1198960minus1119895=1 (119891119895(119909) minus 1) which

appears on the first line of the first displayed inequal-ities on page 70 reads 0 if 1198960 = 1

(iv) The term 1 + 120576 on the right hand side of the secondline of the same inequalities reads as 1 + 1205763

(v) Two 1205762rsquos on the same line read as 1205763(vi) On the next line ((119899 + 1)119899)(1205762) reads as 1205763(vii) For any 1 le 1198960 le 119899 we infer that

1 ge 1 minus 21198960 minus 1119899 ge 1 minus 2119899 minus 1119899 gt minus1 (19)

Hence we have |119891(119909)| le 1 + 120576 if 119909 isin 1198801 by the firstdisplayed inequalities of page 70

(viii) The inequality 119891infin le 120576 on the fifth line on page 70reads as 119891infin le 1 + 120576

Let119870 be a nonempty convex subset of the complex planeand 120593 isin [0 2120587) Put

119888 (119870 120593) = sup 119886 isin R there is a 119887isin R with (119886 + 119894119887) 119890119894120593 isin 119870 (20)

Note that we may write

119888 (119870 120593) = sup Re120582 120582 isin 119890minus119894120593119870 (21)

Let 119860 be a subspace of 119862(119883) for a compact Hausdorff spaceFor 119891 isin 119860 we put 120590(119891) = 119891(119883) and 120590(119891) = co(119891(119883)) whereco(sdot) denotes the closed convex-hull We define the functions

119888119860 119860 times [0 2120587) 997888rarr R119888119860 (119891 120593) = 119888 ( (119891) 120593)

119903119860 119860 times R+ times [0 2120587) 997888rarr R

+119903119860 (119891 119905 120593) = 10038171003817100381710038171003817119891 + 119890119894120593119905110038171003817100381710038171003817infin

(22)

Proof of Theorem 1 Let 119891 isin 119860 First we note that10038171003817100381710038171198911003817100381710038171003817infin = sup 119911 isin 120590 (119891) = sup 119911 isin (119891) (23)

since (119891) is the closed convex-hull of a compact set 120590(119891) =119891(119883) We prove the inequalities

119905 + 119888119860 (119891 120593) le 119903119860 (119891 119905 120593)le radic(119905 + 119888119860 (119891 120593))2 + 100381710038171003817100381711989110038171003817100381710038172infin

119905 ge 0 120593 isin [0 2120587) (24)

which appear on p 68 in [5] Put 119904 = 119888119860(119891 120593) As 120590(119891) iscompact there exists 119887 isin R such that (119904 + 119894119887)119890119894120593 isin 120590(119891)Hence

119890119894120593 (119905 + 119904 + 119894119887) isin 119890119894120593119905 + 120590 (119891) = 120590 (1198901198941205931199051 + 119891) (25)

Journal of Function Spaces 5

As

119905 + 119904 le |119905 + 119904| le |119905 + 119904 + 119894119887| = 10038161003816100381610038161003816119890119894120593 (119905 + 119904 + 119894119887)10038161003816100381610038161003816le 100381710038171003817100381710038171198901198941205931199051 + 11989110038171003817100381710038171003817infin = 119903119860 (119891 119905 120593) (26)

we have

119905 + 119888119860 (119891 120593) le 119903119860 (119891 119905 120593) (27)

Let 119909 isin 119883 By the definition of 119904 = 119888119860(119891 120593) we infer thatRe(119890minus119894120593119891(119909)) le 119904 hence we have Re(119905 + 119890minus119894120593119891(119909)) le 119905 + 119904 forevery 119905 ge 0 Then

Re 119890minus119894120593119891 (119909) le Re (119905 + 119890minus119894120593119891 (119909)) le 119905 + 119904 119905 ge 0 (28)

Letting119872 = max|119905 + 119904| |Re(119890minus119894120593119891(119909))| we have10038161003816100381610038161003816119890119894120593119905 + 119891 (119909)100381610038161003816100381610038162 = 10038161003816100381610038161003816119905 + 119890minus119894120593119891 (119909)100381610038161003816100381610038162

= (Re (119905 + 119890minus119894120593119891 (119909)))2+ (Im (119905 + 119890minus119894120593119891 (119909)))2

le 1198722 + (Im (119905 + 119890minus119894120593119891 (119909)))2le |119905 + 119904|2 + (Re (119890minus119894120593119891 (119909)))2+ (Im (119890minus119894120593119891 (119909)))2

= |119905 + 119904|2 + 10038161003816100381610038161003816119890minus119894120593119891 (119909)100381610038161003816100381610038162

(29)

As 119909 isin 119883 is arbitrary we have

119903119860 (119891 119905 120593) = 10038171003817100381710038171003817119905 + 119890minus11989412059311989110038171003817100381710038171003817infin le radic|119905 + 119904|2 + 100381710038171003817100381711989110038171003817100381710038172infin= radic(119905 + 119888119860 (119891 120593))2 + 100381710038171003817100381711989110038171003817100381710038172infin

(30)

It follows that (24) holds In the same way we have

119905 + 119888119861 (119892 120593) le 119903119861 (119892 119905 120593) le radic(119905 + 119888119861 (119892 120593))2 + 100381710038171003817100381711989210038171003817100381710038172infin119905 ge 0 120593 isin [0 2120587) (31)

for every 119892 isin 119861 By (24) and (31) we infer that

lim119905997888rarrinfin

(119903119860 (119891 119905 120593) minus 119905) = 119888119860 (119891 120593) lim119905997888rarrinfin

(119903119861 (119879119891 119905 120593) minus 119905) = 119888119861 (119879119891 120593) (32)

As lsaquo sdot lsaquo119860 is 1-invariant we have119901 (119903119860 (119891 119905 120593) lsaquo119891lsaquo119860)= 119901 (10038171003817100381710038171003817119891 + 119890119894120593119905110038171003817100381710038171003817infin lsaquo119891 + 1198901198941205931199051lsaquo119860) (33)

As 119879 is an isometry 119879(1) = 1 and lsaquo sdot lsaquo119861 is 1-invariant wehave

119901 (10038171003817100381710038171003817119891 + 119890119894120593119905110038171003817100381710038171003817infin lsaquo119891 + 1198901198941205931199051lsaquo119860)= 119902 (10038171003817100381710038171003817119879 (119891 + 1198901198941205931199051)10038171003817100381710038171003817infin lsaquo119879 (119891 + 1198901198941205931199051)lsaquo119861)= 119902 (10038171003817100381710038171003817119879 (119891) + 119890119894120593119905110038171003817100381710038171003817infin lsaquo119879 (119891) + 1198901198941205931199051lsaquo119861)= 119902 (119903119861 (119879119891 119905 120593) lsaquo119879119891lsaquo119861)

(34)

Thus

119901 (119903119860 (119891 119905 120593) lsaquo119891lsaquo119860) = 119902 (119903119861 (119879119891 119905 120593) lsaquo119879119891lsaquo119861) (35)

It follows that0 = lim

119905997888rarrinfin(119901 (119903119860 (119891 119905 120593) lsaquo119891lsaquo119860)

minus 119902 (119903119861 (119879119891 119905 120593) lsaquo119879119891lsaquo119861))= lim

119905997888rarrinfin(119903119860 (119891 119905 120593) 119901(1 lsaquo119891lsaquo119860119903119860 (119891 119905 120593))

minus 119903119860 (119891 119905 120593)) + lim119905997888rarrinfin

(119903119861 (119879119891 119905 120593)minus 119903119861 (119879119891 119905 120593) 119902(1 lsaquo119879119891lsaquo119861119903119861 (119879119891 119905 120593)))+ lim

119905997888rarrinfin(119903119860 (119891 119905 120593) minus 119905 minus (119903119861 (119879119891 119905 120593) minus 119905))

= lsaquo119891lsaquo119860119863(119901) minus lsaquo119879119891lsaquo119861119863(119902) + 119888119860 (119891 120593)minus 119888119861 (119879119891 120593)

(36)

Recall that 119863(119901) = lim119905997888rarr+0(119901(1 119905) minus 1)119905 and 119863(119902) =lim119905997888rarr+0(119902(1 119905) minus 1)119905 It follows thatlsaquo119891lsaquo119860119863 (119901) minus lsaquo119879119891lsaquo119861119863(119902) = 119888119861 (119879119891 120593) minus 119888119860 (119891 120593)

119891 isin 119860 120593 isin [0 2120587) (37)

Suppose that 119863(119901) = 119863(119902) = 0 Then we have by (37)that 119888119861(119879119891 120593) = 119888119860(119891 120593) for every 119891 isin 119860 and 120593 isin [0 2120587)By Lemma 1 in [5] we infer that 120590(119879119891) = 120590(119891) Thus we have119879119891infin = 119891infin We have proved that 119879 is an isometry from(119860 sdot infin) onto (119861 sdot infin) if119863(119901) = 119863(119902) = 0

Suppose that 119860 and 119861 are regular subspaces of 119862(119883) and119862(119884) respectively Let 119891 isin 119860 PutΔ119891 = lsaquo119891lsaquo119860119863(119901) minus lsaquo119879119891lsaquo119861119863(119902) (38)

Suppose that Δ119891 ge 0 For any 119903 ge 0 and any nonemptycompact convex subset 119870 sub C we have that

119888 (119870 + 119870 (119903) 120593) = 119888 (119870 120593) + 119903 (39)

for all 120593 isin [0 2120587) where 119870(119903) = 119911 isin C |119911| le 119903 Then by(37) we have

119888 (120590 (119879119891) 120593) = 119888119861 (119879119891 120593) = 119888119860 (119891 120593) + Δ119891= 119888 (120590 (119891) 120593) + Δ119891= 119888 (120590 (119891) + 119870 (Δ119891) 120593)

(40)

6 Journal of Function Spaces

for all 120593 isin [0 2120587) It follows by Lemma 1 in [5] that

(119879119891) = 120590 (119891) + 119870 (Δ119891) (41)

and therefore10038171003817100381710038171198791198911003817100381710038171003817infin = 10038171003817100381710038171198911003817100381710038171003817infin + Δ119891 (42)

If Δ119891 le 0 then a similar calculation shows that

(119891) = (119879119891) + 119870 (minusΔ119891) (43)

and10038171003817100381710038171198911003817100381710038171003817infin = 10038171003817100381710038171198791198911003817100381710038171003817infin + (minusΔ119891) (44)

It follows that in any case (Δ119891 ge 0 Δ119891 le 0) we obtainΔ119891 = 10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin 119891 isin 119860 (45)

We will prove that1003817100381710038171003817119879 (119891)1003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin = Δ119891 ge 0 (46)

for all 119891 isin 119860 Once it is proved applying the same argumentfor119879minus1 instead of119879 we see that 119879minus1119892infinminus119892infin ge 0 for every119892 isin 119861 As 119879 is a bijection it follows that 119891infin minus 119879119891infin ge 0for every 119891 isin 119860 It will follow that 119879119891infin = 119891infin for every119891 isin 119860 A proof of (46) is the following For every 120576 gt 0denote

A120576 = 119891 isin 119860 120588 (120590 (119891)) le 120576 (47)

The inequality in (46) is deduced by the following assertionswhich appear in the proof of [5 Theorem]

(1) 119879 is a continuous mapping from (119860 sdot infin) onto (119861 sdotinfin)(2) For each 120576 gt 0 the setA120576 is dense in (119860 sdot infin)(3) For each 120576 gt 0 and each 119891 isin A120576 it holds that119879(119891)infin ge 119891infin minus 120576

Suppose that these assertions are proved Let 119891 isin 119860 By (2)for any 120576 gt 0 there is a sequence 119891119899 of functions inA120576 suchthat 119891119899 minus 119891infin 997888rarr 0 as 119899 997888rarr infin By (3) we have

10038171003817100381710038171198791198911198991003817100381710038171003817infin minus 10038171003817100381710038171198911198991003817100381710038171003817infin ge minus120576 (48)

for every 119899 Letting 119899 997888rarr infin we have10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin ge minus120576 (49)

by (1) As 120576 gt 0 is arbitrary we have that10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin ge 0 (50)

We show proofs of three assertions (1) (2) and (3) aboveprecisely The proof of (1) is slightly different from the corre-sponding one in [5 p 70] This change is rather ambitiousWe also point out that the terms minus1205872 and 1205872 which appearin the formulae (7) and (8) in [5] seem inappropriate theyread for example as 31205874 and 1205874 respectively

We now proceed to prove the first statement Aiming fora contradiction suppose that 119879 is not continuous from (119860 sdotinfin) to (119861 sdot infin) Let 120576 be a positive real number less than1100 Then there is a function 1198910 isin 119860 such that 1198910infin le 120576and 119879(1198910)infin = 1 Then there exist 1199100 isin Ch(119861) such that|119879(1198910)(1199100)| = 1 by [29 Proposition 63] Since 119879 is complex-linear we may suppose that 119879(1198910)(1199100) = 1

By (41) and (45) we deduce that Δ1198910 = 119879(1198910)infin minus1198910infin ge 1 minus 120576 and 120590(119879(1198910)) = 120590(1198910) + 119870(Δ1198910) As 1198910infin le 120576and 119879(1198910)infin = 1 we infer that 120590(1198910) sub 119870(120576) and (119879(1198910) sub119870(1) Let 119911 isin 119870(1 minus 2120576) Let 119909 isin 119883 Then |1198910(119909)| le 120576 assertsthat 119908 = 119911 minus 1198910(119909) isin 119870(1 minus 120576) sub 119870(Δ1198910) Thus

119911 = 1198910 (119909) + 119908 isin 120590 (1198910) + 119870 (Δ1198910) = 120590 (119879 (1198910)) (51)

Hence119870(1 minus 2120576) sub 120590(1198791198910) As 119879(1198910)infin = 1 we have119870(1 minus 2120576) sub 120590 (119879 (1198910)) sub 119870 (1) (52)

Consider the open neighborhood 1198800 of 1199100 in1198832 given by

1198800 = 119910 isin 1198832 1003816100381610038161003816119879 (1198910) (119910) minus 11003816100381610038161003816 lt 120576 (53)

We infer that 1198800 is a proper subset of1198832 by (52) Then by [5Lemma 2] there exists 119892 isin 119861 such that 119892infin le 1 + 120576 119892(1199100) =1 |119892(119910) + 1| lt 120576 for every 119910 isin 1198832 1198800 and |Im119892(119910)| lt 120576 forall 119910 isin 1198832 If 119867 denotes the closed rectangle whose verticesare the four points plusmn(1 + 120576) plusmn 120576119894 we have

120590 (119892) sub 119867 (54)

Consider now the set

119871 = 11989031205871198944119911 |119911| le 1 Re 119911 ge 1 minus 2120576 (55)

We claim that119879(1198910)(1198832)cap119871 = 0 Suppose that119879(1198910)(1198832)cap119871 =0 As 119879(1198910)(1198832) is compact there exists a positive integer119899 such that 119879(1198910)(1198832) cap 119871119899 = 0 where 119871119899 = 11989031205871198944119911 |119911| le 1Re 119911 gt 1 minus 2120576 minus 1119899 Then (52) gives 119879(1198910)(1198832) sub119870(1)119871119899 As (119879(1198910)) is the closed convex-hull of 119879(1198910)(1198832)it is contained in the closed convex set 119870(1) 119871119899 On theother hand (1 minus 2120576)11989031205871198944 isin 119870(1 minus 2120576) sub 120590(119879(1198910)) by (52) As(1minus2120576)11989031205871198944 isin 119871 sub 119871119899 this contradicts 120590(119879(1198910)) sub 119870(1) 119871119899and this proves our claim Hence there is 1199101 isin 1198832 with119879(1198910)(1199101) isin 119871 As 120576 le 1100 it follows that |119879(1198910)(1199101) minus1| ge 120576 and so 1199101 isin 1198832 1198800 Hence |119892(1199101) + 1| lt 120576 Thus119892(1199101) + 119879(1198910)(1199101) is in 119871 minus 1 + K(120576) Thus we have

1 + radic22 minus 3120576 le 119888119861 (119892 + 119879 (1198910) 31205874 ) (56)

We claim that

120590 (119892 + 119879 (1198910)) sub co (119870 (minus1 1) cup 2) + 119870 (3120576) (57)

where119870(1199110 119903) = 119911 isin C |119911 minus 1199110| le 119903 Let 119910 isin 1198832 Supposefirst that |119879(1198910)(119910)minus1| lt 120576 Since 119892(1198832) sub 119867 by (54) we have

119879 (1198910) (119910) + 119892 (119910) isin 119870 (1 120576) + 119867 = (119867 + 1) + 119870 (120576) (58)

Journal of Function Spaces 7

Suppose next that |119879(1198910)(119910)minus1| ge 120576Then 119910 isin 1198832 1198800 and so|119892(119910) + 1| lt 120576 Moreover |119879(1198910)(119910)| le 1 Therefore we have

119892 (119910) + 119879 (1198910) (119910) isin 119870 (1) minus 1 + 119870 (120576)= 119870 (minus1 1) + 119870 (120576) (59)

It follows from (58) and (59) that

(119892 + 119879 (1198910)) (1198832) sub 119870 (minus1 1) cup ((119867 + 1)) + 119870 (120576) (60)

and hence

120590 (119892 + 119879 (1198910)) sub co (119870 (minus1 1) cup 2) + 119870 (3120576) (61)

as is claimed Therefore we have

119888119861 (119892 + 119879 (1198910) 1205874 ) le radic2 + 3120576 (62)

Put 1198911 = 119879minus1(119892) We claim that Δ1198911 le 120576 If Δ1198911 lt 0 thereis nothing to prove Suppose that Δ1198911 ge 0 Then by (41) wehave

120590 (119892) = 120590 (1198911) + 119870 (Δ1198911) (63)

Since 120590(119892) sub 119867 by (54) we have

119867 sup 120590 (1198911) + 119870 (Δ1198911) (64)

As 119867 does not include a closed disk with the radius greaterthan 120576 we conclude that Δ1198911 le 120576

In the following we will consider two cases 0 le Δ1198911 le 120576and Δ1198911 le 0 Suppose first that 0 le Δ1198911 le 120576 Then (64) yields(1198911) sub 119867 From 1198910infin le 120576 we deduce that 120590(1198910) sub 119870(120576)Hence we have

120590 (1198911 + 1198910) sub 120590 (1198911) + 120590 (1198910) sub (1198910) + 120590 (1198910)sub 119867 + 119870 (120576) (65)

Since119867 +119870(120576) is convex we have (1198911 + 1198910) sub 119867 + 119870 (120576) (66)

From (39) we infer that

119888119860 (1198911 + 1198910 31205874 ) = 119888 (120590 (1198911 + 1198910) 31205874 )le 119888 (119867 + 119870 (120576) 31205874 )= 119888 (119867 31205874 ) + 120576= radic22 + (1 + radic2) 120576

(67)

Since 119879(1198911 + 1198910) = 119892 + 119879(1198910) from (56) and (67) we obtainthat

1 minus (4 + radic2) 120576 le 119888119861 (119879 (1198911 + 1198910) 31205874 )minus 119888119860 (1198911 + 1198910 31205874 )

(68)

By (63) and 1 = 119892(1199100) we deduce that 1 isin 120590(1198911) + 119870(Δ1198911)Thus there is 119911 isin 120590(1198911) such that |119911 minus 1| le Δ1198911 It follows thatradic22 minus Δ1198911 le 119888119860(1198911 1205874) hence we have

radic22 minus 2120576 le 119888119860 (1198911 + 1198910 1205874 ) (69)

as 1198910infin le 120576 and 0 le Δ1198911 le 120576 We get by (62) and (69) that

119888119861 (119879 (1198911 + 1198910) 1205874 ) minus 119888119860 (1198911 + 1198910 1205874 ) leradic22 + 5120576 (70)

On the other hand 119888119861(119879(119891) 120593)minus119888119860(119891 120593) is invariant for any 120593by (37) From (68) and (70)we deduce that 120576 ge (2minusradic2)2(9+radic2) and this contradicts that 120576 le 1100

For the second case suppose next that Δ1198911 le 0 Then by(43) we have

120590 (1198911) = 120590 (119892) + 119870 (minusΔ1198911) (71)

and by (54) it follows that (1198911) sub 119867 + 119870(minusΔ1198911) Moreover(1198910) sub 119870(120576) since 1198910infin le 120576 Then

120590 (1198911 + 1198910) sub 120590 (1198911) + 120590 (1198910) sub 120590 (1198911) + 120590 (1198910)sub 119867 + 119870(minusΔ1198911) + 119870 (120576) (72)

Hence (1198911+1198910) sub 119867+119870(minusΔ1198911)+119870(120576) Using (39) we inferthat

119888119860 (1198911 + 1198910 31205874 ) le 119888 (119867 + 119870(minusΔ1198911) + 119870 (120576) 31205874 )= 119888 (119867 31205874 ) + (minusΔ1198911) + 120576= radic22 + (1 + radic2) 120576 + (minusΔ1198911)

(73)

By (71) we obtain that 120590(1198911) sup 119892(1198832) + 119870(minusΔ1198911) and as119892(1199100) = 1 we infer that 120590(1198911) sup 1 + 119870(minusΔ1198911) Hence radic22 +(minusΔ1198911) le 119888119860(1198911 1205874) so thatradic22 + (minusΔ1198911) minus 120576 le 119888119860 (1198911 + 1198910 1205874 ) (74)

as 1198910infin le 120576 Since 119879(1198911 +1198910) = 119892 +119879(1198910) we obtain by (56)and (73) that

1 minus (4 + radic2) 120576 minus (minusΔ1198911)le 119888119861 (119879 (1198911 + 1198910) 31205874 ) minus 119888119860 (1198911 + 1198910 31205874 )

(75)

We also obtain by (62) and (74) that

119888119861 (119879 (1198911 + 1198910) 1205874 ) minus 119888119860 (1198911 + 1198910 1205874 )le radic22 + 4120576 minus (minusΔ1198911)

(76)

Since 119888119861(119879(119891) 120593)minus119888119860(119891 120593) is invariant for any 120593 by (37) from(75) and (76) we deduce that 120576 ge (2 minusradic2)2(8 +radic2) and thisis impossible since 120576 le 1100

8 Journal of Function Spaces

Next we show a proof of the second assertion (2) Let119891 isin 119860 We prove that there exists a sequence 119891119899 sub 119860 whichuniformly converges to 119891 such that 120588(120590(119891119899)) 997888rarr 0 as 119899 997888rarrinfin Without loss of generality we may assume that 119891infin = 1Then there exists 1199100 isin Ch(119860) such that |119891(1199100)| = 119891infinby [29 Proposition 63] We may assume that 119891(1199100) = 1Suppose that 119899 ge 4 Put

119880119899 = 119909 isin 119883 1003816100381610038161003816119891 (119909) minus 11003816100381610038161003816 le 11198992 (77)

and

Ω = co(119870 (1) cup 1 + 1119899) + 119870( 11198992) (78)

(In the following we identify R2 and C that is we identify(119909 119910) and 119909 + 119894119910 for every 119909 119910 isin R) Since we assume that119899 ge 4 we infer by a simple calculation that

[1 minus 11198992 minus 120575119899119899 1 + 1119899 + 120575119899119899 ] times [minus 11198992 minus 120575119899119899 11198992 + 120575119899119899 ]sub Ω

(79)

for 120575119899 with 0 lt 120575119899 le 11198992 We assume that 0 lt 120575119899 le 11198992 By[5 Lemma 2] there exists ℎ119899 isin 119860 such that ℎ119899infin le 1 + 2120575119899ℎ119899(1199100) = 1 |ℎ119899 + 1| le 2120575119899 on119883 119880119899 and |Im ℎ119899| le 2120575119899 on119883Put 119892119899 = (ℎ119899 + 1)2 Then 119892119899 isin 119860 and

10038171003817100381710038171198921198991003817100381710038171003817infin le 1 + 120575119899119892119899 (1199100) = 110038161003816100381610038161198921198991003816100381610038161003816 le 120575119899 on 119883 1198801198991003816100381610038161003816Im1198921198991003816100381610038161003816 le 120575119899 on 119883

minus120575119899 le Re119892119899 le 1 + 120575119899 on 119883

(80)

Let 119909 isin 119880119899 Then we have

1 minus 11198992 le Re119891 (119909) le 1minus 11198992 le Im119891 (119909) le 11198992

(81)

and

minus120575119899119899 le Re119892119899 (119909)119899 le 1119899 + 120575119899119899

minus120575119899119899 le Im119892119899 (119909)119899 le 120575119899119899

(82)

Hence

1 minus 11198992 minus 120575119899119899 le Re(119891 (119909) + 119892119899 (119909)119899 ) le 1 + 1119899 + 120575119899119899 minus 11198992 minus 120575119899119899 le Im(119891 (119909) + 119892119899 (119909)119899 ) le 11198992 + 120575119899119899

(83)

It follows that we have

(119891 + 119892119899119899 ) (119909) isin [1 minus 11198992 minus 120575119899119899 1 + 1119899 + 120575119899119899 ]times [minus 11198992 minus 120575119899119899 11198992 + 120575119899119899 ] sub Ω

(84)

for 119909 isin 119880119899 Suppose that 119909 isin 119883 119880119899 Then10038161003816100381610038161003816100381610038161003816119891 (119909) +

119892119899 (119909)11989910038161003816100381610038161003816100381610038161003816 le 1 +

120575119899119899 le 1 + 11198993 lt 1 + 11198992 (85)

and hence

119891 (119909) + 119892119899 (119909)119899 isin 119870 (1) + 119870( 11198992) sub Ω (86)

for 119909 isin 119883 119880119899 Since 1 + 1119899 = 119891(1199100) + 119892119899(1199100)119899 we have bycombining (84) and (86) that

1 + 1119899 isin (119891 + 119892119899119899 ) (119883) sub Ω (87)

AsΩ is convex we obtain

1 + 1119899 isin 120590 (119891 + 119892119899119899 ) sub Ω (88)

Recall that for 119903 ge 0 and a complex number 1199110119870 (1199110 119903) = 119908 isin C 1003816100381610038161003816119908 minus 11991101003816100381610038161003816 le 119903 (89)

denotes the closed disk with center 1199110 and radius 119903 Weobserve that 120588(Ω 1 + 1119899) = 1119899 + 11198992 Recall that

120588(Ω 1 + 1119899)= sup119903 119911 isin Ω 1 + 1119899 isin 119870 (119911 119903) sub Ω

(90)

Let ℓ1 be the line defined by the equation

119910 = minus119899radic2119899 + 1 (119909 minus (1 +1119899 + 11198992 + 11198993)) (91)

part of which is a part of the boundary ofΩ Let ℓ2 be the linedefined by the equation

119910 = minus119899radic2119899 + 1 (119909 minus (1 minus11198992)) (92)

By some calculation we have that the distance between 1 minus11198992 and 1+1119899 is 1119899+11198992 and it coincides with the distancebetween the point 1 minus 11198992 and the line ℓ1 Hence we see that

1 + 1119899 isin 119870(1 minus 11198992 1119899 + 11198992 ) sub Ω (93)

Thus 1119899 + 11198992 le 120588(Ω 1 + 1119899)Next we prove that 119870(119901 119903) cap Ω119888 = 0 for every 119901 isin Ω

with 119903 = |119901 minus (1 + 1119899)| gt 1119899 + 11198992 It will follow that1119899 + 11198992 ge 120588(Ω 1 + 1119899) and the equality will hold Let

Journal of Function Spaces 9

119901 = (119909119901 119910119901) = 119909119901 + 119894119910119901 isin Ω with 119903 = |119901 minus (1 + 1119899)| gt1119899 + 11198992 We prove the case where 119910119901 ge 0 A proof for thecase where 119910119901 le 0 is the same and we omit it We divideΩ+ =119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902 isin Ω 119910119902 ge 0 into two parts

Ω1 = 119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902isin Ω+ minus119899radic2119899 + 1 (119909119902 minus (1 minus

11198992)) le 119910119902le minus119899radic2119899 + 1 (119909 minus (1 +

1119899 + 11198992 + 11198993))(94)

and

Ω2 = 119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902 isin Ω+ 119910119902le minus119899radic2119899 + 1 (119909119902 minus (1 minus

11198992)) (95)

Suppose that 119901 isin Ω1 and 119903 = |119901minus (1+1119899)| gt 1119899+11198992Thedistance between ℓ1 and ℓ2 is 1119899+11198992 Hence119870(119901 119903)capΩ119888 =0 Suppose that 119901 isin Ω2 and 119903 = |119901 minus (1 + 1119899)| gt 1119899 + 11198992that is119901 = 1minus11198992 Let ℓ1015840 be the line passing through119901whichis parallel to ℓ2 Let 1199011015840 be the unique point in the intersectionof ℓ1015840 and the 119909-axis Then 1199011015840 = 1 minus 11198992 minus 119906 for some 119906 ge 0Then the distance between ℓ1 and ℓ1015840 is 1119899+11198992+119906(1+1119899)which is equal to the distance between the point119901 and the lineℓ1 On the other hand1003816100381610038161003816100381610038161003816119901 minus (1 + 1119899)

1003816100381610038161003816100381610038161003816 ge10038161003816100381610038161003816100381610038161199011015840 minus (1 + 1119899)

1003816100381610038161003816100381610038161003816 = 1119899 + 11198992 + 119906 (96)

It follows that 119870(119901 119903) cap Ω119888 = 0 We conclude that if 119901 isin Ωsatisfies 119903 = |119901minus(1+1119899)| gt 1119899+11198992 then119870(119901 119903)capΩ119888 = 0Thus we have

120588 (Ω 1 + 1119899) = 1119899 + 11198992 (97)

Since 1 + 1119899 isin 120590(119891 + 119892119899119899) sub Ω we have120588 (120590 (119891 + 119892119899119899 )) le 120588 (120590 (119891 + 119892119899119899 ) 1 + 1119899)

le 120588 (Ω 1 + 1119899) = 1119899 + 11198992 (98)

On the other hand 119892119899119899infin 997888rarr 0 as 119899 997888rarr infin ensures that119891 + 119892119899119899 minus 119891infin 997888rarr 0 as 119899 997888rarr infin It follows that for every120576 gt 0A120576 is dense in (119860 sdot infin)Finally we show a proof of the third assertion (3) As is

pointed out in the proof of [5Theorem] 120588(co(119870)+119870(119888)) ge 119888for any 119870 sub C and any 119888 ge 0 Let 120576 gt 0 and 119891 isin A120576 Supposethat Δ119891 le 0 Then by (43) we have (119891) = (119879119891) + 119870(minusΔ119891)Hence we have

120588 ( (119891)) = 120588 (120590 (119879119891) + 119870 (minusΔ119891)) ge minusΔ119891 (99)

As 120576 ge 120588(120590(119891)) we conclude by (45) that10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin = Δ119891 ge minus120576 (100)

This completes the proof of the theorem

4 Hermitian Operators on a BanachAlgebras of Continuous Maps WhoseValues Are in a Uniform Algebras

Let 119883 and 119884 be compact Hausdorff spaces Let 119861 be aunital subalgebra of 119862(119883) which separates the points of 119883Throughout this sectionwe assume119861 is a Banach algebrawiththe norm sdot 119861 and 119860 is a uniform algebra on 119884 Recall thata uniform algebra on 119884 is a uniformly closed subalgebra of119862(119884) which contains constants and separates the points of 119884For functions 119891 isin 119862(119883) and 119892 isin 119862(119884) let 119891 otimes 119892 isin 119862(119883 times 119884)be the function defined by119891otimes119892(119909 119910) = 119891(119909)119892(119910) for (119909 119910) isin119883 times 119884 and for a subspace 119864119883 of 119862(119883) and a subspace 119864119884 of119862(119884) put

119864119883 otimes 119864119884

= 119899sum

119895=1

119891119895 otimes 119892119895 119899 isin N 119891119895 isin 119864119883 119892119895 isin 119864119884

(101)

and

1 otimes 119864119884 = 1 otimes 119892 119892 isin 119864119884 (102)

Throughout the section 119861 is a unital subalgebra of 119862(119883 times 119884)with a Banach algebra norm sdot 119861 We assume that 119861otimes119860 sub 119861Note that 119861 separates the points of119883times119884 since119860 separates thepoints of 119884 and 119861 separates the points of 119883 We assume thatthere exists a compact Hausdorff space M and a complex-linear map 119863 119861 997888rarr 119862(M) such that ker119863 = 1 otimes 119860 Weassume that 119865119861 = 119865infin(119883times119884)+119863(119865)infin(M) for every 119865 isin 119861Hence119863 is continuous Defining

lsaquo119865lsaquo = 119863 (119865)infin(M) 119865 isin 119861 (103)

lsaquo sdot lsaquo is a one-invariant seminorm in the sense of Jarosz lsaquo sdot lsaquois a seminorm on 119861 such that lsaquo119865+ 1lsaquo = lsaquo119865lsaquo for every 119865 isin 119861Hence the norm sdot 119861 is a natural norm (see [5 p67]) Notethat 119861 is a regular subspace of 119862(119883times119884) in the sense of Jarosz[5 Proposition 2]

Lumerrsquos seminal paper [35] opened up a useful methodof finding isometries which is often referred to as Lumerrsquosmethod It involves the notion ofHermitian operators and thefact that 119880119867119880minus1 must be Hermitian if119867 is Hermitian and 119880is a surjective isometry

Definition 6 Let A be a unital Banach algebra We say that119890 isin A is a Hermitian element if1003817100381710038171003817exp (119894119905119890)1003817100381710038171003817A = 1 (104)

for every 119905 isin R The set of all Hermitian elements of A isdenoted by119867(A)

If A is a unital 119862lowast-algebra then 119867(A) is the set of allself-adjoint elements of A Hence 119867(119872119899(C)) is the set of allHermitian matrices and119867(119862(119884)) = 119862R(119884)Definition 7 Let 119864 be a complex Banach space The Banachalgebra of all bounded operators on 119864 is denoted by 119861(119864) Wesay that 119879 isin 119861(119864) is a Hermitian operator if 119879 isin 119867(119861(119864))

10 Journal of Function Spaces

Note that a Hermitian element of a unital Banach algebraand a Hermitian operator are usually defined in terms ofnumerical range or semi-inner product Here we define themby an equivalent form (see [36]) By the definition of aHermitian operator we have the following

Proposition 8 Let 119864119895 be a complex Banach space for 119895 = 1 2Suppose that 119881 1198641 997888rarr 1198642 is a surjective isometry and 119867 1198641 997888rarr 1198641 is a Hermitian operator Then 119881119867119881minus1 1198642 997888rarr 1198642

is a Hermitian operator

Proposition 9 An element 119865 isin 119861 is Hermitian if and only ifthere exists 119891 isin 119860 cap 119862R(119884) such that 119865 = 1 otimes 119891Proof Suppose that 119865 isin 119861 is a Hermitian element Then1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884) + 1003817100381710038171003817119863 (exp (119894119905119865))1003817100381710038171003817infin(M)

= 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817 = 1 (105)

for every 119905 isin R Suppose that there exists a point (119909 119910) isin119883 times 119884 with Im119865(119909 119910) = 0 where Im denotes the imaginarypart of a complex number Suppose that Im119865(119909 119910) gt 0Then1003817100381710038171003817exp (minus119894119865)1003817100381710038171003817infin(119883times119884) ge 1003816100381610038161003816exp (minus119894119865 (119909 119910))1003816100381610038161003816

= exp (Im119865 (119909 119910)) gt 1 (106)

Suppose that Im119865(119909 119910) lt 0 Then1003817100381710038171003817exp (119894119865)1003817100381710038171003817infin(119883times119884) ge 1003816100381610038161003816exp (119894119865 (119909 119910))1003816100381610038161003816= exp (minusIm119865 (119909 119910)) gt 1 (107)

In any case we have there exists 119905 isin R such that1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817119861 ge 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884) gt 1 (108)

which contradicts our assumption We have that

119865 isin 119862R (119883 times 119884) (109)

Thus for every (119904 119905) isin 119883 times 119884 and 119905 isin R |exp(119894119905119865(119904 119905))| = 1Hence exp(119894119905119865)infin(119883times119884) = 1 for every 119905 isin R By (105)we have119863(exp(119894119905119865)infin(M) = 0 which ensures that 119863(exp(119894119905119865)) = 0for every 119905 isin R Thus exp(119894119905119865) isin 1 otimes 119860 for every 119905 isin R Wehave

exp (119894119905119865) minus 1119905 minus 119894119865 = infinsum119899=2

((119894119865)119899 119905119899minus2119899 ) 119905 (110)

and hence for every 119905 isin R with |119905| le 1 we have10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817 le (

infinsum119899=2

119865119899119861|119905|119899minus2119899 ) |119905|

le (infinsum119899=2

119865119899119861119899 ) |119905| le (exp 119865119861) |119905|

(111)

It follows that10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817infin(119883times119884)

le 10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817119861

997888rarr 0(112)

as 119905 997888rarr 0 Since (exp(119894119905119865)minus1)119905 isin 1otimes119860 for each 119905 isin R thereexists 119892119905 isin 119860 such that

exp (119894119905119865) minus 1119905 = 1 otimes 119892119905 (113)

By (112) we have10038171003817100381710038171 otimes 1198921119899 minus 1198941198651003817100381710038171003817infin(119883times119884)

997888rarr 0 (114)

as 119899 997888rarr infin We have that 1 otimes 1198921119899 is a Cauchy sequencein 119862(119883 times 119884) thus we infer that1198921119899 is a Cauchy sequence in119862(119884) Since 119860 is uniformly closed as it is a uniform algebrathere exists 119892 isin 119860 such that

10038171003817100381710038171198921119899 minus 1198921003817100381710038171003817infin(119884)997888rarr 0 (115)

and hence10038171003817100381710038171 otimes 1198921119899 minus 1 otimes 1198921003817100381710038171003817infin(119883times119884)

997888rarr 0 (116)

as 119899 997888rarr infin It follows by (114) that 119894119865 = 1 otimes 119892 thus119865 = 1 otimes (minus119894119892) isin 1 otimes 119860 (117)

By (109) we see that minus119894119892 isin 119862R(119884) thus we have 119891 = minus119894119892 isin119860 cap 119862R(119884) and 119865 = 1 otimes 119891Suppose conversely that 119861 ni 119865 = 1otimes119891 for119891 isin 119860cap119862R(119884)

We infer that 119865 isin 119862R(119883 times 119884) and |exp(119894119905(119865(119909 119910))| = 1 forevery 119905 isin R and (119909 119910) isin 119883 times 119884 Hence exp(119894119905119865)infin(119883times119884) = 1for every 119905 isin R Since

exp (119894119905119865) = exp (119894119905 (1 otimes 119891)) = 1 otimes exp (119894119905119891) isin 1 otimes 119860 (118)

we have 119863(exp(119894119905119865)) = 0 It follows that1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817 = 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884)

+ 1003817100381710038171003817119863 (exp (119894119905119865))1003817100381710038171003817infin(M) = 1 (119)

for every 119905 isin R We conclude that 119865 is a Hermitian elementin 119861

Note that 119891 isin 119860 is Hermitian if and only if119891 isin 119860cap119862R(119884)by [37 Proposition 5] Hence Proposition 9 asserts that 119865 is aHermitian element in119861 if and only if119865 = 1otimes119891 for aHermitianelement 119891 in 119860Proposition 10 Suppose that 119880 119861 997888rarr 119861 is a surjectiveunital isometry Then 119880 is an algebra isomorphism

Proof As we have already mentioned 119861 is a regular subspace(in the sense of Jarosz) with a natural norm Then byTheorem 1 119880 is also an isometry with respect to the supre-mum norm on 119883 times 119884 Then 119880 is uniquely extended to asurjective isometry with respect to the supremum norm from the uniform closure 119861 onto itself Since 119861 is a uniformalgebra a theorem of Nagasawa [32] asserts that is analgebra isomorphism since (1) = 1 Thus 119880 is an algebraisomorphism from 119861 onto itself

Journal of Function Spaces 11

Theorem 11 A bounded operator 119879 119861 997888rarr 119861 is a Hermitianoperator if and only if 119879(1) is a Hermitian element in 119861 and119879 = 119872119879(1) the multiplication operator by 119879(1)Proof By Proposition 10 every surjective unital isometry on119861 is multiplicative Then by [37 Theorem 4] we have theconclusion

5 Banach Algebras of 119862(119884)-Valued Maps

Suppose that119883 is a compact Hausdorff space Suppose that 119861is a unital point separating subalgebra of119862(119884) equipped witha Banach algebra norm Then 119861 is semisimple because 119891 isin119861 119891(119909) = 0 is a maximal ideal of 119861 for every 119909 isin 119883 and theJacobson radical of 119861 vanishes The inequality 119891infin le 119891119861for every 119891 isin 119861 is well known We say that 119861 is natural if themap 119890 119884 997888rarr 119872119861 defined by 119910 997891997888rarr 120601119910 where 120601119910(119891) = 119891(119910)for every 119891 isin 119861 is bijective We say that 119861 is self-adjoint if 119861 isnatural and conjugate-closed in the sense that 119891 isin 119861 impliesthat 119891 isin 119861 for every 119891 isin 119861 where sdot denotes the complexconjugation on 119884Definition 12 Let 119883 and 119884 be compact Hausdorff spacesSuppose that 119861 is a unital point separating subalgebra of119862(119883)equipped with a Banach algebra norm sdot 119861 Suppose that119861 is self-adjoint Suppose that 119861 is a unital point separatingsubalgebra of 119862(119883 times 119884) such that 119861 otimes 119862(119884) sub 119861 equippedwith a Banach algebra norm sdot 119861 Suppose that 119861 is self-adjoint We say that 119861 is a natural 119862(119884)-valuezation of 119861 ifthere exists a compact Hausdorff space M and a complex-linear map 119863 119861 997888rarr 119862(M) such that ker119863 = 1 otimes 119862(119884) and119863(119862R(119883 times 119884) cap 119861) sub 119862R(M) which satisfies

119865 = 119865infin(119883times119884) + 119863 (119865)infin(M) 119865 isin 119861 (120)

The term ldquoa natural 119862(119884)-valuezation of 119861rdquo comes fromthe natural norm defined by Jarosz [5] In fact the norm sdot 119861is a natural norm in the sense of Jarosz [5]

Note that (119883119862(119884) 119861 119861) need not be an admissiblequadruple defined by Nikou and OrsquoFarrell [38] (cf [31]) sincewe do not assume that 119865(sdot 119910) 119865 isin 119861 119910 isin 119884 sub 119861 whichis a requirement for the admissible quadruple On the otherhand if (119883119862(119884) 119861 119861) is an admissible quadruple of type Ldefined in [30] then 119861 is a natural 119862(119884)-valuezation of 119861 dueto Definition 12

Example 13 Let 119861 = 1198621([0 1]) and 119861 = Lip([0 1] 119862(119884))for 119884 = 119901 a singleton Then Lip([0 1]) is algebraically iso-morphic to Lip([0 1] 119862(119884)) Suppose thatM is the maximalideal space of 119871infin([0 1]) and 119863 119861 997888rarr 119862(M) is definedby 119891 997891997888rarr Γ(1198911015840) where Γ denotes the Gelfand transformin 119871infin([0 1]) Then 119861 is a natural 119862(119884)-valuezation of 119861The Banach algebra Lip([0 1]) with the norm 119891infin([01]) +1198911015840infin([01]) is isometrically isomorphic to 119861

Let 119884 be a compact Hausdorff space Note that a closedsubalgebra 119861 of Lip((119870 119889120572) 119862(119884))which appears in Example12 in [30] is an example of a natural 119862(119884)-valuezation of 119861

The Banach algebras1198621([0 1] 119862(119884)) and 1198621(T 119862(119884)))whichappear in Examples 16 and 17 in [30] respectively are alsoexamples of natural 119862(119884)-valuezations of 1198621([0 1])6 Isometries on Natural 119862(119884)-ValuezationsThemain theorem in this paper is the following

Theorem 14 Suppose that 119861119895 is a natural 119862(119884119895)-valuezationof 119861119895 sub 119862(119883119895) for 119895 = 1 2 We assume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (121)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for119895 = 1 2 Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linear isometry Then there exists ℎ isin 119862(1198842) such that |ℎ| = 1on 1198842 a continuous map 120593 1198832 times 1198842 997888rarr 1198831 such that120593(sdot 119910) 1198832 997888rarr 1198831 is a homeomorphism for each 119910 isin 1198842and a homeomorphism 120591 1198842 997888rarr 1198841 which satisfies119880 (119865) (119909 119910) = ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842 (122)

for every 119865 isin 1198611

In short a surjective isometry between 119862(119884)-valuezationsis a weighted composition operator of a specific form thehomeomorphism 1198832 times 1198842 997888rarr 1198831 times 1198841 (119909 119910) 997891997888rarr(120593(119909 119910) 120591(119910)) has the second coordinate that depends onlyon the second variable 119910 isin 1198842 A composition operatorinduced by such a homeomorphism is said to be of type BJin [31 37] after the study of Botelho and Jamison [39]

Quite recently the author of this paper and Oi [30 Theo-rem 8] proved a similar result of Theorem 14 for admissiblequadruples of type L To prove it we apply Proposition 32and the following comments in [31] Instead of this we proveTheorem 14 by Lumerrsquosmethod with which a proof is simplerthan that in [30]

In the following in this section we assume that 119861119895 is anatural 119862(119884119895)-valuezation of 119861 sub 119862(119883119895) for 119895 = 1 2 Weassume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (123)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for 119895 = 1 2Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linearisometry A crucial part of a proof of Theorem 14 is to proveProposition 15

Proposition 15 Suppose that 1198832 is not a singleton Thereexists ℎ isin 119862(1198842) with |ℎ| = 1 on 1198842 such that 119880(11198611) = 11198612 otimesℎ

A similar result for admissible quadruples of type L isproved in [30 Proposition 9] If we assumed that

119865 (sdot 119910) 119865 isin 119861119895 119910 isin 119884119895 sub 119861119895 (124)

then 119861119895 were an admissible quadruple of type L Although119861119895 in this paper need not be an admissible quadruple of type

12 Journal of Function Spaces

L a proof of Proposition 15 is completely the same as that in[30 Proposition 9] since we do not make use of the condition(124) in the proof of [30 Proposition 9] The condition (124)is needed in [30] when we apply Proposition 32 and thefollowing comments in [31]

7 Proof of Theorem 14 An Application ofLumerrsquos Method

Proof of Theorem 14 A proof for the case where 1198831 = 1199091and 1198832 = 1199092 are singletons is the same as the proof ofTheorem 8 in [30]

Suppose that1198832 is not a singleton By Proposition 15 thereexists ℎ isin 119862R(1198842) with |ℎ| = 1 on 1198842 such that 119880(1) =1 otimes ℎ Letting 1198800 1198611 997888rarr 1198612 by 1198800(119865) = (1 otimes ℎ)119880(119865)119865 isin 1198611 we see by the hypothesis (1 otimes ℎ)1198651198612 = 1198651198612 forevery 119865 isin 1198612 that 1198800 is a surjective unital isometry from1198611 onto 1198612 Then Corollary 2 asserts that 1198800 is an algebraisomorphism Let 119891 isin 119862R(1198841) By Proposition 9 1 otimes 119891 isa Hermitian element in 1198611 Then by Theorem 11 1198721otimes119891 is aHermitian operator on 1198611 By Proposition 8 11988001198721otimes119891119880minus1

0 is aHermitian operator on 1198612 Then by Theorem 11 there exists119878(119891) isin 119862R(1198842) such that 11988001198721otimes119891119880minus1

0 = 1198721otimes119878(119891) Hence anoperator 119878 119862R(1198841) 997888rarr 119862R(1198842) is defined Since 1198800 is analgebra isomorphism it is easy to see that 119878 is a real algebraisomorphism from 119862R(1198841) onto 119862R(1198842) Then 119878 119862(1198841) 997888rarr119862(1198842) defined by 119878(119891) = 119878(Re119891) + 119894119878(Im119891) for 119891 isin 119862(1198841)gives a complex algebra isomorphism Gelfand theory assertsthat there is a homeomorphism 120591 1198842 997888rarr 1198841 such that119878(119891) = 119891 ∘ 120591 119891 isin 119862(1198841) It follows that

11988001198721otimes119891119880minus10 = 1198721otimes119891∘120591 119891 isin 119862 (1198841) (125)

Since 119880minus10 (1) = 1 we have

1198800 (1 otimes 119891) = 1 otimes 119891 ∘ 120591 119891 isin 119862 (1198841) (126)

Define Φ 1198611 997888rarr 1198612 by Φ(119886) = 1198800(119886 otimes 1) 119886 isin 1198611Since 1198800 is an algebra isomorphism the map Φ is a unitalhomomorphism Since the maximal ideal space of 1198611 is 1198831

and the maximal ideal space of 1198612 is 1198832 times 1198842 there is acontinuous map 120593 1198832 times 1198842 997888rarr 1198831 such that

Φ(119886) (119909 119910) = 119886 (120593 (119909 119910)) 119886 isin 1198611 (119909 119910) isin 1198832 times 1198842 (127)

It follows by (126) and (127) that

1198800 (119886 otimes 119891) (119909 119910) = 1198800 ((119886 otimes 1) (1 otimes 119891)) (119909 119910)= 1198800 (119886 otimes 1) (119909 119910)1198800 (1 otimes 119891) (119909 119910)= 119886 (120593 (119909 119910))119891 (120591 (119910))= (119886 otimes 119891) (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198832 times 1198842

(128)

for every 119886 isin 1198611 and 119891 isin 119862(1198841) Thus

1198800 (sum(119886119895 otimes 119891119895)) (119909 119910)= (sum(119886119895 otimes 119891119895)) (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842(129)

for every sum(119886119895 otimes 119891119895) isin 1198611 otimes 119862(1198841) By the Stone-Weierstrasstheorem 1198611 otimes 119862(1198841) is uniformly dense in 119862(1198831 times 1198841) henceany element in 1198611 is uniformly approximated by 1198611 otimes 119862(1198841)As 1198800 is also an isometry with respect to the uniform normwe see that

1198800 (119865) (119909 119910) = 119865 (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198831 times 1198841 (130)

for every 119865 isin 1198611 and

119880 (119865) (119909 119910) = (1 otimes ℎ) (119909 119910) 119865 (120593 (119909 119910) 120591 (119910))= ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198831 times 1198841(131)

As 1198800 is an algebra isomorphism the map 1198832 times 1198842 997888rarr1198831 times 1198841 defined by (119909 119910) 997891997888rarr (120593(119909 119910) 120591(119910)) gives ahomeomorphism Therefore for every 119910 isin 1198842 the map

120593 (sdot 119910) 1198832 997888rarr 1198831 (132)

is a homeomorphismSuppose that 1198831 is not a singleton By the same way as

in the last part of the proof of Theorem 8 in [30] we havethat 1198832 is not a singleton Then we have the conclusion bythe previous argument

8 Application of Theorem 14

We exhibit applications of Theorem 14

Corollary 16 ([4 Theorem 33]) Suppose that119880 Lip([0 1]) 997888rarr Lip([0 1]) is a surjective isometrywith respect to the norm defined by 119891infin([01]) + 1198911015840infin([01])

for 119891 isin Lip([0 1]) Then 119880(1) is a constant function of unitmodulus such that

119880(119891) (119909) = 119880 (1) 119891 (119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (133)

or

119880(119891) (119909) = 119880 (1) 119891 (1 minus 119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (134)

The converse statement also holds

Proof By Example 13 we may suppose that Lip([0 1]) is aBanach algebra of 119862(119884)-valuezation Applying Theorem 14we have that119880(1) = 1otimesℎ for ℎ isin 119862(119884)with |ℎ| = 1 Since our

Journal of Function Spaces 13

119884 is a singleton 119880(1) is a constant function of unit modulusWe also see that the corresponding continuous map 120593 [0 1] times 119884 997888rarr [0 1] can be considered as a homeomorphismfrom [0 1] onto [0 1] therefore we have that119880(119891) (119909) = 119880 (1) 119891 (120593 (119909))

119891 isin Lip ([0 1]) 119909 isin [0 1] (135)

The rest is a routine argument to prove that 120593 is an isometryhence 120593(119909) = 119909 119909 isin [0 1] or 120593(119909) = 1 minus 119909 119909 isin [0 1]

The converse statement is trivial

Corollaries 14 15 18 and 19 in [30 Section 6] follow herewith a similar proof

Data Availability

The data used to support the findings of this study areavailable from the corresponding author upon request

Conflicts of Interest

The author declares that they have no conflicts of interest

Acknowledgments

This work was supported by JSPS KAKENHI Grants Num-bers JP16K05172 and JP15K04921

References

[1] K de Leeuw ldquoBanach spaces of Lipschitz functionsrdquo StudiaMathematica vol 21 pp 55ndash66 19611962

[2] A K Roy ldquoExtreme points and linear isometries of the Banachspace of Lipschitz functionsrdquoCanadian Journal of MathematicsJournal Canadien de Mathematiques vol 20 pp 1150ndash11641968

[3] M Cambern ldquoIsometries of certain Banach algebrasrdquo StudiaMathematica vol 25 pp 217ndash225 19641965

[4] N V Rao and A K Roy ldquoLinear isometries of some functionspacesrdquo Pacific Journal of Mathematics vol 38 pp 177ndash192 1971

[5] K Jarosz ldquoIsometries in semisimple commutative BanachalgebrasrdquoProceedings of the AmericanMathematical Society vol94 no 1 pp 65ndash71 1985

[6] K Jarosz and V D Pathak ldquoIsometries between functionspacesrdquo Transactions of the AmericanMathematical Society vol305 no 1 pp 193ndash206 1988

[7] NWeaver ldquoIsometries of noncompact Lipschitz spacesrdquoCana-dian Mathematical Bulletin Bulletin Canadien de Mathemat-iques vol 38 no 2 pp 242ndash249 1995

[8] A Jimenez-Vargas and M Villegas-Vallecillos ldquoLinear isome-tries between spaces of vector-valued Lipschitz functionsrdquoProceedings of the American Mathematical Society vol 137 no4 pp 1381ndash1388 2009

[9] A Jimenez-Vargas and M Villegas-Vallecillos ldquoInto linearisometries between spaces of Lipschitz functionsrdquo HoustonJournal of Mathematics vol 34 no 4 pp 1165ndash1184 2008

[10] F Botelho and J Jamison ldquoSurjective isometries on spacesof differentiable vector-valued functionsrdquo Studia Mathematicavol 192 no 1 pp 39ndash50 2009

[11] E Mayer-Wolf ldquoIsometries between Banach spaces of Lipschitzfunctionsrdquo Israel Journal of Mathematics vol 38 no 1-2 pp 58ndash74 1981

[12] A Jimenez-Vargas M Villegas-Vallecillos and Y-S WangldquoBanach-Stone theorems for vector-valued little Lipschitz func-tionsrdquo Publicationes Mathematicae vol 74 no 1-2 pp 81ndash1002009

[13] J Araujo andLDubarbie ldquoNoncompactness andnoncomplete-ness in isometries of Lipschitz spacesrdquo Journal of MathematicalAnalysis and Applications vol 377 no 1 pp 15ndash29 2011

[14] F Botelho R J Fleming and J Jamison ldquoExtreme points andisometries on vector-valued Lipschitz spacesrdquo Journal of Math-ematical Analysis and Applications vol 381 no 2 pp 821ndash8322011

[15] H Koshimizu ldquoLinear isometries on spaces of continuouslydifferentiable and Lipschitz continuous functionsrdquo NihonkaiMathematical Journal vol 22 no 2 pp 73ndash90 2011

[16] F Botelho J Jamison and B Zheng ldquoIsometries on spaces ofvector valued Lipschitz functionsrdquo Positivity An InternationalMathematics Journal Devoted to Theory and Applications ofPositivity vol 17 no 1 pp 47ndash65 2013

[17] A Ranjbar-Motlagh ldquoA note on isometries of Lipschitz spacesrdquoJournal of Mathematical Analysis and Applications vol 411 no2 pp 555ndash558 2014

[18] F Botelho and J Jamison ldquoSurjective isometries on spaces ofvector valued continuous and Lipschitz functionsrdquo PositivityAn International Mathematics Journal Devoted to Theory andApplications of Positivity vol 17 no 3 pp 395ndash405 2013Erratum to Surjective isometries on spaces of vector valuedcontinuous and Lipschitz functions by F Botelho 20 (2016)757ndash759

[19] T Miura and H Takagi ldquoSurjective isometries on the Banachspace of continuously differentiable functionsrdquo ContemporaryMathematics vol 687 pp 181ndash192 2017

[20] K Kawamura ldquoBanach-Stone type theorems for 1198621-functionspaces over Riemannian manifoldsrdquo Acta Universitatis Szegedi-ensis Acta Scientiarum Mathematicarum vol 83 no 3-4 pp551ndash591 2017

[21] K Kawamura ldquoPerturbations of norms on 1198621-function spacesand associated isometry groupsrdquo Topology Proceedings vol 51pp 169ndash196 2018

[22] K Kawamura ldquoA Banach-Stone type theorem for 1198621-functionspaces over the circlerdquo Topology Proceedings vol 53 pp 15ndash262019

[23] L Li D Chen Q Meng and Y-S Wang ldquoSurjective isometrieson vector-valued differentiable function spacesrdquo Annals ofFunctional Analysis vol 9 no 3 pp 334ndash343 2018

[24] K Kawamura H Koshimizu and T Miura ldquoNorms onC1([01]) and there isometriesrdquo Acta Scientiarum Mathemati-carum vol 84 no 12 pp 239ndash261 2018

[25] L Li A M Peralta L Wang and Y-S Wang ldquoWeak-2-local isometries on uniform algebras and Lipschitz algebrasrdquohttpsarxivorgabs170503619

[26] A Jimenez-Vargas L LiAMPeralta LWang andY-SWangldquo2-local standard isometries on vector-valued Lipschitz func-tion spacesrdquo Journal of Mathematical Analysis and Applicationsvol 461 no 2 pp 1287ndash1298 2018

[27] A Ranjbar-Motlagh ldquoIsometries of Lipschitz type functionspacesrdquoMathematischeNachrichten vol 291 no 11-12 pp 1899ndash1907 2018

14 Journal of Function Spaces

[28] N Weaver Lipschitz Algebras World Scientific Publishing CoInc River Edge NJ USA 1999

[29] R R Phelps Lectures on Choquetrsquos theorem vol 1757 of LectureNotes in Mathematics Springer-Verlag Berlin Germany 2ndedition 2001

[30] O Hatori and S Oi ldquoIsometries on Banach algebras of vector-valued mapsrdquo Acta Scientiarum Mathematicarum vol 84 no12 pp 151ndash183 2018

[31] O Hatori S Oi and H Takagi ldquoPeculiar homomorphismsbetween algebras of vector-valued mapsrdquo Studia Mathematicavol 242 no 2 pp 141ndash163 2018

[32] M Nagasawa ldquoIsomorphisms between commutative Banachalgebras with an application to rings of analytic functionsrdquoKodai Mathematical Seminar Reports vol 11 pp 182ndash188 1959

[33] K de LeeuwW Rudin and J Wermer ldquoThe isometries of somefunction spacesrdquo Proceedings of the American MathematicalSociety vol 11 pp 694ndash698 1960

[34] OHatori A Jimenez-Vargas andMVillegas-Vallecillos ldquoMapswhich preserve norms of non-symmetrical quotients betweengroups of exponentials of Lipschitz functionsrdquo Journal of Math-ematical Analysis and Applications vol 415 no 2 pp 825ndash8452014

[35] G Lumer ldquoOn the isometries of reflexiveOrlicz spacesrdquoAnnalesde lrsquoInstitut Fourier vol 68 pp 99ndash109 1963

[36] R J Fleming and J E Jamison Isometries on Banach SpacesFunction Spaces vol 129 ofMonographs and Surveys in Pure andApplied Mathematics Chapman amp Hall CRC Boca Raton FlaUSA 2003

[37] OHatori and SOi ldquoHermitian operators onBanach algebras ofvector-valued Lipschitzmapsrdquo Journal ofMathematical Analysisand Applications vol 452 no 1 pp 378ndash387 2017 Corrigendumto rdquoHermitian operators on Banach algebras of vector-valuedLipschitz mapsrdquo Journal of Mathematical Analysis and Appli-cations 452 (2017) 378ndash387 MR3628025

[38] A Nikou andA G OrsquoFarrell ldquoBanach algebras of vector-valuedfunctionsrdquo Glasgow Mathematical Journal vol 56 no 2 pp419ndash426 2014

[39] F Botelho and J Jamison ldquoHomomorphisms on a class ofcommutative Banach algebrasrdquo Rocky Mountain Journal ofMathematics vol 43 no 2 pp 395ndash416 2013

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Page 3: Hermitian Operators and Isometries on Banach Algebras of ...downloads.hindawi.com/journals/jfs/2018/8085304.pdfJournalofFunctionSpaces normiseitherthemaxnormorthesumnorm.esituation

Journal of Function Spaces 3

for every 1199050 isin 119870 We denote 1198651015840 = 119866 Put1198621 (119870 119862 (119884)) = 119865isin 119862 (119870119862 (119884)) 119865 is continuously differentiable (9)

Then1198621(119870119862(119884)) with norm 119865 = 119865infin + 1198651015840infin is a unitalsemisimple commutative Banach algebra If 119884 is singleton wemay suppose that 119862(119884) is isometrically isomorphic to C andwe abbreviate 1198621(119870119862(119884)) by 1198621(119870)

By identifying 119862(119870 119862(119884)) with 119862(119870times119884) we may assumethat Lip(119870119862(119884)) (resp lip120572(119870119862(119884))) is a subalgebra of119862(119870 times 119884) by the correspondence

119865 isin Lip (119870 119862 (119884)) larrrarr ((119909 119910) 997891997888rarr (119865 (119909)) (119910)) isin 119862 (119870 times 119884) (10)

Throughout the paper we may suppose that

Lip (119883119862 (119884)) sub 119862 (119883 times 119884) lip120572 (119883119862 (119884)) sub 119862 (119883 times 119884) 1198621 (119870 119862 (119884)) sub 119862 (119870 times 119884)

(11)

