THE ABSOLUTE FUNCTIONAL CALCULUS FOR SECTORIAL OPERATORS

Transcription

1 THE ABSOLUTE FUNCTIONAL CALCULUS FOR SECTORIAL OPERATORS A Dissertation presented to the Faculty of the Graduate School University of Missouri-Columbia In Partial Fulfillment of the Requirements for the Degree Doctor of Philosophy by TAMARA KUCHERENKO Dr. Nigel Kalton, Dissertation Supervisor JULY 005

2 The undersigned, appointed by the Dean of the Graduate School, have examined the dissertation entitled THE ABSOLUTE FUNCTIONAL CALCULUS FOR SECTORIAL OPERATORS presented by Tamara Kucherenko a candidate for the degree of Doctor of Philosophy and hereby certify that in their opinion it is worthy of acceptance.

3 ACKNOWLEDGMENTS I would like to thank the faculty of the University of Missouri Mathematics Department for their efforts in sharing mathematical experience and helping my studies in every possible way. In particular, I would like to express greatest gratitude to my advisor Nigel Kalton for his support and help during my years at Mizzou. His insight and ideas have been crucial to my research. Further, I owe the opportunity of coming to Missouri to my diploma advisor Vladimir Kadets, Kharkov National University, who introduced me to Banach space theory. His guidance as an advisor and friend helped me in many ways. I acknowledge support from Lutz Weis and the University of Karlsruhe where my stay with a DAAD research scholarship was very inspiring. Moreover, I appreciate the welcome of the mathematics faculty at the University of South Carolina where I spent the Spring and Fall semesters of 004. Special thanks go to my loving husband Mark who has always been supportive and encouraging. ii

4 THE ABSOLUTE FUNCTIONAL CALCULUS FOR SECTORIAL OPERATORS Tamara Kucherenko Dr. Nigel Kalton, Dissertation Supervisor ABSTRACT There are four chapters to this dissertation. The first chapter is based on joint work with Mark Hoffmann and Nigel Kalton. It is shown that R-bounded and weakly compact semigroups on L 1 and C(K) can only exist for l 1 and c 0. More generally, R-bounded weakly compact commuting approximating sequences in Banach spaces are discussed. The second chapter is based on joint work with Lutz Weis and was produced while visiting the Universität Karlsruhe, Germany, on a DAAD research fellowship. As in chapter one, we focus on semigroups on L 1 spaces and the lack of H -calculus thereof. Precisely, we consider a sectorial operator A on a non-atomic L p -space, 1 p <, so that its resolvent consists of integral operators, or more generally, has a diffuse representation. Then the fractional domain spaces D(A α ) for α (0, 1) do not coincide with the real interpolation spaces of (L q, D(A)). As a consequence, we iii

5 obtain that no such operator A has a bounded H -calculus for p = 1. The third chapter is demonstrating the pathological properties a sectorial operators on L 1 must have if it generates an R-bounded semigroup. Precisely we show that if A is R-sectorial and ɛ > 0 then there is an invertible operator U : L 1 L 1 with U I < ɛ such that for some positive Borel function w we have U(D(A)) L 1 (w). This actually improves some results from the first chapter. Roughly speaking, this means A is very close to a bounded operator. A central idea in the proof is to associate a family of representing measures to an R- bounded family of operators on L 1 and then show that it forms a weakly compact set. The final chapter introduces absolute functional calculus for sectorial operators. We show that absolute calculus is stronger than H -calculus and establish fundamental facts. We are able to improve a key theorem related to the maximal regularity problem and hence demonstrate the power and usefulness of our new concept. In trying to characterize spaces where sectorial operators have absolute calculus, we find that certain real interpolation spaces play a central role. We are then extending various known results in this setting. The idea of unifying theorems about sectorial operators on real interpolation spaces permeates our work and opens paths for future research on this subject. iv

6 TABLE OF CONTENTS ACKNOWLEDGMENTS ii ABSTRACT iii Chapter 1. R-bounded approximating sequences Introduction The main results Real Interpolation of domains of sectorial operators on L p -spaces Introduction Preliminary results Main results Rademacher bounded families of operators on L Introduction Operators on L Applications to sectorial operators The absolute functional calculus Maximal regularity Real interpolation spaces Sectorial operators Functional calculus Joint functional calculus The definition of absolute H -calculus v

7 4.7 Köthe function spaces The associated spaces X E (f) Absolute calculus and interpolation spaces BIBLIOGRAPHY VITA vi

8 1

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10 Chapter 1 R-bounded approximating sequences and applications to semigroups 1.1 Introduction This chapter is based on joint work with M. Hoffmann and N. Kalton. Our results were published in [8]. It is shown that on certain Banach spaces, including C[0, 1] and L 1 [0, 1], there is no strongly continuous semigroup (T t ) 0<t<1 consisting of weakly compact operators such that (T t ) 0<t<1 is an R-bounded family. More general results concerning approximating sequences are included and some variants of R-boundedness are also discussed. Recent work on semigroup theory ([3], [53]) has highlighted the importance of the concept of R-boundedness. Let us recall the definition of R-bounded families of operators (cf. [6], [15], [11]). Definition A family T of operators in L(X, Y ) is called R-bounded with R-boundedness constant C > 0 if letting (ɛ k ) be a sequence of independent Rademachers on some probability space then for every x 1,..., x n X and T 1,..., T n 3

11 T we have E ɛ k T k x k 1 C E ɛ k x k 1. (1.1) By the Kahane-Khintchine inequality we can replace above by any other exponent 1 p < to obtain an equivalent definition. We will also need the following definition introduced in [3]. Definition A family T of operators in L(X, Y ) is called WR-bounded with WR-boundedness constant C > 0 if for every x 1,..., x n X, y1,..., yn Y and T 1,..., T n T we have 1 T k x k, yk C E ɛ k x k 1 E ɛ k yk. (1.) It is clear by the Cauchy-Schwarz inequality that R-boundedness implies WRboundedness. The converse is not true in general, but it holds for spaces with non-trivial type ([46], [3]). In [3] it was shown that no reasonable differential operator on L 1 can have an H calculus. In this note we consider the related question whether a differentialtype operator on L 1 can generate an R-bounded semigroup. Note that if A is an R-sectorial operator (cf. [3]) with R-sectoriality angle less than π then the semigroup (e ta ) 0<t<1 is necessarily R-bounded. In general, one expects a semigroup generated by a differential operator on a bounded domain to consist of weakly compact operators. We are thus led to consider the question whether one can have a strongly continuous semigroup (T t ) 0<t<1 on L 1 such that each T t is weakly compact (or equivalently compact, since L 1 has the Dunford-Pettis property) and such that the family (T t ) 0<t<1 is R-bounded. In fact this leads to considering versions 4

