Research ArticleOn a Numerical Radius Preserving Onto Isometry on L(119883)
Sun Kwang Kim
Department of Mathematics Kyonggi University Suwon 443-760 Republic of Korea
Correspondence should be addressed to Sun Kwang Kim sunkwangkguackr
Received 17 August 2016 Accepted 25 September 2016
Academic Editor Ajda Fosner
Copyright copy 2016 Sun Kwang Kim This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
We study a numerical radius preserving onto isometry onL(119883) As a main result when119883 is a complex Banach space having bothuniform smoothness and uniform convexity we show that an onto isometry 119879 onL(119883) is numerical radius preserving if and onlyif there exists a scalar 119888119879 of modulus 1 such that 119888119879119879 is numerical range preserving The examples of such spaces are Hilbert spaceand 119871119901 spaces for 1 lt 119901 lt infin
1 Introduction
In this paper we study a numerical radius preserving ontoisometry on the set of bounded linear operators We beginwith some notation to present its definition Let 119883 be aBanach space over the fieldK = R orCWe use119883lowast andL(119883)for the dual of 119883 and the space of bounded linear operatorsfrom 119883 to 119883 respectively We denote by 119861119883 (resp 119878119883) theclosed unit ball of119883 (resp the unit sphere of119883)
The numerical radius and numerical range of an operator119860 isin L(119883) are given by
where Π(119883) = (119909 119909lowast) isin 119878119883 times 119878119883lowast 119909lowast(119909) = 1Note that there are many different definitions of numer-
ical range The above definition of 119881(119860) is sometimes calledBanach algebra numerical range to distinguish from othersIt is worth mentioning that it is also usual to study spatialnumerical range 119909lowast(119860119909) (119909 119909lowast) isin Π(119883) From thedefinition we see that V(119860) le 119860 for every 119860 isin L(119883)
The concept of numerical range was introduced by GLumer and F Bauer in the sixties Later this was extendedto arbitrary continuous functions on a unit sphere of Banachspaces On the other hand the study of numerical radiuswas initiated by Harris for polynomials and holomorphicfunctions We can find more details in [1ndash3]
There has been many different types of research on theseconcepts Among them we can find a lot of works on linearmaps which preserve numerical radius or numerical range[4]
We say that an operator 119879 onL(119883) preserves numericalradius (numerical range) when
V (119879 (119860)) = V (119860)(119881 (119879 (119860)) = 119881 (119860)) (2)
for every 119860 isin L(119883) It is clear that every numerical rangepreserving map is numerical radius preserving
In 1975 Pellegrini [5] studied numerical range preservingoperators on a Banach algebra Particularly when H is acomplex Hilbert space it was shown that an isomorphism 119879on L(H) is 119862lowast-isomorphism if and only if it is numericalrange preserving Later Chan [6] showed that an isomor-phism 119879 onL(H) is numerical radius preserving if and onlyif 119888119879119879 is a 119862lowast-isomorphism for some scalar 119888119879 of modulus 1These results say that for each numerical radius preservingisomorphism 119879 onL(H) there exists a scalar 119888119879 of modulus1 such that 119888119879119879 is a numerical range preserving mapping
In Section 2 we deduce a similar result for an ontoisometry when a complex Banach space 119883 has both uniformconvexity and uniform smoothness Indeed we show that if119879 is a numerical radius preserving onto isometry on L(119883)then there exists a scalar 119888119879 of modulus 1 such that 119888119879119879 isnumerical range preserving It is known that Hilbert spaceand 119871119901 for 1 lt 119901 lt infin have both uniform convexity and
Hindawi Publishing CorporationJournal of Function SpacesVolume 2016 Article ID 9183135 3 pageshttpdxdoiorg10115520169183135
2 Journal of Function Spaces
uniform smoothness However we see that this does not holdwhen119883 = ℓ2infin
2 Results
We first recall the definition of uniform convexityFor every 120598 isin (0 2] themodulus of convexity of a Banach
space (119883 sdot ) is defined by
120575 (120598) = inf 1 minus 1003817100381710038171003817100381710038171003817119909 + 119910
21003817100381710038171003817100381710038171003817 119909 119910 isin 119861119883 1003817100381710038171003817119909 minus 1199101003817100381710038171003817 ge 120598 (3)