We say that a subset 119876 of 119862(119884) is point separating if 119876separates the points of 119884 The unit of commutative Banachalgebra 119861 is denoted by 1 The maximal ideal space of 119861 isdenoted by 119872119861 Suppose that 119861 is a unital point separatingsubalgebra of 119862(119884) equipped with a Banach algebra normThen119861 is semisimple because 119891 isin 119861 119891(119909) = 0 is amaximalideal of 119861 for every 119909 isin 119883 and the Jacobson radical of 119861vanishes

3 A Theorem of Jarosz RevisitedIsometries Preserving Unit

Whether an isometry between unital semisimple commu-tative Banach algebras is of the canonical form dependsnot only on the algebraic structures of these algebras butalso on the norms in these algebra in most cases A simpleexample is a surjective isometry on theWiener algebra whichneed not be canonical Jarosz [5] defined natural norms andprovided a theorem that isometries between a variety ofalgebras equipped with natural norms are of canonical formsFor the sake of completeness we outline the notations and theterminologies which are due to [5]The set of all norms onR2

with 119901(1 0) = 1 is denoted byP For 119901 isin P we put

119863(119901) = lim119905997888rarr+0

119901 (1 119905) minus 1119905 (12)

Recently Tanabe pointed out by a private communication that119863(119901) exists and it is finite for every 119901 isin P (In fact it iseasy to see that (119901(1 119905) minus 1)119905 is increasing since 119901(1 119905) isconvex We also see that inf 119905gt0((119901(1 119905) minus 1)119905) gt minusinfin) Let119883be a compact Hausforff space and 119860 a liner subspace of 119862(119883)which contains constant functions A seminorm lsaquo sdot lsaquo on 119860 iscalled one-invariant (in the sense of Jarosz) if lsaquo119891+ 1lsaquo = lsaquo119891lsaquofor all 119891 isin 119860 Let 119901 isin P A norm sdot on 119860 is called a 119901-norm if there is a one-invariant seminorm lsaquo sdot lsaquo on 119860 suchthat sdot = 119901( sdot infin lsaquo sdot lsaquo) A natural norm is a 119901-norm forsome 119901 isin P

Theorem 1 (Jarosz [5]) Let 119883 and 119884 be compact Hausdorffspaces let 119860 and 119861 be complex-linear subspaces of 119862(119883) and119862(119884) respectively and let 119901 119902 isin P Assume 119860 and 119861 containconstant functions and let sdot119860 sdot119861 be a 119901-norm and 119902-normon 119860 and 119861 respectively Assume next that there is a linearisometry 119879 from (119860 sdot 119860) onto (119861 sdot 119861) with 1198791 = 1 Then if119863(119901) = 119863(119902) = 0 or if 119860 and 119861 are regular subspaces of 119862(119883)and 119862(119884) respectively then 119879 is an isometry from (119860 sdot infin)onto (119861 sdot infin)

In the sequel a unital semisimple commutative Banachalgebra 119860 is identified via the Gelfand transforms with a sub-algebra of 119862(119872119860) A unital semisimple commutative Banachalgebra is regular (in the sense of Jarosz [5]) Hence we haveby a theorem of Nagasawa [32] (cf [33]) that the followingholds

Corollary 2 Let 119860 and 119861 be unital semisimple commutativeBanach algebras Assume they have natural norms respectivelySuppose that 119879 119860 997888rarr 119861 is a surjective complex-linearisometry with 1198791 = 1 Then there exists a homeomorphism120593 119872119861 997888rarr 119872119860 such that

119879 (119891) (119909) = 119891 ∘ 120593 (119909) 119891 isin 119860 119909 isin 119872119861 (13)

Proof A unital semisimple commutative Banach algebra isregular by Proposition 2 in [5]ThenTheorem 1 ensures that119879is a surjective linear isometry from (119860 sdotinfin) onto (119861 sdotinfin) Itis easy to see that 119879 is extended to a surjective linear isometry from the uniform closure 119860 of 119860 onto the uniform closure119861 of 119861 Then a theorem of Nagasawa asserts that there existsa homeomorphism 120593 119872119861 997888rarr 119872119860 such that (119891)(119909) =119891 ∘ 120593(119909) (119891 isin 119860 119909 isin 119872119861) As |119860 = 119879 we have theconclusion

Corollary 3 Let 119870119895 be a compact metric space for 119895 = 1 2Suppose that 119879 Lip(1198701) 997888rarr Lip(1198702) is a surjective complex-linear isometry with respect to the norm sdot 119871 Assume 1198791 = 1Then there exists a surjective isometry 120593 1198832 997888rarr 1198831 such that

119879119891 (119909) = 119891 ∘ 120593 (119909) 119891 isin Lip (1198701) 119909 isin 1198702 (14)

Conversely if 119879 Lip(1198701) 997888rarr Lip(1198702) is of the form as (14)then 119879 is a surjective isometry with respect to both of sdot 119872 and sdot 119871 such that 1198791 = 1

Proof As (Lip(119870119895) sdot 119871) is a unital semisimple commutativeBanach algebra with maximal ideal space 119870119895 Corollary 2asserts that there is a homeomorphism 120593 1198702 997888rarr 1198701 suchthat

119879119891 (119909) = 119891 ∘ 120593 (119909) 119891 isin Lip (1198701) 119909 isin 1198702 (15)

Then by a routine argument we see that 120593 is an isometryConverse statement is trivial

Without assuming 1198791 = 1 we have that 119879 is aweighted composition operator We exhibit a general resultas Theorem 14 (see also [30])(Lip(119870) sdot 119872) need not be a Banach algebra since sdot 119872need not be submultiplicative On the other hand sdot 119872

4 Journal of Function Spaces

is a natural norm in the sense of Jarosz (see [5]) such thatlim119905997888rarr+0((max1 119905 minus 1)119905) = 0 Then by Theorem 1 we havethe following

Corollary 4 Let 119870119895 be a compact metric space for 119895 = 1 2Suppose that 119879 Lip(1198701) 997888rarr Lip(1198702) is a surjective complex-linear isometry with respect to the norm sdot119872 Assume 1198791 = 1Then there exists a surjective isometry 120593 1198832 997888rarr 1198831 such that

119879119891 = 119891 ∘ 120593 119891 isin Lip (1198701) (16)

Conversely if 119879 Lip(1198701) 997888rarr Lip(1198702) is of a similar form as(16) then 119879 is a surjective isometry with respect to both of sdot119872and sdot 119871 such that 1198791 = 1

Proof As sdot 119872 is a natural norm we have by Corollary 2 thatthere is a homeomorphism 120593 1198702 997888rarr 1198701 such that

119879119891 (119909) = 119891 ∘ 120593 (119909) 119891 isin Lip (1198701) 119909 isin 1198702 (17)

Then by a routine argument we see that 120593 is an isometryConverse statement is trivial

Without the assumption that 1198791 = 1 in Corollary 4 onemay expect that 119879 is a weighted composition operator But itis not the case A simple counterexample is given by Weaver[7 p242] (see also [28])

As is pointed out in [34] the original proof of Theorem 1needs a revision in some part and a proof when 119860 and 119861are algebras of Lipschitz functions is revised [34 Proposition7] Although a revised proof for a general case is similar tothat of Proposition 7 in [34] we exhibit it here for the sakeof completeness of this paper To prove Theorem 1 we needLemma 2 in [5] in the sameway as the original proof of JaroszThe following is Lemma 2 in [5]

Lemma5 (Jarosz [5]) Assume119860 is a regular subspace of119862(119883)with 1 isin 119860 and let 1199090 isin 119862ℎ(119860) Then for any 120576 gt 0 and anyopen neighborhood 119880 of 1199090 there is an 119891 isin 119860 such that

10038171003817100381710038171198911003817100381710038171003817infin le 1 + 120576119891 (1199090) = 11003816100381610038161003816119891 (119909) + 11003816100381610038161003816 le 120576

119909 isin 119883 119880(18)

and |Im119891(119909)| le 120576 for all 119909 isin 119883Proof The proof is essentially due to the original proof ofLemma 2 in [5] Several minor changes are needed Weitemize them as follows

(i) Five 1205762rsquos between 11 lines and 5 lines from the bottomof page 69 read as 1205763

(ii) Next 119909 isin 1198831198801 reads as 119909 isin 1198801 on the bottom of page69

(iii) We point out that the term sum1198960minus1119895=1 (119891119895(119909) minus 1) which

appears on the first line of the first displayed inequal-ities on page 70 reads 0 if 1198960 = 1

(iv) The term 1 + 120576 on the right hand side of the secondline of the same inequalities reads as 1 + 1205763

(v) Two 1205762rsquos on the same line read as 1205763(vi) On the next line ((119899 + 1)119899)(1205762) reads as 1205763(vii) For any 1 le 1198960 le 119899 we infer that

1 ge 1 minus 21198960 minus 1119899 ge 1 minus 2119899 minus 1119899 gt minus1 (19)

Hence we have |119891(119909)| le 1 + 120576 if 119909 isin 1198801 by the firstdisplayed inequalities of page 70

(viii) The inequality 119891infin le 120576 on the fifth line on page 70reads as 119891infin le 1 + 120576

Let119870 be a nonempty convex subset of the complex planeand 120593 isin [0 2120587) Put

119888 (119870 120593) = sup 119886 isin R there is a 119887isin R with (119886 + 119894119887) 119890119894120593 isin 119870 (20)

Note that we may write

119888 (119870 120593) = sup Re120582 120582 isin 119890minus119894120593119870 (21)

Let 119860 be a subspace of 119862(119883) for a compact Hausdorff spaceFor 119891 isin 119860 we put 120590(119891) = 119891(119883) and 120590(119891) = co(119891(119883)) whereco(sdot) denotes the closed convex-hull We define the functions

119888119860 119860 times [0 2120587) 997888rarr R119888119860 (119891 120593) = 119888 ( (119891) 120593)

119903119860 119860 times R+ times [0 2120587) 997888rarr R

+119903119860 (119891 119905 120593) = 10038171003817100381710038171003817119891 + 119890119894120593119905110038171003817100381710038171003817infin

(22)

Proof of Theorem 1 Let 119891 isin 119860 First we note that10038171003817100381710038171198911003817100381710038171003817infin = sup 119911 isin 120590 (119891) = sup 119911 isin (119891) (23)

since (119891) is the closed convex-hull of a compact set 120590(119891) =119891(119883) We prove the inequalities

119905 + 119888119860 (119891 120593) le 119903119860 (119891 119905 120593)le radic(119905 + 119888119860 (119891 120593))2 + 100381710038171003817100381711989110038171003817100381710038172infin

119905 ge 0 120593 isin [0 2120587) (24)

which appear on p 68 in [5] Put 119904 = 119888119860(119891 120593) As 120590(119891) iscompact there exists 119887 isin R such that (119904 + 119894119887)119890119894120593 isin 120590(119891)Hence

119890119894120593 (119905 + 119904 + 119894119887) isin 119890119894120593119905 + 120590 (119891) = 120590 (1198901198941205931199051 + 119891) (25)

Journal of Function Spaces 5

As

119905 + 119904 le |119905 + 119904| le |119905 + 119904 + 119894119887| = 10038161003816100381610038161003816119890119894120593 (119905 + 119904 + 119894119887)10038161003816100381610038161003816le 100381710038171003817100381710038171198901198941205931199051 + 11989110038171003817100381710038171003817infin = 119903119860 (119891 119905 120593) (26)

we have

119905 + 119888119860 (119891 120593) le 119903119860 (119891 119905 120593) (27)

Let 119909 isin 119883 By the definition of 119904 = 119888119860(119891 120593) we infer thatRe(119890minus119894120593119891(119909)) le 119904 hence we have Re(119905 + 119890minus119894120593119891(119909)) le 119905 + 119904 forevery 119905 ge 0 Then

Re 119890minus119894120593119891 (119909) le Re (119905 + 119890minus119894120593119891 (119909)) le 119905 + 119904 119905 ge 0 (28)

Letting119872 = max|119905 + 119904| |Re(119890minus119894120593119891(119909))| we have10038161003816100381610038161003816119890119894120593119905 + 119891 (119909)100381610038161003816100381610038162 = 10038161003816100381610038161003816119905 + 119890minus119894120593119891 (119909)100381610038161003816100381610038162

= (Re (119905 + 119890minus119894120593119891 (119909)))2+ (Im (119905 + 119890minus119894120593119891 (119909)))2

le 1198722 + (Im (119905 + 119890minus119894120593119891 (119909)))2le |119905 + 119904|2 + (Re (119890minus119894120593119891 (119909)))2+ (Im (119890minus119894120593119891 (119909)))2

= |119905 + 119904|2 + 10038161003816100381610038161003816119890minus119894120593119891 (119909)100381610038161003816100381610038162

(29)

As 119909 isin 119883 is arbitrary we have

119903119860 (119891 119905 120593) = 10038171003817100381710038171003817119905 + 119890minus11989412059311989110038171003817100381710038171003817infin le radic|119905 + 119904|2 + 100381710038171003817100381711989110038171003817100381710038172infin= radic(119905 + 119888119860 (119891 120593))2 + 100381710038171003817100381711989110038171003817100381710038172infin

(30)

It follows that (24) holds In the same way we have

119905 + 119888119861 (119892 120593) le 119903119861 (119892 119905 120593) le radic(119905 + 119888119861 (119892 120593))2 + 100381710038171003817100381711989210038171003817100381710038172infin119905 ge 0 120593 isin [0 2120587) (31)

for every 119892 isin 119861 By (24) and (31) we infer that

lim119905997888rarrinfin

(119903119860 (119891 119905 120593) minus 119905) = 119888119860 (119891 120593) lim119905997888rarrinfin

(119903119861 (119879119891 119905 120593) minus 119905) = 119888119861 (119879119891 120593) (32)

As lsaquo sdot lsaquo119860 is 1-invariant we have119901 (119903119860 (119891 119905 120593) lsaquo119891lsaquo119860)= 119901 (10038171003817100381710038171003817119891 + 119890119894120593119905110038171003817100381710038171003817infin lsaquo119891 + 1198901198941205931199051lsaquo119860) (33)

As 119879 is an isometry 119879(1) = 1 and lsaquo sdot lsaquo119861 is 1-invariant wehave

119901 (10038171003817100381710038171003817119891 + 119890119894120593119905110038171003817100381710038171003817infin lsaquo119891 + 1198901198941205931199051lsaquo119860)= 119902 (10038171003817100381710038171003817119879 (119891 + 1198901198941205931199051)10038171003817100381710038171003817infin lsaquo119879 (119891 + 1198901198941205931199051)lsaquo119861)= 119902 (10038171003817100381710038171003817119879 (119891) + 119890119894120593119905110038171003817100381710038171003817infin lsaquo119879 (119891) + 1198901198941205931199051lsaquo119861)= 119902 (119903119861 (119879119891 119905 120593) lsaquo119879119891lsaquo119861)

(34)

Thus

119901 (119903119860 (119891 119905 120593) lsaquo119891lsaquo119860) = 119902 (119903119861 (119879119891 119905 120593) lsaquo119879119891lsaquo119861) (35)

It follows that0 = lim

119905997888rarrinfin(119901 (119903119860 (119891 119905 120593) lsaquo119891lsaquo119860)

minus 119902 (119903119861 (119879119891 119905 120593) lsaquo119879119891lsaquo119861))= lim

119905997888rarrinfin(119903119860 (119891 119905 120593) 119901(1 lsaquo119891lsaquo119860119903119860 (119891 119905 120593))

minus 119903119860 (119891 119905 120593)) + lim119905997888rarrinfin

(119903119861 (119879119891 119905 120593)minus 119903119861 (119879119891 119905 120593) 119902(1 lsaquo119879119891lsaquo119861119903119861 (119879119891 119905 120593)))+ lim

119905997888rarrinfin(119903119860 (119891 119905 120593) minus 119905 minus (119903119861 (119879119891 119905 120593) minus 119905))

= lsaquo119891lsaquo119860119863(119901) minus lsaquo119879119891lsaquo119861119863(119902) + 119888119860 (119891 120593)minus 119888119861 (119879119891 120593)

(36)

Recall that 119863(119901) = lim119905997888rarr+0(119901(1 119905) minus 1)119905 and 119863(119902) =lim119905997888rarr+0(119902(1 119905) minus 1)119905 It follows thatlsaquo119891lsaquo119860119863 (119901) minus lsaquo119879119891lsaquo119861119863(119902) = 119888119861 (119879119891 120593) minus 119888119860 (119891 120593)

119891 isin 119860 120593 isin [0 2120587) (37)

Suppose that 119863(119901) = 119863(119902) = 0 Then we have by (37)that 119888119861(119879119891 120593) = 119888119860(119891 120593) for every 119891 isin 119860 and 120593 isin [0 2120587)By Lemma 1 in [5] we infer that 120590(119879119891) = 120590(119891) Thus we have119879119891infin = 119891infin We have proved that 119879 is an isometry from(119860 sdot infin) onto (119861 sdot infin) if119863(119901) = 119863(119902) = 0

Suppose that 119860 and 119861 are regular subspaces of 119862(119883) and119862(119884) respectively Let 119891 isin 119860 PutΔ119891 = lsaquo119891lsaquo119860119863(119901) minus lsaquo119879119891lsaquo119861119863(119902) (38)

Suppose that Δ119891 ge 0 For any 119903 ge 0 and any nonemptycompact convex subset 119870 sub C we have that

119888 (119870 + 119870 (119903) 120593) = 119888 (119870 120593) + 119903 (39)

for all 120593 isin [0 2120587) where 119870(119903) = 119911 isin C |119911| le 119903 Then by(37) we have

119888 (120590 (119879119891) 120593) = 119888119861 (119879119891 120593) = 119888119860 (119891 120593) + Δ119891= 119888 (120590 (119891) 120593) + Δ119891= 119888 (120590 (119891) + 119870 (Δ119891) 120593)

(40)

6 Journal of Function Spaces

for all 120593 isin [0 2120587) It follows by Lemma 1 in [5] that

(119879119891) = 120590 (119891) + 119870 (Δ119891) (41)

and therefore10038171003817100381710038171198791198911003817100381710038171003817infin = 10038171003817100381710038171198911003817100381710038171003817infin + Δ119891 (42)

If Δ119891 le 0 then a similar calculation shows that

(119891) = (119879119891) + 119870 (minusΔ119891) (43)

and10038171003817100381710038171198911003817100381710038171003817infin = 10038171003817100381710038171198791198911003817100381710038171003817infin + (minusΔ119891) (44)

It follows that in any case (Δ119891 ge 0 Δ119891 le 0) we obtainΔ119891 = 10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin 119891 isin 119860 (45)

We will prove that1003817100381710038171003817119879 (119891)1003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin = Δ119891 ge 0 (46)

for all 119891 isin 119860 Once it is proved applying the same argumentfor119879minus1 instead of119879 we see that 119879minus1119892infinminus119892infin ge 0 for every119892 isin 119861 As 119879 is a bijection it follows that 119891infin minus 119879119891infin ge 0for every 119891 isin 119860 It will follow that 119879119891infin = 119891infin for every119891 isin 119860 A proof of (46) is the following For every 120576 gt 0denote

A120576 = 119891 isin 119860 120588 (120590 (119891)) le 120576 (47)

The inequality in (46) is deduced by the following assertionswhich appear in the proof of [5 Theorem]

(1) 119879 is a continuous mapping from (119860 sdot infin) onto (119861 sdotinfin)(2) For each 120576 gt 0 the setA120576 is dense in (119860 sdot infin)(3) For each 120576 gt 0 and each 119891 isin A120576 it holds that119879(119891)infin ge 119891infin minus 120576

Suppose that these assertions are proved Let 119891 isin 119860 By (2)for any 120576 gt 0 there is a sequence 119891119899 of functions inA120576 suchthat 119891119899 minus 119891infin 997888rarr 0 as 119899 997888rarr infin By (3) we have

10038171003817100381710038171198791198911198991003817100381710038171003817infin minus 10038171003817100381710038171198911198991003817100381710038171003817infin ge minus120576 (48)

for every 119899 Letting 119899 997888rarr infin we have10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin ge minus120576 (49)

by (1) As 120576 gt 0 is arbitrary we have that10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin ge 0 (50)

We show proofs of three assertions (1) (2) and (3) aboveprecisely The proof of (1) is slightly different from the corre-sponding one in [5 p 70] This change is rather ambitiousWe also point out that the terms minus1205872 and 1205872 which appearin the formulae (7) and (8) in [5] seem inappropriate theyread for example as 31205874 and 1205874 respectively

We now proceed to prove the first statement Aiming fora contradiction suppose that 119879 is not continuous from (119860 sdotinfin) to (119861 sdot infin) Let 120576 be a positive real number less than1100 Then there is a function 1198910 isin 119860 such that 1198910infin le 120576and 119879(1198910)infin = 1 Then there exist 1199100 isin Ch(119861) such that|119879(1198910)(1199100)| = 1 by [29 Proposition 63] Since 119879 is complex-linear we may suppose that 119879(1198910)(1199100) = 1

By (41) and (45) we deduce that Δ1198910 = 119879(1198910)infin minus1198910infin ge 1 minus 120576 and 120590(119879(1198910)) = 120590(1198910) + 119870(Δ1198910) As 1198910infin le 120576and 119879(1198910)infin = 1 we infer that 120590(1198910) sub 119870(120576) and (119879(1198910) sub119870(1) Let 119911 isin 119870(1 minus 2120576) Let 119909 isin 119883 Then |1198910(119909)| le 120576 assertsthat 119908 = 119911 minus 1198910(119909) isin 119870(1 minus 120576) sub 119870(Δ1198910) Thus

119911 = 1198910 (119909) + 119908 isin 120590 (1198910) + 119870 (Δ1198910) = 120590 (119879 (1198910)) (51)

Hence119870(1 minus 2120576) sub 120590(1198791198910) As 119879(1198910)infin = 1 we have119870(1 minus 2120576) sub 120590 (119879 (1198910)) sub 119870 (1) (52)

Consider the open neighborhood 1198800 of 1199100 in1198832 given by

1198800 = 119910 isin 1198832 1003816100381610038161003816119879 (1198910) (119910) minus 11003816100381610038161003816 lt 120576 (53)

We infer that 1198800 is a proper subset of1198832 by (52) Then by [5Lemma 2] there exists 119892 isin 119861 such that 119892infin le 1 + 120576 119892(1199100) =1 |119892(119910) + 1| lt 120576 for every 119910 isin 1198832 1198800 and |Im119892(119910)| lt 120576 forall 119910 isin 1198832 If 119867 denotes the closed rectangle whose verticesare the four points plusmn(1 + 120576) plusmn 120576119894 we have

120590 (119892) sub 119867 (54)

Consider now the set

119871 = 11989031205871198944119911 |119911| le 1 Re 119911 ge 1 minus 2120576 (55)

We claim that119879(1198910)(1198832)cap119871 = 0 Suppose that119879(1198910)(1198832)cap119871 =0 As 119879(1198910)(1198832) is compact there exists a positive integer119899 such that 119879(1198910)(1198832) cap 119871119899 = 0 where 119871119899 = 11989031205871198944119911 |119911| le 1Re 119911 gt 1 minus 2120576 minus 1119899 Then (52) gives 119879(1198910)(1198832) sub119870(1)119871119899 As (119879(1198910)) is the closed convex-hull of 119879(1198910)(1198832)it is contained in the closed convex set 119870(1) 119871119899 On theother hand (1 minus 2120576)11989031205871198944 isin 119870(1 minus 2120576) sub 120590(119879(1198910)) by (52) As(1minus2120576)11989031205871198944 isin 119871 sub 119871119899 this contradicts 120590(119879(1198910)) sub 119870(1) 119871119899and this proves our claim Hence there is 1199101 isin 1198832 with119879(1198910)(1199101) isin 119871 As 120576 le 1100 it follows that |119879(1198910)(1199101) minus1| ge 120576 and so 1199101 isin 1198832 1198800 Hence |119892(1199101) + 1| lt 120576 Thus119892(1199101) + 119879(1198910)(1199101) is in 119871 minus 1 + K(120576) Thus we have

1 + radic22 minus 3120576 le 119888119861 (119892 + 119879 (1198910) 31205874 ) (56)

We claim that

120590 (119892 + 119879 (1198910)) sub co (119870 (minus1 1) cup 2) + 119870 (3120576) (57)

where119870(1199110 119903) = 119911 isin C |119911 minus 1199110| le 119903 Let 119910 isin 1198832 Supposefirst that |119879(1198910)(119910)minus1| lt 120576 Since 119892(1198832) sub 119867 by (54) we have

119879 (1198910) (119910) + 119892 (119910) isin 119870 (1 120576) + 119867 = (119867 + 1) + 119870 (120576) (58)

Journal of Function Spaces 7

Suppose next that |119879(1198910)(119910)minus1| ge 120576Then 119910 isin 1198832 1198800 and so|119892(119910) + 1| lt 120576 Moreover |119879(1198910)(119910)| le 1 Therefore we have

119892 (119910) + 119879 (1198910) (119910) isin 119870 (1) minus 1 + 119870 (120576)= 119870 (minus1 1) + 119870 (120576) (59)

It follows from (58) and (59) that

(119892 + 119879 (1198910)) (1198832) sub 119870 (minus1 1) cup ((119867 + 1)) + 119870 (120576) (60)

and hence

120590 (119892 + 119879 (1198910)) sub co (119870 (minus1 1) cup 2) + 119870 (3120576) (61)

as is claimed Therefore we have

119888119861 (119892 + 119879 (1198910) 1205874 ) le radic2 + 3120576 (62)

Put 1198911 = 119879minus1(119892) We claim that Δ1198911 le 120576 If Δ1198911 lt 0 thereis nothing to prove Suppose that Δ1198911 ge 0 Then by (41) wehave

120590 (119892) = 120590 (1198911) + 119870 (Δ1198911) (63)

Since 120590(119892) sub 119867 by (54) we have

119867 sup 120590 (1198911) + 119870 (Δ1198911) (64)

As 119867 does not include a closed disk with the radius greaterthan 120576 we conclude that Δ1198911 le 120576

In the following we will consider two cases 0 le Δ1198911 le 120576and Δ1198911 le 0 Suppose first that 0 le Δ1198911 le 120576 Then (64) yields(1198911) sub 119867 From 1198910infin le 120576 we deduce that 120590(1198910) sub 119870(120576)Hence we have

120590 (1198911 + 1198910) sub 120590 (1198911) + 120590 (1198910) sub (1198910) + 120590 (1198910)sub 119867 + 119870 (120576) (65)

Since119867 +119870(120576) is convex we have (1198911 + 1198910) sub 119867 + 119870 (120576) (66)

From (39) we infer that

119888119860 (1198911 + 1198910 31205874 ) = 119888 (120590 (1198911 + 1198910) 31205874 )le 119888 (119867 + 119870 (120576) 31205874 )= 119888 (119867 31205874 ) + 120576= radic22 + (1 + radic2) 120576

(67)

Since 119879(1198911 + 1198910) = 119892 + 119879(1198910) from (56) and (67) we obtainthat

1 minus (4 + radic2) 120576 le 119888119861 (119879 (1198911 + 1198910) 31205874 )minus 119888119860 (1198911 + 1198910 31205874 )

(68)

By (63) and 1 = 119892(1199100) we deduce that 1 isin 120590(1198911) + 119870(Δ1198911)Thus there is 119911 isin 120590(1198911) such that |119911 minus 1| le Δ1198911 It follows thatradic22 minus Δ1198911 le 119888119860(1198911 1205874) hence we have

radic22 minus 2120576 le 119888119860 (1198911 + 1198910 1205874 ) (69)

as 1198910infin le 120576 and 0 le Δ1198911 le 120576 We get by (62) and (69) that

119888119861 (119879 (1198911 + 1198910) 1205874 ) minus 119888119860 (1198911 + 1198910 1205874 ) leradic22 + 5120576 (70)

On the other hand 119888119861(119879(119891) 120593)minus119888119860(119891 120593) is invariant for any 120593by (37) From (68) and (70)we deduce that 120576 ge (2minusradic2)2(9+radic2) and this contradicts that 120576 le 1100

For the second case suppose next that Δ1198911 le 0 Then by(43) we have

120590 (1198911) = 120590 (119892) + 119870 (minusΔ1198911) (71)

and by (54) it follows that (1198911) sub 119867 + 119870(minusΔ1198911) Moreover(1198910) sub 119870(120576) since 1198910infin le 120576 Then

120590 (1198911 + 1198910) sub 120590 (1198911) + 120590 (1198910) sub 120590 (1198911) + 120590 (1198910)sub 119867 + 119870(minusΔ1198911) + 119870 (120576) (72)

Hence (1198911+1198910) sub 119867+119870(minusΔ1198911)+119870(120576) Using (39) we inferthat

119888119860 (1198911 + 1198910 31205874 ) le 119888 (119867 + 119870(minusΔ1198911) + 119870 (120576) 31205874 )= 119888 (119867 31205874 ) + (minusΔ1198911) + 120576= radic22 + (1 + radic2) 120576 + (minusΔ1198911)

(73)

By (71) we obtain that 120590(1198911) sup 119892(1198832) + 119870(minusΔ1198911) and as119892(1199100) = 1 we infer that 120590(1198911) sup 1 + 119870(minusΔ1198911) Hence radic22 +(minusΔ1198911) le 119888119860(1198911 1205874) so thatradic22 + (minusΔ1198911) minus 120576 le 119888119860 (1198911 + 1198910 1205874 ) (74)

as 1198910infin le 120576 Since 119879(1198911 +1198910) = 119892 +119879(1198910) we obtain by (56)and (73) that

1 minus (4 + radic2) 120576 minus (minusΔ1198911)le 119888119861 (119879 (1198911 + 1198910) 31205874 ) minus 119888119860 (1198911 + 1198910 31205874 )

(75)

We also obtain by (62) and (74) that

119888119861 (119879 (1198911 + 1198910) 1205874 ) minus 119888119860 (1198911 + 1198910 1205874 )le radic22 + 4120576 minus (minusΔ1198911)

(76)

Since 119888119861(119879(119891) 120593)minus119888119860(119891 120593) is invariant for any 120593 by (37) from(75) and (76) we deduce that 120576 ge (2 minusradic2)2(8 +radic2) and thisis impossible since 120576 le 1100

8 Journal of Function Spaces

Next we show a proof of the second assertion (2) Let119891 isin 119860 We prove that there exists a sequence 119891119899 sub 119860 whichuniformly converges to 119891 such that 120588(120590(119891119899)) 997888rarr 0 as 119899 997888rarrinfin Without loss of generality we may assume that 119891infin = 1Then there exists 1199100 isin Ch(119860) such that |119891(1199100)| = 119891infinby [29 Proposition 63] We may assume that 119891(1199100) = 1Suppose that 119899 ge 4 Put

119880119899 = 119909 isin 119883 1003816100381610038161003816119891 (119909) minus 11003816100381610038161003816 le 11198992 (77)

and

Ω = co(119870 (1) cup 1 + 1119899) + 119870( 11198992) (78)

(In the following we identify R2 and C that is we identify(119909 119910) and 119909 + 119894119910 for every 119909 119910 isin R) Since we assume that119899 ge 4 we infer by a simple calculation that

[1 minus 11198992 minus 120575119899119899 1 + 1119899 + 120575119899119899 ] times [minus 11198992 minus 120575119899119899 11198992 + 120575119899119899 ]sub Ω

(79)

for 120575119899 with 0 lt 120575119899 le 11198992 We assume that 0 lt 120575119899 le 11198992 By[5 Lemma 2] there exists ℎ119899 isin 119860 such that ℎ119899infin le 1 + 2120575119899ℎ119899(1199100) = 1 |ℎ119899 + 1| le 2120575119899 on119883 119880119899 and |Im ℎ119899| le 2120575119899 on119883Put 119892119899 = (ℎ119899 + 1)2 Then 119892119899 isin 119860 and

10038171003817100381710038171198921198991003817100381710038171003817infin le 1 + 120575119899119892119899 (1199100) = 110038161003816100381610038161198921198991003816100381610038161003816 le 120575119899 on 119883 1198801198991003816100381610038161003816Im1198921198991003816100381610038161003816 le 120575119899 on 119883

minus120575119899 le Re119892119899 le 1 + 120575119899 on 119883

(80)

Let 119909 isin 119880119899 Then we have

1 minus 11198992 le Re119891 (119909) le 1minus 11198992 le Im119891 (119909) le 11198992

(81)

and

minus120575119899119899 le Re119892119899 (119909)119899 le 1119899 + 120575119899119899

minus120575119899119899 le Im119892119899 (119909)119899 le 120575119899119899

(82)

Hence

1 minus 11198992 minus 120575119899119899 le Re(119891 (119909) + 119892119899 (119909)119899 ) le 1 + 1119899 + 120575119899119899 minus 11198992 minus 120575119899119899 le Im(119891 (119909) + 119892119899 (119909)119899 ) le 11198992 + 120575119899119899

(83)

It follows that we have

(119891 + 119892119899119899 ) (119909) isin [1 minus 11198992 minus 120575119899119899 1 + 1119899 + 120575119899119899 ]times [minus 11198992 minus 120575119899119899 11198992 + 120575119899119899 ] sub Ω

(84)

for 119909 isin 119880119899 Suppose that 119909 isin 119883 119880119899 Then10038161003816100381610038161003816100381610038161003816119891 (119909) +

119892119899 (119909)11989910038161003816100381610038161003816100381610038161003816 le 1 +

120575119899119899 le 1 + 11198993 lt 1 + 11198992 (85)

and hence

119891 (119909) + 119892119899 (119909)119899 isin 119870 (1) + 119870( 11198992) sub Ω (86)

for 119909 isin 119883 119880119899 Since 1 + 1119899 = 119891(1199100) + 119892119899(1199100)119899 we have bycombining (84) and (86) that

1 + 1119899 isin (119891 + 119892119899119899 ) (119883) sub Ω (87)

AsΩ is convex we obtain

1 + 1119899 isin 120590 (119891 + 119892119899119899 ) sub Ω (88)

Recall that for 119903 ge 0 and a complex number 1199110119870 (1199110 119903) = 119908 isin C 1003816100381610038161003816119908 minus 11991101003816100381610038161003816 le 119903 (89)

denotes the closed disk with center 1199110 and radius 119903 Weobserve that 120588(Ω 1 + 1119899) = 1119899 + 11198992 Recall that

120588(Ω 1 + 1119899)= sup119903 119911 isin Ω 1 + 1119899 isin 119870 (119911 119903) sub Ω

(90)

Let ℓ1 be the line defined by the equation

119910 = minus119899radic2119899 + 1 (119909 minus (1 +1119899 + 11198992 + 11198993)) (91)

part of which is a part of the boundary ofΩ Let ℓ2 be the linedefined by the equation

119910 = minus119899radic2119899 + 1 (119909 minus (1 minus11198992)) (92)

By some calculation we have that the distance between 1 minus11198992 and 1+1119899 is 1119899+11198992 and it coincides with the distancebetween the point 1 minus 11198992 and the line ℓ1 Hence we see that

1 + 1119899 isin 119870(1 minus 11198992 1119899 + 11198992 ) sub Ω (93)

Thus 1119899 + 11198992 le 120588(Ω 1 + 1119899)Next we prove that 119870(119901 119903) cap Ω119888 = 0 for every 119901 isin Ω

with 119903 = |119901 minus (1 + 1119899)| gt 1119899 + 11198992 It will follow that1119899 + 11198992 ge 120588(Ω 1 + 1119899) and the equality will hold Let

Journal of Function Spaces 9

119901 = (119909119901 119910119901) = 119909119901 + 119894119910119901 isin Ω with 119903 = |119901 minus (1 + 1119899)| gt1119899 + 11198992 We prove the case where 119910119901 ge 0 A proof for thecase where 119910119901 le 0 is the same and we omit it We divideΩ+ =119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902 isin Ω 119910119902 ge 0 into two parts

Ω1 = 119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902isin Ω+ minus119899radic2119899 + 1 (119909119902 minus (1 minus

11198992)) le 119910119902le minus119899radic2119899 + 1 (119909 minus (1 +

1119899 + 11198992 + 11198993))(94)

and

Ω2 = 119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902 isin Ω+ 119910119902le minus119899radic2119899 + 1 (119909119902 minus (1 minus

11198992)) (95)

Suppose that 119901 isin Ω1 and 119903 = |119901minus (1+1119899)| gt 1119899+11198992Thedistance between ℓ1 and ℓ2 is 1119899+11198992 Hence119870(119901 119903)capΩ119888 =0 Suppose that 119901 isin Ω2 and 119903 = |119901 minus (1 + 1119899)| gt 1119899 + 11198992that is119901 = 1minus11198992 Let ℓ1015840 be the line passing through119901whichis parallel to ℓ2 Let 1199011015840 be the unique point in the intersectionof ℓ1015840 and the 119909-axis Then 1199011015840 = 1 minus 11198992 minus 119906 for some 119906 ge 0Then the distance between ℓ1 and ℓ1015840 is 1119899+11198992+119906(1+1119899)which is equal to the distance between the point119901 and the lineℓ1 On the other hand1003816100381610038161003816100381610038161003816119901 minus (1 + 1119899)

1003816100381610038161003816100381610038161003816 ge10038161003816100381610038161003816100381610038161199011015840 minus (1 + 1119899)

1003816100381610038161003816100381610038161003816 = 1119899 + 11198992 + 119906 (96)

It follows that 119870(119901 119903) cap Ω119888 = 0 We conclude that if 119901 isin Ωsatisfies 119903 = |119901minus(1+1119899)| gt 1119899+11198992 then119870(119901 119903)capΩ119888 = 0Thus we have

120588 (Ω 1 + 1119899) = 1119899 + 11198992 (97)

Since 1 + 1119899 isin 120590(119891 + 119892119899119899) sub Ω we have120588 (120590 (119891 + 119892119899119899 )) le 120588 (120590 (119891 + 119892119899119899 ) 1 + 1119899)

le 120588 (Ω 1 + 1119899) = 1119899 + 11198992 (98)

On the other hand 119892119899119899infin 997888rarr 0 as 119899 997888rarr infin ensures that119891 + 119892119899119899 minus 119891infin 997888rarr 0 as 119899 997888rarr infin It follows that for every120576 gt 0A120576 is dense in (119860 sdot infin)Finally we show a proof of the third assertion (3) As is

pointed out in the proof of [5Theorem] 120588(co(119870)+119870(119888)) ge 119888for any 119870 sub C and any 119888 ge 0 Let 120576 gt 0 and 119891 isin A120576 Supposethat Δ119891 le 0 Then by (43) we have (119891) = (119879119891) + 119870(minusΔ119891)Hence we have

120588 ( (119891)) = 120588 (120590 (119879119891) + 119870 (minusΔ119891)) ge minusΔ119891 (99)

As 120576 ge 120588(120590(119891)) we conclude by (45) that10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin = Δ119891 ge minus120576 (100)

This completes the proof of the theorem

4 Hermitian Operators on a BanachAlgebras of Continuous Maps WhoseValues Are in a Uniform Algebras

Let 119883 and 119884 be compact Hausdorff spaces Let 119861 be aunital subalgebra of 119862(119883) which separates the points of 119883Throughout this sectionwe assume119861 is a Banach algebrawiththe norm sdot 119861 and 119860 is a uniform algebra on 119884 Recall thata uniform algebra on 119884 is a uniformly closed subalgebra of119862(119884) which contains constants and separates the points of 119884For functions 119891 isin 119862(119883) and 119892 isin 119862(119884) let 119891 otimes 119892 isin 119862(119883 times 119884)be the function defined by119891otimes119892(119909 119910) = 119891(119909)119892(119910) for (119909 119910) isin119883 times 119884 and for a subspace 119864119883 of 119862(119883) and a subspace 119864119884 of119862(119884) put

119864119883 otimes 119864119884

= 119899sum

119895=1

119891119895 otimes 119892119895 119899 isin N 119891119895 isin 119864119883 119892119895 isin 119864119884

(101)

and

1 otimes 119864119884 = 1 otimes 119892 119892 isin 119864119884 (102)

Throughout the section 119861 is a unital subalgebra of 119862(119883 times 119884)with a Banach algebra norm sdot 119861 We assume that 119861otimes119860 sub 119861Note that 119861 separates the points of119883times119884 since119860 separates thepoints of 119884 and 119861 separates the points of 119883 We assume thatthere exists a compact Hausdorff space M and a complex-linear map 119863 119861 997888rarr 119862(M) such that ker119863 = 1 otimes 119860 Weassume that 119865119861 = 119865infin(119883times119884)+119863(119865)infin(M) for every 119865 isin 119861Hence119863 is continuous Defining

lsaquo119865lsaquo = 119863 (119865)infin(M) 119865 isin 119861 (103)

lsaquo sdot lsaquo is a one-invariant seminorm in the sense of Jarosz lsaquo sdot lsaquois a seminorm on 119861 such that lsaquo119865+ 1lsaquo = lsaquo119865lsaquo for every 119865 isin 119861Hence the norm sdot 119861 is a natural norm (see [5 p67]) Notethat 119861 is a regular subspace of 119862(119883times119884) in the sense of Jarosz[5 Proposition 2]

Lumerrsquos seminal paper [35] opened up a useful methodof finding isometries which is often referred to as Lumerrsquosmethod It involves the notion ofHermitian operators and thefact that 119880119867119880minus1 must be Hermitian if119867 is Hermitian and 119880is a surjective isometry

Definition 6 Let A be a unital Banach algebra We say that119890 isin A is a Hermitian element if1003817100381710038171003817exp (119894119905119890)1003817100381710038171003817A = 1 (104)

for every 119905 isin R The set of all Hermitian elements of A isdenoted by119867(A)

If A is a unital 119862lowast-algebra then 119867(A) is the set of allself-adjoint elements of A Hence 119867(119872119899(C)) is the set of allHermitian matrices and119867(119862(119884)) = 119862R(119884)Definition 7 Let 119864 be a complex Banach space The Banachalgebra of all bounded operators on 119864 is denoted by 119861(119864) Wesay that 119879 isin 119861(119864) is a Hermitian operator if 119879 isin 119867(119861(119864))

10 Journal of Function Spaces

Note that a Hermitian element of a unital Banach algebraand a Hermitian operator are usually defined in terms ofnumerical range or semi-inner product Here we define themby an equivalent form (see [36]) By the definition of aHermitian operator we have the following

Proposition 8 Let 119864119895 be a complex Banach space for 119895 = 1 2Suppose that 119881 1198641 997888rarr 1198642 is a surjective isometry and 119867 1198641 997888rarr 1198641 is a Hermitian operator Then 119881119867119881minus1 1198642 997888rarr 1198642

is a Hermitian operator

Proposition 9 An element 119865 isin 119861 is Hermitian if and only ifthere exists 119891 isin 119860 cap 119862R(119884) such that 119865 = 1 otimes 119891Proof Suppose that 119865 isin 119861 is a Hermitian element Then1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884) + 1003817100381710038171003817119863 (exp (119894119905119865))1003817100381710038171003817infin(M)

= 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817 = 1 (105)

for every 119905 isin R Suppose that there exists a point (119909 119910) isin119883 times 119884 with Im119865(119909 119910) = 0 where Im denotes the imaginarypart of a complex number Suppose that Im119865(119909 119910) gt 0Then1003817100381710038171003817exp (minus119894119865)1003817100381710038171003817infin(119883times119884) ge 1003816100381610038161003816exp (minus119894119865 (119909 119910))1003816100381610038161003816

= exp (Im119865 (119909 119910)) gt 1 (106)

Suppose that Im119865(119909 119910) lt 0 Then1003817100381710038171003817exp (119894119865)1003817100381710038171003817infin(119883times119884) ge 1003816100381610038161003816exp (119894119865 (119909 119910))1003816100381610038161003816= exp (minusIm119865 (119909 119910)) gt 1 (107)

In any case we have there exists 119905 isin R such that1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817119861 ge 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884) gt 1 (108)

which contradicts our assumption We have that

119865 isin 119862R (119883 times 119884) (109)

Thus for every (119904 119905) isin 119883 times 119884 and 119905 isin R |exp(119894119905119865(119904 119905))| = 1Hence exp(119894119905119865)infin(119883times119884) = 1 for every 119905 isin R By (105)we have119863(exp(119894119905119865)infin(M) = 0 which ensures that 119863(exp(119894119905119865)) = 0for every 119905 isin R Thus exp(119894119905119865) isin 1 otimes 119860 for every 119905 isin R Wehave

exp (119894119905119865) minus 1119905 minus 119894119865 = infinsum119899=2

((119894119865)119899 119905119899minus2119899 ) 119905 (110)

and hence for every 119905 isin R with |119905| le 1 we have10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817 le (

infinsum119899=2

119865119899119861|119905|119899minus2119899 ) |119905|

le (infinsum119899=2

119865119899119861119899 ) |119905| le (exp 119865119861) |119905|

(111)

It follows that10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817infin(119883times119884)

le 10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817119861

997888rarr 0(112)

as 119905 997888rarr 0 Since (exp(119894119905119865)minus1)119905 isin 1otimes119860 for each 119905 isin R thereexists 119892119905 isin 119860 such that

exp (119894119905119865) minus 1119905 = 1 otimes 119892119905 (113)

By (112) we have10038171003817100381710038171 otimes 1198921119899 minus 1198941198651003817100381710038171003817infin(119883times119884)

997888rarr 0 (114)

as 119899 997888rarr infin We have that 1 otimes 1198921119899 is a Cauchy sequencein 119862(119883 times 119884) thus we infer that1198921119899 is a Cauchy sequence in119862(119884) Since 119860 is uniformly closed as it is a uniform algebrathere exists 119892 isin 119860 such that

10038171003817100381710038171198921119899 minus 1198921003817100381710038171003817infin(119884)997888rarr 0 (115)

and hence10038171003817100381710038171 otimes 1198921119899 minus 1 otimes 1198921003817100381710038171003817infin(119883times119884)

997888rarr 0 (116)

as 119899 997888rarr infin It follows by (114) that 119894119865 = 1 otimes 119892 thus119865 = 1 otimes (minus119894119892) isin 1 otimes 119860 (117)

By (109) we see that minus119894119892 isin 119862R(119884) thus we have 119891 = minus119894119892 isin119860 cap 119862R(119884) and 119865 = 1 otimes 119891Suppose conversely that 119861 ni 119865 = 1otimes119891 for119891 isin 119860cap119862R(119884)

We infer that 119865 isin 119862R(119883 times 119884) and |exp(119894119905(119865(119909 119910))| = 1 forevery 119905 isin R and (119909 119910) isin 119883 times 119884 Hence exp(119894119905119865)infin(119883times119884) = 1for every 119905 isin R Since

exp (119894119905119865) = exp (119894119905 (1 otimes 119891)) = 1 otimes exp (119894119905119891) isin 1 otimes 119860 (118)

we have 119863(exp(119894119905119865)) = 0 It follows that1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817 = 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884)

+ 1003817100381710038171003817119863 (exp (119894119905119865))1003817100381710038171003817infin(M) = 1 (119)

for every 119905 isin R We conclude that 119865 is a Hermitian elementin 119861

Note that 119891 isin 119860 is Hermitian if and only if119891 isin 119860cap119862R(119884)by [37 Proposition 5] Hence Proposition 9 asserts that 119865 is aHermitian element in119861 if and only if119865 = 1otimes119891 for aHermitianelement 119891 in 119860Proposition 10 Suppose that 119880 119861 997888rarr 119861 is a surjectiveunital isometry Then 119880 is an algebra isomorphism

Proof As we have already mentioned 119861 is a regular subspace(in the sense of Jarosz) with a natural norm Then byTheorem 1 119880 is also an isometry with respect to the supre-mum norm on 119883 times 119884 Then 119880 is uniquely extended to asurjective isometry with respect to the supremum norm from the uniform closure 119861 onto itself Since 119861 is a uniformalgebra a theorem of Nagasawa [32] asserts that is analgebra isomorphism since (1) = 1 Thus 119880 is an algebraisomorphism from 119861 onto itself

Journal of Function Spaces 11

Theorem 11 A bounded operator 119879 119861 997888rarr 119861 is a Hermitianoperator if and only if 119879(1) is a Hermitian element in 119861 and119879 = 119872119879(1) the multiplication operator by 119879(1)Proof By Proposition 10 every surjective unital isometry on119861 is multiplicative Then by [37 Theorem 4] we have theconclusion

5 Banach Algebras of 119862(119884)-Valued Maps

Suppose that119883 is a compact Hausdorff space Suppose that 119861is a unital point separating subalgebra of119862(119884) equipped witha Banach algebra norm Then 119861 is semisimple because 119891 isin119861 119891(119909) = 0 is a maximal ideal of 119861 for every 119909 isin 119883 and theJacobson radical of 119861 vanishes The inequality 119891infin le 119891119861for every 119891 isin 119861 is well known We say that 119861 is natural if themap 119890 119884 997888rarr 119872119861 defined by 119910 997891997888rarr 120601119910 where 120601119910(119891) = 119891(119910)for every 119891 isin 119861 is bijective We say that 119861 is self-adjoint if 119861 isnatural and conjugate-closed in the sense that 119891 isin 119861 impliesthat 119891 isin 119861 for every 119891 isin 119861 where sdot denotes the complexconjugation on 119884Definition 12 Let 119883 and 119884 be compact Hausdorff spacesSuppose that 119861 is a unital point separating subalgebra of119862(119883)equipped with a Banach algebra norm sdot 119861 Suppose that119861 is self-adjoint Suppose that 119861 is a unital point separatingsubalgebra of 119862(119883 times 119884) such that 119861 otimes 119862(119884) sub 119861 equippedwith a Banach algebra norm sdot 119861 Suppose that 119861 is self-adjoint We say that 119861 is a natural 119862(119884)-valuezation of 119861 ifthere exists a compact Hausdorff space M and a complex-linear map 119863 119861 997888rarr 119862(M) such that ker119863 = 1 otimes 119862(119884) and119863(119862R(119883 times 119884) cap 119861) sub 119862R(M) which satisfies

119865 = 119865infin(119883times119884) + 119863 (119865)infin(M) 119865 isin 119861 (120)

The term ldquoa natural 119862(119884)-valuezation of 119861rdquo comes fromthe natural norm defined by Jarosz [5] In fact the norm sdot 119861is a natural norm in the sense of Jarosz [5]

Note that (119883119862(119884) 119861 119861) need not be an admissiblequadruple defined by Nikou and OrsquoFarrell [38] (cf [31]) sincewe do not assume that 119865(sdot 119910) 119865 isin 119861 119910 isin 119884 sub 119861 whichis a requirement for the admissible quadruple On the otherhand if (119883119862(119884) 119861 119861) is an admissible quadruple of type Ldefined in [30] then 119861 is a natural 119862(119884)-valuezation of 119861 dueto Definition 12

Example 13 Let 119861 = 1198621([0 1]) and 119861 = Lip([0 1] 119862(119884))for 119884 = 119901 a singleton Then Lip([0 1]) is algebraically iso-morphic to Lip([0 1] 119862(119884)) Suppose thatM is the maximalideal space of 119871infin([0 1]) and 119863 119861 997888rarr 119862(M) is definedby 119891 997891997888rarr Γ(1198911015840) where Γ denotes the Gelfand transformin 119871infin([0 1]) Then 119861 is a natural 119862(119884)-valuezation of 119861The Banach algebra Lip([0 1]) with the norm 119891infin([01]) +1198911015840infin([01]) is isometrically isomorphic to 119861

Let 119884 be a compact Hausdorff space Note that a closedsubalgebra 119861 of Lip((119870 119889120572) 119862(119884))which appears in Example12 in [30] is an example of a natural 119862(119884)-valuezation of 119861

The Banach algebras1198621([0 1] 119862(119884)) and 1198621(T 119862(119884)))whichappear in Examples 16 and 17 in [30] respectively are alsoexamples of natural 119862(119884)-valuezations of 1198621([0 1])6 Isometries on Natural 119862(119884)-ValuezationsThemain theorem in this paper is the following

Theorem 14 Suppose that 119861119895 is a natural 119862(119884119895)-valuezationof 119861119895 sub 119862(119883119895) for 119895 = 1 2 We assume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (121)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for119895 = 1 2 Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linear isometry Then there exists ℎ isin 119862(1198842) such that |ℎ| = 1on 1198842 a continuous map 120593 1198832 times 1198842 997888rarr 1198831 such that120593(sdot 119910) 1198832 997888rarr 1198831 is a homeomorphism for each 119910 isin 1198842and a homeomorphism 120591 1198842 997888rarr 1198841 which satisfies119880 (119865) (119909 119910) = ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842 (122)

for every 119865 isin 1198611

In short a surjective isometry between 119862(119884)-valuezationsis a weighted composition operator of a specific form thehomeomorphism 1198832 times 1198842 997888rarr 1198831 times 1198841 (119909 119910) 997891997888rarr(120593(119909 119910) 120591(119910)) has the second coordinate that depends onlyon the second variable 119910 isin 1198842 A composition operatorinduced by such a homeomorphism is said to be of type BJin [31 37] after the study of Botelho and Jamison [39]

Quite recently the author of this paper and Oi [30 Theo-rem 8] proved a similar result of Theorem 14 for admissiblequadruples of type L To prove it we apply Proposition 32and the following comments in [31] Instead of this we proveTheorem 14 by Lumerrsquosmethod with which a proof is simplerthan that in [30]

In the following in this section we assume that 119861119895 is anatural 119862(119884119895)-valuezation of 119861 sub 119862(119883119895) for 119895 = 1 2 Weassume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (123)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for 119895 = 1 2Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linearisometry A crucial part of a proof of Theorem 14 is to proveProposition 15

Proposition 15 Suppose that 1198832 is not a singleton Thereexists ℎ isin 119862(1198842) with |ℎ| = 1 on 1198842 such that 119880(11198611) = 11198612 otimesℎ

A similar result for admissible quadruples of type L isproved in [30 Proposition 9] If we assumed that

119865 (sdot 119910) 119865 isin 119861119895 119910 isin 119884119895 sub 119861119895 (124)

then 119861119895 were an admissible quadruple of type L Although119861119895 in this paper need not be an admissible quadruple of type

12 Journal of Function Spaces

L a proof of Proposition 15 is completely the same as that in[30 Proposition 9] since we do not make use of the condition(124) in the proof of [30 Proposition 9] The condition (124)is needed in [30] when we apply Proposition 32 and thefollowing comments in [31]

7 Proof of Theorem 14 An Application ofLumerrsquos Method

Proof of Theorem 14 A proof for the case where 1198831 = 1199091and 1198832 = 1199092 are singletons is the same as the proof ofTheorem 8 in [30]

Suppose that1198832 is not a singleton By Proposition 15 thereexists ℎ isin 119862R(1198842) with |ℎ| = 1 on 1198842 such that 119880(1) =1 otimes ℎ Letting 1198800 1198611 997888rarr 1198612 by 1198800(119865) = (1 otimes ℎ)119880(119865)119865 isin 1198611 we see by the hypothesis (1 otimes ℎ)1198651198612 = 1198651198612 forevery 119865 isin 1198612 that 1198800 is a surjective unital isometry from1198611 onto 1198612 Then Corollary 2 asserts that 1198800 is an algebraisomorphism Let 119891 isin 119862R(1198841) By Proposition 9 1 otimes 119891 isa Hermitian element in 1198611 Then by Theorem 11 1198721otimes119891 is aHermitian operator on 1198611 By Proposition 8 11988001198721otimes119891119880minus1

0 is aHermitian operator on 1198612 Then by Theorem 11 there exists119878(119891) isin 119862R(1198842) such that 11988001198721otimes119891119880minus1

0 = 1198721otimes119878(119891) Hence anoperator 119878 119862R(1198841) 997888rarr 119862R(1198842) is defined Since 1198800 is analgebra isomorphism it is easy to see that 119878 is a real algebraisomorphism from 119862R(1198841) onto 119862R(1198842) Then 119878 119862(1198841) 997888rarr119862(1198842) defined by 119878(119891) = 119878(Re119891) + 119894119878(Im119891) for 119891 isin 119862(1198841)gives a complex algebra isomorphism Gelfand theory assertsthat there is a homeomorphism 120591 1198842 997888rarr 1198841 such that119878(119891) = 119891 ∘ 120591 119891 isin 119862(1198841) It follows that

11988001198721otimes119891119880minus10 = 1198721otimes119891∘120591 119891 isin 119862 (1198841) (125)

Since 119880minus10 (1) = 1 we have

1198800 (1 otimes 119891) = 1 otimes 119891 ∘ 120591 119891 isin 119862 (1198841) (126)

Define Φ 1198611 997888rarr 1198612 by Φ(119886) = 1198800(119886 otimes 1) 119886 isin 1198611Since 1198800 is an algebra isomorphism the map Φ is a unitalhomomorphism Since the maximal ideal space of 1198611 is 1198831

and the maximal ideal space of 1198612 is 1198832 times 1198842 there is acontinuous map 120593 1198832 times 1198842 997888rarr 1198831 such that

Φ(119886) (119909 119910) = 119886 (120593 (119909 119910)) 119886 isin 1198611 (119909 119910) isin 1198832 times 1198842 (127)

It follows by (126) and (127) that

1198800 (119886 otimes 119891) (119909 119910) = 1198800 ((119886 otimes 1) (1 otimes 119891)) (119909 119910)= 1198800 (119886 otimes 1) (119909 119910)1198800 (1 otimes 119891) (119909 119910)= 119886 (120593 (119909 119910))119891 (120591 (119910))= (119886 otimes 119891) (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198832 times 1198842

(128)

for every 119886 isin 1198611 and 119891 isin 119862(1198841) Thus

1198800 (sum(119886119895 otimes 119891119895)) (119909 119910)= (sum(119886119895 otimes 119891119895)) (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842(129)

for every sum(119886119895 otimes 119891119895) isin 1198611 otimes 119862(1198841) By the Stone-Weierstrasstheorem 1198611 otimes 119862(1198841) is uniformly dense in 119862(1198831 times 1198841) henceany element in 1198611 is uniformly approximated by 1198611 otimes 119862(1198841)As 1198800 is also an isometry with respect to the uniform normwe see that

1198800 (119865) (119909 119910) = 119865 (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198831 times 1198841 (130)

for every 119865 isin 1198611 and

119880 (119865) (119909 119910) = (1 otimes ℎ) (119909 119910) 119865 (120593 (119909 119910) 120591 (119910))= ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198831 times 1198841(131)

As 1198800 is an algebra isomorphism the map 1198832 times 1198842 997888rarr1198831 times 1198841 defined by (119909 119910) 997891997888rarr (120593(119909 119910) 120591(119910)) gives ahomeomorphism Therefore for every 119910 isin 1198842 the map

120593 (sdot 119910) 1198832 997888rarr 1198831 (132)

is a homeomorphismSuppose that 1198831 is not a singleton By the same way as

in the last part of the proof of Theorem 8 in [30] we havethat 1198832 is not a singleton Then we have the conclusion bythe previous argument

8 Application of Theorem 14

We exhibit applications of Theorem 14

Corollary 16 ([4 Theorem 33]) Suppose that119880 Lip([0 1]) 997888rarr Lip([0 1]) is a surjective isometrywith respect to the norm defined by 119891infin([01]) + 1198911015840infin([01])

for 119891 isin Lip([0 1]) Then 119880(1) is a constant function of unitmodulus such that

119880(119891) (119909) = 119880 (1) 119891 (119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (133)

or

119880(119891) (119909) = 119880 (1) 119891 (1 minus 119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (134)

The converse statement also holds

Proof By Example 13 we may suppose that Lip([0 1]) is aBanach algebra of 119862(119884)-valuezation Applying Theorem 14we have that119880(1) = 1otimesℎ for ℎ isin 119862(119884)with |ℎ| = 1 Since our

Journal of Function Spaces 13

119884 is a singleton 119880(1) is a constant function of unit modulusWe also see that the corresponding continuous map 120593 [0 1] times 119884 997888rarr [0 1] can be considered as a homeomorphismfrom [0 1] onto [0 1] therefore we have that119880(119891) (119909) = 119880 (1) 119891 (120593 (119909))

119891 isin Lip ([0 1]) 119909 isin [0 1] (135)

The rest is a routine argument to prove that 120593 is an isometryhence 120593(119909) = 119909 119909 isin [0 1] or 120593(119909) = 1 minus 119909 119909 isin [0 1]

The converse statement is trivial

Corollaries 14 15 18 and 19 in [30 Section 6] follow herewith a similar proof

Data Availability

The data used to support the findings of this study areavailable from the corresponding author upon request

Conflicts of Interest

The author declares that they have no conflicts of interest

Acknowledgments

This work was supported by JSPS KAKENHI Grants Num-bers JP16K05172 and JP15K04921

References

[1] K de Leeuw ldquoBanach spaces of Lipschitz functionsrdquo StudiaMathematica vol 21 pp 55ndash66 19611962

[2] A K Roy ldquoExtreme points and linear isometries of the Banachspace of Lipschitz functionsrdquoCanadian Journal of MathematicsJournal Canadien de Mathematiques vol 20 pp 1150ndash11641968

[3] M Cambern ldquoIsometries of certain Banach algebrasrdquo StudiaMathematica vol 25 pp 217ndash225 19641965

[4] N V Rao and A K Roy ldquoLinear isometries of some functionspacesrdquo Pacific Journal of Mathematics vol 38 pp 177ndash192 1971

[5] K Jarosz ldquoIsometries in semisimple commutative BanachalgebrasrdquoProceedings of the AmericanMathematical Society vol94 no 1 pp 65ndash71 1985

[6] K Jarosz and V D Pathak ldquoIsometries between functionspacesrdquo Transactions of the AmericanMathematical Society vol305 no 1 pp 193ndash206 1988

[7] NWeaver ldquoIsometries of noncompact Lipschitz spacesrdquoCana-dian Mathematical Bulletin Bulletin Canadien de Mathemat-iques vol 38 no 2 pp 242ndash249 1995

[8] A Jimenez-Vargas and M Villegas-Vallecillos ldquoLinear isome-tries between spaces of vector-valued Lipschitz functionsrdquoProceedings of the American Mathematical Society vol 137 no4 pp 1381ndash1388 2009