12 of the approximation property; the only property of the semigroup needed is commutativity. We consider the general question whether on a given separable Banach space one can find an R-bounded sequence (T n ) n N of commuting weakly compact operators such that lim n T n x = x for all x X. Our main results show that for the spaces L 1 [0, 1], C(K) (except c 0 ) and the disk algebra A(D) this is impossible. These results may be regarded as extensions of classical results that the spaces L 1, C(K) do not have unconditional bases [38]. In the case of L 1 we are led to consider a natural weakening of R-boundedness, where we use the definition (1.1) but only for single vectors. Definition A family T of operators in L(X, Y ) is called semi-r-bounded if there is a constant C > 0 such that for every x X, a 1,..., a n C and T 1,..., T n T we have E ɛ k a k T k x 1 C ( a k ) 1 x. (1.3) We note that semi-r-boundedness is equivalent to R-boundedness for operators on L 1. In Theorem 1.. we actually characterize all spaces where semi-rboundedness is equivalent to R-boundedness as spaces which are either Hilbert spaces or GT-spaces of cotype in the terminology of Pisier [45]. The authors would like to thank Dale Alspach for drawing their attention to the notion of the l 1 index of sequences. 1. R-boundedness and WR-boundedness In this section, we make some remarks about R-boundedness and related notions. 5

13 Note that in a space of type, any uniformly bounded collection T L(X, X) is semi-r-bounded. The converse is also true: Proposition A Banach space X has type if and only if uniform boundedness is equivalent to semi-r-boundedness. Proof. Suppose that every uniformly bounded family of operators is already semi- R-bounded. Pick any x X and x X such that x = x = 1 and x (x) = 1. Notice that the family T = {x u : u = 1} is uniformly bounded with constant one and hence semi-r-bounded by assumption. Let C be the semi-r-boundedness constant of T. Select any x 1,..., x n X and write x k = x k u k where u k = 1. Then {x u k : k = 1,..., n} T and E ɛ k x k 1 = E ɛ k x k (x u k )x C E ɛ k x k x = C x = C ( ( x k ) 1 x k ) Thus, X has type. For some spaces, semi-r-boundedness is equivalent to R-boundedness and we are able to completely characterize these spaces in the next theorem. Let us recall that a Banach space X is called a GT-space if every bounded operator T : X l 6

14 is absolutely summing. Examples of GT-spaces of cotype are L 1, the quotient of L 1 by a reflexive subspace [45],[33], and L 1 /H 1 [13]. It is unknown whether every GT-space has cotype. Theorem 1... Suppose X is separable. Then the following are equivalent : (i) Every semi-r-bounded family of operators on X is R-bounded. (ii) X is isomorphic to l or X is a GT-space of cotype. Proof. First we prove that (i) implies (ii). Suppose that every semi-r-bounded family of operators on X is R-bounded. Let us note that this implies the existence of a constant K so that if T has semi-r-boundedness constant C then it has R- boundedness constant KC; for otherwise we could find a sequence T n of families with semi-r-boundedness constant 1 and R-boundedness constant at least 4 n ; then the family n 1 n T n contradicts our assumption. Fix M > 1 and take x X. Choose n N. By Dvoretzky s theorem [40] we can find e 1,..., e n X such that for any a 1,..., a n C we have M 1 ( a k ) 1 ( a k e k M a k ) 1. Consider the family of operators T n = {u e k : u = 1, k = 1,..., n}. Then each T n is semi-r-bounded with constant M as follows. A finite subfamily of T n is of the form {u kj e k : 1 k n, 1 j m k } for some m 1,..., m n N. Then for 7

15 every a 11,..., a nmn C we have, (letting ɛ kj denote independent Rademachers), m k E ɛ kj a kj u kj(x)e k j=1 M E 1 m k ɛ kj u kj(x)a kj j=1 M m k E ɛ kj u kj(x)a kj j=1 j=1 ( ) 1 m k M a kj x Our assumption implies that each T n is R-bounded with constant KM. Let x 1,..., x n X and write x k = x k u k where u k = 1. Choose u k X such that u k (u k) = 1 and u k = 1. Now we have ( ) 1 x k M E This shows that X has cotype. ɛ k x k e k = M E ɛ k x k u k(u k )e k = M E ɛ k x k (u k e k )(u k ) KM E ɛ k x k u k = KM E ɛ k x k Let us assume that X has non-trivial type. Then by results of Pisier [45] and also by Figiel and Tomczak-Jaegermann [5], l n is uniformly complemented in X

16 Thus, for some constant C, for every n N we can choose a biorthogonal system {(e k, e k ) : k = 1,..., n} in X X such that and ( ) 1 a k e k C a k ( ) 1 a k e k C a k for all a 1,..., a n C. Note that for any x X and a 1,..., a n C, and so ( ) 1 a k e k(x) C a k x ( e k(x) ) 1 C x. Consider the family of operators T n = {e k u : u = 1; k = 1,..., n}. Let x X. Then for any a 1,..., a n C and every u 1,..., u n X of norm one we have E ɛ k a k (e k u k )(x) = E ɛ k a k e k(x)u k a k e k(x)u k = a k e k(x) ( ) 1 ( ) 1 a k e k(x) C ( a k ) 1 x. We conclude that T n is semi-r-bounded with constant C and hence T n is R- bounded for constant KC independent of n. This implies that X has type as 9

17 follows. Choose any x 1,..., x n X and write x k = x k u k where u k = 1. Then E ɛ k x k 1 = E ɛ k x k u k = E ɛ k x k e k(e k )u k = E ɛ k x k (e k u k )(e k ) 1 KC E ɛ k x k e k KC ( x k ) (1.4) Now, X has type and cotype and is therefore isomorphic to l by Kwapien s theorem [56]. Now suppose on the contrary that X has trivial type. We will show that X is a GT-space, i.e. any T : X l is 1-summing. Fix T : X l of norm one. Since X has cotype we can equivalently show that any such T is -summing [19]. It suffices to check that for any n N and operator S : l n X such that S 1 we have π (T S) C where C does not depend on n [56]. One can assume that T S : l n l n and that T S is diagonal with respect to the canonical orthonormal basis (e k ) in l n, i.e. T Se k = λ k e k for some λ 1,..., λ n. Then it suffices to show uniform boundedness of the Hilbert-Schmidt norms T S HS = ( n T Se k ) 1. Write f k = T e k X and f k = Se k X. Consider {f k u : k = 1,..., n; u = 1}. We will show that this family is semi-r-bounded with constant one. Take 10

