On the Borel Complexity of some classes of Banach spaces

Transcription

1 Kent State University On the Borel Complexity of some classes of Banach spaces Bruno de Mendonça Braga Kent, Ohio July 2013

2 To anyone that, somehow, enjoy it. i

3 Acknowledgements: First of all, I would like to thank my adviser Joe Diestel. I thank him not only for all the beers we had together and the amazing stories he shared but also (and mainly) for all the mathematical help he gave me. Without Joe I would still not even know about the existence of this amazing branch of mathematics, i.e., descriptive set theory and its applications to the geometry of Banach spaces. His advices and guru-like guidance were essential for this dissertation. I am really happy to say that Kent State University provided me with a great study environment and I thank all the professors I had while a student here. Specially, I would like to thank Prof. Richard Aron for being one of the friendliest people I have ever met, for being always available to help me not only with math but also with random advices, and also for being one of the only people I could practice my (beautiful) Portuguese with. I would also like to express my sincere gratitude to Kent State University and its department of mathematics for the financial support they provided me. I also thank all members of my committee for spending their time reading this dissertation. I cannot forget to thank all my monkey friends and family. Although faraway their support and mere existence were of great importance to me. My vacations back home worked to recharge my body with both love and caipirinhas. I specially thank Pedro Igor and Caio Guimares for all the personal assistance they gave me. Also, the friends I made here in Kent were essential for my sanity and I sincerely thank them for that. I was able to find some really amazing people in this tiny university town. I thank my officemate, Willian Franca, for all the mathematical brainstorms we had toii

4 gether and for not carrying about an (annoying) officemate listening to loud music all day long. At last, I thank Masha, the best girlfriend in the world, for all the support and help. iii

5 A guy goes to a psychiatrist and says: Doc, my brother s crazy; he thinks he s a chicken. And the doctor says, Well, why don t you turn him in? The guy says, I would, but I need the eggs. Alvy Singer quoting Groucho Marx. iv

6 Contents 1 Background Banach spaces Basis and Sequences Operators Some Classes of Banach Spaces Local Theory Descriptive set theory Standard Borel spaces Trees Complementability of some ideals in L(X) Unconditionally Converging Operators Weakly Compact Operators Geometry of Banach spaces Banach-Saks Property Alternating Banach-Saks Property Weak Banach-Saks property Complementability of some ideals in L(X), Part II. 40 1

7 CONTENTS 4.1 Banach-Saks Operators Geometry of Banach spaces, Part II Schur Property Dunford-Pettis Property B-convex Banach Spaces Daugavet Property Complete Continuous Property Radon-Nikodym property Analytic Radon-Nikodym property The Equivalence Relation of Isomorphisms Local Structure of Banach Spaces Finite Representability Super Reflexibility Local Unconditional Structure On the Borel complexity of C P Pure classes not cointaining some l p Spaces containing a minimal subspace Non-Universality Results Non-Universality Results Hausdorff Distance and Openings Between Banach Spaces The Hausdorff Distance on F(X) Openings Between Banach Spaces

8 Preface. Happiness is a Banach space. (Dewitt-Morette and Choquet Bruhat) Our goal for this dissertation is to study the Borel complexity of certain classes of Banach spaces, hence, these notes lie in the intersection of descriptive set theory and the theory of Banach spaces. With this dissertation we intend to introduce and familiarize the reader with general techniques to compute the Borel complexity of classes of Banach spaces. Therefore, we do not intend to be concise and we take our time making the text as complete as possible for the understanding of the general ideas behind the proofs. With this in mind, we start this dissertation with chapter 1, which is a brief introduction to all the background the reader may need regarding descriptive set theory and the theory of Banach spaces. In chapter 2, we study two problems related to special classes of operators on separable Banach spaces being complemented in the space of its bounded operators or not. Specifically, we will show that both the set of Banach spaces with its unconditionally converging operators complemented in its bounded operators, and the set of Banach spaces with its weakly compact operators complemented in its bounded operators, are non Borel. The first one is actually complete coanalytic. In both of these problems, we will be using re- 3

9 CONTENTS sults of [BBG] concerning the complementability of those ideals in its space of all bounded operators and the fact that the space itself contains c 0. Next, in chapter 3, we study the Borel complexity of other classes of Banach spaces, namely, Banach spaces with the so called Banach-Saks property, alternating Banach-Saks property, and weak Banach-Saks property. We show that the first two of them are complete coanalytic sets in the class of separable Banach spaces, and that the third one is at least non Borel (it is also shown that the weak Banach-Saks property is at most Σ 1 2). In order to show some of those results we use the geometric sequential characterizations of Banach spaces with the Banach-Saks property and the alternating Banach-Saks property given by B. Beauzamy (see [Be]). The stability under l 2 -sums of the Banach-Saks property shown by J. R. Partington ([P]) will also be of great importance in our proofs. In chapter 4 we revisit the main concern of chapter 2 by studying another problem related to the complementability of an ideal of L(X). Precisely, it is shown that the set of Banach spaces whose set of Banach-Saks operators is complemented in its bounded operators is non Borel. For this, a result by J. Diestel and C. J. Seifert ([DiSe]) that says that weakly compact operators T : C(K) X, where K is a compact Hausdorff space, are Banach-Saks operators, will be essential. In chapter 5 we study the Borel complexity of more geometric classes of separable Banach spaces. In order to show that the class of Banach spaces with the Schur property is non Borel we will rely on the stability of this property shown by B. Tanbay ([T]), and, when dealing with the Dunford-Pettis property, the same will be shown using one of its characterizations (see [FLR], and [Fa]) and Tanbay s result. It is also shown that the Schur property is at least Σ 1 2. After that, we show that both the set of B-convex Banach 4

10 CONTENTS spaces and the set of Banach spaces with the Daugavet property are Borel sets. For the latter, our proof uses the geometric characterization of the Daugavet property given in [KaSSiW], and [KaW]. Following, we show that the set of separable Banach spaces with the complete continuous property (CCP), and the set of Banach spaces with the Radon-Nikodym property (RNP), are complete coanalytic. For this we use characterizations of these properties in terms of the existence of a special kind of bush on the space ([G], [J3]). Also, we show that the analytic Radon-Nikodym property is non Borel. Chapter 6 is dedicated to the local structure of separable Banach spaces. We show that the set of Banach spaces Y that are finitely representable in a given space X, the set of super reflexive spaces and spaces with local unconditional structure are all Borel. It is also shown that if a property P is Borel then property super-p is at least coanalytic. Chapter 7 deals with the Borel complexity of C P = {Y SB Z P, Z Y }, for some specific classes of Banach spaces P SB. Specifically, we show that C P is non Borel if P is the class of minimal spaces, tight spaces, HI spaces, Banach-Saks spaces, alternating Banach-Saks spaces, Schur spaces, B-convex spaces, and infinite dimentional super reflexive Banach spaces. In chapter 8 we give several applications of the theorems obtained along these notes to non-universality like results. In all the results proven in these notes we will be applying techniques related to descriptive set theory and its applications to the geometry of Banach spaces that can be found in [D], and [S]. 5