[9] A Jimenez-Vargas and M Villegas-Vallecillos ldquoInto linearisometries between spaces of Lipschitz functionsrdquo HoustonJournal of Mathematics vol 34 no 4 pp 1165ndash1184 2008

[10] F Botelho and J Jamison ldquoSurjective isometries on spacesof differentiable vector-valued functionsrdquo Studia Mathematicavol 192 no 1 pp 39ndash50 2009

[11] E Mayer-Wolf ldquoIsometries between Banach spaces of Lipschitzfunctionsrdquo Israel Journal of Mathematics vol 38 no 1-2 pp 58ndash74 1981

[12] A Jimenez-Vargas M Villegas-Vallecillos and Y-S WangldquoBanach-Stone theorems for vector-valued little Lipschitz func-tionsrdquo Publicationes Mathematicae vol 74 no 1-2 pp 81ndash1002009

[13] J Araujo andLDubarbie ldquoNoncompactness andnoncomplete-ness in isometries of Lipschitz spacesrdquo Journal of MathematicalAnalysis and Applications vol 377 no 1 pp 15ndash29 2011

[14] F Botelho R J Fleming and J Jamison ldquoExtreme points andisometries on vector-valued Lipschitz spacesrdquo Journal of Math-ematical Analysis and Applications vol 381 no 2 pp 821ndash8322011

[15] H Koshimizu ldquoLinear isometries on spaces of continuouslydifferentiable and Lipschitz continuous functionsrdquo NihonkaiMathematical Journal vol 22 no 2 pp 73ndash90 2011

[16] F Botelho J Jamison and B Zheng ldquoIsometries on spaces ofvector valued Lipschitz functionsrdquo Positivity An InternationalMathematics Journal Devoted to Theory and Applications ofPositivity vol 17 no 1 pp 47ndash65 2013

[17] A Ranjbar-Motlagh ldquoA note on isometries of Lipschitz spacesrdquoJournal of Mathematical Analysis and Applications vol 411 no2 pp 555ndash558 2014

[18] F Botelho and J Jamison ldquoSurjective isometries on spaces ofvector valued continuous and Lipschitz functionsrdquo PositivityAn International Mathematics Journal Devoted to Theory andApplications of Positivity vol 17 no 3 pp 395ndash405 2013Erratum to Surjective isometries on spaces of vector valuedcontinuous and Lipschitz functions by F Botelho 20 (2016)757ndash759

[19] T Miura and H Takagi ldquoSurjective isometries on the Banachspace of continuously differentiable functionsrdquo ContemporaryMathematics vol 687 pp 181ndash192 2017

[20] K Kawamura ldquoBanach-Stone type theorems for 1198621-functionspaces over Riemannian manifoldsrdquo Acta Universitatis Szegedi-ensis Acta Scientiarum Mathematicarum vol 83 no 3-4 pp551ndash591 2017

[21] K Kawamura ldquoPerturbations of norms on 1198621-function spacesand associated isometry groupsrdquo Topology Proceedings vol 51pp 169ndash196 2018

[22] K Kawamura ldquoA Banach-Stone type theorem for 1198621-functionspaces over the circlerdquo Topology Proceedings vol 53 pp 15ndash262019

[23] L Li D Chen Q Meng and Y-S Wang ldquoSurjective isometrieson vector-valued differentiable function spacesrdquo Annals ofFunctional Analysis vol 9 no 3 pp 334ndash343 2018

[24] K Kawamura H Koshimizu and T Miura ldquoNorms onC1([01]) and there isometriesrdquo Acta Scientiarum Mathemati-carum vol 84 no 12 pp 239ndash261 2018

[25] L Li A M Peralta L Wang and Y-S Wang ldquoWeak-2-local isometries on uniform algebras and Lipschitz algebrasrdquohttpsarxivorgabs170503619

[26] A Jimenez-Vargas L LiAMPeralta LWang andY-SWangldquo2-local standard isometries on vector-valued Lipschitz func-tion spacesrdquo Journal of Mathematical Analysis and Applicationsvol 461 no 2 pp 1287ndash1298 2018

[27] A Ranjbar-Motlagh ldquoIsometries of Lipschitz type functionspacesrdquoMathematischeNachrichten vol 291 no 11-12 pp 1899ndash1907 2018

14 Journal of Function Spaces

[28] N Weaver Lipschitz Algebras World Scientific Publishing CoInc River Edge NJ USA 1999

[29] R R Phelps Lectures on Choquetrsquos theorem vol 1757 of LectureNotes in Mathematics Springer-Verlag Berlin Germany 2ndedition 2001

[30] O Hatori and S Oi ldquoIsometries on Banach algebras of vector-valued mapsrdquo Acta Scientiarum Mathematicarum vol 84 no12 pp 151ndash183 2018

[31] O Hatori S Oi and H Takagi ldquoPeculiar homomorphismsbetween algebras of vector-valued mapsrdquo Studia Mathematicavol 242 no 2 pp 141ndash163 2018

[32] M Nagasawa ldquoIsomorphisms between commutative Banachalgebras with an application to rings of analytic functionsrdquoKodai Mathematical Seminar Reports vol 11 pp 182ndash188 1959

[33] K de LeeuwW Rudin and J Wermer ldquoThe isometries of somefunction spacesrdquo Proceedings of the American MathematicalSociety vol 11 pp 694ndash698 1960

[34] OHatori A Jimenez-Vargas andMVillegas-Vallecillos ldquoMapswhich preserve norms of non-symmetrical quotients betweengroups of exponentials of Lipschitz functionsrdquo Journal of Math-ematical Analysis and Applications vol 415 no 2 pp 825ndash8452014

[35] G Lumer ldquoOn the isometries of reflexiveOrlicz spacesrdquoAnnalesde lrsquoInstitut Fourier vol 68 pp 99ndash109 1963

[36] R J Fleming and J E Jamison Isometries on Banach SpacesFunction Spaces vol 129 ofMonographs and Surveys in Pure andApplied Mathematics Chapman amp Hall CRC Boca Raton FlaUSA 2003

[37] OHatori and SOi ldquoHermitian operators onBanach algebras ofvector-valued Lipschitzmapsrdquo Journal ofMathematical Analysisand Applications vol 452 no 1 pp 378ndash387 2017 Corrigendumto rdquoHermitian operators on Banach algebras of vector-valuedLipschitz mapsrdquo Journal of Mathematical Analysis and Appli-cations 452 (2017) 378ndash387 MR3628025

[38] A Nikou andA G OrsquoFarrell ldquoBanach algebras of vector-valuedfunctionsrdquo Glasgow Mathematical Journal vol 56 no 2 pp419ndash426 2014

[39] F Botelho and J Jamison ldquoHomomorphisms on a class ofcommutative Banach algebrasrdquo Rocky Mountain Journal ofMathematics vol 43 no 2 pp 395ndash416 2013

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Page 4: Hermitian Operators and Isometries on Banach Algebras of ...downloads.hindawi.com/journals/jfs/2018/8085304.pdfJournalofFunctionSpaces normiseitherthemaxnormorthesumnorm.esituation

4 Journal of Function Spaces

is a natural norm in the sense of Jarosz (see [5]) such thatlim119905997888rarr+0((max1 119905 minus 1)119905) = 0 Then by Theorem 1 we havethe following

Corollary 4 Let 119870119895 be a compact metric space for 119895 = 1 2Suppose that 119879 Lip(1198701) 997888rarr Lip(1198702) is a surjective complex-linear isometry with respect to the norm sdot119872 Assume 1198791 = 1Then there exists a surjective isometry 120593 1198832 997888rarr 1198831 such that

119879119891 = 119891 ∘ 120593 119891 isin Lip (1198701) (16)

Conversely if 119879 Lip(1198701) 997888rarr Lip(1198702) is of a similar form as(16) then 119879 is a surjective isometry with respect to both of sdot119872and sdot 119871 such that 1198791 = 1

Proof As sdot 119872 is a natural norm we have by Corollary 2 thatthere is a homeomorphism 120593 1198702 997888rarr 1198701 such that

119879119891 (119909) = 119891 ∘ 120593 (119909) 119891 isin Lip (1198701) 119909 isin 1198702 (17)

Then by a routine argument we see that 120593 is an isometryConverse statement is trivial

Without the assumption that 1198791 = 1 in Corollary 4 onemay expect that 119879 is a weighted composition operator But itis not the case A simple counterexample is given by Weaver[7 p242] (see also [28])

As is pointed out in [34] the original proof of Theorem 1needs a revision in some part and a proof when 119860 and 119861are algebras of Lipschitz functions is revised [34 Proposition7] Although a revised proof for a general case is similar tothat of Proposition 7 in [34] we exhibit it here for the sakeof completeness of this paper To prove Theorem 1 we needLemma 2 in [5] in the sameway as the original proof of JaroszThe following is Lemma 2 in [5]

Lemma5 (Jarosz [5]) Assume119860 is a regular subspace of119862(119883)with 1 isin 119860 and let 1199090 isin 119862ℎ(119860) Then for any 120576 gt 0 and anyopen neighborhood 119880 of 1199090 there is an 119891 isin 119860 such that

10038171003817100381710038171198911003817100381710038171003817infin le 1 + 120576119891 (1199090) = 11003816100381610038161003816119891 (119909) + 11003816100381610038161003816 le 120576

119909 isin 119883 119880(18)

and |Im119891(119909)| le 120576 for all 119909 isin 119883Proof The proof is essentially due to the original proof ofLemma 2 in [5] Several minor changes are needed Weitemize them as follows

(i) Five 1205762rsquos between 11 lines and 5 lines from the bottomof page 69 read as 1205763

(ii) Next 119909 isin 1198831198801 reads as 119909 isin 1198801 on the bottom of page69

(iii) We point out that the term sum1198960minus1119895=1 (119891119895(119909) minus 1) which

appears on the first line of the first displayed inequal-ities on page 70 reads 0 if 1198960 = 1

(iv) The term 1 + 120576 on the right hand side of the secondline of the same inequalities reads as 1 + 1205763

(v) Two 1205762rsquos on the same line read as 1205763(vi) On the next line ((119899 + 1)119899)(1205762) reads as 1205763(vii) For any 1 le 1198960 le 119899 we infer that

1 ge 1 minus 21198960 minus 1119899 ge 1 minus 2119899 minus 1119899 gt minus1 (19)

Hence we have |119891(119909)| le 1 + 120576 if 119909 isin 1198801 by the firstdisplayed inequalities of page 70

(viii) The inequality 119891infin le 120576 on the fifth line on page 70reads as 119891infin le 1 + 120576

Let119870 be a nonempty convex subset of the complex planeand 120593 isin [0 2120587) Put

119888 (119870 120593) = sup 119886 isin R there is a 119887isin R with (119886 + 119894119887) 119890119894120593 isin 119870 (20)

Note that we may write

119888 (119870 120593) = sup Re120582 120582 isin 119890minus119894120593119870 (21)

Let 119860 be a subspace of 119862(119883) for a compact Hausdorff spaceFor 119891 isin 119860 we put 120590(119891) = 119891(119883) and 120590(119891) = co(119891(119883)) whereco(sdot) denotes the closed convex-hull We define the functions

119888119860 119860 times [0 2120587) 997888rarr R119888119860 (119891 120593) = 119888 ( (119891) 120593)

119903119860 119860 times R+ times [0 2120587) 997888rarr R

+119903119860 (119891 119905 120593) = 10038171003817100381710038171003817119891 + 119890119894120593119905110038171003817100381710038171003817infin

(22)

Proof of Theorem 1 Let 119891 isin 119860 First we note that10038171003817100381710038171198911003817100381710038171003817infin = sup 119911 isin 120590 (119891) = sup 119911 isin (119891) (23)

since (119891) is the closed convex-hull of a compact set 120590(119891) =119891(119883) We prove the inequalities

119905 + 119888119860 (119891 120593) le 119903119860 (119891 119905 120593)le radic(119905 + 119888119860 (119891 120593))2 + 100381710038171003817100381711989110038171003817100381710038172infin

119905 ge 0 120593 isin [0 2120587) (24)

which appear on p 68 in [5] Put 119904 = 119888119860(119891 120593) As 120590(119891) iscompact there exists 119887 isin R such that (119904 + 119894119887)119890119894120593 isin 120590(119891)Hence

119890119894120593 (119905 + 119904 + 119894119887) isin 119890119894120593119905 + 120590 (119891) = 120590 (1198901198941205931199051 + 119891) (25)

Journal of Function Spaces 5

As

119905 + 119904 le |119905 + 119904| le |119905 + 119904 + 119894119887| = 10038161003816100381610038161003816119890119894120593 (119905 + 119904 + 119894119887)10038161003816100381610038161003816le 100381710038171003817100381710038171198901198941205931199051 + 11989110038171003817100381710038171003817infin = 119903119860 (119891 119905 120593) (26)

we have

119905 + 119888119860 (119891 120593) le 119903119860 (119891 119905 120593) (27)

Let 119909 isin 119883 By the definition of 119904 = 119888119860(119891 120593) we infer thatRe(119890minus119894120593119891(119909)) le 119904 hence we have Re(119905 + 119890minus119894120593119891(119909)) le 119905 + 119904 forevery 119905 ge 0 Then

Re 119890minus119894120593119891 (119909) le Re (119905 + 119890minus119894120593119891 (119909)) le 119905 + 119904 119905 ge 0 (28)

Letting119872 = max|119905 + 119904| |Re(119890minus119894120593119891(119909))| we have10038161003816100381610038161003816119890119894120593119905 + 119891 (119909)100381610038161003816100381610038162 = 10038161003816100381610038161003816119905 + 119890minus119894120593119891 (119909)100381610038161003816100381610038162

= (Re (119905 + 119890minus119894120593119891 (119909)))2+ (Im (119905 + 119890minus119894120593119891 (119909)))2

le 1198722 + (Im (119905 + 119890minus119894120593119891 (119909)))2le |119905 + 119904|2 + (Re (119890minus119894120593119891 (119909)))2+ (Im (119890minus119894120593119891 (119909)))2

= |119905 + 119904|2 + 10038161003816100381610038161003816119890minus119894120593119891 (119909)100381610038161003816100381610038162

(29)

As 119909 isin 119883 is arbitrary we have

119903119860 (119891 119905 120593) = 10038171003817100381710038171003817119905 + 119890minus11989412059311989110038171003817100381710038171003817infin le radic|119905 + 119904|2 + 100381710038171003817100381711989110038171003817100381710038172infin= radic(119905 + 119888119860 (119891 120593))2 + 100381710038171003817100381711989110038171003817100381710038172infin

(30)

It follows that (24) holds In the same way we have

119905 + 119888119861 (119892 120593) le 119903119861 (119892 119905 120593) le radic(119905 + 119888119861 (119892 120593))2 + 100381710038171003817100381711989210038171003817100381710038172infin119905 ge 0 120593 isin [0 2120587) (31)

for every 119892 isin 119861 By (24) and (31) we infer that

lim119905997888rarrinfin

(119903119860 (119891 119905 120593) minus 119905) = 119888119860 (119891 120593) lim119905997888rarrinfin

(119903119861 (119879119891 119905 120593) minus 119905) = 119888119861 (119879119891 120593) (32)

As lsaquo sdot lsaquo119860 is 1-invariant we have119901 (119903119860 (119891 119905 120593) lsaquo119891lsaquo119860)= 119901 (10038171003817100381710038171003817119891 + 119890119894120593119905110038171003817100381710038171003817infin lsaquo119891 + 1198901198941205931199051lsaquo119860) (33)

As 119879 is an isometry 119879(1) = 1 and lsaquo sdot lsaquo119861 is 1-invariant wehave

119901 (10038171003817100381710038171003817119891 + 119890119894120593119905110038171003817100381710038171003817infin lsaquo119891 + 1198901198941205931199051lsaquo119860)= 119902 (10038171003817100381710038171003817119879 (119891 + 1198901198941205931199051)10038171003817100381710038171003817infin lsaquo119879 (119891 + 1198901198941205931199051)lsaquo119861)= 119902 (10038171003817100381710038171003817119879 (119891) + 119890119894120593119905110038171003817100381710038171003817infin lsaquo119879 (119891) + 1198901198941205931199051lsaquo119861)= 119902 (119903119861 (119879119891 119905 120593) lsaquo119879119891lsaquo119861)

(34)

Thus

119901 (119903119860 (119891 119905 120593) lsaquo119891lsaquo119860) = 119902 (119903119861 (119879119891 119905 120593) lsaquo119879119891lsaquo119861) (35)

It follows that0 = lim

119905997888rarrinfin(119901 (119903119860 (119891 119905 120593) lsaquo119891lsaquo119860)

minus 119902 (119903119861 (119879119891 119905 120593) lsaquo119879119891lsaquo119861))= lim

119905997888rarrinfin(119903119860 (119891 119905 120593) 119901(1 lsaquo119891lsaquo119860119903119860 (119891 119905 120593))

minus 119903119860 (119891 119905 120593)) + lim119905997888rarrinfin

(119903119861 (119879119891 119905 120593)minus 119903119861 (119879119891 119905 120593) 119902(1 lsaquo119879119891lsaquo119861119903119861 (119879119891 119905 120593)))+ lim

119905997888rarrinfin(119903119860 (119891 119905 120593) minus 119905 minus (119903119861 (119879119891 119905 120593) minus 119905))

= lsaquo119891lsaquo119860119863(119901) minus lsaquo119879119891lsaquo119861119863(119902) + 119888119860 (119891 120593)minus 119888119861 (119879119891 120593)

(36)

Recall that 119863(119901) = lim119905997888rarr+0(119901(1 119905) minus 1)119905 and 119863(119902) =lim119905997888rarr+0(119902(1 119905) minus 1)119905 It follows thatlsaquo119891lsaquo119860119863 (119901) minus lsaquo119879119891lsaquo119861119863(119902) = 119888119861 (119879119891 120593) minus 119888119860 (119891 120593)

119891 isin 119860 120593 isin [0 2120587) (37)

Suppose that 119863(119901) = 119863(119902) = 0 Then we have by (37)that 119888119861(119879119891 120593) = 119888119860(119891 120593) for every 119891 isin 119860 and 120593 isin [0 2120587)By Lemma 1 in [5] we infer that 120590(119879119891) = 120590(119891) Thus we have119879119891infin = 119891infin We have proved that 119879 is an isometry from(119860 sdot infin) onto (119861 sdot infin) if119863(119901) = 119863(119902) = 0

Suppose that 119860 and 119861 are regular subspaces of 119862(119883) and119862(119884) respectively Let 119891 isin 119860 PutΔ119891 = lsaquo119891lsaquo119860119863(119901) minus lsaquo119879119891lsaquo119861119863(119902) (38)

Suppose that Δ119891 ge 0 For any 119903 ge 0 and any nonemptycompact convex subset 119870 sub C we have that

119888 (119870 + 119870 (119903) 120593) = 119888 (119870 120593) + 119903 (39)

for all 120593 isin [0 2120587) where 119870(119903) = 119911 isin C |119911| le 119903 Then by(37) we have

119888 (120590 (119879119891) 120593) = 119888119861 (119879119891 120593) = 119888119860 (119891 120593) + Δ119891= 119888 (120590 (119891) 120593) + Δ119891= 119888 (120590 (119891) + 119870 (Δ119891) 120593)

(40)

6 Journal of Function Spaces

for all 120593 isin [0 2120587) It follows by Lemma 1 in [5] that

(119879119891) = 120590 (119891) + 119870 (Δ119891) (41)

and therefore10038171003817100381710038171198791198911003817100381710038171003817infin = 10038171003817100381710038171198911003817100381710038171003817infin + Δ119891 (42)

If Δ119891 le 0 then a similar calculation shows that

(119891) = (119879119891) + 119870 (minusΔ119891) (43)

and10038171003817100381710038171198911003817100381710038171003817infin = 10038171003817100381710038171198791198911003817100381710038171003817infin + (minusΔ119891) (44)

It follows that in any case (Δ119891 ge 0 Δ119891 le 0) we obtainΔ119891 = 10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin 119891 isin 119860 (45)

We will prove that1003817100381710038171003817119879 (119891)1003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin = Δ119891 ge 0 (46)

for all 119891 isin 119860 Once it is proved applying the same argumentfor119879minus1 instead of119879 we see that 119879minus1119892infinminus119892infin ge 0 for every119892 isin 119861 As 119879 is a bijection it follows that 119891infin minus 119879119891infin ge 0for every 119891 isin 119860 It will follow that 119879119891infin = 119891infin for every119891 isin 119860 A proof of (46) is the following For every 120576 gt 0denote

A120576 = 119891 isin 119860 120588 (120590 (119891)) le 120576 (47)

The inequality in (46) is deduced by the following assertionswhich appear in the proof of [5 Theorem]

(1) 119879 is a continuous mapping from (119860 sdot infin) onto (119861 sdotinfin)(2) For each 120576 gt 0 the setA120576 is dense in (119860 sdot infin)(3) For each 120576 gt 0 and each 119891 isin A120576 it holds that119879(119891)infin ge 119891infin minus 120576

Suppose that these assertions are proved Let 119891 isin 119860 By (2)for any 120576 gt 0 there is a sequence 119891119899 of functions inA120576 suchthat 119891119899 minus 119891infin 997888rarr 0 as 119899 997888rarr infin By (3) we have

10038171003817100381710038171198791198911198991003817100381710038171003817infin minus 10038171003817100381710038171198911198991003817100381710038171003817infin ge minus120576 (48)

for every 119899 Letting 119899 997888rarr infin we have10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin ge minus120576 (49)

by (1) As 120576 gt 0 is arbitrary we have that10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin ge 0 (50)

We show proofs of three assertions (1) (2) and (3) aboveprecisely The proof of (1) is slightly different from the corre-sponding one in [5 p 70] This change is rather ambitiousWe also point out that the terms minus1205872 and 1205872 which appearin the formulae (7) and (8) in [5] seem inappropriate theyread for example as 31205874 and 1205874 respectively

We now proceed to prove the first statement Aiming fora contradiction suppose that 119879 is not continuous from (119860 sdotinfin) to (119861 sdot infin) Let 120576 be a positive real number less than1100 Then there is a function 1198910 isin 119860 such that 1198910infin le 120576and 119879(1198910)infin = 1 Then there exist 1199100 isin Ch(119861) such that|119879(1198910)(1199100)| = 1 by [29 Proposition 63] Since 119879 is complex-linear we may suppose that 119879(1198910)(1199100) = 1

By (41) and (45) we deduce that Δ1198910 = 119879(1198910)infin minus1198910infin ge 1 minus 120576 and 120590(119879(1198910)) = 120590(1198910) + 119870(Δ1198910) As 1198910infin le 120576and 119879(1198910)infin = 1 we infer that 120590(1198910) sub 119870(120576) and (119879(1198910) sub119870(1) Let 119911 isin 119870(1 minus 2120576) Let 119909 isin 119883 Then |1198910(119909)| le 120576 assertsthat 119908 = 119911 minus 1198910(119909) isin 119870(1 minus 120576) sub 119870(Δ1198910) Thus

119911 = 1198910 (119909) + 119908 isin 120590 (1198910) + 119870 (Δ1198910) = 120590 (119879 (1198910)) (51)

Hence119870(1 minus 2120576) sub 120590(1198791198910) As 119879(1198910)infin = 1 we have119870(1 minus 2120576) sub 120590 (119879 (1198910)) sub 119870 (1) (52)

Consider the open neighborhood 1198800 of 1199100 in1198832 given by

1198800 = 119910 isin 1198832 1003816100381610038161003816119879 (1198910) (119910) minus 11003816100381610038161003816 lt 120576 (53)

We infer that 1198800 is a proper subset of1198832 by (52) Then by [5Lemma 2] there exists 119892 isin 119861 such that 119892infin le 1 + 120576 119892(1199100) =1 |119892(119910) + 1| lt 120576 for every 119910 isin 1198832 1198800 and |Im119892(119910)| lt 120576 forall 119910 isin 1198832 If 119867 denotes the closed rectangle whose verticesare the four points plusmn(1 + 120576) plusmn 120576119894 we have

120590 (119892) sub 119867 (54)

Consider now the set

119871 = 11989031205871198944119911 |119911| le 1 Re 119911 ge 1 minus 2120576 (55)

We claim that119879(1198910)(1198832)cap119871 = 0 Suppose that119879(1198910)(1198832)cap119871 =0 As 119879(1198910)(1198832) is compact there exists a positive integer119899 such that 119879(1198910)(1198832) cap 119871119899 = 0 where 119871119899 = 11989031205871198944119911 |119911| le 1Re 119911 gt 1 minus 2120576 minus 1119899 Then (52) gives 119879(1198910)(1198832) sub119870(1)119871119899 As (119879(1198910)) is the closed convex-hull of 119879(1198910)(1198832)it is contained in the closed convex set 119870(1) 119871119899 On theother hand (1 minus 2120576)11989031205871198944 isin 119870(1 minus 2120576) sub 120590(119879(1198910)) by (52) As(1minus2120576)11989031205871198944 isin 119871 sub 119871119899 this contradicts 120590(119879(1198910)) sub 119870(1) 119871119899and this proves our claim Hence there is 1199101 isin 1198832 with119879(1198910)(1199101) isin 119871 As 120576 le 1100 it follows that |119879(1198910)(1199101) minus1| ge 120576 and so 1199101 isin 1198832 1198800 Hence |119892(1199101) + 1| lt 120576 Thus119892(1199101) + 119879(1198910)(1199101) is in 119871 minus 1 + K(120576) Thus we have

1 + radic22 minus 3120576 le 119888119861 (119892 + 119879 (1198910) 31205874 ) (56)

We claim that

120590 (119892 + 119879 (1198910)) sub co (119870 (minus1 1) cup 2) + 119870 (3120576) (57)

where119870(1199110 119903) = 119911 isin C |119911 minus 1199110| le 119903 Let 119910 isin 1198832 Supposefirst that |119879(1198910)(119910)minus1| lt 120576 Since 119892(1198832) sub 119867 by (54) we have

119879 (1198910) (119910) + 119892 (119910) isin 119870 (1 120576) + 119867 = (119867 + 1) + 119870 (120576) (58)

Journal of Function Spaces 7

Suppose next that |119879(1198910)(119910)minus1| ge 120576Then 119910 isin 1198832 1198800 and so|119892(119910) + 1| lt 120576 Moreover |119879(1198910)(119910)| le 1 Therefore we have

119892 (119910) + 119879 (1198910) (119910) isin 119870 (1) minus 1 + 119870 (120576)= 119870 (minus1 1) + 119870 (120576) (59)

It follows from (58) and (59) that

(119892 + 119879 (1198910)) (1198832) sub 119870 (minus1 1) cup ((119867 + 1)) + 119870 (120576) (60)

and hence

120590 (119892 + 119879 (1198910)) sub co (119870 (minus1 1) cup 2) + 119870 (3120576) (61)

as is claimed Therefore we have

119888119861 (119892 + 119879 (1198910) 1205874 ) le radic2 + 3120576 (62)

Put 1198911 = 119879minus1(119892) We claim that Δ1198911 le 120576 If Δ1198911 lt 0 thereis nothing to prove Suppose that Δ1198911 ge 0 Then by (41) wehave

120590 (119892) = 120590 (1198911) + 119870 (Δ1198911) (63)

Since 120590(119892) sub 119867 by (54) we have

119867 sup 120590 (1198911) + 119870 (Δ1198911) (64)

As 119867 does not include a closed disk with the radius greaterthan 120576 we conclude that Δ1198911 le 120576

In the following we will consider two cases 0 le Δ1198911 le 120576and Δ1198911 le 0 Suppose first that 0 le Δ1198911 le 120576 Then (64) yields(1198911) sub 119867 From 1198910infin le 120576 we deduce that 120590(1198910) sub 119870(120576)Hence we have

120590 (1198911 + 1198910) sub 120590 (1198911) + 120590 (1198910) sub (1198910) + 120590 (1198910)sub 119867 + 119870 (120576) (65)

Since119867 +119870(120576) is convex we have (1198911 + 1198910) sub 119867 + 119870 (120576) (66)

From (39) we infer that

119888119860 (1198911 + 1198910 31205874 ) = 119888 (120590 (1198911 + 1198910) 31205874 )le 119888 (119867 + 119870 (120576) 31205874 )= 119888 (119867 31205874 ) + 120576= radic22 + (1 + radic2) 120576

(67)

Since 119879(1198911 + 1198910) = 119892 + 119879(1198910) from (56) and (67) we obtainthat

1 minus (4 + radic2) 120576 le 119888119861 (119879 (1198911 + 1198910) 31205874 )minus 119888119860 (1198911 + 1198910 31205874 )

(68)

By (63) and 1 = 119892(1199100) we deduce that 1 isin 120590(1198911) + 119870(Δ1198911)Thus there is 119911 isin 120590(1198911) such that |119911 minus 1| le Δ1198911 It follows thatradic22 minus Δ1198911 le 119888119860(1198911 1205874) hence we have

radic22 minus 2120576 le 119888119860 (1198911 + 1198910 1205874 ) (69)

as 1198910infin le 120576 and 0 le Δ1198911 le 120576 We get by (62) and (69) that

119888119861 (119879 (1198911 + 1198910) 1205874 ) minus 119888119860 (1198911 + 1198910 1205874 ) leradic22 + 5120576 (70)

On the other hand 119888119861(119879(119891) 120593)minus119888119860(119891 120593) is invariant for any 120593by (37) From (68) and (70)we deduce that 120576 ge (2minusradic2)2(9+radic2) and this contradicts that 120576 le 1100

For the second case suppose next that Δ1198911 le 0 Then by(43) we have

120590 (1198911) = 120590 (119892) + 119870 (minusΔ1198911) (71)

and by (54) it follows that (1198911) sub 119867 + 119870(minusΔ1198911) Moreover(1198910) sub 119870(120576) since 1198910infin le 120576 Then

120590 (1198911 + 1198910) sub 120590 (1198911) + 120590 (1198910) sub 120590 (1198911) + 120590 (1198910)sub 119867 + 119870(minusΔ1198911) + 119870 (120576) (72)

Hence (1198911+1198910) sub 119867+119870(minusΔ1198911)+119870(120576) Using (39) we inferthat

119888119860 (1198911 + 1198910 31205874 ) le 119888 (119867 + 119870(minusΔ1198911) + 119870 (120576) 31205874 )= 119888 (119867 31205874 ) + (minusΔ1198911) + 120576= radic22 + (1 + radic2) 120576 + (minusΔ1198911)

(73)

By (71) we obtain that 120590(1198911) sup 119892(1198832) + 119870(minusΔ1198911) and as119892(1199100) = 1 we infer that 120590(1198911) sup 1 + 119870(minusΔ1198911) Hence radic22 +(minusΔ1198911) le 119888119860(1198911 1205874) so thatradic22 + (minusΔ1198911) minus 120576 le 119888119860 (1198911 + 1198910 1205874 ) (74)

as 1198910infin le 120576 Since 119879(1198911 +1198910) = 119892 +119879(1198910) we obtain by (56)and (73) that

1 minus (4 + radic2) 120576 minus (minusΔ1198911)le 119888119861 (119879 (1198911 + 1198910) 31205874 ) minus 119888119860 (1198911 + 1198910 31205874 )

(75)

We also obtain by (62) and (74) that

119888119861 (119879 (1198911 + 1198910) 1205874 ) minus 119888119860 (1198911 + 1198910 1205874 )le radic22 + 4120576 minus (minusΔ1198911)

(76)

Since 119888119861(119879(119891) 120593)minus119888119860(119891 120593) is invariant for any 120593 by (37) from(75) and (76) we deduce that 120576 ge (2 minusradic2)2(8 +radic2) and thisis impossible since 120576 le 1100

8 Journal of Function Spaces

Next we show a proof of the second assertion (2) Let119891 isin 119860 We prove that there exists a sequence 119891119899 sub 119860 whichuniformly converges to 119891 such that 120588(120590(119891119899)) 997888rarr 0 as 119899 997888rarrinfin Without loss of generality we may assume that 119891infin = 1Then there exists 1199100 isin Ch(119860) such that |119891(1199100)| = 119891infinby [29 Proposition 63] We may assume that 119891(1199100) = 1Suppose that 119899 ge 4 Put

119880119899 = 119909 isin 119883 1003816100381610038161003816119891 (119909) minus 11003816100381610038161003816 le 11198992 (77)

and

Ω = co(119870 (1) cup 1 + 1119899) + 119870( 11198992) (78)

(In the following we identify R2 and C that is we identify(119909 119910) and 119909 + 119894119910 for every 119909 119910 isin R) Since we assume that119899 ge 4 we infer by a simple calculation that

[1 minus 11198992 minus 120575119899119899 1 + 1119899 + 120575119899119899 ] times [minus 11198992 minus 120575119899119899 11198992 + 120575119899119899 ]sub Ω

(79)

for 120575119899 with 0 lt 120575119899 le 11198992 We assume that 0 lt 120575119899 le 11198992 By[5 Lemma 2] there exists ℎ119899 isin 119860 such that ℎ119899infin le 1 + 2120575119899ℎ119899(1199100) = 1 |ℎ119899 + 1| le 2120575119899 on119883 119880119899 and |Im ℎ119899| le 2120575119899 on119883Put 119892119899 = (ℎ119899 + 1)2 Then 119892119899 isin 119860 and

10038171003817100381710038171198921198991003817100381710038171003817infin le 1 + 120575119899119892119899 (1199100) = 110038161003816100381610038161198921198991003816100381610038161003816 le 120575119899 on 119883 1198801198991003816100381610038161003816Im1198921198991003816100381610038161003816 le 120575119899 on 119883

minus120575119899 le Re119892119899 le 1 + 120575119899 on 119883

(80)

Let 119909 isin 119880119899 Then we have

1 minus 11198992 le Re119891 (119909) le 1minus 11198992 le Im119891 (119909) le 11198992

(81)

and

minus120575119899119899 le Re119892119899 (119909)119899 le 1119899 + 120575119899119899

minus120575119899119899 le Im119892119899 (119909)119899 le 120575119899119899

(82)

Hence

1 minus 11198992 minus 120575119899119899 le Re(119891 (119909) + 119892119899 (119909)119899 ) le 1 + 1119899 + 120575119899119899 minus 11198992 minus 120575119899119899 le Im(119891 (119909) + 119892119899 (119909)119899 ) le 11198992 + 120575119899119899

(83)

It follows that we have

(119891 + 119892119899119899 ) (119909) isin [1 minus 11198992 minus 120575119899119899 1 + 1119899 + 120575119899119899 ]times [minus 11198992 minus 120575119899119899 11198992 + 120575119899119899 ] sub Ω

(84)

for 119909 isin 119880119899 Suppose that 119909 isin 119883 119880119899 Then10038161003816100381610038161003816100381610038161003816119891 (119909) +

119892119899 (119909)11989910038161003816100381610038161003816100381610038161003816 le 1 +

120575119899119899 le 1 + 11198993 lt 1 + 11198992 (85)

and hence

119891 (119909) + 119892119899 (119909)119899 isin 119870 (1) + 119870( 11198992) sub Ω (86)

for 119909 isin 119883 119880119899 Since 1 + 1119899 = 119891(1199100) + 119892119899(1199100)119899 we have bycombining (84) and (86) that

1 + 1119899 isin (119891 + 119892119899119899 ) (119883) sub Ω (87)

AsΩ is convex we obtain

1 + 1119899 isin 120590 (119891 + 119892119899119899 ) sub Ω (88)

Recall that for 119903 ge 0 and a complex number 1199110119870 (1199110 119903) = 119908 isin C 1003816100381610038161003816119908 minus 11991101003816100381610038161003816 le 119903 (89)

denotes the closed disk with center 1199110 and radius 119903 Weobserve that 120588(Ω 1 + 1119899) = 1119899 + 11198992 Recall that

120588(Ω 1 + 1119899)= sup119903 119911 isin Ω 1 + 1119899 isin 119870 (119911 119903) sub Ω

(90)

Let ℓ1 be the line defined by the equation

119910 = minus119899radic2119899 + 1 (119909 minus (1 +1119899 + 11198992 + 11198993)) (91)

part of which is a part of the boundary ofΩ Let ℓ2 be the linedefined by the equation

119910 = minus119899radic2119899 + 1 (119909 minus (1 minus11198992)) (92)

By some calculation we have that the distance between 1 minus11198992 and 1+1119899 is 1119899+11198992 and it coincides with the distancebetween the point 1 minus 11198992 and the line ℓ1 Hence we see that

1 + 1119899 isin 119870(1 minus 11198992 1119899 + 11198992 ) sub Ω (93)

Thus 1119899 + 11198992 le 120588(Ω 1 + 1119899)Next we prove that 119870(119901 119903) cap Ω119888 = 0 for every 119901 isin Ω

with 119903 = |119901 minus (1 + 1119899)| gt 1119899 + 11198992 It will follow that1119899 + 11198992 ge 120588(Ω 1 + 1119899) and the equality will hold Let

Journal of Function Spaces 9

119901 = (119909119901 119910119901) = 119909119901 + 119894119910119901 isin Ω with 119903 = |119901 minus (1 + 1119899)| gt1119899 + 11198992 We prove the case where 119910119901 ge 0 A proof for thecase where 119910119901 le 0 is the same and we omit it We divideΩ+ =119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902 isin Ω 119910119902 ge 0 into two parts

Ω1 = 119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902isin Ω+ minus119899radic2119899 + 1 (119909119902 minus (1 minus

11198992)) le 119910119902le minus119899radic2119899 + 1 (119909 minus (1 +

1119899 + 11198992 + 11198993))(94)

and

Ω2 = 119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902 isin Ω+ 119910119902le minus119899radic2119899 + 1 (119909119902 minus (1 minus

11198992)) (95)

Suppose that 119901 isin Ω1 and 119903 = |119901minus (1+1119899)| gt 1119899+11198992Thedistance between ℓ1 and ℓ2 is 1119899+11198992 Hence119870(119901 119903)capΩ119888 =0 Suppose that 119901 isin Ω2 and 119903 = |119901 minus (1 + 1119899)| gt 1119899 + 11198992that is119901 = 1minus11198992 Let ℓ1015840 be the line passing through119901whichis parallel to ℓ2 Let 1199011015840 be the unique point in the intersectionof ℓ1015840 and the 119909-axis Then 1199011015840 = 1 minus 11198992 minus 119906 for some 119906 ge 0Then the distance between ℓ1 and ℓ1015840 is 1119899+11198992+119906(1+1119899)which is equal to the distance between the point119901 and the lineℓ1 On the other hand1003816100381610038161003816100381610038161003816119901 minus (1 + 1119899)

1003816100381610038161003816100381610038161003816 ge10038161003816100381610038161003816100381610038161199011015840 minus (1 + 1119899)

1003816100381610038161003816100381610038161003816 = 1119899 + 11198992 + 119906 (96)

It follows that 119870(119901 119903) cap Ω119888 = 0 We conclude that if 119901 isin Ωsatisfies 119903 = |119901minus(1+1119899)| gt 1119899+11198992 then119870(119901 119903)capΩ119888 = 0Thus we have

120588 (Ω 1 + 1119899) = 1119899 + 11198992 (97)

Since 1 + 1119899 isin 120590(119891 + 119892119899119899) sub Ω we have120588 (120590 (119891 + 119892119899119899 )) le 120588 (120590 (119891 + 119892119899119899 ) 1 + 1119899)

le 120588 (Ω 1 + 1119899) = 1119899 + 11198992 (98)

On the other hand 119892119899119899infin 997888rarr 0 as 119899 997888rarr infin ensures that119891 + 119892119899119899 minus 119891infin 997888rarr 0 as 119899 997888rarr infin It follows that for every120576 gt 0A120576 is dense in (119860 sdot infin)Finally we show a proof of the third assertion (3) As is

pointed out in the proof of [5Theorem] 120588(co(119870)+119870(119888)) ge 119888for any 119870 sub C and any 119888 ge 0 Let 120576 gt 0 and 119891 isin A120576 Supposethat Δ119891 le 0 Then by (43) we have (119891) = (119879119891) + 119870(minusΔ119891)Hence we have

120588 ( (119891)) = 120588 (120590 (119879119891) + 119870 (minusΔ119891)) ge minusΔ119891 (99)

As 120576 ge 120588(120590(119891)) we conclude by (45) that10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin = Δ119891 ge minus120576 (100)

This completes the proof of the theorem

4 Hermitian Operators on a BanachAlgebras of Continuous Maps WhoseValues Are in a Uniform Algebras

Let 119883 and 119884 be compact Hausdorff spaces Let 119861 be aunital subalgebra of 119862(119883) which separates the points of 119883Throughout this sectionwe assume119861 is a Banach algebrawiththe norm sdot 119861 and 119860 is a uniform algebra on 119884 Recall thata uniform algebra on 119884 is a uniformly closed subalgebra of119862(119884) which contains constants and separates the points of 119884For functions 119891 isin 119862(119883) and 119892 isin 119862(119884) let 119891 otimes 119892 isin 119862(119883 times 119884)be the function defined by119891otimes119892(119909 119910) = 119891(119909)119892(119910) for (119909 119910) isin119883 times 119884 and for a subspace 119864119883 of 119862(119883) and a subspace 119864119884 of119862(119884) put

119864119883 otimes 119864119884

= 119899sum

119895=1

119891119895 otimes 119892119895 119899 isin N 119891119895 isin 119864119883 119892119895 isin 119864119884

(101)

and

1 otimes 119864119884 = 1 otimes 119892 119892 isin 119864119884 (102)

Throughout the section 119861 is a unital subalgebra of 119862(119883 times 119884)with a Banach algebra norm sdot 119861 We assume that 119861otimes119860 sub 119861Note that 119861 separates the points of119883times119884 since119860 separates thepoints of 119884 and 119861 separates the points of 119883 We assume thatthere exists a compact Hausdorff space M and a complex-linear map 119863 119861 997888rarr 119862(M) such that ker119863 = 1 otimes 119860 Weassume that 119865119861 = 119865infin(119883times119884)+119863(119865)infin(M) for every 119865 isin 119861Hence119863 is continuous Defining

lsaquo119865lsaquo = 119863 (119865)infin(M) 119865 isin 119861 (103)

lsaquo sdot lsaquo is a one-invariant seminorm in the sense of Jarosz lsaquo sdot lsaquois a seminorm on 119861 such that lsaquo119865+ 1lsaquo = lsaquo119865lsaquo for every 119865 isin 119861Hence the norm sdot 119861 is a natural norm (see [5 p67]) Notethat 119861 is a regular subspace of 119862(119883times119884) in the sense of Jarosz[5 Proposition 2]

Lumerrsquos seminal paper [35] opened up a useful methodof finding isometries which is often referred to as Lumerrsquosmethod It involves the notion ofHermitian operators and thefact that 119880119867119880minus1 must be Hermitian if119867 is Hermitian and 119880is a surjective isometry

Definition 6 Let A be a unital Banach algebra We say that119890 isin A is a Hermitian element if1003817100381710038171003817exp (119894119905119890)1003817100381710038171003817A = 1 (104)

for every 119905 isin R The set of all Hermitian elements of A isdenoted by119867(A)

If A is a unital 119862lowast-algebra then 119867(A) is the set of allself-adjoint elements of A Hence 119867(119872119899(C)) is the set of allHermitian matrices and119867(119862(119884)) = 119862R(119884)Definition 7 Let 119864 be a complex Banach space The Banachalgebra of all bounded operators on 119864 is denoted by 119861(119864) Wesay that 119879 isin 119861(119864) is a Hermitian operator if 119879 isin 119867(119861(119864))

10 Journal of Function Spaces

Note that a Hermitian element of a unital Banach algebraand a Hermitian operator are usually defined in terms ofnumerical range or semi-inner product Here we define themby an equivalent form (see [36]) By the definition of aHermitian operator we have the following

Proposition 8 Let 119864119895 be a complex Banach space for 119895 = 1 2Suppose that 119881 1198641 997888rarr 1198642 is a surjective isometry and 119867 1198641 997888rarr 1198641 is a Hermitian operator Then 119881119867119881minus1 1198642 997888rarr 1198642

is a Hermitian operator

Proposition 9 An element 119865 isin 119861 is Hermitian if and only ifthere exists 119891 isin 119860 cap 119862R(119884) such that 119865 = 1 otimes 119891Proof Suppose that 119865 isin 119861 is a Hermitian element Then1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884) + 1003817100381710038171003817119863 (exp (119894119905119865))1003817100381710038171003817infin(M)

= 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817 = 1 (105)

for every 119905 isin R Suppose that there exists a point (119909 119910) isin119883 times 119884 with Im119865(119909 119910) = 0 where Im denotes the imaginarypart of a complex number Suppose that Im119865(119909 119910) gt 0Then1003817100381710038171003817exp (minus119894119865)1003817100381710038171003817infin(119883times119884) ge 1003816100381610038161003816exp (minus119894119865 (119909 119910))1003816100381610038161003816

= exp (Im119865 (119909 119910)) gt 1 (106)

Suppose that Im119865(119909 119910) lt 0 Then1003817100381710038171003817exp (119894119865)1003817100381710038171003817infin(119883times119884) ge 1003816100381610038161003816exp (119894119865 (119909 119910))1003816100381610038161003816= exp (minusIm119865 (119909 119910)) gt 1 (107)

In any case we have there exists 119905 isin R such that1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817119861 ge 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884) gt 1 (108)

which contradicts our assumption We have that

119865 isin 119862R (119883 times 119884) (109)

Thus for every (119904 119905) isin 119883 times 119884 and 119905 isin R |exp(119894119905119865(119904 119905))| = 1Hence exp(119894119905119865)infin(119883times119884) = 1 for every 119905 isin R By (105)we have119863(exp(119894119905119865)infin(M) = 0 which ensures that 119863(exp(119894119905119865)) = 0for every 119905 isin R Thus exp(119894119905119865) isin 1 otimes 119860 for every 119905 isin R Wehave

exp (119894119905119865) minus 1119905 minus 119894119865 = infinsum119899=2

((119894119865)119899 119905119899minus2119899 ) 119905 (110)

and hence for every 119905 isin R with |119905| le 1 we have10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817 le (

infinsum119899=2

119865119899119861|119905|119899minus2119899 ) |119905|

le (infinsum119899=2

119865119899119861119899 ) |119905| le (exp 119865119861) |119905|

(111)

It follows that10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817infin(119883times119884)

le 10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817119861

997888rarr 0(112)

as 119905 997888rarr 0 Since (exp(119894119905119865)minus1)119905 isin 1otimes119860 for each 119905 isin R thereexists 119892119905 isin 119860 such that

exp (119894119905119865) minus 1119905 = 1 otimes 119892119905 (113)

By (112) we have10038171003817100381710038171 otimes 1198921119899 minus 1198941198651003817100381710038171003817infin(119883times119884)

997888rarr 0 (114)

as 119899 997888rarr infin We have that 1 otimes 1198921119899 is a Cauchy sequencein 119862(119883 times 119884) thus we infer that1198921119899 is a Cauchy sequence in119862(119884) Since 119860 is uniformly closed as it is a uniform algebrathere exists 119892 isin 119860 such that

10038171003817100381710038171198921119899 minus 1198921003817100381710038171003817infin(119884)997888rarr 0 (115)

and hence10038171003817100381710038171 otimes 1198921119899 minus 1 otimes 1198921003817100381710038171003817infin(119883times119884)

997888rarr 0 (116)

as 119899 997888rarr infin It follows by (114) that 119894119865 = 1 otimes 119892 thus119865 = 1 otimes (minus119894119892) isin 1 otimes 119860 (117)

By (109) we see that minus119894119892 isin 119862R(119884) thus we have 119891 = minus119894119892 isin119860 cap 119862R(119884) and 119865 = 1 otimes 119891Suppose conversely that 119861 ni 119865 = 1otimes119891 for119891 isin 119860cap119862R(119884)

We infer that 119865 isin 119862R(119883 times 119884) and |exp(119894119905(119865(119909 119910))| = 1 forevery 119905 isin R and (119909 119910) isin 119883 times 119884 Hence exp(119894119905119865)infin(119883times119884) = 1for every 119905 isin R Since

exp (119894119905119865) = exp (119894119905 (1 otimes 119891)) = 1 otimes exp (119894119905119891) isin 1 otimes 119860 (118)

we have 119863(exp(119894119905119865)) = 0 It follows that1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817 = 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884)

+ 1003817100381710038171003817119863 (exp (119894119905119865))1003817100381710038171003817infin(M) = 1 (119)

for every 119905 isin R We conclude that 119865 is a Hermitian elementin 119861

Note that 119891 isin 119860 is Hermitian if and only if119891 isin 119860cap119862R(119884)by [37 Proposition 5] Hence Proposition 9 asserts that 119865 is aHermitian element in119861 if and only if119865 = 1otimes119891 for aHermitianelement 119891 in 119860Proposition 10 Suppose that 119880 119861 997888rarr 119861 is a surjectiveunital isometry Then 119880 is an algebra isomorphism

Proof As we have already mentioned 119861 is a regular subspace(in the sense of Jarosz) with a natural norm Then byTheorem 1 119880 is also an isometry with respect to the supre-mum norm on 119883 times 119884 Then 119880 is uniquely extended to asurjective isometry with respect to the supremum norm from the uniform closure 119861 onto itself Since 119861 is a uniformalgebra a theorem of Nagasawa [32] asserts that is analgebra isomorphism since (1) = 1 Thus 119880 is an algebraisomorphism from 119861 onto itself

Journal of Function Spaces 11

Theorem 11 A bounded operator 119879 119861 997888rarr 119861 is a Hermitianoperator if and only if 119879(1) is a Hermitian element in 119861 and119879 = 119872119879(1) the multiplication operator by 119879(1)Proof By Proposition 10 every surjective unital isometry on119861 is multiplicative Then by [37 Theorem 4] we have theconclusion

5 Banach Algebras of 119862(119884)-Valued Maps

Suppose that119883 is a compact Hausdorff space Suppose that 119861is a unital point separating subalgebra of119862(119884) equipped witha Banach algebra norm Then 119861 is semisimple because 119891 isin119861 119891(119909) = 0 is a maximal ideal of 119861 for every 119909 isin 119883 and theJacobson radical of 119861 vanishes The inequality 119891infin le 119891119861for every 119891 isin 119861 is well known We say that 119861 is natural if themap 119890 119884 997888rarr 119872119861 defined by 119910 997891997888rarr 120601119910 where 120601119910(119891) = 119891(119910)for every 119891 isin 119861 is bijective We say that 119861 is self-adjoint if 119861 isnatural and conjugate-closed in the sense that 119891 isin 119861 impliesthat 119891 isin 119861 for every 119891 isin 119861 where sdot denotes the complexconjugation on 119884Definition 12 Let 119883 and 119884 be compact Hausdorff spacesSuppose that 119861 is a unital point separating subalgebra of119862(119883)equipped with a Banach algebra norm sdot 119861 Suppose that119861 is self-adjoint Suppose that 119861 is a unital point separatingsubalgebra of 119862(119883 times 119884) such that 119861 otimes 119862(119884) sub 119861 equippedwith a Banach algebra norm sdot 119861 Suppose that 119861 is self-adjoint We say that 119861 is a natural 119862(119884)-valuezation of 119861 ifthere exists a compact Hausdorff space M and a complex-linear map 119863 119861 997888rarr 119862(M) such that ker119863 = 1 otimes 119862(119884) and119863(119862R(119883 times 119884) cap 119861) sub 119862R(M) which satisfies

119865 = 119865infin(119883times119884) + 119863 (119865)infin(M) 119865 isin 119861 (120)

The term ldquoa natural 119862(119884)-valuezation of 119861rdquo comes fromthe natural norm defined by Jarosz [5] In fact the norm sdot 119861is a natural norm in the sense of Jarosz [5]

Note that (119883119862(119884) 119861 119861) need not be an admissiblequadruple defined by Nikou and OrsquoFarrell [38] (cf [31]) sincewe do not assume that 119865(sdot 119910) 119865 isin 119861 119910 isin 119884 sub 119861 whichis a requirement for the admissible quadruple On the otherhand if (119883119862(119884) 119861 119861) is an admissible quadruple of type Ldefined in [30] then 119861 is a natural 119862(119884)-valuezation of 119861 dueto Definition 12

Example 13 Let 119861 = 1198621([0 1]) and 119861 = Lip([0 1] 119862(119884))for 119884 = 119901 a singleton Then Lip([0 1]) is algebraically iso-morphic to Lip([0 1] 119862(119884)) Suppose thatM is the maximalideal space of 119871infin([0 1]) and 119863 119861 997888rarr 119862(M) is definedby 119891 997891997888rarr Γ(1198911015840) where Γ denotes the Gelfand transformin 119871infin([0 1]) Then 119861 is a natural 119862(119884)-valuezation of 119861The Banach algebra Lip([0 1]) with the norm 119891infin([01]) +1198911015840infin([01]) is isometrically isomorphic to 119861

Let 119884 be a compact Hausdorff space Note that a closedsubalgebra 119861 of Lip((119870 119889120572) 119862(119884))which appears in Example12 in [30] is an example of a natural 119862(119884)-valuezation of 119861

The Banach algebras1198621([0 1] 119862(119884)) and 1198621(T 119862(119884)))whichappear in Examples 16 and 17 in [30] respectively are alsoexamples of natural 119862(119884)-valuezations of 1198621([0 1])6 Isometries on Natural 119862(119884)-ValuezationsThemain theorem in this paper is the following

Theorem 14 Suppose that 119861119895 is a natural 119862(119884119895)-valuezationof 119861119895 sub 119862(119883119895) for 119895 = 1 2 We assume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (121)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for119895 = 1 2 Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linear isometry Then there exists ℎ isin 119862(1198842) such that |ℎ| = 1on 1198842 a continuous map 120593 1198832 times 1198842 997888rarr 1198831 such that120593(sdot 119910) 1198832 997888rarr 1198831 is a homeomorphism for each 119910 isin 1198842and a homeomorphism 120591 1198842 997888rarr 1198841 which satisfies119880 (119865) (119909 119910) = ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842 (122)

for every 119865 isin 1198611

In short a surjective isometry between 119862(119884)-valuezationsis a weighted composition operator of a specific form thehomeomorphism 1198832 times 1198842 997888rarr 1198831 times 1198841 (119909 119910) 997891997888rarr(120593(119909 119910) 120591(119910)) has the second coordinate that depends onlyon the second variable 119910 isin 1198842 A composition operatorinduced by such a homeomorphism is said to be of type BJin [31 37] after the study of Botelho and Jamison [39]

Quite recently the author of this paper and Oi [30 Theo-rem 8] proved a similar result of Theorem 14 for admissiblequadruples of type L To prove it we apply Proposition 32and the following comments in [31] Instead of this we proveTheorem 14 by Lumerrsquosmethod with which a proof is simplerthan that in [30]

In the following in this section we assume that 119861119895 is anatural 119862(119884119895)-valuezation of 119861 sub 119862(119883119895) for 119895 = 1 2 Weassume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (123)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for 119895 = 1 2Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linearisometry A crucial part of a proof of Theorem 14 is to proveProposition 15

Proposition 15 Suppose that 1198832 is not a singleton Thereexists ℎ isin 119862(1198842) with |ℎ| = 1 on 1198842 such that 119880(11198611) = 11198612 otimesℎ

A similar result for admissible quadruples of type L isproved in [30 Proposition 9] If we assumed that

119865 (sdot 119910) 119865 isin 119861119895 119910 isin 119884119895 sub 119861119895 (124)

then 119861119895 were an admissible quadruple of type L Although119861119895 in this paper need not be an admissible quadruple of type

12 Journal of Function Spaces

L a proof of Proposition 15 is completely the same as that in[30 Proposition 9] since we do not make use of the condition(124) in the proof of [30 Proposition 9] The condition (124)is needed in [30] when we apply Proposition 32 and thefollowing comments in [31]

7 Proof of Theorem 14 An Application ofLumerrsquos Method

Proof of Theorem 14 A proof for the case where 1198831 = 1199091and 1198832 = 1199092 are singletons is the same as the proof ofTheorem 8 in [30]

Suppose that1198832 is not a singleton By Proposition 15 thereexists ℎ isin 119862R(1198842) with |ℎ| = 1 on 1198842 such that 119880(1) =1 otimes ℎ Letting 1198800 1198611 997888rarr 1198612 by 1198800(119865) = (1 otimes ℎ)119880(119865)119865 isin 1198611 we see by the hypothesis (1 otimes ℎ)1198651198612 = 1198651198612 forevery 119865 isin 1198612 that 1198800 is a surjective unital isometry from1198611 onto 1198612 Then Corollary 2 asserts that 1198800 is an algebraisomorphism Let 119891 isin 119862R(1198841) By Proposition 9 1 otimes 119891 isa Hermitian element in 1198611 Then by Theorem 11 1198721otimes119891 is aHermitian operator on 1198611 By Proposition 8 11988001198721otimes119891119880minus1

0 is aHermitian operator on 1198612 Then by Theorem 11 there exists119878(119891) isin 119862R(1198842) such that 11988001198721otimes119891119880minus1

0 = 1198721otimes119878(119891) Hence anoperator 119878 119862R(1198841) 997888rarr 119862R(1198842) is defined Since 1198800 is analgebra isomorphism it is easy to see that 119878 is a real algebraisomorphism from 119862R(1198841) onto 119862R(1198842) Then 119878 119862(1198841) 997888rarr119862(1198842) defined by 119878(119891) = 119878(Re119891) + 119894119878(Im119891) for 119891 isin 119862(1198841)gives a complex algebra isomorphism Gelfand theory assertsthat there is a homeomorphism 120591 1198842 997888rarr 1198841 such that119878(119891) = 119891 ∘ 120591 119891 isin 119862(1198841) It follows that

11988001198721otimes119891119880minus10 = 1198721otimes119891∘120591 119891 isin 119862 (1198841) (125)

Since 119880minus10 (1) = 1 we have

1198800 (1 otimes 119891) = 1 otimes 119891 ∘ 120591 119891 isin 119862 (1198841) (126)

Define Φ 1198611 997888rarr 1198612 by Φ(119886) = 1198800(119886 otimes 1) 119886 isin 1198611Since 1198800 is an algebra isomorphism the map Φ is a unitalhomomorphism Since the maximal ideal space of 1198611 is 1198831

and the maximal ideal space of 1198612 is 1198832 times 1198842 there is acontinuous map 120593 1198832 times 1198842 997888rarr 1198831 such that

Φ(119886) (119909 119910) = 119886 (120593 (119909 119910)) 119886 isin 1198611 (119909 119910) isin 1198832 times 1198842 (127)

It follows by (126) and (127) that

1198800 (119886 otimes 119891) (119909 119910) = 1198800 ((119886 otimes 1) (1 otimes 119891)) (119909 119910)= 1198800 (119886 otimes 1) (119909 119910)1198800 (1 otimes 119891) (119909 119910)= 119886 (120593 (119909 119910))119891 (120591 (119910))= (119886 otimes 119891) (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198832 times 1198842

(128)

for every 119886 isin 1198611 and 119891 isin 119862(1198841) Thus

1198800 (sum(119886119895 otimes 119891119895)) (119909 119910)= (sum(119886119895 otimes 119891119895)) (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842(129)

for every sum(119886119895 otimes 119891119895) isin 1198611 otimes 119862(1198841) By the Stone-Weierstrasstheorem 1198611 otimes 119862(1198841) is uniformly dense in 119862(1198831 times 1198841) henceany element in 1198611 is uniformly approximated by 1198611 otimes 119862(1198841)As 1198800 is also an isometry with respect to the uniform normwe see that

1198800 (119865) (119909 119910) = 119865 (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198831 times 1198841 (130)

for every 119865 isin 1198611 and

119880 (119865) (119909 119910) = (1 otimes ℎ) (119909 119910) 119865 (120593 (119909 119910) 120591 (119910))= ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198831 times 1198841(131)

As 1198800 is an algebra isomorphism the map 1198832 times 1198842 997888rarr1198831 times 1198841 defined by (119909 119910) 997891997888rarr (120593(119909 119910) 120591(119910)) gives ahomeomorphism Therefore for every 119910 isin 1198842 the map

120593 (sdot 119910) 1198832 997888rarr 1198831 (132)

is a homeomorphismSuppose that 1198831 is not a singleton By the same way as

in the last part of the proof of Theorem 8 in [30] we havethat 1198832 is not a singleton Then we have the conclusion bythe previous argument

8 Application of Theorem 14

We exhibit applications of Theorem 14

Corollary 16 ([4 Theorem 33]) Suppose that119880 Lip([0 1]) 997888rarr Lip([0 1]) is a surjective isometrywith respect to the norm defined by 119891infin([01]) + 1198911015840infin([01])

for 119891 isin Lip([0 1]) Then 119880(1) is a constant function of unitmodulus such that

119880(119891) (119909) = 119880 (1) 119891 (119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (133)

or

119880(119891) (119909) = 119880 (1) 119891 (1 minus 119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (134)

The converse statement also holds

Proof By Example 13 we may suppose that Lip([0 1]) is aBanach algebra of 119862(119884)-valuezation Applying Theorem 14we have that119880(1) = 1otimesℎ for ℎ isin 119862(119884)with |ℎ| = 1 Since our

Journal of Function Spaces 13

119884 is a singleton 119880(1) is a constant function of unit modulusWe also see that the corresponding continuous map 120593 [0 1] times 119884 997888rarr [0 1] can be considered as a homeomorphismfrom [0 1] onto [0 1] therefore we have that119880(119891) (119909) = 119880 (1) 119891 (120593 (119909))

119891 isin Lip ([0 1]) 119909 isin [0 1] (135)

The rest is a routine argument to prove that 120593 is an isometryhence 120593(119909) = 119909 119909 isin [0 1] or 120593(119909) = 1 minus 119909 119909 isin [0 1]

The converse statement is trivial

Corollaries 14 15 18 and 19 in [30 Section 6] follow herewith a similar proof

Data Availability

The data used to support the findings of this study areavailable from the corresponding author upon request

Conflicts of Interest

The author declares that they have no conflicts of interest

Acknowledgments

This work was supported by JSPS KAKENHI Grants Num-bers JP16K05172 and JP15K04921

References

[1] K de Leeuw ldquoBanach spaces of Lipschitz functionsrdquo StudiaMathematica vol 21 pp 55ndash66 19611962