18 u 1,..., u n X of norm one and a 1,..., a n C. Then for x X we have E ɛ k a k fk (x)u k a k fk (x) ( ) 1 ( ) 1 a k e k(t x) = ( ( a k ) 1 T x a k ) 1 x Therefore, {f k u : k = 1,..., n; u = 1} is R-bounded with constant K. Since X has trivial type, it contains l n 1 uniformly [45]. Hence, for fixed M > 1 and every n N there are y 1,..., y n X with y k = 1 for 1 k n such that a k M a k y k (1.5) Choose any scalars b 1,..., b n. Now using R-boundedness and Kahane s inequality for p = 1 with constant A we have 11

19 Thus, b k λ k = b k fk (f k ) M E ɛ k b k fk (f k )y k = M E ɛ k b k (fk y k )(f k ) KM E ɛ k b k f k KM E ɛ k b k e k KM ( ( λ k ) 1 b k ) 1 KM and so T S HS KM. Therefore, any operator T : X l is -summing. This completes the proof of (i) implies (ii). Now we will show that (ii) implies (i). Suppose that X is a GT-space of cotype, and that T is a family of semi-r-bounded operators. We will show that T is R-bounded. Since X is separable, there is a quotient map Q : l 1 X. First, we show that any semi-r-bounded family of operators from l 1 into X is already R- bounded. Let S be such a family with semi-r-boundedness constant one. Suppose S 1,..., S n S and x 1,..., x n l 1. Then x k = j=1 ξ jke j where (e j ) is the canonical basis of l 1. Let us denote by C the constant in the Kahane-Khintchine inequality for any

20 Banach space: Thus Then E ɛ k x k 1 E ɛ k S k x k 1 CE ɛ k x k. CE ɛ k S k x k E ɛ k S k x k = E ɛ k S k ξ jk e j j=1 E ɛ k ξ jk S k e j j=1 E ɛ k ξ jk S k e j j=1 j=1 ( ) 1 ξ jk. 1 Combining and using the Khintchine inequality again we obtain E ɛ k S k x k 1 C E ɛ k ξ jk j=1 ( ) = C E ɛ k ξ jk j=1 = C E ɛ k ξ jk e j j=1 l1 = C E ɛ k ξ jk e j j=1 = C E ɛ k x k Combining the previous two computations gives that S is R-bounded. 13

21 Now let T be a family of operators on X with semi-boundedness constant one. Let Q : l 1 X be a quotient map and note that the family S = {T Q : T T } is R-bounded with some constant B by the above calculation. We will apply a characterization of GT-spaces of cotype due to Pisier [45]. Proposition 1..3 (Pisier). X is a GT-space of cotype if and only if there is a constant C > 0 such that for any n N, x 1,..., x n X there are y 1,..., y n l 1 such that Qy k = x k, k = 1,..., n and E ɛ k y k CE ɛ k x k (1.6) Now take n N, T 1,..., T n T and x 1,..., x n X. Choose y 1,..., y n l 1 according to Then E ɛ k T k x k = E ɛ k T k Qy k BE ɛ k y k CBE ɛ k x k Thus, T is R-bounded. The proof is complete. For a set T of bounded linear operators we will use the notation T = {T : T T }. Lemma (i) If T is R-bounded then T is R-bounded (with the same constant). 14

22 (ii) If T is WR-bounded then T and T are WR-bounded (with the same constant). (iii) If T is semi-r-bounded then T is semi-r-bounded (with the same constant). Proof. The proofs of (i) and (iii) are similar. For (i) suppose T 1,..., T n T and that T has R-boundedness constant one. Let Ω = { 1, 1} n with P normalized counting measure on Ω. Let ɛ k be the sequence of coordinate maps on Ω. Let Rad (Ω; X) be the subspace of L (Ω, P; X) generated by the functions ɛ k x for 1 k n and x X (this space is isomorphic to X n ). Then Rad (Ω; X ) can be identified naturally with a subspace of Rad (Ω; X). Consider the map T : Rad (Ω; X) Rad (Ω; X) defined by ( ) T ɛ k x k = ɛ k T k x k. Then T 1 and so T 1 and (i) follows. Let us now prove (ii). Suppose T is WR-bounded with constant one and T 1,..., T n T. Suppose x 1,..., x n X are such that 1 E ɛ k x k 1. Then, using the identification of Rad (Ω, X ) as the bidual of Rad (Ω, X) we observe that the set of functions of the form n ɛ kx k in Rad (Ω, X ) such that T k x k, x k 1 is weak -closed and contains the unit ball of Rad (Ω, X). By Goldstine s theorem it contains the unit ball of Rad (Ω, X ) and this implies that T is WR-bounded with constant one. 15

23 Now it is time to give an example of a family of operators that is uniformly bounded but not WR-bounded. The previous lemma will imply that the corresponding dual family is semi-r-bounded but not WR-bounded. Example. Let X = l p, 1 p <. Pick any non-zero element x X and choose u X of norm one such that u (x) 0. Define T k = u e k where (e k ) is the canonical basis of X. The family {T k } is uniformly bounded, T k = 1, but we will show that it is not wr-bounded. Consider the dual basis (e k ) in (l p). Then T k x, e k = u (x)e k, e k = n u (x) (1.7) On the other hand, we have 1 E ɛ k x E ɛ k e k 1 = x n 1 n 1/q (1.8) Here q satisfies 1/p+1/q = 1. If p < then q > and 1 +1/q < 1, so for 1 p < the family {T k } can not be wr-bounded. We have T k = e k u on X = l q where < q. Consider q. Since by reflexivity Tk = T k and using 1..4 we see that {Tk } is not WR-bounded. However, X has type and hence {Tk } is semi-r-bounded by The main results Suppose X is any Banach space. We shall say that a sequence T = (T k ) is an approximating sequence if lim k x T k x = 0 for every x X. We will say that T is compact (relatively, weakly compact) if each T k is compact (relatively, weakly compact). We will say that T is commuting if we have T k T l = T l T k for l, k N. 16

24 If T is a commuting approximating sequence, let us define the subspace E T of X to be the closed linear span of k T k (X ). The following Lemma is trivial. Lemma If T is a commuting approximating sequence then E T is a norming subspace of X, i.e. for some C we have x C sup x (x) x X x B ET and T is weakly compact, lim n T nx = x weakly for x E T. (T n ET ) n=1 is an approximating sequence for E T. Let us recall that a Banach space X has property (V ) of Pe lczyński if every unconditionally converging operator T : X Y is weakly compact. The spaces C(K) have property (V) [4] and more generally any C -algebra has property (V) [44]. The disk algebra A(D) also has property (V) [33], [17]; see also [49]. We also recall that a Banach space X is said to have property (V ) if whenever (x n ) is a bounded sequence in X then either: (i) (x n ) has a subsequence which is weakly Cauchy, or (ii) (x n ) has a subsequence (y n ) such that for some sequence (yn) in X and δ > 0 we have yn(y n ) δ and a k yk max a k a 1,..., a n C, n N. 1 k n Property (V ) was introduced by Pe lczyński [4]. We note that Bombal [8] shows that every Banach lattice not containing c 0 has property (V ). Any subspace of a space with property (V ) also has property (V ). 17