11 CONTENTS At last, in chapter 9, we turn ourselves to a completely different kind of problem, therefore, this chapter can be read independently from the rest of this dissertation. In this chapter we study the Hausdorff metric in SB 2 and several openings (or gaps) between Banach spaces. Our goal in this chapter is to notice that those functions are Borel in relation to the Effros-Borel structure and to study the Borel complexity of other distances that are commonly defined on SB. We would like to point out that this work was highly motivated by two papers. One of them is D. Puglisi s paper on the position of K(X, Y ) in L(X, Y ), in this paper Puglisi shows that the set of pairs of separable Banach spaces (X, Y ) such that the ideal of compact operators from X to Y is complemented in the bounded operators from X to Y is non Borel (see [Pu]). The second one is M. Ostrovskii s paper on topologies generated by several openings between Banach spaces and their applications to the geometry of Banach spaces (see [O]). 6

12 Chapter 1 Background. My brain? That s my second favorite organ. (Woody Allen) In this chapter we give all the background the reader may need regarding the theory of Banach spaces and descriptive set theory. For more details on Banach space theory we refer to Sequences and Series in Banach Spaces, by J. Diestel ([Di2]), and Topics in Banach Space Theory, by F. Albiac and N. J. Kalton ([AK]). For more on descriptive set theory we refer to Classical Descriptive Set Theory, by A. Kechris ([Ke]), and we refer to Notes on Descriptive set theory and Applications to Banach Spaces, by Th. Schlumprecht ([S]), for a really good introduction to descriptive set theory focusing on its applications to the geometry of Banach spaces. 7

13 1.1. BANACH SPACES. 1.1 Banach spaces Basis and Sequences. Given a Banach space X we say that a sequence (x n ) n N of elements of X is a Schauder basis of X if every element of X can be uniquely written as an infinite linear combination of (x n ) n N, i.e., for all x X there exist an unique (a n ) n N R N such that x = n N a nx n. If X has a Schauder basis (x n ) n N we can define, for all n N, natural projections P n ( i N a ix i ) = n a ix i. It can be shown that the norm of those projections are uniformly bounded ([AK]). If the supremum of these norms is K we say the Schauder basis (x n ) n N has basis constant K. The Schauder basis is called monotone if its basis constant is 1, and bimonotone if it is monotone and Id P n = 1, for all n N, where Id is the identity on X. A sequence (x n ) n N is called a basic sequence if it is a Schauder basis for its closed span. Equivalently (see [AK]), (x n ) n N is a basic sequence if its elements are not zero and there exists K > 0 such that k n a i x i K a i x i, for all k, n N (k n), and for all a 1,..., a n R. The infimum of the constants K for which the inequality above holds is called the basic constant, if this minimum is 1 the sequence is called monotone. If a basis (resp. basic sequence) has the property that it remains a basis (resp. basic sequence) no matter how one reorders it, then the basis (resp. basic sequence) is called an unconditional basis (resp. unconditional basic sequence). Equivalently (see [AK]), a sequence (x n ) n N is unconditional if its elements are not zero and there exists K > 0 such 8

14 CHAPTER 1. BACKGROUND. that n n a i x i K b i x i, for all n N, and all a 1,..., a n, b 1,..., b n R such that a i b i, for all i {1,..., n}. The infimum of this constants is called the unconditional constant of the sequence. A sequence (x n ) n N is called unconditionally converging if the limit n N x π(n) exists for all permutation (reordering) π : N N. The sequence (x n ) n N is called weakly unconditionally Cauchy (or sometimes weakly unconditionally convergent) if n N f(x π(n)) exists for all permutation π and all linear functional f : X R. Given two sequences (x n ) n N and (y n ) n N we say that they are equivalent if there exists C 1 such that 1 k k k a i x i a i y i C a i x i, C for all k N, and all a 1,..., a k R Operators. We denote by L(X) the space of bounded linear operators from X to itself endowed with the supremum norm. It is standard that L(X) is a Banach space with this norm. In this dissertation we will be interested in operator ideals of L(X) and when those ideals are complemented in L(X). Precisely, a subset I L(X) is called an operator ideal (or just an ideal) if it contains the compact operators (i.e., operators that carries bounded sets into relatively compact sets) and if, for all T I and all A L(X), we have A T I and T A I. It is easy to see that the set of compact operators (we denote it by K(X)) is the smallest of the ideals and L(X) the biggest. 9

15 1.1. BANACH SPACES. We say that an operator T : X Y is unconditionally converging (see [Pe]) if it maps weakly unconditionally Cauchy series into unconditionally converging series, T is said to be weakly compact if it maps bounded sets into relatively weakly compact sets. We denote by U(X) and W(X) the set of unconditionally converging and weakly compact operators from X to itself, respectively. Both of these sets lie in L(X), and it is easy to see that both U(X) and W(X) are ideals of L(X). Another important operator ideal is the ideal of strictly singular operators. Those operators will be essential in some of the main proofs in this dissertation. An operator T : X Y is said to be strictly singular if T E is not an isomorphism onto its image, for all infinite dimensional subspace E X Some Classes of Banach Spaces. A Banach space X is said to have the Banach-Saks property if every bounded sequence (x n ) n N in X has a subsequence (x nk ) k N such that its Cesaro mean n 1 n k=1 x n k is norm convergent. A Banach space X is said to have the alternating Banach-Saks property if every bounded sequence (x n ) n N in X has a subsequence (x nk ) k N such that its alternatingsigns Cesaro mean n 1 n k=1 ( 1)k x nk is norm convergent. A Banach space is said to have the weak Banach-Saks property if every weakly null sequence has a subsequence such that its Cesaro mean is norm convergent to zero. The weak Banach-Saks property if often called Banach-Saks-Rosenthal property. Given two isomorphic Banach spaces X and Y, we define the Banach-Mazur distance D(X, Y ) between X and Y as the infimum of all the quantities T T 1, where T varies among all the isomorphisms between X and Y. A Banach space X is called B-convex if 10