A Banach space (119883 sdot ) is said to be uniformly convex if120575(120598) gt 0 for all 120598 isin (0 2]
It is well known that every uniformly convex space isstrictly convex and 119871119901 is uniformly convex when 1 lt 119901 lt infin
Very recently the following characterization of uniformconvexity was shown [7]
Theorem 1 A Banach space119883 is uniformly convex if and onlyif for each 120576 gt 0 there is 120578(120576) gt 0 such that for any (119909 119909lowast) isin119878119883 times 119878119883lowast with |119909lowast(119909)| gt 1 minus 120578(120576) there exists 119910 isin 119878119884 satisfying
119909lowast (119910) = 11003817100381710038171003817119910 minus 1199091003817100381710038171003817 lt 120576
(4)
From this theorem we show the following lemma
Lemma 2 Let 119883 be a uniformly convex space For any 120576 gt 0there exists 0 lt 120573(120576) lt 120576 such that if two pairs (119909 119909lowast) (119910 119910lowast) isinΠ(119883) satisfy 119909lowast minus 119910lowast lt 120573(120576) then 119909 minus 119910 lt 120576Proof Let 120578(sdot) be the function inTheorem 1 and assume that(119909 119909lowast) (119910 119910lowast) isin Π(119883) satisfy 119909lowast minus 119910lowast lt 120578(120576)
Note that 119909 is the unique element in 119861119883 such that 119909lowast(119909) =1 by the strict convexity of119883 Since |119909lowast(119910)| ge |119910lowast(119910)|minus |(119910lowast minus119909lowast)(119909)| gt 1 minus 120578(120576) we see that 119909 minus 119910 lt 120576 from Theorem 1Hence we may take 120573(120576) = min120578(120576) 1205762
It is worth remarking that the converse of Lemma 2 is truewhen the space 119883 is finite dimensional Indeed if 119883 is notuniformly convex then there exist 120576 gt 0 and a sequence ofpairs (119909119894 119910119894) isin 119861119883 times 119861119883 such that 119909119894 minus 119910119894 gt 120576 and 119909119894 +119910119894 converge to 2 Since 119861119883 is compact we may assume that(119909119894) and (119910119894) converge to 119909 and 119910 respectively It is clear that119909+119910 = 2 and so there exists a functional 119891 isin 119878119883lowast such that119891(119909 + 119910) = 2 Since (119909 119891) (119910 119891) isin Π(119883) and 119909 minus 119910 gt 120576 weget the desired contradiction
In paper [6] of Chan it was shown that if 119879 is a numericalradius preserving isomorphism on L(H) then 119879(119868) = 119888119879119868for some constant 119888119879 of modulus 1 where 119868 is the identitymap on H We show that the same result holds for anonto isometry on L(119883) when 119883 has uniform convexity anduniform smoothness Before that we first see the following
Lemma 3 If 119879 is a numerical radius preserving map onL(119883) then for each 119860 isin L(119883) there exists a scalar 120572119860 ofmodulus 1 such that V(119879(119868)) + V(119879(119860)) = V(119879(119868) + 120572119860119879(119860))Proof We first see that for any 119860 isin L(119883) there is a scalar120572119860 of modulus 1 satisfying V(119868 + 120572119860119860) = 1 + V(119860) Indeed
for a sequence (119909119894 119909lowast119894 ) isin Π(119883) such that |119909lowast119894 (119860119909119894)| convergesto V(119860) we may assume that 119909lowast119894 (119860119909119894) converges to someconstant120573 Let120572119860 = 120573|120573|Thenwe have that119909lowast119894 ((119868+120572119860119860)119909119894)converges to 1+|120573| = 1+V(119860) Since it is clear that V(119868+120572119860119860) le1 + V(119860) we deduce V(119868 + 120572119860119860) = 1 + V(119860)
The fact that 119879 preserves numerical radius gives V(119879(119868))+V(119879(119860)) = V(119868) + V(119860) = V(119868 + 120572119860119860) = V(119879(119868 + 120572119860119860)) =V(119879(119868) + 120572119860119879(119860))Lemma4 Assume119883 is a strictly convex space and an operator119860 isin L(119883) satisfies 119860 = 1 If absolute value of every elementin 119881(119860) is 1 then 119860 = 119888119860119868 for some constant 119888119860 of modulus 1
Proof From the assumption we see that 119860119909 = 1 =|119909lowast(119860119909)| for arbitrary 119909 isin 119878119883 and 119909lowast satisfying 119909lowast(119909) = 1Since 119883 is strictly convex 119909lowast attains its norm only at 120582119909 |120582| = 1 This implies that 119860119909 = 120582119909119909 for some constant 120582119909 ofmodulus 1