25 Lemma Let X, Y be Banach spaces and let T = (T k ) be any sequence of operators in L(X, Y ). Suppose either (i) T is semi-r-bounded or (ii) T is WRbounded and Y has property (V ). Then for every x X the sequence (T k x) has a weakly Cauchy subsequence. Proof. If not, by passing to a subsequence we can suppose (T k x) is equivalent to the canonical l 1 -basis ([48], [41]). If T is semi-r-bounded we observe that for some C we have ( ) 1 E ɛ k a k T k x C a k x, a 1,..., a n C, n N. This gives a contradiction. In case (ii), we can pass to a subsequence and assume the existence of y n Y such that a k yk max a k 1 k n and y n(t x n ) δ > 0 for all n. Then nδ yk(t k x) a 1,..., a n C, n N 1 C E ɛ k x E ɛ k yk C n. 1 This also yields a contradiction. Theorem Let X be a Banach space with a commuting weakly compact approximating sequence T. Suppose either that (i) T is semi-r-bounded and X is weakly sequentially complete or (ii) T is WR-bounded and X has property (V ). Then X is isomorphic to a dual space. 18

26 Proof. In either case we consider the family T L(X, X). By Lemma 4. for each x X we can find a subsequence T k n x so that T k n (x ) is weakly convergent to some y X. Then for x X x (T k y) = lim n x (T k T n x ) = lim n x (T n T k x ) so that T k y = Tk x. Hence lim k y Tk x = 0. We now show that E T can be identified with X. Clearly X canonically embeds in E T since E T is norming. If f E T then by the Hahn-Banach theorem there exists x X with x = f and x (x ) = f (x ) for x E T. Let y = lim k T k x. Then for x E T, Hence E T = X. x (y) = lim x (Tk x ) = lim x (Tk x ) = f (x ). k k Theorem The space L 1 (0, 1) does not have a commuting weakly compact approximating sequence which is either semi-r-bounded or WR-bounded. Proof. L 1 is not a dual space [56]. Of course a semi-r-bounded sequence in L 1 is actually R-bounded. Theorem Let X be a separable Banach space with property (V). If X has a commuting weakly compact approximating sequence (T n ) n=1 which is WR-bounded, then X is separable and has a WR-bounded commuting weakly compact approximating sequence. Proof. Since X has (V), it follows that X has property (V ). We show that lim n T nx = x weakly for x X. Indeed T nx converges weak to x and 19

27 it must have a weakly convergent subsequence by Lemma 4.. Hence x E T so X = E T. Now Tn(B X ) is weakly compact by Gantmacher s theorem also and weak -metrizable, hence norm separable. Thus X is separable, and so by Mazur s theorem and a diagonal argument we can find a sequence of convex combinations (Sn) n=1 of (Tn) n=1 which is an approximating sequence. Corollary If K is an uncountable compact metric space then C(K) has no WR-bounded commuting weakly compact approximating sequence. The disk algebra has no WR-bounded weakly compact approximating sequence. We now consider C(K) when K is countable. In this case C(K) is homeomorphic to a space C(α) = C([1, α]) where α is a countable ordinal. There is a characterization of such C(K) due to Bessaga and Pe lczyński [7]. Theorem (Bessaga-Pe lczyński). If α < β, C(ω α k) is isomorphic to C(ω β n) if and only if β < α ω. Consequently, C(ω ωγ ), 0 γ < ω 1, is a complete list of representatives of the isomorphism classes of C(K) for K a countable compact metric space. The following lemma can be obtained as an applications of l 1 -indices ([1], [9], [10]). However, for convenience of the reader we will give a direct proof by construction. Lemma Let α be a countable ordinal with α ω ω. Then there exists f C(α) so that whenever f n C(α) converges to f C(α) weak then for any m N there exist n 1,..., n m N such that m ɛ k f nk 1 m ɛ k = ±1, k = 1,,..., m. 0

28 Proof. In this case C(α) can be identified with l (α). It is easy to see that it suffices to consider the case α = ω ω. Consider f X defined by f( N k=0 ωk l k ) = ( 1) N k=0 l k and f(ω ω ) = 1. Writing K for the space [1, ω ω ] let K (p) denote the p th derived set of K. Then K (p) consists of all ordinals of the form n k=p ωk l k together with ω ω. For each p N, K (p) is nonempty. Furthermore for each α K (p) and every open neighborhood V of α we have that f takes both values ±1 on V K (p 1). Let f n C(K) be any sequence such that (f n ) converges to f weak. Fix 0 < δ < 1 and m N. We construct (f n 1,..., f nm ) inductively. We start from K (m). By definition of f we can pick α 1 1, α 1 K (m) such that f(α 1 j) = ( 1) j for j = 1,. Then find n 1 N such that f n1 (α 1 j) ( 1) j < δ. Since f n1 is continuous we can choose open neighborhoods U 1 j of α 1 j such that f n1 (α) ( 1) j < δ for all α U 1 j. For the inductive step, suppose that (n j ) k j=1, (α k j ) k j=1 and open sets (U k j ) k j=1 have been chosen so that αj k Uj k. Then for i = 1,..., k find points α k+1 i 1, αk+1 i Ui k K (m k+1) with f(α k+1 j ) = ( 1) j. By pointwise convergence, we can select n k+1 > n k such that f nk+1 (α k+1 j ) ( 1) j < δ. Since f nk+1 is continuous, there are neighborhoods U k+1 i 1, U k+1 i Ui k, i = 1,..., k such that for all α U k+1 j f nk+1 (α) ( 1) j < δ. we have In the m-th iteration this will give m neighborhoods and m functions f n1,..., f nm so that for any ɛ 1,..., ɛ m { 1, +1} there is an α contained in one of these neigh- 1

29 borhoods such that f k (α) ɛ k < δ for all k = 1,..., m. Hence m ɛ k f nk (1 δ)m. Theorem Let K be a compact metric space. Suppose there is an R-bounded commuting weakly compact approximating sequence in C(K). Then C(K) is isomorphic to c 0. Proof. By Corollary we need only consider the case when K is countable. By Theorem it suffices to consider the case when K = [1, α] where α ω ω. Pick f C(K) satisfying the hypotheses of Lemma Suppose (T n ) is an R-bounded weakly compact approximating sequence for C(K). Then (T n) is an approximating sequence for C(K) by Theorem and hence T n f converges to f weak. It follows that for any m we can choose n 1,..., n m so that Hence m ɛ k T n k f m E ɛ k T 1m ɛ k = ±1. n k f 1 1 m. This contradicts the fact that T n is R-bounded (or even semi-r-bounded). Remark. We can replace the assumption of R-boundedness by the assumption that (T n ) and (Tn) are both semi-r-bounded. By Theorem 1.. this hypothesis would imply that (Tn) is actually R-bounded and hence that (T n ) is WR-bounded. We only used the fact that (T n ) is both semi-r-bounded and WR-bounded.