16 CHAPTER 1. BACKGROUND. lim inf{d(x n, l n 1) X n is a n-dimensional subspace of X} =, n where l n 1 is the n-dimensional space R n endowed with the l 1 -norm (x k ) n k=1 1 = n k=1 x k. We denote the set of Banach spaces with the Banach-Saks property, the set of Banach spaces with the alternating Banach-Saks property, and the set of Banach spaces with the weak Banach-Saks property, by BS, ABS, and WBS, respectively. One can see that the inclusions BS ABS WBS hold (see [Be]). We denote the set of B-convex spaces by BC. An operator T L(X) that carries bounded sequences into sequences with convergent Cesaro mean subsequences is called a Banach-Saks operator. We denote by BS(X) the set of all Banach-Saks operators from X to itself. It is easy to see that BS(X) is an operator ideal. An operator T : X Y between two Banach spaces is said to be completely continuous (or Dunford-Pettis) if it carries weak-compact sets into norm-compact sets. A Banach space X is said to have the complete continuous property if every operator from L 1 [0, 1] to X is completely continuous. A Banach space X is said to have the Dunford-Pettis property if for all Banach spaces Y every weakly compact operator T : X Y is completely continuous. A Banach space X has the Daugavet property if every rank-1 operator T : X X satisfies Id + T = 1 + T. We say that a Banach space X has the Schur property if every weakly convergent sequence of X is norm convergent. A classical example of a Schur space is l 1 (see [AK]). A Banach space X is said to have the Radon-Nikodym property if for any totally finite 11

17 1.1. BANACH SPACES. positive measure space (Y, Σ, µ) and any X-valued µ-continuous measure m on Σ, with m (X) <, there exists f L 1 X (Y, Σ, µ) such that m(e) = fdµ, for all E Σ. E A complex Banach space X has the analytic Radon-Nikodym property if every X-valued measure of bounded variation, defined on the Borel subsets of T = {z C z = 1}, whose negative Fourier coefficients vanish, has a Randon-Nikodym Bochner integral derivative with respect to the Lebesgue measure on T Local Theory. A Banach space Y is said to be finitely representable in a Banach space X if for all ε > 0, and each finite dimensional subspace E Y, there exists a finite dimensional subspace F X and an isomorphism u : E F such that u u 1 < 1 + ε. In other words, Y is finitely representable in X if for all ε > 0, and each finite dimensional subspace E Y, one can find a subspace F X such that the Banach-Mazur distance between E and F is smaller than 1 + ε. A Banach space X is called super reflexive if every Banach space Y that is finitely representable in X is reflexive. P. Enflo had shown in [E] that a Banach space X is super reflexive if, and only if, there exists an equivalent uniformly convex norm on X, i.e., if for all ε > 0 there exists δ > 0 such that x y ε implies x+y /2 1 δ, for all x, y S X. A Banach space X is said to have local unconditional structure (or l.u.st.) if there exists λ > 0 such that for each finite dimensional Banach space E X there exists a finite dimensional space F with an unconditional basis and operators u : E F, and w : F X such that w u = Id E, and ub(f ) u w λ, where ub(f ) is an unconditional constant for F. Notice that a complemented subspace of a space with l.u.st. clearly has l.u.st.. 12

18 CHAPTER 1. BACKGROUND. 1.2 Descriptive set theory Standard Borel spaces. A separable metric space is said to be a Polish space if there exists an equivalent metric in which it is complete. A continuous image of a Polish space into another Polish space is called an analytic set and a set whose complement is analytic is called coanalytic. A measure space (X, A), where X is a set and A is a σ-algebra of subsets of X, is called a standard Borel space if there exists a Polish topology on this set whose Borel σ-algebra coincides with A. We define Borel, analytic and coanalytic sets in standard Borel spaces by saying that these are the sets that, by considering a compatible Polish topology, are Borel, analytic, and coanalytic, respectively. Observe that this is well defined, i.e., this definition does not depend on the Polish topology itself but on its Borel structure. A function between two standard Borel spaces is called Borel measurable if the inverse image of each Borel subset of its codomain is Borel in its domain. We usually refer to Borel measurable functions just as Borel functions. Notice that, if you consider a Borel function between two standard Borel spaces, its inverse image of analytic sets (resp. coanalytic) is analytic (resp. coanalytic) (see [S]). Given a Polish space X the set of analytic (resp. coanalytic) subsets of X is denoted by Σ 1 1(X) (resp. Π 1 1(X)). Hence, the terminology Σ 1 1-set (resp. Π 1 1-set) is used to refer to analytic sets (resp. coanalytic sets). Let X be a standard Borel space. An analytic (resp. coanalytic) subset A X is said 13

19 1.2. DESCRIPTIVE SET THEORY. to be complete analytic (resp. complete coanalytic) if for all standard Borel space Y and all B Y analytic (resp. coanalytic), there exists a Borel function f : Y X such that f 1 (A) = B. This function is called a Borel reduction from B to A, and B is said to be Borel reducible to A. Let X be a standard Borel space. We call a subset A X Σ 1 1-hard (resp. Π 1 1-hard) if for all standard Borel space Y and all B Y analytic (resp. coanalytic) there exists a Borel reduction from B to A. Therefore, to say that a set A X is Σ 1 1-hard (resp. Π 1 1-hard) means that A is at least as complex as Σ 1 1-sets (resp. Π 1 1-sets) in the projective hierarchy. With this terminology we have that A X is complete analytic (resp. complete coanalytic) if, and only if, A is Σ 1 1-hard (resp. Π 1 1-hard) and analytic (resp. coanalytic). As there exist analytic non Borel (resp. coanalytic non Borel) sets we have that Σ 1 1-hard (resp. Π 1 1-hard) sets are non Borel. Also, if X is a standard Borel space, A X, and there exists a Borel reduction from a Σ 1 1-hard (resp. Π 1 1-hard) subset B of a standard Borel space Y to A, then A is Σ 1 1-hard (resp. Π 1 1-hard). If A is also analytic (resp. coanalytic), then A is complete analytic (resp. complete coanalytic). We refer to [S] and [Ke] for more on complete analytic and coanalytic sets. Complete analytic sets (resp. complete coanalytic sets) are also called Σ 1 1-complete sets (resp. Π 1 1-complete). Consider a Polish space X and let F(X) be the set of all its non empty closed sets. We endow F(X) with the Effros-Borel structure, i.e., the σ-algebra generated by {F X F U }, where U varies among the open sets of X. It can be shown that F(X) with the Effros- Borel structure is a standard Borel space ([Ke]). The following well-known lemma (see 14

