Functional calculus on analytic UMD Banach spaces · 2020-05-19 · J. Aust. Math. Soc. 74 (2003),...

J. Aust. Math. Soc.74 (2003), 351–378

TWO RESULTS ABOUT H ∞ FUNCTIONAL CALCULUS ONANALYTIC UMD BANACH SPACES

CHRISTIAN LE MERDY

(Received 3 December 2000; revised 26 March 2002)

Communicated by A. Pryde

Abstract

Let X be a Banach space with the analytic UMD property, and letA andB be two commuting sectorialoperators onX which admit boundedH∞ functional calculi with respect to angles�1 and�2 satisfying�1 + �2 < ³ . It was proved by Kalton and Weis that in this case,A + B is closed. The first result ofthis paper is that under the same conditions,A + B actually admits a boundedH∞ functional calculus.Our second result is that given a Banach spaceX and a number 1≤ p < ∞, the derivation operator onthe vector valued Hardy spaceH p.R; X/ admits a boundedH∞ functional calculus if and only ifX hasthe analytic UMD property. This is an ‘analytic’ version of the well-known characterization of UMD bythe boundedness of theH ∞ functional calculus of the derivation operator on vector valuedL p-spacesL p.R; X/ for 1< p < ∞ (Dore-Venni, Hieber-Pruss, Pruss).

2000Mathematics subject classification: primary 47A60.

1. Introduction and main statements

This paper deals with two questions concerningH∞ functional calculus of sectorialoperators, as introduced by McIntosh on Hilbert spaces (see [23]) and then developedin the Banach space setting by Cowling, Doust, McIntosh and Yagi in [6]. Thesequestions are both closely related to the pioneering work of Dore and Venni [9]concerning the sum of commuting operators with bounded imaginary powers onUMD Banach spaces.

Let X be a Banach space and letA and B be two commuting sectorial operatorson X, with respective types!1 and!2, and with respective domainsD.A/ andD.B/.Then their sumA+ B : x 7→ A.x/+ B.x/, with domainD.A+ B/ = D.A/∩ D.B/,

c© 2003 Australian Mathematical Society 1446-8107/03$A2:00+ 0:00

351

http://www.austms.org.au/Publ/JAustMS/V74P3/n18.html

352 Christian Le Merdy [2]

is a closable operator. Assume that!1 + !2 < ³ . Then according to some earlierwork of Da Prato and Grisvard [7, Section 3], the closureA + B is in turn a sectorialoperator of type max{!1; !2}. Now assume the stronger condition thatA andB admitbounded imaginary powers, with the following estimates:

∀ s ∈ R;∥∥Ais

∥∥ ≤ K1e�1|s| and∥∥Bis

∥∥ ≤ K2e�2|s|;(1.1)

for some constantsK1; K2 > 0 and�1; �2 in .0; ³/ such that�1 + �2 < ³ . It wasproved in [9] (in the invertible case) and then in [13] and [26] (in the general case)that if X is a UMD Banach space, thenA+ B is closed. Furthermore, it was proved in[26] and [10] that under these conditions,A + B admits bounded imaginary powers.

This led to the following two natural questions. Assume that

(H) A has a boundedH∞.6�1/ functional calculus,B has a boundedH∞.6�2/

functional calculus, and�1 + �2 < ³ .

For which Banach spacesX does this imply thatA + B is closed and for whichones does this imply thatA + B admits a boundedH∞ functional calculus? Thisamounts to consider the following two possible properties (P1) and (P2) of a BanachspaceX.

(P1) WheneverA andB are commuting sectorial operators onX satisfying (H) forsome�1; �2 ∈ .0; ³/, the sumA + B is closed.(P2) WheneverA andB are commuting sectorial operators onX satisfying (H) for

some�1; �2 ∈ .0; ³/, the operatorA + B admits a boundedH∞ functional calculus.

The above questions were first tackled in [19] where it is shown for example thatBanach lattices, or Banach spaces with Pisier’s property.Þ/ satisfy (P1) and (P2). Onthe other hand, UMD Banach spaces obviously satisfy (P1) by the above mentionedDore-Venni Theorem. However the problem whether (P2) is satisfied by all UMDBanach spaces was left open in [19]. Our first result (Theorem1.1below) solves thisquestion. We will actually be able to consider the larger class of Banach spaces withthe so-called property.1/, defined by the inequality (1.4) below.

Let."i /i ≥1 be a Rademacher sequence on a probability space.�;P/. That is, the"i ’sare pairwise independent random variables on� andP."i = 1/ = P."i = −1/ = 1=2for anyi ≥ 1. Then for any finite familyx1; : : : ; xn in X, we let

∥∥∥∥∥n∑

i =1

"i xi

∥∥∥∥∥Rad.X/

=∫�

∥∥∥∥∥n∑

i =1

"i .!/xi

∥∥∥∥∥X

dP.!/ :(1.2)

Now let ."′j / j ≥1 be another Rademacher sequence on.�;P/ and assume that."i /i ≥1

and."′j / j ≥1 are mutually independent. Then for any finite family.xi j /1≤i; j ≤n in X, we

[3] Results aboutH ∞ calculus 353

let ∥∥∥∥∥n∑

i; j =1

"i "′j xi j

∥∥∥∥∥Rad2.X/

=∫�

∥∥∥∥∥n∑

i; j =1

"i .!/"′j .!/xi j

∥∥∥∥∥X

dP.!/ :(1.3)

By definition we say thatX satisfies.1/ if there is a constantC > 0 such that for anyfinite family .xi j /1≤i; j ≤n in X, we have∥∥∥∥∥ ∑

1≤i ≤ j ≤n

"i "′j xi j

∥∥∥∥∥Rad2.X/

≤ C

∥∥∥∥∥n∑

i; j =1

"i "′j xi j

∥∥∥∥∥Rad2.X/

:(1.4)

This property was explicitly defined by Kalton and Weis in [18]. It goes back to apaper of Haagerup and Pisier [15] where it is implicitly shown that any analytic UMDBanach space (AUMD in short), hence any UMD Banach space satisfies.1/ (seealso [18, Proposition 3.2]). We refer to [11], [15, Section 4] and Section2 below forthe definition of AUMD Banach spaces and relevant information. Kalton and Weisshowed in [18, Corollary 6.4] that any Banach space with property.1/ satisfies (P1)above. Our first result says that (P2) is satisfied as well for this class.

THEOREM 1.1. Let A and B be two commuting sectorial operators on a BanachspaceX and let�1; �2 in .0; ³/ such that�1 + �2 < ³ . Assume thatA has a boundedH∞.6�1/ functional calculus,B has a boundedH∞.6�2/ functional calculus andXhas property.1/. ThenA + B is closed and for any� > max{�1; �2}, A + B has aboundedH∞.6�/ functional calculus.

In [9], Dore and Venni were mainly interested in applications toL p-maximalregularity for generatorsof bounded analytic semigroups on UMD Banach spaces. Forthat purpose, they needed one of the operatorsAor B in (1.1) to be a derivation operator.Indeed, they proved the following result [9, Theorem 3.1]: given a number 1< p < ∞and a UMD Banach spaceX, the derivation operatord=dt on L p.R; X/, with domainW1;p.R; X/, admits bounded imaginary powers. This result was strengthened in [16]where it is proved that in this case,d=dt has a boundedH∞.6�/ functional calculuson L p.R; X/ for any � > ³=2. Shortly after the Dore-Venni paper appeared, Prüssshowed the following converse to their result. If the derivation operatord=dt onL p.R; X/ admits bounded imaginary powers, thenX is UMD (see [25, Section 8.1]).

Our second result (Theorem1.2 below) says that similarly, the AUMD propertycharacterizes those Banach spacesX such that the derivation operator has a boundedH∞ functional calculus (or bounded imaginary powers) onX-valued Hardy spacesH p.R; X/. We note that contrary to the above mentioned results, the valuep = 1 canbe included in our analytic setting. We will assume that the reader is familiar with


classical (= scalar valued) Hardy spaces on the real lineRand on the torusT = R=2³Z,and we refer to the monographs [17] and [12] for the necessary background.

Vector-valued Hardy spaces on the real line are defined as follows. We letX bea Banach space. Given anyf ∈ L1.R; X/, we let f : R → X denote its Fouriertransform defined by

f .¾/ =∫

f .t/e−i ¾ t dt; ¾ ∈ R:

By definition, H 1.R; X/ is the (closed) subspace ofL1.R; X/ of all functions f suchthat f .¾/ = 0 for any¾ ≤ 0. Now let 1< p < ∞. Then we defineH p.R; X/ ⊂L p.R; X/as the closure ofH1.R; X/∩L p.R; X/ in L p.R; X/. Equivalently,H p.R; X/is the subspace of all functionsf ∈ L p.R; X/ whose Poisson integral on the upperhalf-plane ofC is analytic. In the case whenX = C, these spaces coincide with theclassical Hardy spacesH p.R/.