Now assume that there exist two elements 1199091 1199092 isin 119878119883such that 120582119909
1
= 1205821199092
Note that 1199091 and 1199092 are linearly inde-pendent Consider 119911 = (1199091 + 1199092)1199091 + 1199092 and let 119860119911 = 120582119911119911This means that
minus 120582119911)1199092 which is acontradiction
Theorem 5 Assume 119883 has both uniform convexity anduniform smoothness If 119879 is a numerical radius preserving ontoisometry on L(119883) then 119879(119868) = 119888119879119868 for some constant 119888119879 ofmodulus 1
Proof From Lemma 3 for each119860 isin L(119883) there is 120572119860 so thatV(119879(119868)) + V(119879(119860)) = V(119879(119868) + 120572119860119879(119860))
We now show that |119909lowast(119879(119868)(119909))| = V(119879(119868)) = V(119868) = 1 forany (119909 119909lowast) isin Π(119883) If this is not true there exists (1199110 119911lowast0 ) isinΠ(119883) such that |119911lowast0 (119879(119868)(1199110))| lt 1 Take an operator 119861 isinL(119883) so that 119879(119861)(119909) = 119911lowast0 (119909)1199110 for each 119909 isin 119883 and take0 lt 120574 lt 1 such that |119911lowast0 (119879(119868)(1199110))| lt 120574 lt 1
Let 120575 gt 0 be a number satisfying that every (119911 119911lowast) isin Π(119883)with 119911 minus 1199110 lt 120575 and 119911lowast minus 119911lowast0 lt 120575 implies |119911lowast(119879(119868))119911| lt 120574Since119883 is uniformly convex we may take 120573(sdot) in Lemma 2
For a pair (119910 119910lowast) isin Π(119883) if 119910lowast minus 119911lowast0 lt 120573(120575) lt 120575 then119910 minus 1199110 lt 120575 This gives that
On the other hand there exists 120576 gt 0 such that if (119910 119910lowast) isinΠ(119883) satisfies 119910lowast minus 119911lowast0 ge 120573(120575) then |119910lowast(1199110)| le 1 minus 120576 To seethat this is true we note that dual of uniformly smooth spaceis uniformly convex Hence119883lowast is uniformly convex We mayuse again Lemma 2 and then for some small enough 120576 gt 0 if(119910 119910lowast) isin Π(119883) satisfies |119910lowast(1199110)| gt 1minus120576 then 119910lowastminus119911lowast0 lt 120573(120575)
Journal of Function Spaces 3
Therefore we see that if (119910 119910lowast) isin Π(119883) satisfies that 119910lowastminus119911lowast0 ge 120573(120575) then |119910lowast(119879(119861)(119910))| = |119910lowast(1199110)119911lowast0 (119910)| le 1 minus 120576 Wehave
This shows that V(119879(119868 + 120572119861119861)) le max2 minus 120576 120574 + 1 lt 2 =V(119879(119868)) + V(119879(119861)) and so we have desired contradiction
From Lemma 4 we see that 119879(119868) = 119888119879119868 for somemodulus1 constant 119888119879Remark 6 From Theorem 5 we can easily construct ontoisometries which are not numerical radius preserving Forexample consider 2-dimensional Hilbert spaceR2 we definean operator 119861 R2 rarr R2 by 119861(1199091 1199092) = (1199091 minus1199092) forevery (1199091 1199092) isin R2 It is clear that the operator 119879 on L(R2)given by 119879(119860) = 119861 ∘ 119860 for every 119860 isin L(R2) is an ontoisometry However this does not preserve numerical radiussince 119879(119868) = 119888119879119868 for any 119888119879
We denote a state space S of a Banach algebraL(119883) byS = 119891 isin L (119883)lowast 10038171003817100381710038171198911003817100381710038171003817 = 1 = 119891 (119868) (8)
When 119883 is a complex Banach space according to [5Theorem 22] if the adjoint operator 119878lowast of operator 119878 L(119883) rarr L(119883) satisfies that 119878lowast(S) sub S then 119881(119878(119860)) sub119881(119860) For119879 inTheorem5 since1198790 = 119888minus1119879 119879 is an onto isometryand so satisfies 119879lowast0 (S) = S we see that 119881(1198790(119860)) = 119881(119860)Now we deduce the following main result
Theorem 7 Assume a complex Banach space 119883 has bothuniform convexity and uniform smoothness An onto isometry119879 on L(119883) is numerical radius preserving if and only if thereexists a scalar 119888119879 of modulus 1 such that 119888119879119879 is a numericalrange preserving mapping
Remark 8 Theorem 7 does not hold for some Banach spacesIn order to see this let us recall numerical index 119899(119883) of aBanach space
119899 (119883) = inf V (119879) 119879 isin 119878L(119883) (9)