20 CHAPTER 1. BACKGROUND. [Ke]) will be crucial in some of our proofs. Lemma 1. Let X be a Polish space. There exists a sequence of Borel functions (S n ) n N : F(X) X such that {S n (F )} n N is dense in F, for all closed F X. In this dissertation we will only work with separable Banach spaces. We denote the closed unit ball of a Banach space X and its unit sphere by B X and S X, respectively. It is well known that every separable Banach space is isometrically isomorphic to a closed linear subspace of C( ) (see [Ke]), where denotes the Cantor set. Therefore, C( ) is called universal for the class of separable Banach spaces and we can code the class of separable Banach spaces, denoting it by SB, by SB = {X C( ) X is a closed linear subspace of C( )}. As C( ) is clearly a Polish space we can endow F(C( )) with the Effros-Borel structure. It can be shown that SB is a Borel set in F(C( )) and hence it is also a standard Borel space (see [S]). It now makes sense to wonder if specific sets of separable Banach spaces are Borel or not. Throughout these notes we will denote by {S n } n N the sequence of Borel functions S n : SB C( ) given by lemma 1 (more precisely, the restriction of those functions to SB). Hence, for all X SB, {S n (X)} n N is dense in X. Also, letting S n(x) = S n (X), if S n (X) 0 S min{m N Sm(X) 0}(X), if S n (X) = 0, we obtain that { Sn (X) = S n(x)/ S n(x) } n N is dense in S X for each X SB. Clearly, each S n : SB S C( ) is also Borel (see [S]). 15

21 1.2. DESCRIPTIVE SET THEORY Trees. Denote by N <N the set of all finite tuples of natural numbers plus the empty set. Given s = (s 0,..., s n 1 ), t = (t 0,..., t m 1 ) N <N we say that the length of s is s = n, s i = (s 0,..., s i 1 ), for all i {1,..., n}, s 0 = { }, s t iff n m and s i = t i, for all i {0,..., n 1}, i.e., if t is an extension of s. We define s < t analogously. Define the concatenation of s and t as s t = (s 0,..., s n 1, t 0,..., t m 1 ). A subset T of N <N is called a tree if t T implies t i T, for all i {0,..., t }. We denote the set of trees on N by Tr. A subset I of a tree T is called a segment if I is completely ordered and if s, t I with s t, then l I, for all l T such that s l t. Two segments I 1, I 2 are called completely incomparable if neither s t nor t s hold if s I 1 and t I 2. As N <N is countable, 2 N<N (the power set of N <N ) is Polish with its standard product topology. If we think about Tr as a subset of 2 N<N it is easy to see that Tr is a G δ set in 2 N<N ([Ke]). Thus, it is also Borel in 2 N<N. As Tr is Borel in the Polish space 2 N<N we know that Tr is a standard Borel space ([Ke]). A β N N is called a branch of a tree T if β i T, for all i N, where β i is defined analogously as above. We call a tree T well-founded if T has no branches and ill-founded otherwise, we denote the set of well-founded and ill-founded trees by WF and IF, respectively. It is well known that WF is a complete coanalytic set of Tr, hence IF is complete analytic (see [S]). We denote the set of trees with finite length by FTr, i.e., FTr = {T Tr n N, s T implies s n}. This set is easily seen to be Borel. Indeed, we have FTr = n s N <N s >n {T Tr s / T }. There is a really important index that can be defined on the set of trees called the order index of a tree. For a given tree T Tr we define the derived tree of T as 16

22 CHAPTER 1. BACKGROUND. T = {s T t T, s < t}. By transfinite induction we define (T ξ ) ξ ON, where ON denotes the ordinal numbers, as follows T α = (T β ), if α = β + 1, for some β ON, T α = β<α T β, if α is a limit ordinal. We now define the order index on Tr. If there exists an ordinal number α < ω 1, where ω 1 denotes the smallest uncountable ordinal, such that T α = we say the order index of T is o(t ) = min{α < ω 1 T α = }. If there is no such countable ordinal we set o(t ) = ω 1. The reason why we introduce this index is because of the way it interacts with the notion of well-founded and ill-founded trees. We have the following easy proposition. Proposition 2. A tree T Tr on the natural numbers is well-founded if, and only if, its order index is countable, i.e., if, and only if, o(t ) < ω 1. For T Tr and k N, let T (k) = {s N <N (k) s T } and T k = {s T (k) s}. We have another useful application of the order index to well-founded trees. Proposition 3. Let T WF with o(t ) > 1, then o(t (k)) < o(t ), for all k N. Let θ Tr. We define c 00 (θ) = {(a s ) s θ R θ a s 0 for finitely many s θ}. We denote by (e s ) s θ the standard unit basis of c 00 (θ). Now we introduce a very useful way to construct a Borel ϕ : Tr SB. Let (x n ) n N be a basic sequence in a separable Banach space X. Pick θ Tr and p [1, ). We define a norm on c 00 (θ) as follows: For each x c 00 (θ), let 17

23 1.2. DESCRIPTIVE SET THEORY. {( n x θ = sup s Ii x(s)x s p X ) 1 } p n N, I 1,..., I n incomparable segments of θ, where. X is the X norm. Define ϕ(θ) to be the completion of c 00 (θ) under the norm. θ. Pick Y C( ) such that ϕ(n <N ) = Y. If we consider the natural isometries of ϕ(θ) into ϕ(n <N ), we can see ϕ as a function from Tr into SB. With this in mind, we have (see [S]): Proposition 4. ϕ : Tr SB is a Borel function. The same is true if we define. θ as { } x θ = sup x(s)x X s I segment of θ. s I For now on, every time we define a function on Tr as the completion of c 00 (θ) under one of those norms we will be considering an isometry of ϕ(n <N ) into C( ) and the natural isometries of ϕ(θ) into ϕ(n <N ). Now that we ve seen all the background we need in order to understand our results and their proofs let s start with the real math. 18