30 Let us conclude by stating our main result with respect to semigroups. (Actually our results are somewhat stronger than stated below.) Theorem Let X be a separable Banach space with an R-bounded strongly continuous semigroup (T t ) t>0 consisting of weakly compact operators. Then if 1. X = L 1 (µ) for some measure µ then X is isomorphic to l 1 (i.e. µ is purely atomic).. X = C(K) then X is isomorphic to c 0. 3

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34 Chapter Real interpolation of domains of sectorial operators on L p -spaces.1 Introduction This chapter is based on joint work with L. Weis. Let A be a sectorial operator on a non-atomic L p -space, 1 p <, whose resolvent consists of integral operators, or more generally, has a diffuse representation. Then the fractional domain spaces D(A α ) for α (0, 1) do not coincide with the real interpolation spaces of (L q, D(A)). As a consequence, we obtain that no such operator A has a bounded H -calculus if p = 1. It is not uncommon that properties of the Laplace operator extend to a sectorial operator A which satisfies a pointwise kernel bound of the kind (λ + A) 1 f(u) k λ (u, v) f(v) dv, u R n (.1) Ω for f L q and λ in a sector about R +. Here, k λ is the kernel of (λ ) 1 or a more general Poisson bound. In the case of 1 < q <, (.1) implies that ( A) has maximal L p -regularity for 1 < p < (see e.g. [7], [35, section 5] ), or that A has a H -functional calculus on L q if A has one on L ( [3], [35, section 5]). In 7

35 this paper we exhibit two more examples of such phenomena. It is well known that Laplace operator on L 1 (R n ) does not have a bounded H -calculus. In Corollary.3.3 we show that if q=1 then (.1) implies that A does not have a bounded H -functional calculus. This is still true if k λ is the kernel of any positive integral operator on L 1 (Ω) or if (λ + A) 1 has a diffuse representation (see the definition below). If ( A) generates a weakly compact semigroup this result is already contained in [8]. It seems remarkable that the very same estimate (.1) that guarantees the boundedness of the H -calculus in so many cases if q (1, ), absolutely excludes it if q = 1. It is also well known that for on L q (R n ), 1 < q <, q the fractional domains D((1 ) α ) are isomorphic to the Bessel potential spaces Wq α (R n ). So they do not coincide with the real interpolation spaces (L q, D( )) α,r which are isomorphic to the Besov potential spaces B α q,r(r n ) (of course, they are the same for q=). In Theorem 3.1 we will show that (.1) implies such a result for any sectorial operator A on L q with 0 ρ(a) and 1 < q <, q, i.e. D(A α ) (L q, D(A)) α,r, 0 < α < 1 Again, it is enough that k λ is the kernel of a positive integral operator on L q (Ω), or that (λ + A) 1 has a diffuse representation. If we assume in addition that A has bounded imaginary powers it follows that the complex and real interpolation methods yield different results for the interpolation pair (L q, D(A)) (see Corollary.3.). Let us recall now some definitions. A closed operator A with domain D(A) is 8

36 called sectorial of type ω if the spectrum σ(a) is contained in a sector {z C : arg(z) < ω} {0} and we have λr(λ, A) C ω for arg(λ) > ω. We will write ρ(a) = C\σ(A) for the resolvent set of A and R(λ, A) for the resolvent at λ ρ(a). Suppose that A is a sectorial operator of type ω and f is a holomorphic function on Σ σ where σ > ω. Given that f satisfies the condition Σ δ f(λ) 1 dλ <, λ we can define f(a) = Σ δ f(λ)r(λ, A)dλ, ω < δ < σ We say that A has bounded H (Σ σ )-functional calculus if the map f f(a) can be extended to a bounded map from the space H (Σ σ ) of bounded holomorphic functions on Σ σ to the space of bounded linear operators on X (see [3] for details). For the definition of fractional powers in terms of the H -calculus see e.g. [35] and if 0 ρ(a) see also [5]. A sectorial operator A has bounded imaginary powers if A it for t R define bounded operators on X. Clearly, a bounded H -calculus implies bounded imaginary powers. For the most part we consider L q -spaces on σ-finite non-atomic measure spaces (K, B, m) and (Ω, Σ, µ). We recall that a bounded operator T on L q is positive if the image of every non-negative function is again a non-negative function. If an operator can be split into a difference of two positive operators then it is called regular. Regular operators between L p spaces have a particularly useful representation (see [30, 54, 50]). Given a regular operator T : L p (K, m) L q (Ω, µ) there is a family of regular Borel measures (ν y (x)) y Ω on K such that for every f L p (K, m) 9

37 we have T f(y) = f(x) dν y (x) K µ a.e. Note that if all measures ν y are absolutely continuous with respect to m then by the Radon-Nikodym theorem we obtain classical integral operators, T f(y) = f(x)k(y, x) dm(x), K k(y, ) = dν y /dm In case that all measures ν y are non-atomic we say that the operator has a diffuse representation. While resolvents of second order elliptic operators are typically classical integral operators, the resolvents of first order differential operators have usually a diffuse representation. As an example, consider the operator A : D(A) L 1 (R ) L 1 (R ) given by Af(x 1, x ) = x 1 f(x 1, x ) Its resolvent (R(A, λ)f)(x 1, x ) = has representing measure 0 e λt f(x 1 + t, x ) dt µ λ (x 1,x ) = η λ x 1 δ x where δ x is the Dirac measure and dηx λ 1 = χ [x1, )(t)e λ(t x1) dt. Therefore, R(A, λ) is not an integral operator but has a diffuse representation. However, given a diffuse operator T we can always pass to a sub-σ-algebra for which T is integral [55]. 30

38 . Preliminary results The following lemma is a vector-valued version of a classical result about uniform integrability in L 1. Lemma..1. Let X be a Banach space and T be an isomorphic embedding from X into L p (X) (1 p < ). Assume that for some subspace Y X the set { T y(t) p X : y Y, y X = 1} is not uniformly integrable as a subset of L 1. Then there exist a sequence (y n ) in Y isomorphic to a unit vector basis of l p. Proof. Since { T y(t) p X : y Y, y X = 1} is not uniformly integrable in L 1 we can find a sequence (y n ) in Y with y n 1 such that T y n (t) p Xdt = 1 and T y n (t) p X 0 (n ) almost everywhere. To see this, assume the contrary, i.e. every sequence from T (Y ) converging to zero almost everywhere is converging to zero in L p (X)-norm. Then for all 0 < q < p there exists C > 0 such that T y(t) p dt C T y(t) q dt for all y Y. Hence, we have lim sup M ( y =1 T y(t) >M C lim M sup T y(t) p dt) 1/p ( C lim sup ( M y =1 y =1 T y(t) >M T y(t) q dt) 1/q T y(t) p M q p dt) 1/p = 0 This contradicts the fact that { T y(t) p X : y Y, y X = 1} is not uniformly integrable in L 1. For convenience define f n (t) = T y n (t) p X. Then (f n) are functions in L 1 of norm one. We will use a subsequence splitting lemma. 31