[2] A K Roy ldquoExtreme points and linear isometries of the Banachspace of Lipschitz functionsrdquoCanadian Journal of MathematicsJournal Canadien de Mathematiques vol 20 pp 1150ndash11641968

[3] M Cambern ldquoIsometries of certain Banach algebrasrdquo StudiaMathematica vol 25 pp 217ndash225 19641965

[4] N V Rao and A K Roy ldquoLinear isometries of some functionspacesrdquo Pacific Journal of Mathematics vol 38 pp 177ndash192 1971

[5] K Jarosz ldquoIsometries in semisimple commutative BanachalgebrasrdquoProceedings of the AmericanMathematical Society vol94 no 1 pp 65ndash71 1985

[6] K Jarosz and V D Pathak ldquoIsometries between functionspacesrdquo Transactions of the AmericanMathematical Society vol305 no 1 pp 193ndash206 1988

[7] NWeaver ldquoIsometries of noncompact Lipschitz spacesrdquoCana-dian Mathematical Bulletin Bulletin Canadien de Mathemat-iques vol 38 no 2 pp 242ndash249 1995

[8] A Jimenez-Vargas and M Villegas-Vallecillos ldquoLinear isome-tries between spaces of vector-valued Lipschitz functionsrdquoProceedings of the American Mathematical Society vol 137 no4 pp 1381ndash1388 2009

[9] A Jimenez-Vargas and M Villegas-Vallecillos ldquoInto linearisometries between spaces of Lipschitz functionsrdquo HoustonJournal of Mathematics vol 34 no 4 pp 1165ndash1184 2008

[10] F Botelho and J Jamison ldquoSurjective isometries on spacesof differentiable vector-valued functionsrdquo Studia Mathematicavol 192 no 1 pp 39ndash50 2009

[11] E Mayer-Wolf ldquoIsometries between Banach spaces of Lipschitzfunctionsrdquo Israel Journal of Mathematics vol 38 no 1-2 pp 58ndash74 1981

[12] A Jimenez-Vargas M Villegas-Vallecillos and Y-S WangldquoBanach-Stone theorems for vector-valued little Lipschitz func-tionsrdquo Publicationes Mathematicae vol 74 no 1-2 pp 81ndash1002009

[13] J Araujo andLDubarbie ldquoNoncompactness andnoncomplete-ness in isometries of Lipschitz spacesrdquo Journal of MathematicalAnalysis and Applications vol 377 no 1 pp 15ndash29 2011

[14] F Botelho R J Fleming and J Jamison ldquoExtreme points andisometries on vector-valued Lipschitz spacesrdquo Journal of Math-ematical Analysis and Applications vol 381 no 2 pp 821ndash8322011

[15] H Koshimizu ldquoLinear isometries on spaces of continuouslydifferentiable and Lipschitz continuous functionsrdquo NihonkaiMathematical Journal vol 22 no 2 pp 73ndash90 2011

[16] F Botelho J Jamison and B Zheng ldquoIsometries on spaces ofvector valued Lipschitz functionsrdquo Positivity An InternationalMathematics Journal Devoted to Theory and Applications ofPositivity vol 17 no 1 pp 47ndash65 2013

[17] A Ranjbar-Motlagh ldquoA note on isometries of Lipschitz spacesrdquoJournal of Mathematical Analysis and Applications vol 411 no2 pp 555ndash558 2014

[18] F Botelho and J Jamison ldquoSurjective isometries on spaces ofvector valued continuous and Lipschitz functionsrdquo PositivityAn International Mathematics Journal Devoted to Theory andApplications of Positivity vol 17 no 3 pp 395ndash405 2013Erratum to Surjective isometries on spaces of vector valuedcontinuous and Lipschitz functions by F Botelho 20 (2016)757ndash759

[19] T Miura and H Takagi ldquoSurjective isometries on the Banachspace of continuously differentiable functionsrdquo ContemporaryMathematics vol 687 pp 181ndash192 2017

[20] K Kawamura ldquoBanach-Stone type theorems for 1198621-functionspaces over Riemannian manifoldsrdquo Acta Universitatis Szegedi-ensis Acta Scientiarum Mathematicarum vol 83 no 3-4 pp551ndash591 2017

[21] K Kawamura ldquoPerturbations of norms on 1198621-function spacesand associated isometry groupsrdquo Topology Proceedings vol 51pp 169ndash196 2018

[22] K Kawamura ldquoA Banach-Stone type theorem for 1198621-functionspaces over the circlerdquo Topology Proceedings vol 53 pp 15ndash262019

[23] L Li D Chen Q Meng and Y-S Wang ldquoSurjective isometrieson vector-valued differentiable function spacesrdquo Annals ofFunctional Analysis vol 9 no 3 pp 334ndash343 2018

[24] K Kawamura H Koshimizu and T Miura ldquoNorms onC1([01]) and there isometriesrdquo Acta Scientiarum Mathemati-carum vol 84 no 12 pp 239ndash261 2018

[25] L Li A M Peralta L Wang and Y-S Wang ldquoWeak-2-local isometries on uniform algebras and Lipschitz algebrasrdquohttpsarxivorgabs170503619

[26] A Jimenez-Vargas L LiAMPeralta LWang andY-SWangldquo2-local standard isometries on vector-valued Lipschitz func-tion spacesrdquo Journal of Mathematical Analysis and Applicationsvol 461 no 2 pp 1287ndash1298 2018

[27] A Ranjbar-Motlagh ldquoIsometries of Lipschitz type functionspacesrdquoMathematischeNachrichten vol 291 no 11-12 pp 1899ndash1907 2018

14 Journal of Function Spaces

[28] N Weaver Lipschitz Algebras World Scientific Publishing CoInc River Edge NJ USA 1999

[29] R R Phelps Lectures on Choquetrsquos theorem vol 1757 of LectureNotes in Mathematics Springer-Verlag Berlin Germany 2ndedition 2001

[30] O Hatori and S Oi ldquoIsometries on Banach algebras of vector-valued mapsrdquo Acta Scientiarum Mathematicarum vol 84 no12 pp 151ndash183 2018

[31] O Hatori S Oi and H Takagi ldquoPeculiar homomorphismsbetween algebras of vector-valued mapsrdquo Studia Mathematicavol 242 no 2 pp 141ndash163 2018

[32] M Nagasawa ldquoIsomorphisms between commutative Banachalgebras with an application to rings of analytic functionsrdquoKodai Mathematical Seminar Reports vol 11 pp 182ndash188 1959

[33] K de LeeuwW Rudin and J Wermer ldquoThe isometries of somefunction spacesrdquo Proceedings of the American MathematicalSociety vol 11 pp 694ndash698 1960

[34] OHatori A Jimenez-Vargas andMVillegas-Vallecillos ldquoMapswhich preserve norms of non-symmetrical quotients betweengroups of exponentials of Lipschitz functionsrdquo Journal of Math-ematical Analysis and Applications vol 415 no 2 pp 825ndash8452014

[35] G Lumer ldquoOn the isometries of reflexiveOrlicz spacesrdquoAnnalesde lrsquoInstitut Fourier vol 68 pp 99ndash109 1963

[36] R J Fleming and J E Jamison Isometries on Banach SpacesFunction Spaces vol 129 ofMonographs and Surveys in Pure andApplied Mathematics Chapman amp Hall CRC Boca Raton FlaUSA 2003

[37] OHatori and SOi ldquoHermitian operators onBanach algebras ofvector-valued Lipschitzmapsrdquo Journal ofMathematical Analysisand Applications vol 452 no 1 pp 378ndash387 2017 Corrigendumto rdquoHermitian operators on Banach algebras of vector-valuedLipschitz mapsrdquo Journal of Mathematical Analysis and Appli-cations 452 (2017) 378ndash387 MR3628025

[38] A Nikou andA G OrsquoFarrell ldquoBanach algebras of vector-valuedfunctionsrdquo Glasgow Mathematical Journal vol 56 no 2 pp419ndash426 2014

[39] F Botelho and J Jamison ldquoHomomorphisms on a class ofcommutative Banach algebrasrdquo Rocky Mountain Journal ofMathematics vol 43 no 2 pp 395ndash416 2013

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Page 5: Hermitian Operators and Isometries on Banach Algebras of ...downloads.hindawi.com/journals/jfs/2018/8085304.pdfJournalofFunctionSpaces normiseitherthemaxnormorthesumnorm.esituation

Journal of Function Spaces 5

As

119905 + 119904 le |119905 + 119904| le |119905 + 119904 + 119894119887| = 10038161003816100381610038161003816119890119894120593 (119905 + 119904 + 119894119887)10038161003816100381610038161003816le 100381710038171003817100381710038171198901198941205931199051 + 11989110038171003817100381710038171003817infin = 119903119860 (119891 119905 120593) (26)

we have

119905 + 119888119860 (119891 120593) le 119903119860 (119891 119905 120593) (27)

Let 119909 isin 119883 By the definition of 119904 = 119888119860(119891 120593) we infer thatRe(119890minus119894120593119891(119909)) le 119904 hence we have Re(119905 + 119890minus119894120593119891(119909)) le 119905 + 119904 forevery 119905 ge 0 Then

Re 119890minus119894120593119891 (119909) le Re (119905 + 119890minus119894120593119891 (119909)) le 119905 + 119904 119905 ge 0 (28)

Letting119872 = max|119905 + 119904| |Re(119890minus119894120593119891(119909))| we have10038161003816100381610038161003816119890119894120593119905 + 119891 (119909)100381610038161003816100381610038162 = 10038161003816100381610038161003816119905 + 119890minus119894120593119891 (119909)100381610038161003816100381610038162

= (Re (119905 + 119890minus119894120593119891 (119909)))2+ (Im (119905 + 119890minus119894120593119891 (119909)))2

le 1198722 + (Im (119905 + 119890minus119894120593119891 (119909)))2le |119905 + 119904|2 + (Re (119890minus119894120593119891 (119909)))2+ (Im (119890minus119894120593119891 (119909)))2

= |119905 + 119904|2 + 10038161003816100381610038161003816119890minus119894120593119891 (119909)100381610038161003816100381610038162

(29)

As 119909 isin 119883 is arbitrary we have

119903119860 (119891 119905 120593) = 10038171003817100381710038171003817119905 + 119890minus11989412059311989110038171003817100381710038171003817infin le radic|119905 + 119904|2 + 100381710038171003817100381711989110038171003817100381710038172infin= radic(119905 + 119888119860 (119891 120593))2 + 100381710038171003817100381711989110038171003817100381710038172infin

(30)

It follows that (24) holds In the same way we have

119905 + 119888119861 (119892 120593) le 119903119861 (119892 119905 120593) le radic(119905 + 119888119861 (119892 120593))2 + 100381710038171003817100381711989210038171003817100381710038172infin119905 ge 0 120593 isin [0 2120587) (31)

for every 119892 isin 119861 By (24) and (31) we infer that

lim119905997888rarrinfin

(119903119860 (119891 119905 120593) minus 119905) = 119888119860 (119891 120593) lim119905997888rarrinfin

(119903119861 (119879119891 119905 120593) minus 119905) = 119888119861 (119879119891 120593) (32)

As lsaquo sdot lsaquo119860 is 1-invariant we have119901 (119903119860 (119891 119905 120593) lsaquo119891lsaquo119860)= 119901 (10038171003817100381710038171003817119891 + 119890119894120593119905110038171003817100381710038171003817infin lsaquo119891 + 1198901198941205931199051lsaquo119860) (33)

As 119879 is an isometry 119879(1) = 1 and lsaquo sdot lsaquo119861 is 1-invariant wehave

119901 (10038171003817100381710038171003817119891 + 119890119894120593119905110038171003817100381710038171003817infin lsaquo119891 + 1198901198941205931199051lsaquo119860)= 119902 (10038171003817100381710038171003817119879 (119891 + 1198901198941205931199051)10038171003817100381710038171003817infin lsaquo119879 (119891 + 1198901198941205931199051)lsaquo119861)= 119902 (10038171003817100381710038171003817119879 (119891) + 119890119894120593119905110038171003817100381710038171003817infin lsaquo119879 (119891) + 1198901198941205931199051lsaquo119861)= 119902 (119903119861 (119879119891 119905 120593) lsaquo119879119891lsaquo119861)

(34)

Thus

119901 (119903119860 (119891 119905 120593) lsaquo119891lsaquo119860) = 119902 (119903119861 (119879119891 119905 120593) lsaquo119879119891lsaquo119861) (35)

It follows that0 = lim

119905997888rarrinfin(119901 (119903119860 (119891 119905 120593) lsaquo119891lsaquo119860)

minus 119902 (119903119861 (119879119891 119905 120593) lsaquo119879119891lsaquo119861))= lim

119905997888rarrinfin(119903119860 (119891 119905 120593) 119901(1 lsaquo119891lsaquo119860119903119860 (119891 119905 120593))

minus 119903119860 (119891 119905 120593)) + lim119905997888rarrinfin

(119903119861 (119879119891 119905 120593)minus 119903119861 (119879119891 119905 120593) 119902(1 lsaquo119879119891lsaquo119861119903119861 (119879119891 119905 120593)))+ lim

119905997888rarrinfin(119903119860 (119891 119905 120593) minus 119905 minus (119903119861 (119879119891 119905 120593) minus 119905))

= lsaquo119891lsaquo119860119863(119901) minus lsaquo119879119891lsaquo119861119863(119902) + 119888119860 (119891 120593)minus 119888119861 (119879119891 120593)

(36)

Recall that 119863(119901) = lim119905997888rarr+0(119901(1 119905) minus 1)119905 and 119863(119902) =lim119905997888rarr+0(119902(1 119905) minus 1)119905 It follows thatlsaquo119891lsaquo119860119863 (119901) minus lsaquo119879119891lsaquo119861119863(119902) = 119888119861 (119879119891 120593) minus 119888119860 (119891 120593)

119891 isin 119860 120593 isin [0 2120587) (37)

Suppose that 119863(119901) = 119863(119902) = 0 Then we have by (37)that 119888119861(119879119891 120593) = 119888119860(119891 120593) for every 119891 isin 119860 and 120593 isin [0 2120587)By Lemma 1 in [5] we infer that 120590(119879119891) = 120590(119891) Thus we have119879119891infin = 119891infin We have proved that 119879 is an isometry from(119860 sdot infin) onto (119861 sdot infin) if119863(119901) = 119863(119902) = 0

Suppose that 119860 and 119861 are regular subspaces of 119862(119883) and119862(119884) respectively Let 119891 isin 119860 PutΔ119891 = lsaquo119891lsaquo119860119863(119901) minus lsaquo119879119891lsaquo119861119863(119902) (38)

Suppose that Δ119891 ge 0 For any 119903 ge 0 and any nonemptycompact convex subset 119870 sub C we have that

119888 (119870 + 119870 (119903) 120593) = 119888 (119870 120593) + 119903 (39)

for all 120593 isin [0 2120587) where 119870(119903) = 119911 isin C |119911| le 119903 Then by(37) we have

119888 (120590 (119879119891) 120593) = 119888119861 (119879119891 120593) = 119888119860 (119891 120593) + Δ119891= 119888 (120590 (119891) 120593) + Δ119891= 119888 (120590 (119891) + 119870 (Δ119891) 120593)

(40)

6 Journal of Function Spaces

for all 120593 isin [0 2120587) It follows by Lemma 1 in [5] that

(119879119891) = 120590 (119891) + 119870 (Δ119891) (41)

and therefore10038171003817100381710038171198791198911003817100381710038171003817infin = 10038171003817100381710038171198911003817100381710038171003817infin + Δ119891 (42)

If Δ119891 le 0 then a similar calculation shows that

(119891) = (119879119891) + 119870 (minusΔ119891) (43)

and10038171003817100381710038171198911003817100381710038171003817infin = 10038171003817100381710038171198791198911003817100381710038171003817infin + (minusΔ119891) (44)

It follows that in any case (Δ119891 ge 0 Δ119891 le 0) we obtainΔ119891 = 10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin 119891 isin 119860 (45)

We will prove that1003817100381710038171003817119879 (119891)1003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin = Δ119891 ge 0 (46)

for all 119891 isin 119860 Once it is proved applying the same argumentfor119879minus1 instead of119879 we see that 119879minus1119892infinminus119892infin ge 0 for every119892 isin 119861 As 119879 is a bijection it follows that 119891infin minus 119879119891infin ge 0for every 119891 isin 119860 It will follow that 119879119891infin = 119891infin for every119891 isin 119860 A proof of (46) is the following For every 120576 gt 0denote

A120576 = 119891 isin 119860 120588 (120590 (119891)) le 120576 (47)

The inequality in (46) is deduced by the following assertionswhich appear in the proof of [5 Theorem]

(1) 119879 is a continuous mapping from (119860 sdot infin) onto (119861 sdotinfin)(2) For each 120576 gt 0 the setA120576 is dense in (119860 sdot infin)(3) For each 120576 gt 0 and each 119891 isin A120576 it holds that119879(119891)infin ge 119891infin minus 120576

Suppose that these assertions are proved Let 119891 isin 119860 By (2)for any 120576 gt 0 there is a sequence 119891119899 of functions inA120576 suchthat 119891119899 minus 119891infin 997888rarr 0 as 119899 997888rarr infin By (3) we have

10038171003817100381710038171198791198911198991003817100381710038171003817infin minus 10038171003817100381710038171198911198991003817100381710038171003817infin ge minus120576 (48)

for every 119899 Letting 119899 997888rarr infin we have10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin ge minus120576 (49)

by (1) As 120576 gt 0 is arbitrary we have that10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin ge 0 (50)

We show proofs of three assertions (1) (2) and (3) aboveprecisely The proof of (1) is slightly different from the corre-sponding one in [5 p 70] This change is rather ambitiousWe also point out that the terms minus1205872 and 1205872 which appearin the formulae (7) and (8) in [5] seem inappropriate theyread for example as 31205874 and 1205874 respectively

We now proceed to prove the first statement Aiming fora contradiction suppose that 119879 is not continuous from (119860 sdotinfin) to (119861 sdot infin) Let 120576 be a positive real number less than1100 Then there is a function 1198910 isin 119860 such that 1198910infin le 120576and 119879(1198910)infin = 1 Then there exist 1199100 isin Ch(119861) such that|119879(1198910)(1199100)| = 1 by [29 Proposition 63] Since 119879 is complex-linear we may suppose that 119879(1198910)(1199100) = 1

By (41) and (45) we deduce that Δ1198910 = 119879(1198910)infin minus1198910infin ge 1 minus 120576 and 120590(119879(1198910)) = 120590(1198910) + 119870(Δ1198910) As 1198910infin le 120576and 119879(1198910)infin = 1 we infer that 120590(1198910) sub 119870(120576) and (119879(1198910) sub119870(1) Let 119911 isin 119870(1 minus 2120576) Let 119909 isin 119883 Then |1198910(119909)| le 120576 assertsthat 119908 = 119911 minus 1198910(119909) isin 119870(1 minus 120576) sub 119870(Δ1198910) Thus

119911 = 1198910 (119909) + 119908 isin 120590 (1198910) + 119870 (Δ1198910) = 120590 (119879 (1198910)) (51)

Hence119870(1 minus 2120576) sub 120590(1198791198910) As 119879(1198910)infin = 1 we have119870(1 minus 2120576) sub 120590 (119879 (1198910)) sub 119870 (1) (52)

Consider the open neighborhood 1198800 of 1199100 in1198832 given by

1198800 = 119910 isin 1198832 1003816100381610038161003816119879 (1198910) (119910) minus 11003816100381610038161003816 lt 120576 (53)

We infer that 1198800 is a proper subset of1198832 by (52) Then by [5Lemma 2] there exists 119892 isin 119861 such that 119892infin le 1 + 120576 119892(1199100) =1 |119892(119910) + 1| lt 120576 for every 119910 isin 1198832 1198800 and |Im119892(119910)| lt 120576 forall 119910 isin 1198832 If 119867 denotes the closed rectangle whose verticesare the four points plusmn(1 + 120576) plusmn 120576119894 we have

120590 (119892) sub 119867 (54)

Consider now the set

119871 = 11989031205871198944119911 |119911| le 1 Re 119911 ge 1 minus 2120576 (55)

We claim that119879(1198910)(1198832)cap119871 = 0 Suppose that119879(1198910)(1198832)cap119871 =0 As 119879(1198910)(1198832) is compact there exists a positive integer119899 such that 119879(1198910)(1198832) cap 119871119899 = 0 where 119871119899 = 11989031205871198944119911 |119911| le 1Re 119911 gt 1 minus 2120576 minus 1119899 Then (52) gives 119879(1198910)(1198832) sub119870(1)119871119899 As (119879(1198910)) is the closed convex-hull of 119879(1198910)(1198832)it is contained in the closed convex set 119870(1) 119871119899 On theother hand (1 minus 2120576)11989031205871198944 isin 119870(1 minus 2120576) sub 120590(119879(1198910)) by (52) As(1minus2120576)11989031205871198944 isin 119871 sub 119871119899 this contradicts 120590(119879(1198910)) sub 119870(1) 119871119899and this proves our claim Hence there is 1199101 isin 1198832 with119879(1198910)(1199101) isin 119871 As 120576 le 1100 it follows that |119879(1198910)(1199101) minus1| ge 120576 and so 1199101 isin 1198832 1198800 Hence |119892(1199101) + 1| lt 120576 Thus119892(1199101) + 119879(1198910)(1199101) is in 119871 minus 1 + K(120576) Thus we have

1 + radic22 minus 3120576 le 119888119861 (119892 + 119879 (1198910) 31205874 ) (56)

We claim that

120590 (119892 + 119879 (1198910)) sub co (119870 (minus1 1) cup 2) + 119870 (3120576) (57)

where119870(1199110 119903) = 119911 isin C |119911 minus 1199110| le 119903 Let 119910 isin 1198832 Supposefirst that |119879(1198910)(119910)minus1| lt 120576 Since 119892(1198832) sub 119867 by (54) we have

119879 (1198910) (119910) + 119892 (119910) isin 119870 (1 120576) + 119867 = (119867 + 1) + 119870 (120576) (58)

Journal of Function Spaces 7

Suppose next that |119879(1198910)(119910)minus1| ge 120576Then 119910 isin 1198832 1198800 and so|119892(119910) + 1| lt 120576 Moreover |119879(1198910)(119910)| le 1 Therefore we have

119892 (119910) + 119879 (1198910) (119910) isin 119870 (1) minus 1 + 119870 (120576)= 119870 (minus1 1) + 119870 (120576) (59)

It follows from (58) and (59) that

(119892 + 119879 (1198910)) (1198832) sub 119870 (minus1 1) cup ((119867 + 1)) + 119870 (120576) (60)

and hence

120590 (119892 + 119879 (1198910)) sub co (119870 (minus1 1) cup 2) + 119870 (3120576) (61)

as is claimed Therefore we have

119888119861 (119892 + 119879 (1198910) 1205874 ) le radic2 + 3120576 (62)

Put 1198911 = 119879minus1(119892) We claim that Δ1198911 le 120576 If Δ1198911 lt 0 thereis nothing to prove Suppose that Δ1198911 ge 0 Then by (41) wehave

120590 (119892) = 120590 (1198911) + 119870 (Δ1198911) (63)

Since 120590(119892) sub 119867 by (54) we have

119867 sup 120590 (1198911) + 119870 (Δ1198911) (64)

As 119867 does not include a closed disk with the radius greaterthan 120576 we conclude that Δ1198911 le 120576

In the following we will consider two cases 0 le Δ1198911 le 120576and Δ1198911 le 0 Suppose first that 0 le Δ1198911 le 120576 Then (64) yields(1198911) sub 119867 From 1198910infin le 120576 we deduce that 120590(1198910) sub 119870(120576)Hence we have

120590 (1198911 + 1198910) sub 120590 (1198911) + 120590 (1198910) sub (1198910) + 120590 (1198910)sub 119867 + 119870 (120576) (65)

Since119867 +119870(120576) is convex we have (1198911 + 1198910) sub 119867 + 119870 (120576) (66)

From (39) we infer that

119888119860 (1198911 + 1198910 31205874 ) = 119888 (120590 (1198911 + 1198910) 31205874 )le 119888 (119867 + 119870 (120576) 31205874 )= 119888 (119867 31205874 ) + 120576= radic22 + (1 + radic2) 120576

(67)

Since 119879(1198911 + 1198910) = 119892 + 119879(1198910) from (56) and (67) we obtainthat

1 minus (4 + radic2) 120576 le 119888119861 (119879 (1198911 + 1198910) 31205874 )minus 119888119860 (1198911 + 1198910 31205874 )

(68)

By (63) and 1 = 119892(1199100) we deduce that 1 isin 120590(1198911) + 119870(Δ1198911)Thus there is 119911 isin 120590(1198911) such that |119911 minus 1| le Δ1198911 It follows thatradic22 minus Δ1198911 le 119888119860(1198911 1205874) hence we have

radic22 minus 2120576 le 119888119860 (1198911 + 1198910 1205874 ) (69)

as 1198910infin le 120576 and 0 le Δ1198911 le 120576 We get by (62) and (69) that

119888119861 (119879 (1198911 + 1198910) 1205874 ) minus 119888119860 (1198911 + 1198910 1205874 ) leradic22 + 5120576 (70)

On the other hand 119888119861(119879(119891) 120593)minus119888119860(119891 120593) is invariant for any 120593by (37) From (68) and (70)we deduce that 120576 ge (2minusradic2)2(9+radic2) and this contradicts that 120576 le 1100

For the second case suppose next that Δ1198911 le 0 Then by(43) we have

120590 (1198911) = 120590 (119892) + 119870 (minusΔ1198911) (71)

and by (54) it follows that (1198911) sub 119867 + 119870(minusΔ1198911) Moreover(1198910) sub 119870(120576) since 1198910infin le 120576 Then

120590 (1198911 + 1198910) sub 120590 (1198911) + 120590 (1198910) sub 120590 (1198911) + 120590 (1198910)sub 119867 + 119870(minusΔ1198911) + 119870 (120576) (72)

Hence (1198911+1198910) sub 119867+119870(minusΔ1198911)+119870(120576) Using (39) we inferthat

119888119860 (1198911 + 1198910 31205874 ) le 119888 (119867 + 119870(minusΔ1198911) + 119870 (120576) 31205874 )= 119888 (119867 31205874 ) + (minusΔ1198911) + 120576= radic22 + (1 + radic2) 120576 + (minusΔ1198911)

(73)

By (71) we obtain that 120590(1198911) sup 119892(1198832) + 119870(minusΔ1198911) and as119892(1199100) = 1 we infer that 120590(1198911) sup 1 + 119870(minusΔ1198911) Hence radic22 +(minusΔ1198911) le 119888119860(1198911 1205874) so thatradic22 + (minusΔ1198911) minus 120576 le 119888119860 (1198911 + 1198910 1205874 ) (74)

as 1198910infin le 120576 Since 119879(1198911 +1198910) = 119892 +119879(1198910) we obtain by (56)and (73) that

1 minus (4 + radic2) 120576 minus (minusΔ1198911)le 119888119861 (119879 (1198911 + 1198910) 31205874 ) minus 119888119860 (1198911 + 1198910 31205874 )

(75)

We also obtain by (62) and (74) that

119888119861 (119879 (1198911 + 1198910) 1205874 ) minus 119888119860 (1198911 + 1198910 1205874 )le radic22 + 4120576 minus (minusΔ1198911)

(76)

Since 119888119861(119879(119891) 120593)minus119888119860(119891 120593) is invariant for any 120593 by (37) from(75) and (76) we deduce that 120576 ge (2 minusradic2)2(8 +radic2) and thisis impossible since 120576 le 1100

8 Journal of Function Spaces

Next we show a proof of the second assertion (2) Let119891 isin 119860 We prove that there exists a sequence 119891119899 sub 119860 whichuniformly converges to 119891 such that 120588(120590(119891119899)) 997888rarr 0 as 119899 997888rarrinfin Without loss of generality we may assume that 119891infin = 1Then there exists 1199100 isin Ch(119860) such that |119891(1199100)| = 119891infinby [29 Proposition 63] We may assume that 119891(1199100) = 1Suppose that 119899 ge 4 Put

119880119899 = 119909 isin 119883 1003816100381610038161003816119891 (119909) minus 11003816100381610038161003816 le 11198992 (77)

and

Ω = co(119870 (1) cup 1 + 1119899) + 119870( 11198992) (78)

(In the following we identify R2 and C that is we identify(119909 119910) and 119909 + 119894119910 for every 119909 119910 isin R) Since we assume that119899 ge 4 we infer by a simple calculation that

[1 minus 11198992 minus 120575119899119899 1 + 1119899 + 120575119899119899 ] times [minus 11198992 minus 120575119899119899 11198992 + 120575119899119899 ]sub Ω

(79)

for 120575119899 with 0 lt 120575119899 le 11198992 We assume that 0 lt 120575119899 le 11198992 By[5 Lemma 2] there exists ℎ119899 isin 119860 such that ℎ119899infin le 1 + 2120575119899ℎ119899(1199100) = 1 |ℎ119899 + 1| le 2120575119899 on119883 119880119899 and |Im ℎ119899| le 2120575119899 on119883Put 119892119899 = (ℎ119899 + 1)2 Then 119892119899 isin 119860 and

10038171003817100381710038171198921198991003817100381710038171003817infin le 1 + 120575119899119892119899 (1199100) = 110038161003816100381610038161198921198991003816100381610038161003816 le 120575119899 on 119883 1198801198991003816100381610038161003816Im1198921198991003816100381610038161003816 le 120575119899 on 119883

minus120575119899 le Re119892119899 le 1 + 120575119899 on 119883

(80)

Let 119909 isin 119880119899 Then we have

1 minus 11198992 le Re119891 (119909) le 1minus 11198992 le Im119891 (119909) le 11198992

(81)

and

minus120575119899119899 le Re119892119899 (119909)119899 le 1119899 + 120575119899119899

minus120575119899119899 le Im119892119899 (119909)119899 le 120575119899119899

(82)

Hence

1 minus 11198992 minus 120575119899119899 le Re(119891 (119909) + 119892119899 (119909)119899 ) le 1 + 1119899 + 120575119899119899 minus 11198992 minus 120575119899119899 le Im(119891 (119909) + 119892119899 (119909)119899 ) le 11198992 + 120575119899119899

(83)

It follows that we have

(119891 + 119892119899119899 ) (119909) isin [1 minus 11198992 minus 120575119899119899 1 + 1119899 + 120575119899119899 ]times [minus 11198992 minus 120575119899119899 11198992 + 120575119899119899 ] sub Ω

(84)

for 119909 isin 119880119899 Suppose that 119909 isin 119883 119880119899 Then10038161003816100381610038161003816100381610038161003816119891 (119909) +

119892119899 (119909)11989910038161003816100381610038161003816100381610038161003816 le 1 +

120575119899119899 le 1 + 11198993 lt 1 + 11198992 (85)

and hence

119891 (119909) + 119892119899 (119909)119899 isin 119870 (1) + 119870( 11198992) sub Ω (86)

for 119909 isin 119883 119880119899 Since 1 + 1119899 = 119891(1199100) + 119892119899(1199100)119899 we have bycombining (84) and (86) that

1 + 1119899 isin (119891 + 119892119899119899 ) (119883) sub Ω (87)

AsΩ is convex we obtain

1 + 1119899 isin 120590 (119891 + 119892119899119899 ) sub Ω (88)

Recall that for 119903 ge 0 and a complex number 1199110119870 (1199110 119903) = 119908 isin C 1003816100381610038161003816119908 minus 11991101003816100381610038161003816 le 119903 (89)

denotes the closed disk with center 1199110 and radius 119903 Weobserve that 120588(Ω 1 + 1119899) = 1119899 + 11198992 Recall that

120588(Ω 1 + 1119899)= sup119903 119911 isin Ω 1 + 1119899 isin 119870 (119911 119903) sub Ω

(90)

Let ℓ1 be the line defined by the equation

119910 = minus119899radic2119899 + 1 (119909 minus (1 +1119899 + 11198992 + 11198993)) (91)

part of which is a part of the boundary ofΩ Let ℓ2 be the linedefined by the equation

119910 = minus119899radic2119899 + 1 (119909 minus (1 minus11198992)) (92)

By some calculation we have that the distance between 1 minus11198992 and 1+1119899 is 1119899+11198992 and it coincides with the distancebetween the point 1 minus 11198992 and the line ℓ1 Hence we see that

1 + 1119899 isin 119870(1 minus 11198992 1119899 + 11198992 ) sub Ω (93)

Thus 1119899 + 11198992 le 120588(Ω 1 + 1119899)Next we prove that 119870(119901 119903) cap Ω119888 = 0 for every 119901 isin Ω

with 119903 = |119901 minus (1 + 1119899)| gt 1119899 + 11198992 It will follow that1119899 + 11198992 ge 120588(Ω 1 + 1119899) and the equality will hold Let

Journal of Function Spaces 9

119901 = (119909119901 119910119901) = 119909119901 + 119894119910119901 isin Ω with 119903 = |119901 minus (1 + 1119899)| gt1119899 + 11198992 We prove the case where 119910119901 ge 0 A proof for thecase where 119910119901 le 0 is the same and we omit it We divideΩ+ =119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902 isin Ω 119910119902 ge 0 into two parts

Ω1 = 119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902isin Ω+ minus119899radic2119899 + 1 (119909119902 minus (1 minus

11198992)) le 119910119902le minus119899radic2119899 + 1 (119909 minus (1 +

1119899 + 11198992 + 11198993))(94)

and

Ω2 = 119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902 isin Ω+ 119910119902le minus119899radic2119899 + 1 (119909119902 minus (1 minus

11198992)) (95)

Suppose that 119901 isin Ω1 and 119903 = |119901minus (1+1119899)| gt 1119899+11198992Thedistance between ℓ1 and ℓ2 is 1119899+11198992 Hence119870(119901 119903)capΩ119888 =0 Suppose that 119901 isin Ω2 and 119903 = |119901 minus (1 + 1119899)| gt 1119899 + 11198992that is119901 = 1minus11198992 Let ℓ1015840 be the line passing through119901whichis parallel to ℓ2 Let 1199011015840 be the unique point in the intersectionof ℓ1015840 and the 119909-axis Then 1199011015840 = 1 minus 11198992 minus 119906 for some 119906 ge 0Then the distance between ℓ1 and ℓ1015840 is 1119899+11198992+119906(1+1119899)which is equal to the distance between the point119901 and the lineℓ1 On the other hand1003816100381610038161003816100381610038161003816119901 minus (1 + 1119899)

1003816100381610038161003816100381610038161003816 ge10038161003816100381610038161003816100381610038161199011015840 minus (1 + 1119899)

1003816100381610038161003816100381610038161003816 = 1119899 + 11198992 + 119906 (96)

It follows that 119870(119901 119903) cap Ω119888 = 0 We conclude that if 119901 isin Ωsatisfies 119903 = |119901minus(1+1119899)| gt 1119899+11198992 then119870(119901 119903)capΩ119888 = 0Thus we have

120588 (Ω 1 + 1119899) = 1119899 + 11198992 (97)

Since 1 + 1119899 isin 120590(119891 + 119892119899119899) sub Ω we have120588 (120590 (119891 + 119892119899119899 )) le 120588 (120590 (119891 + 119892119899119899 ) 1 + 1119899)

le 120588 (Ω 1 + 1119899) = 1119899 + 11198992 (98)

On the other hand 119892119899119899infin 997888rarr 0 as 119899 997888rarr infin ensures that119891 + 119892119899119899 minus 119891infin 997888rarr 0 as 119899 997888rarr infin It follows that for every120576 gt 0A120576 is dense in (119860 sdot infin)Finally we show a proof of the third assertion (3) As is

pointed out in the proof of [5Theorem] 120588(co(119870)+119870(119888)) ge 119888for any 119870 sub C and any 119888 ge 0 Let 120576 gt 0 and 119891 isin A120576 Supposethat Δ119891 le 0 Then by (43) we have (119891) = (119879119891) + 119870(minusΔ119891)Hence we have

120588 ( (119891)) = 120588 (120590 (119879119891) + 119870 (minusΔ119891)) ge minusΔ119891 (99)

As 120576 ge 120588(120590(119891)) we conclude by (45) that10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin = Δ119891 ge minus120576 (100)

This completes the proof of the theorem

4 Hermitian Operators on a BanachAlgebras of Continuous Maps WhoseValues Are in a Uniform Algebras

Let 119883 and 119884 be compact Hausdorff spaces Let 119861 be aunital subalgebra of 119862(119883) which separates the points of 119883Throughout this sectionwe assume119861 is a Banach algebrawiththe norm sdot 119861 and 119860 is a uniform algebra on 119884 Recall thata uniform algebra on 119884 is a uniformly closed subalgebra of119862(119884) which contains constants and separates the points of 119884For functions 119891 isin 119862(119883) and 119892 isin 119862(119884) let 119891 otimes 119892 isin 119862(119883 times 119884)be the function defined by119891otimes119892(119909 119910) = 119891(119909)119892(119910) for (119909 119910) isin119883 times 119884 and for a subspace 119864119883 of 119862(119883) and a subspace 119864119884 of119862(119884) put

119864119883 otimes 119864119884

= 119899sum

119895=1

119891119895 otimes 119892119895 119899 isin N 119891119895 isin 119864119883 119892119895 isin 119864119884

(101)

and

1 otimes 119864119884 = 1 otimes 119892 119892 isin 119864119884 (102)

Throughout the section 119861 is a unital subalgebra of 119862(119883 times 119884)with a Banach algebra norm sdot 119861 We assume that 119861otimes119860 sub 119861Note that 119861 separates the points of119883times119884 since119860 separates thepoints of 119884 and 119861 separates the points of 119883 We assume thatthere exists a compact Hausdorff space M and a complex-linear map 119863 119861 997888rarr 119862(M) such that ker119863 = 1 otimes 119860 Weassume that 119865119861 = 119865infin(119883times119884)+119863(119865)infin(M) for every 119865 isin 119861Hence119863 is continuous Defining

lsaquo119865lsaquo = 119863 (119865)infin(M) 119865 isin 119861 (103)

lsaquo sdot lsaquo is a one-invariant seminorm in the sense of Jarosz lsaquo sdot lsaquois a seminorm on 119861 such that lsaquo119865+ 1lsaquo = lsaquo119865lsaquo for every 119865 isin 119861Hence the norm sdot 119861 is a natural norm (see [5 p67]) Notethat 119861 is a regular subspace of 119862(119883times119884) in the sense of Jarosz[5 Proposition 2]

Lumerrsquos seminal paper [35] opened up a useful methodof finding isometries which is often referred to as Lumerrsquosmethod It involves the notion ofHermitian operators and thefact that 119880119867119880minus1 must be Hermitian if119867 is Hermitian and 119880is a surjective isometry

Definition 6 Let A be a unital Banach algebra We say that119890 isin A is a Hermitian element if1003817100381710038171003817exp (119894119905119890)1003817100381710038171003817A = 1 (104)

for every 119905 isin R The set of all Hermitian elements of A isdenoted by119867(A)

If A is a unital 119862lowast-algebra then 119867(A) is the set of allself-adjoint elements of A Hence 119867(119872119899(C)) is the set of allHermitian matrices and119867(119862(119884)) = 119862R(119884)Definition 7 Let 119864 be a complex Banach space The Banachalgebra of all bounded operators on 119864 is denoted by 119861(119864) Wesay that 119879 isin 119861(119864) is a Hermitian operator if 119879 isin 119867(119861(119864))

10 Journal of Function Spaces

Note that a Hermitian element of a unital Banach algebraand a Hermitian operator are usually defined in terms ofnumerical range or semi-inner product Here we define themby an equivalent form (see [36]) By the definition of aHermitian operator we have the following

Proposition 8 Let 119864119895 be a complex Banach space for 119895 = 1 2Suppose that 119881 1198641 997888rarr 1198642 is a surjective isometry and 119867 1198641 997888rarr 1198641 is a Hermitian operator Then 119881119867119881minus1 1198642 997888rarr 1198642

is a Hermitian operator

Proposition 9 An element 119865 isin 119861 is Hermitian if and only ifthere exists 119891 isin 119860 cap 119862R(119884) such that 119865 = 1 otimes 119891Proof Suppose that 119865 isin 119861 is a Hermitian element Then1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884) + 1003817100381710038171003817119863 (exp (119894119905119865))1003817100381710038171003817infin(M)

= 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817 = 1 (105)

for every 119905 isin R Suppose that there exists a point (119909 119910) isin119883 times 119884 with Im119865(119909 119910) = 0 where Im denotes the imaginarypart of a complex number Suppose that Im119865(119909 119910) gt 0Then1003817100381710038171003817exp (minus119894119865)1003817100381710038171003817infin(119883times119884) ge 1003816100381610038161003816exp (minus119894119865 (119909 119910))1003816100381610038161003816

= exp (Im119865 (119909 119910)) gt 1 (106)

Suppose that Im119865(119909 119910) lt 0 Then1003817100381710038171003817exp (119894119865)1003817100381710038171003817infin(119883times119884) ge 1003816100381610038161003816exp (119894119865 (119909 119910))1003816100381610038161003816= exp (minusIm119865 (119909 119910)) gt 1 (107)

In any case we have there exists 119905 isin R such that1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817119861 ge 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884) gt 1 (108)

which contradicts our assumption We have that

119865 isin 119862R (119883 times 119884) (109)

Thus for every (119904 119905) isin 119883 times 119884 and 119905 isin R |exp(119894119905119865(119904 119905))| = 1Hence exp(119894119905119865)infin(119883times119884) = 1 for every 119905 isin R By (105)we have119863(exp(119894119905119865)infin(M) = 0 which ensures that 119863(exp(119894119905119865)) = 0for every 119905 isin R Thus exp(119894119905119865) isin 1 otimes 119860 for every 119905 isin R Wehave

exp (119894119905119865) minus 1119905 minus 119894119865 = infinsum119899=2

((119894119865)119899 119905119899minus2119899 ) 119905 (110)

and hence for every 119905 isin R with |119905| le 1 we have10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817 le (

infinsum119899=2

119865119899119861|119905|119899minus2119899 ) |119905|

le (infinsum119899=2

119865119899119861119899 ) |119905| le (exp 119865119861) |119905|

(111)

It follows that10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817infin(119883times119884)

le 10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817119861

997888rarr 0(112)

as 119905 997888rarr 0 Since (exp(119894119905119865)minus1)119905 isin 1otimes119860 for each 119905 isin R thereexists 119892119905 isin 119860 such that

exp (119894119905119865) minus 1119905 = 1 otimes 119892119905 (113)

By (112) we have10038171003817100381710038171 otimes 1198921119899 minus 1198941198651003817100381710038171003817infin(119883times119884)

997888rarr 0 (114)

as 119899 997888rarr infin We have that 1 otimes 1198921119899 is a Cauchy sequencein 119862(119883 times 119884) thus we infer that1198921119899 is a Cauchy sequence in119862(119884) Since 119860 is uniformly closed as it is a uniform algebrathere exists 119892 isin 119860 such that

10038171003817100381710038171198921119899 minus 1198921003817100381710038171003817infin(119884)997888rarr 0 (115)

and hence10038171003817100381710038171 otimes 1198921119899 minus 1 otimes 1198921003817100381710038171003817infin(119883times119884)

997888rarr 0 (116)

as 119899 997888rarr infin It follows by (114) that 119894119865 = 1 otimes 119892 thus119865 = 1 otimes (minus119894119892) isin 1 otimes 119860 (117)

By (109) we see that minus119894119892 isin 119862R(119884) thus we have 119891 = minus119894119892 isin119860 cap 119862R(119884) and 119865 = 1 otimes 119891Suppose conversely that 119861 ni 119865 = 1otimes119891 for119891 isin 119860cap119862R(119884)

We infer that 119865 isin 119862R(119883 times 119884) and |exp(119894119905(119865(119909 119910))| = 1 forevery 119905 isin R and (119909 119910) isin 119883 times 119884 Hence exp(119894119905119865)infin(119883times119884) = 1for every 119905 isin R Since

exp (119894119905119865) = exp (119894119905 (1 otimes 119891)) = 1 otimes exp (119894119905119891) isin 1 otimes 119860 (118)

we have 119863(exp(119894119905119865)) = 0 It follows that1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817 = 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884)

+ 1003817100381710038171003817119863 (exp (119894119905119865))1003817100381710038171003817infin(M) = 1 (119)

for every 119905 isin R We conclude that 119865 is a Hermitian elementin 119861

Note that 119891 isin 119860 is Hermitian if and only if119891 isin 119860cap119862R(119884)by [37 Proposition 5] Hence Proposition 9 asserts that 119865 is aHermitian element in119861 if and only if119865 = 1otimes119891 for aHermitianelement 119891 in 119860Proposition 10 Suppose that 119880 119861 997888rarr 119861 is a surjectiveunital isometry Then 119880 is an algebra isomorphism

Proof As we have already mentioned 119861 is a regular subspace(in the sense of Jarosz) with a natural norm Then byTheorem 1 119880 is also an isometry with respect to the supre-mum norm on 119883 times 119884 Then 119880 is uniquely extended to asurjective isometry with respect to the supremum norm from the uniform closure 119861 onto itself Since 119861 is a uniformalgebra a theorem of Nagasawa [32] asserts that is analgebra isomorphism since (1) = 1 Thus 119880 is an algebraisomorphism from 119861 onto itself

Journal of Function Spaces 11

Theorem 11 A bounded operator 119879 119861 997888rarr 119861 is a Hermitianoperator if and only if 119879(1) is a Hermitian element in 119861 and119879 = 119872119879(1) the multiplication operator by 119879(1)Proof By Proposition 10 every surjective unital isometry on119861 is multiplicative Then by [37 Theorem 4] we have theconclusion

5 Banach Algebras of 119862(119884)-Valued Maps

Suppose that119883 is a compact Hausdorff space Suppose that 119861is a unital point separating subalgebra of119862(119884) equipped witha Banach algebra norm Then 119861 is semisimple because 119891 isin119861 119891(119909) = 0 is a maximal ideal of 119861 for every 119909 isin 119883 and theJacobson radical of 119861 vanishes The inequality 119891infin le 119891119861for every 119891 isin 119861 is well known We say that 119861 is natural if themap 119890 119884 997888rarr 119872119861 defined by 119910 997891997888rarr 120601119910 where 120601119910(119891) = 119891(119910)for every 119891 isin 119861 is bijective We say that 119861 is self-adjoint if 119861 isnatural and conjugate-closed in the sense that 119891 isin 119861 impliesthat 119891 isin 119861 for every 119891 isin 119861 where sdot denotes the complexconjugation on 119884Definition 12 Let 119883 and 119884 be compact Hausdorff spacesSuppose that 119861 is a unital point separating subalgebra of119862(119883)equipped with a Banach algebra norm sdot 119861 Suppose that119861 is self-adjoint Suppose that 119861 is a unital point separatingsubalgebra of 119862(119883 times 119884) such that 119861 otimes 119862(119884) sub 119861 equippedwith a Banach algebra norm sdot 119861 Suppose that 119861 is self-adjoint We say that 119861 is a natural 119862(119884)-valuezation of 119861 ifthere exists a compact Hausdorff space M and a complex-linear map 119863 119861 997888rarr 119862(M) such that ker119863 = 1 otimes 119862(119884) and119863(119862R(119883 times 119884) cap 119861) sub 119862R(M) which satisfies

119865 = 119865infin(119883times119884) + 119863 (119865)infin(M) 119865 isin 119861 (120)

The term ldquoa natural 119862(119884)-valuezation of 119861rdquo comes fromthe natural norm defined by Jarosz [5] In fact the norm sdot 119861is a natural norm in the sense of Jarosz [5]

Note that (119883119862(119884) 119861 119861) need not be an admissiblequadruple defined by Nikou and OrsquoFarrell [38] (cf [31]) sincewe do not assume that 119865(sdot 119910) 119865 isin 119861 119910 isin 119884 sub 119861 whichis a requirement for the admissible quadruple On the otherhand if (119883119862(119884) 119861 119861) is an admissible quadruple of type Ldefined in [30] then 119861 is a natural 119862(119884)-valuezation of 119861 dueto Definition 12

Example 13 Let 119861 = 1198621([0 1]) and 119861 = Lip([0 1] 119862(119884))for 119884 = 119901 a singleton Then Lip([0 1]) is algebraically iso-morphic to Lip([0 1] 119862(119884)) Suppose thatM is the maximalideal space of 119871infin([0 1]) and 119863 119861 997888rarr 119862(M) is definedby 119891 997891997888rarr Γ(1198911015840) where Γ denotes the Gelfand transformin 119871infin([0 1]) Then 119861 is a natural 119862(119884)-valuezation of 119861The Banach algebra Lip([0 1]) with the norm 119891infin([01]) +1198911015840infin([01]) is isometrically isomorphic to 119861

Let 119884 be a compact Hausdorff space Note that a closedsubalgebra 119861 of Lip((119870 119889120572) 119862(119884))which appears in Example12 in [30] is an example of a natural 119862(119884)-valuezation of 119861

The Banach algebras1198621([0 1] 119862(119884)) and 1198621(T 119862(119884)))whichappear in Examples 16 and 17 in [30] respectively are alsoexamples of natural 119862(119884)-valuezations of 1198621([0 1])6 Isometries on Natural 119862(119884)-ValuezationsThemain theorem in this paper is the following

Theorem 14 Suppose that 119861119895 is a natural 119862(119884119895)-valuezationof 119861119895 sub 119862(119883119895) for 119895 = 1 2 We assume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (121)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for119895 = 1 2 Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linear isometry Then there exists ℎ isin 119862(1198842) such that |ℎ| = 1on 1198842 a continuous map 120593 1198832 times 1198842 997888rarr 1198831 such that120593(sdot 119910) 1198832 997888rarr 1198831 is a homeomorphism for each 119910 isin 1198842and a homeomorphism 120591 1198842 997888rarr 1198841 which satisfies119880 (119865) (119909 119910) = ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842 (122)

for every 119865 isin 1198611

In short a surjective isometry between 119862(119884)-valuezationsis a weighted composition operator of a specific form thehomeomorphism 1198832 times 1198842 997888rarr 1198831 times 1198841 (119909 119910) 997891997888rarr(120593(119909 119910) 120591(119910)) has the second coordinate that depends onlyon the second variable 119910 isin 1198842 A composition operatorinduced by such a homeomorphism is said to be of type BJin [31 37] after the study of Botelho and Jamison [39]

Quite recently the author of this paper and Oi [30 Theo-rem 8] proved a similar result of Theorem 14 for admissiblequadruples of type L To prove it we apply Proposition 32and the following comments in [31] Instead of this we proveTheorem 14 by Lumerrsquosmethod with which a proof is simplerthan that in [30]

In the following in this section we assume that 119861119895 is anatural 119862(119884119895)-valuezation of 119861 sub 119862(119883119895) for 119895 = 1 2 Weassume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (123)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for 119895 = 1 2Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linearisometry A crucial part of a proof of Theorem 14 is to proveProposition 15

Proposition 15 Suppose that 1198832 is not a singleton Thereexists ℎ isin 119862(1198842) with |ℎ| = 1 on 1198842 such that 119880(11198611) = 11198612 otimesℎ

A similar result for admissible quadruples of type L isproved in [30 Proposition 9] If we assumed that

119865 (sdot 119910) 119865 isin 119861119895 119910 isin 119884119895 sub 119861119895 (124)

then 119861119895 were an admissible quadruple of type L Although119861119895 in this paper need not be an admissible quadruple of type

12 Journal of Function Spaces

L a proof of Proposition 15 is completely the same as that in[30 Proposition 9] since we do not make use of the condition(124) in the proof of [30 Proposition 9] The condition (124)is needed in [30] when we apply Proposition 32 and thefollowing comments in [31]

7 Proof of Theorem 14 An Application ofLumerrsquos Method

Proof of Theorem 14 A proof for the case where 1198831 = 1199091and 1198832 = 1199092 are singletons is the same as the proof ofTheorem 8 in [30]

Suppose that1198832 is not a singleton By Proposition 15 thereexists ℎ isin 119862R(1198842) with |ℎ| = 1 on 1198842 such that 119880(1) =1 otimes ℎ Letting 1198800 1198611 997888rarr 1198612 by 1198800(119865) = (1 otimes ℎ)119880(119865)119865 isin 1198611 we see by the hypothesis (1 otimes ℎ)1198651198612 = 1198651198612 forevery 119865 isin 1198612 that 1198800 is a surjective unital isometry from1198611 onto 1198612 Then Corollary 2 asserts that 1198800 is an algebraisomorphism Let 119891 isin 119862R(1198841) By Proposition 9 1 otimes 119891 isa Hermitian element in 1198611 Then by Theorem 11 1198721otimes119891 is aHermitian operator on 1198611 By Proposition 8 11988001198721otimes119891119880minus1

0 is aHermitian operator on 1198612 Then by Theorem 11 there exists119878(119891) isin 119862R(1198842) such that 11988001198721otimes119891119880minus1

0 = 1198721otimes119878(119891) Hence anoperator 119878 119862R(1198841) 997888rarr 119862R(1198842) is defined Since 1198800 is analgebra isomorphism it is easy to see that 119878 is a real algebraisomorphism from 119862R(1198841) onto 119862R(1198842) Then 119878 119862(1198841) 997888rarr119862(1198842) defined by 119878(119891) = 119878(Re119891) + 119894119878(Im119891) for 119891 isin 119862(1198841)gives a complex algebra isomorphism Gelfand theory assertsthat there is a homeomorphism 120591 1198842 997888rarr 1198841 such that119878(119891) = 119891 ∘ 120591 119891 isin 119862(1198841) It follows that

11988001198721otimes119891119880minus10 = 1198721otimes119891∘120591 119891 isin 119862 (1198841) (125)

Since 119880minus10 (1) = 1 we have

1198800 (1 otimes 119891) = 1 otimes 119891 ∘ 120591 119891 isin 119862 (1198841) (126)

Define Φ 1198611 997888rarr 1198612 by Φ(119886) = 1198800(119886 otimes 1) 119886 isin 1198611Since 1198800 is an algebra isomorphism the map Φ is a unitalhomomorphism Since the maximal ideal space of 1198611 is 1198831

and the maximal ideal space of 1198612 is 1198832 times 1198842 there is acontinuous map 120593 1198832 times 1198842 997888rarr 1198831 such that

Φ(119886) (119909 119910) = 119886 (120593 (119909 119910)) 119886 isin 1198611 (119909 119910) isin 1198832 times 1198842 (127)

It follows by (126) and (127) that

1198800 (119886 otimes 119891) (119909 119910) = 1198800 ((119886 otimes 1) (1 otimes 119891)) (119909 119910)= 1198800 (119886 otimes 1) (119909 119910)1198800 (1 otimes 119891) (119909 119910)= 119886 (120593 (119909 119910))119891 (120591 (119910))= (119886 otimes 119891) (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198832 times 1198842

(128)

for every 119886 isin 1198611 and 119891 isin 119862(1198841) Thus

1198800 (sum(119886119895 otimes 119891119895)) (119909 119910)= (sum(119886119895 otimes 119891119895)) (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842(129)

for every sum(119886119895 otimes 119891119895) isin 1198611 otimes 119862(1198841) By the Stone-Weierstrasstheorem 1198611 otimes 119862(1198841) is uniformly dense in 119862(1198831 times 1198841) henceany element in 1198611 is uniformly approximated by 1198611 otimes 119862(1198841)As 1198800 is also an isometry with respect to the uniform normwe see that

1198800 (119865) (119909 119910) = 119865 (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198831 times 1198841 (130)

for every 119865 isin 1198611 and

119880 (119865) (119909 119910) = (1 otimes ℎ) (119909 119910) 119865 (120593 (119909 119910) 120591 (119910))= ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198831 times 1198841(131)

As 1198800 is an algebra isomorphism the map 1198832 times 1198842 997888rarr1198831 times 1198841 defined by (119909 119910) 997891997888rarr (120593(119909 119910) 120591(119910)) gives ahomeomorphism Therefore for every 119910 isin 1198842 the map

120593 (sdot 119910) 1198832 997888rarr 1198831 (132)

is a homeomorphismSuppose that 1198831 is not a singleton By the same way as

in the last part of the proof of Theorem 8 in [30] we havethat 1198832 is not a singleton Then we have the conclusion bythe previous argument

8 Application of Theorem 14

We exhibit applications of Theorem 14

Corollary 16 ([4 Theorem 33]) Suppose that119880 Lip([0 1]) 997888rarr Lip([0 1]) is a surjective isometrywith respect to the norm defined by 119891infin([01]) + 1198911015840infin([01])

for 119891 isin Lip([0 1]) Then 119880(1) is a constant function of unitmodulus such that

119880(119891) (119909) = 119880 (1) 119891 (119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (133)

or

119880(119891) (119909) = 119880 (1) 119891 (1 minus 119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (134)

The converse statement also holds

Proof By Example 13 we may suppose that Lip([0 1]) is aBanach algebra of 119862(119884)-valuezation Applying Theorem 14we have that119880(1) = 1otimesℎ for ℎ isin 119862(119884)with |ℎ| = 1 Since our

Journal of Function Spaces 13

119884 is a singleton 119880(1) is a constant function of unit modulusWe also see that the corresponding continuous map 120593 [0 1] times 119884 997888rarr [0 1] can be considered as a homeomorphismfrom [0 1] onto [0 1] therefore we have that119880(119891) (119909) = 119880 (1) 119891 (120593 (119909))

119891 isin Lip ([0 1]) 119909 isin [0 1] (135)

The rest is a routine argument to prove that 120593 is an isometryhence 120593(119909) = 119909 119909 isin [0 1] or 120593(119909) = 1 minus 119909 119909 isin [0 1]

The converse statement is trivial

Corollaries 14 15 18 and 19 in [30 Section 6] follow herewith a similar proof

Data Availability

The data used to support the findings of this study areavailable from the corresponding author upon request

Conflicts of Interest

The author declares that they have no conflicts of interest

Acknowledgments

This work was supported by JSPS KAKENHI Grants Num-bers JP16K05172 and JP15K04921

References

[1] K de Leeuw ldquoBanach spaces of Lipschitz functionsrdquo StudiaMathematica vol 21 pp 55ndash66 19611962

[2] A K Roy ldquoExtreme points and linear isometries of the Banachspace of Lipschitz functionsrdquoCanadian Journal of MathematicsJournal Canadien de Mathematiques vol 20 pp 1150ndash11641968

[3] M Cambern ldquoIsometries of certain Banach algebrasrdquo StudiaMathematica vol 25 pp 217ndash225 19641965

[4] N V Rao and A K Roy ldquoLinear isometries of some functionspacesrdquo Pacific Journal of Mathematics vol 38 pp 177ndash192 1971

[5] K Jarosz ldquoIsometries in semisimple commutative BanachalgebrasrdquoProceedings of the AmericanMathematical Society vol94 no 1 pp 65ndash71 1985

[6] K Jarosz and V D Pathak ldquoIsometries between functionspacesrdquo Transactions of the AmericanMathematical Society vol305 no 1 pp 193ndash206 1988

[7] NWeaver ldquoIsometries of noncompact Lipschitz spacesrdquoCana-dian Mathematical Bulletin Bulletin Canadien de Mathemat-iques vol 38 no 2 pp 242ndash249 1995

[8] A Jimenez-Vargas and M Villegas-Vallecillos ldquoLinear isome-tries between spaces of vector-valued Lipschitz functionsrdquoProceedings of the American Mathematical Society vol 137 no4 pp 1381ndash1388 2009

[9] A Jimenez-Vargas and M Villegas-Vallecillos ldquoInto linearisometries between spaces of Lipschitz functionsrdquo HoustonJournal of Mathematics vol 34 no 4 pp 1165ndash1184 2008

[10] F Botelho and J Jamison ldquoSurjective isometries on spacesof differentiable vector-valued functionsrdquo Studia Mathematicavol 192 no 1 pp 39ndash50 2009

[11] E Mayer-Wolf ldquoIsometries between Banach spaces of Lipschitzfunctionsrdquo Israel Journal of Mathematics vol 38 no 1-2 pp 58ndash74 1981

[12] A Jimenez-Vargas M Villegas-Vallecillos and Y-S WangldquoBanach-Stone theorems for vector-valued little Lipschitz func-tionsrdquo Publicationes Mathematicae vol 74 no 1-2 pp 81ndash1002009

[13] J Araujo andLDubarbie ldquoNoncompactness andnoncomplete-ness in isometries of Lipschitz spacesrdquo Journal of MathematicalAnalysis and Applications vol 377 no 1 pp 15ndash29 2011

[14] F Botelho R J Fleming and J Jamison ldquoExtreme points andisometries on vector-valued Lipschitz spacesrdquo Journal of Math-ematical Analysis and Applications vol 381 no 2 pp 821ndash8322011

[15] H Koshimizu ldquoLinear isometries on spaces of continuouslydifferentiable and Lipschitz continuous functionsrdquo NihonkaiMathematical Journal vol 22 no 2 pp 73ndash90 2011

[16] F Botelho J Jamison and B Zheng ldquoIsometries on spaces ofvector valued Lipschitz functionsrdquo Positivity An InternationalMathematics Journal Devoted to Theory and Applications ofPositivity vol 17 no 1 pp 47ndash65 2013

[17] A Ranjbar-Motlagh ldquoA note on isometries of Lipschitz spacesrdquoJournal of Mathematical Analysis and Applications vol 411 no2 pp 555ndash558 2014

[18] F Botelho and J Jamison ldquoSurjective isometries on spaces ofvector valued continuous and Lipschitz functionsrdquo PositivityAn International Mathematics Journal Devoted to Theory andApplications of Positivity vol 17 no 3 pp 395ndash405 2013Erratum to Surjective isometries on spaces of vector valuedcontinuous and Lipschitz functions by F Botelho 20 (2016)757ndash759

[19] T Miura and H Takagi ldquoSurjective isometries on the Banachspace of continuously differentiable functionsrdquo ContemporaryMathematics vol 687 pp 181ndash192 2017