We see that 119899(119883) is the greatest constant 119896 ge 0 such that119896119879 le V(119879) for every119879 isin L(119883) and so V is equivalent to theoperator norm if and only if 119899(119883) gt 0 For more informationwe give [8ndash11]
From the definition of numerical index if 119899(119883) = 1then every isometry on L(119883) preserves numerical radiusThere are many classes of Banach spaces having numericalindex 1 like 1198880 ℓ1 and ℓinfin [12] Among them one of thesimplest examples having numerical index 1 is R2 as thetypical subspace of ℓinfin Since the same operator 119879 defined inRemark 6 is also an onto isometry on this space 119879 preservesnumerical radius However this does not preserve numericalrange The reason is that we have 119881(119868) = 1 and minus1 =(0 1)119879(119868)(0 1) isin 119881(119879(119868))
Remark 9 It is not possible to say that Theorem 7 is true forevery isometry Consider shift operators 1198611 and 1198612 on ℓ119901 (1 lt119901 lt infin) defined by 1198611((119909119894)infin119894=1) = (119909119894)infin119894=2 and 1198612((119909119894)infin119894=1) =(119909119894)infin119894=0 for (119909119894)infin119894=1 isin ℓ119901 where 1199090 = 0 Then it is easy to seethat an operator 119879 given by 119879(119860) = 1198612 ∘ 119860 ∘ 1198611 for each 119860 isin119871(ℓ119901) is a numerical radius preserving isometry However 119879does not preserve numerical range since 119881(119868) = 1 and 0 =(1 0 0 0 )119879(119868)(1 0 0 0 ) isin 119881(119879(119868))
Competing Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work was supported by Kyonggi University ResearchGrant 2015
References
[1] F F Bonsall and J Duncan Numerical Ranges I vol 2 ofLecture Notes in Mathematics London Mathematical SocietyCambridge UK 1971
[2] F F Bonsall and J DuncanNumerical Ranges II LondonMath-ematical Society Lecture Note Series 10 Cambridge UniversityPress Cambridge Uk 1973
[3] L A Harris ldquoThe numerical range of holomorphic functions inBanach spacesrdquo The American Journal of Mathematics vol 93pp 1005ndash1019 1971
[4] C-K Li ldquoA survey on linear preservers of numerical ranges andradiirdquo Taiwanese Journal of Mathematics vol 5 no 3 pp 477ndash496 2001
[5] V J Pellegrini ldquoNumerical range preserving operators on aBanach algebrardquo Polska Akademia Nauk Instytut Matematy-czny Studia Mathematica vol 54 no 2 pp 143ndash147 1975
[6] J-T Chan ldquoNumerical radius perserving operators on B(H)rdquoProceedings of the American Mathematical Society vol 123 no5 pp 1437ndash1439 1995
[7] S K Kim andH J Lee ldquoUniform convexity and Bishop-Phelps-Bollobas propertyrdquo Canadian Journal of Mathematics vol 66no 2 pp 373ndash386 2014
[8] K Boyko V Kadets M Martın and D Werner ldquoNumericalindex of Banach spaces and dualityrdquo Mathematical Proceedingsof the Cambridge Philosophical Society vol 142 no 1 pp 93ndash1022007
[9] C Finet M Martın and R Paya ldquoNumerical index andrenormingrdquo Proceedings of the American Mathematical Societyvol 131 no 3 pp 871ndash877 2003
[10] V Kadets M Martn and R Paya ldquoRecent progress and openquestions on the numerical index of Banach spacesrdquo Revista dela Real Academia de Ciencias Exactas Fısicas y Naturales SerieA Matematicas vol 100 pp 155ndash182 2006
[11] M Martın J Merı M Popov and B RandrianantoaninaldquoNumerical index of absolute sums of Banach spacesrdquo Journalof Mathematical Analysis and Applications vol 375 no 1 pp207ndash222 2011
[12] K Boyko V Kadets M Martın and J Merı ldquoProperties of lushspaces and applications to Banach spaces with numerical index1rdquo Studia Mathematica vol 190 no 2 pp 117ndash133 2009
uniform smoothness However we see that this does not holdwhen119883 = ℓ2infin
2 Results
We first recall the definition of uniform convexityFor every 120598 isin (0 2] themodulus of convexity of a Banach
space (119883 sdot ) is defined by
120575 (120598) = inf 1 minus 1003817100381710038171003817100381710038171003817119909 + 119910
21003817100381710038171003817100381710038171003817 119909 119910 isin 119861119883 1003817100381710038171003817119909 minus 1199101003817100381710038171003817 ge 120598 (3)
A Banach space (119883 sdot ) is said to be uniformly convex if120575(120598) gt 0 for all 120598 isin (0 2]
It is well known that every uniformly convex space isstrictly convex and 119871119901 is uniformly convex when 1 lt 119901 lt infin
Very recently the following characterization of uniformconvexity was shown [7]