39 Lemma... [9] If (f n ) is a sequence in the unit ball of L 1 then there exist a subsequence (f nk ) and disjoint sets (A k ) with their complements B k such that f nk Bk are uniformly integrable. Since the sequence (f nk Bk ) is uniformly integrable and still goes to zero almost everywhere when k is approaching infinity we get that f nk Bk goes to zero in L 1 - norm. So f nk Ak is bounded in norm from below. Now T y nk = T y nk Bk + T y nk Ak where T y nk Ak Lp(X) = f nk Ak L1 is bounded from below. Thus the sequence (T y nk Ak ) is isomorphic to the unit vector basis of l p since it has disjoint support and bounded from below in L p (X). On the other hand T y nk T y nk Ak Lp(X) = T y nk Bk Lp(X) = f nk Bk L1 0 (k ) It follows by perturbation of basis that some subsequence of (T y nk ) is equivalent to the unit vector basis of l p. Denote this subsequence again by (T y nk ). Then (y nk ) is also equivalent to the unit vector basis of l p since T is an isomorphism. The next proposition is related to a result in [3]. The expression appearing in the statement will be applied to the setting of interpolation spaces between X and D(A). Proposition..3. Suppose X is a Banach space and A is a sectorial operator on X. Assume there is a constant C > 0, 1 p < and α (0, 1) such that for every x X 0 C 1 x ( t α 1/p A α R(t, A)x p dt) 1/p C x (.) 3

40 Then if Y is an infinite-dimensional closed subspace of D(A) (with a graph norm) and Y does not contain a copy of l p then A is bounded on Y. Proof. We will consider an operator T : X L p (R, dt, X) given by T x(t) = t α 1/p A α R(t, A)x It follows from (.) that T is an isomorphic embedding. Since α < 1 we can find a natural number m such that α (m 1)/m. Fix s < 0. Then R(s, A) maps X isomorphically onto D(A) (with a graph norm). Let Y 0 = R(s, A) 1 Y. Then Y 0 is an infinite-dimensional subspace of X that does not contain a copy of l p. By lemma..1 the set { T y(t) p X : y Y 0, y X = 1} is uniformly integrable. The operator A α R(s, A) has a lower bound on Y 0 since otherwise, there would exist a sequence y n in Y 0 of elements of norm one such that A α R(s, A)y n 0. However, the resolvent equation yields for any t < 0 A α R(t, A)y n = A α R(s, A)y n + (s t)r(t, A)(A α R(s, A)y n ) Therefore A α R(t, A)y n 0 pointwise. Now by uniform integrability and., we have y n 0 which gives a contradiction. The operator A α R(s, A) is an isomorphism on Y 0. Thus the subspace Y 1 = A α R(s, A)(Y 0 ) does not contain a copy of l p and by the same argument we get that A α R(s, A) is bounded from below on Y 1. This gives us a lower bound for the operator A α R(s, A) on Y 0. Repeating the same procedure m times we get that the operator A mα R(s, A) m is bounded from below on Y 0 by some constant C > 0. It follows from the boundedness of the operator A mα R(s, A) m 1 33 (α (m 1)/m)

41 and the simple computation C y 0 A mα R(s, A) m y 0 A mα R(s, A) m 1 R(s, A)y 0 y 0 Y 0 that the resolvent R(s, A) is bounded from below on Y 0. Now we see that A is bounded on Y = R(s, A)Y 0. Take any y in Y and find y 0 in Y 0 such that y = R(s, A)y 0. Then Ay AR(s, A) y 0 (1/C) AR(s, A) A mα R(s, A) m 1 R(s, A)y 0 = C 1 y Remark..4. The proposition cannot be applied for p =. In this case X is isomorphic to L (R, dt, X). Thus there is no subspace in X and hence in D(A) which does not contain a copy of l. We assume that zero is contained in the resolvent set, then (, 0] ρ(a) and we have an estimate R(t, A) C for all t (, 0]. This allows us to apply 1+ t a theorem from [5] which yields that an equivalent norm on the real interpolation space (X, D(A)) α,p for 0 < α < 1 and 1 p is given by for x (X, D(A)) α,p. x (Lq,D(A)) α,p ( 0 t α A(A + t) 1 x p dt t )1/p (.3) In [3] it was shown that if A has an H -calculus on L 1 then x L1 A s R(t, A)x dt t. 34

42 Formula.3 allows us to reformulate this statement as follows. Proposition..5. If A has a bounded H -calculus on L 1 (Ω, µ) then (L 1, D(A)) α,1 = D(A) α with equivalence of norms for 0 < α < 1..3 Main results In general we have the following inclusions between the domain D(A α ) of a fractional power of A and real interpolation spaces (X, D(A)) α,1 and (X, D(A)) α, (X, D(A)) α,1 D(A α ) (X, D(A)) α,. If a sectorial operator A has a bounded H -calculus on X = L (K, B, m) then we have D(A α ) = (X, D(A)) α,. This result can be found in [?]. As we will see now this statement is wrong for L q with q. Theorem.3.1. Let A be a sectorial operator on L q (K, B, m) for a non-atomic measure space (K, B, m) and 1 q <, q. Assume that 0 ρ(a) and there exists s < 0 such that R(s, A) is a regular operator with a diffuse representation. Then for any α (0, 1) and 1 p D(A α ) (L q, D(A)) α,p Proof. We will assume that D(A α ) = (L q, D(A α )) α,p and derive a contradiction. It follows from [5] that there exists a constant C > 0 such that for any y 35