24 Chapter 2 Complementability of some ideals in L(X). (M. Gromov s answer to my ) In this chapter we deal with operator ideals and whether they are complemented or not in L(X). 2.1 Unconditionally Converging Operators. Let X and Y be Banach spaces. We write Y subspace of X. Theorem 5. Let U = {X SB U(X) L(X)}, where U(X) is the set of unconditionally convergent operators on X. Then U is complete coanalytic. X if Y is isomorphic to a complemented Proof. In order to show this we only need to use that U(X) is complemented in L(X) if, and only if, c 0 does not embed in X (see [BBG], pag. 452). Therefore, U = NC c0 (where 19

25 2.1. UNCONDITIONALLY CONVERGING OPERATORS. NC X = {Y SB X Y }, for X SB), which is well known to be non Borel (complete coanalytic actually, see [Bo]), so there is nothing to be done. However, as the proof of this fact will give us a lemma that will be crucial in the proof of many other theorems in these notes, we believe it is worth revisiting this result. Let s define a Borel function ϕ : Tr SB such that ϕ 1 (U) = WF. Let (e n ) n N be the standart basis of c 0. For each θ Tr and x c 00 (θ) we define {( n x θ = sup x(s)e s 2 s Ii ) 1 2 c 0 } n N, I 1,..., I n incomparable segments of θ, where. c0 is the norm of c 0. Now we define ϕ(θ) as the completion of c 00 (θ) under the norm. θ. Let s now show that ϕ works as we want. If θ IF it s clear that c 0 ϕ(θ). Indeed, let β be a branch of θ, then c 0 = ϕ(β) ϕ(θ), where by ϕ(β) we mean ϕ applied to the tree {s N <N s < β}. The only thing left to show is that if θ WF, then c 0 does not embed in ϕ(θ). For this we will proceed by induction on the order of θ. If o(θ) = 1 the result is clear. Indeed, in this case θ = { }, so ϕ(θ) is one dimensional. Assume c 0 ϕ(θ) for all θ WF with o(θ) < α, for some α < ω 1. Fix θ WF with o(θ) = α. Let Λ = {λ N (λ) θ} and enumerate Λ, say Λ = {λ i i N}. For each λ Λ, let θ λ = {s θ (λ) s}. As θ WF Proposition 3 gives us ( n o j=1 ) θ(λ j ) < o(θ) = α, n N. 20

26 CHAPTER 2. COMPLEMENTABILITY OF SOME IDEALS IN L(X). Our induction hypothesis implies that c 0 ϕ( n j=1 θ(λ j)), for all n N. Therefore, as ϕ( n j=1 θ λ j ) is a direct sum of ϕ( n j=1 θ(λ j)) with a finite dimensional space, c 0 ϕ( n j=1 θ λ j ), for all n N. Consider now the projections ( n P λn : ϕ(θ) ϕ j=1 θ λj ) (a s ) s θ (a s ) s n j=1 θ λ j. Say E is a subspace of ϕ(θ). Assume E = c 0 and let s find a contradiction. Case 1: j N such that P λj : E ϕ( j θ λ i ) is not strictly singular. Then there exists an infinite dimensional subspace Ẽ E such that P λ j Ẽ is an isomorphism. As E is minimal, this implies c 0 ϕ( j θ λ i ), absurd. Case 2: P λj : E ϕ( j θ λ i ) is strictly singular, for all j N. Claim: (x n ) n N a normalized sequence in E such that P λj (x n ) 0, as n, j N. Indeed, by a well-known consequence of the definition of strictly singular operators, for all j N, there exists a normalized sequence (x j n) n N such that P λj (x j n) < 1/n, for all n N. Let (x n ) n N be the diagonal sequence of the sequences (x j n) n N, i.e., x n = x n n, for all n N. As, i j implies P λi (x) P λj (x), for all x E, (x n ) n N has the required property. Say (ε i ) i N is a sequence of positive numbers converging to zero. Using the claim above 21

27 2.1. UNCONDITIONALLY CONVERGING OPERATORS. and the fact that P λj (x) x, as n N, for all x ϕ(θ), we can pick increasing sequences of natural numbers (n k ) k N and (l k ) k N such that i) P λlk (x nk ) x nk θ < ε k, for all k N, and ii) P λlk (x nk+1 ) θ < ε k, for all k N. For all k N, let y k = P λlk (x nk ) P λlk 1 (x nk ). Choosing ε k small enough we can assume y k 2 θ ( 1 2, 2). It s easy to see that (y k) k N is equivalent to (ẽ k ) k N, where (ẽ k ) k N is the standard l 2 -basis. Indeed, pick a 1,..., a N R, then 1 2 N a i 2 N N a i y i 2 θ = a i y i 2 θ = N N a i y i 2 θ 2 a i 2, where the equalities above only hold because the supports of (y k ) k N are completely incomparable. Therefore, by choosing (ε k ) k N converging to zero fast enough, the principle of small perturbations (see [AK]) gives us that (x nk ) k N is equivalent to (y k ) k N (ẽ k ) k N, absurd. Indeed, this would imply l 2 E = c 0. The proof above actually gives us a stronger result, which will be essential in many of the proofs in these notes. We have the following: Lemma 6. Let X SB be a minimal Banach space and assume X does not contain l p, p [1, ). Say Y X is a subspace with basis (e n ) n N. For each θ Tr, and each x c 00 (θ), define {( n x θ = sup s Ii x(s)e s p Y ) 1 } p n N, I 1,..., I n incomparable segments of θ, 22

28 CHAPTER 2. COMPLEMENTABILITY OF SOME IDEALS IN L(X). where. Y stands for the norm of Y. If we define a function ϕ : Tr SB by letting ϕ(θ) be the completion of c 00 (θ) under the norm. θ we have that X ϕ(θ) if, and only if, θ IF. An analogous result holds if. θ is defined as { } x θ = sup x(s)e Y s I segment of θ, s I and we assume c 0 X. If we forget the condition Y X we still have that X ϕ(θ), for all θ WF. 2.2 Weakly Compact Operators. We say that an operator T : X Y is weakly compact if it maps bounded sets into relatively weakly compact sets. Theorem 7. Let W = {X SB W(X) L(X)}, where W(X) is the set of weakly compact operators on X. Then W is Π 1 1-hard. In particular, W is non Borel. Proof. In order to show this we will use another result of [BBG] (pag. 450). In this paper it is shown that if c 0 X, then W(X) is not complemented in L(X). Let ϕ : Tr SB be as in the proof of theorem 5 for X = c 0 and (e n ) n N the standard c 0 -basis. Let s observe that ϕ 1 (W) = WF, and we will be done. If θ IF we saw that c 0 ϕ(θ), hence ϕ(θ) / W. Let s show that if θ WF, then ϕ(θ) is reflexive, which implies ϕ(θ) W. Indeed, a Banach space is reflexive if and only if its unit ball is weakly compact, therefore W(X) = L(X). The statement that ϕ carries WF into reflexive spaces can be found (as well as part of its proof) in [S]. But for the seek of completeness of these notes and as this result will give us a lemma that will be crucial when computing the Borel complexity of both the 23