Given 1 ≤ p < ∞, let .Tt/t≥0 denote the isometric translation semigroup onL p.R; X/, defined by

Tt. f /.s/ = f .s − t/; f ∈ L p.R; X/; t ≥ 0; s ∈ R:

Then H p.R; X/ is an invariant subspace of.Tt /t≥0. Indeed, for anyf ∈ H 1.R; X/,for anyt ≥ 0, and for any¾ ≤ 0,

[Tt. f /.¾/ = e−i ¾ t f .¾/ = 0:

The negative generator of.Tt/t≥0 onL p.R; X/ is equal to the derivation operatord=dt,with domainW1;p.R; X/. We will use the same notationd=dt to denote its restrictionto H p.R; X/, with domainW1;p.R; X/ ∩ H p.R; X/. Of course the latter coincideswith the negative generator of the restriction of.Tt /t≥0 to H p.R; X/. We do not referto eitherp or X in this notation, but the space on which we considerd=dt should beclear from the context.

We now turn to analogous definitions on the torusT = R=2³Z. We assume thatTis equipped with its normalized Haar measure. That is, if we identifyT with [−³;³/in the usual way, then the associated measure on this interval isdt=2³ . Given aBanach spaceX and f ∈ L1.T; X/, we define its (X-valued) Fourier coefficients by

f .k/ = 1

2³

∫ ³

−³f .t/e−i kt dt; k ∈ Z:

By definition, H p.T; X/ (respectivelyH p0 .T; X/) is the subspace ofL p.T; X/ of all

functions f such that f .k/ = 0 for any k < 0 (respectivelyk ≤ 0). We simplywrite H p.T/ and H p

0 .T/ in the case whenX = C. Again we may define derivation


operators in this context. Indeed, for any 1≤ p < ∞, let .Tt /t≥0 be the isometricsemigroup onL p.T; X/ defined by lettingTt. f /.s/ = f .s − t/ for any t ≥ 0 andany s ∈ R=2³Z. ThenH p.T; X/ and H p

0 .T; X/ are both invariant subspaces underthe action of.Tt/t≥0 and we letd=dt denote the negative generator of.Tt/t≥0 either onL p.T; X/, or onH p.T; X/, or onH p

0 .T; X/.Before stating our result, we note that for anyX and any 1≤ p < ∞, d=dt is

a sectorial operator of type³=2 (in the sense of (2.1) below) on eitherH p.R; X/ orH p

0 .T; X/.

THEOREM 1.2. Given a Banach spaceX, the following assertions are equiva-lent.

.i/ X is an AUMD Banach space..ii/ For all 1 ≤ p < ∞, and for all � > ³=2, d=dt has a boundedH∞.6�/

functional calculus onH p.R; X/..iii / There exist1 ≤ p < ∞ and s ∈ R

∗ such that.d=dt/i s is bounded onH p.R; X/..iv/ For all 1 ≤ p < ∞, and for all � > ³=2, d=dt has a boundedH∞.6�/

functional calculus onH p0 .T; X/.

.v/ There exist1 ≤ p < ∞ ands ∈ R∗ such that.d=dt/i s is bounded onH p

0 .T; X/.

Section2contains the necessary background onH∞ functional calculus and AUMDBanach spaces. Section3 is mainly devoted to the proof of Theorem1.1. The latterrelies on remarkable recent results of Kalton and Weis [18] connectingH∞ functionalcalculus andR-bounded sets of operators in the sense of [3]. Section4 is mainlydevoted to the proof of Theorem1.2. The latter reduces to the study of certain Fouriermultipliers on vector valued Hardy spaces and we include several results on this topic.

NOTE. The reader should notice that we use the same notationf 7→ f for all sortsof Fourier transforms.

2. Preliminaries and notation

Given a Banach spaceX, we let B.X/ denote the Banach algebra of all boundedoperators onX. If A is a linear operator onX, we let D.A/ and R.A/ denote thedomain and the range ofA respectively. Furthermore, we denote by¦.A/ the spectrumof A and by².A/ the resolvent set ofA. For½ ∈ ².A/, we let R.½; A/ = .½− A/−1

denote the corresponding resolvent operator. For! ∈ .0; ³/, let6! be the open sectorof all z ∈ C \ {0} such that| Arg.z/| < !. By definition, A is a sectorial operator oftype! ∈ .0; ³/ if A is closed and densely defined,A is one-to-one,A has a dense


range,¦.A/ ⊂ 6!, and for any� ∈ .!; ³/ there is a constantC > 0 such that

∀½ ∈ 6�

c; ‖½R.½; A/‖ ≤ C:(2.1)

We note the classical fact that if.Tt/t≥0 is a boundedC0-semigroup onX and if −Adenotes its infinitesimal generator, thenA is a sectorial operator of type³=2 providedthat A is one-to-one and has dense range.

For � ∈ .0; ³/, and for a Banach spaceE, we let H∞.6� ; E/ be the space of allbounded analytic functionsF : 6� → E. This is a Banach space for the norm

‖F‖∞;� = sup{‖F.z/‖E : z ∈ 6�}:Then we letH∞

0 .6� ; E/ be the subspace of allF ∈ H∞.6� ; E/ for which there existtwo positive numberss;C > 0 such that

‖F.z/‖E ≤ C|z|s

.1 + |z|/2s; z ∈ 6�:(2.2)

We will simply denoteH∞.6� ;C/ andH∞0 .6� ;C/ by H∞.6�/ andH∞

0 .6�/.We now come toH∞ functional calculus for sectorial operators. The definitions

and basic facts below essentially go back to [23] and [6]. The reader may also consult[20] or [18] for more information. Given a sectorial operatorA of type! ∈ .0; ³/ ona Banach spaceX, we define its commutant by

EA = {T ∈ B.X/ : T R.½; A/ = R.½; A/T; ½ ∈ ².A/} :Clearly EA is a closed subalgebra ofB.X/. Let! < < � < ³ , and let0 be theoriented contour defined by

0 .t/ ={

−tei ; t ∈ R−;

te−i ; t ∈ R+:(2.3)

Then for any functionF ∈ H ∞0 .6� ; EA/, we set

F.A/ = 1

2³ i

∫0

F.½/R.½; A/d½ :(2.4)

Since A satisfies (2.1) and F satisfies (2.2), F.A/ is well defined and belongs toB.X/. By Cauchy’s Theorem, the definition (2.4) does not depend on the choice of ∈ .!; �/. Furthermore,H∞

0 .6� ; EA/ is an algebra and the mappingF 7→ F.A/ isan algebra homomorphism. Note that the latter is unbounded in general. Now let' bethe scalar valued function defined by'.z/ = z=.1 + z/2. Then the bounded operator'.A/ = A.1 + A/−2 is one-to-one and its range is equal toD.A/ ∩ R.A/, which is


dense inX. Given anyF ∈ H∞.6� ; EA/, we may therefore defineF.A/ as follows.We note that the function'F belongs toH ∞

0 .6� ; EA/ and we set

F.A/ = '.A/−1.'F/.A/:

This possibly unbounded operator has domain equal to the space of allx ∈ X suchthat

[.'F/.A/

].x/ ∈ R.'.A//. The latter is dense andF.A/ is closed. We record for

further the following well-known lemma.

LEMMA 2.1. Let0 < ! < � < ³ be two numbers and letA be a sectorial operatorof type! on X.

(1) Let F ∈ H ∞.6� ; EA/ and let .Fn/n≥1 be the uniformly bounded sequence ofH∞

0 .6� ; EA/ defined by lettingFn.z/ = (n=.n+z/−1=.1+nz/

)F.z/ for anyz ∈ 6� .

ThenF.A/ is bounded if and only if the sequence.Fn.A//n≥1 ⊂ B.X/ is bounded andin this case,F.A/ is the strong limit of.Fn.A//n≥1.(2) Let F ∈ H∞.6� ; EA/ be a function with the following property: there exist two

positive numberss;C > 0 such that

‖F.z/‖EA≤ C

1

|z|s; z ∈ 6�:(2.5)

Assume moreover thatA is invertible. ThenF.A/ is bounded and for any ∈ .!; �/,

F.A/ = 1

2³ i

∫0

F.½/R.½; A/d½ :(2.6)

PROOF. Part (1) is a variant of the so-called ‘convergence lemma’ [23, 6]. Turningto (2), note that since 0∈ ².A/, the mapping½ 7→ F.½/R.½; A/ is bounded on0 .Hence (2.1) and (2.5) ensure that the integral in the right-hand side of (2.6) convergesand defines an element ofB.X/. It is easy to conclude from (1) and Lebesgue’sTheorem that this element ofB.X/ equalsF.A/.

We finally recall two major definitions. First, letA be a sectorial operator onX oftype! ∈ .0; ³/, and let� > !. We say thatA has a boundedH∞.6�/ functionalcalculus ifF.A/ is bounded for anyF ∈ H∞.6�/. Second, letB be another sectorialoperator onX. We say thatA andB commute ifR.½; A/R.¼; B/ = R.¼; B/R.½; A/for any½ ∈ ².A/, ¼ ∈ ².B/.