43 (L q, D(A)) α,p we have C 1 ( 0 t α A(A + t) 1 y p dt t )1/p y (Lq,D(A)) α,p C( 0 t α A(A + t) 1 y p dt t )1/p Since D(A α ) = (L q, D(A)) α,p, we obtain for any y D(A α ) that the quantities A α y, y D(A α ), and y (Lq,D(A)) α,p are equivalent. Pick x from the range of A α and take y D(A α ) such that x = A α y. Then using t α A(A+t) 1 A α x p dt 0 t α 1/p A 1 α R(t, A)x p dt we obtain for some 0 suitable constant C > 0 that = t 0 C 1 ( t α 1/p A 1 α R(t, A)x p dt) 1/p x Lq(K,B,m) 0 C( t α 1/p A 1 α R(t, A)x p dt) 1/p The range of A α is dense in L q (K, B, m) and therefore condition (.) is fulfilled. We will use Proposition..3. Since the resolvent R(s, A) is a regular operator with a diffuse representation there is a non-atomic sub σ-algebra B 1 of B such that R(s, A) Lq(K,B1,m) is a compact operator ([54]). Let Y 1 be a closed infinitedimensional subspace of L q (K, B 1, m) which does not contain a copy of l p, for instance, take the span of a sequence equivalent to the Rademacher functions. Consider Y = R(s, A)Y 1. Since R(s, A) is an isomorphism from L q (K, B, m) onto D(A) (with the graph norm), Y is a closed infinite-dimensional subspace of D(A) and does not contain l p. By Proposition..3 A is bounded on Y and therefore 36

44 si A is also bounded on Y. We consider the bounded operator J : (D(A),. A ) L q (K, B, m), J = R(s, A)(sI A) Then J(Y ) = Y. On the other hand, J Y = R(s, A)(sI A) Y = R(s, A) Y1 is a compact operator since Y 1 L q (K, B 1, m). This is impossible since J is onto Y and Y is infinite-dimensional. We hence obtain a contradiction. It is well known that if A has bounded imaginary powers on X then D(A α ) coincides with the complex interpolation spaces [X, D(A)] α = D(A) α (see e.g. [35] [5]). Hence our theorem implies Corollary.3.. Assume in addition to the assumption of Theorem.3.1 that A has bounded imaginary powers. Then (L p, D(A)) α,p [L p, D(A)] α for all 1 p and α (0, 1). Our next results will show that no reasonable differential operator on L 1 (Ω, µ) can have a bounded H -calculus. Corollary.3.3. Let A be a sectorial operator on L 1 (Ω, Σ, µ). Assume there is a point λ ρ(a) such that the resolvent R(λ, A) has a diffuse representation. Then A does not have a bounded H -calculus. Proof. Combine Proposition..5 and Theorem.3.1 noticing that all operators on L 1 are regular. 37

45 For a variant of our assumption recall the Sobolev spaces defined for s R and 1 p as H s p = {u S : F 1 {(1 + ξ ) s/ Fu(ξ)} Lp <. where F : S S denotes the Fourier transform for tempered distributions (see [5], [5]). Corollary.3.4. Let Ω R n with piecewise smooth boundary. Suppose that A : L 1 (Ω) D(A) L 1 (Ω) is a sectorial operator such that D(A) H s 1(Ω) for some s > 0. Then A does not have an H -calculus. Proof. To apply Theorem.3.3 we need to show that R(λ, A) has a diffuse representation for some λ ρ(a). Pick any λ ρ(a). Then by Sobolev s theorem we have a continuous inclusion H1(Ω) s L p (Ω) for some p > 1. Hence, for any bounded set U Ω with piecewise smooth boundary we obtain that χ U R(λ, A) factors through L p (U), L 1 (Ω) χ U R(λ,A) D(A) L 1 (U) H s 1(U) L p (U) L 1 (U). Consequently, χ U R(λ, A) is a weakly compact operator. Notice that µ(u) is finite. Therefore, χ U R(λ, A) is an integral operator [0]. This argument works for all bounded U Ω with piecewise smooth boundary and thus R(λ, A) has a diffuse representation. According to Corollary.3.3, A does not have an H -calculus. 38

46 39

47 40

48 Chapter 3 Rademacher bounded families of operators on L Introduction Recall that a closed operator A on a Banach space X is called sectorial if The domain D(A) and range R(A) are dense, A is one-to-one The spectrum σ(a) is contained in a sector Σ ω = {ζ : arg ζ ω}. The resolvent R(ζ, A) satisfies the estimate ζr(ζ, A) C φ, arg(ζ) φ whenever ω < φ < π. If ω < π than the operator A generates a bounded analytic semigroup, T u = e ua, u [0, 1]. Conversely if A is the generator of a bounded analytic semigroup then A is sectorial, provided it is one-one with ω < π/. A is called R-sectorial with angle of R-sectoriality ω R = ω R (A) if for every φ > ω R the collection of operators {R(ζ, A) : arg ζ φ} is R-bounded. We recall 41

49 here that a collection of operators T on a Banach space X is called R-bounded if there is a constant C so that (E ε j T j x j ) 1/ C(E ε j x j ) 1/ x 1,..., x n X, T 1,..., T n T. j=1 j=1 Here (ε j ) j=1 is a sequence of independent Rademacher functions. R-sectoriality has become a very important property for sectorial operators because of the theorem of Weis [53] which shows that in a Banach space with (UMD) this property is closely related to maximal regularity. This note is concerned with the structure of R-sectorial operators on the Banach space L 1 = L 1 (K, λ) where K is a Polish space and λ is a nonatomic σ finite Borel measure. All such spaces are isometric to L 1 = L 1 [0, 1], and so we will assume that K is a compact metric space and λ is a probability measure. If A is a sectorial operator on L 1 which has H -calculus (for some angle) then A is R-sectorial (for some angle). We refer to [3] for the definition and discussion of the H calculus. In [3] it was shown that if A has an H calculus then A is bounded on any reflexive subspace of D(A) (with the graph norm); this had the implication there are very few examples of sectorial operators with an H calculus on L 1 and in particular essentially no reasonable differential operator can have this property. In [8] it was shown that there are no R-bounded strongly continuous semigroups on L 1 consisting of weakly compact operators; it also follows from the results of [8] that if A is an R-sectorial operator on L 1 then the resolvent R(ζ, A) can never be a weakly compact operator. 4

50 The simplest example of a sectorial operator on L 1 (K, λ) which has an H calculus and hence is R-sectorial is the operator Af(s) = b(s)f(s) where b > 0 a.e. and with domain D(A) = { f : } f(s) b(s) 1 dλ(s) <. Note here that the domain is very large indeed; in fact for any ɛ > 0 we can find a Borel set B with λ(b) > 1 ɛ and such that L 1 (B) D(A). Of course one can get further examples by considering A = UAU 1 for U any invertible operator with D(A ) = U(D(A)). In this note, we show that this example is typical. Precisely we show that if A is R-sectorial and ɛ > 0 then there is an invertible operator U : L 1 L 1 with U I < ɛ such that for some positive Borel function w we have U(D(A)) L 1 (w). This improves both the results of [8] and [3]. 3. Operators on L 1 Let K be a compact metric space and suppose λ is a probability measure on K. We will utilize the so-called random measure representation of operators on L 1, developed in [30], [4] and [54]. A random measure on K is a map s µ s from K into M(K) which is Borel for the weak -topology on M(K). If the random measure satisfies the condition µ s (B)dλ(s) Cλ(B) B B (3.1) K 43