Theorem 1 A Banach space119883 is uniformly convex if and onlyif for each 120576 gt 0 there is 120578(120576) gt 0 such that for any (119909 119909lowast) isin119878119883 times 119878119883lowast with |119909lowast(119909)| gt 1 minus 120578(120576) there exists 119910 isin 119878119884 satisfying
119909lowast (119910) = 11003817100381710038171003817119910 minus 1199091003817100381710038171003817 lt 120576
(4)
From this theorem we show the following lemma
Lemma 2 Let 119883 be a uniformly convex space For any 120576 gt 0there exists 0 lt 120573(120576) lt 120576 such that if two pairs (119909 119909lowast) (119910 119910lowast) isinΠ(119883) satisfy 119909lowast minus 119910lowast lt 120573(120576) then 119909 minus 119910 lt 120576Proof Let 120578(sdot) be the function inTheorem 1 and assume that(119909 119909lowast) (119910 119910lowast) isin Π(119883) satisfy 119909lowast minus 119910lowast lt 120578(120576)
Note that 119909 is the unique element in 119861119883 such that 119909lowast(119909) =1 by the strict convexity of119883 Since |119909lowast(119910)| ge |119910lowast(119910)|minus |(119910lowast minus119909lowast)(119909)| gt 1 minus 120578(120576) we see that 119909 minus 119910 lt 120576 from Theorem 1Hence we may take 120573(120576) = min120578(120576) 1205762
It is worth remarking that the converse of Lemma 2 is truewhen the space 119883 is finite dimensional Indeed if 119883 is notuniformly convex then there exist 120576 gt 0 and a sequence ofpairs (119909119894 119910119894) isin 119861119883 times 119861119883 such that 119909119894 minus 119910119894 gt 120576 and 119909119894 +119910119894 converge to 2 Since 119861119883 is compact we may assume that(119909119894) and (119910119894) converge to 119909 and 119910 respectively It is clear that119909+119910 = 2 and so there exists a functional 119891 isin 119878119883lowast such that119891(119909 + 119910) = 2 Since (119909 119891) (119910 119891) isin Π(119883) and 119909 minus 119910 gt 120576 weget the desired contradiction
In paper [6] of Chan it was shown that if 119879 is a numericalradius preserving isomorphism on L(H) then 119879(119868) = 119888119879119868for some constant 119888119879 of modulus 1 where 119868 is the identitymap on H We show that the same result holds for anonto isometry on L(119883) when 119883 has uniform convexity anduniform smoothness Before that we first see the following
Lemma 3 If 119879 is a numerical radius preserving map onL(119883) then for each 119860 isin L(119883) there exists a scalar 120572119860 ofmodulus 1 such that V(119879(119868)) + V(119879(119860)) = V(119879(119868) + 120572119860119879(119860))Proof We first see that for any 119860 isin L(119883) there is a scalar120572119860 of modulus 1 satisfying V(119868 + 120572119860119860) = 1 + V(119860) Indeed
for a sequence (119909119894 119909lowast119894 ) isin Π(119883) such that |119909lowast119894 (119860119909119894)| convergesto V(119860) we may assume that 119909lowast119894 (119860119909119894) converges to someconstant120573 Let120572119860 = 120573|120573|Thenwe have that119909lowast119894 ((119868+120572119860119860)119909119894)converges to 1+|120573| = 1+V(119860) Since it is clear that V(119868+120572119860119860) le1 + V(119860) we deduce V(119868 + 120572119860119860) = 1 + V(119860)
The fact that 119879 preserves numerical radius gives V(119879(119868))+V(119879(119860)) = V(119868) + V(119860) = V(119868 + 120572119860119860) = V(119879(119868 + 120572119860119860)) =V(119879(119868) + 120572119860119879(119860))Lemma4 Assume119883 is a strictly convex space and an operator119860 isin L(119883) satisfies 119860 = 1 If absolute value of every elementin 119881(119860) is 1 then 119860 = 119888119860119868 for some constant 119888119860 of modulus 1
Proof From the assumption we see that 119860119909 = 1 =|119909lowast(119860119909)| for arbitrary 119909 isin 119878119883 and 119909lowast satisfying 119909lowast(119909) = 1Since 119883 is strictly convex 119909lowast attains its norm only at 120582119909 |120582| = 1 This implies that 119860119909 = 120582119909119909 for some constant 120582119909 ofmodulus 1
Now assume that there exist two elements 1199091 1199092 isin 119878119883such that 120582119909
1
= 1205821199092
Note that 1199091 and 1199092 are linearly inde-pendent Consider 119911 = (1199091 + 1199092)1199091 + 1199092 and let 119860119911 = 120582119911119911This means that
minus 120582119911)1199092 which is acontradiction