We now turn to some background and notation on AUMD Banach spaces for whichwe refer to [11] and [15, Section 4]. Equip the compact spaceTN with its productmeasure and let.t1; : : : ; tn; : : : / denote a typical element ofTN. For any integern ≥ 1, letFn denote the¦ -field generated by the firstn variablest1; : : : ; tn. Let.gn/n≥1 be anX-valued martingale with respect to the filtration.Fn/n≥1, that is, each


gn : TN → X is anFn-measurable function and lettingdn = gn − gn−1, we haveE.dn|Fn−1/ = 0 for anyn ≥ 1. As usual the convention is thatg0 = 0 andF0 isthe trivial ¦ -field. We say that.gn/n≥1 is analytic if for anyn ≥ 1, there exists ameasurable function8n : Tn−1 → X such that

dn.t1; : : : ; tn/ = 8n.t1; : : : ; tn−1/eitn; t1; : : : ; tn ∈ T:(2.7)

Let 1 ≤ p < ∞ be a number. By definitionX is an AUMD Banach space if there isa constantK p such that whenever.gn/n≥1 is anX-valued analytic martingale,N ≥ 1is an integer, and"1; : : : ; "N ∈ {−1;1}, we have an estimate∥∥∥∥∥

N∑n=1

"ndn

∥∥∥∥∥p

≤ K p

∥∥∥∥∥N∑

n=1

dn

∥∥∥∥∥p

;(2.8)

where the norms are computed inL p.TN ; X/. This property does not depend onp.Furthermore, to prove that a given Banach space is AUMD, it suffices to show (2.8)in the case when the martingale is finite (that is,dn is eventually 0), and each8n in(2.7) is anX-valued trigonometric polynomial, that is, a sum of elements of the formxeiq1t1 · · · eiqn−1tn−1, wherex ∈ X andq1; : : : ;qn−1 ∈ Z.

The class of AUMD spaces includes UMD Banach spaces,L1-spaces, and quotientsL1=R of an L1-space by one of its reflexive subspacesR. Also, it is stable undertaking subspaces. Conversely,C.�/-spaces (where� is an infinite compact set), thequotient spaceL1.T/=H1.T/, and the Schatten spaceS1.H / of trace class operatorson an infinite dimensional Hilbert spaceH are not AUMD.

3. Perturbation of R-sectorial operators and proof of Theorem1.1

The main purpose of this section is the proof of Theorem1.1. The latter relieson some recent work of Kalton and Weis [18] involving R-boundedness, and ona perturbation result (Proposition3.2 below) of independent interest, which is thekey ingredient of the proof. We shall first give the necessary background onR-boundedness. Our main reference for this notion is [3], see also [18].

Let X be a Banach space and letT ⊂ B.X/ be a set of bounded operators onX.By definition, we say thatT is R-bounded if there is a constantC > 0 such that forany finite familiesT1; : : : ;Tn in T , andx1; : : : ; xn in X, we have∥∥∥∥∥

n∑i =1

"i Ti .xi /

∥∥∥∥∥Rad.X/

≤ C

∥∥∥∥∥n∑

i =1

"i xi

∥∥∥∥∥Rad.X/

:(3.1)

In this definition, the norms‖ ‖Rad.X/ are defined by (1.2). The least constantCsatisfying (3.1) is called theR-boundedness constant ofT and is denoted byR.T /.


Obviously anyR-bounded setT is bounded and‖T‖ ≤ R.T / for any T ∈ T ,but the converse does not hold on non-Hilbertian Banach spaces. Given any two setsT1;T2 ⊂ B.X/, we letT1 +T2 = {T1 + T2 : T1 ∈ T1;T2 ∈ T2} andT1T2 = {T1T2 :T1 ∈ T1;T2 ∈ T2}. In the next lemma, we record some well-known stability resultsconcerningR-bounded sets.

LEMMA 3.1. (1) If T ⊂ B.X/ is R-bounded, then its closureST is R-boundedand R.ST / = R.T /.(2) If T1;T2 ⊂ B.X/ are R-bounded, thenT1+T2 is R-boundedandR.T1+T2/ ≤

R.T1/+ R.T2/.(3) If T1;T2 ⊂ B.X/ are R-bounded, thenT1T2 is R-bounded andR.T1T2/ ≤

R.T1/R.T2/.(4) If T ⊂ B.X/ is R-bounded, then its absolute convex hullaco.T / is R-bounded

and R.aco.T // ≤ 2R.T /.(5) Let T ⊂ B.X/ be R-bounded and letC > 0 be a constant. Then the

set{ ∫ ∞

0 f .t/T.t/dt∣∣ T : R∗

+ → T and f : R∗+ → C are continuous, and∫ ∞

0 | f .t/| dt ≤ C}

is R-bounded and itsR-boundedness constant is less than orequal to2C R.T /.

PROOF. The first three assertions are more or less obvious. The assertion (4) is [3,Lemma 3.2]. To prove (5), it therefore suffices to check that ifT : R∗

+ → T andf : R∗

+ → C are two continuous functions and if∫ ∞

0 | f .t/| dt ≤ C, then∫ ∞

0

f .t/T.t/dt ∈ C aco.T /:

For this it suffices to show that for any 0< Þ < þ < ∞,∫ þ

Þ

f .t/T.t/dt ∈ C aco.T /:

The latter property clearly follows from an approximation of the integral by Riemannsums.

PROPOSITION3.2. Let A be a sectorial operator on a Banach spaceX, and let¼ ∈ .0; ³/ be such that the set{

½R.½; A/ : ½ =∈ 6¼

}(3.2)

is R-bounded. Let¹; ' ∈ .0; ³/be two numbers such that¼+¹ < ³ andmax{¼; ¹} <'. Suppose thatF ∈ H ∞.6'/ and thatF.A/ is bounded. Then the set

{F.A + z/ : z ∈ 6¹}is R-bounded.


Note that all the operators considered in the previous statement make sense. Indeed,our assumption on (3.2) implies thatA is sectorial of type< ¼. (According to theterminology of [18], A is actually anR-sectorial operator ofR-sectorial type< ¼.)Since¼ + ¹ < ³ , A + z is therefore a sectorial operator of type max{¼; ¹} for anyz ∈ 6¹ , which allows us to defineF.A + z/ for any F ∈ H∞.6'/.

ReplacingR-boundedness by boundedness, our statement corresponds to the fol-lowing perturbation result established by Uiterdijk in his Ph.D. thesis [28]: if A is asectorial operator of type< ¼, if ¼+ ¹ < ³ and max{¼; ¹} < ' and if F ∈ H∞.6'/

is a function such thatF.A/ is bounded, thenF.A + z/ is bounded for anyz ∈ 6¹

and the resulting family of operators is bounded. Strictly speaking, Uiterdijk provedthat result only for¹ = 0 but it is possible to extend his proof to the general case. Itturns out that his arguments also yield our Proposition3.2, up to some estimates onthe R-boundedness of certain sets of operators, as explained in the proof below.

PROOF OFPROPOSITION3.2. We let A be as in Proposition3.2 and we letF ∈ H∞.6'/be such thatF.A/ is bounded. Following Uiterdijk’s idea in [28, 2.3], wedecomposeF.A+z/ for anyz ∈ 6¹ as follows. WritingI X = z.A+z/−1+A.A+z/−1,we haveF.A + z/ = z.A + z/−1F.A + z/+ A.A + z/−1F.A + z/ hence

F.A + z/ = z.A + z/−1F.A + z/+ A.A + z/−1F.A/

+ A.A + z/−1.F.A + z/ − F.A//:

Since¹ < ³ − ¼, the R-boundedness of (3.2) implies that{z.A + z/−1 : z ∈ 6¹} isR-bounded. Applying Lemma3.1(1), we deduce that{A.A + z/−1F.A/ : z ∈ 6¹} isR-bounded and that it suffices to show that{

z.A + z/−1F.A + z/ : z ∈ 6¹

}is R-bounded;(3.3)

and {A.A + z/−1

(F.A + z/− F.A/

) : z ∈ 6¹

}is R-bounded:(3.4)

We first prove (3.3). Let ¼′ > ¼ and¹ ′ > ¹ be such that¼′ + ¹ ′ < ³ and let' > max{¼′; ¹ ′}. Then there exists a positive numberr > 0 (depending on¹ and¹ ′)such that

∀ z ∈ 6¹; z− 2r |z| ∈ 6¹ ′:(3.5)

Fix somez ∈ 6¹. For any½ ∈ 6¼′, the complex number½ + z − r |z| belongs to6max{¼′;¹′}, hence to6'. Indeed,z − r |z| ∈ 6¹ ′ by (3.5) and¼′ + ¹ ′ < ³ . We maytherefore definehz ∈ H∞.6¼′/ by letting

hz.½/ = z

½+ z− r |z| F.½ + z− r |z|/; ½ ∈ 6¼′:


Let 0 = 0 be defined by (2.3), for some ∈ .¼;¼′/. Clearly½hz.½/ is boundedaway from 0 on6¼′ andA + r |z| is an invertible sectorial operator of type¼. Henceby Lemma2.1(2), hz.A + r |z|/ is bounded and

hz.A + r |z|/ = 1

2³ i

∫0

h.½/R.½; A + r |z|/d½ :

However,hz.A+r |z|/ = z.A+z/−1F.A+z/hence we have the integral representation

z.A + z/−1F.A + z/ = 1

2³ i

∫0

zF.½+ z − r |z|/½+ z − r |z| R.½− r |z|; A/d½ :(3.6)

Let I .z/ be the integral in the right-hand side of (3.6). Letting0+ = 0 ∩{Im.½/ > 0}and0− = 0∩{Im.½/ < 0}, we writeI .z/ = I+.z/+ I−.z/, whereI+.z/ = 1

2³ i

∫0+

· · ·and I−.z/ = 1

2³ i

∫0−

· · · . Then

I+.z/ = − ei

2³ i

∫ ∞

0

zF.tei + z − r |z|/tei + z − r |z| R.tei − r |z|; A/dt

= − ei

2³ i

∫ ∞

0

zF.tei + z − r |z|/.tei + z − r |z|/.tei − r |z|/

(.tei − r |z|/R.tei − r |z|; A/

)dt :

We will prove below that

supz∈6¹

∫ ∞

0

∣∣∣∣ zF.tei + z − r |z|/.tei + z − r |z|/.tei − r |z|/

∣∣∣∣ dt < ∞:(3.7)

Now recall that we chose > ¼. Thus for anyt > 0, tei =∈ 6¼ hencetei −r |z| =∈ 6¼.Hence the continuous functionT.t/ = .tei − r |z|/R.tei − r |z|; A/ is valued in theR-bounded set (3.2). Thus according to Lemma3.1 (5), and (3.7), we obtain that{I+.z/ : z ∈ 6¹} is R-bounded. Similarly, the set{I−.z/ : z ∈ 6¹} is R-bounded, andso the first required result (3.3) follows using (3.6) and Lemma3.1(1).

We now prove the crucial estimate (3.7). Write anyz ∈ 6¹ as z = |z|ei �, with|� | < ¹. Then the integral in (3.7) is

≤ ‖F‖∞;'

∫ ∞

0

|z|∣∣tei + |z|.ei � − r /∣∣∣∣tei − r |z|∣∣ dt :

Changingt into |z|t , the latter is equal to

‖F‖∞;'

∫ ∞

0

1∣∣tei + ei � − r∣∣∣∣tei − r

∣∣ dt :(3.8)


Now observe thatei � ∈ 6¹ , henceei � − 2r ∈ 6¹ ′ by (3.5). Hence 2r − ei � belongs to6³−¹ ′

c, hence to6¼′

c. Consequently, we have∣∣tei + ei � − r

∣∣ = ∣∣.tei + r /− .2r − ei �/∣∣

≥ dist(

tei + r;6¼′c)

≥ K |tei + r |;

whereK > 0 is a constant not depending ont > 0. Therefore the integral in (3.8) is

≤ ‖F‖∞;'

K

∫ ∞

0

1

|tei + r ||tei − r | dt :

This shows (3.7).We now turn to the proof of (3.4). Let 0 = 0 be defined by (2.3), for some

satisfying max{¼; ¹} < < '. We will need the following integral representation,which is essentially due to [28]. For anyz ∈ 6¹ ,

A.A + z/−1(F.A + z/− F.A/

)(3.9)

= 1

2³ i

∫0

F.½/z(

AR.½; A/.A + z/−1)R.½− z; A/d½ :

To prove this, first note from the boundedness of the operatorA.A + z/−1 and thesectoriality ofA that there exists a constantK > 0 (depending onz) such that∥∥(AR.½; A/.z + A/−1

)R.½− z; A/

∥∥ ≤ K min

{1;

1

|½||½− z|}; ½ ∈ 0:(3.10)

Hence the right-hand side of (3.9) makes sense. Assume thatF ∈ H ∞0 .6'/. Then

applying (2.4), we have

F.A + z/ = 1

2³ i

∫0

F.½/R.½ − z; A/d½ and F.A/ = 1

2³ i

∫0

F.½/R.½; A/d½;

hence (3.9) follows by applying the identity

R.½− z; A/− R.½; A/ = zR.½− z; A/R.½; A/:

Now for an arbitraryF ∈ H∞.6'/, we let .Fn/n≥1 be the sequence ofH∞0 .6'/

defined in Lemma2.1 (1) so thatA.A + z/−1(F.A + z/ − F.A/

)is the strong limit

of A.A + z/−1(Fn.A + z/ − Fn.A/

). Thanks to (3.10), we may apply Lebesgue’s

Theorem to deduce that since (3.9) holds for eachFn, it holds as well forF .We let0z

1 = 0 ∩{|½| ≤ |z|} and0z2 = 0 ∩{|½| > |z|}. According to (3.9), we have

that for anyz ∈ 6¹

A.A + z/−1(F.A + z/ − F.A/

) = z.A + z/−1B1.z/+ A.A + z/−1B2.z/;(3.11)


with

B1.z/ = 1

2³ i

∫0z

1

F.½/AR.½; A/R.½ − z; A/d½ ;

B2.z/ = 1

2³ i

∫0z

2

F.½/zR.½; A/R.½ − z; A/d½ :

Then we decompose again asB1.z/ = B+1 .z/ + B−

1 .z/ andB2.z/ = B+2 .z/ + B−

2 .z/,where B+

i .z/ corresponds to the integration along0zi ∩ {Im.½/ > 0} and B−

i .z/corresponds to the integration along0z

i ∩ {Im.½/ < 0}. Let us show that

{B+i .z/ : z ∈ 6¹} is R-bounded; i = 1;2:(3.12)

For anyz ∈ 6¹, we have

B+1 .z/ = − ei

2³ i

∫ |z|

0

F.tei /AR.tei ; A/R.tei − z; A/dt

= − ei

2³ i

∫ |z|

0

F.tei /

tei − z

(AR.tei ; A/.tei − z/R.tei − z; A/

)dt :

By assumption, the set{¦ R.¦; A/ : ¦ =∈ 6¼} is R-bounded hence the set{AR.¦; A/ :¦ =∈ 6¼} is R-bounded as well. Hence applying Lemma3.1(2) we find that{

¦ R.¦; A/!R.!; A/ : ¦;! =∈ 6¼

}(3.13)

is R-bounded. Now for any 0< t < |z|, the operatorAR.tei ; A/.tei − z/R.tei −z; A/ belongs to the set (3.13). Hence by Lemma3.1(5), the set{B+

1 .z/ : z ∈ 6¹} isR-bounded provided that

supz∈6¹

∫ |z|

0

∣∣F.tei /∣∣∣∣tei − z∣∣ dt < ∞ :(3.14)

To check (3.14), we letz = |z|ei � be an arbitrary element of6¹ with |� | < ¹. Thenchangingt into |z|t , the integral in (3.14) is

≤ ‖F‖∞;'

∫ 1

0

dt

|tei − ei �| :Hence it remains to observe that the latter integral is less than or equal to the inverseof the distance between the two disjoint compact sets{tei : t ∈ [0;1]} and {ei � :� ∈ [−¹; ¹]}. This concludes the proof of theR-boundedness of{B+

1 .z/ : z ∈ 6¹}.Similarly we write

B+2 .z/ = − ei

2³ i

∫ ∞

|z|F.tei /zR.tei ; A/R.tei − z; A/dt

= − ei

2³ i

∫ ∞

|z|

zF.tei /

.tei /.tei − z/

(tei R.tei ; A/.tei − z/R.tei − z; A/

)dt :


The proof that{B+2 .z/ : z ∈ 6¹} is R-bounded reduces to showing that

supz∈6¹

∫ ∞

|z|

|z||F.tei /||tei ||tei − z| dt < ∞ :(3.15)

We let z = |z|ei � with |� | < ¹ and changingt into |z|t , we find that the integral in(3.15) is

≤ ‖F‖∞;'

∫ ∞

1

dt

t |tei − ei � | :

There is a constantK > 0 such that|tei −ei � | ≥ K t for anyt ≥ 1 and any� ∈ [−¹; ¹].Consequently the latter integral is less than or equal toK −1 ‖F‖∞;'

∫ ∞1 dt=t2. This

completes the proof of (3.12).We now conclude the proof by applying Lemma3.1. Arguing as above we find

that the sets{B−i .z/ : z ∈ 6¹} are R-bounded hence the sets{Bi .z/ : z ∈ 6¹} are

R-bounded. Since−6¹ = 6³−¹c ⊂ 6¼

c, our assumption implies that{z.A + z/−1 :

z ∈ 6¹} is R-bounded. We therefore deduce that{z.A + z/−1B1.z/ : z ∈ 6¹} isR-bounded. Likewise{A.A + z/−1 : z ∈ 6¹} hence{A.A + z/−1B2.z/ : z ∈ 6¹} isR-bounded. Now the result (3.4) follows from the decomposition (3.11).