[20] K Kawamura ldquoBanach-Stone type theorems for 1198621-functionspaces over Riemannian manifoldsrdquo Acta Universitatis Szegedi-ensis Acta Scientiarum Mathematicarum vol 83 no 3-4 pp551ndash591 2017

[21] K Kawamura ldquoPerturbations of norms on 1198621-function spacesand associated isometry groupsrdquo Topology Proceedings vol 51pp 169ndash196 2018

[22] K Kawamura ldquoA Banach-Stone type theorem for 1198621-functionspaces over the circlerdquo Topology Proceedings vol 53 pp 15ndash262019

[23] L Li D Chen Q Meng and Y-S Wang ldquoSurjective isometrieson vector-valued differentiable function spacesrdquo Annals ofFunctional Analysis vol 9 no 3 pp 334ndash343 2018

[24] K Kawamura H Koshimizu and T Miura ldquoNorms onC1([01]) and there isometriesrdquo Acta Scientiarum Mathemati-carum vol 84 no 12 pp 239ndash261 2018

[25] L Li A M Peralta L Wang and Y-S Wang ldquoWeak-2-local isometries on uniform algebras and Lipschitz algebrasrdquohttpsarxivorgabs170503619

[26] A Jimenez-Vargas L LiAMPeralta LWang andY-SWangldquo2-local standard isometries on vector-valued Lipschitz func-tion spacesrdquo Journal of Mathematical Analysis and Applicationsvol 461 no 2 pp 1287ndash1298 2018

[27] A Ranjbar-Motlagh ldquoIsometries of Lipschitz type functionspacesrdquoMathematischeNachrichten vol 291 no 11-12 pp 1899ndash1907 2018

14 Journal of Function Spaces

[28] N Weaver Lipschitz Algebras World Scientific Publishing CoInc River Edge NJ USA 1999

[29] R R Phelps Lectures on Choquetrsquos theorem vol 1757 of LectureNotes in Mathematics Springer-Verlag Berlin Germany 2ndedition 2001

[30] O Hatori and S Oi ldquoIsometries on Banach algebras of vector-valued mapsrdquo Acta Scientiarum Mathematicarum vol 84 no12 pp 151ndash183 2018

[31] O Hatori S Oi and H Takagi ldquoPeculiar homomorphismsbetween algebras of vector-valued mapsrdquo Studia Mathematicavol 242 no 2 pp 141ndash163 2018

[32] M Nagasawa ldquoIsomorphisms between commutative Banachalgebras with an application to rings of analytic functionsrdquoKodai Mathematical Seminar Reports vol 11 pp 182ndash188 1959

[33] K de LeeuwW Rudin and J Wermer ldquoThe isometries of somefunction spacesrdquo Proceedings of the American MathematicalSociety vol 11 pp 694ndash698 1960

[34] OHatori A Jimenez-Vargas andMVillegas-Vallecillos ldquoMapswhich preserve norms of non-symmetrical quotients betweengroups of exponentials of Lipschitz functionsrdquo Journal of Math-ematical Analysis and Applications vol 415 no 2 pp 825ndash8452014

[35] G Lumer ldquoOn the isometries of reflexiveOrlicz spacesrdquoAnnalesde lrsquoInstitut Fourier vol 68 pp 99ndash109 1963

[36] R J Fleming and J E Jamison Isometries on Banach SpacesFunction Spaces vol 129 ofMonographs and Surveys in Pure andApplied Mathematics Chapman amp Hall CRC Boca Raton FlaUSA 2003

[37] OHatori and SOi ldquoHermitian operators onBanach algebras ofvector-valued Lipschitzmapsrdquo Journal ofMathematical Analysisand Applications vol 452 no 1 pp 378ndash387 2017 Corrigendumto rdquoHermitian operators on Banach algebras of vector-valuedLipschitz mapsrdquo Journal of Mathematical Analysis and Appli-cations 452 (2017) 378ndash387 MR3628025

[38] A Nikou andA G OrsquoFarrell ldquoBanach algebras of vector-valuedfunctionsrdquo Glasgow Mathematical Journal vol 56 no 2 pp419ndash426 2014

[39] F Botelho and J Jamison ldquoHomomorphisms on a class ofcommutative Banach algebrasrdquo Rocky Mountain Journal ofMathematics vol 43 no 2 pp 395ndash416 2013

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Page 6: Hermitian Operators and Isometries on Banach Algebras of ...downloads.hindawi.com/journals/jfs/2018/8085304.pdfJournalofFunctionSpaces normiseitherthemaxnormorthesumnorm.esituation

6 Journal of Function Spaces

for all 120593 isin [0 2120587) It follows by Lemma 1 in [5] that

(119879119891) = 120590 (119891) + 119870 (Δ119891) (41)

and therefore10038171003817100381710038171198791198911003817100381710038171003817infin = 10038171003817100381710038171198911003817100381710038171003817infin + Δ119891 (42)

If Δ119891 le 0 then a similar calculation shows that

(119891) = (119879119891) + 119870 (minusΔ119891) (43)

and10038171003817100381710038171198911003817100381710038171003817infin = 10038171003817100381710038171198791198911003817100381710038171003817infin + (minusΔ119891) (44)

It follows that in any case (Δ119891 ge 0 Δ119891 le 0) we obtainΔ119891 = 10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin 119891 isin 119860 (45)

We will prove that1003817100381710038171003817119879 (119891)1003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin = Δ119891 ge 0 (46)

for all 119891 isin 119860 Once it is proved applying the same argumentfor119879minus1 instead of119879 we see that 119879minus1119892infinminus119892infin ge 0 for every119892 isin 119861 As 119879 is a bijection it follows that 119891infin minus 119879119891infin ge 0for every 119891 isin 119860 It will follow that 119879119891infin = 119891infin for every119891 isin 119860 A proof of (46) is the following For every 120576 gt 0denote

A120576 = 119891 isin 119860 120588 (120590 (119891)) le 120576 (47)

The inequality in (46) is deduced by the following assertionswhich appear in the proof of [5 Theorem]

(1) 119879 is a continuous mapping from (119860 sdot infin) onto (119861 sdotinfin)(2) For each 120576 gt 0 the setA120576 is dense in (119860 sdot infin)(3) For each 120576 gt 0 and each 119891 isin A120576 it holds that119879(119891)infin ge 119891infin minus 120576

Suppose that these assertions are proved Let 119891 isin 119860 By (2)for any 120576 gt 0 there is a sequence 119891119899 of functions inA120576 suchthat 119891119899 minus 119891infin 997888rarr 0 as 119899 997888rarr infin By (3) we have

10038171003817100381710038171198791198911198991003817100381710038171003817infin minus 10038171003817100381710038171198911198991003817100381710038171003817infin ge minus120576 (48)

for every 119899 Letting 119899 997888rarr infin we have10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin ge minus120576 (49)

by (1) As 120576 gt 0 is arbitrary we have that10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin ge 0 (50)

We show proofs of three assertions (1) (2) and (3) aboveprecisely The proof of (1) is slightly different from the corre-sponding one in [5 p 70] This change is rather ambitiousWe also point out that the terms minus1205872 and 1205872 which appearin the formulae (7) and (8) in [5] seem inappropriate theyread for example as 31205874 and 1205874 respectively

We now proceed to prove the first statement Aiming fora contradiction suppose that 119879 is not continuous from (119860 sdotinfin) to (119861 sdot infin) Let 120576 be a positive real number less than1100 Then there is a function 1198910 isin 119860 such that 1198910infin le 120576and 119879(1198910)infin = 1 Then there exist 1199100 isin Ch(119861) such that|119879(1198910)(1199100)| = 1 by [29 Proposition 63] Since 119879 is complex-linear we may suppose that 119879(1198910)(1199100) = 1

By (41) and (45) we deduce that Δ1198910 = 119879(1198910)infin minus1198910infin ge 1 minus 120576 and 120590(119879(1198910)) = 120590(1198910) + 119870(Δ1198910) As 1198910infin le 120576and 119879(1198910)infin = 1 we infer that 120590(1198910) sub 119870(120576) and (119879(1198910) sub119870(1) Let 119911 isin 119870(1 minus 2120576) Let 119909 isin 119883 Then |1198910(119909)| le 120576 assertsthat 119908 = 119911 minus 1198910(119909) isin 119870(1 minus 120576) sub 119870(Δ1198910) Thus

119911 = 1198910 (119909) + 119908 isin 120590 (1198910) + 119870 (Δ1198910) = 120590 (119879 (1198910)) (51)

Hence119870(1 minus 2120576) sub 120590(1198791198910) As 119879(1198910)infin = 1 we have119870(1 minus 2120576) sub 120590 (119879 (1198910)) sub 119870 (1) (52)

Consider the open neighborhood 1198800 of 1199100 in1198832 given by

1198800 = 119910 isin 1198832 1003816100381610038161003816119879 (1198910) (119910) minus 11003816100381610038161003816 lt 120576 (53)

We infer that 1198800 is a proper subset of1198832 by (52) Then by [5Lemma 2] there exists 119892 isin 119861 such that 119892infin le 1 + 120576 119892(1199100) =1 |119892(119910) + 1| lt 120576 for every 119910 isin 1198832 1198800 and |Im119892(119910)| lt 120576 forall 119910 isin 1198832 If 119867 denotes the closed rectangle whose verticesare the four points plusmn(1 + 120576) plusmn 120576119894 we have

120590 (119892) sub 119867 (54)

Consider now the set

119871 = 11989031205871198944119911 |119911| le 1 Re 119911 ge 1 minus 2120576 (55)

We claim that119879(1198910)(1198832)cap119871 = 0 Suppose that119879(1198910)(1198832)cap119871 =0 As 119879(1198910)(1198832) is compact there exists a positive integer119899 such that 119879(1198910)(1198832) cap 119871119899 = 0 where 119871119899 = 11989031205871198944119911 |119911| le 1Re 119911 gt 1 minus 2120576 minus 1119899 Then (52) gives 119879(1198910)(1198832) sub119870(1)119871119899 As (119879(1198910)) is the closed convex-hull of 119879(1198910)(1198832)it is contained in the closed convex set 119870(1) 119871119899 On theother hand (1 minus 2120576)11989031205871198944 isin 119870(1 minus 2120576) sub 120590(119879(1198910)) by (52) As(1minus2120576)11989031205871198944 isin 119871 sub 119871119899 this contradicts 120590(119879(1198910)) sub 119870(1) 119871119899and this proves our claim Hence there is 1199101 isin 1198832 with119879(1198910)(1199101) isin 119871 As 120576 le 1100 it follows that |119879(1198910)(1199101) minus1| ge 120576 and so 1199101 isin 1198832 1198800 Hence |119892(1199101) + 1| lt 120576 Thus119892(1199101) + 119879(1198910)(1199101) is in 119871 minus 1 + K(120576) Thus we have

1 + radic22 minus 3120576 le 119888119861 (119892 + 119879 (1198910) 31205874 ) (56)

We claim that

120590 (119892 + 119879 (1198910)) sub co (119870 (minus1 1) cup 2) + 119870 (3120576) (57)

where119870(1199110 119903) = 119911 isin C |119911 minus 1199110| le 119903 Let 119910 isin 1198832 Supposefirst that |119879(1198910)(119910)minus1| lt 120576 Since 119892(1198832) sub 119867 by (54) we have

119879 (1198910) (119910) + 119892 (119910) isin 119870 (1 120576) + 119867 = (119867 + 1) + 119870 (120576) (58)

Journal of Function Spaces 7

Suppose next that |119879(1198910)(119910)minus1| ge 120576Then 119910 isin 1198832 1198800 and so|119892(119910) + 1| lt 120576 Moreover |119879(1198910)(119910)| le 1 Therefore we have

119892 (119910) + 119879 (1198910) (119910) isin 119870 (1) minus 1 + 119870 (120576)= 119870 (minus1 1) + 119870 (120576) (59)

It follows from (58) and (59) that

(119892 + 119879 (1198910)) (1198832) sub 119870 (minus1 1) cup ((119867 + 1)) + 119870 (120576) (60)

and hence

120590 (119892 + 119879 (1198910)) sub co (119870 (minus1 1) cup 2) + 119870 (3120576) (61)

as is claimed Therefore we have

119888119861 (119892 + 119879 (1198910) 1205874 ) le radic2 + 3120576 (62)

Put 1198911 = 119879minus1(119892) We claim that Δ1198911 le 120576 If Δ1198911 lt 0 thereis nothing to prove Suppose that Δ1198911 ge 0 Then by (41) wehave

120590 (119892) = 120590 (1198911) + 119870 (Δ1198911) (63)

Since 120590(119892) sub 119867 by (54) we have

119867 sup 120590 (1198911) + 119870 (Δ1198911) (64)

As 119867 does not include a closed disk with the radius greaterthan 120576 we conclude that Δ1198911 le 120576

In the following we will consider two cases 0 le Δ1198911 le 120576and Δ1198911 le 0 Suppose first that 0 le Δ1198911 le 120576 Then (64) yields(1198911) sub 119867 From 1198910infin le 120576 we deduce that 120590(1198910) sub 119870(120576)Hence we have

120590 (1198911 + 1198910) sub 120590 (1198911) + 120590 (1198910) sub (1198910) + 120590 (1198910)sub 119867 + 119870 (120576) (65)

Since119867 +119870(120576) is convex we have (1198911 + 1198910) sub 119867 + 119870 (120576) (66)

From (39) we infer that

119888119860 (1198911 + 1198910 31205874 ) = 119888 (120590 (1198911 + 1198910) 31205874 )le 119888 (119867 + 119870 (120576) 31205874 )= 119888 (119867 31205874 ) + 120576= radic22 + (1 + radic2) 120576

(67)

Since 119879(1198911 + 1198910) = 119892 + 119879(1198910) from (56) and (67) we obtainthat

1 minus (4 + radic2) 120576 le 119888119861 (119879 (1198911 + 1198910) 31205874 )minus 119888119860 (1198911 + 1198910 31205874 )

(68)

By (63) and 1 = 119892(1199100) we deduce that 1 isin 120590(1198911) + 119870(Δ1198911)Thus there is 119911 isin 120590(1198911) such that |119911 minus 1| le Δ1198911 It follows thatradic22 minus Δ1198911 le 119888119860(1198911 1205874) hence we have

radic22 minus 2120576 le 119888119860 (1198911 + 1198910 1205874 ) (69)

as 1198910infin le 120576 and 0 le Δ1198911 le 120576 We get by (62) and (69) that

119888119861 (119879 (1198911 + 1198910) 1205874 ) minus 119888119860 (1198911 + 1198910 1205874 ) leradic22 + 5120576 (70)

On the other hand 119888119861(119879(119891) 120593)minus119888119860(119891 120593) is invariant for any 120593by (37) From (68) and (70)we deduce that 120576 ge (2minusradic2)2(9+radic2) and this contradicts that 120576 le 1100

For the second case suppose next that Δ1198911 le 0 Then by(43) we have

120590 (1198911) = 120590 (119892) + 119870 (minusΔ1198911) (71)

and by (54) it follows that (1198911) sub 119867 + 119870(minusΔ1198911) Moreover(1198910) sub 119870(120576) since 1198910infin le 120576 Then

120590 (1198911 + 1198910) sub 120590 (1198911) + 120590 (1198910) sub 120590 (1198911) + 120590 (1198910)sub 119867 + 119870(minusΔ1198911) + 119870 (120576) (72)

Hence (1198911+1198910) sub 119867+119870(minusΔ1198911)+119870(120576) Using (39) we inferthat

119888119860 (1198911 + 1198910 31205874 ) le 119888 (119867 + 119870(minusΔ1198911) + 119870 (120576) 31205874 )= 119888 (119867 31205874 ) + (minusΔ1198911) + 120576= radic22 + (1 + radic2) 120576 + (minusΔ1198911)

(73)

By (71) we obtain that 120590(1198911) sup 119892(1198832) + 119870(minusΔ1198911) and as119892(1199100) = 1 we infer that 120590(1198911) sup 1 + 119870(minusΔ1198911) Hence radic22 +(minusΔ1198911) le 119888119860(1198911 1205874) so thatradic22 + (minusΔ1198911) minus 120576 le 119888119860 (1198911 + 1198910 1205874 ) (74)

as 1198910infin le 120576 Since 119879(1198911 +1198910) = 119892 +119879(1198910) we obtain by (56)and (73) that

1 minus (4 + radic2) 120576 minus (minusΔ1198911)le 119888119861 (119879 (1198911 + 1198910) 31205874 ) minus 119888119860 (1198911 + 1198910 31205874 )

(75)

We also obtain by (62) and (74) that

119888119861 (119879 (1198911 + 1198910) 1205874 ) minus 119888119860 (1198911 + 1198910 1205874 )le radic22 + 4120576 minus (minusΔ1198911)

(76)

Since 119888119861(119879(119891) 120593)minus119888119860(119891 120593) is invariant for any 120593 by (37) from(75) and (76) we deduce that 120576 ge (2 minusradic2)2(8 +radic2) and thisis impossible since 120576 le 1100

8 Journal of Function Spaces

Next we show a proof of the second assertion (2) Let119891 isin 119860 We prove that there exists a sequence 119891119899 sub 119860 whichuniformly converges to 119891 such that 120588(120590(119891119899)) 997888rarr 0 as 119899 997888rarrinfin Without loss of generality we may assume that 119891infin = 1Then there exists 1199100 isin Ch(119860) such that |119891(1199100)| = 119891infinby [29 Proposition 63] We may assume that 119891(1199100) = 1Suppose that 119899 ge 4 Put

119880119899 = 119909 isin 119883 1003816100381610038161003816119891 (119909) minus 11003816100381610038161003816 le 11198992 (77)

and

Ω = co(119870 (1) cup 1 + 1119899) + 119870( 11198992) (78)

(In the following we identify R2 and C that is we identify(119909 119910) and 119909 + 119894119910 for every 119909 119910 isin R) Since we assume that119899 ge 4 we infer by a simple calculation that

[1 minus 11198992 minus 120575119899119899 1 + 1119899 + 120575119899119899 ] times [minus 11198992 minus 120575119899119899 11198992 + 120575119899119899 ]sub Ω

(79)

for 120575119899 with 0 lt 120575119899 le 11198992 We assume that 0 lt 120575119899 le 11198992 By[5 Lemma 2] there exists ℎ119899 isin 119860 such that ℎ119899infin le 1 + 2120575119899ℎ119899(1199100) = 1 |ℎ119899 + 1| le 2120575119899 on119883 119880119899 and |Im ℎ119899| le 2120575119899 on119883Put 119892119899 = (ℎ119899 + 1)2 Then 119892119899 isin 119860 and

10038171003817100381710038171198921198991003817100381710038171003817infin le 1 + 120575119899119892119899 (1199100) = 110038161003816100381610038161198921198991003816100381610038161003816 le 120575119899 on 119883 1198801198991003816100381610038161003816Im1198921198991003816100381610038161003816 le 120575119899 on 119883

minus120575119899 le Re119892119899 le 1 + 120575119899 on 119883

(80)

Let 119909 isin 119880119899 Then we have

1 minus 11198992 le Re119891 (119909) le 1minus 11198992 le Im119891 (119909) le 11198992

(81)

and

minus120575119899119899 le Re119892119899 (119909)119899 le 1119899 + 120575119899119899

minus120575119899119899 le Im119892119899 (119909)119899 le 120575119899119899

(82)

Hence

1 minus 11198992 minus 120575119899119899 le Re(119891 (119909) + 119892119899 (119909)119899 ) le 1 + 1119899 + 120575119899119899 minus 11198992 minus 120575119899119899 le Im(119891 (119909) + 119892119899 (119909)119899 ) le 11198992 + 120575119899119899

(83)

It follows that we have

(119891 + 119892119899119899 ) (119909) isin [1 minus 11198992 minus 120575119899119899 1 + 1119899 + 120575119899119899 ]times [minus 11198992 minus 120575119899119899 11198992 + 120575119899119899 ] sub Ω

(84)

for 119909 isin 119880119899 Suppose that 119909 isin 119883 119880119899 Then10038161003816100381610038161003816100381610038161003816119891 (119909) +

119892119899 (119909)11989910038161003816100381610038161003816100381610038161003816 le 1 +

120575119899119899 le 1 + 11198993 lt 1 + 11198992 (85)

and hence

119891 (119909) + 119892119899 (119909)119899 isin 119870 (1) + 119870( 11198992) sub Ω (86)

for 119909 isin 119883 119880119899 Since 1 + 1119899 = 119891(1199100) + 119892119899(1199100)119899 we have bycombining (84) and (86) that

1 + 1119899 isin (119891 + 119892119899119899 ) (119883) sub Ω (87)

AsΩ is convex we obtain

1 + 1119899 isin 120590 (119891 + 119892119899119899 ) sub Ω (88)

Recall that for 119903 ge 0 and a complex number 1199110119870 (1199110 119903) = 119908 isin C 1003816100381610038161003816119908 minus 11991101003816100381610038161003816 le 119903 (89)

denotes the closed disk with center 1199110 and radius 119903 Weobserve that 120588(Ω 1 + 1119899) = 1119899 + 11198992 Recall that

120588(Ω 1 + 1119899)= sup119903 119911 isin Ω 1 + 1119899 isin 119870 (119911 119903) sub Ω

(90)

Let ℓ1 be the line defined by the equation

119910 = minus119899radic2119899 + 1 (119909 minus (1 +1119899 + 11198992 + 11198993)) (91)

part of which is a part of the boundary ofΩ Let ℓ2 be the linedefined by the equation

119910 = minus119899radic2119899 + 1 (119909 minus (1 minus11198992)) (92)

By some calculation we have that the distance between 1 minus11198992 and 1+1119899 is 1119899+11198992 and it coincides with the distancebetween the point 1 minus 11198992 and the line ℓ1 Hence we see that

1 + 1119899 isin 119870(1 minus 11198992 1119899 + 11198992 ) sub Ω (93)

Thus 1119899 + 11198992 le 120588(Ω 1 + 1119899)Next we prove that 119870(119901 119903) cap Ω119888 = 0 for every 119901 isin Ω

with 119903 = |119901 minus (1 + 1119899)| gt 1119899 + 11198992 It will follow that1119899 + 11198992 ge 120588(Ω 1 + 1119899) and the equality will hold Let

Journal of Function Spaces 9

119901 = (119909119901 119910119901) = 119909119901 + 119894119910119901 isin Ω with 119903 = |119901 minus (1 + 1119899)| gt1119899 + 11198992 We prove the case where 119910119901 ge 0 A proof for thecase where 119910119901 le 0 is the same and we omit it We divideΩ+ =119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902 isin Ω 119910119902 ge 0 into two parts

Ω1 = 119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902isin Ω+ minus119899radic2119899 + 1 (119909119902 minus (1 minus

11198992)) le 119910119902le minus119899radic2119899 + 1 (119909 minus (1 +

1119899 + 11198992 + 11198993))(94)

and

Ω2 = 119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902 isin Ω+ 119910119902le minus119899radic2119899 + 1 (119909119902 minus (1 minus

11198992)) (95)

Suppose that 119901 isin Ω1 and 119903 = |119901minus (1+1119899)| gt 1119899+11198992Thedistance between ℓ1 and ℓ2 is 1119899+11198992 Hence119870(119901 119903)capΩ119888 =0 Suppose that 119901 isin Ω2 and 119903 = |119901 minus (1 + 1119899)| gt 1119899 + 11198992that is119901 = 1minus11198992 Let ℓ1015840 be the line passing through119901whichis parallel to ℓ2 Let 1199011015840 be the unique point in the intersectionof ℓ1015840 and the 119909-axis Then 1199011015840 = 1 minus 11198992 minus 119906 for some 119906 ge 0Then the distance between ℓ1 and ℓ1015840 is 1119899+11198992+119906(1+1119899)which is equal to the distance between the point119901 and the lineℓ1 On the other hand1003816100381610038161003816100381610038161003816119901 minus (1 + 1119899)

1003816100381610038161003816100381610038161003816 ge10038161003816100381610038161003816100381610038161199011015840 minus (1 + 1119899)

1003816100381610038161003816100381610038161003816 = 1119899 + 11198992 + 119906 (96)

It follows that 119870(119901 119903) cap Ω119888 = 0 We conclude that if 119901 isin Ωsatisfies 119903 = |119901minus(1+1119899)| gt 1119899+11198992 then119870(119901 119903)capΩ119888 = 0Thus we have

120588 (Ω 1 + 1119899) = 1119899 + 11198992 (97)

Since 1 + 1119899 isin 120590(119891 + 119892119899119899) sub Ω we have120588 (120590 (119891 + 119892119899119899 )) le 120588 (120590 (119891 + 119892119899119899 ) 1 + 1119899)

le 120588 (Ω 1 + 1119899) = 1119899 + 11198992 (98)

On the other hand 119892119899119899infin 997888rarr 0 as 119899 997888rarr infin ensures that119891 + 119892119899119899 minus 119891infin 997888rarr 0 as 119899 997888rarr infin It follows that for every120576 gt 0A120576 is dense in (119860 sdot infin)Finally we show a proof of the third assertion (3) As is

pointed out in the proof of [5Theorem] 120588(co(119870)+119870(119888)) ge 119888for any 119870 sub C and any 119888 ge 0 Let 120576 gt 0 and 119891 isin A120576 Supposethat Δ119891 le 0 Then by (43) we have (119891) = (119879119891) + 119870(minusΔ119891)Hence we have

120588 ( (119891)) = 120588 (120590 (119879119891) + 119870 (minusΔ119891)) ge minusΔ119891 (99)

As 120576 ge 120588(120590(119891)) we conclude by (45) that10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin = Δ119891 ge minus120576 (100)

This completes the proof of the theorem

4 Hermitian Operators on a BanachAlgebras of Continuous Maps WhoseValues Are in a Uniform Algebras

Let 119883 and 119884 be compact Hausdorff spaces Let 119861 be aunital subalgebra of 119862(119883) which separates the points of 119883Throughout this sectionwe assume119861 is a Banach algebrawiththe norm sdot 119861 and 119860 is a uniform algebra on 119884 Recall thata uniform algebra on 119884 is a uniformly closed subalgebra of119862(119884) which contains constants and separates the points of 119884For functions 119891 isin 119862(119883) and 119892 isin 119862(119884) let 119891 otimes 119892 isin 119862(119883 times 119884)be the function defined by119891otimes119892(119909 119910) = 119891(119909)119892(119910) for (119909 119910) isin119883 times 119884 and for a subspace 119864119883 of 119862(119883) and a subspace 119864119884 of119862(119884) put

119864119883 otimes 119864119884

= 119899sum

119895=1

119891119895 otimes 119892119895 119899 isin N 119891119895 isin 119864119883 119892119895 isin 119864119884

(101)

and

1 otimes 119864119884 = 1 otimes 119892 119892 isin 119864119884 (102)

Throughout the section 119861 is a unital subalgebra of 119862(119883 times 119884)with a Banach algebra norm sdot 119861 We assume that 119861otimes119860 sub 119861Note that 119861 separates the points of119883times119884 since119860 separates thepoints of 119884 and 119861 separates the points of 119883 We assume thatthere exists a compact Hausdorff space M and a complex-linear map 119863 119861 997888rarr 119862(M) such that ker119863 = 1 otimes 119860 Weassume that 119865119861 = 119865infin(119883times119884)+119863(119865)infin(M) for every 119865 isin 119861Hence119863 is continuous Defining

lsaquo119865lsaquo = 119863 (119865)infin(M) 119865 isin 119861 (103)

lsaquo sdot lsaquo is a one-invariant seminorm in the sense of Jarosz lsaquo sdot lsaquois a seminorm on 119861 such that lsaquo119865+ 1lsaquo = lsaquo119865lsaquo for every 119865 isin 119861Hence the norm sdot 119861 is a natural norm (see [5 p67]) Notethat 119861 is a regular subspace of 119862(119883times119884) in the sense of Jarosz[5 Proposition 2]

Lumerrsquos seminal paper [35] opened up a useful methodof finding isometries which is often referred to as Lumerrsquosmethod It involves the notion ofHermitian operators and thefact that 119880119867119880minus1 must be Hermitian if119867 is Hermitian and 119880is a surjective isometry

Definition 6 Let A be a unital Banach algebra We say that119890 isin A is a Hermitian element if1003817100381710038171003817exp (119894119905119890)1003817100381710038171003817A = 1 (104)

for every 119905 isin R The set of all Hermitian elements of A isdenoted by119867(A)

If A is a unital 119862lowast-algebra then 119867(A) is the set of allself-adjoint elements of A Hence 119867(119872119899(C)) is the set of allHermitian matrices and119867(119862(119884)) = 119862R(119884)Definition 7 Let 119864 be a complex Banach space The Banachalgebra of all bounded operators on 119864 is denoted by 119861(119864) Wesay that 119879 isin 119861(119864) is a Hermitian operator if 119879 isin 119867(119861(119864))

10 Journal of Function Spaces

Note that a Hermitian element of a unital Banach algebraand a Hermitian operator are usually defined in terms ofnumerical range or semi-inner product Here we define themby an equivalent form (see [36]) By the definition of aHermitian operator we have the following

Proposition 8 Let 119864119895 be a complex Banach space for 119895 = 1 2Suppose that 119881 1198641 997888rarr 1198642 is a surjective isometry and 119867 1198641 997888rarr 1198641 is a Hermitian operator Then 119881119867119881minus1 1198642 997888rarr 1198642

is a Hermitian operator

Proposition 9 An element 119865 isin 119861 is Hermitian if and only ifthere exists 119891 isin 119860 cap 119862R(119884) such that 119865 = 1 otimes 119891Proof Suppose that 119865 isin 119861 is a Hermitian element Then1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884) + 1003817100381710038171003817119863 (exp (119894119905119865))1003817100381710038171003817infin(M)

= 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817 = 1 (105)

for every 119905 isin R Suppose that there exists a point (119909 119910) isin119883 times 119884 with Im119865(119909 119910) = 0 where Im denotes the imaginarypart of a complex number Suppose that Im119865(119909 119910) gt 0Then1003817100381710038171003817exp (minus119894119865)1003817100381710038171003817infin(119883times119884) ge 1003816100381610038161003816exp (minus119894119865 (119909 119910))1003816100381610038161003816

= exp (Im119865 (119909 119910)) gt 1 (106)

Suppose that Im119865(119909 119910) lt 0 Then1003817100381710038171003817exp (119894119865)1003817100381710038171003817infin(119883times119884) ge 1003816100381610038161003816exp (119894119865 (119909 119910))1003816100381610038161003816= exp (minusIm119865 (119909 119910)) gt 1 (107)

In any case we have there exists 119905 isin R such that1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817119861 ge 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884) gt 1 (108)

which contradicts our assumption We have that

119865 isin 119862R (119883 times 119884) (109)

Thus for every (119904 119905) isin 119883 times 119884 and 119905 isin R |exp(119894119905119865(119904 119905))| = 1Hence exp(119894119905119865)infin(119883times119884) = 1 for every 119905 isin R By (105)we have119863(exp(119894119905119865)infin(M) = 0 which ensures that 119863(exp(119894119905119865)) = 0for every 119905 isin R Thus exp(119894119905119865) isin 1 otimes 119860 for every 119905 isin R Wehave

exp (119894119905119865) minus 1119905 minus 119894119865 = infinsum119899=2

((119894119865)119899 119905119899minus2119899 ) 119905 (110)

and hence for every 119905 isin R with |119905| le 1 we have10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817 le (

infinsum119899=2

119865119899119861|119905|119899minus2119899 ) |119905|

le (infinsum119899=2

119865119899119861119899 ) |119905| le (exp 119865119861) |119905|

(111)

It follows that10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817infin(119883times119884)

le 10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817119861

997888rarr 0(112)

as 119905 997888rarr 0 Since (exp(119894119905119865)minus1)119905 isin 1otimes119860 for each 119905 isin R thereexists 119892119905 isin 119860 such that

exp (119894119905119865) minus 1119905 = 1 otimes 119892119905 (113)

By (112) we have10038171003817100381710038171 otimes 1198921119899 minus 1198941198651003817100381710038171003817infin(119883times119884)

997888rarr 0 (114)

as 119899 997888rarr infin We have that 1 otimes 1198921119899 is a Cauchy sequencein 119862(119883 times 119884) thus we infer that1198921119899 is a Cauchy sequence in119862(119884) Since 119860 is uniformly closed as it is a uniform algebrathere exists 119892 isin 119860 such that

10038171003817100381710038171198921119899 minus 1198921003817100381710038171003817infin(119884)997888rarr 0 (115)

and hence10038171003817100381710038171 otimes 1198921119899 minus 1 otimes 1198921003817100381710038171003817infin(119883times119884)

997888rarr 0 (116)

as 119899 997888rarr infin It follows by (114) that 119894119865 = 1 otimes 119892 thus119865 = 1 otimes (minus119894119892) isin 1 otimes 119860 (117)

By (109) we see that minus119894119892 isin 119862R(119884) thus we have 119891 = minus119894119892 isin119860 cap 119862R(119884) and 119865 = 1 otimes 119891Suppose conversely that 119861 ni 119865 = 1otimes119891 for119891 isin 119860cap119862R(119884)

We infer that 119865 isin 119862R(119883 times 119884) and |exp(119894119905(119865(119909 119910))| = 1 forevery 119905 isin R and (119909 119910) isin 119883 times 119884 Hence exp(119894119905119865)infin(119883times119884) = 1for every 119905 isin R Since

exp (119894119905119865) = exp (119894119905 (1 otimes 119891)) = 1 otimes exp (119894119905119891) isin 1 otimes 119860 (118)

we have 119863(exp(119894119905119865)) = 0 It follows that1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817 = 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884)

+ 1003817100381710038171003817119863 (exp (119894119905119865))1003817100381710038171003817infin(M) = 1 (119)

for every 119905 isin R We conclude that 119865 is a Hermitian elementin 119861

Note that 119891 isin 119860 is Hermitian if and only if119891 isin 119860cap119862R(119884)by [37 Proposition 5] Hence Proposition 9 asserts that 119865 is aHermitian element in119861 if and only if119865 = 1otimes119891 for aHermitianelement 119891 in 119860Proposition 10 Suppose that 119880 119861 997888rarr 119861 is a surjectiveunital isometry Then 119880 is an algebra isomorphism

Proof As we have already mentioned 119861 is a regular subspace(in the sense of Jarosz) with a natural norm Then byTheorem 1 119880 is also an isometry with respect to the supre-mum norm on 119883 times 119884 Then 119880 is uniquely extended to asurjective isometry with respect to the supremum norm from the uniform closure 119861 onto itself Since 119861 is a uniformalgebra a theorem of Nagasawa [32] asserts that is analgebra isomorphism since (1) = 1 Thus 119880 is an algebraisomorphism from 119861 onto itself

Journal of Function Spaces 11

Theorem 11 A bounded operator 119879 119861 997888rarr 119861 is a Hermitianoperator if and only if 119879(1) is a Hermitian element in 119861 and119879 = 119872119879(1) the multiplication operator by 119879(1)Proof By Proposition 10 every surjective unital isometry on119861 is multiplicative Then by [37 Theorem 4] we have theconclusion

5 Banach Algebras of 119862(119884)-Valued Maps

Suppose that119883 is a compact Hausdorff space Suppose that 119861is a unital point separating subalgebra of119862(119884) equipped witha Banach algebra norm Then 119861 is semisimple because 119891 isin119861 119891(119909) = 0 is a maximal ideal of 119861 for every 119909 isin 119883 and theJacobson radical of 119861 vanishes The inequality 119891infin le 119891119861for every 119891 isin 119861 is well known We say that 119861 is natural if themap 119890 119884 997888rarr 119872119861 defined by 119910 997891997888rarr 120601119910 where 120601119910(119891) = 119891(119910)for every 119891 isin 119861 is bijective We say that 119861 is self-adjoint if 119861 isnatural and conjugate-closed in the sense that 119891 isin 119861 impliesthat 119891 isin 119861 for every 119891 isin 119861 where sdot denotes the complexconjugation on 119884Definition 12 Let 119883 and 119884 be compact Hausdorff spacesSuppose that 119861 is a unital point separating subalgebra of119862(119883)equipped with a Banach algebra norm sdot 119861 Suppose that119861 is self-adjoint Suppose that 119861 is a unital point separatingsubalgebra of 119862(119883 times 119884) such that 119861 otimes 119862(119884) sub 119861 equippedwith a Banach algebra norm sdot 119861 Suppose that 119861 is self-adjoint We say that 119861 is a natural 119862(119884)-valuezation of 119861 ifthere exists a compact Hausdorff space M and a complex-linear map 119863 119861 997888rarr 119862(M) such that ker119863 = 1 otimes 119862(119884) and119863(119862R(119883 times 119884) cap 119861) sub 119862R(M) which satisfies

119865 = 119865infin(119883times119884) + 119863 (119865)infin(M) 119865 isin 119861 (120)

The term ldquoa natural 119862(119884)-valuezation of 119861rdquo comes fromthe natural norm defined by Jarosz [5] In fact the norm sdot 119861is a natural norm in the sense of Jarosz [5]

Note that (119883119862(119884) 119861 119861) need not be an admissiblequadruple defined by Nikou and OrsquoFarrell [38] (cf [31]) sincewe do not assume that 119865(sdot 119910) 119865 isin 119861 119910 isin 119884 sub 119861 whichis a requirement for the admissible quadruple On the otherhand if (119883119862(119884) 119861 119861) is an admissible quadruple of type Ldefined in [30] then 119861 is a natural 119862(119884)-valuezation of 119861 dueto Definition 12

Example 13 Let 119861 = 1198621([0 1]) and 119861 = Lip([0 1] 119862(119884))for 119884 = 119901 a singleton Then Lip([0 1]) is algebraically iso-morphic to Lip([0 1] 119862(119884)) Suppose thatM is the maximalideal space of 119871infin([0 1]) and 119863 119861 997888rarr 119862(M) is definedby 119891 997891997888rarr Γ(1198911015840) where Γ denotes the Gelfand transformin 119871infin([0 1]) Then 119861 is a natural 119862(119884)-valuezation of 119861The Banach algebra Lip([0 1]) with the norm 119891infin([01]) +1198911015840infin([01]) is isometrically isomorphic to 119861

Let 119884 be a compact Hausdorff space Note that a closedsubalgebra 119861 of Lip((119870 119889120572) 119862(119884))which appears in Example12 in [30] is an example of a natural 119862(119884)-valuezation of 119861

The Banach algebras1198621([0 1] 119862(119884)) and 1198621(T 119862(119884)))whichappear in Examples 16 and 17 in [30] respectively are alsoexamples of natural 119862(119884)-valuezations of 1198621([0 1])6 Isometries on Natural 119862(119884)-ValuezationsThemain theorem in this paper is the following

Theorem 14 Suppose that 119861119895 is a natural 119862(119884119895)-valuezationof 119861119895 sub 119862(119883119895) for 119895 = 1 2 We assume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (121)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for119895 = 1 2 Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linear isometry Then there exists ℎ isin 119862(1198842) such that |ℎ| = 1on 1198842 a continuous map 120593 1198832 times 1198842 997888rarr 1198831 such that120593(sdot 119910) 1198832 997888rarr 1198831 is a homeomorphism for each 119910 isin 1198842and a homeomorphism 120591 1198842 997888rarr 1198841 which satisfies119880 (119865) (119909 119910) = ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842 (122)

for every 119865 isin 1198611

In short a surjective isometry between 119862(119884)-valuezationsis a weighted composition operator of a specific form thehomeomorphism 1198832 times 1198842 997888rarr 1198831 times 1198841 (119909 119910) 997891997888rarr(120593(119909 119910) 120591(119910)) has the second coordinate that depends onlyon the second variable 119910 isin 1198842 A composition operatorinduced by such a homeomorphism is said to be of type BJin [31 37] after the study of Botelho and Jamison [39]

Quite recently the author of this paper and Oi [30 Theo-rem 8] proved a similar result of Theorem 14 for admissiblequadruples of type L To prove it we apply Proposition 32and the following comments in [31] Instead of this we proveTheorem 14 by Lumerrsquosmethod with which a proof is simplerthan that in [30]

In the following in this section we assume that 119861119895 is anatural 119862(119884119895)-valuezation of 119861 sub 119862(119883119895) for 119895 = 1 2 Weassume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (123)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for 119895 = 1 2Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linearisometry A crucial part of a proof of Theorem 14 is to proveProposition 15

Proposition 15 Suppose that 1198832 is not a singleton Thereexists ℎ isin 119862(1198842) with |ℎ| = 1 on 1198842 such that 119880(11198611) = 11198612 otimesℎ

A similar result for admissible quadruples of type L isproved in [30 Proposition 9] If we assumed that

119865 (sdot 119910) 119865 isin 119861119895 119910 isin 119884119895 sub 119861119895 (124)

then 119861119895 were an admissible quadruple of type L Although119861119895 in this paper need not be an admissible quadruple of type

12 Journal of Function Spaces

L a proof of Proposition 15 is completely the same as that in[30 Proposition 9] since we do not make use of the condition(124) in the proof of [30 Proposition 9] The condition (124)is needed in [30] when we apply Proposition 32 and thefollowing comments in [31]

7 Proof of Theorem 14 An Application ofLumerrsquos Method

Proof of Theorem 14 A proof for the case where 1198831 = 1199091and 1198832 = 1199092 are singletons is the same as the proof ofTheorem 8 in [30]

Suppose that1198832 is not a singleton By Proposition 15 thereexists ℎ isin 119862R(1198842) with |ℎ| = 1 on 1198842 such that 119880(1) =1 otimes ℎ Letting 1198800 1198611 997888rarr 1198612 by 1198800(119865) = (1 otimes ℎ)119880(119865)119865 isin 1198611 we see by the hypothesis (1 otimes ℎ)1198651198612 = 1198651198612 forevery 119865 isin 1198612 that 1198800 is a surjective unital isometry from1198611 onto 1198612 Then Corollary 2 asserts that 1198800 is an algebraisomorphism Let 119891 isin 119862R(1198841) By Proposition 9 1 otimes 119891 isa Hermitian element in 1198611 Then by Theorem 11 1198721otimes119891 is aHermitian operator on 1198611 By Proposition 8 11988001198721otimes119891119880minus1

0 is aHermitian operator on 1198612 Then by Theorem 11 there exists119878(119891) isin 119862R(1198842) such that 11988001198721otimes119891119880minus1

0 = 1198721otimes119878(119891) Hence anoperator 119878 119862R(1198841) 997888rarr 119862R(1198842) is defined Since 1198800 is analgebra isomorphism it is easy to see that 119878 is a real algebraisomorphism from 119862R(1198841) onto 119862R(1198842) Then 119878 119862(1198841) 997888rarr119862(1198842) defined by 119878(119891) = 119878(Re119891) + 119894119878(Im119891) for 119891 isin 119862(1198841)gives a complex algebra isomorphism Gelfand theory assertsthat there is a homeomorphism 120591 1198842 997888rarr 1198841 such that119878(119891) = 119891 ∘ 120591 119891 isin 119862(1198841) It follows that

11988001198721otimes119891119880minus10 = 1198721otimes119891∘120591 119891 isin 119862 (1198841) (125)

Since 119880minus10 (1) = 1 we have

1198800 (1 otimes 119891) = 1 otimes 119891 ∘ 120591 119891 isin 119862 (1198841) (126)

Define Φ 1198611 997888rarr 1198612 by Φ(119886) = 1198800(119886 otimes 1) 119886 isin 1198611Since 1198800 is an algebra isomorphism the map Φ is a unitalhomomorphism Since the maximal ideal space of 1198611 is 1198831

and the maximal ideal space of 1198612 is 1198832 times 1198842 there is acontinuous map 120593 1198832 times 1198842 997888rarr 1198831 such that

Φ(119886) (119909 119910) = 119886 (120593 (119909 119910)) 119886 isin 1198611 (119909 119910) isin 1198832 times 1198842 (127)

It follows by (126) and (127) that

1198800 (119886 otimes 119891) (119909 119910) = 1198800 ((119886 otimes 1) (1 otimes 119891)) (119909 119910)= 1198800 (119886 otimes 1) (119909 119910)1198800 (1 otimes 119891) (119909 119910)= 119886 (120593 (119909 119910))119891 (120591 (119910))= (119886 otimes 119891) (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198832 times 1198842

(128)

for every 119886 isin 1198611 and 119891 isin 119862(1198841) Thus

1198800 (sum(119886119895 otimes 119891119895)) (119909 119910)= (sum(119886119895 otimes 119891119895)) (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842(129)

for every sum(119886119895 otimes 119891119895) isin 1198611 otimes 119862(1198841) By the Stone-Weierstrasstheorem 1198611 otimes 119862(1198841) is uniformly dense in 119862(1198831 times 1198841) henceany element in 1198611 is uniformly approximated by 1198611 otimes 119862(1198841)As 1198800 is also an isometry with respect to the uniform normwe see that

1198800 (119865) (119909 119910) = 119865 (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198831 times 1198841 (130)

for every 119865 isin 1198611 and

119880 (119865) (119909 119910) = (1 otimes ℎ) (119909 119910) 119865 (120593 (119909 119910) 120591 (119910))= ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198831 times 1198841(131)

As 1198800 is an algebra isomorphism the map 1198832 times 1198842 997888rarr1198831 times 1198841 defined by (119909 119910) 997891997888rarr (120593(119909 119910) 120591(119910)) gives ahomeomorphism Therefore for every 119910 isin 1198842 the map

120593 (sdot 119910) 1198832 997888rarr 1198831 (132)

is a homeomorphismSuppose that 1198831 is not a singleton By the same way as

in the last part of the proof of Theorem 8 in [30] we havethat 1198832 is not a singleton Then we have the conclusion bythe previous argument

8 Application of Theorem 14

We exhibit applications of Theorem 14

Corollary 16 ([4 Theorem 33]) Suppose that119880 Lip([0 1]) 997888rarr Lip([0 1]) is a surjective isometrywith respect to the norm defined by 119891infin([01]) + 1198911015840infin([01])

for 119891 isin Lip([0 1]) Then 119880(1) is a constant function of unitmodulus such that

119880(119891) (119909) = 119880 (1) 119891 (119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (133)

or

119880(119891) (119909) = 119880 (1) 119891 (1 minus 119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (134)

The converse statement also holds

Proof By Example 13 we may suppose that Lip([0 1]) is aBanach algebra of 119862(119884)-valuezation Applying Theorem 14we have that119880(1) = 1otimesℎ for ℎ isin 119862(119884)with |ℎ| = 1 Since our

Journal of Function Spaces 13

119884 is a singleton 119880(1) is a constant function of unit modulusWe also see that the corresponding continuous map 120593 [0 1] times 119884 997888rarr [0 1] can be considered as a homeomorphismfrom [0 1] onto [0 1] therefore we have that119880(119891) (119909) = 119880 (1) 119891 (120593 (119909))

119891 isin Lip ([0 1]) 119909 isin [0 1] (135)

The rest is a routine argument to prove that 120593 is an isometryhence 120593(119909) = 119909 119909 isin [0 1] or 120593(119909) = 1 minus 119909 119909 isin [0 1]

The converse statement is trivial

Corollaries 14 15 18 and 19 in [30 Section 6] follow herewith a similar proof

Data Availability

The data used to support the findings of this study areavailable from the corresponding author upon request

Conflicts of Interest

The author declares that they have no conflicts of interest

Acknowledgments

This work was supported by JSPS KAKENHI Grants Num-bers JP16K05172 and JP15K04921

References

[1] K de Leeuw ldquoBanach spaces of Lipschitz functionsrdquo StudiaMathematica vol 21 pp 55ndash66 19611962

[2] A K Roy ldquoExtreme points and linear isometries of the Banachspace of Lipschitz functionsrdquoCanadian Journal of MathematicsJournal Canadien de Mathematiques vol 20 pp 1150ndash11641968

[3] M Cambern ldquoIsometries of certain Banach algebrasrdquo StudiaMathematica vol 25 pp 217ndash225 19641965

[4] N V Rao and A K Roy ldquoLinear isometries of some functionspacesrdquo Pacific Journal of Mathematics vol 38 pp 177ndash192 1971

[5] K Jarosz ldquoIsometries in semisimple commutative BanachalgebrasrdquoProceedings of the AmericanMathematical Society vol94 no 1 pp 65ndash71 1985

[6] K Jarosz and V D Pathak ldquoIsometries between functionspacesrdquo Transactions of the AmericanMathematical Society vol305 no 1 pp 193ndash206 1988

[7] NWeaver ldquoIsometries of noncompact Lipschitz spacesrdquoCana-dian Mathematical Bulletin Bulletin Canadien de Mathemat-iques vol 38 no 2 pp 242ndash249 1995

[8] A Jimenez-Vargas and M Villegas-Vallecillos ldquoLinear isome-tries between spaces of vector-valued Lipschitz functionsrdquoProceedings of the American Mathematical Society vol 137 no4 pp 1381ndash1388 2009

[9] A Jimenez-Vargas and M Villegas-Vallecillos ldquoInto linearisometries between spaces of Lipschitz functionsrdquo HoustonJournal of Mathematics vol 34 no 4 pp 1165ndash1184 2008

[10] F Botelho and J Jamison ldquoSurjective isometries on spacesof differentiable vector-valued functionsrdquo Studia Mathematicavol 192 no 1 pp 39ndash50 2009

[11] E Mayer-Wolf ldquoIsometries between Banach spaces of Lipschitzfunctionsrdquo Israel Journal of Mathematics vol 38 no 1-2 pp 58ndash74 1981

[12] A Jimenez-Vargas M Villegas-Vallecillos and Y-S WangldquoBanach-Stone theorems for vector-valued little Lipschitz func-tionsrdquo Publicationes Mathematicae vol 74 no 1-2 pp 81ndash1002009

[13] J Araujo andLDubarbie ldquoNoncompactness andnoncomplete-ness in isometries of Lipschitz spacesrdquo Journal of MathematicalAnalysis and Applications vol 377 no 1 pp 15ndash29 2011

[14] F Botelho R J Fleming and J Jamison ldquoExtreme points andisometries on vector-valued Lipschitz spacesrdquo Journal of Math-ematical Analysis and Applications vol 381 no 2 pp 821ndash8322011

[15] H Koshimizu ldquoLinear isometries on spaces of continuouslydifferentiable and Lipschitz continuous functionsrdquo NihonkaiMathematical Journal vol 22 no 2 pp 73ndash90 2011

[16] F Botelho J Jamison and B Zheng ldquoIsometries on spaces ofvector valued Lipschitz functionsrdquo Positivity An InternationalMathematics Journal Devoted to Theory and Applications ofPositivity vol 17 no 1 pp 47ndash65 2013

[17] A Ranjbar-Motlagh ldquoA note on isometries of Lipschitz spacesrdquoJournal of Mathematical Analysis and Applications vol 411 no2 pp 555ndash558 2014

[18] F Botelho and J Jamison ldquoSurjective isometries on spaces ofvector valued continuous and Lipschitz functionsrdquo PositivityAn International Mathematics Journal Devoted to Theory andApplications of Positivity vol 17 no 3 pp 395ndash405 2013Erratum to Surjective isometries on spaces of vector valuedcontinuous and Lipschitz functions by F Botelho 20 (2016)757ndash759

[19] T Miura and H Takagi ldquoSurjective isometries on the Banachspace of continuously differentiable functionsrdquo ContemporaryMathematics vol 687 pp 181ndash192 2017

[20] K Kawamura ldquoBanach-Stone type theorems for 1198621-functionspaces over Riemannian manifoldsrdquo Acta Universitatis Szegedi-ensis Acta Scientiarum Mathematicarum vol 83 no 3-4 pp551ndash591 2017

[21] K Kawamura ldquoPerturbations of norms on 1198621-function spacesand associated isometry groupsrdquo Topology Proceedings vol 51pp 169ndash196 2018

[22] K Kawamura ldquoA Banach-Stone type theorem for 1198621-functionspaces over the circlerdquo Topology Proceedings vol 53 pp 15ndash262019

[23] L Li D Chen Q Meng and Y-S Wang ldquoSurjective isometrieson vector-valued differentiable function spacesrdquo Annals ofFunctional Analysis vol 9 no 3 pp 334ndash343 2018

[24] K Kawamura H Koshimizu and T Miura ldquoNorms onC1([01]) and there isometriesrdquo Acta Scientiarum Mathemati-carum vol 84 no 12 pp 239ndash261 2018

[25] L Li A M Peralta L Wang and Y-S Wang ldquoWeak-2-local isometries on uniform algebras and Lipschitz algebrasrdquohttpsarxivorgabs170503619

[26] A Jimenez-Vargas L LiAMPeralta LWang andY-SWangldquo2-local standard isometries on vector-valued Lipschitz func-tion spacesrdquo Journal of Mathematical Analysis and Applicationsvol 461 no 2 pp 1287ndash1298 2018

[27] A Ranjbar-Motlagh ldquoIsometries of Lipschitz type functionspacesrdquoMathematischeNachrichten vol 291 no 11-12 pp 1899ndash1907 2018

14 Journal of Function Spaces

[28] N Weaver Lipschitz Algebras World Scientific Publishing CoInc River Edge NJ USA 1999

[29] R R Phelps Lectures on Choquetrsquos theorem vol 1757 of LectureNotes in Mathematics Springer-Verlag Berlin Germany 2ndedition 2001

[30] O Hatori and S Oi ldquoIsometries on Banach algebras of vector-valued mapsrdquo Acta Scientiarum Mathematicarum vol 84 no12 pp 151ndash183 2018

[31] O Hatori S Oi and H Takagi ldquoPeculiar homomorphismsbetween algebras of vector-valued mapsrdquo Studia Mathematicavol 242 no 2 pp 141ndash163 2018

[32] M Nagasawa ldquoIsomorphisms between commutative Banachalgebras with an application to rings of analytic functionsrdquoKodai Mathematical Seminar Reports vol 11 pp 182ndash188 1959

[33] K de LeeuwW Rudin and J Wermer ldquoThe isometries of somefunction spacesrdquo Proceedings of the American MathematicalSociety vol 11 pp 694ndash698 1960

[34] OHatori A Jimenez-Vargas andMVillegas-Vallecillos ldquoMapswhich preserve norms of non-symmetrical quotients betweengroups of exponentials of Lipschitz functionsrdquo Journal of Math-ematical Analysis and Applications vol 415 no 2 pp 825ndash8452014

[35] G Lumer ldquoOn the isometries of reflexiveOrlicz spacesrdquoAnnalesde lrsquoInstitut Fourier vol 68 pp 99ndash109 1963

[36] R J Fleming and J E Jamison Isometries on Banach SpacesFunction Spaces vol 129 ofMonographs and Surveys in Pure andApplied Mathematics Chapman amp Hall CRC Boca Raton FlaUSA 2003

[37] OHatori and SOi ldquoHermitian operators onBanach algebras ofvector-valued Lipschitzmapsrdquo Journal ofMathematical Analysisand Applications vol 452 no 1 pp 378ndash387 2017 Corrigendumto rdquoHermitian operators on Banach algebras of vector-valuedLipschitz mapsrdquo Journal of Mathematical Analysis and Appli-cations 452 (2017) 378ndash387 MR3628025

[38] A Nikou andA G OrsquoFarrell ldquoBanach algebras of vector-valuedfunctionsrdquo Glasgow Mathematical Journal vol 56 no 2 pp419ndash426 2014

[39] F Botelho and J Jamison ldquoHomomorphisms on a class ofcommutative Banach algebrasrdquo Rocky Mountain Journal ofMathematics vol 43 no 2 pp 395ndash416 2013

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Page 7: Hermitian Operators and Isometries on Banach Algebras of ...downloads.hindawi.com/journals/jfs/2018/8085304.pdfJournalofFunctionSpaces normiseitherthemaxnormorthesumnorm.esituation

Journal of Function Spaces 7

Suppose next that |119879(1198910)(119910)minus1| ge 120576Then 119910 isin 1198832 1198800 and so|119892(119910) + 1| lt 120576 Moreover |119879(1198910)(119910)| le 1 Therefore we have

119892 (119910) + 119879 (1198910) (119910) isin 119870 (1) minus 1 + 119870 (120576)= 119870 (minus1 1) + 119870 (120576) (59)

It follows from (58) and (59) that

(119892 + 119879 (1198910)) (1198832) sub 119870 (minus1 1) cup ((119867 + 1)) + 119870 (120576) (60)

and hence

120590 (119892 + 119879 (1198910)) sub co (119870 (minus1 1) cup 2) + 119870 (3120576) (61)

as is claimed Therefore we have

119888119861 (119892 + 119879 (1198910) 1205874 ) le radic2 + 3120576 (62)

Put 1198911 = 119879minus1(119892) We claim that Δ1198911 le 120576 If Δ1198911 lt 0 thereis nothing to prove Suppose that Δ1198911 ge 0 Then by (41) wehave

120590 (119892) = 120590 (1198911) + 119870 (Δ1198911) (63)

Since 120590(119892) sub 119867 by (54) we have

119867 sup 120590 (1198911) + 119870 (Δ1198911) (64)

As 119867 does not include a closed disk with the radius greaterthan 120576 we conclude that Δ1198911 le 120576

In the following we will consider two cases 0 le Δ1198911 le 120576and Δ1198911 le 0 Suppose first that 0 le Δ1198911 le 120576 Then (64) yields(1198911) sub 119867 From 1198910infin le 120576 we deduce that 120590(1198910) sub 119870(120576)Hence we have

120590 (1198911 + 1198910) sub 120590 (1198911) + 120590 (1198910) sub (1198910) + 120590 (1198910)sub 119867 + 119870 (120576) (65)

Since119867 +119870(120576) is convex we have (1198911 + 1198910) sub 119867 + 119870 (120576) (66)

From (39) we infer that

119888119860 (1198911 + 1198910 31205874 ) = 119888 (120590 (1198911 + 1198910) 31205874 )le 119888 (119867 + 119870 (120576) 31205874 )= 119888 (119867 31205874 ) + 120576= radic22 + (1 + radic2) 120576

(67)

Since 119879(1198911 + 1198910) = 119892 + 119879(1198910) from (56) and (67) we obtainthat

1 minus (4 + radic2) 120576 le 119888119861 (119879 (1198911 + 1198910) 31205874 )minus 119888119860 (1198911 + 1198910 31205874 )

(68)

By (63) and 1 = 119892(1199100) we deduce that 1 isin 120590(1198911) + 119870(Δ1198911)Thus there is 119911 isin 120590(1198911) such that |119911 minus 1| le Δ1198911 It follows thatradic22 minus Δ1198911 le 119888119860(1198911 1205874) hence we have

radic22 minus 2120576 le 119888119860 (1198911 + 1198910 1205874 ) (69)

as 1198910infin le 120576 and 0 le Δ1198911 le 120576 We get by (62) and (69) that

119888119861 (119879 (1198911 + 1198910) 1205874 ) minus 119888119860 (1198911 + 1198910 1205874 ) leradic22 + 5120576 (70)

On the other hand 119888119861(119879(119891) 120593)minus119888119860(119891 120593) is invariant for any 120593by (37) From (68) and (70)we deduce that 120576 ge (2minusradic2)2(9+radic2) and this contradicts that 120576 le 1100

For the second case suppose next that Δ1198911 le 0 Then by(43) we have

120590 (1198911) = 120590 (119892) + 119870 (minusΔ1198911) (71)

and by (54) it follows that (1198911) sub 119867 + 119870(minusΔ1198911) Moreover(1198910) sub 119870(120576) since 1198910infin le 120576 Then

120590 (1198911 + 1198910) sub 120590 (1198911) + 120590 (1198910) sub 120590 (1198911) + 120590 (1198910)sub 119867 + 119870(minusΔ1198911) + 119870 (120576) (72)

Hence (1198911+1198910) sub 119867+119870(minusΔ1198911)+119870(120576) Using (39) we inferthat

119888119860 (1198911 + 1198910 31205874 ) le 119888 (119867 + 119870(minusΔ1198911) + 119870 (120576) 31205874 )= 119888 (119867 31205874 ) + (minusΔ1198911) + 120576= radic22 + (1 + radic2) 120576 + (minusΔ1198911)

(73)

By (71) we obtain that 120590(1198911) sup 119892(1198832) + 119870(minusΔ1198911) and as119892(1199100) = 1 we infer that 120590(1198911) sup 1 + 119870(minusΔ1198911) Hence radic22 +(minusΔ1198911) le 119888119860(1198911 1205874) so thatradic22 + (minusΔ1198911) minus 120576 le 119888119860 (1198911 + 1198910 1205874 ) (74)

as 1198910infin le 120576 Since 119879(1198911 +1198910) = 119892 +119879(1198910) we obtain by (56)and (73) that

1 minus (4 + radic2) 120576 minus (minusΔ1198911)le 119888119861 (119879 (1198911 + 1198910) 31205874 ) minus 119888119860 (1198911 + 1198910 31205874 )

(75)

We also obtain by (62) and (74) that

119888119861 (119879 (1198911 + 1198910) 1205874 ) minus 119888119860 (1198911 + 1198910 1205874 )le radic22 + 4120576 minus (minusΔ1198911)

(76)

Since 119888119861(119879(119891) 120593)minus119888119860(119891 120593) is invariant for any 120593 by (37) from(75) and (76) we deduce that 120576 ge (2 minusradic2)2(8 +radic2) and thisis impossible since 120576 le 1100

8 Journal of Function Spaces

Next we show a proof of the second assertion (2) Let119891 isin 119860 We prove that there exists a sequence 119891119899 sub 119860 whichuniformly converges to 119891 such that 120588(120590(119891119899)) 997888rarr 0 as 119899 997888rarrinfin Without loss of generality we may assume that 119891infin = 1Then there exists 1199100 isin Ch(119860) such that |119891(1199100)| = 119891infinby [29 Proposition 63] We may assume that 119891(1199100) = 1Suppose that 119899 ge 4 Put

119880119899 = 119909 isin 119883 1003816100381610038161003816119891 (119909) minus 11003816100381610038161003816 le 11198992 (77)

and

Ω = co(119870 (1) cup 1 + 1119899) + 119870( 11198992) (78)

(In the following we identify R2 and C that is we identify(119909 119910) and 119909 + 119894119910 for every 119909 119910 isin R) Since we assume that119899 ge 4 we infer by a simple calculation that