29 2.2. WEAKLY COMPACT OPERATORS. Radon-Nikodym property and the complete continuous property, we believe it is worth showing its complete proof here. Let (s i ) i N be a compatible enumeration of θ, i.e., s i s j implies i j. Notice that (e si ) i N is a basis for ϕ(θ), where each e si denotes the vector in ϕ(θ) with coordinate s i being 1 and 0 elsewhere. This base is clearly bimonotone, which implies the biorthogonal functionals (e s i ) i N are also a bimonotone basis for their closed span (see [AK]). By a now-classical result of R. C. James ([J]), a Banach space with a Schauder basis is reflexive if, and only if, it has a basis that is both boundedly complete and shrinking. Hence, it is enough to show that the basis (e si ) si θ of ϕ(θ) is both boundedly complete and shrinking. We start by showing that (e si ) si θ is boundedly complete. For this let s proceed by transfinite induction on the order of θ. If o(θ) = 1 the result is trivial. Assume (e si ) si θ is boundedly complete for all θ WF of order smaller than α < ω 1. Fix θ WF with o(θ) = α. Consider (a s ) s θ c 00 (θ) with a = 0, then a s e s 2 θ = { N = sup } a s e s 2 c 0 n N, I 1,..., I n incomparable segments of θ s Ii = k = k { N sup } a s e s 2 c 0 n N, I 1,..., I n incomparable segments of θ s I i θ k a s e s 2 θ s θk Now say (a si ) i N is a sequence of real numbers with the property that a s1 = a = 0 and sup N N a s i e si θ <. As θ is well-founded Proposition 3 implies that o(θ(k)) < o(θ) = α, for all k N. Hence, our inductive hypothesis together with the equation above 24

30 CHAPTER 2. COMPLEMENTABILITY OF SOME IDEALS IN L(X). implies that a e si si i N s i θ k exists for all k N. The equation above also implies that, for all ε > 0, there exists n 0 N such that, for n > n 0, k=n a e si si s i θ k There exists n 1 N such that m n n 1 implies 2 < ε2 θ 2, m a e si si i=n s i θ k 2 θ < ε2 2n 0, for all k n 0. Putting all of this together we have that, for all m n n 1, m i=n a si e si 2 = θ k N m n 0 m k=1 a e si si i=n s i θ k a e si si i=n s i θ k ε2 2 + ε2 2 = ε2. 2 θ 2 θ + k=n 0 +1 a e si si s i θ k 2 θ Notice that here we used the fact that our basis is bimonotone. We just showed that, for all ε > 0, there exists n 1 N such that m n n 1 implies m i=n a s i e si θ < ε, thus (e si ) si θ is boundedly complete. We now show that (e si ) si θ is shrinking. Again, let s proceed by transfinite induction on the order of θ. If o(θ) = 1 the result is trivial. Assume (e si ) si θ is shrinking for all θ WF of order smaller than α < ω 1. Fix θ WF 25

31 2.2. WEAKLY COMPACT OPERATORS. with o(θ) = α. Consider (a s ) s θ c 00 (θ) with a = 0. a s e s θ = sup ( ) θ a s e s bs e s b se s 1 = sup a s b s a se s 1 = sup a s b s a se s 1 s θ k k N (c n) B l2 k N (c n) B l2 = sup sup a s b s b se s c k s θ k = sup c k a s e s θ k N s θ k ( = ) 1 a s e s 2 2 θ s θ k k It s well known that (e si ) si θ is shrinking if, and only if, (e s i ) si θ is boundedly complete (see [AK]). We show (e s i ) si θ is boundedly complete. Consider (a si ) i N a sequence of real numbers such that a s1 = a = 0 and sup N a s i e s i θ <. The equation above implies N a s e 2 s <. θ k N i N s i θ k So, for all ε > 0, there exists n 0 N such that a si e 2 s < ε2 i θ 2, for all n n 0. k=n i Ns i θ k By our inductive hypothesis, there exists n 1 N such that m n n 1 implies m a e si i=n s i θ k Therefore, if m n n 1, we have s 2 < ε2 i θ 2n 0, for all k n 0. 26

32 CHAPTER 2. COMPLEMENTABILITY OF SOME IDEALS IN L(X). m a si e s 2 = i θ i=n k a s e 2 s θ i=n s θ k n 0 m a s e 2 s + θ k=1 i=n s θ k m ε2 2 + ε2 2 = ε2, k=n 0 +1 a s e s θ k 2 s θ here we use the fact that (e s) s θ is bimonotone. We are now done. The proof above actually gives us the following lemma, that will be really useful for us later on. Lemma 8. Let X be a Banach space with basis (e n ) n N, and norm. X. Let p (1, ). For each θ Tr and x c 00 (θ) we define {( n x θ = sup s Ii x(s)e s p X ) 1 } p n N, I 1,..., I n incomparable segments of θ, and let ϕ(θ) be the completion of c 00 (θ) under the norm. θ. Then, if θ WF, we have that ϕ(θ) is reflexive. Problem 9. Is W coanalytic? If yes, we had shown that W is complete coanalytic. 27

33 Chapter 3 Geometry of Banach spaces. With great power comes great responsibility. (Ben Parker) In this chapter we study the Borel complexity of the Banach-Saks property, the alternating Banach-Saks property and the weak Banach-Saks property. 3.1 Banach-Saks Property. A Banach space X is said to have the Banach-Saks property if every bounded sequence (x n ) n N in X has a subsequence (x nk ) k N such that its Cesaro mean n 1 n k=1 x n k is norm convergent. In this section we show that the class of separable Banach spaces having the Banach-Saks property is complete coanalytic. For this we use a characterization of this property given by B. Beauzamy in [Be] (pag. 373). Beauzamy characterized not having the Banach-Saks property in terms of the existence of a sequence satisfying some geometrical inequality. 28