PROOF OFTHEOREM 1.1. Suppose thatA, B satisfy the assumptions of Theorem1.1on a Banach spaceX with property.1/. We already know thatA + B is closed by[18, Corollary 6.4], hence is sectorial of type max{�1; �2}. We let � > max{�1; �2}and F ∈ H∞.6�/. Then we choose¼ > �1 and¹ > �2 such that¼ + ¹ < ³ and� > max{¼; ¹}. SinceA has a boundedH∞.6�1/ functional calculus andX satisfies.1/, the set{½R.½; A/ : z =∈ 6¼} is R-bounded by [18, Theorem 5.3, (3)]. Hence theset

{F.A + z/ : z ∈ 6¹}(3.16)

is R-bounded by Proposition3.2 applied with' = � . Since A and B commute,the functionF.A + ·/ : z 7→ F.A + z/ takes values in the commutant algebraEB,and hence belongs toH∞.6� ; EB/. Now the R-boundedness of (3.16) implies that.F.A + ·//.B/ is bounded by [18, Theorem 4.4]. Hence it remains to check that

.F.A + ·//.B/ = F.A + B/;(3.17)

which shows thatF.A + B/ is a bounded operator.To check this equality, we use the function'.z/ = z=.1 + z/2 considered in

Section2. We let!1 and!2 denote the respective types ofA andB. We let01; 02; 03

be three contours as defined by (2.3) corresponding to three numbers 1; 2; 3 such


that!1 < 1 < �1, !2 < 2 < �2, and max{�1; �2} < 3 < � . It follows from the firstpart of the proof of [19, Theorem 4.1] that for any½ =∈ 6max{�1;�2}, we have

'.A/'.B/R.½; A + B/ =(

1

2³ i

)2 ∫01

∫02

'.z′/'.z/½− .z′ + z/

R.z′; A/R.z; B/dzdz′ :

(3.18)

Hence ifF ∈ H∞0 .6�/, we have

F.A + B/'.A/'.B/

= 1

2³ i

∫03

F.½/'.A/'.B/R.½; A + B/d½ by (2.4)

=(

1

2³ i

)3 ∫03

∫01

∫02

F.½/'.z′/'.z/½− .z′ + z/

R.z′; A/R.z; B/d½dzdz′ by (3.18)

=(

1

2³ i

)2 ∫01

∫02

F.z′ + z/'.z′/'.z/R.z′; A/R.z; B/dzdz′

by Cauchy’s Theorem

= 1

2³ i

∫02

F.A + z/'.A/'.z/R.z; B/dz = (F.A + ·/).B/'.A/'.B/:

Since'.A/'.B/ has a dense range this shows (3.17) in the case whenF ∈ H ∞0 .6�/.

The general case now follows from Lemma2.1(1).

We note that Theorem1.1 does not remain true if sums are replaced by products,even on UMD Banach spaces. Namely, letA; B be two commuting sectorial operatorson a UMD Banach space, and assume thatA has a boundedH∞.6�1/ functionalcalculus,B has a boundedH∞.6�2/ functional calculus, and�1 + �2 < ³ . Then theoperatorAB, with domain equal to the space of allx ∈ D.B/ such thatB.x/ ∈ D.A/,is closable and it is proved in [26] that its closureAB is a sectorial operator of type�1 + �2 which admits bounded imaginary powers. HoweverAB does not admit aboundedH∞ functional calculus in general. Indeed given 1≤ p < ∞, let Sp denotethe Schatten space of all compact operatorsT : `2 → `2 such that|T |p has a finitetrace, equipped with the norm‖T‖p = .tr.|T |p//1=p. Then Sp is a UMD Banachspace if 1< p < ∞ and the example given in the proof of [19, Theorem 3.9] toshow thatSp fails the so-called joint calculus property shows as well that ifp 6= 2,there exist commuting operatorsA; B on Sp so thatAB does not admit a boundedH∞ functional calculus although for any�1; �2 > 0, A (respectivelyB) has a boundedH∞.6�1/ (respectivelyH∞.6�2/) functional calculus.

We do not know any Banach space satisfying (P1) without satisfying (P2) or satis-fying (P2) without satisfying (P1). The example below [18, Corollary 6.4] showing


that if X is a Banach space such that Rad2.X/ satisfies (P1), thenX has property.1/,can be easily adapted to show that if Rad2.X/ satisfies (P2), thenX has property.1/.On the other hand we notice that the Banach spaceS∞ of compact operators on2

does not satisfy (P2) by [28, Chapter 7]. The same argument shows thatS1 does notsatisfy (P2). We conclude this section by a remark and an open question.

REMARK 3.3. Up to now, we know two classes of Banachspaces satisfying (P1) and(P2), namely Banach spaces with property (1) and Banach spaces with the so-calledproperty.A/ introduced in [19]. It is natural to consider the following property, whichis both weaker than (1) and.A/. Let us say that a Banach spaceX satisfies (W1)(for weak (1)) if there is a constantC > 0 such that for any finite families.xi j /1≤i; j ≤n

in X and.x∗i j /1≤i; j ≤n in X∗ we have

∣∣∣∣∣ ∑1≤i ≤ j ≤n

〈xi j ; x∗i j 〉∣∣∣∣∣ ≤ C

∥∥∥∥∥n∑

i; j =1

"i "′j xi j

∥∥∥∥∥Rad2.X/

∥∥∥∥∥n∑

i; j =1

"i "′j x

∗i j

∥∥∥∥∥Rad2.X∗/

:

Any Banach space with property (W1) satisfies (P1). Indeed the argument in theproof of [18, Theorem 5.3, (3)] can be easily adapted to show that ifA has a boundedH∞.6�/ functional calculus onX, thenA is W R-sectorial (in the sense of [18]) withrespect to any! > � . The result therefore follows from [18, Theorem 4.5]. Howeverwe do not know whether any Banach space with property (W1) satisfies (P2). Inparticular we do not know whether our Proposition3.2 remains true if ‘R-bounded’is replaced by ‘W R-bounded’.

4. Fourier multipliers on vector valued Hardy spacesand proof of Theorem1.2

In this section, we shall first explain how to define Fourier multipliers on spacesH p.R; X/ or H p

0 .T; X/ and their link with theH∞ functional calculus of the derivationoperators. We shall then study the relationships between multipliers onH p.R; X/ andmultipliers onH p

0 .T; X/ and establish the proof of Theorem1.2.Let 1 ≤ p < ∞ be a number and letX be an arbitrary Banach space. LetP denote

the space of all complex trigonometric polynomials, that is, the linear span of thefunctionsek.t/ = eikt , k ∈ Z. Then we letP A (respectivelyP A

0 ) denote the subspaceofP spanned by{ek : k ≥ 0} (respectively{ek : k ≥ 1}). Using for example Fejér’sapproximation, we see thatP A ⊗ X andP A

0 ⊗ X are dense subspaces ofH p.T; X/andH p

0 .T; X/ respectively. Let.mk/k≥1 be a bounded sequence of complex numbers.We say that.mk/k≥1 is a bounded Fourier multiplier onH p

0 .T; X/ if there exists a


constantC > 0 such that for anyf = ∑k≥1 f .k/ ⊗ ek ∈P A

0 ⊗ X,∥∥∥∥∥∑k≥1

m.k/ f .k/ ⊗ ek

∥∥∥∥∥p

≤ C

∥∥∥∥∥∑k≥1

f .k/⊗ ek

∥∥∥∥∥p

:(4.1)

In that case there is a (necessarily unique) bounded operator onH p0 .T; X/ mapping

x ⊗ ek to m.k/x ⊗ ek for anyx ∈ X and anyk ≥ 1. Its norm is the least constantCsatisfying (4.1) and is called the norm of the Fourier multiplier.mk/k≥1 on H p

0 .T; X/.Of course, similar definitions can be given onH p.T; X/.

We now proceed to Fourier multipliers on H p.R; X/. We first note that the tensorproductH p.R/ ⊗ X is a dense subspace ofH p.R; X/. Indeed the arguments in [17,Chapter 8] showing thatH p.R/ andH p.T/ are isometric extend almost verbatim to thevector-valued case giving an isometric isomorphismJp;X : H p.R; X/ → H p.T; X/.MoreoverJp;X mapsH p.R/⊗X ontoH p.T/⊗X. Since the latter is dense inH p.T; X/(by Fejer’s approximation), we deduce the density ofH p.R/ ⊗ X in H p.R; X/.Consequently,.H2.R/ ∩ H p.R// ⊗ X is a dense subspace ofH p.R; X/.