[1 minus 11198992 minus 120575119899119899 1 + 1119899 + 120575119899119899 ] times [minus 11198992 minus 120575119899119899 11198992 + 120575119899119899 ]sub Ω

(79)

for 120575119899 with 0 lt 120575119899 le 11198992 We assume that 0 lt 120575119899 le 11198992 By[5 Lemma 2] there exists ℎ119899 isin 119860 such that ℎ119899infin le 1 + 2120575119899ℎ119899(1199100) = 1 |ℎ119899 + 1| le 2120575119899 on119883 119880119899 and |Im ℎ119899| le 2120575119899 on119883Put 119892119899 = (ℎ119899 + 1)2 Then 119892119899 isin 119860 and

10038171003817100381710038171198921198991003817100381710038171003817infin le 1 + 120575119899119892119899 (1199100) = 110038161003816100381610038161198921198991003816100381610038161003816 le 120575119899 on 119883 1198801198991003816100381610038161003816Im1198921198991003816100381610038161003816 le 120575119899 on 119883

minus120575119899 le Re119892119899 le 1 + 120575119899 on 119883

(80)

Let 119909 isin 119880119899 Then we have

1 minus 11198992 le Re119891 (119909) le 1minus 11198992 le Im119891 (119909) le 11198992

(81)

and

minus120575119899119899 le Re119892119899 (119909)119899 le 1119899 + 120575119899119899

minus120575119899119899 le Im119892119899 (119909)119899 le 120575119899119899

(82)

Hence

1 minus 11198992 minus 120575119899119899 le Re(119891 (119909) + 119892119899 (119909)119899 ) le 1 + 1119899 + 120575119899119899 minus 11198992 minus 120575119899119899 le Im(119891 (119909) + 119892119899 (119909)119899 ) le 11198992 + 120575119899119899

(83)

It follows that we have

(119891 + 119892119899119899 ) (119909) isin [1 minus 11198992 minus 120575119899119899 1 + 1119899 + 120575119899119899 ]times [minus 11198992 minus 120575119899119899 11198992 + 120575119899119899 ] sub Ω

(84)

for 119909 isin 119880119899 Suppose that 119909 isin 119883 119880119899 Then10038161003816100381610038161003816100381610038161003816119891 (119909) +

119892119899 (119909)11989910038161003816100381610038161003816100381610038161003816 le 1 +

120575119899119899 le 1 + 11198993 lt 1 + 11198992 (85)

and hence

119891 (119909) + 119892119899 (119909)119899 isin 119870 (1) + 119870( 11198992) sub Ω (86)

for 119909 isin 119883 119880119899 Since 1 + 1119899 = 119891(1199100) + 119892119899(1199100)119899 we have bycombining (84) and (86) that

1 + 1119899 isin (119891 + 119892119899119899 ) (119883) sub Ω (87)

AsΩ is convex we obtain

1 + 1119899 isin 120590 (119891 + 119892119899119899 ) sub Ω (88)

Recall that for 119903 ge 0 and a complex number 1199110119870 (1199110 119903) = 119908 isin C 1003816100381610038161003816119908 minus 11991101003816100381610038161003816 le 119903 (89)

denotes the closed disk with center 1199110 and radius 119903 Weobserve that 120588(Ω 1 + 1119899) = 1119899 + 11198992 Recall that

120588(Ω 1 + 1119899)= sup119903 119911 isin Ω 1 + 1119899 isin 119870 (119911 119903) sub Ω

(90)

Let ℓ1 be the line defined by the equation

119910 = minus119899radic2119899 + 1 (119909 minus (1 +1119899 + 11198992 + 11198993)) (91)

part of which is a part of the boundary ofΩ Let ℓ2 be the linedefined by the equation

119910 = minus119899radic2119899 + 1 (119909 minus (1 minus11198992)) (92)

By some calculation we have that the distance between 1 minus11198992 and 1+1119899 is 1119899+11198992 and it coincides with the distancebetween the point 1 minus 11198992 and the line ℓ1 Hence we see that

1 + 1119899 isin 119870(1 minus 11198992 1119899 + 11198992 ) sub Ω (93)

Thus 1119899 + 11198992 le 120588(Ω 1 + 1119899)Next we prove that 119870(119901 119903) cap Ω119888 = 0 for every 119901 isin Ω

with 119903 = |119901 minus (1 + 1119899)| gt 1119899 + 11198992 It will follow that1119899 + 11198992 ge 120588(Ω 1 + 1119899) and the equality will hold Let

Journal of Function Spaces 9

119901 = (119909119901 119910119901) = 119909119901 + 119894119910119901 isin Ω with 119903 = |119901 minus (1 + 1119899)| gt1119899 + 11198992 We prove the case where 119910119901 ge 0 A proof for thecase where 119910119901 le 0 is the same and we omit it We divideΩ+ =119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902 isin Ω 119910119902 ge 0 into two parts

Ω1 = 119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902isin Ω+ minus119899radic2119899 + 1 (119909119902 minus (1 minus

11198992)) le 119910119902le minus119899radic2119899 + 1 (119909 minus (1 +

1119899 + 11198992 + 11198993))(94)

and

Ω2 = 119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902 isin Ω+ 119910119902le minus119899radic2119899 + 1 (119909119902 minus (1 minus

11198992)) (95)

Suppose that 119901 isin Ω1 and 119903 = |119901minus (1+1119899)| gt 1119899+11198992Thedistance between ℓ1 and ℓ2 is 1119899+11198992 Hence119870(119901 119903)capΩ119888 =0 Suppose that 119901 isin Ω2 and 119903 = |119901 minus (1 + 1119899)| gt 1119899 + 11198992that is119901 = 1minus11198992 Let ℓ1015840 be the line passing through119901whichis parallel to ℓ2 Let 1199011015840 be the unique point in the intersectionof ℓ1015840 and the 119909-axis Then 1199011015840 = 1 minus 11198992 minus 119906 for some 119906 ge 0Then the distance between ℓ1 and ℓ1015840 is 1119899+11198992+119906(1+1119899)which is equal to the distance between the point119901 and the lineℓ1 On the other hand1003816100381610038161003816100381610038161003816119901 minus (1 + 1119899)

1003816100381610038161003816100381610038161003816 ge10038161003816100381610038161003816100381610038161199011015840 minus (1 + 1119899)

1003816100381610038161003816100381610038161003816 = 1119899 + 11198992 + 119906 (96)

It follows that 119870(119901 119903) cap Ω119888 = 0 We conclude that if 119901 isin Ωsatisfies 119903 = |119901minus(1+1119899)| gt 1119899+11198992 then119870(119901 119903)capΩ119888 = 0Thus we have

120588 (Ω 1 + 1119899) = 1119899 + 11198992 (97)

Since 1 + 1119899 isin 120590(119891 + 119892119899119899) sub Ω we have120588 (120590 (119891 + 119892119899119899 )) le 120588 (120590 (119891 + 119892119899119899 ) 1 + 1119899)

le 120588 (Ω 1 + 1119899) = 1119899 + 11198992 (98)

On the other hand 119892119899119899infin 997888rarr 0 as 119899 997888rarr infin ensures that119891 + 119892119899119899 minus 119891infin 997888rarr 0 as 119899 997888rarr infin It follows that for every120576 gt 0A120576 is dense in (119860 sdot infin)Finally we show a proof of the third assertion (3) As is

pointed out in the proof of [5Theorem] 120588(co(119870)+119870(119888)) ge 119888for any 119870 sub C and any 119888 ge 0 Let 120576 gt 0 and 119891 isin A120576 Supposethat Δ119891 le 0 Then by (43) we have (119891) = (119879119891) + 119870(minusΔ119891)Hence we have

120588 ( (119891)) = 120588 (120590 (119879119891) + 119870 (minusΔ119891)) ge minusΔ119891 (99)

As 120576 ge 120588(120590(119891)) we conclude by (45) that10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin = Δ119891 ge minus120576 (100)

This completes the proof of the theorem

4 Hermitian Operators on a BanachAlgebras of Continuous Maps WhoseValues Are in a Uniform Algebras

Let 119883 and 119884 be compact Hausdorff spaces Let 119861 be aunital subalgebra of 119862(119883) which separates the points of 119883Throughout this sectionwe assume119861 is a Banach algebrawiththe norm sdot 119861 and 119860 is a uniform algebra on 119884 Recall thata uniform algebra on 119884 is a uniformly closed subalgebra of119862(119884) which contains constants and separates the points of 119884For functions 119891 isin 119862(119883) and 119892 isin 119862(119884) let 119891 otimes 119892 isin 119862(119883 times 119884)be the function defined by119891otimes119892(119909 119910) = 119891(119909)119892(119910) for (119909 119910) isin119883 times 119884 and for a subspace 119864119883 of 119862(119883) and a subspace 119864119884 of119862(119884) put

119864119883 otimes 119864119884

= 119899sum

119895=1

119891119895 otimes 119892119895 119899 isin N 119891119895 isin 119864119883 119892119895 isin 119864119884

(101)

and

1 otimes 119864119884 = 1 otimes 119892 119892 isin 119864119884 (102)

Throughout the section 119861 is a unital subalgebra of 119862(119883 times 119884)with a Banach algebra norm sdot 119861 We assume that 119861otimes119860 sub 119861Note that 119861 separates the points of119883times119884 since119860 separates thepoints of 119884 and 119861 separates the points of 119883 We assume thatthere exists a compact Hausdorff space M and a complex-linear map 119863 119861 997888rarr 119862(M) such that ker119863 = 1 otimes 119860 Weassume that 119865119861 = 119865infin(119883times119884)+119863(119865)infin(M) for every 119865 isin 119861Hence119863 is continuous Defining

lsaquo119865lsaquo = 119863 (119865)infin(M) 119865 isin 119861 (103)

lsaquo sdot lsaquo is a one-invariant seminorm in the sense of Jarosz lsaquo sdot lsaquois a seminorm on 119861 such that lsaquo119865+ 1lsaquo = lsaquo119865lsaquo for every 119865 isin 119861Hence the norm sdot 119861 is a natural norm (see [5 p67]) Notethat 119861 is a regular subspace of 119862(119883times119884) in the sense of Jarosz[5 Proposition 2]

Lumerrsquos seminal paper [35] opened up a useful methodof finding isometries which is often referred to as Lumerrsquosmethod It involves the notion ofHermitian operators and thefact that 119880119867119880minus1 must be Hermitian if119867 is Hermitian and 119880is a surjective isometry

Definition 6 Let A be a unital Banach algebra We say that119890 isin A is a Hermitian element if1003817100381710038171003817exp (119894119905119890)1003817100381710038171003817A = 1 (104)

for every 119905 isin R The set of all Hermitian elements of A isdenoted by119867(A)

If A is a unital 119862lowast-algebra then 119867(A) is the set of allself-adjoint elements of A Hence 119867(119872119899(C)) is the set of allHermitian matrices and119867(119862(119884)) = 119862R(119884)Definition 7 Let 119864 be a complex Banach space The Banachalgebra of all bounded operators on 119864 is denoted by 119861(119864) Wesay that 119879 isin 119861(119864) is a Hermitian operator if 119879 isin 119867(119861(119864))

10 Journal of Function Spaces

Note that a Hermitian element of a unital Banach algebraand a Hermitian operator are usually defined in terms ofnumerical range or semi-inner product Here we define themby an equivalent form (see [36]) By the definition of aHermitian operator we have the following

Proposition 8 Let 119864119895 be a complex Banach space for 119895 = 1 2Suppose that 119881 1198641 997888rarr 1198642 is a surjective isometry and 119867 1198641 997888rarr 1198641 is a Hermitian operator Then 119881119867119881minus1 1198642 997888rarr 1198642

is a Hermitian operator

Proposition 9 An element 119865 isin 119861 is Hermitian if and only ifthere exists 119891 isin 119860 cap 119862R(119884) such that 119865 = 1 otimes 119891Proof Suppose that 119865 isin 119861 is a Hermitian element Then1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884) + 1003817100381710038171003817119863 (exp (119894119905119865))1003817100381710038171003817infin(M)

= 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817 = 1 (105)

for every 119905 isin R Suppose that there exists a point (119909 119910) isin119883 times 119884 with Im119865(119909 119910) = 0 where Im denotes the imaginarypart of a complex number Suppose that Im119865(119909 119910) gt 0Then1003817100381710038171003817exp (minus119894119865)1003817100381710038171003817infin(119883times119884) ge 1003816100381610038161003816exp (minus119894119865 (119909 119910))1003816100381610038161003816

= exp (Im119865 (119909 119910)) gt 1 (106)

Suppose that Im119865(119909 119910) lt 0 Then1003817100381710038171003817exp (119894119865)1003817100381710038171003817infin(119883times119884) ge 1003816100381610038161003816exp (119894119865 (119909 119910))1003816100381610038161003816= exp (minusIm119865 (119909 119910)) gt 1 (107)

In any case we have there exists 119905 isin R such that1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817119861 ge 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884) gt 1 (108)

which contradicts our assumption We have that

119865 isin 119862R (119883 times 119884) (109)

Thus for every (119904 119905) isin 119883 times 119884 and 119905 isin R |exp(119894119905119865(119904 119905))| = 1Hence exp(119894119905119865)infin(119883times119884) = 1 for every 119905 isin R By (105)we have119863(exp(119894119905119865)infin(M) = 0 which ensures that 119863(exp(119894119905119865)) = 0for every 119905 isin R Thus exp(119894119905119865) isin 1 otimes 119860 for every 119905 isin R Wehave

exp (119894119905119865) minus 1119905 minus 119894119865 = infinsum119899=2

((119894119865)119899 119905119899minus2119899 ) 119905 (110)

and hence for every 119905 isin R with |119905| le 1 we have10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817 le (

infinsum119899=2

119865119899119861|119905|119899minus2119899 ) |119905|

le (infinsum119899=2

119865119899119861119899 ) |119905| le (exp 119865119861) |119905|

(111)

It follows that10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817infin(119883times119884)

le 10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817119861

997888rarr 0(112)

as 119905 997888rarr 0 Since (exp(119894119905119865)minus1)119905 isin 1otimes119860 for each 119905 isin R thereexists 119892119905 isin 119860 such that

exp (119894119905119865) minus 1119905 = 1 otimes 119892119905 (113)

By (112) we have10038171003817100381710038171 otimes 1198921119899 minus 1198941198651003817100381710038171003817infin(119883times119884)

997888rarr 0 (114)

as 119899 997888rarr infin We have that 1 otimes 1198921119899 is a Cauchy sequencein 119862(119883 times 119884) thus we infer that1198921119899 is a Cauchy sequence in119862(119884) Since 119860 is uniformly closed as it is a uniform algebrathere exists 119892 isin 119860 such that

10038171003817100381710038171198921119899 minus 1198921003817100381710038171003817infin(119884)997888rarr 0 (115)

and hence10038171003817100381710038171 otimes 1198921119899 minus 1 otimes 1198921003817100381710038171003817infin(119883times119884)

997888rarr 0 (116)

as 119899 997888rarr infin It follows by (114) that 119894119865 = 1 otimes 119892 thus119865 = 1 otimes (minus119894119892) isin 1 otimes 119860 (117)

By (109) we see that minus119894119892 isin 119862R(119884) thus we have 119891 = minus119894119892 isin119860 cap 119862R(119884) and 119865 = 1 otimes 119891Suppose conversely that 119861 ni 119865 = 1otimes119891 for119891 isin 119860cap119862R(119884)

We infer that 119865 isin 119862R(119883 times 119884) and |exp(119894119905(119865(119909 119910))| = 1 forevery 119905 isin R and (119909 119910) isin 119883 times 119884 Hence exp(119894119905119865)infin(119883times119884) = 1for every 119905 isin R Since

exp (119894119905119865) = exp (119894119905 (1 otimes 119891)) = 1 otimes exp (119894119905119891) isin 1 otimes 119860 (118)

we have 119863(exp(119894119905119865)) = 0 It follows that1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817 = 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884)

+ 1003817100381710038171003817119863 (exp (119894119905119865))1003817100381710038171003817infin(M) = 1 (119)

for every 119905 isin R We conclude that 119865 is a Hermitian elementin 119861

Note that 119891 isin 119860 is Hermitian if and only if119891 isin 119860cap119862R(119884)by [37 Proposition 5] Hence Proposition 9 asserts that 119865 is aHermitian element in119861 if and only if119865 = 1otimes119891 for aHermitianelement 119891 in 119860Proposition 10 Suppose that 119880 119861 997888rarr 119861 is a surjectiveunital isometry Then 119880 is an algebra isomorphism

Proof As we have already mentioned 119861 is a regular subspace(in the sense of Jarosz) with a natural norm Then byTheorem 1 119880 is also an isometry with respect to the supre-mum norm on 119883 times 119884 Then 119880 is uniquely extended to asurjective isometry with respect to the supremum norm from the uniform closure 119861 onto itself Since 119861 is a uniformalgebra a theorem of Nagasawa [32] asserts that is analgebra isomorphism since (1) = 1 Thus 119880 is an algebraisomorphism from 119861 onto itself

Journal of Function Spaces 11

Theorem 11 A bounded operator 119879 119861 997888rarr 119861 is a Hermitianoperator if and only if 119879(1) is a Hermitian element in 119861 and119879 = 119872119879(1) the multiplication operator by 119879(1)Proof By Proposition 10 every surjective unital isometry on119861 is multiplicative Then by [37 Theorem 4] we have theconclusion

5 Banach Algebras of 119862(119884)-Valued Maps

Suppose that119883 is a compact Hausdorff space Suppose that 119861is a unital point separating subalgebra of119862(119884) equipped witha Banach algebra norm Then 119861 is semisimple because 119891 isin119861 119891(119909) = 0 is a maximal ideal of 119861 for every 119909 isin 119883 and theJacobson radical of 119861 vanishes The inequality 119891infin le 119891119861for every 119891 isin 119861 is well known We say that 119861 is natural if themap 119890 119884 997888rarr 119872119861 defined by 119910 997891997888rarr 120601119910 where 120601119910(119891) = 119891(119910)for every 119891 isin 119861 is bijective We say that 119861 is self-adjoint if 119861 isnatural and conjugate-closed in the sense that 119891 isin 119861 impliesthat 119891 isin 119861 for every 119891 isin 119861 where sdot denotes the complexconjugation on 119884Definition 12 Let 119883 and 119884 be compact Hausdorff spacesSuppose that 119861 is a unital point separating subalgebra of119862(119883)equipped with a Banach algebra norm sdot 119861 Suppose that119861 is self-adjoint Suppose that 119861 is a unital point separatingsubalgebra of 119862(119883 times 119884) such that 119861 otimes 119862(119884) sub 119861 equippedwith a Banach algebra norm sdot 119861 Suppose that 119861 is self-adjoint We say that 119861 is a natural 119862(119884)-valuezation of 119861 ifthere exists a compact Hausdorff space M and a complex-linear map 119863 119861 997888rarr 119862(M) such that ker119863 = 1 otimes 119862(119884) and119863(119862R(119883 times 119884) cap 119861) sub 119862R(M) which satisfies

119865 = 119865infin(119883times119884) + 119863 (119865)infin(M) 119865 isin 119861 (120)

The term ldquoa natural 119862(119884)-valuezation of 119861rdquo comes fromthe natural norm defined by Jarosz [5] In fact the norm sdot 119861is a natural norm in the sense of Jarosz [5]

Note that (119883119862(119884) 119861 119861) need not be an admissiblequadruple defined by Nikou and OrsquoFarrell [38] (cf [31]) sincewe do not assume that 119865(sdot 119910) 119865 isin 119861 119910 isin 119884 sub 119861 whichis a requirement for the admissible quadruple On the otherhand if (119883119862(119884) 119861 119861) is an admissible quadruple of type Ldefined in [30] then 119861 is a natural 119862(119884)-valuezation of 119861 dueto Definition 12

Example 13 Let 119861 = 1198621([0 1]) and 119861 = Lip([0 1] 119862(119884))for 119884 = 119901 a singleton Then Lip([0 1]) is algebraically iso-morphic to Lip([0 1] 119862(119884)) Suppose thatM is the maximalideal space of 119871infin([0 1]) and 119863 119861 997888rarr 119862(M) is definedby 119891 997891997888rarr Γ(1198911015840) where Γ denotes the Gelfand transformin 119871infin([0 1]) Then 119861 is a natural 119862(119884)-valuezation of 119861The Banach algebra Lip([0 1]) with the norm 119891infin([01]) +1198911015840infin([01]) is isometrically isomorphic to 119861

Let 119884 be a compact Hausdorff space Note that a closedsubalgebra 119861 of Lip((119870 119889120572) 119862(119884))which appears in Example12 in [30] is an example of a natural 119862(119884)-valuezation of 119861

The Banach algebras1198621([0 1] 119862(119884)) and 1198621(T 119862(119884)))whichappear in Examples 16 and 17 in [30] respectively are alsoexamples of natural 119862(119884)-valuezations of 1198621([0 1])6 Isometries on Natural 119862(119884)-ValuezationsThemain theorem in this paper is the following

Theorem 14 Suppose that 119861119895 is a natural 119862(119884119895)-valuezationof 119861119895 sub 119862(119883119895) for 119895 = 1 2 We assume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (121)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for119895 = 1 2 Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linear isometry Then there exists ℎ isin 119862(1198842) such that |ℎ| = 1on 1198842 a continuous map 120593 1198832 times 1198842 997888rarr 1198831 such that120593(sdot 119910) 1198832 997888rarr 1198831 is a homeomorphism for each 119910 isin 1198842and a homeomorphism 120591 1198842 997888rarr 1198841 which satisfies119880 (119865) (119909 119910) = ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842 (122)

for every 119865 isin 1198611

In short a surjective isometry between 119862(119884)-valuezationsis a weighted composition operator of a specific form thehomeomorphism 1198832 times 1198842 997888rarr 1198831 times 1198841 (119909 119910) 997891997888rarr(120593(119909 119910) 120591(119910)) has the second coordinate that depends onlyon the second variable 119910 isin 1198842 A composition operatorinduced by such a homeomorphism is said to be of type BJin [31 37] after the study of Botelho and Jamison [39]

Quite recently the author of this paper and Oi [30 Theo-rem 8] proved a similar result of Theorem 14 for admissiblequadruples of type L To prove it we apply Proposition 32and the following comments in [31] Instead of this we proveTheorem 14 by Lumerrsquosmethod with which a proof is simplerthan that in [30]

In the following in this section we assume that 119861119895 is anatural 119862(119884119895)-valuezation of 119861 sub 119862(119883119895) for 119895 = 1 2 Weassume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (123)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for 119895 = 1 2Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linearisometry A crucial part of a proof of Theorem 14 is to proveProposition 15

Proposition 15 Suppose that 1198832 is not a singleton Thereexists ℎ isin 119862(1198842) with |ℎ| = 1 on 1198842 such that 119880(11198611) = 11198612 otimesℎ

A similar result for admissible quadruples of type L isproved in [30 Proposition 9] If we assumed that

119865 (sdot 119910) 119865 isin 119861119895 119910 isin 119884119895 sub 119861119895 (124)

then 119861119895 were an admissible quadruple of type L Although119861119895 in this paper need not be an admissible quadruple of type

12 Journal of Function Spaces

L a proof of Proposition 15 is completely the same as that in[30 Proposition 9] since we do not make use of the condition(124) in the proof of [30 Proposition 9] The condition (124)is needed in [30] when we apply Proposition 32 and thefollowing comments in [31]

7 Proof of Theorem 14 An Application ofLumerrsquos Method

Proof of Theorem 14 A proof for the case where 1198831 = 1199091and 1198832 = 1199092 are singletons is the same as the proof ofTheorem 8 in [30]

Suppose that1198832 is not a singleton By Proposition 15 thereexists ℎ isin 119862R(1198842) with |ℎ| = 1 on 1198842 such that 119880(1) =1 otimes ℎ Letting 1198800 1198611 997888rarr 1198612 by 1198800(119865) = (1 otimes ℎ)119880(119865)119865 isin 1198611 we see by the hypothesis (1 otimes ℎ)1198651198612 = 1198651198612 forevery 119865 isin 1198612 that 1198800 is a surjective unital isometry from1198611 onto 1198612 Then Corollary 2 asserts that 1198800 is an algebraisomorphism Let 119891 isin 119862R(1198841) By Proposition 9 1 otimes 119891 isa Hermitian element in 1198611 Then by Theorem 11 1198721otimes119891 is aHermitian operator on 1198611 By Proposition 8 11988001198721otimes119891119880minus1

0 is aHermitian operator on 1198612 Then by Theorem 11 there exists119878(119891) isin 119862R(1198842) such that 11988001198721otimes119891119880minus1

0 = 1198721otimes119878(119891) Hence anoperator 119878 119862R(1198841) 997888rarr 119862R(1198842) is defined Since 1198800 is analgebra isomorphism it is easy to see that 119878 is a real algebraisomorphism from 119862R(1198841) onto 119862R(1198842) Then 119878 119862(1198841) 997888rarr119862(1198842) defined by 119878(119891) = 119878(Re119891) + 119894119878(Im119891) for 119891 isin 119862(1198841)gives a complex algebra isomorphism Gelfand theory assertsthat there is a homeomorphism 120591 1198842 997888rarr 1198841 such that119878(119891) = 119891 ∘ 120591 119891 isin 119862(1198841) It follows that

11988001198721otimes119891119880minus10 = 1198721otimes119891∘120591 119891 isin 119862 (1198841) (125)

Since 119880minus10 (1) = 1 we have

1198800 (1 otimes 119891) = 1 otimes 119891 ∘ 120591 119891 isin 119862 (1198841) (126)

Define Φ 1198611 997888rarr 1198612 by Φ(119886) = 1198800(119886 otimes 1) 119886 isin 1198611Since 1198800 is an algebra isomorphism the map Φ is a unitalhomomorphism Since the maximal ideal space of 1198611 is 1198831

and the maximal ideal space of 1198612 is 1198832 times 1198842 there is acontinuous map 120593 1198832 times 1198842 997888rarr 1198831 such that

Φ(119886) (119909 119910) = 119886 (120593 (119909 119910)) 119886 isin 1198611 (119909 119910) isin 1198832 times 1198842 (127)

It follows by (126) and (127) that

1198800 (119886 otimes 119891) (119909 119910) = 1198800 ((119886 otimes 1) (1 otimes 119891)) (119909 119910)= 1198800 (119886 otimes 1) (119909 119910)1198800 (1 otimes 119891) (119909 119910)= 119886 (120593 (119909 119910))119891 (120591 (119910))= (119886 otimes 119891) (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198832 times 1198842

(128)

for every 119886 isin 1198611 and 119891 isin 119862(1198841) Thus

1198800 (sum(119886119895 otimes 119891119895)) (119909 119910)= (sum(119886119895 otimes 119891119895)) (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842(129)

for every sum(119886119895 otimes 119891119895) isin 1198611 otimes 119862(1198841) By the Stone-Weierstrasstheorem 1198611 otimes 119862(1198841) is uniformly dense in 119862(1198831 times 1198841) henceany element in 1198611 is uniformly approximated by 1198611 otimes 119862(1198841)As 1198800 is also an isometry with respect to the uniform normwe see that

1198800 (119865) (119909 119910) = 119865 (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198831 times 1198841 (130)

for every 119865 isin 1198611 and

119880 (119865) (119909 119910) = (1 otimes ℎ) (119909 119910) 119865 (120593 (119909 119910) 120591 (119910))= ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198831 times 1198841(131)

As 1198800 is an algebra isomorphism the map 1198832 times 1198842 997888rarr1198831 times 1198841 defined by (119909 119910) 997891997888rarr (120593(119909 119910) 120591(119910)) gives ahomeomorphism Therefore for every 119910 isin 1198842 the map

120593 (sdot 119910) 1198832 997888rarr 1198831 (132)

is a homeomorphismSuppose that 1198831 is not a singleton By the same way as

in the last part of the proof of Theorem 8 in [30] we havethat 1198832 is not a singleton Then we have the conclusion bythe previous argument

8 Application of Theorem 14

We exhibit applications of Theorem 14

Corollary 16 ([4 Theorem 33]) Suppose that119880 Lip([0 1]) 997888rarr Lip([0 1]) is a surjective isometrywith respect to the norm defined by 119891infin([01]) + 1198911015840infin([01])

for 119891 isin Lip([0 1]) Then 119880(1) is a constant function of unitmodulus such that

119880(119891) (119909) = 119880 (1) 119891 (119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (133)

or

119880(119891) (119909) = 119880 (1) 119891 (1 minus 119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (134)

The converse statement also holds

Proof By Example 13 we may suppose that Lip([0 1]) is aBanach algebra of 119862(119884)-valuezation Applying Theorem 14we have that119880(1) = 1otimesℎ for ℎ isin 119862(119884)with |ℎ| = 1 Since our

Journal of Function Spaces 13

119884 is a singleton 119880(1) is a constant function of unit modulusWe also see that the corresponding continuous map 120593 [0 1] times 119884 997888rarr [0 1] can be considered as a homeomorphismfrom [0 1] onto [0 1] therefore we have that119880(119891) (119909) = 119880 (1) 119891 (120593 (119909))

119891 isin Lip ([0 1]) 119909 isin [0 1] (135)

The rest is a routine argument to prove that 120593 is an isometryhence 120593(119909) = 119909 119909 isin [0 1] or 120593(119909) = 1 minus 119909 119909 isin [0 1]

The converse statement is trivial

Corollaries 14 15 18 and 19 in [30 Section 6] follow herewith a similar proof

Data Availability

The data used to support the findings of this study areavailable from the corresponding author upon request

Conflicts of Interest

The author declares that they have no conflicts of interest

Acknowledgments

This work was supported by JSPS KAKENHI Grants Num-bers JP16K05172 and JP15K04921

References

[1] K de Leeuw ldquoBanach spaces of Lipschitz functionsrdquo StudiaMathematica vol 21 pp 55ndash66 19611962

[2] A K Roy ldquoExtreme points and linear isometries of the Banachspace of Lipschitz functionsrdquoCanadian Journal of MathematicsJournal Canadien de Mathematiques vol 20 pp 1150ndash11641968

[3] M Cambern ldquoIsometries of certain Banach algebrasrdquo StudiaMathematica vol 25 pp 217ndash225 19641965

[4] N V Rao and A K Roy ldquoLinear isometries of some functionspacesrdquo Pacific Journal of Mathematics vol 38 pp 177ndash192 1971

[5] K Jarosz ldquoIsometries in semisimple commutative BanachalgebrasrdquoProceedings of the AmericanMathematical Society vol94 no 1 pp 65ndash71 1985

[6] K Jarosz and V D Pathak ldquoIsometries between functionspacesrdquo Transactions of the AmericanMathematical Society vol305 no 1 pp 193ndash206 1988

[7] NWeaver ldquoIsometries of noncompact Lipschitz spacesrdquoCana-dian Mathematical Bulletin Bulletin Canadien de Mathemat-iques vol 38 no 2 pp 242ndash249 1995

[8] A Jimenez-Vargas and M Villegas-Vallecillos ldquoLinear isome-tries between spaces of vector-valued Lipschitz functionsrdquoProceedings of the American Mathematical Society vol 137 no4 pp 1381ndash1388 2009

[9] A Jimenez-Vargas and M Villegas-Vallecillos ldquoInto linearisometries between spaces of Lipschitz functionsrdquo HoustonJournal of Mathematics vol 34 no 4 pp 1165ndash1184 2008

[10] F Botelho and J Jamison ldquoSurjective isometries on spacesof differentiable vector-valued functionsrdquo Studia Mathematicavol 192 no 1 pp 39ndash50 2009

[11] E Mayer-Wolf ldquoIsometries between Banach spaces of Lipschitzfunctionsrdquo Israel Journal of Mathematics vol 38 no 1-2 pp 58ndash74 1981

[12] A Jimenez-Vargas M Villegas-Vallecillos and Y-S WangldquoBanach-Stone theorems for vector-valued little Lipschitz func-tionsrdquo Publicationes Mathematicae vol 74 no 1-2 pp 81ndash1002009

[13] J Araujo andLDubarbie ldquoNoncompactness andnoncomplete-ness in isometries of Lipschitz spacesrdquo Journal of MathematicalAnalysis and Applications vol 377 no 1 pp 15ndash29 2011

[14] F Botelho R J Fleming and J Jamison ldquoExtreme points andisometries on vector-valued Lipschitz spacesrdquo Journal of Math-ematical Analysis and Applications vol 381 no 2 pp 821ndash8322011

[15] H Koshimizu ldquoLinear isometries on spaces of continuouslydifferentiable and Lipschitz continuous functionsrdquo NihonkaiMathematical Journal vol 22 no 2 pp 73ndash90 2011

[16] F Botelho J Jamison and B Zheng ldquoIsometries on spaces ofvector valued Lipschitz functionsrdquo Positivity An InternationalMathematics Journal Devoted to Theory and Applications ofPositivity vol 17 no 1 pp 47ndash65 2013

[17] A Ranjbar-Motlagh ldquoA note on isometries of Lipschitz spacesrdquoJournal of Mathematical Analysis and Applications vol 411 no2 pp 555ndash558 2014

[18] F Botelho and J Jamison ldquoSurjective isometries on spaces ofvector valued continuous and Lipschitz functionsrdquo PositivityAn International Mathematics Journal Devoted to Theory andApplications of Positivity vol 17 no 3 pp 395ndash405 2013Erratum to Surjective isometries on spaces of vector valuedcontinuous and Lipschitz functions by F Botelho 20 (2016)757ndash759

[19] T Miura and H Takagi ldquoSurjective isometries on the Banachspace of continuously differentiable functionsrdquo ContemporaryMathematics vol 687 pp 181ndash192 2017

[20] K Kawamura ldquoBanach-Stone type theorems for 1198621-functionspaces over Riemannian manifoldsrdquo Acta Universitatis Szegedi-ensis Acta Scientiarum Mathematicarum vol 83 no 3-4 pp551ndash591 2017

[21] K Kawamura ldquoPerturbations of norms on 1198621-function spacesand associated isometry groupsrdquo Topology Proceedings vol 51pp 169ndash196 2018

[22] K Kawamura ldquoA Banach-Stone type theorem for 1198621-functionspaces over the circlerdquo Topology Proceedings vol 53 pp 15ndash262019

[23] L Li D Chen Q Meng and Y-S Wang ldquoSurjective isometrieson vector-valued differentiable function spacesrdquo Annals ofFunctional Analysis vol 9 no 3 pp 334ndash343 2018

[24] K Kawamura H Koshimizu and T Miura ldquoNorms onC1([01]) and there isometriesrdquo Acta Scientiarum Mathemati-carum vol 84 no 12 pp 239ndash261 2018

[25] L Li A M Peralta L Wang and Y-S Wang ldquoWeak-2-local isometries on uniform algebras and Lipschitz algebrasrdquohttpsarxivorgabs170503619

[26] A Jimenez-Vargas L LiAMPeralta LWang andY-SWangldquo2-local standard isometries on vector-valued Lipschitz func-tion spacesrdquo Journal of Mathematical Analysis and Applicationsvol 461 no 2 pp 1287ndash1298 2018

[27] A Ranjbar-Motlagh ldquoIsometries of Lipschitz type functionspacesrdquoMathematischeNachrichten vol 291 no 11-12 pp 1899ndash1907 2018

14 Journal of Function Spaces

[28] N Weaver Lipschitz Algebras World Scientific Publishing CoInc River Edge NJ USA 1999

[29] R R Phelps Lectures on Choquetrsquos theorem vol 1757 of LectureNotes in Mathematics Springer-Verlag Berlin Germany 2ndedition 2001

[30] O Hatori and S Oi ldquoIsometries on Banach algebras of vector-valued mapsrdquo Acta Scientiarum Mathematicarum vol 84 no12 pp 151ndash183 2018

[31] O Hatori S Oi and H Takagi ldquoPeculiar homomorphismsbetween algebras of vector-valued mapsrdquo Studia Mathematicavol 242 no 2 pp 141ndash163 2018

[32] M Nagasawa ldquoIsomorphisms between commutative Banachalgebras with an application to rings of analytic functionsrdquoKodai Mathematical Seminar Reports vol 11 pp 182ndash188 1959

[33] K de LeeuwW Rudin and J Wermer ldquoThe isometries of somefunction spacesrdquo Proceedings of the American MathematicalSociety vol 11 pp 694ndash698 1960

[34] OHatori A Jimenez-Vargas andMVillegas-Vallecillos ldquoMapswhich preserve norms of non-symmetrical quotients betweengroups of exponentials of Lipschitz functionsrdquo Journal of Math-ematical Analysis and Applications vol 415 no 2 pp 825ndash8452014

[35] G Lumer ldquoOn the isometries of reflexiveOrlicz spacesrdquoAnnalesde lrsquoInstitut Fourier vol 68 pp 99ndash109 1963

[36] R J Fleming and J E Jamison Isometries on Banach SpacesFunction Spaces vol 129 ofMonographs and Surveys in Pure andApplied Mathematics Chapman amp Hall CRC Boca Raton FlaUSA 2003

[37] OHatori and SOi ldquoHermitian operators onBanach algebras ofvector-valued Lipschitzmapsrdquo Journal ofMathematical Analysisand Applications vol 452 no 1 pp 378ndash387 2017 Corrigendumto rdquoHermitian operators on Banach algebras of vector-valuedLipschitz mapsrdquo Journal of Mathematical Analysis and Appli-cations 452 (2017) 378ndash387 MR3628025

[38] A Nikou andA G OrsquoFarrell ldquoBanach algebras of vector-valuedfunctionsrdquo Glasgow Mathematical Journal vol 56 no 2 pp419ndash426 2014

[39] F Botelho and J Jamison ldquoHomomorphisms on a class ofcommutative Banach algebrasrdquo Rocky Mountain Journal ofMathematics vol 43 no 2 pp 395ndash416 2013

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Page 8: Hermitian Operators and Isometries on Banach Algebras of ...downloads.hindawi.com/journals/jfs/2018/8085304.pdfJournalofFunctionSpaces normiseitherthemaxnormorthesumnorm.esituation

8 Journal of Function Spaces

Next we show a proof of the second assertion (2) Let119891 isin 119860 We prove that there exists a sequence 119891119899 sub 119860 whichuniformly converges to 119891 such that 120588(120590(119891119899)) 997888rarr 0 as 119899 997888rarrinfin Without loss of generality we may assume that 119891infin = 1Then there exists 1199100 isin Ch(119860) such that |119891(1199100)| = 119891infinby [29 Proposition 63] We may assume that 119891(1199100) = 1Suppose that 119899 ge 4 Put

119880119899 = 119909 isin 119883 1003816100381610038161003816119891 (119909) minus 11003816100381610038161003816 le 11198992 (77)

and

Ω = co(119870 (1) cup 1 + 1119899) + 119870( 11198992) (78)

(In the following we identify R2 and C that is we identify(119909 119910) and 119909 + 119894119910 for every 119909 119910 isin R) Since we assume that119899 ge 4 we infer by a simple calculation that

[1 minus 11198992 minus 120575119899119899 1 + 1119899 + 120575119899119899 ] times [minus 11198992 minus 120575119899119899 11198992 + 120575119899119899 ]sub Ω

(79)

for 120575119899 with 0 lt 120575119899 le 11198992 We assume that 0 lt 120575119899 le 11198992 By[5 Lemma 2] there exists ℎ119899 isin 119860 such that ℎ119899infin le 1 + 2120575119899ℎ119899(1199100) = 1 |ℎ119899 + 1| le 2120575119899 on119883 119880119899 and |Im ℎ119899| le 2120575119899 on119883Put 119892119899 = (ℎ119899 + 1)2 Then 119892119899 isin 119860 and

10038171003817100381710038171198921198991003817100381710038171003817infin le 1 + 120575119899119892119899 (1199100) = 110038161003816100381610038161198921198991003816100381610038161003816 le 120575119899 on 119883 1198801198991003816100381610038161003816Im1198921198991003816100381610038161003816 le 120575119899 on 119883

minus120575119899 le Re119892119899 le 1 + 120575119899 on 119883

(80)

Let 119909 isin 119880119899 Then we have

1 minus 11198992 le Re119891 (119909) le 1minus 11198992 le Im119891 (119909) le 11198992

(81)

and

minus120575119899119899 le Re119892119899 (119909)119899 le 1119899 + 120575119899119899

minus120575119899119899 le Im119892119899 (119909)119899 le 120575119899119899

(82)

Hence

1 minus 11198992 minus 120575119899119899 le Re(119891 (119909) + 119892119899 (119909)119899 ) le 1 + 1119899 + 120575119899119899 minus 11198992 minus 120575119899119899 le Im(119891 (119909) + 119892119899 (119909)119899 ) le 11198992 + 120575119899119899

(83)

It follows that we have

(119891 + 119892119899119899 ) (119909) isin [1 minus 11198992 minus 120575119899119899 1 + 1119899 + 120575119899119899 ]times [minus 11198992 minus 120575119899119899 11198992 + 120575119899119899 ] sub Ω

(84)

for 119909 isin 119880119899 Suppose that 119909 isin 119883 119880119899 Then10038161003816100381610038161003816100381610038161003816119891 (119909) +

119892119899 (119909)11989910038161003816100381610038161003816100381610038161003816 le 1 +

120575119899119899 le 1 + 11198993 lt 1 + 11198992 (85)

and hence

119891 (119909) + 119892119899 (119909)119899 isin 119870 (1) + 119870( 11198992) sub Ω (86)

for 119909 isin 119883 119880119899 Since 1 + 1119899 = 119891(1199100) + 119892119899(1199100)119899 we have bycombining (84) and (86) that

1 + 1119899 isin (119891 + 119892119899119899 ) (119883) sub Ω (87)

AsΩ is convex we obtain

1 + 1119899 isin 120590 (119891 + 119892119899119899 ) sub Ω (88)

Recall that for 119903 ge 0 and a complex number 1199110119870 (1199110 119903) = 119908 isin C 1003816100381610038161003816119908 minus 11991101003816100381610038161003816 le 119903 (89)

denotes the closed disk with center 1199110 and radius 119903 Weobserve that 120588(Ω 1 + 1119899) = 1119899 + 11198992 Recall that

120588(Ω 1 + 1119899)= sup119903 119911 isin Ω 1 + 1119899 isin 119870 (119911 119903) sub Ω

(90)

Let ℓ1 be the line defined by the equation

119910 = minus119899radic2119899 + 1 (119909 minus (1 +1119899 + 11198992 + 11198993)) (91)

part of which is a part of the boundary ofΩ Let ℓ2 be the linedefined by the equation

119910 = minus119899radic2119899 + 1 (119909 minus (1 minus11198992)) (92)

By some calculation we have that the distance between 1 minus11198992 and 1+1119899 is 1119899+11198992 and it coincides with the distancebetween the point 1 minus 11198992 and the line ℓ1 Hence we see that

1 + 1119899 isin 119870(1 minus 11198992 1119899 + 11198992 ) sub Ω (93)

Thus 1119899 + 11198992 le 120588(Ω 1 + 1119899)Next we prove that 119870(119901 119903) cap Ω119888 = 0 for every 119901 isin Ω

with 119903 = |119901 minus (1 + 1119899)| gt 1119899 + 11198992 It will follow that1119899 + 11198992 ge 120588(Ω 1 + 1119899) and the equality will hold Let

Journal of Function Spaces 9

119901 = (119909119901 119910119901) = 119909119901 + 119894119910119901 isin Ω with 119903 = |119901 minus (1 + 1119899)| gt1119899 + 11198992 We prove the case where 119910119901 ge 0 A proof for thecase where 119910119901 le 0 is the same and we omit it We divideΩ+ =119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902 isin Ω 119910119902 ge 0 into two parts

Ω1 = 119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902isin Ω+ minus119899radic2119899 + 1 (119909119902 minus (1 minus

11198992)) le 119910119902le minus119899radic2119899 + 1 (119909 minus (1 +

1119899 + 11198992 + 11198993))(94)

and

Ω2 = 119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902 isin Ω+ 119910119902le minus119899radic2119899 + 1 (119909119902 minus (1 minus

11198992)) (95)

Suppose that 119901 isin Ω1 and 119903 = |119901minus (1+1119899)| gt 1119899+11198992Thedistance between ℓ1 and ℓ2 is 1119899+11198992 Hence119870(119901 119903)capΩ119888 =0 Suppose that 119901 isin Ω2 and 119903 = |119901 minus (1 + 1119899)| gt 1119899 + 11198992that is119901 = 1minus11198992 Let ℓ1015840 be the line passing through119901whichis parallel to ℓ2 Let 1199011015840 be the unique point in the intersectionof ℓ1015840 and the 119909-axis Then 1199011015840 = 1 minus 11198992 minus 119906 for some 119906 ge 0Then the distance between ℓ1 and ℓ1015840 is 1119899+11198992+119906(1+1119899)which is equal to the distance between the point119901 and the lineℓ1 On the other hand1003816100381610038161003816100381610038161003816119901 minus (1 + 1119899)

1003816100381610038161003816100381610038161003816 ge10038161003816100381610038161003816100381610038161199011015840 minus (1 + 1119899)

1003816100381610038161003816100381610038161003816 = 1119899 + 11198992 + 119906 (96)

It follows that 119870(119901 119903) cap Ω119888 = 0 We conclude that if 119901 isin Ωsatisfies 119903 = |119901minus(1+1119899)| gt 1119899+11198992 then119870(119901 119903)capΩ119888 = 0Thus we have

120588 (Ω 1 + 1119899) = 1119899 + 11198992 (97)

Since 1 + 1119899 isin 120590(119891 + 119892119899119899) sub Ω we have120588 (120590 (119891 + 119892119899119899 )) le 120588 (120590 (119891 + 119892119899119899 ) 1 + 1119899)

le 120588 (Ω 1 + 1119899) = 1119899 + 11198992 (98)

On the other hand 119892119899119899infin 997888rarr 0 as 119899 997888rarr infin ensures that119891 + 119892119899119899 minus 119891infin 997888rarr 0 as 119899 997888rarr infin It follows that for every120576 gt 0A120576 is dense in (119860 sdot infin)Finally we show a proof of the third assertion (3) As is

pointed out in the proof of [5Theorem] 120588(co(119870)+119870(119888)) ge 119888for any 119870 sub C and any 119888 ge 0 Let 120576 gt 0 and 119891 isin A120576 Supposethat Δ119891 le 0 Then by (43) we have (119891) = (119879119891) + 119870(minusΔ119891)Hence we have

120588 ( (119891)) = 120588 (120590 (119879119891) + 119870 (minusΔ119891)) ge minusΔ119891 (99)

As 120576 ge 120588(120590(119891)) we conclude by (45) that10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin = Δ119891 ge minus120576 (100)

This completes the proof of the theorem

4 Hermitian Operators on a BanachAlgebras of Continuous Maps WhoseValues Are in a Uniform Algebras

Let 119883 and 119884 be compact Hausdorff spaces Let 119861 be aunital subalgebra of 119862(119883) which separates the points of 119883Throughout this sectionwe assume119861 is a Banach algebrawiththe norm sdot 119861 and 119860 is a uniform algebra on 119884 Recall thata uniform algebra on 119884 is a uniformly closed subalgebra of119862(119884) which contains constants and separates the points of 119884For functions 119891 isin 119862(119883) and 119892 isin 119862(119884) let 119891 otimes 119892 isin 119862(119883 times 119884)be the function defined by119891otimes119892(119909 119910) = 119891(119909)119892(119910) for (119909 119910) isin119883 times 119884 and for a subspace 119864119883 of 119862(119883) and a subspace 119864119884 of119862(119884) put

119864119883 otimes 119864119884

= 119899sum

119895=1

119891119895 otimes 119892119895 119899 isin N 119891119895 isin 119864119883 119892119895 isin 119864119884

(101)

and

1 otimes 119864119884 = 1 otimes 119892 119892 isin 119864119884 (102)

Throughout the section 119861 is a unital subalgebra of 119862(119883 times 119884)with a Banach algebra norm sdot 119861 We assume that 119861otimes119860 sub 119861Note that 119861 separates the points of119883times119884 since119860 separates thepoints of 119884 and 119861 separates the points of 119883 We assume thatthere exists a compact Hausdorff space M and a complex-linear map 119863 119861 997888rarr 119862(M) such that ker119863 = 1 otimes 119860 Weassume that 119865119861 = 119865infin(119883times119884)+119863(119865)infin(M) for every 119865 isin 119861Hence119863 is continuous Defining

lsaquo119865lsaquo = 119863 (119865)infin(M) 119865 isin 119861 (103)

lsaquo sdot lsaquo is a one-invariant seminorm in the sense of Jarosz lsaquo sdot lsaquois a seminorm on 119861 such that lsaquo119865+ 1lsaquo = lsaquo119865lsaquo for every 119865 isin 119861Hence the norm sdot 119861 is a natural norm (see [5 p67]) Notethat 119861 is a regular subspace of 119862(119883times119884) in the sense of Jarosz[5 Proposition 2]

Lumerrsquos seminal paper [35] opened up a useful methodof finding isometries which is often referred to as Lumerrsquosmethod It involves the notion ofHermitian operators and thefact that 119880119867119880minus1 must be Hermitian if119867 is Hermitian and 119880is a surjective isometry

Definition 6 Let A be a unital Banach algebra We say that119890 isin A is a Hermitian element if1003817100381710038171003817exp (119894119905119890)1003817100381710038171003817A = 1 (104)

for every 119905 isin R The set of all Hermitian elements of A isdenoted by119867(A)

If A is a unital 119862lowast-algebra then 119867(A) is the set of allself-adjoint elements of A Hence 119867(119872119899(C)) is the set of allHermitian matrices and119867(119862(119884)) = 119862R(119884)Definition 7 Let 119864 be a complex Banach space The Banachalgebra of all bounded operators on 119864 is denoted by 119861(119864) Wesay that 119879 isin 119861(119864) is a Hermitian operator if 119879 isin 119867(119861(119864))

10 Journal of Function Spaces

Note that a Hermitian element of a unital Banach algebraand a Hermitian operator are usually defined in terms ofnumerical range or semi-inner product Here we define themby an equivalent form (see [36]) By the definition of aHermitian operator we have the following

Proposition 8 Let 119864119895 be a complex Banach space for 119895 = 1 2Suppose that 119881 1198641 997888rarr 1198642 is a surjective isometry and 119867 1198641 997888rarr 1198641 is a Hermitian operator Then 119881119867119881minus1 1198642 997888rarr 1198642

is a Hermitian operator

Proposition 9 An element 119865 isin 119861 is Hermitian if and only ifthere exists 119891 isin 119860 cap 119862R(119884) such that 119865 = 1 otimes 119891Proof Suppose that 119865 isin 119861 is a Hermitian element Then1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884) + 1003817100381710038171003817119863 (exp (119894119905119865))1003817100381710038171003817infin(M)

= 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817 = 1 (105)

for every 119905 isin R Suppose that there exists a point (119909 119910) isin119883 times 119884 with Im119865(119909 119910) = 0 where Im denotes the imaginarypart of a complex number Suppose that Im119865(119909 119910) gt 0Then1003817100381710038171003817exp (minus119894119865)1003817100381710038171003817infin(119883times119884) ge 1003816100381610038161003816exp (minus119894119865 (119909 119910))1003816100381610038161003816

= exp (Im119865 (119909 119910)) gt 1 (106)

Suppose that Im119865(119909 119910) lt 0 Then1003817100381710038171003817exp (119894119865)1003817100381710038171003817infin(119883times119884) ge 1003816100381610038161003816exp (119894119865 (119909 119910))1003816100381610038161003816= exp (minusIm119865 (119909 119910)) gt 1 (107)

In any case we have there exists 119905 isin R such that1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817119861 ge 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884) gt 1 (108)

which contradicts our assumption We have that

119865 isin 119862R (119883 times 119884) (109)

Thus for every (119904 119905) isin 119883 times 119884 and 119905 isin R |exp(119894119905119865(119904 119905))| = 1Hence exp(119894119905119865)infin(119883times119884) = 1 for every 119905 isin R By (105)we have119863(exp(119894119905119865)infin(M) = 0 which ensures that 119863(exp(119894119905119865)) = 0for every 119905 isin R Thus exp(119894119905119865) isin 1 otimes 119860 for every 119905 isin R Wehave

exp (119894119905119865) minus 1119905 minus 119894119865 = infinsum119899=2

((119894119865)119899 119905119899minus2119899 ) 119905 (110)

and hence for every 119905 isin R with |119905| le 1 we have10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817 le (

infinsum119899=2

119865119899119861|119905|119899minus2119899 ) |119905|

le (infinsum119899=2

119865119899119861119899 ) |119905| le (exp 119865119861) |119905|

(111)

It follows that10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817infin(119883times119884)

le 10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817119861

997888rarr 0(112)

as 119905 997888rarr 0 Since (exp(119894119905119865)minus1)119905 isin 1otimes119860 for each 119905 isin R thereexists 119892119905 isin 119860 such that

exp (119894119905119865) minus 1119905 = 1 otimes 119892119905 (113)

By (112) we have10038171003817100381710038171 otimes 1198921119899 minus 1198941198651003817100381710038171003817infin(119883times119884)

997888rarr 0 (114)

as 119899 997888rarr infin We have that 1 otimes 1198921119899 is a Cauchy sequencein 119862(119883 times 119884) thus we infer that1198921119899 is a Cauchy sequence in119862(119884) Since 119860 is uniformly closed as it is a uniform algebrathere exists 119892 isin 119860 such that

10038171003817100381710038171198921119899 minus 1198921003817100381710038171003817infin(119884)997888rarr 0 (115)

and hence10038171003817100381710038171 otimes 1198921119899 minus 1 otimes 1198921003817100381710038171003817infin(119883times119884)

997888rarr 0 (116)

as 119899 997888rarr infin It follows by (114) that 119894119865 = 1 otimes 119892 thus119865 = 1 otimes (minus119894119892) isin 1 otimes 119860 (117)

By (109) we see that minus119894119892 isin 119862R(119884) thus we have 119891 = minus119894119892 isin119860 cap 119862R(119884) and 119865 = 1 otimes 119891Suppose conversely that 119861 ni 119865 = 1otimes119891 for119891 isin 119860cap119862R(119884)

We infer that 119865 isin 119862R(119883 times 119884) and |exp(119894119905(119865(119909 119910))| = 1 forevery 119905 isin R and (119909 119910) isin 119883 times 119884 Hence exp(119894119905119865)infin(119883times119884) = 1for every 119905 isin R Since

exp (119894119905119865) = exp (119894119905 (1 otimes 119891)) = 1 otimes exp (119894119905119891) isin 1 otimes 119860 (118)

we have 119863(exp(119894119905119865)) = 0 It follows that1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817 = 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884)

+ 1003817100381710038171003817119863 (exp (119894119905119865))1003817100381710038171003817infin(M) = 1 (119)

for every 119905 isin R We conclude that 119865 is a Hermitian elementin 119861

Note that 119891 isin 119860 is Hermitian if and only if119891 isin 119860cap119862R(119884)by [37 Proposition 5] Hence Proposition 9 asserts that 119865 is aHermitian element in119861 if and only if119865 = 1otimes119891 for aHermitianelement 119891 in 119860Proposition 10 Suppose that 119880 119861 997888rarr 119861 is a surjectiveunital isometry Then 119880 is an algebra isomorphism

Proof As we have already mentioned 119861 is a regular subspace(in the sense of Jarosz) with a natural norm Then byTheorem 1 119880 is also an isometry with respect to the supre-mum norm on 119883 times 119884 Then 119880 is uniquely extended to asurjective isometry with respect to the supremum norm from the uniform closure 119861 onto itself Since 119861 is a uniformalgebra a theorem of Nagasawa [32] asserts that is analgebra isomorphism since (1) = 1 Thus 119880 is an algebraisomorphism from 119861 onto itself

Journal of Function Spaces 11

Theorem 11 A bounded operator 119879 119861 997888rarr 119861 is a Hermitianoperator if and only if 119879(1) is a Hermitian element in 119861 and119879 = 119872119879(1) the multiplication operator by 119879(1)Proof By Proposition 10 every surjective unital isometry on119861 is multiplicative Then by [37 Theorem 4] we have theconclusion

5 Banach Algebras of 119862(119884)-Valued Maps

Suppose that119883 is a compact Hausdorff space Suppose that 119861is a unital point separating subalgebra of119862(119884) equipped witha Banach algebra norm Then 119861 is semisimple because 119891 isin119861 119891(119909) = 0 is a maximal ideal of 119861 for every 119909 isin 119883 and theJacobson radical of 119861 vanishes The inequality 119891infin le 119891119861for every 119891 isin 119861 is well known We say that 119861 is natural if themap 119890 119884 997888rarr 119872119861 defined by 119910 997891997888rarr 120601119910 where 120601119910(119891) = 119891(119910)for every 119891 isin 119861 is bijective We say that 119861 is self-adjoint if 119861 isnatural and conjugate-closed in the sense that 119891 isin 119861 impliesthat 119891 isin 119861 for every 119891 isin 119861 where sdot denotes the complexconjugation on 119884Definition 12 Let 119883 and 119884 be compact Hausdorff spacesSuppose that 119861 is a unital point separating subalgebra of119862(119883)equipped with a Banach algebra norm sdot 119861 Suppose that119861 is self-adjoint Suppose that 119861 is a unital point separatingsubalgebra of 119862(119883 times 119884) such that 119861 otimes 119862(119884) sub 119861 equippedwith a Banach algebra norm sdot 119861 Suppose that 119861 is self-adjoint We say that 119861 is a natural 119862(119884)-valuezation of 119861 ifthere exists a compact Hausdorff space M and a complex-linear map 119863 119861 997888rarr 119862(M) such that ker119863 = 1 otimes 119862(119884) and119863(119862R(119883 times 119884) cap 119861) sub 119862R(M) which satisfies

119865 = 119865infin(119883times119884) + 119863 (119865)infin(M) 119865 isin 119861 (120)

The term ldquoa natural 119862(119884)-valuezation of 119861rdquo comes fromthe natural norm defined by Jarosz [5] In fact the norm sdot 119861is a natural norm in the sense of Jarosz [5]

Note that (119883119862(119884) 119861 119861) need not be an admissiblequadruple defined by Nikou and OrsquoFarrell [38] (cf [31]) sincewe do not assume that 119865(sdot 119910) 119865 isin 119861 119910 isin 119884 sub 119861 whichis a requirement for the admissible quadruple On the otherhand if (119883119862(119884) 119861 119861) is an admissible quadruple of type Ldefined in [30] then 119861 is a natural 119862(119884)-valuezation of 119861 dueto Definition 12

Example 13 Let 119861 = 1198621([0 1]) and 119861 = Lip([0 1] 119862(119884))for 119884 = 119901 a singleton Then Lip([0 1]) is algebraically iso-morphic to Lip([0 1] 119862(119884)) Suppose thatM is the maximalideal space of 119871infin([0 1]) and 119863 119861 997888rarr 119862(M) is definedby 119891 997891997888rarr Γ(1198911015840) where Γ denotes the Gelfand transformin 119871infin([0 1]) Then 119861 is a natural 119862(119884)-valuezation of 119861The Banach algebra Lip([0 1]) with the norm 119891infin([01]) +1198911015840infin([01]) is isometrically isomorphic to 119861

Let 119884 be a compact Hausdorff space Note that a closedsubalgebra 119861 of Lip((119870 119889120572) 119862(119884))which appears in Example12 in [30] is an example of a natural 119862(119884)-valuezation of 119861

The Banach algebras1198621([0 1] 119862(119884)) and 1198621(T 119862(119884)))whichappear in Examples 16 and 17 in [30] respectively are alsoexamples of natural 119862(119884)-valuezations of 1198621([0 1])6 Isometries on Natural 119862(119884)-ValuezationsThemain theorem in this paper is the following

Theorem 14 Suppose that 119861119895 is a natural 119862(119884119895)-valuezationof 119861119895 sub 119862(119883119895) for 119895 = 1 2 We assume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (121)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for119895 = 1 2 Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linear isometry Then there exists ℎ isin 119862(1198842) such that |ℎ| = 1on 1198842 a continuous map 120593 1198832 times 1198842 997888rarr 1198831 such that120593(sdot 119910) 1198832 997888rarr 1198831 is a homeomorphism for each 119910 isin 1198842and a homeomorphism 120591 1198842 997888rarr 1198841 which satisfies119880 (119865) (119909 119910) = ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842 (122)

for every 119865 isin 1198611

In short a surjective isometry between 119862(119884)-valuezationsis a weighted composition operator of a specific form thehomeomorphism 1198832 times 1198842 997888rarr 1198831 times 1198841 (119909 119910) 997891997888rarr(120593(119909 119910) 120591(119910)) has the second coordinate that depends onlyon the second variable 119910 isin 1198842 A composition operatorinduced by such a homeomorphism is said to be of type BJin [31 37] after the study of Botelho and Jamison [39]

Quite recently the author of this paper and Oi [30 Theo-rem 8] proved a similar result of Theorem 14 for admissiblequadruples of type L To prove it we apply Proposition 32and the following comments in [31] Instead of this we proveTheorem 14 by Lumerrsquosmethod with which a proof is simplerthan that in [30]

In the following in this section we assume that 119861119895 is anatural 119862(119884119895)-valuezation of 119861 sub 119862(119883119895) for 119895 = 1 2 Weassume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (123)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for 119895 = 1 2Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linearisometry A crucial part of a proof of Theorem 14 is to proveProposition 15

Proposition 15 Suppose that 1198832 is not a singleton Thereexists ℎ isin 119862(1198842) with |ℎ| = 1 on 1198842 such that 119880(11198611) = 11198612 otimesℎ

A similar result for admissible quadruples of type L isproved in [30 Proposition 9] If we assumed that

119865 (sdot 119910) 119865 isin 119861119895 119910 isin 119884119895 sub 119861119895 (124)

then 119861119895 were an admissible quadruple of type L Although119861119895 in this paper need not be an admissible quadruple of type

12 Journal of Function Spaces

L a proof of Proposition 15 is completely the same as that in[30 Proposition 9] since we do not make use of the condition(124) in the proof of [30 Proposition 9] The condition (124)is needed in [30] when we apply Proposition 32 and thefollowing comments in [31]