Theorem 5 Assume 119883 has both uniform convexity anduniform smoothness If 119879 is a numerical radius preserving ontoisometry on L(119883) then 119879(119868) = 119888119879119868 for some constant 119888119879 ofmodulus 1
Proof From Lemma 3 for each119860 isin L(119883) there is 120572119860 so thatV(119879(119868)) + V(119879(119860)) = V(119879(119868) + 120572119860119879(119860))
We now show that |119909lowast(119879(119868)(119909))| = V(119879(119868)) = V(119868) = 1 forany (119909 119909lowast) isin Π(119883) If this is not true there exists (1199110 119911lowast0 ) isinΠ(119883) such that |119911lowast0 (119879(119868)(1199110))| lt 1 Take an operator 119861 isinL(119883) so that 119879(119861)(119909) = 119911lowast0 (119909)1199110 for each 119909 isin 119883 and take0 lt 120574 lt 1 such that |119911lowast0 (119879(119868)(1199110))| lt 120574 lt 1
Let 120575 gt 0 be a number satisfying that every (119911 119911lowast) isin Π(119883)with 119911 minus 1199110 lt 120575 and 119911lowast minus 119911lowast0 lt 120575 implies |119911lowast(119879(119868))119911| lt 120574Since119883 is uniformly convex we may take 120573(sdot) in Lemma 2
For a pair (119910 119910lowast) isin Π(119883) if 119910lowast minus 119911lowast0 lt 120573(120575) lt 120575 then119910 minus 1199110 lt 120575 This gives that
On the other hand there exists 120576 gt 0 such that if (119910 119910lowast) isinΠ(119883) satisfies 119910lowast minus 119911lowast0 ge 120573(120575) then |119910lowast(1199110)| le 1 minus 120576 To seethat this is true we note that dual of uniformly smooth spaceis uniformly convex Hence119883lowast is uniformly convex We mayuse again Lemma 2 and then for some small enough 120576 gt 0 if(119910 119910lowast) isin Π(119883) satisfies |119910lowast(1199110)| gt 1minus120576 then 119910lowastminus119911lowast0 lt 120573(120575)
Journal of Function Spaces 3
Therefore we see that if (119910 119910lowast) isin Π(119883) satisfies that 119910lowastminus119911lowast0 ge 120573(120575) then |119910lowast(119879(119861)(119910))| = |119910lowast(1199110)119911lowast0 (119910)| le 1 minus 120576 Wehave
This shows that V(119879(119868 + 120572119861119861)) le max2 minus 120576 120574 + 1 lt 2 =V(119879(119868)) + V(119879(119861)) and so we have desired contradiction
From Lemma 4 we see that 119879(119868) = 119888119879119868 for somemodulus1 constant 119888119879Remark 6 From Theorem 5 we can easily construct ontoisometries which are not numerical radius preserving Forexample consider 2-dimensional Hilbert spaceR2 we definean operator 119861 R2 rarr R2 by 119861(1199091 1199092) = (1199091 minus1199092) forevery (1199091 1199092) isin R2 It is clear that the operator 119879 on L(R2)given by 119879(119860) = 119861 ∘ 119860 for every 119860 isin L(R2) is an ontoisometry However this does not preserve numerical radiussince 119879(119868) = 119888119879119868 for any 119888119879
We denote a state space S of a Banach algebraL(119883) byS = 119891 isin L (119883)lowast 10038171003817100381710038171198911003817100381710038171003817 = 1 = 119891 (119868) (8)
When 119883 is a complex Banach space according to [5Theorem 22] if the adjoint operator 119878lowast of operator 119878 L(119883) rarr L(119883) satisfies that 119878lowast(S) sub S then 119881(119878(119860)) sub119881(119860) For119879 inTheorem5 since1198790 = 119888minus1119879 119879 is an onto isometryand so satisfies 119879lowast0 (S) = S we see that 119881(1198790(119860)) = 119881(119860)Now we deduce the following main result
Theorem 7 Assume a complex Banach space 119883 has bothuniform convexity and uniform smoothness An onto isometry119879 on L(119883) is numerical radius preserving if and only if thereexists a scalar 119888119879 of modulus 1 such that 119888119879119879 is a numericalrange preserving mapping
Remark 8 Theorem 7 does not hold for some Banach spacesIn order to see this let us recall numerical index 119899(119883) of aBanach space
119899 (119883) = inf V (119879) 119879 isin 119878L(119883) (9)
We see that 119899(119883) is the greatest constant 119896 ge 0 such that119896119879 le V(119879) for every119879 isin L(119883) and so V is equivalent to theoperator norm if and only if 119899(119883) gt 0 For more informationwe give [8ndash11]