Let m : R∗+ → C be a bounded measurable function, and let−m : H 2.R/ → H2.R/

be the associated Fourier multiplier defined by letting\−m. f / = m f on R∗+ for any

f ∈ H 2.R/. We say thatm is a bounded Fourier multiplier onH p.R; X/ if −m mapsH 2.R/ ∩ H p.R/ into itself and if there exists a constantC > 0 such that for any∑

k fk ⊗ xk in(H 2.R/ ∩ H p.R/

)⊗ X,∥∥∥∥∥∑k

−m. fk/⊗ xk

∥∥∥∥∥p

≤ C

∥∥∥∥∥∑k

fk ⊗ xk

∥∥∥∥∥p

:(4.2)

In other words,m is a bounded Fourier multiplier onH p.R; X/ if −m ⊗ I X is boundedwith respect to theH p.R; X/-norm. In that case,−m ⊗ I X uniquely extends to abounded operator onH p.R; X/ whose norm is the least constantC satisfying (4.2).This norm will be called the norm of the Fourier multiplierm on H p.R; X/.

LEMMA 4.1. Let 1 ≤ p < ∞ and consider a functionF ∈ H∞.6�/ for some� > ³=2.

(1) Let A = d=dt on the spaceH p.R; X/. ThenF.A/ is bounded if and only if thefunctionm : R∗

+ → C defined bym.¾/ = F.i ¾/ is a bounded Fourier multiplier onH p.R; X/.(2) Let A = d=dt on the spaceH p

0 .T; X/. ThenF.A/ is bounded if and only if thesequence.F.ik//k≥1 is a bounded Fourier multiplier onH p

0 .T; X/.

PROOF. This result is elementary. Using the definitions above, one can reduce tothe scalar case. Then it suffices to apply the formulaf ′.¾/ = i ¾ f .¾/ (in case (1)) orf ′.k/ = ik f .k/ (in case (2)) for suitable functions.


LEMMA 4.2. Let f : R → C be a continuous2³-periodic function, and let' ∈L1.R/ with

∫'.t/dt = 1. Then

1

2³

∫ ³

−³f .t/dt = lim

�→0�

∫f .t/'.�t/dt:(4.3)

PROOF. By equicontinuity, we may reduce to the case whenf ∈ P hence bylinearity, we may assume thatf = ek for somek ∈ Z. Since we have

�

∫ek.t/'.�t/dt = '

(−k

�

);

the result follows at once.

The next result extends the well-known fact that ifm is a bounded and continuousfunction onR, thenm is a bounded Fourier multiplier onL p.R/ if and only if thesequences.m."k//k∈Z are uniformly bounded Fourier multipliers onL p.T/. In thelatter result, the ‘if’ part is due to Stein and Weiss [27, VII, Theorem 3.18] whereasthe ‘only if’ part goes back to de Leeuw [21]. Note that de Leeuw’s Theoremcan be regarded as a consequence of the Coifman-Weiss transference principle (see[4, Chapter 3]). The above mentioned equivalence extends to vector-valuedL p-spaces with identical proofs. To deal with vector-valuedH p-spaces, we will needsome substantial modifications that are indicated below. Note for example that theCoifman-Weiss transference principle is no longer available for Hardy spaces.

PROPOSITION4.3. Let1 ≤ p < ∞ be a number, letm : R∗+ → C be a bounded and

continuous function, and letC ≥ 0 be a constant. Then the following two assertionsare equivalent:

.i/ m is a bounded Fourier multiplier onH p.R; X/ whose norm is less than orequal toC;.ii/ for any " ∈ .0;1], the sequence.m."k//k≥1 is a bounded Fourier multiplier

on H p0 .T; X/ whose norm is less than or equal toC.

PROOF. We first show that (i) implies (ii), by adapting the proof of [21, Theo-rem 2.3]. In proving (ii), we may clearly assume that" = 1. Indeed ifm satisfies (i),then the function¾ 7→ m."¾/ satisfies it as well for any" > 0. We let 1< q ≤ ∞ bethe conjugate number ofp, that is,p−1 + q−1 = 1. Suppose that

P.t/ =∑k≥1

P.k/eikt and Q.t/ =∑

n

Q.n/eint :(4.4)

are two vector-valued trigonometric polynomials withP ∈P A0 ⊗X andQ ∈P⊗X∗.

Let us consider two (scalar-valued) functions 1 ∈ L1.R/ ∩ L p.R/ and 2 ∈ L1.R/ ∩


Lq.R/ which satisfy

‖ 1‖p = ‖ 2‖q = 1 and Supp. 1/ ⊂ [−1;1]:(4.5)

Then for any� ∈ .0;1/, let us defineP� ∈ L1.R; X/∩L p.R; X/andQ� ∈ L1.R; X∗/∩Lq.R; X∗/ by letting P�.t/ = P.t/ 1.�t/ and Q�.t/ = Q.t/ 2.�t/ for any t ∈ R.Givenk ≥ 1, � ∈ .0;1/, and¾ ∈ R, the Fourier transform oft 7→ eikt 1.�t/ at thepoint ¾ equals.1=�/ 1

(.¾ − k/=�

). Hence we have

P�.¾/ = 1

�

∑k≥1

P.k/ 1

(¾ − k

�

); ¾ ∈ R:(4.6)

For any¾ ≤ 0 andk ≥ 1, we have¾ − k ≤ −1. Hence.¾ − k/=� ≤ −1 andhence 1

(.¾ − k/=�

) = 0. It therefore follows from (4.6) that P� actually belongs toH 1.R; X/ ∩ L p.R; X/ ⊂ H p.R; X/.

Applying the assumption (i), letT : H p.R; X/ → H p.R; X/ be the boundedFourier multiplier operator induced bym, which we may apply toP�. We will showthat

lim�→0

�〈T.P�/;Q�〉 = 〈 1; 2〉⟨∑

k≥1

m.k/P.k/ ⊗ ek;Q

⟩;(4.7)

where the brackets in the left-hand side stand for the duality betweenL p.R; X/ andLq.R; X∗/, the first brackets in the right-hand side stand for the duality betweenL p.R/

andLq.R/, and the second brackets in the right-hand side stand for the duality betweenL p.T; X/ andLq.T; X∗/.

Once this is established, one can conclude as follows. On the one hand, we notethat

�∥∥P�

∥∥p

p= �

∫‖P.t/‖p

X | 1.�t/|p dt → 1

2³

∫ ³

−³‖P.t/‖p

X dt = ‖P‖pp

by Lemma4.2. Similarly, �∥∥Q�

∥∥q

q→ ‖Q‖q

q if q < ∞, and in the case whenq = ∞,

we have an estimate∥∥Q�

∥∥∞ ≤ ‖Q‖∞. Hence, we finally have that

lim sup�→0

�∥∥P�

∥∥p

∥∥Q�

∥∥q

≤ ‖P‖p ‖Q‖q ;

and hence

lim sup�→0

∣∣�〈T.P�/;Q�〉∣∣ ≤ C ‖P‖p ‖Q‖q :(4.8)

On the other hand, recall that the natural embeddingL p.T; X/ ,→ Lq.T; X∗/∗ is anisometry (see for example [8, IV.1]). We easily deduce that∥∥∥∥∥∑

k≥1

m.k/P.k/ ⊗ ek

∥∥∥∥∥p

= sup

{∣∣∣∣∣⟨∑

k≥1

m.k/P.k/ ⊗ ek;Q

⟩∣∣∣∣∣};


where the supremum runs over allQ ∈ P ⊗ X∗ with ‖Q‖q ≤ 1. Combiningwith (4.8) and (4.7), we obtain that|〈 1; 2〉|

∥∥∑k≥1 m.k/P.k/ ⊗ ek

∥∥p

≤ C ‖P‖p.Since the supremum of|〈 1; 2〉| over all 1 and 2 satisfying (4.5) equals 1, we finallyobtain that.m.k//k≥1 is a bounded Fourier multiplier onH p

0 .T; X/ with norm≤ C.It therefore remains to prove (4.7). Applying (4.6) we see that

\T.P�/.¾/ = 1

�

∑k≥1

P.k/m.¾/ 1

(¾ − k

�

); ¾ > 0:

Furthermore arguing as in the proof of (4.6), we find that

Q�.¾/ = 1

�

∑n

Q.n/ 2

(¾ − n

�

); ¾ ∈ R:

Hence we have for any� ∈ .0;1/

〈T.P�/;Q�〉 = 1

2³

⟨\T.P�/; Q�.−·/⟩

= 1

2³�2

∫ ⟨∑k≥1

P.k/m.¾/ 1

(¾ − k

�

);∑

n

Q.n/ 2

(−¾ − n

�

)⟩d¾

= 1

2³�2

∑k;n

⟨P.k/; Q.n/

⟩ ∫m.¾/ 1

(¾ − k

�

) 2

(−¾ − n

�

)d¾:

Let us denote byIk;n.�/ the integral on the right-hand side of this equality. By thechange of variables = .¾ + n/=�, we see that

Ik;n.�/ = �

∫m.�s − n/ 1

(s − n + k

�

) 2.−s/ds:

In the case whenn = −k, this integral equals�∫

m.k +�s/ 1.s/ 2.−s/ds hence thecontinuity ofm at the pointk yields

lim�→0

1

�Ik;−k.�/ = m.k/

∫ 1.s/ 2.−s/ds = 2³m.k/〈 1; 2〉:

Assume now thatn + k 6= 0. Since Supp. 1/ ⊂ [−1;1], we have∣∣∣∣1� Ik;n.�/

∣∣∣∣ ≤ ‖m‖∞

∫ ∣∣∣∣ 1

(s − n + k

�

)∣∣∣∣ ∣∣ 2.−s/∣∣ ds

≤ ‖m‖∞ ‖ 1‖1 sup

{∣∣ 2.−s/∣∣ : −1 + n + k

�≤ s ≤ 1 + n + k

�

}→ 0 when � → 0 since lim

s→∞ 2.s/ = 0:


Combining these estimates, we finally obtain that

lim�→0

�〈T.P�/;Q�〉 = 〈 1; 2〉∑k≥1

m.k/⟨P.k/; Q.−k/

⟩= 〈 1; 2〉 1

2³

∫ ³

−³

⟨∑k≥1

m.k/P.k/eikt ;∑

n

Q.n/eint

⟩dt

= 〈 1; 2〉⟨∑

k≥1

m.k/P.k/⊗ ek;Q

⟩;

which completes the proof of (4.7).We now prove that (ii) implies (i), by adapting the proof of [27, VII, Theorem 3.18]

to our analytic setting. In particular we will use the Poisson summation principle. Welet U be the space of allC∞ functions f : R → C belonging toH 1.R/ such thatlim |t |→∞ |t2 f .t/| = 0. Then according to [12, Chapter II, Corollary 3.3],U is a densesubspace ofH p.R/ for any 1≤ p < ∞. Therefore,U ⊗ X is dense inH p.R; X/.Hence to prove (i), it suffices to prove (4.2) onU ⊗ X.

Let T = −m ⊗ I X be the Fourier multiplier operator corresponding tom defined onU ⊗ X and let f be an arbitrary element ofU ⊗ X. Since any function inU is C∞

henceC2, we have lim|¾ |→∞ |¾ |2 ∥∥ f .¾/∥∥ = 0 hence lim|¾ |→∞ |¾ |2|m.¾/|∥∥ f .¾/

∥∥ = 0.By Fourier’s inversion formula we deduce that for anyt ∈ R,

T f .t/ = 1

2³

∫ ∞

0

m.¾/ f .¾/eit ¾ d¾ :

Sincem is continuous, we deduce by means of Riemann sums that

T f .t/ = lim"→0

("

2³

∞∑k=1

m."k/ f ."k/eit "k

); t ∈ R:(4.9)

For any" ∈ .0;1/, let F" be defined by

F".t/ = 2³

"

∑j ∈Z

f

(t + 2³ j

"

); t ∈ R:

Since f ∈ U ⊗ X, F" is a well-defined continuous function onR. MoreoverF" is2³-periodic. Then regarding it as an element ofC.T; X/, we see that

∀ k ∈ Z; F".k/ = f ."k/:(4.10)

Indeed, this follows from the standard proof of the Poisson summation formula. Thisshows in particular thatF" ∈ H p

0 .T; X/. We now claim that

lim"→0

"p−1

.2³/p

∫ ³

−³‖F".t/‖p dt = ‖ f ‖p

p :(4.11)


Indeed, by our assumption thatf ∈ U ⊗ X, there is a constantK > 0 such that‖ f .s/‖ ≤ K |s|−2 for anys satisfying|s| ≥ ³ . Let t ∈ [−³;³]. For any non-zerointeger j and any" ∈ .0;1/, we have

∣∣.t + 2³ j /="∣∣ ≥ ³ and hence∥∥∥∥ f

(t + 2³ j

"

)∥∥∥∥ ≤ K "2

.t + 2³ j /2≤ K "2

³2.2| j | − 1/2:

We deduce that for some absolute constantK ′ > 0, we have∥∥∥∥F".t/ − 2³

"f

(t

"

)∥∥∥∥ ≤ K ′"; t ∈ [−³;³]; " ∈ .0;1/:

Integrating this inequality and passing to the limit, we obtain that

lim"→0

"p−1

∫ ³

−³

∥∥∥∥F".t/ − 2³

"f

(t

"

)∥∥∥∥p

dt = 0:

Our claim (4.11) now follows from the fact that

"p−1

∫ ³

−³

∥∥∥∥2³

"f

(t

"

)∥∥∥∥p

dt = .2³/p

∫ ³="

−³="‖ f .s/‖p ds

→ .2³/p ‖ f ‖pp when " → 0:

For any" ∈ .0;1/, we let T" : H p0 .T; X/ → H p

0 .T; X/ be the bounded Fouriermultiplier operator induced by the sequence.m."k//k≥1. Then we let� : R → R+ bea compactly supported continuous function such that�.0/ = 1 and∑

j ∈Z�.t + 2³ j /p = 1; t ∈ R:

(See [27, VII, Lemma 3.21].) Arguing as in [27, page 266], we find that

"p

∫‖T"F"."t/�."t/‖ p dt = "p−1

∫ ³

−³‖T"F".y/‖p dy:

Hence applying our assumption, we have

"p

∫‖T"F"."t/�."t/‖ p dt ≤ Cp"p−1

∫ ³

−³‖F".y/‖p dy:(4.12)

Combining (4.10) and (4.9), we infer that

T f .t/ = lim"→0

"

2³T"F"."t/�."t/; t ∈ R:

It therefore follows from Fatou’s Lemma, (4.12), and (4.11) that∫‖T f .t/‖p dt ≤ Cp

∫‖ f .t/‖p dt:

This concludes the proof thatm satisfies (i).


In proving Theorem1.2, we will need the following elementary result whose proofis left to the reader.

LEMMA 4.4. Let A ⊂ Z be a finite set of integers, lets; > 0 be two positivenumbers, and let" ∈ {−1;1} be a sign. Then there exists an integerk ≥ 1 such thatfor anya ∈ A, a + k ≥ 1 and|.a + k/i s − "| < .

PROOF OFTHEOREM 1.2. Clearly (ii) implies (iii) and (iv) implies (v). More-over, (ii) implies (iv) and (iii) implies (v) by Lemma4.1 and Proposition4.3. Thuswe only have to show that (i) implies (ii) and (v) implies (i).

Assume (i), that is,X is an AUMD Banach space, and let.mk/k≥1 be a boundedsequence of complex numbers. Then according to Blower’s extension of Mikhlin’sTheorem in [1], .mk/k≥1 is a bounded Fourier multiplier onH 1

0 .T; X/ provided that

C1 = supk≥1

k|mk+1 − mk| < ∞ and

C2 = supk≥1

k2|mk+2 − 2mk+1 + mk| < ∞:(4.13)

Moreover, lettingC0 = supk≥1 |mk|, the norm of the Fourier multiplier.mk/k≥1 onH 1

0 .T; X/ only depends onC0, C1, C2 and on the ‘AUMD constant’K1 appearingin (2.8) for p = 1. Furthermore it is easy to check that for any 1< p < ∞, Blower’sTheorem extends to Fourier multipliers onH p

0 .T; X/ with the same proof. Of coursein this case,K p replacesK1 in the estimate of the norm of the Fourier multiplier.mk/k≥1 on H p

0 .T; X/.We now prove (ii). Let 1≤ p < ∞ and � > ³=2 be two numbers, and let

F ∈ H∞.6�/. Then according to Lemma4.1, Proposition4.3, and the precedingdiscussion, it suffices to show that for any" ∈ .0;1/, the complex numbersmk =F.i "k/ satisfy (4.13) and that the resulting constantsC1, C2 are uniformly boundedwith respect to" ∈ .0;1/. To prove this, we first note the well-known fact that foranyz ∈ iR, |zF′.z/| ≤ K�‖F‖∞;� and|z2F ′′.z/| ≤ K�‖F‖∞;� , for some constantK�

only depending on� . Indeed, this follows from Cauchy’s Theorem and probably goesback to [5]. Now for anyk ≥ 1 and any" ∈ .0;1/, we have

mk+1 − mk = F.i ".k + 1//− F.i "k/ = i "∫ 1

0

F ′.i ".k + t//dt

hence

k|mk+1 − mk| ≤ k"∫ 1

0

|F ′.i ".k + t//| dt

≤ k"K�‖F‖∞;�

∫ 1

0

dt

".k + t/≤ K�‖F‖∞;� :