51 then it induces a bounded operator T : L 1 (λ) L 1 (λ) given by the formula T f(s) = f(t)dµ s (t) λ a.e. (3.) K and then T C. Conversely every bounded linear operator T : L 1 (λ) L 1 (λ) has an essentially unique random measure representation and T is the least constant C so that (3.1) holds. We may also associate to T a unique measure ρ T on K K given by ρ T (A) = K ( K ) χ A (s, t)dµ T s (t) dλ(s) A B(K K). Thus ρ T (A B) = T χ B dλ. A The map T ρ T maps L(L 1 ) onto an order-ideal in M(K K) consisting of all measures ρ such that ρ (A B) Cλ(B), A, B B(K K). The space L(L 1 (K, λ)) is a complex Banach lattice and it is easily checked that if T L(L 1 ) then µ T s = µ T s (λ-a.e.) and that ρ T = ρ T. Since it is a Banach lattice we can define as usual an operator ( n j=1 T j ) 1 for any T 1,..., T n X. The following result is implicitly contained in ideas of [30], and more explicitly in [31] : Proposition Let T n : L 1 L 1 be a uniformly bounded sequence of operators such that lim n ρ Tn = 0. Then given any ɛ > 0 there is a Borel subset B of K 44

52 with λ(b) > 1 ɛ and n N so that we have T n f ɛ f f L 1 (B). Proof. Consider the measures on K given by ν n (A) = ρ Tn (A K). Then ν n is absolutely continuous with respect to λ. Let w n be the Radon-Nikodym derivatives. Then w n dλ = ρ Tn (K K) 0. Therefore, w n 0 in measure. Hence there exists n N and B with λ(b) > 1 ɛ so that w n < ɛ on B. However T n = w n. This follows from T n = sup λ(b)>0 v n(b) λ(b), v n(b) = B w n dλ, and v n (B) = ρ Tn (B K) = χ B K d µ Tn s (t) = K µtn s (B) dλ. If f L 1 (B) we have T n f f(s) d ρ T (s, t) K K B f(s) w n (s)dλ(s) ɛ f. If T L(L 1 ) then we can write µ s as given in (3.) in the form µ s = a(s)δ s + µ s λ a.e. where µ s{s} = 0 λ a.e. and a is a bounded Borel function. (See for example [30]). Thus T f(s) = a(s)f(s) + f(t)dµ s(t) K 45 λ a.e.

53 If we define the diagonal part of T by Π(T )f = a(s)f(s) then ρ Π(T ) is the restriction of ρ T to the diagonal subset = {(s, s) : s K}. Thus ρ Π(T ) (B) = ρ T (B ). Theorem 3... Let T be a family of operators in L(L 1 ). Then the following are equivalent: (i) T is R bounded. (ii) {( n j=1 a j T j ) 1/ : n j=1 a j 1, T k T, n N} is uniformly bounded. Proof. Assume T is R-bounded so that E ɛ j T j x j ME j=1 m ɛ j x j for any T 1,..., T n T and x 1,..., x n X. Suppose T 1,... T n T and a 1,..., a n C are such that n j=1 a j 1. Then, by Khintchine s inequality for lattices, ( a j T j ) 1 CE ɛ j a j T j j=1 where C is an absolute constant. Now choose any sequence of partitions A m = (A mj ) Nm j=1 so that each A m+1 refines A m and lim m j=1 j=1 sup diama mj = 0. 1 j N m Then for any positive function f L 1 (K, λ) and any T L(L 1 (K, λ)) we have T f = lim m N m j=1 T (fχ Amj ) λ a.e. 46

54 Thus ɛ k a k T k f = lim m N m j=1 ɛ k a k T k (fχ Amj ) λ a.e. Now by R-boundedness E K N m j=1 ɛ k a k T k (fχ Amj ) dλ = N m j=1 M M( E N m j=1 ɛ k a k T k (fχ Amj ) E ɛ k a k fχ Amj a k ) 1 M f. N m j=1 fχ Amj It follows from Fatou s Lemma that E ɛ k a k T k M and hence ( a k T k ) 1 CM. Now let us prove that (ii) = (i). First suppose f L 1 (K, λ) is positive and T 1,..., T n L(L 1 (K, λ)). Then if a 1,..., a n 0 and a a n = 1 we have a j T j f ( T j ) 1 f. j=1 Now the least upper bound of the left hand-side over all choices of a 1,..., a n is ( n j=1 ( T j f) ) 1 and so j=1 j=1 ( ( T j f) ) 1 ( T j ) 1 f. 47 j=1

55 Let from our assumption (ii) there is a constant M so that ( a j T j ) 1 M T1,..., T n T, a a n = 1. j=1 Suppose f L 1 and T 1,..., T n T. Then E ɛ j a j T j f ( a j T j f ) 1 j=1 ( ( j=1 a j ( T j f ) ) 1 j=1 a j T j ) 1 f j=1 M f. Now by Theorem. of [8] this means T is R-bounded. Proposition Suppose T is an R-bounded family of operators on L 1 (K, λ). Then the family of measures {ρ T : T T } is relatively weakly compact in M(K K). Proof. Note that if T 1,..., T n T then max 1 k n T k ( T k ) 1 Mn 1/ where M = sup{ a j T j ) 1 : T1,..., T n T, j=1 a j 1} j=1 which is finite by Theorem 3... The maximum here is computed in the lattice L(L 1 ). 48

56 Hence max 1 k n ρ T k M(K K) Mn 1. Assume the set {ρ T : T T } is not relatively weakly compact. Then by Dieudonne-Grothendieck characterization [18] there is a δ > 0, a sequence (T k ) n and a sequence of disjoint open sets U k in K K so that ρ Tk (U k ) δ for all k. Then max 1 k n ρ T k M(K K) which gives a contradiction. ρ Tk (U k ) δn n = 1,, Applications to sectorial operators Now we will give some applications of this theorem to sectorial operators. Proposition If A is R-sectorial and ω R (A) < π/ then {e ta : 0 < t < } is an R-bounded semigroup. Conversely if A is sectorial and A generates an R-bounded semigroup then A is R-sectorial with ω R (A) π/. If A is a sectorial operator which generates a semigroup {e ta : 0 < t < } with the property that {e ta : 0 < t 1} is R-bounded then for any φ > π/ there exists M so that the set {ζr(ζ, A) : arg(ζ + M) φ} is R-bounded. Proof. These are simple deductions from the formulas: ζr(ζ, A) = 0 ζe ζt e ta dt and e ta (1 + ta) 1 = 1 πi Γ ν (e tζ (1 + tζ) 1 )R(ζ, A)dζ 49