From the definition of numerical index if 119899(119883) = 1then every isometry on L(119883) preserves numerical radiusThere are many classes of Banach spaces having numericalindex 1 like 1198880 ℓ1 and ℓinfin [12] Among them one of thesimplest examples having numerical index 1 is R2 as thetypical subspace of ℓinfin Since the same operator 119879 defined inRemark 6 is also an onto isometry on this space 119879 preservesnumerical radius However this does not preserve numericalrange The reason is that we have 119881(119868) = 1 and minus1 =(0 1)119879(119868)(0 1) isin 119881(119879(119868))
Remark 9 It is not possible to say that Theorem 7 is true forevery isometry Consider shift operators 1198611 and 1198612 on ℓ119901 (1 lt119901 lt infin) defined by 1198611((119909119894)infin119894=1) = (119909119894)infin119894=2 and 1198612((119909119894)infin119894=1) =(119909119894)infin119894=0 for (119909119894)infin119894=1 isin ℓ119901 where 1199090 = 0 Then it is easy to seethat an operator 119879 given by 119879(119860) = 1198612 ∘ 119860 ∘ 1198611 for each 119860 isin119871(ℓ119901) is a numerical radius preserving isometry However 119879does not preserve numerical range since 119881(119868) = 1 and 0 =(1 0 0 0 )119879(119868)(1 0 0 0 ) isin 119881(119879(119868))
Competing Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work was supported by Kyonggi University ResearchGrant 2015
References
[1] F F Bonsall and J Duncan Numerical Ranges I vol 2 ofLecture Notes in Mathematics London Mathematical SocietyCambridge UK 1971
[2] F F Bonsall and J DuncanNumerical Ranges II LondonMath-ematical Society Lecture Note Series 10 Cambridge UniversityPress Cambridge Uk 1973
[3] L A Harris ldquoThe numerical range of holomorphic functions inBanach spacesrdquo The American Journal of Mathematics vol 93pp 1005ndash1019 1971
[4] C-K Li ldquoA survey on linear preservers of numerical ranges andradiirdquo Taiwanese Journal of Mathematics vol 5 no 3 pp 477ndash496 2001
[5] V J Pellegrini ldquoNumerical range preserving operators on aBanach algebrardquo Polska Akademia Nauk Instytut Matematy-czny Studia Mathematica vol 54 no 2 pp 143ndash147 1975
[6] J-T Chan ldquoNumerical radius perserving operators on B(H)rdquoProceedings of the American Mathematical Society vol 123 no5 pp 1437ndash1439 1995
[7] S K Kim andH J Lee ldquoUniform convexity and Bishop-Phelps-Bollobas propertyrdquo Canadian Journal of Mathematics vol 66no 2 pp 373ndash386 2014
[8] K Boyko V Kadets M Martın and D Werner ldquoNumericalindex of Banach spaces and dualityrdquo Mathematical Proceedingsof the Cambridge Philosophical Society vol 142 no 1 pp 93ndash1022007
[9] C Finet M Martın and R Paya ldquoNumerical index andrenormingrdquo Proceedings of the American Mathematical Societyvol 131 no 3 pp 871ndash877 2003
[10] V Kadets M Martn and R Paya ldquoRecent progress and openquestions on the numerical index of Banach spacesrdquo Revista dela Real Academia de Ciencias Exactas Fısicas y Naturales SerieA Matematicas vol 100 pp 155ndash182 2006
[11] M Martın J Merı M Popov and B RandrianantoaninaldquoNumerical index of absolute sums of Banach spacesrdquo Journalof Mathematical Analysis and Applications vol 375 no 1 pp207ndash222 2011
[12] K Boyko V Kadets M Martın and J Merı ldquoProperties of lushspaces and applications to Banach spaces with numerical index1rdquo Studia Mathematica vol 190 no 2 pp 117ndash133 2009
Therefore we see that if (119910 119910lowast) isin Π(119883) satisfies that 119910lowastminus119911lowast0 ge 120573(120575) then |119910lowast(119879(119861)(119910))| = |119910lowast(1199110)119911lowast0 (119910)| le 1 minus 120576 Wehave
This shows that V(119879(119868 + 120572119861119861)) le max2 minus 120576 120574 + 1 lt 2 =V(119879(119868)) + V(119879(119861)) and so we have desired contradiction
From Lemma 4 we see that 119879(119868) = 119888119879119868 for somemodulus1 constant 119888119879Remark 6 From Theorem 5 we can easily construct ontoisometries which are not numerical radius preserving Forexample consider 2-dimensional Hilbert spaceR2 we definean operator 119861 R2 rarr R2 by 119861(1199091 1199092) = (1199091 minus1199092) forevery (1199091 1199092) isin R2 It is clear that the operator 119879 on L(R2)given by 119879(119860) = 119861 ∘ 119860 for every 119860 isin L(R2) is an ontoisometry However this does not preserve numerical radiussince 119879(119868) = 119888119879119868 for any 119888119879
We denote a state space S of a Banach algebraL(119883) byS = 119891 isin L (119883)lowast 10038171003817100381710038171198911003817100381710038171003817 = 1 = 119891 (119868) (8)