Similarly, we have

mk+2−2mk+1+mk = (F.i ".k+2//− F.i ".k+1//

)−(F.i ".k+1//− F.i "k/)

= .i "/2∫ 1

0

∫ 1

0

F ′′.i ".k + t + s//dt ds

hencek2|mk+2 − 2mk+1 + mk| ≤ K�‖F‖∞;� . This completes the proof of (ii).We now assume (v) and prove (i). We will use Bourgain’s transference technique

introduced in [2]. We note that our proof is close to the Guerre-Delabrière char-acterization of UMD spaces [14]. Let p and s be given by (v). For any integerk ≥ 1, .ik/i s = kise−s³=2 hence by Lemma4.1, the sequence.kis/k≥1 is a boundedFourier multiplier onH p

0 .T; X/. We denote byT : H p0 .T; X/ → H p

0 .T; X/ theresulting bounded operator. LetN ≥ 1 be an integer, and for any 1≤ n ≤ N, let8n : Tn−1 → X be anX-valued trigonometric polynomial (81 ∈ X being a constant),and letdn : Tn → X be defined by (2.7). Given "1; : : : ; "N ∈ {−1;1}, we aim toprove that ∥∥∥∥∥

N∑n=1

"ndn

∥∥∥∥∥p

≤ 21−1=p ‖T‖∥∥∥∥∥

N∑n=1

dn

∥∥∥∥∥p

;(4.14)

where the norms are computed inL p.TN ; X/. This will show thatX is an AUMDBanach space.

For 1 ≤ n ≤ N, let3.8n/ ⊂ Zn−1 be the spectrum of8n, that is, the support of

the Fourier transform of8n. Then3.8n/ is a finite set and

8n.t1; : : : ; tn−1/ =∑

q∈3.8n/

8n.q/eiq1t1 · · · eiqn−1tn−1; t1; : : : ; tn−1 ∈ T:(4.15)

We let Ž > 0 be an arbitrary positive number. Then we define (by induction) asequencek1; : : : ; kN of positive integers as follows. We letCn = ∑

q∈3.8n/

∥∥8n.q/∥∥

for 1 ≤ n ≤ N. We first choosek1 ≥ 1 such that

|kis1 − "1| ≤ Ž

NC1:

Then we assume that 2≤ n ≤ N and thatk1; : : : ; kn−1 have been chosen. We let

An ={

n−1∑j =1

qj kj : q = .q1; : : : ;qn−1/ ∈ 3.8n/

}:

ThenAn is a finite subset ofZ and so applying Lemma4.4with A = An and" = "n,we choosekn ≥ 1 such that

ifn−1∑j =1

qj kj ∈ An; then kn +n−1∑j =1

qj kj ≥ 1(4.16)


and ∣∣∣∣∣∣(

kn +n−1∑j =1

qj kj

)i s

− "n

∣∣∣∣∣∣ ≤ Ž

NCn

:

Thus for any 1≤ n ≤ N, we have the following estimate

∑q∈3.8n/

∣∣∣∣∣∣(

kn +n−1∑j =1

qj kj

)i s

− "n

∣∣∣∣∣∣ ∥∥8n.q/∥∥ ≤ Ž

n:(4.17)

We fix t1; : : : ; tN ∈ T and, for any 1≤ n ≤ N, we introduce1n : T → X byletting1n.t/ = dn.t1 + k1t; : : : ; tn + knt/ for anyt ∈ T. Then we have

1n.t/ = 8n.t1 + k1t; : : : ; tn−1 + kn−1t/ei .tn+knt/ by (2.7)

=∑

q∈3.8n/

8n.q/eiq1.t1+k1t/ · · · eiqn−1.tn−1+kn−1t/ei .tn+knt/ by (4.15)

=∑

q∈3.8n/

8n.q/eiq1t1 · · · eiqn−1tn−1eitnei(

kn+∑n−1j=1 qj kj

)t:

Looking at (4.16), we see that1n is an analytic polynomial without constant term(that is,1n ∈P A

0 ⊗ X) and applying the Fourier multiplier operatorT , we obtain that

T1n.t/ =∑

q∈3.8n /

8n.q/eiq1t1 · · · eiqn−1tn−1eitn

(kn +

n−1∑j =1

qj kj

)i s

ei(

kn+∑n−1j=1 qj kj

)t; t ∈ T:

It therefore follows from (4.17) that for any 1≤ n ≤ N and anyt ∈ T,

‖T1n.t/ − "n1n.t/‖ ≤ Ž=N

whence∥∥T(∑N

n=11n

).t/ − ∑N

n=1 "n1n.t/∥∥ ≤ Ž, t ∈ T. Integrating onT yields∥∥T

(∑Nn=11n

)−∑Nn=1 "n1n

∥∥p

≤ Ž, hence∥∥∑N

n=1 "n1n

∥∥p

≤ Ž + ‖T‖∥∥∑Nn=11n

∥∥p,

whence ∥∥∥∥∥N∑

n=1

"n1n

∥∥∥∥∥p

p

≤ 2p−1

Ž p + ‖T‖p

∥∥∥∥∥N∑

n=1

1n

∥∥∥∥∥p

p

:More explicitly,∫ ³

−³

∥∥∥∥∥N∑

n=1

"ndn.t1 + k1t; : : : ; tn + knt/

∥∥∥∥∥p

dt

2³

≤ 2p−1

(Ž p + ‖T‖p

∫ ³

−³

∥∥∥∥∥N∑

n=1

dn.t1 + k1t; : : : ; tn + knt/

∥∥∥∥∥p

dt

2³

):


Now by integrating the latter inequality with respect to.t1; : : : ; tN/ ∈ TN and applying

Fubini’s Theorem, we deduce that∥∥∥∥∥N∑

n=1

"ndn

∥∥∥∥∥p

p

≤ 2p−1

Ž p + ‖T‖p

∥∥∥∥∥N∑

n=1

dn

∥∥∥∥∥p

p

:SinceŽ > 0 is arbitrary, we finally obtain (4.14), which completes the proof.

REMARK 4.5. Let X be any Banach space and consider the sectorial operatord=dteither onH p.R; X/ or on H p

0 .T; X/. Then it is easy to see that for any! > 0,the operator−id=dt is sectorial of type!. Moreover it is clear from the proof ofTheorem1.2 that in the case whenX is AUMD, the operator−id=dt has a boundedH∞.6�/ functional calculus for any� > 0.

REMARK 4.6. It was first established by Lust-Piquard and independently by Drury(see [22]) that there exist bounded Fourier multipliers onH 1

0 .T/which are not boundedon H 1

0 .T; S1/. Since S1 is not AUMD [15], Theorem1.2 yields several explicitexamples. In particular we recover Drury’s example, namely for anys ∈ R

∗, thesequence.kis/k≥1 is not a bounded Fourier multiplier onH 1

0 .T; S1/. We note that thelatter fact can also be easily deduced from [24, Theorem 6.2]. More generally, if.mk/k≥1 is a bounded non-converging sequence and if limk→∞.mk+1 − mk/ = 0, then.mk/k≥1 cannot be a bounded Fourier multiplier onH 1

0 .T; S1/. Indeed, let.mkp/p≥1 and

.mlq/q≥1 be two converging subsequences of.mk/k≥1, with distinct limitsþ1 andþ2.

Assume that.mk/k≥1 is a bounded Fourier multiplier onH 10 .T; S1/. Then according to

[24, Theorem 6.2], there exist a Hilbert spaceH and two bounded sequences.xk/k≥1

and .yl /l≥1 in H such thatmk+l = 〈xk; yl 〉 for any k; l ≥ 1. In particular, we maywrite mkp+lq

= 〈xkp; ylq

〉 for any p;q ≥ 1. For anyq ≥ 1, limp→∞.mkp+lq− mkp

/ = 0hence limp→∞ mkp+lq

= þ1. Similarly, limq→∞ mkp+lq= þ2 for any p ≥ 1. Now if x

andy are weak cluster points of the bounded sequences.xkp/p≥1 and.ylq

/q≥1, we seethat for anyq ≥ 1, þ1 = lim p→∞〈xkp

; ylq〉 = 〈x; ylq

〉 henceþ1 = 〈x; y〉. Likewise,þ2 = 〈x; y〉, which gives a contradiction.

REMARK 4.7. Let X be a Banach space, and letH1at.R; X/ be theX-valued atomic

H 1-space. Then the translation semigroup is well-defined and isometric onH 1at.R; X/.

We may then defined=dt on H 1at.R; X/ as its negative generator, and consider the

question whether it admits a boundedH∞ functional calculus. Using Calderon-Zygmund Theory for vector-valuedL p-spaces, one can show thatd=dt has a boundedH∞.6�/ functional calculus for any� > ³=2 if and only if d=dt has boundedimaginary powers if and only ifX is UMD. We omit the details.


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Departement de MathématiquesUniversite de Franche-Comté25030 Besançon CedexFrancee-mail: [email protected]

mailto:[email protected]

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