34 CHAPTER 3. GEOMETRY OF BANACH SPACES. Precisely: Theorem 10. A X SB does not have the Banach-Saks property if, and only if, there exist ε > 0 and a sequence (x n ) n N in B X such that, for all subsequences (x nk ) k N, m N, and l {1,..., m}, the following holds Theorem 11. BS is coanalytic in SB. 1 l m m ( x nk x nk ) ε. k=1 k=l+1 Proof. In order to show that BS is coanalytic we construct a Borel function ϕ : SB Tr such that ϕ 1 (WF) = BS, as WF is coanalytic we will be done. Pick ε > 0 and X SB. Let s associate a tree T BS (X, ε) on X to the pair (X, ε) in the following manner, for all finite sequences (x k ) n k=1 in B X, (x k ) n k=1 T BS(X, ε) n 1 <... < n m n, l {1,..., m}, 1 l m m ( x nk x nk ) ε k=1 k=l+1 It is clear, by theorem 10, that X does not have the BS property if, and only if, T BS (X, ε) is ill-founded for some ε. Now we use lemma 1 to make T BS (X, ε) into a tree on N<N. Let (S n ) n N be the sequence of Borel maps given by lemma 1 for X = C( ). Given X SB and ε > 0 we define a tree T BS (X, ε) on N <N as, for all finite sequences (n k ) n k=1 N<N, (n k ) n k=1 T BS (X, ε) (S nk (B X )) n k=1 T BS(X, ε). As {S n (B X )} n N is dense in B X, for all X SB, standard approximation arguments imply that X does not have the BS property if, and only if, T BS (X, ε) is ill-founded for some ε. 29

35 3.1. BANACH-SAKS PROPERTY. We now glue the trees (T BS (X, ε)) ε together in order to get a single tree T BS (X) on N <N. We define T BS (X) as follows, for all (m) (n k ) n k=1 N<N, ( (m) (n k ) n k=1 T BS (X) (n k ) n 1 ) k=1 T BS X,. m It is clear that X / BS if, and only if, T BS (X) is ill-founded. Therefore, we had just defined the ϕ : SB Tr we wanted. We only need to observe ϕ is Borel and we will be done. For this we pick s N <N, say s = (m) (n k ) n k=1, and show that {X SB s T (X)} is Borel. Clearly, {X SB s T (X)} = n 1 <...<n m n l {1,...,m} {X SB 1 l m ( S nk (B X ) k=1 m k=l+1 S nk (B X )) ε }, and we are done. Indeed, as S n is Borel, n N, the set {X SB 1 m ( l k=1 S n k (B X ) m k=l+1 S n k (B X )) ε} is Borel. The previous theorem shows that BS is at least coanalytic in SB, but it doesn t say anything about BS being Borel or not. The next theorem takes care of this by showing that coanalyticity is the most we can get of BS in relation to its Borel complexity. Theorem 12. BS is Π 1 1-hard. In particular, BS is non Borel. Proof. For this we construct a Borel function ϕ : Tr SB such that ϕ 1 (BS) = WF, as WF is Π 1 1-hard this will suffice. For each θ Tr and x c 00 (θ) define {( n x θ = sup x(s)e s 2 s Ii ) 1 2 l 1 } n N, I 1,..., I n incomparable segments of θ, where (e n ) n N is the standard l 1 -basis and. l1 is its norm. We define ϕ(θ) as the com- 30

36 CHAPTER 3. GEOMETRY OF BANACH SPACES. pletion of c 00 (θ) under the norm. θ. If θ IF we clearly have l 1 ϕ(θ). Indeed, if β is a branch of θ we have ϕ(β) = l 1. As l 1 ϕ(θ) and l 1 is clearly not in BS (take its standard basis for example, it clearly doesn t have a subsequence with norm converging Cesaro mean) we conclude that ϕ(θ) / BS. Let s show that if θ WF, then ϕ(θ) BS. We proceed by transfinite induction on the order of θ WF. Say o(θ) = 1, then ϕ(θ) is 1-dimensional and we are clearly done. Assume ϕ(θ) BS, for all θ WF with o(θ) < α, for some α < ω 1. Pick θ WF with o(θ) = α, let s show that ϕ(θ) BS. Let Λ = {λ N (λ) θ}. As θ WF, Proposition 3 gives us o ( θ(λ) ) < o(θ) = α, λ Λ. Our induction hypothesis implies that ϕ(θ(λ)) BS, for all λ Λ. Now, notice that ϕ(θ) ( ) = R ϕ(θ(λ)), l 2 λ Λ where we get the R above because of the coordinate related to s = θ. By J. R. Partington s result in [P] (pag. 370), we have that the l 2 -sum of spaces in BS is also in BS. Hence, ( λ Λ ϕ(θ(λ))) l 2 is in BS and conclude that ϕ(θ) BS. The transfinite induction is now over, and so is our proof. Theorem 13. BS is complete coanalytic. 3.2 Alternating Banach-Saks Property. A Banach space X is said to have the alternating Banach-Saks property if every bounded sequence (x n ) n N in X has a subsequence (x nk ) k N such that its alternating-signs Cesaro 31

37 3.2. ALTERNATING BANACH-SAKS PROPERTY. mean n 1 n k=1 ( 1)k x nk is norm convergent. In this section, instead of looking at the Banach-Saks property, we study Banach spaces that have the alternating Banach-Saks property. More precisely, we show that the class of separable Banach spaces having the alternating Banach-Saks property is complete coanalytic. For this we use a characterization of this property given by B. Beauzamy in [Be] (pag. 369). Beauzamy characterized not having the alternating Banach-Saks property in terms of the existence of a sequence satisfying some geometrical relation. Precisely: Theorem 14. A X SB does not have the alternating Banach-Saks property if, and only if, there exist ε > 0 and a sequence (x n ) n N in B X such that for all l N, if l n(1) <... < n(2 l ), where n(i) N, i {1,..., 2 l }, then for all c 1,..., c 2 l R. 2l c i x n(i) 2l ε c i, Theorem 15. ABS is coanalytic in SB. Proof. We proceed exactly as in the proof of theorem 11. Let s construct a Borel function ϕ : SB Tr such that ϕ 1 (WF) = ABS. Pick ε > 0 and X SB. We associate a tree T ABS (X, ε) on X to the pair (X, ε) in the following manner, for all finite sequences (x k) n k=1 in B X, 32