7 Proof of Theorem 14 An Application ofLumerrsquos Method

Proof of Theorem 14 A proof for the case where 1198831 = 1199091and 1198832 = 1199092 are singletons is the same as the proof ofTheorem 8 in [30]

Suppose that1198832 is not a singleton By Proposition 15 thereexists ℎ isin 119862R(1198842) with |ℎ| = 1 on 1198842 such that 119880(1) =1 otimes ℎ Letting 1198800 1198611 997888rarr 1198612 by 1198800(119865) = (1 otimes ℎ)119880(119865)119865 isin 1198611 we see by the hypothesis (1 otimes ℎ)1198651198612 = 1198651198612 forevery 119865 isin 1198612 that 1198800 is a surjective unital isometry from1198611 onto 1198612 Then Corollary 2 asserts that 1198800 is an algebraisomorphism Let 119891 isin 119862R(1198841) By Proposition 9 1 otimes 119891 isa Hermitian element in 1198611 Then by Theorem 11 1198721otimes119891 is aHermitian operator on 1198611 By Proposition 8 11988001198721otimes119891119880minus1

0 is aHermitian operator on 1198612 Then by Theorem 11 there exists119878(119891) isin 119862R(1198842) such that 11988001198721otimes119891119880minus1

0 = 1198721otimes119878(119891) Hence anoperator 119878 119862R(1198841) 997888rarr 119862R(1198842) is defined Since 1198800 is analgebra isomorphism it is easy to see that 119878 is a real algebraisomorphism from 119862R(1198841) onto 119862R(1198842) Then 119878 119862(1198841) 997888rarr119862(1198842) defined by 119878(119891) = 119878(Re119891) + 119894119878(Im119891) for 119891 isin 119862(1198841)gives a complex algebra isomorphism Gelfand theory assertsthat there is a homeomorphism 120591 1198842 997888rarr 1198841 such that119878(119891) = 119891 ∘ 120591 119891 isin 119862(1198841) It follows that

11988001198721otimes119891119880minus10 = 1198721otimes119891∘120591 119891 isin 119862 (1198841) (125)

Since 119880minus10 (1) = 1 we have

1198800 (1 otimes 119891) = 1 otimes 119891 ∘ 120591 119891 isin 119862 (1198841) (126)

Define Φ 1198611 997888rarr 1198612 by Φ(119886) = 1198800(119886 otimes 1) 119886 isin 1198611Since 1198800 is an algebra isomorphism the map Φ is a unitalhomomorphism Since the maximal ideal space of 1198611 is 1198831

and the maximal ideal space of 1198612 is 1198832 times 1198842 there is acontinuous map 120593 1198832 times 1198842 997888rarr 1198831 such that

Φ(119886) (119909 119910) = 119886 (120593 (119909 119910)) 119886 isin 1198611 (119909 119910) isin 1198832 times 1198842 (127)

It follows by (126) and (127) that

1198800 (119886 otimes 119891) (119909 119910) = 1198800 ((119886 otimes 1) (1 otimes 119891)) (119909 119910)= 1198800 (119886 otimes 1) (119909 119910)1198800 (1 otimes 119891) (119909 119910)= 119886 (120593 (119909 119910))119891 (120591 (119910))= (119886 otimes 119891) (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198832 times 1198842

(128)

for every 119886 isin 1198611 and 119891 isin 119862(1198841) Thus

1198800 (sum(119886119895 otimes 119891119895)) (119909 119910)= (sum(119886119895 otimes 119891119895)) (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842(129)

for every sum(119886119895 otimes 119891119895) isin 1198611 otimes 119862(1198841) By the Stone-Weierstrasstheorem 1198611 otimes 119862(1198841) is uniformly dense in 119862(1198831 times 1198841) henceany element in 1198611 is uniformly approximated by 1198611 otimes 119862(1198841)As 1198800 is also an isometry with respect to the uniform normwe see that

1198800 (119865) (119909 119910) = 119865 (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198831 times 1198841 (130)

for every 119865 isin 1198611 and

119880 (119865) (119909 119910) = (1 otimes ℎ) (119909 119910) 119865 (120593 (119909 119910) 120591 (119910))= ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198831 times 1198841(131)

As 1198800 is an algebra isomorphism the map 1198832 times 1198842 997888rarr1198831 times 1198841 defined by (119909 119910) 997891997888rarr (120593(119909 119910) 120591(119910)) gives ahomeomorphism Therefore for every 119910 isin 1198842 the map

120593 (sdot 119910) 1198832 997888rarr 1198831 (132)

is a homeomorphismSuppose that 1198831 is not a singleton By the same way as

in the last part of the proof of Theorem 8 in [30] we havethat 1198832 is not a singleton Then we have the conclusion bythe previous argument

8 Application of Theorem 14

We exhibit applications of Theorem 14

Corollary 16 ([4 Theorem 33]) Suppose that119880 Lip([0 1]) 997888rarr Lip([0 1]) is a surjective isometrywith respect to the norm defined by 119891infin([01]) + 1198911015840infin([01])

for 119891 isin Lip([0 1]) Then 119880(1) is a constant function of unitmodulus such that

119880(119891) (119909) = 119880 (1) 119891 (119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (133)

or

119880(119891) (119909) = 119880 (1) 119891 (1 minus 119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (134)

The converse statement also holds

Proof By Example 13 we may suppose that Lip([0 1]) is aBanach algebra of 119862(119884)-valuezation Applying Theorem 14we have that119880(1) = 1otimesℎ for ℎ isin 119862(119884)with |ℎ| = 1 Since our

Journal of Function Spaces 13

119884 is a singleton 119880(1) is a constant function of unit modulusWe also see that the corresponding continuous map 120593 [0 1] times 119884 997888rarr [0 1] can be considered as a homeomorphismfrom [0 1] onto [0 1] therefore we have that119880(119891) (119909) = 119880 (1) 119891 (120593 (119909))

119891 isin Lip ([0 1]) 119909 isin [0 1] (135)

The rest is a routine argument to prove that 120593 is an isometryhence 120593(119909) = 119909 119909 isin [0 1] or 120593(119909) = 1 minus 119909 119909 isin [0 1]

The converse statement is trivial

Corollaries 14 15 18 and 19 in [30 Section 6] follow herewith a similar proof

Data Availability

The data used to support the findings of this study areavailable from the corresponding author upon request

Conflicts of Interest

The author declares that they have no conflicts of interest

Acknowledgments

This work was supported by JSPS KAKENHI Grants Num-bers JP16K05172 and JP15K04921

References

[1] K de Leeuw ldquoBanach spaces of Lipschitz functionsrdquo StudiaMathematica vol 21 pp 55ndash66 19611962

[2] A K Roy ldquoExtreme points and linear isometries of the Banachspace of Lipschitz functionsrdquoCanadian Journal of MathematicsJournal Canadien de Mathematiques vol 20 pp 1150ndash11641968

[3] M Cambern ldquoIsometries of certain Banach algebrasrdquo StudiaMathematica vol 25 pp 217ndash225 19641965

[4] N V Rao and A K Roy ldquoLinear isometries of some functionspacesrdquo Pacific Journal of Mathematics vol 38 pp 177ndash192 1971

[5] K Jarosz ldquoIsometries in semisimple commutative BanachalgebrasrdquoProceedings of the AmericanMathematical Society vol94 no 1 pp 65ndash71 1985

[6] K Jarosz and V D Pathak ldquoIsometries between functionspacesrdquo Transactions of the AmericanMathematical Society vol305 no 1 pp 193ndash206 1988

[7] NWeaver ldquoIsometries of noncompact Lipschitz spacesrdquoCana-dian Mathematical Bulletin Bulletin Canadien de Mathemat-iques vol 38 no 2 pp 242ndash249 1995

[8] A Jimenez-Vargas and M Villegas-Vallecillos ldquoLinear isome-tries between spaces of vector-valued Lipschitz functionsrdquoProceedings of the American Mathematical Society vol 137 no4 pp 1381ndash1388 2009

[9] A Jimenez-Vargas and M Villegas-Vallecillos ldquoInto linearisometries between spaces of Lipschitz functionsrdquo HoustonJournal of Mathematics vol 34 no 4 pp 1165ndash1184 2008

[10] F Botelho and J Jamison ldquoSurjective isometries on spacesof differentiable vector-valued functionsrdquo Studia Mathematicavol 192 no 1 pp 39ndash50 2009

[11] E Mayer-Wolf ldquoIsometries between Banach spaces of Lipschitzfunctionsrdquo Israel Journal of Mathematics vol 38 no 1-2 pp 58ndash74 1981

[12] A Jimenez-Vargas M Villegas-Vallecillos and Y-S WangldquoBanach-Stone theorems for vector-valued little Lipschitz func-tionsrdquo Publicationes Mathematicae vol 74 no 1-2 pp 81ndash1002009

[13] J Araujo andLDubarbie ldquoNoncompactness andnoncomplete-ness in isometries of Lipschitz spacesrdquo Journal of MathematicalAnalysis and Applications vol 377 no 1 pp 15ndash29 2011

[14] F Botelho R J Fleming and J Jamison ldquoExtreme points andisometries on vector-valued Lipschitz spacesrdquo Journal of Math-ematical Analysis and Applications vol 381 no 2 pp 821ndash8322011

[15] H Koshimizu ldquoLinear isometries on spaces of continuouslydifferentiable and Lipschitz continuous functionsrdquo NihonkaiMathematical Journal vol 22 no 2 pp 73ndash90 2011

[16] F Botelho J Jamison and B Zheng ldquoIsometries on spaces ofvector valued Lipschitz functionsrdquo Positivity An InternationalMathematics Journal Devoted to Theory and Applications ofPositivity vol 17 no 1 pp 47ndash65 2013

[17] A Ranjbar-Motlagh ldquoA note on isometries of Lipschitz spacesrdquoJournal of Mathematical Analysis and Applications vol 411 no2 pp 555ndash558 2014

[18] F Botelho and J Jamison ldquoSurjective isometries on spaces ofvector valued continuous and Lipschitz functionsrdquo PositivityAn International Mathematics Journal Devoted to Theory andApplications of Positivity vol 17 no 3 pp 395ndash405 2013Erratum to Surjective isometries on spaces of vector valuedcontinuous and Lipschitz functions by F Botelho 20 (2016)757ndash759

[19] T Miura and H Takagi ldquoSurjective isometries on the Banachspace of continuously differentiable functionsrdquo ContemporaryMathematics vol 687 pp 181ndash192 2017

[20] K Kawamura ldquoBanach-Stone type theorems for 1198621-functionspaces over Riemannian manifoldsrdquo Acta Universitatis Szegedi-ensis Acta Scientiarum Mathematicarum vol 83 no 3-4 pp551ndash591 2017

[21] K Kawamura ldquoPerturbations of norms on 1198621-function spacesand associated isometry groupsrdquo Topology Proceedings vol 51pp 169ndash196 2018

[22] K Kawamura ldquoA Banach-Stone type theorem for 1198621-functionspaces over the circlerdquo Topology Proceedings vol 53 pp 15ndash262019

[23] L Li D Chen Q Meng and Y-S Wang ldquoSurjective isometrieson vector-valued differentiable function spacesrdquo Annals ofFunctional Analysis vol 9 no 3 pp 334ndash343 2018

[24] K Kawamura H Koshimizu and T Miura ldquoNorms onC1([01]) and there isometriesrdquo Acta Scientiarum Mathemati-carum vol 84 no 12 pp 239ndash261 2018

[25] L Li A M Peralta L Wang and Y-S Wang ldquoWeak-2-local isometries on uniform algebras and Lipschitz algebrasrdquohttpsarxivorgabs170503619

[26] A Jimenez-Vargas L LiAMPeralta LWang andY-SWangldquo2-local standard isometries on vector-valued Lipschitz func-tion spacesrdquo Journal of Mathematical Analysis and Applicationsvol 461 no 2 pp 1287ndash1298 2018

[27] A Ranjbar-Motlagh ldquoIsometries of Lipschitz type functionspacesrdquoMathematischeNachrichten vol 291 no 11-12 pp 1899ndash1907 2018

14 Journal of Function Spaces

[28] N Weaver Lipschitz Algebras World Scientific Publishing CoInc River Edge NJ USA 1999

[29] R R Phelps Lectures on Choquetrsquos theorem vol 1757 of LectureNotes in Mathematics Springer-Verlag Berlin Germany 2ndedition 2001

[30] O Hatori and S Oi ldquoIsometries on Banach algebras of vector-valued mapsrdquo Acta Scientiarum Mathematicarum vol 84 no12 pp 151ndash183 2018

[31] O Hatori S Oi and H Takagi ldquoPeculiar homomorphismsbetween algebras of vector-valued mapsrdquo Studia Mathematicavol 242 no 2 pp 141ndash163 2018

[32] M Nagasawa ldquoIsomorphisms between commutative Banachalgebras with an application to rings of analytic functionsrdquoKodai Mathematical Seminar Reports vol 11 pp 182ndash188 1959

[33] K de LeeuwW Rudin and J Wermer ldquoThe isometries of somefunction spacesrdquo Proceedings of the American MathematicalSociety vol 11 pp 694ndash698 1960

[34] OHatori A Jimenez-Vargas andMVillegas-Vallecillos ldquoMapswhich preserve norms of non-symmetrical quotients betweengroups of exponentials of Lipschitz functionsrdquo Journal of Math-ematical Analysis and Applications vol 415 no 2 pp 825ndash8452014

[35] G Lumer ldquoOn the isometries of reflexiveOrlicz spacesrdquoAnnalesde lrsquoInstitut Fourier vol 68 pp 99ndash109 1963

[36] R J Fleming and J E Jamison Isometries on Banach SpacesFunction Spaces vol 129 ofMonographs and Surveys in Pure andApplied Mathematics Chapman amp Hall CRC Boca Raton FlaUSA 2003

[37] OHatori and SOi ldquoHermitian operators onBanach algebras ofvector-valued Lipschitzmapsrdquo Journal ofMathematical Analysisand Applications vol 452 no 1 pp 378ndash387 2017 Corrigendumto rdquoHermitian operators on Banach algebras of vector-valuedLipschitz mapsrdquo Journal of Mathematical Analysis and Appli-cations 452 (2017) 378ndash387 MR3628025

[38] A Nikou andA G OrsquoFarrell ldquoBanach algebras of vector-valuedfunctionsrdquo Glasgow Mathematical Journal vol 56 no 2 pp419ndash426 2014

[39] F Botelho and J Jamison ldquoHomomorphisms on a class ofcommutative Banach algebrasrdquo Rocky Mountain Journal ofMathematics vol 43 no 2 pp 395ndash416 2013

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Page 9: Hermitian Operators and Isometries on Banach Algebras of ...downloads.hindawi.com/journals/jfs/2018/8085304.pdfJournalofFunctionSpaces normiseitherthemaxnormorthesumnorm.esituation

Journal of Function Spaces 9

119901 = (119909119901 119910119901) = 119909119901 + 119894119910119901 isin Ω with 119903 = |119901 minus (1 + 1119899)| gt1119899 + 11198992 We prove the case where 119910119901 ge 0 A proof for thecase where 119910119901 le 0 is the same and we omit it We divideΩ+ =119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902 isin Ω 119910119902 ge 0 into two parts

Ω1 = 119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902isin Ω+ minus119899radic2119899 + 1 (119909119902 minus (1 minus

11198992)) le 119910119902le minus119899radic2119899 + 1 (119909 minus (1 +

1119899 + 11198992 + 11198993))(94)

and

Ω2 = 119902 = (119909119902 119910119902) = 119909119902 + 119894119910119902 isin Ω+ 119910119902le minus119899radic2119899 + 1 (119909119902 minus (1 minus

11198992)) (95)

Suppose that 119901 isin Ω1 and 119903 = |119901minus (1+1119899)| gt 1119899+11198992Thedistance between ℓ1 and ℓ2 is 1119899+11198992 Hence119870(119901 119903)capΩ119888 =0 Suppose that 119901 isin Ω2 and 119903 = |119901 minus (1 + 1119899)| gt 1119899 + 11198992that is119901 = 1minus11198992 Let ℓ1015840 be the line passing through119901whichis parallel to ℓ2 Let 1199011015840 be the unique point in the intersectionof ℓ1015840 and the 119909-axis Then 1199011015840 = 1 minus 11198992 minus 119906 for some 119906 ge 0Then the distance between ℓ1 and ℓ1015840 is 1119899+11198992+119906(1+1119899)which is equal to the distance between the point119901 and the lineℓ1 On the other hand1003816100381610038161003816100381610038161003816119901 minus (1 + 1119899)

1003816100381610038161003816100381610038161003816 ge10038161003816100381610038161003816100381610038161199011015840 minus (1 + 1119899)

1003816100381610038161003816100381610038161003816 = 1119899 + 11198992 + 119906 (96)

It follows that 119870(119901 119903) cap Ω119888 = 0 We conclude that if 119901 isin Ωsatisfies 119903 = |119901minus(1+1119899)| gt 1119899+11198992 then119870(119901 119903)capΩ119888 = 0Thus we have

120588 (Ω 1 + 1119899) = 1119899 + 11198992 (97)

Since 1 + 1119899 isin 120590(119891 + 119892119899119899) sub Ω we have120588 (120590 (119891 + 119892119899119899 )) le 120588 (120590 (119891 + 119892119899119899 ) 1 + 1119899)

le 120588 (Ω 1 + 1119899) = 1119899 + 11198992 (98)

On the other hand 119892119899119899infin 997888rarr 0 as 119899 997888rarr infin ensures that119891 + 119892119899119899 minus 119891infin 997888rarr 0 as 119899 997888rarr infin It follows that for every120576 gt 0A120576 is dense in (119860 sdot infin)Finally we show a proof of the third assertion (3) As is

pointed out in the proof of [5Theorem] 120588(co(119870)+119870(119888)) ge 119888for any 119870 sub C and any 119888 ge 0 Let 120576 gt 0 and 119891 isin A120576 Supposethat Δ119891 le 0 Then by (43) we have (119891) = (119879119891) + 119870(minusΔ119891)Hence we have

120588 ( (119891)) = 120588 (120590 (119879119891) + 119870 (minusΔ119891)) ge minusΔ119891 (99)

As 120576 ge 120588(120590(119891)) we conclude by (45) that10038171003817100381710038171198791198911003817100381710038171003817infin minus 10038171003817100381710038171198911003817100381710038171003817infin = Δ119891 ge minus120576 (100)

This completes the proof of the theorem

4 Hermitian Operators on a BanachAlgebras of Continuous Maps WhoseValues Are in a Uniform Algebras

Let 119883 and 119884 be compact Hausdorff spaces Let 119861 be aunital subalgebra of 119862(119883) which separates the points of 119883Throughout this sectionwe assume119861 is a Banach algebrawiththe norm sdot 119861 and 119860 is a uniform algebra on 119884 Recall thata uniform algebra on 119884 is a uniformly closed subalgebra of119862(119884) which contains constants and separates the points of 119884For functions 119891 isin 119862(119883) and 119892 isin 119862(119884) let 119891 otimes 119892 isin 119862(119883 times 119884)be the function defined by119891otimes119892(119909 119910) = 119891(119909)119892(119910) for (119909 119910) isin119883 times 119884 and for a subspace 119864119883 of 119862(119883) and a subspace 119864119884 of119862(119884) put

119864119883 otimes 119864119884

= 119899sum

119895=1

119891119895 otimes 119892119895 119899 isin N 119891119895 isin 119864119883 119892119895 isin 119864119884

(101)

and

1 otimes 119864119884 = 1 otimes 119892 119892 isin 119864119884 (102)

Throughout the section 119861 is a unital subalgebra of 119862(119883 times 119884)with a Banach algebra norm sdot 119861 We assume that 119861otimes119860 sub 119861Note that 119861 separates the points of119883times119884 since119860 separates thepoints of 119884 and 119861 separates the points of 119883 We assume thatthere exists a compact Hausdorff space M and a complex-linear map 119863 119861 997888rarr 119862(M) such that ker119863 = 1 otimes 119860 Weassume that 119865119861 = 119865infin(119883times119884)+119863(119865)infin(M) for every 119865 isin 119861Hence119863 is continuous Defining

lsaquo119865lsaquo = 119863 (119865)infin(M) 119865 isin 119861 (103)

lsaquo sdot lsaquo is a one-invariant seminorm in the sense of Jarosz lsaquo sdot lsaquois a seminorm on 119861 such that lsaquo119865+ 1lsaquo = lsaquo119865lsaquo for every 119865 isin 119861Hence the norm sdot 119861 is a natural norm (see [5 p67]) Notethat 119861 is a regular subspace of 119862(119883times119884) in the sense of Jarosz[5 Proposition 2]

Lumerrsquos seminal paper [35] opened up a useful methodof finding isometries which is often referred to as Lumerrsquosmethod It involves the notion ofHermitian operators and thefact that 119880119867119880minus1 must be Hermitian if119867 is Hermitian and 119880is a surjective isometry

Definition 6 Let A be a unital Banach algebra We say that119890 isin A is a Hermitian element if1003817100381710038171003817exp (119894119905119890)1003817100381710038171003817A = 1 (104)

for every 119905 isin R The set of all Hermitian elements of A isdenoted by119867(A)

If A is a unital 119862lowast-algebra then 119867(A) is the set of allself-adjoint elements of A Hence 119867(119872119899(C)) is the set of allHermitian matrices and119867(119862(119884)) = 119862R(119884)Definition 7 Let 119864 be a complex Banach space The Banachalgebra of all bounded operators on 119864 is denoted by 119861(119864) Wesay that 119879 isin 119861(119864) is a Hermitian operator if 119879 isin 119867(119861(119864))

10 Journal of Function Spaces

Note that a Hermitian element of a unital Banach algebraand a Hermitian operator are usually defined in terms ofnumerical range or semi-inner product Here we define themby an equivalent form (see [36]) By the definition of aHermitian operator we have the following

Proposition 8 Let 119864119895 be a complex Banach space for 119895 = 1 2Suppose that 119881 1198641 997888rarr 1198642 is a surjective isometry and 119867 1198641 997888rarr 1198641 is a Hermitian operator Then 119881119867119881minus1 1198642 997888rarr 1198642

is a Hermitian operator

Proposition 9 An element 119865 isin 119861 is Hermitian if and only ifthere exists 119891 isin 119860 cap 119862R(119884) such that 119865 = 1 otimes 119891Proof Suppose that 119865 isin 119861 is a Hermitian element Then1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884) + 1003817100381710038171003817119863 (exp (119894119905119865))1003817100381710038171003817infin(M)

= 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817 = 1 (105)

for every 119905 isin R Suppose that there exists a point (119909 119910) isin119883 times 119884 with Im119865(119909 119910) = 0 where Im denotes the imaginarypart of a complex number Suppose that Im119865(119909 119910) gt 0Then1003817100381710038171003817exp (minus119894119865)1003817100381710038171003817infin(119883times119884) ge 1003816100381610038161003816exp (minus119894119865 (119909 119910))1003816100381610038161003816

= exp (Im119865 (119909 119910)) gt 1 (106)

Suppose that Im119865(119909 119910) lt 0 Then1003817100381710038171003817exp (119894119865)1003817100381710038171003817infin(119883times119884) ge 1003816100381610038161003816exp (119894119865 (119909 119910))1003816100381610038161003816= exp (minusIm119865 (119909 119910)) gt 1 (107)

In any case we have there exists 119905 isin R such that1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817119861 ge 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884) gt 1 (108)

which contradicts our assumption We have that

119865 isin 119862R (119883 times 119884) (109)

Thus for every (119904 119905) isin 119883 times 119884 and 119905 isin R |exp(119894119905119865(119904 119905))| = 1Hence exp(119894119905119865)infin(119883times119884) = 1 for every 119905 isin R By (105)we have119863(exp(119894119905119865)infin(M) = 0 which ensures that 119863(exp(119894119905119865)) = 0for every 119905 isin R Thus exp(119894119905119865) isin 1 otimes 119860 for every 119905 isin R Wehave

exp (119894119905119865) minus 1119905 minus 119894119865 = infinsum119899=2

((119894119865)119899 119905119899minus2119899 ) 119905 (110)

and hence for every 119905 isin R with |119905| le 1 we have10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817 le (

infinsum119899=2

119865119899119861|119905|119899minus2119899 ) |119905|

le (infinsum119899=2

119865119899119861119899 ) |119905| le (exp 119865119861) |119905|

(111)

It follows that10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817infin(119883times119884)

le 10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817119861

997888rarr 0(112)

as 119905 997888rarr 0 Since (exp(119894119905119865)minus1)119905 isin 1otimes119860 for each 119905 isin R thereexists 119892119905 isin 119860 such that

exp (119894119905119865) minus 1119905 = 1 otimes 119892119905 (113)

By (112) we have10038171003817100381710038171 otimes 1198921119899 minus 1198941198651003817100381710038171003817infin(119883times119884)

997888rarr 0 (114)

as 119899 997888rarr infin We have that 1 otimes 1198921119899 is a Cauchy sequencein 119862(119883 times 119884) thus we infer that1198921119899 is a Cauchy sequence in119862(119884) Since 119860 is uniformly closed as it is a uniform algebrathere exists 119892 isin 119860 such that

10038171003817100381710038171198921119899 minus 1198921003817100381710038171003817infin(119884)997888rarr 0 (115)

and hence10038171003817100381710038171 otimes 1198921119899 minus 1 otimes 1198921003817100381710038171003817infin(119883times119884)

997888rarr 0 (116)

as 119899 997888rarr infin It follows by (114) that 119894119865 = 1 otimes 119892 thus119865 = 1 otimes (minus119894119892) isin 1 otimes 119860 (117)

By (109) we see that minus119894119892 isin 119862R(119884) thus we have 119891 = minus119894119892 isin119860 cap 119862R(119884) and 119865 = 1 otimes 119891Suppose conversely that 119861 ni 119865 = 1otimes119891 for119891 isin 119860cap119862R(119884)

We infer that 119865 isin 119862R(119883 times 119884) and |exp(119894119905(119865(119909 119910))| = 1 forevery 119905 isin R and (119909 119910) isin 119883 times 119884 Hence exp(119894119905119865)infin(119883times119884) = 1for every 119905 isin R Since

exp (119894119905119865) = exp (119894119905 (1 otimes 119891)) = 1 otimes exp (119894119905119891) isin 1 otimes 119860 (118)

we have 119863(exp(119894119905119865)) = 0 It follows that1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817 = 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884)

+ 1003817100381710038171003817119863 (exp (119894119905119865))1003817100381710038171003817infin(M) = 1 (119)

for every 119905 isin R We conclude that 119865 is a Hermitian elementin 119861

Note that 119891 isin 119860 is Hermitian if and only if119891 isin 119860cap119862R(119884)by [37 Proposition 5] Hence Proposition 9 asserts that 119865 is aHermitian element in119861 if and only if119865 = 1otimes119891 for aHermitianelement 119891 in 119860Proposition 10 Suppose that 119880 119861 997888rarr 119861 is a surjectiveunital isometry Then 119880 is an algebra isomorphism

Proof As we have already mentioned 119861 is a regular subspace(in the sense of Jarosz) with a natural norm Then byTheorem 1 119880 is also an isometry with respect to the supre-mum norm on 119883 times 119884 Then 119880 is uniquely extended to asurjective isometry with respect to the supremum norm from the uniform closure 119861 onto itself Since 119861 is a uniformalgebra a theorem of Nagasawa [32] asserts that is analgebra isomorphism since (1) = 1 Thus 119880 is an algebraisomorphism from 119861 onto itself

Journal of Function Spaces 11

Theorem 11 A bounded operator 119879 119861 997888rarr 119861 is a Hermitianoperator if and only if 119879(1) is a Hermitian element in 119861 and119879 = 119872119879(1) the multiplication operator by 119879(1)Proof By Proposition 10 every surjective unital isometry on119861 is multiplicative Then by [37 Theorem 4] we have theconclusion

5 Banach Algebras of 119862(119884)-Valued Maps

Suppose that119883 is a compact Hausdorff space Suppose that 119861is a unital point separating subalgebra of119862(119884) equipped witha Banach algebra norm Then 119861 is semisimple because 119891 isin119861 119891(119909) = 0 is a maximal ideal of 119861 for every 119909 isin 119883 and theJacobson radical of 119861 vanishes The inequality 119891infin le 119891119861for every 119891 isin 119861 is well known We say that 119861 is natural if themap 119890 119884 997888rarr 119872119861 defined by 119910 997891997888rarr 120601119910 where 120601119910(119891) = 119891(119910)for every 119891 isin 119861 is bijective We say that 119861 is self-adjoint if 119861 isnatural and conjugate-closed in the sense that 119891 isin 119861 impliesthat 119891 isin 119861 for every 119891 isin 119861 where sdot denotes the complexconjugation on 119884Definition 12 Let 119883 and 119884 be compact Hausdorff spacesSuppose that 119861 is a unital point separating subalgebra of119862(119883)equipped with a Banach algebra norm sdot 119861 Suppose that119861 is self-adjoint Suppose that 119861 is a unital point separatingsubalgebra of 119862(119883 times 119884) such that 119861 otimes 119862(119884) sub 119861 equippedwith a Banach algebra norm sdot 119861 Suppose that 119861 is self-adjoint We say that 119861 is a natural 119862(119884)-valuezation of 119861 ifthere exists a compact Hausdorff space M and a complex-linear map 119863 119861 997888rarr 119862(M) such that ker119863 = 1 otimes 119862(119884) and119863(119862R(119883 times 119884) cap 119861) sub 119862R(M) which satisfies

119865 = 119865infin(119883times119884) + 119863 (119865)infin(M) 119865 isin 119861 (120)

The term ldquoa natural 119862(119884)-valuezation of 119861rdquo comes fromthe natural norm defined by Jarosz [5] In fact the norm sdot 119861is a natural norm in the sense of Jarosz [5]

Note that (119883119862(119884) 119861 119861) need not be an admissiblequadruple defined by Nikou and OrsquoFarrell [38] (cf [31]) sincewe do not assume that 119865(sdot 119910) 119865 isin 119861 119910 isin 119884 sub 119861 whichis a requirement for the admissible quadruple On the otherhand if (119883119862(119884) 119861 119861) is an admissible quadruple of type Ldefined in [30] then 119861 is a natural 119862(119884)-valuezation of 119861 dueto Definition 12

Example 13 Let 119861 = 1198621([0 1]) and 119861 = Lip([0 1] 119862(119884))for 119884 = 119901 a singleton Then Lip([0 1]) is algebraically iso-morphic to Lip([0 1] 119862(119884)) Suppose thatM is the maximalideal space of 119871infin([0 1]) and 119863 119861 997888rarr 119862(M) is definedby 119891 997891997888rarr Γ(1198911015840) where Γ denotes the Gelfand transformin 119871infin([0 1]) Then 119861 is a natural 119862(119884)-valuezation of 119861The Banach algebra Lip([0 1]) with the norm 119891infin([01]) +1198911015840infin([01]) is isometrically isomorphic to 119861

Let 119884 be a compact Hausdorff space Note that a closedsubalgebra 119861 of Lip((119870 119889120572) 119862(119884))which appears in Example12 in [30] is an example of a natural 119862(119884)-valuezation of 119861

The Banach algebras1198621([0 1] 119862(119884)) and 1198621(T 119862(119884)))whichappear in Examples 16 and 17 in [30] respectively are alsoexamples of natural 119862(119884)-valuezations of 1198621([0 1])6 Isometries on Natural 119862(119884)-ValuezationsThemain theorem in this paper is the following

Theorem 14 Suppose that 119861119895 is a natural 119862(119884119895)-valuezationof 119861119895 sub 119862(119883119895) for 119895 = 1 2 We assume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (121)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for119895 = 1 2 Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linear isometry Then there exists ℎ isin 119862(1198842) such that |ℎ| = 1on 1198842 a continuous map 120593 1198832 times 1198842 997888rarr 1198831 such that120593(sdot 119910) 1198832 997888rarr 1198831 is a homeomorphism for each 119910 isin 1198842and a homeomorphism 120591 1198842 997888rarr 1198841 which satisfies119880 (119865) (119909 119910) = ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842 (122)

for every 119865 isin 1198611

In short a surjective isometry between 119862(119884)-valuezationsis a weighted composition operator of a specific form thehomeomorphism 1198832 times 1198842 997888rarr 1198831 times 1198841 (119909 119910) 997891997888rarr(120593(119909 119910) 120591(119910)) has the second coordinate that depends onlyon the second variable 119910 isin 1198842 A composition operatorinduced by such a homeomorphism is said to be of type BJin [31 37] after the study of Botelho and Jamison [39]

Quite recently the author of this paper and Oi [30 Theo-rem 8] proved a similar result of Theorem 14 for admissiblequadruples of type L To prove it we apply Proposition 32and the following comments in [31] Instead of this we proveTheorem 14 by Lumerrsquosmethod with which a proof is simplerthan that in [30]

In the following in this section we assume that 119861119895 is anatural 119862(119884119895)-valuezation of 119861 sub 119862(119883119895) for 119895 = 1 2 Weassume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (123)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for 119895 = 1 2Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linearisometry A crucial part of a proof of Theorem 14 is to proveProposition 15

Proposition 15 Suppose that 1198832 is not a singleton Thereexists ℎ isin 119862(1198842) with |ℎ| = 1 on 1198842 such that 119880(11198611) = 11198612 otimesℎ

A similar result for admissible quadruples of type L isproved in [30 Proposition 9] If we assumed that

119865 (sdot 119910) 119865 isin 119861119895 119910 isin 119884119895 sub 119861119895 (124)

then 119861119895 were an admissible quadruple of type L Although119861119895 in this paper need not be an admissible quadruple of type

12 Journal of Function Spaces

L a proof of Proposition 15 is completely the same as that in[30 Proposition 9] since we do not make use of the condition(124) in the proof of [30 Proposition 9] The condition (124)is needed in [30] when we apply Proposition 32 and thefollowing comments in [31]

7 Proof of Theorem 14 An Application ofLumerrsquos Method

Proof of Theorem 14 A proof for the case where 1198831 = 1199091and 1198832 = 1199092 are singletons is the same as the proof ofTheorem 8 in [30]

Suppose that1198832 is not a singleton By Proposition 15 thereexists ℎ isin 119862R(1198842) with |ℎ| = 1 on 1198842 such that 119880(1) =1 otimes ℎ Letting 1198800 1198611 997888rarr 1198612 by 1198800(119865) = (1 otimes ℎ)119880(119865)119865 isin 1198611 we see by the hypothesis (1 otimes ℎ)1198651198612 = 1198651198612 forevery 119865 isin 1198612 that 1198800 is a surjective unital isometry from1198611 onto 1198612 Then Corollary 2 asserts that 1198800 is an algebraisomorphism Let 119891 isin 119862R(1198841) By Proposition 9 1 otimes 119891 isa Hermitian element in 1198611 Then by Theorem 11 1198721otimes119891 is aHermitian operator on 1198611 By Proposition 8 11988001198721otimes119891119880minus1

0 is aHermitian operator on 1198612 Then by Theorem 11 there exists119878(119891) isin 119862R(1198842) such that 11988001198721otimes119891119880minus1

0 = 1198721otimes119878(119891) Hence anoperator 119878 119862R(1198841) 997888rarr 119862R(1198842) is defined Since 1198800 is analgebra isomorphism it is easy to see that 119878 is a real algebraisomorphism from 119862R(1198841) onto 119862R(1198842) Then 119878 119862(1198841) 997888rarr119862(1198842) defined by 119878(119891) = 119878(Re119891) + 119894119878(Im119891) for 119891 isin 119862(1198841)gives a complex algebra isomorphism Gelfand theory assertsthat there is a homeomorphism 120591 1198842 997888rarr 1198841 such that119878(119891) = 119891 ∘ 120591 119891 isin 119862(1198841) It follows that

11988001198721otimes119891119880minus10 = 1198721otimes119891∘120591 119891 isin 119862 (1198841) (125)

Since 119880minus10 (1) = 1 we have

1198800 (1 otimes 119891) = 1 otimes 119891 ∘ 120591 119891 isin 119862 (1198841) (126)

Define Φ 1198611 997888rarr 1198612 by Φ(119886) = 1198800(119886 otimes 1) 119886 isin 1198611Since 1198800 is an algebra isomorphism the map Φ is a unitalhomomorphism Since the maximal ideal space of 1198611 is 1198831

and the maximal ideal space of 1198612 is 1198832 times 1198842 there is acontinuous map 120593 1198832 times 1198842 997888rarr 1198831 such that

Φ(119886) (119909 119910) = 119886 (120593 (119909 119910)) 119886 isin 1198611 (119909 119910) isin 1198832 times 1198842 (127)

It follows by (126) and (127) that

1198800 (119886 otimes 119891) (119909 119910) = 1198800 ((119886 otimes 1) (1 otimes 119891)) (119909 119910)= 1198800 (119886 otimes 1) (119909 119910)1198800 (1 otimes 119891) (119909 119910)= 119886 (120593 (119909 119910))119891 (120591 (119910))= (119886 otimes 119891) (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198832 times 1198842

(128)

for every 119886 isin 1198611 and 119891 isin 119862(1198841) Thus

1198800 (sum(119886119895 otimes 119891119895)) (119909 119910)= (sum(119886119895 otimes 119891119895)) (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842(129)

for every sum(119886119895 otimes 119891119895) isin 1198611 otimes 119862(1198841) By the Stone-Weierstrasstheorem 1198611 otimes 119862(1198841) is uniformly dense in 119862(1198831 times 1198841) henceany element in 1198611 is uniformly approximated by 1198611 otimes 119862(1198841)As 1198800 is also an isometry with respect to the uniform normwe see that

1198800 (119865) (119909 119910) = 119865 (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198831 times 1198841 (130)

for every 119865 isin 1198611 and

119880 (119865) (119909 119910) = (1 otimes ℎ) (119909 119910) 119865 (120593 (119909 119910) 120591 (119910))= ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198831 times 1198841(131)

As 1198800 is an algebra isomorphism the map 1198832 times 1198842 997888rarr1198831 times 1198841 defined by (119909 119910) 997891997888rarr (120593(119909 119910) 120591(119910)) gives ahomeomorphism Therefore for every 119910 isin 1198842 the map

120593 (sdot 119910) 1198832 997888rarr 1198831 (132)

is a homeomorphismSuppose that 1198831 is not a singleton By the same way as

in the last part of the proof of Theorem 8 in [30] we havethat 1198832 is not a singleton Then we have the conclusion bythe previous argument

8 Application of Theorem 14

We exhibit applications of Theorem 14

Corollary 16 ([4 Theorem 33]) Suppose that119880 Lip([0 1]) 997888rarr Lip([0 1]) is a surjective isometrywith respect to the norm defined by 119891infin([01]) + 1198911015840infin([01])

for 119891 isin Lip([0 1]) Then 119880(1) is a constant function of unitmodulus such that

119880(119891) (119909) = 119880 (1) 119891 (119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (133)

or

119880(119891) (119909) = 119880 (1) 119891 (1 minus 119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (134)

The converse statement also holds

Proof By Example 13 we may suppose that Lip([0 1]) is aBanach algebra of 119862(119884)-valuezation Applying Theorem 14we have that119880(1) = 1otimesℎ for ℎ isin 119862(119884)with |ℎ| = 1 Since our

Journal of Function Spaces 13

119884 is a singleton 119880(1) is a constant function of unit modulusWe also see that the corresponding continuous map 120593 [0 1] times 119884 997888rarr [0 1] can be considered as a homeomorphismfrom [0 1] onto [0 1] therefore we have that119880(119891) (119909) = 119880 (1) 119891 (120593 (119909))

119891 isin Lip ([0 1]) 119909 isin [0 1] (135)

The rest is a routine argument to prove that 120593 is an isometryhence 120593(119909) = 119909 119909 isin [0 1] or 120593(119909) = 1 minus 119909 119909 isin [0 1]

The converse statement is trivial

Corollaries 14 15 18 and 19 in [30 Section 6] follow herewith a similar proof

Data Availability

The data used to support the findings of this study areavailable from the corresponding author upon request

Conflicts of Interest

The author declares that they have no conflicts of interest

Acknowledgments

This work was supported by JSPS KAKENHI Grants Num-bers JP16K05172 and JP15K04921

References

[1] K de Leeuw ldquoBanach spaces of Lipschitz functionsrdquo StudiaMathematica vol 21 pp 55ndash66 19611962

[2] A K Roy ldquoExtreme points and linear isometries of the Banachspace of Lipschitz functionsrdquoCanadian Journal of MathematicsJournal Canadien de Mathematiques vol 20 pp 1150ndash11641968

[3] M Cambern ldquoIsometries of certain Banach algebrasrdquo StudiaMathematica vol 25 pp 217ndash225 19641965

[4] N V Rao and A K Roy ldquoLinear isometries of some functionspacesrdquo Pacific Journal of Mathematics vol 38 pp 177ndash192 1971

[5] K Jarosz ldquoIsometries in semisimple commutative BanachalgebrasrdquoProceedings of the AmericanMathematical Society vol94 no 1 pp 65ndash71 1985

[6] K Jarosz and V D Pathak ldquoIsometries between functionspacesrdquo Transactions of the AmericanMathematical Society vol305 no 1 pp 193ndash206 1988

[7] NWeaver ldquoIsometries of noncompact Lipschitz spacesrdquoCana-dian Mathematical Bulletin Bulletin Canadien de Mathemat-iques vol 38 no 2 pp 242ndash249 1995

[8] A Jimenez-Vargas and M Villegas-Vallecillos ldquoLinear isome-tries between spaces of vector-valued Lipschitz functionsrdquoProceedings of the American Mathematical Society vol 137 no4 pp 1381ndash1388 2009

[9] A Jimenez-Vargas and M Villegas-Vallecillos ldquoInto linearisometries between spaces of Lipschitz functionsrdquo HoustonJournal of Mathematics vol 34 no 4 pp 1165ndash1184 2008

[10] F Botelho and J Jamison ldquoSurjective isometries on spacesof differentiable vector-valued functionsrdquo Studia Mathematicavol 192 no 1 pp 39ndash50 2009

[11] E Mayer-Wolf ldquoIsometries between Banach spaces of Lipschitzfunctionsrdquo Israel Journal of Mathematics vol 38 no 1-2 pp 58ndash74 1981

[12] A Jimenez-Vargas M Villegas-Vallecillos and Y-S WangldquoBanach-Stone theorems for vector-valued little Lipschitz func-tionsrdquo Publicationes Mathematicae vol 74 no 1-2 pp 81ndash1002009

[13] J Araujo andLDubarbie ldquoNoncompactness andnoncomplete-ness in isometries of Lipschitz spacesrdquo Journal of MathematicalAnalysis and Applications vol 377 no 1 pp 15ndash29 2011

[14] F Botelho R J Fleming and J Jamison ldquoExtreme points andisometries on vector-valued Lipschitz spacesrdquo Journal of Math-ematical Analysis and Applications vol 381 no 2 pp 821ndash8322011

[15] H Koshimizu ldquoLinear isometries on spaces of continuouslydifferentiable and Lipschitz continuous functionsrdquo NihonkaiMathematical Journal vol 22 no 2 pp 73ndash90 2011

[16] F Botelho J Jamison and B Zheng ldquoIsometries on spaces ofvector valued Lipschitz functionsrdquo Positivity An InternationalMathematics Journal Devoted to Theory and Applications ofPositivity vol 17 no 1 pp 47ndash65 2013

[17] A Ranjbar-Motlagh ldquoA note on isometries of Lipschitz spacesrdquoJournal of Mathematical Analysis and Applications vol 411 no2 pp 555ndash558 2014

[18] F Botelho and J Jamison ldquoSurjective isometries on spaces ofvector valued continuous and Lipschitz functionsrdquo PositivityAn International Mathematics Journal Devoted to Theory andApplications of Positivity vol 17 no 3 pp 395ndash405 2013Erratum to Surjective isometries on spaces of vector valuedcontinuous and Lipschitz functions by F Botelho 20 (2016)757ndash759

[19] T Miura and H Takagi ldquoSurjective isometries on the Banachspace of continuously differentiable functionsrdquo ContemporaryMathematics vol 687 pp 181ndash192 2017

[20] K Kawamura ldquoBanach-Stone type theorems for 1198621-functionspaces over Riemannian manifoldsrdquo Acta Universitatis Szegedi-ensis Acta Scientiarum Mathematicarum vol 83 no 3-4 pp551ndash591 2017

[21] K Kawamura ldquoPerturbations of norms on 1198621-function spacesand associated isometry groupsrdquo Topology Proceedings vol 51pp 169ndash196 2018

[22] K Kawamura ldquoA Banach-Stone type theorem for 1198621-functionspaces over the circlerdquo Topology Proceedings vol 53 pp 15ndash262019

[23] L Li D Chen Q Meng and Y-S Wang ldquoSurjective isometrieson vector-valued differentiable function spacesrdquo Annals ofFunctional Analysis vol 9 no 3 pp 334ndash343 2018

[24] K Kawamura H Koshimizu and T Miura ldquoNorms onC1([01]) and there isometriesrdquo Acta Scientiarum Mathemati-carum vol 84 no 12 pp 239ndash261 2018

[25] L Li A M Peralta L Wang and Y-S Wang ldquoWeak-2-local isometries on uniform algebras and Lipschitz algebrasrdquohttpsarxivorgabs170503619

[26] A Jimenez-Vargas L LiAMPeralta LWang andY-SWangldquo2-local standard isometries on vector-valued Lipschitz func-tion spacesrdquo Journal of Mathematical Analysis and Applicationsvol 461 no 2 pp 1287ndash1298 2018

[27] A Ranjbar-Motlagh ldquoIsometries of Lipschitz type functionspacesrdquoMathematischeNachrichten vol 291 no 11-12 pp 1899ndash1907 2018

14 Journal of Function Spaces

[28] N Weaver Lipschitz Algebras World Scientific Publishing CoInc River Edge NJ USA 1999

[29] R R Phelps Lectures on Choquetrsquos theorem vol 1757 of LectureNotes in Mathematics Springer-Verlag Berlin Germany 2ndedition 2001

[30] O Hatori and S Oi ldquoIsometries on Banach algebras of vector-valued mapsrdquo Acta Scientiarum Mathematicarum vol 84 no12 pp 151ndash183 2018

[31] O Hatori S Oi and H Takagi ldquoPeculiar homomorphismsbetween algebras of vector-valued mapsrdquo Studia Mathematicavol 242 no 2 pp 141ndash163 2018

[32] M Nagasawa ldquoIsomorphisms between commutative Banachalgebras with an application to rings of analytic functionsrdquoKodai Mathematical Seminar Reports vol 11 pp 182ndash188 1959

[33] K de LeeuwW Rudin and J Wermer ldquoThe isometries of somefunction spacesrdquo Proceedings of the American MathematicalSociety vol 11 pp 694ndash698 1960

[34] OHatori A Jimenez-Vargas andMVillegas-Vallecillos ldquoMapswhich preserve norms of non-symmetrical quotients betweengroups of exponentials of Lipschitz functionsrdquo Journal of Math-ematical Analysis and Applications vol 415 no 2 pp 825ndash8452014

[35] G Lumer ldquoOn the isometries of reflexiveOrlicz spacesrdquoAnnalesde lrsquoInstitut Fourier vol 68 pp 99ndash109 1963

[36] R J Fleming and J E Jamison Isometries on Banach SpacesFunction Spaces vol 129 ofMonographs and Surveys in Pure andApplied Mathematics Chapman amp Hall CRC Boca Raton FlaUSA 2003

[37] OHatori and SOi ldquoHermitian operators onBanach algebras ofvector-valued Lipschitzmapsrdquo Journal ofMathematical Analysisand Applications vol 452 no 1 pp 378ndash387 2017 Corrigendumto rdquoHermitian operators on Banach algebras of vector-valuedLipschitz mapsrdquo Journal of Mathematical Analysis and Appli-cations 452 (2017) 378ndash387 MR3628025

[38] A Nikou andA G OrsquoFarrell ldquoBanach algebras of vector-valuedfunctionsrdquo Glasgow Mathematical Journal vol 56 no 2 pp419ndash426 2014

[39] F Botelho and J Jamison ldquoHomomorphisms on a class ofcommutative Banach algebrasrdquo Rocky Mountain Journal ofMathematics vol 43 no 2 pp 395ndash416 2013

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Page 10: Hermitian Operators and Isometries on Banach Algebras of ...downloads.hindawi.com/journals/jfs/2018/8085304.pdfJournalofFunctionSpaces normiseitherthemaxnormorthesumnorm.esituation

10 Journal of Function Spaces

Note that a Hermitian element of a unital Banach algebraand a Hermitian operator are usually defined in terms ofnumerical range or semi-inner product Here we define themby an equivalent form (see [36]) By the definition of aHermitian operator we have the following

Proposition 8 Let 119864119895 be a complex Banach space for 119895 = 1 2Suppose that 119881 1198641 997888rarr 1198642 is a surjective isometry and 119867 1198641 997888rarr 1198641 is a Hermitian operator Then 119881119867119881minus1 1198642 997888rarr 1198642

is a Hermitian operator

Proposition 9 An element 119865 isin 119861 is Hermitian if and only ifthere exists 119891 isin 119860 cap 119862R(119884) such that 119865 = 1 otimes 119891Proof Suppose that 119865 isin 119861 is a Hermitian element Then1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884) + 1003817100381710038171003817119863 (exp (119894119905119865))1003817100381710038171003817infin(M)

= 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817 = 1 (105)

for every 119905 isin R Suppose that there exists a point (119909 119910) isin119883 times 119884 with Im119865(119909 119910) = 0 where Im denotes the imaginarypart of a complex number Suppose that Im119865(119909 119910) gt 0Then1003817100381710038171003817exp (minus119894119865)1003817100381710038171003817infin(119883times119884) ge 1003816100381610038161003816exp (minus119894119865 (119909 119910))1003816100381610038161003816

= exp (Im119865 (119909 119910)) gt 1 (106)

Suppose that Im119865(119909 119910) lt 0 Then1003817100381710038171003817exp (119894119865)1003817100381710038171003817infin(119883times119884) ge 1003816100381610038161003816exp (119894119865 (119909 119910))1003816100381610038161003816= exp (minusIm119865 (119909 119910)) gt 1 (107)

In any case we have there exists 119905 isin R such that1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817119861 ge 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884) gt 1 (108)

which contradicts our assumption We have that

119865 isin 119862R (119883 times 119884) (109)

Thus for every (119904 119905) isin 119883 times 119884 and 119905 isin R |exp(119894119905119865(119904 119905))| = 1Hence exp(119894119905119865)infin(119883times119884) = 1 for every 119905 isin R By (105)we have119863(exp(119894119905119865)infin(M) = 0 which ensures that 119863(exp(119894119905119865)) = 0for every 119905 isin R Thus exp(119894119905119865) isin 1 otimes 119860 for every 119905 isin R Wehave

exp (119894119905119865) minus 1119905 minus 119894119865 = infinsum119899=2

((119894119865)119899 119905119899minus2119899 ) 119905 (110)

and hence for every 119905 isin R with |119905| le 1 we have10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817 le (

infinsum119899=2

119865119899119861|119905|119899minus2119899 ) |119905|

le (infinsum119899=2

119865119899119861119899 ) |119905| le (exp 119865119861) |119905|

(111)

It follows that10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817infin(119883times119884)

le 10038171003817100381710038171003817100381710038171003817exp (119894119905119865) minus 1119905 minus 11989411986510038171003817100381710038171003817100381710038171003817119861

997888rarr 0(112)

as 119905 997888rarr 0 Since (exp(119894119905119865)minus1)119905 isin 1otimes119860 for each 119905 isin R thereexists 119892119905 isin 119860 such that

exp (119894119905119865) minus 1119905 = 1 otimes 119892119905 (113)

By (112) we have10038171003817100381710038171 otimes 1198921119899 minus 1198941198651003817100381710038171003817infin(119883times119884)

997888rarr 0 (114)

as 119899 997888rarr infin We have that 1 otimes 1198921119899 is a Cauchy sequencein 119862(119883 times 119884) thus we infer that1198921119899 is a Cauchy sequence in119862(119884) Since 119860 is uniformly closed as it is a uniform algebrathere exists 119892 isin 119860 such that

10038171003817100381710038171198921119899 minus 1198921003817100381710038171003817infin(119884)997888rarr 0 (115)

and hence10038171003817100381710038171 otimes 1198921119899 minus 1 otimes 1198921003817100381710038171003817infin(119883times119884)

997888rarr 0 (116)

as 119899 997888rarr infin It follows by (114) that 119894119865 = 1 otimes 119892 thus119865 = 1 otimes (minus119894119892) isin 1 otimes 119860 (117)

By (109) we see that minus119894119892 isin 119862R(119884) thus we have 119891 = minus119894119892 isin119860 cap 119862R(119884) and 119865 = 1 otimes 119891Suppose conversely that 119861 ni 119865 = 1otimes119891 for119891 isin 119860cap119862R(119884)

We infer that 119865 isin 119862R(119883 times 119884) and |exp(119894119905(119865(119909 119910))| = 1 forevery 119905 isin R and (119909 119910) isin 119883 times 119884 Hence exp(119894119905119865)infin(119883times119884) = 1for every 119905 isin R Since

exp (119894119905119865) = exp (119894119905 (1 otimes 119891)) = 1 otimes exp (119894119905119891) isin 1 otimes 119860 (118)

we have 119863(exp(119894119905119865)) = 0 It follows that1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817 = 1003817100381710038171003817exp (119894119905119865)1003817100381710038171003817infin(119883times119884)

+ 1003817100381710038171003817119863 (exp (119894119905119865))1003817100381710038171003817infin(M) = 1 (119)

for every 119905 isin R We conclude that 119865 is a Hermitian elementin 119861

Note that 119891 isin 119860 is Hermitian if and only if119891 isin 119860cap119862R(119884)by [37 Proposition 5] Hence Proposition 9 asserts that 119865 is aHermitian element in119861 if and only if119865 = 1otimes119891 for aHermitianelement 119891 in 119860Proposition 10 Suppose that 119880 119861 997888rarr 119861 is a surjectiveunital isometry Then 119880 is an algebra isomorphism

Proof As we have already mentioned 119861 is a regular subspace(in the sense of Jarosz) with a natural norm Then byTheorem 1 119880 is also an isometry with respect to the supre-mum norm on 119883 times 119884 Then 119880 is uniquely extended to asurjective isometry with respect to the supremum norm from the uniform closure 119861 onto itself Since 119861 is a uniformalgebra a theorem of Nagasawa [32] asserts that is analgebra isomorphism since (1) = 1 Thus 119880 is an algebraisomorphism from 119861 onto itself

Journal of Function Spaces 11

Theorem 11 A bounded operator 119879 119861 997888rarr 119861 is a Hermitianoperator if and only if 119879(1) is a Hermitian element in 119861 and119879 = 119872119879(1) the multiplication operator by 119879(1)Proof By Proposition 10 every surjective unital isometry on119861 is multiplicative Then by [37 Theorem 4] we have theconclusion

5 Banach Algebras of 119862(119884)-Valued Maps

Suppose that119883 is a compact Hausdorff space Suppose that 119861is a unital point separating subalgebra of119862(119884) equipped witha Banach algebra norm Then 119861 is semisimple because 119891 isin119861 119891(119909) = 0 is a maximal ideal of 119861 for every 119909 isin 119883 and theJacobson radical of 119861 vanishes The inequality 119891infin le 119891119861for every 119891 isin 119861 is well known We say that 119861 is natural if themap 119890 119884 997888rarr 119872119861 defined by 119910 997891997888rarr 120601119910 where 120601119910(119891) = 119891(119910)for every 119891 isin 119861 is bijective We say that 119861 is self-adjoint if 119861 isnatural and conjugate-closed in the sense that 119891 isin 119861 impliesthat 119891 isin 119861 for every 119891 isin 119861 where sdot denotes the complexconjugation on 119884Definition 12 Let 119883 and 119884 be compact Hausdorff spacesSuppose that 119861 is a unital point separating subalgebra of119862(119883)equipped with a Banach algebra norm sdot 119861 Suppose that119861 is self-adjoint Suppose that 119861 is a unital point separatingsubalgebra of 119862(119883 times 119884) such that 119861 otimes 119862(119884) sub 119861 equippedwith a Banach algebra norm sdot 119861 Suppose that 119861 is self-adjoint We say that 119861 is a natural 119862(119884)-valuezation of 119861 ifthere exists a compact Hausdorff space M and a complex-linear map 119863 119861 997888rarr 119862(M) such that ker119863 = 1 otimes 119862(119884) and119863(119862R(119883 times 119884) cap 119861) sub 119862R(M) which satisfies

119865 = 119865infin(119883times119884) + 119863 (119865)infin(M) 119865 isin 119861 (120)

The term ldquoa natural 119862(119884)-valuezation of 119861rdquo comes fromthe natural norm defined by Jarosz [5] In fact the norm sdot 119861is a natural norm in the sense of Jarosz [5]

Note that (119883119862(119884) 119861 119861) need not be an admissiblequadruple defined by Nikou and OrsquoFarrell [38] (cf [31]) sincewe do not assume that 119865(sdot 119910) 119865 isin 119861 119910 isin 119884 sub 119861 whichis a requirement for the admissible quadruple On the otherhand if (119883119862(119884) 119861 119861) is an admissible quadruple of type Ldefined in [30] then 119861 is a natural 119862(119884)-valuezation of 119861 dueto Definition 12

Example 13 Let 119861 = 1198621([0 1]) and 119861 = Lip([0 1] 119862(119884))for 119884 = 119901 a singleton Then Lip([0 1]) is algebraically iso-morphic to Lip([0 1] 119862(119884)) Suppose thatM is the maximalideal space of 119871infin([0 1]) and 119863 119861 997888rarr 119862(M) is definedby 119891 997891997888rarr Γ(1198911015840) where Γ denotes the Gelfand transformin 119871infin([0 1]) Then 119861 is a natural 119862(119884)-valuezation of 119861The Banach algebra Lip([0 1]) with the norm 119891infin([01]) +1198911015840infin([01]) is isometrically isomorphic to 119861

Let 119884 be a compact Hausdorff space Note that a closedsubalgebra 119861 of Lip((119870 119889120572) 119862(119884))which appears in Example12 in [30] is an example of a natural 119862(119884)-valuezation of 119861

The Banach algebras1198621([0 1] 119862(119884)) and 1198621(T 119862(119884)))whichappear in Examples 16 and 17 in [30] respectively are alsoexamples of natural 119862(119884)-valuezations of 1198621([0 1])6 Isometries on Natural 119862(119884)-ValuezationsThemain theorem in this paper is the following

Theorem 14 Suppose that 119861119895 is a natural 119862(119884119895)-valuezationof 119861119895 sub 119862(119883119895) for 119895 = 1 2 We assume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (121)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for119895 = 1 2 Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linear isometry Then there exists ℎ isin 119862(1198842) such that |ℎ| = 1on 1198842 a continuous map 120593 1198832 times 1198842 997888rarr 1198831 such that120593(sdot 119910) 1198832 997888rarr 1198831 is a homeomorphism for each 119910 isin 1198842and a homeomorphism 120591 1198842 997888rarr 1198841 which satisfies119880 (119865) (119909 119910) = ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842 (122)

for every 119865 isin 1198611

In short a surjective isometry between 119862(119884)-valuezationsis a weighted composition operator of a specific form thehomeomorphism 1198832 times 1198842 997888rarr 1198831 times 1198841 (119909 119910) 997891997888rarr(120593(119909 119910) 120591(119910)) has the second coordinate that depends onlyon the second variable 119910 isin 1198842 A composition operatorinduced by such a homeomorphism is said to be of type BJin [31 37] after the study of Botelho and Jamison [39]

Quite recently the author of this paper and Oi [30 Theo-rem 8] proved a similar result of Theorem 14 for admissiblequadruples of type L To prove it we apply Proposition 32and the following comments in [31] Instead of this we proveTheorem 14 by Lumerrsquosmethod with which a proof is simplerthan that in [30]

In the following in this section we assume that 119861119895 is anatural 119862(119884119895)-valuezation of 119861 sub 119862(119883119895) for 119895 = 1 2 Weassume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (123)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for 119895 = 1 2Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linearisometry A crucial part of a proof of Theorem 14 is to proveProposition 15

Proposition 15 Suppose that 1198832 is not a singleton Thereexists ℎ isin 119862(1198842) with |ℎ| = 1 on 1198842 such that 119880(11198611) = 11198612 otimesℎ

A similar result for admissible quadruples of type L isproved in [30 Proposition 9] If we assumed that

119865 (sdot 119910) 119865 isin 119861119895 119910 isin 119884119895 sub 119861119895 (124)

then 119861119895 were an admissible quadruple of type L Although119861119895 in this paper need not be an admissible quadruple of type

12 Journal of Function Spaces

L a proof of Proposition 15 is completely the same as that in[30 Proposition 9] since we do not make use of the condition(124) in the proof of [30 Proposition 9] The condition (124)is needed in [30] when we apply Proposition 32 and thefollowing comments in [31]

7 Proof of Theorem 14 An Application ofLumerrsquos Method

Proof of Theorem 14 A proof for the case where 1198831 = 1199091and 1198832 = 1199092 are singletons is the same as the proof ofTheorem 8 in [30]

Suppose that1198832 is not a singleton By Proposition 15 thereexists ℎ isin 119862R(1198842) with |ℎ| = 1 on 1198842 such that 119880(1) =1 otimes ℎ Letting 1198800 1198611 997888rarr 1198612 by 1198800(119865) = (1 otimes ℎ)119880(119865)119865 isin 1198611 we see by the hypothesis (1 otimes ℎ)1198651198612 = 1198651198612 forevery 119865 isin 1198612 that 1198800 is a surjective unital isometry from1198611 onto 1198612 Then Corollary 2 asserts that 1198800 is an algebraisomorphism Let 119891 isin 119862R(1198841) By Proposition 9 1 otimes 119891 isa Hermitian element in 1198611 Then by Theorem 11 1198721otimes119891 is aHermitian operator on 1198611 By Proposition 8 11988001198721otimes119891119880minus1