When 119883 is a complex Banach space according to [5Theorem 22] if the adjoint operator 119878lowast of operator 119878 L(119883) rarr L(119883) satisfies that 119878lowast(S) sub S then 119881(119878(119860)) sub119881(119860) For119879 inTheorem5 since1198790 = 119888minus1119879 119879 is an onto isometryand so satisfies 119879lowast0 (S) = S we see that 119881(1198790(119860)) = 119881(119860)Now we deduce the following main result
Theorem 7 Assume a complex Banach space 119883 has bothuniform convexity and uniform smoothness An onto isometry119879 on L(119883) is numerical radius preserving if and only if thereexists a scalar 119888119879 of modulus 1 such that 119888119879119879 is a numericalrange preserving mapping
Remark 8 Theorem 7 does not hold for some Banach spacesIn order to see this let us recall numerical index 119899(119883) of aBanach space
119899 (119883) = inf V (119879) 119879 isin 119878L(119883) (9)
We see that 119899(119883) is the greatest constant 119896 ge 0 such that119896119879 le V(119879) for every119879 isin L(119883) and so V is equivalent to theoperator norm if and only if 119899(119883) gt 0 For more informationwe give [8ndash11]
From the definition of numerical index if 119899(119883) = 1then every isometry on L(119883) preserves numerical radiusThere are many classes of Banach spaces having numericalindex 1 like 1198880 ℓ1 and ℓinfin [12] Among them one of thesimplest examples having numerical index 1 is R2 as thetypical subspace of ℓinfin Since the same operator 119879 defined inRemark 6 is also an onto isometry on this space 119879 preservesnumerical radius However this does not preserve numericalrange The reason is that we have 119881(119868) = 1 and minus1 =(0 1)119879(119868)(0 1) isin 119881(119879(119868))
Remark 9 It is not possible to say that Theorem 7 is true forevery isometry Consider shift operators 1198611 and 1198612 on ℓ119901 (1 lt119901 lt infin) defined by 1198611((119909119894)infin119894=1) = (119909119894)infin119894=2 and 1198612((119909119894)infin119894=1) =(119909119894)infin119894=0 for (119909119894)infin119894=1 isin ℓ119901 where 1199090 = 0 Then it is easy to seethat an operator 119879 given by 119879(119860) = 1198612 ∘ 119860 ∘ 1198611 for each 119860 isin119871(ℓ119901) is a numerical radius preserving isometry However 119879does not preserve numerical range since 119881(119868) = 1 and 0 =(1 0 0 0 )119879(119868)(1 0 0 0 ) isin 119881(119879(119868))
Competing Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work was supported by Kyonggi University ResearchGrant 2015
References
[1] F F Bonsall and J Duncan Numerical Ranges I vol 2 ofLecture Notes in Mathematics London Mathematical SocietyCambridge UK 1971
[2] F F Bonsall and J DuncanNumerical Ranges II LondonMath-ematical Society Lecture Note Series 10 Cambridge UniversityPress Cambridge Uk 1973
[3] L A Harris ldquoThe numerical range of holomorphic functions inBanach spacesrdquo The American Journal of Mathematics vol 93pp 1005ndash1019 1971
[4] C-K Li ldquoA survey on linear preservers of numerical ranges andradiirdquo Taiwanese Journal of Mathematics vol 5 no 3 pp 477ndash496 2001
[5] V J Pellegrini ldquoNumerical range preserving operators on aBanach algebrardquo Polska Akademia Nauk Instytut Matematy-czny Studia Mathematica vol 54 no 2 pp 143ndash147 1975
[6] J-T Chan ldquoNumerical radius perserving operators on B(H)rdquoProceedings of the American Mathematical Society vol 123 no5 pp 1437ndash1439 1995
[7] S K Kim andH J Lee ldquoUniform convexity and Bishop-Phelps-Bollobas propertyrdquo Canadian Journal of Mathematics vol 66no 2 pp 373ndash386 2014
[8] K Boyko V Kadets M Martın and D Werner ldquoNumericalindex of Banach spaces and dualityrdquo Mathematical Proceedingsof the Cambridge Philosophical Society vol 142 no 1 pp 93ndash1022007
[9] C Finet M Martın and R Paya ldquoNumerical index andrenormingrdquo Proceedings of the American Mathematical Societyvol 131 no 3 pp 871ndash877 2003
[10] V Kadets M Martn and R Paya ldquoRecent progress and openquestions on the numerical index of Banach spacesrdquo Revista dela Real Academia de Ciencias Exactas Fısicas y Naturales SerieA Matematicas vol 100 pp 155ndash182 2006
[11] M Martın J Merı M Popov and B RandrianantoaninaldquoNumerical index of absolute sums of Banach spacesrdquo Journalof Mathematical Analysis and Applications vol 375 no 1 pp207ndash222 2011
[12] K Boyko V Kadets M Martın and J Merı ldquoProperties of lushspaces and applications to Banach spaces with numerical index1rdquo Studia Mathematica vol 190 no 2 pp 117ndash133 2009