38 CHAPTER 3. GEOMETRY OF BANACH SPACES. (x k ) n k=1 T ABS(X, ε) l N, n(1) <... < n(2 l ) (n(k) N, k {1,..., 2 l }, & n(1) l & n(2 l ) n), c 1,..., c 2 l R, 2 l 2 l c k x n(k) ε c k k=1 k=1 It is clear, by theorem 14, that X does not have the ABS property if, and only if, T ABS (X, ε) is ill-founded for some ε > 0. Now we use lemma 1 to make T ABS (X, ε) into a tree on N <N. Let (S n ) n N be the sequence of Borel maps given by lemma 1 for X = C( ). Given X SB and ε > 0 we define a tree T ABS (X, ε) on N <N as, for all finite sequences (n k ) n k=1 N<N, (n k ) n k=1 T ABS (X, ε) (S nk (B X )) n k=1 T ABS(X, ε). By standard approximation arguments we see that X does not have the ABS property if, and only if, T ABS (X, ε) is ill-founded for some ε. We now glue the trees (T ABS (X, ε)) ε together in order to get a single tree T ABS (X) on N <N. We define T ABS (X) as follows, for all (m) (n k ) n k=1 N<N, (m) (n k ) n k=1 T ABS (X) (n k ) n k=1 T ABS (X, 1 m ). It is clear that X / ABS if, and only if, T ABS (X) is ill-founded. Therefore, we have just defined the ϕ : SB Tr we wanted. We only need to observe ϕ is Borel and we will be done. For this we pick s N <N, say s = (m) (n k ) n k=1, and we show that {X SB s T (X)} is Borel. Clearly, 33

39 3.2. ALTERNATING BANACH-SAKS PROPERTY. {X SB s T (X)} = l N c 1,...,c n(1)<...<n(2 l 2 l Q + ) N n(1) l & n(2 l ) n {X SB 2l c k S n(k) (B X ) 1 m k=1 2 l k=1 } c k, and we are done. Indeed, as S n is Borel, n N, the set {X SB 2 l k=1 c ks n(k) (B X ) 1 m 2 l k=1 c k } is Borel. Now we show that coanalyticity is the most we can get of ABS in relation to its Borel complexity. Theorem 16. ABS is Π 1 1-hard. Moreover, ABS is complete coanalytic. Proof. Let ϕ : Tr SB be define exactly as in the proof of theorem 12. We will show that ϕ 1 (ABS) = WF. If θ IF, we have l 1 ϕ(θ). As l 1 is not in ABS (we can take its standard basis again, it clearly doesn t have a subsequence with norm converging alternating-signs Cesaro mean) we conclude that ϕ(θ) / ABS. Let s show that if θ WF, then ϕ(θ) ABS. We proceed by transfinite induction on the order of θ WF. Say o(θ) = 1, then ϕ(θ) is 1-dimensional and we are clearly done. Assume ϕ(θ) ABS for all θ WF with o(θ) < α, for some α < ω 1. Pick θ WF with o(θ) = α. Using the same notation as in the proof of theorem 12, we have ϕ(θ) ( ) = R ϕ(θ(λ)). l 2 λ Λ By lemma 6, l 1 ϕ(θ). B. Beauzamy showed in [Be] (pag. 386) that a Banach space not 34

40 CHAPTER 3. GEOMETRY OF BANACH SPACES. containing l 1 has the alternating Banach-Saks property if, and only if, it has the weak Banach-Saks property. So, we only need to show that ϕ(θ) is in WBS. As ϕ(θ(λ)) ABS, for all λ Λ, we have ϕ(θ(λ)) WBS, for all λ Λ. By a corollary of J. R. Partington (see [P], pag. 373), ( λ Λ ϕ(θ(λ))) l 2 is also in WBS. Thus, we conclude that ϕ(θ) WBS, and we are done. 3.3 Weak Banach-Saks property. A Banach space is said to have the weak Banach-Saks property if every weakly null sequence has a subsequence such that its Cesaro mean is norm convergent to zero. Let s study its complexity. Theorem 17. WBS is Π 1 1-hard. In particular, WBS is non Borel. Proof. First we notice that we cannot use the same function we constructed in theorem 12, this because, as l 1 has the Schur property, l 1 is clearly in WBS. We construct a similar function, for each θ Tr and x c 00 (θ) define {( n x θ = sup s Ii x(s)e s 2 C( ) ) 1 } 2 I 1,..., I n incomparable segments of θ, where (e n ) n N is a basis of C( ) and. C( ) is its norm. In the same fashion, we define ϕ(θ) as the completion of c 00 (θ) under the norm. θ. If θ IF we clearly have C( ) ϕ(θ). As C( ) is not in WBS (see [F]) we have ϕ(θ) / WBS. We now show that if θ WF, then ϕ(θ) WBS. We proceed by transfinite induction on the order of θ WF. Say o(θ) = 1, then ϕ(θ) is 1-dimensional and we are clearly done. Assume ϕ(θ) WBS, for all θ WF with o(θ) < α, for some α < ω 1. Pick θ WF with o(θ) = α, let s show that ϕ(θ) WBS. 35

41 3.3. WEAK BANACH-SAKS PROPERTY. Using the same notation as in the proof of theorem 12, we have ϕ(θ) ( ) = R ϕ(θ(λ)). l 2 λ Λ By our inductive hypothesis, ϕ(θ(λ)) WBS, for all λ Λ. Therefore, by a corollary of J. R. Partington (see [P], pag. 373), ( λ Λ ϕ(θ(λ))) l 2 is also in WBS, and we are done. Remark: It is worth noticing that the same ϕ constructed above could be used to proof theorem 12, and theorem 16. With that being said, let s try to obtain more information about the Borel complexity of WBS. For this we use the following lemma. Lemma 18. Let (x n ) n N be a bounded sequence in a Banach space X. (x n ) n N is weakly null if, and only if, every subsequence of (x n ) n N has a convex block subsequence converging to zero in norm. In particular, if (x n ) n N is a weakly null sequence in a Banach space X, and if X embeds into another Banach space Y, then (x n ) n N is weaklly null in Y. Proof. Say every subsequence of (x n ) n N has a convex block subsequence converging to zero in norm. First we show that (x n ) n N has a weakly null subsequence. As (x n ) n N is bounded, Rosenthal s l 1 -theorem (see [R2]) says that we can find a subsequence that is either weak-cauchy or equivalent to the usual l 1 -basis. As l 1 s usual basis has no subsequence with a convex block sequence converging to zero in norm, we conclude that (x n ) n N must have a weak-cauchy subsequence. By hypothesis, this sequence must have a convex block subsequence converging to zero in norm, say (y k = l k+1 i=l k +1 a ix ni ) k N, for some subsequence (n k ) of natural numbers. 36