0 is aHermitian operator on 1198612 Then by Theorem 11 there exists119878(119891) isin 119862R(1198842) such that 11988001198721otimes119891119880minus1

0 = 1198721otimes119878(119891) Hence anoperator 119878 119862R(1198841) 997888rarr 119862R(1198842) is defined Since 1198800 is analgebra isomorphism it is easy to see that 119878 is a real algebraisomorphism from 119862R(1198841) onto 119862R(1198842) Then 119878 119862(1198841) 997888rarr119862(1198842) defined by 119878(119891) = 119878(Re119891) + 119894119878(Im119891) for 119891 isin 119862(1198841)gives a complex algebra isomorphism Gelfand theory assertsthat there is a homeomorphism 120591 1198842 997888rarr 1198841 such that119878(119891) = 119891 ∘ 120591 119891 isin 119862(1198841) It follows that

11988001198721otimes119891119880minus10 = 1198721otimes119891∘120591 119891 isin 119862 (1198841) (125)

Since 119880minus10 (1) = 1 we have

1198800 (1 otimes 119891) = 1 otimes 119891 ∘ 120591 119891 isin 119862 (1198841) (126)

Define Φ 1198611 997888rarr 1198612 by Φ(119886) = 1198800(119886 otimes 1) 119886 isin 1198611Since 1198800 is an algebra isomorphism the map Φ is a unitalhomomorphism Since the maximal ideal space of 1198611 is 1198831

and the maximal ideal space of 1198612 is 1198832 times 1198842 there is acontinuous map 120593 1198832 times 1198842 997888rarr 1198831 such that

Φ(119886) (119909 119910) = 119886 (120593 (119909 119910)) 119886 isin 1198611 (119909 119910) isin 1198832 times 1198842 (127)

It follows by (126) and (127) that

1198800 (119886 otimes 119891) (119909 119910) = 1198800 ((119886 otimes 1) (1 otimes 119891)) (119909 119910)= 1198800 (119886 otimes 1) (119909 119910)1198800 (1 otimes 119891) (119909 119910)= 119886 (120593 (119909 119910))119891 (120591 (119910))= (119886 otimes 119891) (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198832 times 1198842

(128)

for every 119886 isin 1198611 and 119891 isin 119862(1198841) Thus

1198800 (sum(119886119895 otimes 119891119895)) (119909 119910)= (sum(119886119895 otimes 119891119895)) (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842(129)

for every sum(119886119895 otimes 119891119895) isin 1198611 otimes 119862(1198841) By the Stone-Weierstrasstheorem 1198611 otimes 119862(1198841) is uniformly dense in 119862(1198831 times 1198841) henceany element in 1198611 is uniformly approximated by 1198611 otimes 119862(1198841)As 1198800 is also an isometry with respect to the uniform normwe see that

1198800 (119865) (119909 119910) = 119865 (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198831 times 1198841 (130)

for every 119865 isin 1198611 and

119880 (119865) (119909 119910) = (1 otimes ℎ) (119909 119910) 119865 (120593 (119909 119910) 120591 (119910))= ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198831 times 1198841(131)

As 1198800 is an algebra isomorphism the map 1198832 times 1198842 997888rarr1198831 times 1198841 defined by (119909 119910) 997891997888rarr (120593(119909 119910) 120591(119910)) gives ahomeomorphism Therefore for every 119910 isin 1198842 the map

120593 (sdot 119910) 1198832 997888rarr 1198831 (132)

is a homeomorphismSuppose that 1198831 is not a singleton By the same way as

in the last part of the proof of Theorem 8 in [30] we havethat 1198832 is not a singleton Then we have the conclusion bythe previous argument

8 Application of Theorem 14

We exhibit applications of Theorem 14

Corollary 16 ([4 Theorem 33]) Suppose that119880 Lip([0 1]) 997888rarr Lip([0 1]) is a surjective isometrywith respect to the norm defined by 119891infin([01]) + 1198911015840infin([01])

for 119891 isin Lip([0 1]) Then 119880(1) is a constant function of unitmodulus such that

119880(119891) (119909) = 119880 (1) 119891 (119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (133)

or

119880(119891) (119909) = 119880 (1) 119891 (1 minus 119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (134)

The converse statement also holds

Proof By Example 13 we may suppose that Lip([0 1]) is aBanach algebra of 119862(119884)-valuezation Applying Theorem 14we have that119880(1) = 1otimesℎ for ℎ isin 119862(119884)with |ℎ| = 1 Since our

Journal of Function Spaces 13

119884 is a singleton 119880(1) is a constant function of unit modulusWe also see that the corresponding continuous map 120593 [0 1] times 119884 997888rarr [0 1] can be considered as a homeomorphismfrom [0 1] onto [0 1] therefore we have that119880(119891) (119909) = 119880 (1) 119891 (120593 (119909))

119891 isin Lip ([0 1]) 119909 isin [0 1] (135)

The rest is a routine argument to prove that 120593 is an isometryhence 120593(119909) = 119909 119909 isin [0 1] or 120593(119909) = 1 minus 119909 119909 isin [0 1]

The converse statement is trivial

Corollaries 14 15 18 and 19 in [30 Section 6] follow herewith a similar proof

Data Availability

The data used to support the findings of this study areavailable from the corresponding author upon request

Conflicts of Interest

The author declares that they have no conflicts of interest

Acknowledgments

This work was supported by JSPS KAKENHI Grants Num-bers JP16K05172 and JP15K04921

References

[1] K de Leeuw ldquoBanach spaces of Lipschitz functionsrdquo StudiaMathematica vol 21 pp 55ndash66 19611962

[2] A K Roy ldquoExtreme points and linear isometries of the Banachspace of Lipschitz functionsrdquoCanadian Journal of MathematicsJournal Canadien de Mathematiques vol 20 pp 1150ndash11641968

[3] M Cambern ldquoIsometries of certain Banach algebrasrdquo StudiaMathematica vol 25 pp 217ndash225 19641965

[4] N V Rao and A K Roy ldquoLinear isometries of some functionspacesrdquo Pacific Journal of Mathematics vol 38 pp 177ndash192 1971

[5] K Jarosz ldquoIsometries in semisimple commutative BanachalgebrasrdquoProceedings of the AmericanMathematical Society vol94 no 1 pp 65ndash71 1985

[6] K Jarosz and V D Pathak ldquoIsometries between functionspacesrdquo Transactions of the AmericanMathematical Society vol305 no 1 pp 193ndash206 1988

[7] NWeaver ldquoIsometries of noncompact Lipschitz spacesrdquoCana-dian Mathematical Bulletin Bulletin Canadien de Mathemat-iques vol 38 no 2 pp 242ndash249 1995

[8] A Jimenez-Vargas and M Villegas-Vallecillos ldquoLinear isome-tries between spaces of vector-valued Lipschitz functionsrdquoProceedings of the American Mathematical Society vol 137 no4 pp 1381ndash1388 2009

[9] A Jimenez-Vargas and M Villegas-Vallecillos ldquoInto linearisometries between spaces of Lipschitz functionsrdquo HoustonJournal of Mathematics vol 34 no 4 pp 1165ndash1184 2008

[10] F Botelho and J Jamison ldquoSurjective isometries on spacesof differentiable vector-valued functionsrdquo Studia Mathematicavol 192 no 1 pp 39ndash50 2009

[11] E Mayer-Wolf ldquoIsometries between Banach spaces of Lipschitzfunctionsrdquo Israel Journal of Mathematics vol 38 no 1-2 pp 58ndash74 1981

[12] A Jimenez-Vargas M Villegas-Vallecillos and Y-S WangldquoBanach-Stone theorems for vector-valued little Lipschitz func-tionsrdquo Publicationes Mathematicae vol 74 no 1-2 pp 81ndash1002009

[13] J Araujo andLDubarbie ldquoNoncompactness andnoncomplete-ness in isometries of Lipschitz spacesrdquo Journal of MathematicalAnalysis and Applications vol 377 no 1 pp 15ndash29 2011

[14] F Botelho R J Fleming and J Jamison ldquoExtreme points andisometries on vector-valued Lipschitz spacesrdquo Journal of Math-ematical Analysis and Applications vol 381 no 2 pp 821ndash8322011

[15] H Koshimizu ldquoLinear isometries on spaces of continuouslydifferentiable and Lipschitz continuous functionsrdquo NihonkaiMathematical Journal vol 22 no 2 pp 73ndash90 2011

[16] F Botelho J Jamison and B Zheng ldquoIsometries on spaces ofvector valued Lipschitz functionsrdquo Positivity An InternationalMathematics Journal Devoted to Theory and Applications ofPositivity vol 17 no 1 pp 47ndash65 2013

[17] A Ranjbar-Motlagh ldquoA note on isometries of Lipschitz spacesrdquoJournal of Mathematical Analysis and Applications vol 411 no2 pp 555ndash558 2014

[18] F Botelho and J Jamison ldquoSurjective isometries on spaces ofvector valued continuous and Lipschitz functionsrdquo PositivityAn International Mathematics Journal Devoted to Theory andApplications of Positivity vol 17 no 3 pp 395ndash405 2013Erratum to Surjective isometries on spaces of vector valuedcontinuous and Lipschitz functions by F Botelho 20 (2016)757ndash759

[19] T Miura and H Takagi ldquoSurjective isometries on the Banachspace of continuously differentiable functionsrdquo ContemporaryMathematics vol 687 pp 181ndash192 2017

[20] K Kawamura ldquoBanach-Stone type theorems for 1198621-functionspaces over Riemannian manifoldsrdquo Acta Universitatis Szegedi-ensis Acta Scientiarum Mathematicarum vol 83 no 3-4 pp551ndash591 2017

[21] K Kawamura ldquoPerturbations of norms on 1198621-function spacesand associated isometry groupsrdquo Topology Proceedings vol 51pp 169ndash196 2018

[22] K Kawamura ldquoA Banach-Stone type theorem for 1198621-functionspaces over the circlerdquo Topology Proceedings vol 53 pp 15ndash262019

[23] L Li D Chen Q Meng and Y-S Wang ldquoSurjective isometrieson vector-valued differentiable function spacesrdquo Annals ofFunctional Analysis vol 9 no 3 pp 334ndash343 2018

[24] K Kawamura H Koshimizu and T Miura ldquoNorms onC1([01]) and there isometriesrdquo Acta Scientiarum Mathemati-carum vol 84 no 12 pp 239ndash261 2018

[25] L Li A M Peralta L Wang and Y-S Wang ldquoWeak-2-local isometries on uniform algebras and Lipschitz algebrasrdquohttpsarxivorgabs170503619

[26] A Jimenez-Vargas L LiAMPeralta LWang andY-SWangldquo2-local standard isometries on vector-valued Lipschitz func-tion spacesrdquo Journal of Mathematical Analysis and Applicationsvol 461 no 2 pp 1287ndash1298 2018

[27] A Ranjbar-Motlagh ldquoIsometries of Lipschitz type functionspacesrdquoMathematischeNachrichten vol 291 no 11-12 pp 1899ndash1907 2018

14 Journal of Function Spaces

[28] N Weaver Lipschitz Algebras World Scientific Publishing CoInc River Edge NJ USA 1999

[29] R R Phelps Lectures on Choquetrsquos theorem vol 1757 of LectureNotes in Mathematics Springer-Verlag Berlin Germany 2ndedition 2001

[30] O Hatori and S Oi ldquoIsometries on Banach algebras of vector-valued mapsrdquo Acta Scientiarum Mathematicarum vol 84 no12 pp 151ndash183 2018

[31] O Hatori S Oi and H Takagi ldquoPeculiar homomorphismsbetween algebras of vector-valued mapsrdquo Studia Mathematicavol 242 no 2 pp 141ndash163 2018

[32] M Nagasawa ldquoIsomorphisms between commutative Banachalgebras with an application to rings of analytic functionsrdquoKodai Mathematical Seminar Reports vol 11 pp 182ndash188 1959

[33] K de LeeuwW Rudin and J Wermer ldquoThe isometries of somefunction spacesrdquo Proceedings of the American MathematicalSociety vol 11 pp 694ndash698 1960

[34] OHatori A Jimenez-Vargas andMVillegas-Vallecillos ldquoMapswhich preserve norms of non-symmetrical quotients betweengroups of exponentials of Lipschitz functionsrdquo Journal of Math-ematical Analysis and Applications vol 415 no 2 pp 825ndash8452014

[35] G Lumer ldquoOn the isometries of reflexiveOrlicz spacesrdquoAnnalesde lrsquoInstitut Fourier vol 68 pp 99ndash109 1963

[36] R J Fleming and J E Jamison Isometries on Banach SpacesFunction Spaces vol 129 ofMonographs and Surveys in Pure andApplied Mathematics Chapman amp Hall CRC Boca Raton FlaUSA 2003

[37] OHatori and SOi ldquoHermitian operators onBanach algebras ofvector-valued Lipschitzmapsrdquo Journal ofMathematical Analysisand Applications vol 452 no 1 pp 378ndash387 2017 Corrigendumto rdquoHermitian operators on Banach algebras of vector-valuedLipschitz mapsrdquo Journal of Mathematical Analysis and Appli-cations 452 (2017) 378ndash387 MR3628025

[38] A Nikou andA G OrsquoFarrell ldquoBanach algebras of vector-valuedfunctionsrdquo Glasgow Mathematical Journal vol 56 no 2 pp419ndash426 2014

[39] F Botelho and J Jamison ldquoHomomorphisms on a class ofcommutative Banach algebrasrdquo Rocky Mountain Journal ofMathematics vol 43 no 2 pp 395ndash416 2013

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Page 11: Hermitian Operators and Isometries on Banach Algebras of ...downloads.hindawi.com/journals/jfs/2018/8085304.pdfJournalofFunctionSpaces normiseitherthemaxnormorthesumnorm.esituation

Journal of Function Spaces 11

Theorem 11 A bounded operator 119879 119861 997888rarr 119861 is a Hermitianoperator if and only if 119879(1) is a Hermitian element in 119861 and119879 = 119872119879(1) the multiplication operator by 119879(1)Proof By Proposition 10 every surjective unital isometry on119861 is multiplicative Then by [37 Theorem 4] we have theconclusion

5 Banach Algebras of 119862(119884)-Valued Maps

Suppose that119883 is a compact Hausdorff space Suppose that 119861is a unital point separating subalgebra of119862(119884) equipped witha Banach algebra norm Then 119861 is semisimple because 119891 isin119861 119891(119909) = 0 is a maximal ideal of 119861 for every 119909 isin 119883 and theJacobson radical of 119861 vanishes The inequality 119891infin le 119891119861for every 119891 isin 119861 is well known We say that 119861 is natural if themap 119890 119884 997888rarr 119872119861 defined by 119910 997891997888rarr 120601119910 where 120601119910(119891) = 119891(119910)for every 119891 isin 119861 is bijective We say that 119861 is self-adjoint if 119861 isnatural and conjugate-closed in the sense that 119891 isin 119861 impliesthat 119891 isin 119861 for every 119891 isin 119861 where sdot denotes the complexconjugation on 119884Definition 12 Let 119883 and 119884 be compact Hausdorff spacesSuppose that 119861 is a unital point separating subalgebra of119862(119883)equipped with a Banach algebra norm sdot 119861 Suppose that119861 is self-adjoint Suppose that 119861 is a unital point separatingsubalgebra of 119862(119883 times 119884) such that 119861 otimes 119862(119884) sub 119861 equippedwith a Banach algebra norm sdot 119861 Suppose that 119861 is self-adjoint We say that 119861 is a natural 119862(119884)-valuezation of 119861 ifthere exists a compact Hausdorff space M and a complex-linear map 119863 119861 997888rarr 119862(M) such that ker119863 = 1 otimes 119862(119884) and119863(119862R(119883 times 119884) cap 119861) sub 119862R(M) which satisfies

119865 = 119865infin(119883times119884) + 119863 (119865)infin(M) 119865 isin 119861 (120)

The term ldquoa natural 119862(119884)-valuezation of 119861rdquo comes fromthe natural norm defined by Jarosz [5] In fact the norm sdot 119861is a natural norm in the sense of Jarosz [5]

Note that (119883119862(119884) 119861 119861) need not be an admissiblequadruple defined by Nikou and OrsquoFarrell [38] (cf [31]) sincewe do not assume that 119865(sdot 119910) 119865 isin 119861 119910 isin 119884 sub 119861 whichis a requirement for the admissible quadruple On the otherhand if (119883119862(119884) 119861 119861) is an admissible quadruple of type Ldefined in [30] then 119861 is a natural 119862(119884)-valuezation of 119861 dueto Definition 12

Example 13 Let 119861 = 1198621([0 1]) and 119861 = Lip([0 1] 119862(119884))for 119884 = 119901 a singleton Then Lip([0 1]) is algebraically iso-morphic to Lip([0 1] 119862(119884)) Suppose thatM is the maximalideal space of 119871infin([0 1]) and 119863 119861 997888rarr 119862(M) is definedby 119891 997891997888rarr Γ(1198911015840) where Γ denotes the Gelfand transformin 119871infin([0 1]) Then 119861 is a natural 119862(119884)-valuezation of 119861The Banach algebra Lip([0 1]) with the norm 119891infin([01]) +1198911015840infin([01]) is isometrically isomorphic to 119861

Let 119884 be a compact Hausdorff space Note that a closedsubalgebra 119861 of Lip((119870 119889120572) 119862(119884))which appears in Example12 in [30] is an example of a natural 119862(119884)-valuezation of 119861

The Banach algebras1198621([0 1] 119862(119884)) and 1198621(T 119862(119884)))whichappear in Examples 16 and 17 in [30] respectively are alsoexamples of natural 119862(119884)-valuezations of 1198621([0 1])6 Isometries on Natural 119862(119884)-ValuezationsThemain theorem in this paper is the following

Theorem 14 Suppose that 119861119895 is a natural 119862(119884119895)-valuezationof 119861119895 sub 119862(119883119895) for 119895 = 1 2 We assume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (121)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for119895 = 1 2 Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linear isometry Then there exists ℎ isin 119862(1198842) such that |ℎ| = 1on 1198842 a continuous map 120593 1198832 times 1198842 997888rarr 1198831 such that120593(sdot 119910) 1198832 997888rarr 1198831 is a homeomorphism for each 119910 isin 1198842and a homeomorphism 120591 1198842 997888rarr 1198841 which satisfies119880 (119865) (119909 119910) = ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842 (122)

for every 119865 isin 1198611

In short a surjective isometry between 119862(119884)-valuezationsis a weighted composition operator of a specific form thehomeomorphism 1198832 times 1198842 997888rarr 1198831 times 1198841 (119909 119910) 997891997888rarr(120593(119909 119910) 120591(119910)) has the second coordinate that depends onlyon the second variable 119910 isin 1198842 A composition operatorinduced by such a homeomorphism is said to be of type BJin [31 37] after the study of Botelho and Jamison [39]

Quite recently the author of this paper and Oi [30 Theo-rem 8] proved a similar result of Theorem 14 for admissiblequadruples of type L To prove it we apply Proposition 32and the following comments in [31] Instead of this we proveTheorem 14 by Lumerrsquosmethod with which a proof is simplerthan that in [30]

In the following in this section we assume that 119861119895 is anatural 119862(119884119895)-valuezation of 119861 sub 119862(119883119895) for 119895 = 1 2 Weassume that

(1 otimes ℎ) 119865119861119895 = 119865119861119895 (123)

for every 119865 isin 119861119895 and ℎ isin 119862(119884119895) with |ℎ| = 1 on 119884119895 for 119895 = 1 2Suppose that 119880 1198611 997888rarr 1198612 is a surjective complex-linearisometry A crucial part of a proof of Theorem 14 is to proveProposition 15

Proposition 15 Suppose that 1198832 is not a singleton Thereexists ℎ isin 119862(1198842) with |ℎ| = 1 on 1198842 such that 119880(11198611) = 11198612 otimesℎ

A similar result for admissible quadruples of type L isproved in [30 Proposition 9] If we assumed that

119865 (sdot 119910) 119865 isin 119861119895 119910 isin 119884119895 sub 119861119895 (124)

then 119861119895 were an admissible quadruple of type L Although119861119895 in this paper need not be an admissible quadruple of type

12 Journal of Function Spaces

L a proof of Proposition 15 is completely the same as that in[30 Proposition 9] since we do not make use of the condition(124) in the proof of [30 Proposition 9] The condition (124)is needed in [30] when we apply Proposition 32 and thefollowing comments in [31]

7 Proof of Theorem 14 An Application ofLumerrsquos Method

Proof of Theorem 14 A proof for the case where 1198831 = 1199091and 1198832 = 1199092 are singletons is the same as the proof ofTheorem 8 in [30]

Suppose that1198832 is not a singleton By Proposition 15 thereexists ℎ isin 119862R(1198842) with |ℎ| = 1 on 1198842 such that 119880(1) =1 otimes ℎ Letting 1198800 1198611 997888rarr 1198612 by 1198800(119865) = (1 otimes ℎ)119880(119865)119865 isin 1198611 we see by the hypothesis (1 otimes ℎ)1198651198612 = 1198651198612 forevery 119865 isin 1198612 that 1198800 is a surjective unital isometry from1198611 onto 1198612 Then Corollary 2 asserts that 1198800 is an algebraisomorphism Let 119891 isin 119862R(1198841) By Proposition 9 1 otimes 119891 isa Hermitian element in 1198611 Then by Theorem 11 1198721otimes119891 is aHermitian operator on 1198611 By Proposition 8 11988001198721otimes119891119880minus1

0 is aHermitian operator on 1198612 Then by Theorem 11 there exists119878(119891) isin 119862R(1198842) such that 11988001198721otimes119891119880minus1

0 = 1198721otimes119878(119891) Hence anoperator 119878 119862R(1198841) 997888rarr 119862R(1198842) is defined Since 1198800 is analgebra isomorphism it is easy to see that 119878 is a real algebraisomorphism from 119862R(1198841) onto 119862R(1198842) Then 119878 119862(1198841) 997888rarr119862(1198842) defined by 119878(119891) = 119878(Re119891) + 119894119878(Im119891) for 119891 isin 119862(1198841)gives a complex algebra isomorphism Gelfand theory assertsthat there is a homeomorphism 120591 1198842 997888rarr 1198841 such that119878(119891) = 119891 ∘ 120591 119891 isin 119862(1198841) It follows that

11988001198721otimes119891119880minus10 = 1198721otimes119891∘120591 119891 isin 119862 (1198841) (125)

Since 119880minus10 (1) = 1 we have

1198800 (1 otimes 119891) = 1 otimes 119891 ∘ 120591 119891 isin 119862 (1198841) (126)

Define Φ 1198611 997888rarr 1198612 by Φ(119886) = 1198800(119886 otimes 1) 119886 isin 1198611Since 1198800 is an algebra isomorphism the map Φ is a unitalhomomorphism Since the maximal ideal space of 1198611 is 1198831

and the maximal ideal space of 1198612 is 1198832 times 1198842 there is acontinuous map 120593 1198832 times 1198842 997888rarr 1198831 such that

Φ(119886) (119909 119910) = 119886 (120593 (119909 119910)) 119886 isin 1198611 (119909 119910) isin 1198832 times 1198842 (127)

It follows by (126) and (127) that

1198800 (119886 otimes 119891) (119909 119910) = 1198800 ((119886 otimes 1) (1 otimes 119891)) (119909 119910)= 1198800 (119886 otimes 1) (119909 119910)1198800 (1 otimes 119891) (119909 119910)= 119886 (120593 (119909 119910))119891 (120591 (119910))= (119886 otimes 119891) (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198832 times 1198842

(128)

for every 119886 isin 1198611 and 119891 isin 119862(1198841) Thus

1198800 (sum(119886119895 otimes 119891119895)) (119909 119910)= (sum(119886119895 otimes 119891119895)) (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842(129)

for every sum(119886119895 otimes 119891119895) isin 1198611 otimes 119862(1198841) By the Stone-Weierstrasstheorem 1198611 otimes 119862(1198841) is uniformly dense in 119862(1198831 times 1198841) henceany element in 1198611 is uniformly approximated by 1198611 otimes 119862(1198841)As 1198800 is also an isometry with respect to the uniform normwe see that

1198800 (119865) (119909 119910) = 119865 (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198831 times 1198841 (130)

for every 119865 isin 1198611 and

119880 (119865) (119909 119910) = (1 otimes ℎ) (119909 119910) 119865 (120593 (119909 119910) 120591 (119910))= ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198831 times 1198841(131)

As 1198800 is an algebra isomorphism the map 1198832 times 1198842 997888rarr1198831 times 1198841 defined by (119909 119910) 997891997888rarr (120593(119909 119910) 120591(119910)) gives ahomeomorphism Therefore for every 119910 isin 1198842 the map

120593 (sdot 119910) 1198832 997888rarr 1198831 (132)

is a homeomorphismSuppose that 1198831 is not a singleton By the same way as

in the last part of the proof of Theorem 8 in [30] we havethat 1198832 is not a singleton Then we have the conclusion bythe previous argument

8 Application of Theorem 14

We exhibit applications of Theorem 14

Corollary 16 ([4 Theorem 33]) Suppose that119880 Lip([0 1]) 997888rarr Lip([0 1]) is a surjective isometrywith respect to the norm defined by 119891infin([01]) + 1198911015840infin([01])

for 119891 isin Lip([0 1]) Then 119880(1) is a constant function of unitmodulus such that

119880(119891) (119909) = 119880 (1) 119891 (119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (133)

or

119880(119891) (119909) = 119880 (1) 119891 (1 minus 119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (134)

The converse statement also holds

Proof By Example 13 we may suppose that Lip([0 1]) is aBanach algebra of 119862(119884)-valuezation Applying Theorem 14we have that119880(1) = 1otimesℎ for ℎ isin 119862(119884)with |ℎ| = 1 Since our

Journal of Function Spaces 13

119884 is a singleton 119880(1) is a constant function of unit modulusWe also see that the corresponding continuous map 120593 [0 1] times 119884 997888rarr [0 1] can be considered as a homeomorphismfrom [0 1] onto [0 1] therefore we have that119880(119891) (119909) = 119880 (1) 119891 (120593 (119909))

119891 isin Lip ([0 1]) 119909 isin [0 1] (135)

The rest is a routine argument to prove that 120593 is an isometryhence 120593(119909) = 119909 119909 isin [0 1] or 120593(119909) = 1 minus 119909 119909 isin [0 1]

The converse statement is trivial

Corollaries 14 15 18 and 19 in [30 Section 6] follow herewith a similar proof

Data Availability

The data used to support the findings of this study areavailable from the corresponding author upon request

Conflicts of Interest

The author declares that they have no conflicts of interest

Acknowledgments

This work was supported by JSPS KAKENHI Grants Num-bers JP16K05172 and JP15K04921

References

[1] K de Leeuw ldquoBanach spaces of Lipschitz functionsrdquo StudiaMathematica vol 21 pp 55ndash66 19611962

[2] A K Roy ldquoExtreme points and linear isometries of the Banachspace of Lipschitz functionsrdquoCanadian Journal of MathematicsJournal Canadien de Mathematiques vol 20 pp 1150ndash11641968

[3] M Cambern ldquoIsometries of certain Banach algebrasrdquo StudiaMathematica vol 25 pp 217ndash225 19641965

[4] N V Rao and A K Roy ldquoLinear isometries of some functionspacesrdquo Pacific Journal of Mathematics vol 38 pp 177ndash192 1971

[5] K Jarosz ldquoIsometries in semisimple commutative BanachalgebrasrdquoProceedings of the AmericanMathematical Society vol94 no 1 pp 65ndash71 1985

[6] K Jarosz and V D Pathak ldquoIsometries between functionspacesrdquo Transactions of the AmericanMathematical Society vol305 no 1 pp 193ndash206 1988

[7] NWeaver ldquoIsometries of noncompact Lipschitz spacesrdquoCana-dian Mathematical Bulletin Bulletin Canadien de Mathemat-iques vol 38 no 2 pp 242ndash249 1995

[8] A Jimenez-Vargas and M Villegas-Vallecillos ldquoLinear isome-tries between spaces of vector-valued Lipschitz functionsrdquoProceedings of the American Mathematical Society vol 137 no4 pp 1381ndash1388 2009

[9] A Jimenez-Vargas and M Villegas-Vallecillos ldquoInto linearisometries between spaces of Lipschitz functionsrdquo HoustonJournal of Mathematics vol 34 no 4 pp 1165ndash1184 2008

[10] F Botelho and J Jamison ldquoSurjective isometries on spacesof differentiable vector-valued functionsrdquo Studia Mathematicavol 192 no 1 pp 39ndash50 2009

[11] E Mayer-Wolf ldquoIsometries between Banach spaces of Lipschitzfunctionsrdquo Israel Journal of Mathematics vol 38 no 1-2 pp 58ndash74 1981

[12] A Jimenez-Vargas M Villegas-Vallecillos and Y-S WangldquoBanach-Stone theorems for vector-valued little Lipschitz func-tionsrdquo Publicationes Mathematicae vol 74 no 1-2 pp 81ndash1002009

[13] J Araujo andLDubarbie ldquoNoncompactness andnoncomplete-ness in isometries of Lipschitz spacesrdquo Journal of MathematicalAnalysis and Applications vol 377 no 1 pp 15ndash29 2011

[14] F Botelho R J Fleming and J Jamison ldquoExtreme points andisometries on vector-valued Lipschitz spacesrdquo Journal of Math-ematical Analysis and Applications vol 381 no 2 pp 821ndash8322011

[15] H Koshimizu ldquoLinear isometries on spaces of continuouslydifferentiable and Lipschitz continuous functionsrdquo NihonkaiMathematical Journal vol 22 no 2 pp 73ndash90 2011

[16] F Botelho J Jamison and B Zheng ldquoIsometries on spaces ofvector valued Lipschitz functionsrdquo Positivity An InternationalMathematics Journal Devoted to Theory and Applications ofPositivity vol 17 no 1 pp 47ndash65 2013

[17] A Ranjbar-Motlagh ldquoA note on isometries of Lipschitz spacesrdquoJournal of Mathematical Analysis and Applications vol 411 no2 pp 555ndash558 2014

[18] F Botelho and J Jamison ldquoSurjective isometries on spaces ofvector valued continuous and Lipschitz functionsrdquo PositivityAn International Mathematics Journal Devoted to Theory andApplications of Positivity vol 17 no 3 pp 395ndash405 2013Erratum to Surjective isometries on spaces of vector valuedcontinuous and Lipschitz functions by F Botelho 20 (2016)757ndash759

[19] T Miura and H Takagi ldquoSurjective isometries on the Banachspace of continuously differentiable functionsrdquo ContemporaryMathematics vol 687 pp 181ndash192 2017

[20] K Kawamura ldquoBanach-Stone type theorems for 1198621-functionspaces over Riemannian manifoldsrdquo Acta Universitatis Szegedi-ensis Acta Scientiarum Mathematicarum vol 83 no 3-4 pp551ndash591 2017

[21] K Kawamura ldquoPerturbations of norms on 1198621-function spacesand associated isometry groupsrdquo Topology Proceedings vol 51pp 169ndash196 2018

[22] K Kawamura ldquoA Banach-Stone type theorem for 1198621-functionspaces over the circlerdquo Topology Proceedings vol 53 pp 15ndash262019

[23] L Li D Chen Q Meng and Y-S Wang ldquoSurjective isometrieson vector-valued differentiable function spacesrdquo Annals ofFunctional Analysis vol 9 no 3 pp 334ndash343 2018

[24] K Kawamura H Koshimizu and T Miura ldquoNorms onC1([01]) and there isometriesrdquo Acta Scientiarum Mathemati-carum vol 84 no 12 pp 239ndash261 2018

[25] L Li A M Peralta L Wang and Y-S Wang ldquoWeak-2-local isometries on uniform algebras and Lipschitz algebrasrdquohttpsarxivorgabs170503619

[26] A Jimenez-Vargas L LiAMPeralta LWang andY-SWangldquo2-local standard isometries on vector-valued Lipschitz func-tion spacesrdquo Journal of Mathematical Analysis and Applicationsvol 461 no 2 pp 1287ndash1298 2018

[27] A Ranjbar-Motlagh ldquoIsometries of Lipschitz type functionspacesrdquoMathematischeNachrichten vol 291 no 11-12 pp 1899ndash1907 2018

14 Journal of Function Spaces

[28] N Weaver Lipschitz Algebras World Scientific Publishing CoInc River Edge NJ USA 1999

[29] R R Phelps Lectures on Choquetrsquos theorem vol 1757 of LectureNotes in Mathematics Springer-Verlag Berlin Germany 2ndedition 2001

[30] O Hatori and S Oi ldquoIsometries on Banach algebras of vector-valued mapsrdquo Acta Scientiarum Mathematicarum vol 84 no12 pp 151ndash183 2018

[31] O Hatori S Oi and H Takagi ldquoPeculiar homomorphismsbetween algebras of vector-valued mapsrdquo Studia Mathematicavol 242 no 2 pp 141ndash163 2018

[32] M Nagasawa ldquoIsomorphisms between commutative Banachalgebras with an application to rings of analytic functionsrdquoKodai Mathematical Seminar Reports vol 11 pp 182ndash188 1959

[33] K de LeeuwW Rudin and J Wermer ldquoThe isometries of somefunction spacesrdquo Proceedings of the American MathematicalSociety vol 11 pp 694ndash698 1960

[34] OHatori A Jimenez-Vargas andMVillegas-Vallecillos ldquoMapswhich preserve norms of non-symmetrical quotients betweengroups of exponentials of Lipschitz functionsrdquo Journal of Math-ematical Analysis and Applications vol 415 no 2 pp 825ndash8452014

[35] G Lumer ldquoOn the isometries of reflexiveOrlicz spacesrdquoAnnalesde lrsquoInstitut Fourier vol 68 pp 99ndash109 1963

[36] R J Fleming and J E Jamison Isometries on Banach SpacesFunction Spaces vol 129 ofMonographs and Surveys in Pure andApplied Mathematics Chapman amp Hall CRC Boca Raton FlaUSA 2003

[37] OHatori and SOi ldquoHermitian operators onBanach algebras ofvector-valued Lipschitzmapsrdquo Journal ofMathematical Analysisand Applications vol 452 no 1 pp 378ndash387 2017 Corrigendumto rdquoHermitian operators on Banach algebras of vector-valuedLipschitz mapsrdquo Journal of Mathematical Analysis and Appli-cations 452 (2017) 378ndash387 MR3628025

[38] A Nikou andA G OrsquoFarrell ldquoBanach algebras of vector-valuedfunctionsrdquo Glasgow Mathematical Journal vol 56 no 2 pp419ndash426 2014

[39] F Botelho and J Jamison ldquoHomomorphisms on a class ofcommutative Banach algebrasrdquo Rocky Mountain Journal ofMathematics vol 43 no 2 pp 395ndash416 2013

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Page 12: Hermitian Operators and Isometries on Banach Algebras of ...downloads.hindawi.com/journals/jfs/2018/8085304.pdfJournalofFunctionSpaces normiseitherthemaxnormorthesumnorm.esituation

12 Journal of Function Spaces

L a proof of Proposition 15 is completely the same as that in[30 Proposition 9] since we do not make use of the condition(124) in the proof of [30 Proposition 9] The condition (124)is needed in [30] when we apply Proposition 32 and thefollowing comments in [31]

7 Proof of Theorem 14 An Application ofLumerrsquos Method

Proof of Theorem 14 A proof for the case where 1198831 = 1199091and 1198832 = 1199092 are singletons is the same as the proof ofTheorem 8 in [30]

Suppose that1198832 is not a singleton By Proposition 15 thereexists ℎ isin 119862R(1198842) with |ℎ| = 1 on 1198842 such that 119880(1) =1 otimes ℎ Letting 1198800 1198611 997888rarr 1198612 by 1198800(119865) = (1 otimes ℎ)119880(119865)119865 isin 1198611 we see by the hypothesis (1 otimes ℎ)1198651198612 = 1198651198612 forevery 119865 isin 1198612 that 1198800 is a surjective unital isometry from1198611 onto 1198612 Then Corollary 2 asserts that 1198800 is an algebraisomorphism Let 119891 isin 119862R(1198841) By Proposition 9 1 otimes 119891 isa Hermitian element in 1198611 Then by Theorem 11 1198721otimes119891 is aHermitian operator on 1198611 By Proposition 8 11988001198721otimes119891119880minus1

0 is aHermitian operator on 1198612 Then by Theorem 11 there exists119878(119891) isin 119862R(1198842) such that 11988001198721otimes119891119880minus1

0 = 1198721otimes119878(119891) Hence anoperator 119878 119862R(1198841) 997888rarr 119862R(1198842) is defined Since 1198800 is analgebra isomorphism it is easy to see that 119878 is a real algebraisomorphism from 119862R(1198841) onto 119862R(1198842) Then 119878 119862(1198841) 997888rarr119862(1198842) defined by 119878(119891) = 119878(Re119891) + 119894119878(Im119891) for 119891 isin 119862(1198841)gives a complex algebra isomorphism Gelfand theory assertsthat there is a homeomorphism 120591 1198842 997888rarr 1198841 such that119878(119891) = 119891 ∘ 120591 119891 isin 119862(1198841) It follows that

11988001198721otimes119891119880minus10 = 1198721otimes119891∘120591 119891 isin 119862 (1198841) (125)

Since 119880minus10 (1) = 1 we have

1198800 (1 otimes 119891) = 1 otimes 119891 ∘ 120591 119891 isin 119862 (1198841) (126)

Define Φ 1198611 997888rarr 1198612 by Φ(119886) = 1198800(119886 otimes 1) 119886 isin 1198611Since 1198800 is an algebra isomorphism the map Φ is a unitalhomomorphism Since the maximal ideal space of 1198611 is 1198831

and the maximal ideal space of 1198612 is 1198832 times 1198842 there is acontinuous map 120593 1198832 times 1198842 997888rarr 1198831 such that

Φ(119886) (119909 119910) = 119886 (120593 (119909 119910)) 119886 isin 1198611 (119909 119910) isin 1198832 times 1198842 (127)

It follows by (126) and (127) that

1198800 (119886 otimes 119891) (119909 119910) = 1198800 ((119886 otimes 1) (1 otimes 119891)) (119909 119910)= 1198800 (119886 otimes 1) (119909 119910)1198800 (1 otimes 119891) (119909 119910)= 119886 (120593 (119909 119910))119891 (120591 (119910))= (119886 otimes 119891) (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198832 times 1198842

(128)

for every 119886 isin 1198611 and 119891 isin 119862(1198841) Thus

1198800 (sum(119886119895 otimes 119891119895)) (119909 119910)= (sum(119886119895 otimes 119891119895)) (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198832 times 1198842(129)

for every sum(119886119895 otimes 119891119895) isin 1198611 otimes 119862(1198841) By the Stone-Weierstrasstheorem 1198611 otimes 119862(1198841) is uniformly dense in 119862(1198831 times 1198841) henceany element in 1198611 is uniformly approximated by 1198611 otimes 119862(1198841)As 1198800 is also an isometry with respect to the uniform normwe see that

1198800 (119865) (119909 119910) = 119865 (120593 (119909 119910) 120591 (119910)) (119909 119910) isin 1198831 times 1198841 (130)

for every 119865 isin 1198611 and

119880 (119865) (119909 119910) = (1 otimes ℎ) (119909 119910) 119865 (120593 (119909 119910) 120591 (119910))= ℎ (119910) 119865 (120593 (119909 119910) 120591 (119910))

(119909 119910) isin 1198831 times 1198841(131)

As 1198800 is an algebra isomorphism the map 1198832 times 1198842 997888rarr1198831 times 1198841 defined by (119909 119910) 997891997888rarr (120593(119909 119910) 120591(119910)) gives ahomeomorphism Therefore for every 119910 isin 1198842 the map

120593 (sdot 119910) 1198832 997888rarr 1198831 (132)

is a homeomorphismSuppose that 1198831 is not a singleton By the same way as

in the last part of the proof of Theorem 8 in [30] we havethat 1198832 is not a singleton Then we have the conclusion bythe previous argument

8 Application of Theorem 14

We exhibit applications of Theorem 14

Corollary 16 ([4 Theorem 33]) Suppose that119880 Lip([0 1]) 997888rarr Lip([0 1]) is a surjective isometrywith respect to the norm defined by 119891infin([01]) + 1198911015840infin([01])

for 119891 isin Lip([0 1]) Then 119880(1) is a constant function of unitmodulus such that

119880(119891) (119909) = 119880 (1) 119891 (119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (133)

or

119880(119891) (119909) = 119880 (1) 119891 (1 minus 119909) 119891 isin Lip ([0 1]) 119909 isin [0 1] (134)

The converse statement also holds

Proof By Example 13 we may suppose that Lip([0 1]) is aBanach algebra of 119862(119884)-valuezation Applying Theorem 14we have that119880(1) = 1otimesℎ for ℎ isin 119862(119884)with |ℎ| = 1 Since our

Journal of Function Spaces 13

119884 is a singleton 119880(1) is a constant function of unit modulusWe also see that the corresponding continuous map 120593 [0 1] times 119884 997888rarr [0 1] can be considered as a homeomorphismfrom [0 1] onto [0 1] therefore we have that119880(119891) (119909) = 119880 (1) 119891 (120593 (119909))

119891 isin Lip ([0 1]) 119909 isin [0 1] (135)

The rest is a routine argument to prove that 120593 is an isometryhence 120593(119909) = 119909 119909 isin [0 1] or 120593(119909) = 1 minus 119909 119909 isin [0 1]

The converse statement is trivial

Corollaries 14 15 18 and 19 in [30 Section 6] follow herewith a similar proof

Data Availability

The data used to support the findings of this study areavailable from the corresponding author upon request

Conflicts of Interest

The author declares that they have no conflicts of interest

Acknowledgments

This work was supported by JSPS KAKENHI Grants Num-bers JP16K05172 and JP15K04921

References

[1] K de Leeuw ldquoBanach spaces of Lipschitz functionsrdquo StudiaMathematica vol 21 pp 55ndash66 19611962

[2] A K Roy ldquoExtreme points and linear isometries of the Banachspace of Lipschitz functionsrdquoCanadian Journal of MathematicsJournal Canadien de Mathematiques vol 20 pp 1150ndash11641968

[3] M Cambern ldquoIsometries of certain Banach algebrasrdquo StudiaMathematica vol 25 pp 217ndash225 19641965

[4] N V Rao and A K Roy ldquoLinear isometries of some functionspacesrdquo Pacific Journal of Mathematics vol 38 pp 177ndash192 1971

[5] K Jarosz ldquoIsometries in semisimple commutative BanachalgebrasrdquoProceedings of the AmericanMathematical Society vol94 no 1 pp 65ndash71 1985

[6] K Jarosz and V D Pathak ldquoIsometries between functionspacesrdquo Transactions of the AmericanMathematical Society vol305 no 1 pp 193ndash206 1988

[7] NWeaver ldquoIsometries of noncompact Lipschitz spacesrdquoCana-dian Mathematical Bulletin Bulletin Canadien de Mathemat-iques vol 38 no 2 pp 242ndash249 1995

[8] A Jimenez-Vargas and M Villegas-Vallecillos ldquoLinear isome-tries between spaces of vector-valued Lipschitz functionsrdquoProceedings of the American Mathematical Society vol 137 no4 pp 1381ndash1388 2009

[9] A Jimenez-Vargas and M Villegas-Vallecillos ldquoInto linearisometries between spaces of Lipschitz functionsrdquo HoustonJournal of Mathematics vol 34 no 4 pp 1165ndash1184 2008

[10] F Botelho and J Jamison ldquoSurjective isometries on spacesof differentiable vector-valued functionsrdquo Studia Mathematicavol 192 no 1 pp 39ndash50 2009

[11] E Mayer-Wolf ldquoIsometries between Banach spaces of Lipschitzfunctionsrdquo Israel Journal of Mathematics vol 38 no 1-2 pp 58ndash74 1981

[12] A Jimenez-Vargas M Villegas-Vallecillos and Y-S WangldquoBanach-Stone theorems for vector-valued little Lipschitz func-tionsrdquo Publicationes Mathematicae vol 74 no 1-2 pp 81ndash1002009

[13] J Araujo andLDubarbie ldquoNoncompactness andnoncomplete-ness in isometries of Lipschitz spacesrdquo Journal of MathematicalAnalysis and Applications vol 377 no 1 pp 15ndash29 2011

[14] F Botelho R J Fleming and J Jamison ldquoExtreme points andisometries on vector-valued Lipschitz spacesrdquo Journal of Math-ematical Analysis and Applications vol 381 no 2 pp 821ndash8322011

[15] H Koshimizu ldquoLinear isometries on spaces of continuouslydifferentiable and Lipschitz continuous functionsrdquo NihonkaiMathematical Journal vol 22 no 2 pp 73ndash90 2011

[16] F Botelho J Jamison and B Zheng ldquoIsometries on spaces ofvector valued Lipschitz functionsrdquo Positivity An InternationalMathematics Journal Devoted to Theory and Applications ofPositivity vol 17 no 1 pp 47ndash65 2013

[17] A Ranjbar-Motlagh ldquoA note on isometries of Lipschitz spacesrdquoJournal of Mathematical Analysis and Applications vol 411 no2 pp 555ndash558 2014

[18] F Botelho and J Jamison ldquoSurjective isometries on spaces ofvector valued continuous and Lipschitz functionsrdquo PositivityAn International Mathematics Journal Devoted to Theory andApplications of Positivity vol 17 no 3 pp 395ndash405 2013Erratum to Surjective isometries on spaces of vector valuedcontinuous and Lipschitz functions by F Botelho 20 (2016)757ndash759

[19] T Miura and H Takagi ldquoSurjective isometries on the Banachspace of continuously differentiable functionsrdquo ContemporaryMathematics vol 687 pp 181ndash192 2017

[20] K Kawamura ldquoBanach-Stone type theorems for 1198621-functionspaces over Riemannian manifoldsrdquo Acta Universitatis Szegedi-ensis Acta Scientiarum Mathematicarum vol 83 no 3-4 pp551ndash591 2017

[21] K Kawamura ldquoPerturbations of norms on 1198621-function spacesand associated isometry groupsrdquo Topology Proceedings vol 51pp 169ndash196 2018

[22] K Kawamura ldquoA Banach-Stone type theorem for 1198621-functionspaces over the circlerdquo Topology Proceedings vol 53 pp 15ndash262019

[23] L Li D Chen Q Meng and Y-S Wang ldquoSurjective isometrieson vector-valued differentiable function spacesrdquo Annals ofFunctional Analysis vol 9 no 3 pp 334ndash343 2018

[24] K Kawamura H Koshimizu and T Miura ldquoNorms onC1([01]) and there isometriesrdquo Acta Scientiarum Mathemati-carum vol 84 no 12 pp 239ndash261 2018

[25] L Li A M Peralta L Wang and Y-S Wang ldquoWeak-2-local isometries on uniform algebras and Lipschitz algebrasrdquohttpsarxivorgabs170503619

[26] A Jimenez-Vargas L LiAMPeralta LWang andY-SWangldquo2-local standard isometries on vector-valued Lipschitz func-tion spacesrdquo Journal of Mathematical Analysis and Applicationsvol 461 no 2 pp 1287ndash1298 2018

[27] A Ranjbar-Motlagh ldquoIsometries of Lipschitz type functionspacesrdquoMathematischeNachrichten vol 291 no 11-12 pp 1899ndash1907 2018

14 Journal of Function Spaces

[28] N Weaver Lipschitz Algebras World Scientific Publishing CoInc River Edge NJ USA 1999

[29] R R Phelps Lectures on Choquetrsquos theorem vol 1757 of LectureNotes in Mathematics Springer-Verlag Berlin Germany 2ndedition 2001

[30] O Hatori and S Oi ldquoIsometries on Banach algebras of vector-valued mapsrdquo Acta Scientiarum Mathematicarum vol 84 no12 pp 151ndash183 2018

[31] O Hatori S Oi and H Takagi ldquoPeculiar homomorphismsbetween algebras of vector-valued mapsrdquo Studia Mathematicavol 242 no 2 pp 141ndash163 2018

[32] M Nagasawa ldquoIsomorphisms between commutative Banachalgebras with an application to rings of analytic functionsrdquoKodai Mathematical Seminar Reports vol 11 pp 182ndash188 1959

[33] K de LeeuwW Rudin and J Wermer ldquoThe isometries of somefunction spacesrdquo Proceedings of the American MathematicalSociety vol 11 pp 694ndash698 1960

[34] OHatori A Jimenez-Vargas andMVillegas-Vallecillos ldquoMapswhich preserve norms of non-symmetrical quotients betweengroups of exponentials of Lipschitz functionsrdquo Journal of Math-ematical Analysis and Applications vol 415 no 2 pp 825ndash8452014

[35] G Lumer ldquoOn the isometries of reflexiveOrlicz spacesrdquoAnnalesde lrsquoInstitut Fourier vol 68 pp 99ndash109 1963

[36] R J Fleming and J E Jamison Isometries on Banach SpacesFunction Spaces vol 129 ofMonographs and Surveys in Pure andApplied Mathematics Chapman amp Hall CRC Boca Raton FlaUSA 2003

[37] OHatori and SOi ldquoHermitian operators onBanach algebras ofvector-valued Lipschitzmapsrdquo Journal ofMathematical Analysisand Applications vol 452 no 1 pp 378ndash387 2017 Corrigendumto rdquoHermitian operators on Banach algebras of vector-valuedLipschitz mapsrdquo Journal of Mathematical Analysis and Appli-cations 452 (2017) 378ndash387 MR3628025

[38] A Nikou andA G OrsquoFarrell ldquoBanach algebras of vector-valuedfunctionsrdquo Glasgow Mathematical Journal vol 56 no 2 pp419ndash426 2014

[39] F Botelho and J Jamison ldquoHomomorphisms on a class ofcommutative Banach algebrasrdquo Rocky Mountain Journal ofMathematics vol 43 no 2 pp 395ndash416 2013

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

Page 13: Hermitian Operators and Isometries on Banach Algebras of ...downloads.hindawi.com/journals/jfs/2018/8085304.pdfJournalofFunctionSpaces normiseitherthemaxnormorthesumnorm.esituation

Journal of Function Spaces 13

119884 is a singleton 119880(1) is a constant function of unit modulusWe also see that the corresponding continuous map 120593 [0 1] times 119884 997888rarr [0 1] can be considered as a homeomorphismfrom [0 1] onto [0 1] therefore we have that119880(119891) (119909) = 119880 (1) 119891 (120593 (119909))

119891 isin Lip ([0 1]) 119909 isin [0 1] (135)

The rest is a routine argument to prove that 120593 is an isometryhence 120593(119909) = 119909 119909 isin [0 1] or 120593(119909) = 1 minus 119909 119909 isin [0 1]

The converse statement is trivial

Corollaries 14 15 18 and 19 in [30 Section 6] follow herewith a similar proof

Data Availability

The data used to support the findings of this study areavailable from the corresponding author upon request

Conflicts of Interest

The author declares that they have no conflicts of interest

Acknowledgments

This work was supported by JSPS KAKENHI Grants Num-bers JP16K05172 and JP15K04921

References

[1] K de Leeuw ldquoBanach spaces of Lipschitz functionsrdquo StudiaMathematica vol 21 pp 55ndash66 19611962

[2] A K Roy ldquoExtreme points and linear isometries of the Banachspace of Lipschitz functionsrdquoCanadian Journal of MathematicsJournal Canadien de Mathematiques vol 20 pp 1150ndash11641968

[3] M Cambern ldquoIsometries of certain Banach algebrasrdquo StudiaMathematica vol 25 pp 217ndash225 19641965

[4] N V Rao and A K Roy ldquoLinear isometries of some functionspacesrdquo Pacific Journal of Mathematics vol 38 pp 177ndash192 1971

[5] K Jarosz ldquoIsometries in semisimple commutative BanachalgebrasrdquoProceedings of the AmericanMathematical Society vol94 no 1 pp 65ndash71 1985

[6] K Jarosz and V D Pathak ldquoIsometries between functionspacesrdquo Transactions of the AmericanMathematical Society vol305 no 1 pp 193ndash206 1988

[7] NWeaver ldquoIsometries of noncompact Lipschitz spacesrdquoCana-dian Mathematical Bulletin Bulletin Canadien de Mathemat-iques vol 38 no 2 pp 242ndash249 1995

[8] A Jimenez-Vargas and M Villegas-Vallecillos ldquoLinear isome-tries between spaces of vector-valued Lipschitz functionsrdquoProceedings of the American Mathematical Society vol 137 no4 pp 1381ndash1388 2009

[9] A Jimenez-Vargas and M Villegas-Vallecillos ldquoInto linearisometries between spaces of Lipschitz functionsrdquo HoustonJournal of Mathematics vol 34 no 4 pp 1165ndash1184 2008

[10] F Botelho and J Jamison ldquoSurjective isometries on spacesof differentiable vector-valued functionsrdquo Studia Mathematicavol 192 no 1 pp 39ndash50 2009

[11] E Mayer-Wolf ldquoIsometries between Banach spaces of Lipschitzfunctionsrdquo Israel Journal of Mathematics vol 38 no 1-2 pp 58ndash74 1981

[12] A Jimenez-Vargas M Villegas-Vallecillos and Y-S WangldquoBanach-Stone theorems for vector-valued little Lipschitz func-tionsrdquo Publicationes Mathematicae vol 74 no 1-2 pp 81ndash1002009

[13] J Araujo andLDubarbie ldquoNoncompactness andnoncomplete-ness in isometries of Lipschitz spacesrdquo Journal of MathematicalAnalysis and Applications vol 377 no 1 pp 15ndash29 2011

[14] F Botelho R J Fleming and J Jamison ldquoExtreme points andisometries on vector-valued Lipschitz spacesrdquo Journal of Math-ematical Analysis and Applications vol 381 no 2 pp 821ndash8322011

[15] H Koshimizu ldquoLinear isometries on spaces of continuouslydifferentiable and Lipschitz continuous functionsrdquo NihonkaiMathematical Journal vol 22 no 2 pp 73ndash90 2011

[16] F Botelho J Jamison and B Zheng ldquoIsometries on spaces ofvector valued Lipschitz functionsrdquo Positivity An InternationalMathematics Journal Devoted to Theory and Applications ofPositivity vol 17 no 1 pp 47ndash65 2013

[17] A Ranjbar-Motlagh ldquoA note on isometries of Lipschitz spacesrdquoJournal of Mathematical Analysis and Applications vol 411 no2 pp 555ndash558 2014

[18] F Botelho and J Jamison ldquoSurjective isometries on spaces ofvector valued continuous and Lipschitz functionsrdquo PositivityAn International Mathematics Journal Devoted to Theory andApplications of Positivity vol 17 no 3 pp 395ndash405 2013Erratum to Surjective isometries on spaces of vector valuedcontinuous and Lipschitz functions by F Botelho 20 (2016)757ndash759

[19] T Miura and H Takagi ldquoSurjective isometries on the Banachspace of continuously differentiable functionsrdquo ContemporaryMathematics vol 687 pp 181ndash192 2017

[20] K Kawamura ldquoBanach-Stone type theorems for 1198621-functionspaces over Riemannian manifoldsrdquo Acta Universitatis Szegedi-ensis Acta Scientiarum Mathematicarum vol 83 no 3-4 pp551ndash591 2017

[21] K Kawamura ldquoPerturbations of norms on 1198621-function spacesand associated isometry groupsrdquo Topology Proceedings vol 51pp 169ndash196 2018

[22] K Kawamura ldquoA Banach-Stone type theorem for 1198621-functionspaces over the circlerdquo Topology Proceedings vol 53 pp 15ndash262019

[23] L Li D Chen Q Meng and Y-S Wang ldquoSurjective isometrieson vector-valued differentiable function spacesrdquo Annals ofFunctional Analysis vol 9 no 3 pp 334ndash343 2018

[24] K Kawamura H Koshimizu and T Miura ldquoNorms onC1([01]) and there isometriesrdquo Acta Scientiarum Mathemati-carum vol 84 no 12 pp 239ndash261 2018

[25] L Li A M Peralta L Wang and Y-S Wang ldquoWeak-2-local isometries on uniform algebras and Lipschitz algebrasrdquohttpsarxivorgabs170503619

[26] A Jimenez-Vargas L LiAMPeralta LWang andY-SWangldquo2-local standard isometries on vector-valued Lipschitz func-tion spacesrdquo Journal of Mathematical Analysis and Applicationsvol 461 no 2 pp 1287ndash1298 2018

[27] A Ranjbar-Motlagh ldquoIsometries of Lipschitz type functionspacesrdquoMathematischeNachrichten vol 291 no 11-12 pp 1899ndash1907 2018

14 Journal of Function Spaces

[28] N Weaver Lipschitz Algebras World Scientific Publishing CoInc River Edge NJ USA 1999

[29] R R Phelps Lectures on Choquetrsquos theorem vol 1757 of LectureNotes in Mathematics Springer-Verlag Berlin Germany 2ndedition 2001

[30] O Hatori and S Oi ldquoIsometries on Banach algebras of vector-valued mapsrdquo Acta Scientiarum Mathematicarum vol 84 no12 pp 151ndash183 2018

[31] O Hatori S Oi and H Takagi ldquoPeculiar homomorphismsbetween algebras of vector-valued mapsrdquo Studia Mathematicavol 242 no 2 pp 141ndash163 2018

[32] M Nagasawa ldquoIsomorphisms between commutative Banachalgebras with an application to rings of analytic functionsrdquoKodai Mathematical Seminar Reports vol 11 pp 182ndash188 1959

[33] K de LeeuwW Rudin and J Wermer ldquoThe isometries of somefunction spacesrdquo Proceedings of the American MathematicalSociety vol 11 pp 694ndash698 1960

[34] OHatori A Jimenez-Vargas andMVillegas-Vallecillos ldquoMapswhich preserve norms of non-symmetrical quotients betweengroups of exponentials of Lipschitz functionsrdquo Journal of Math-ematical Analysis and Applications vol 415 no 2 pp 825ndash8452014

[35] G Lumer ldquoOn the isometries of reflexiveOrlicz spacesrdquoAnnalesde lrsquoInstitut Fourier vol 68 pp 99ndash109 1963

[36] R J Fleming and J E Jamison Isometries on Banach SpacesFunction Spaces vol 129 ofMonographs and Surveys in Pure andApplied Mathematics Chapman amp Hall CRC Boca Raton FlaUSA 2003

[37] OHatori and SOi ldquoHermitian operators onBanach algebras ofvector-valued Lipschitzmapsrdquo Journal ofMathematical Analysisand Applications vol 452 no 1 pp 378ndash387 2017 Corrigendumto rdquoHermitian operators on Banach algebras of vector-valuedLipschitz mapsrdquo Journal of Mathematical Analysis and Appli-cations 452 (2017) 378ndash387 MR3628025

[38] A Nikou andA G OrsquoFarrell ldquoBanach algebras of vector-valuedfunctionsrdquo Glasgow Mathematical Journal vol 56 no 2 pp419ndash426 2014

[39] F Botelho and J Jamison ldquoHomomorphisms on a class ofcommutative Banach algebrasrdquo Rocky Mountain Journal ofMathematics vol 43 no 2 pp 395ndash416 2013

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

Page 14: Hermitian Operators and Isometries on Banach Algebras of ...downloads.hindawi.com/journals/jfs/2018/8085304.pdfJournalofFunctionSpaces normiseitherthemaxnormorthesumnorm.esituation

14 Journal of Function Spaces

[28] N Weaver Lipschitz Algebras World Scientific Publishing CoInc River Edge NJ USA 1999

[29] R R Phelps Lectures on Choquetrsquos theorem vol 1757 of LectureNotes in Mathematics Springer-Verlag Berlin Germany 2ndedition 2001

[30] O Hatori and S Oi ldquoIsometries on Banach algebras of vector-valued mapsrdquo Acta Scientiarum Mathematicarum vol 84 no12 pp 151ndash183 2018

[31] O Hatori S Oi and H Takagi ldquoPeculiar homomorphismsbetween algebras of vector-valued mapsrdquo Studia Mathematicavol 242 no 2 pp 141ndash163 2018

[32] M Nagasawa ldquoIsomorphisms between commutative Banachalgebras with an application to rings of analytic functionsrdquoKodai Mathematical Seminar Reports vol 11 pp 182ndash188 1959

[33] K de LeeuwW Rudin and J Wermer ldquoThe isometries of somefunction spacesrdquo Proceedings of the American MathematicalSociety vol 11 pp 694ndash698 1960

[34] OHatori A Jimenez-Vargas andMVillegas-Vallecillos ldquoMapswhich preserve norms of non-symmetrical quotients betweengroups of exponentials of Lipschitz functionsrdquo Journal of Math-ematical Analysis and Applications vol 415 no 2 pp 825ndash8452014

[35] G Lumer ldquoOn the isometries of reflexiveOrlicz spacesrdquoAnnalesde lrsquoInstitut Fourier vol 68 pp 99ndash109 1963

[36] R J Fleming and J E Jamison Isometries on Banach SpacesFunction Spaces vol 129 ofMonographs and Surveys in Pure andApplied Mathematics Chapman amp Hall CRC Boca Raton FlaUSA 2003

[37] OHatori and SOi ldquoHermitian operators onBanach algebras ofvector-valued Lipschitzmapsrdquo Journal ofMathematical Analysisand Applications vol 452 no 1 pp 378ndash387 2017 Corrigendumto rdquoHermitian operators on Banach algebras of vector-valuedLipschitz mapsrdquo Journal of Mathematical Analysis and Appli-cations 452 (2017) 378ndash387 MR3628025

[38] A Nikou andA G OrsquoFarrell ldquoBanach algebras of vector-valuedfunctionsrdquo Glasgow Mathematical Journal vol 56 no 2 pp419ndash426 2014

[39] F Botelho and J Jamison ldquoHomomorphisms on a class ofcommutative Banach algebrasrdquo Rocky Mountain Journal ofMathematics vol 43 no 2 pp 395ndash416 2013

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

Page 15: Hermitian Operators and Isometries on Banach Algebras of ...downloads.hindawi.com/journals/jfs/2018/8085304.pdfJournalofFunctionSpaces normiseitherthemaxnormorthesumnorm.esituation

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom


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