On the Borel Complexity of Some Classes of Banach Spaces

Kent State University

On the Borel Complexity of some classes of Banach spaces

Bruno de Mendon¸caBraga

Kent, Ohio

July 2013 To anyone that, somehow, enjoy it.

i 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 depart- ment 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 to-

ii 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 “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 Contents

1 Background. 7

1.1 Banach spaces...... 8

1.1.1 Basis and Sequences...... 8

1.1.2 Operators...... 9

1.1.3 Some Classes of Banach Spaces...... 10

1.1.4 Local Theory...... 12

1.2 Descriptive set theory...... 13

1.2.1 Standard Borel spaces...... 13

1.2.2 Trees...... 16

2 Complementability of some ideals in L(X). 19

2.1 Unconditionally Converging Operators...... 19

2.2 Weakly Compact Operators...... 23

3 Geometry of Banach spaces. 28

3.1 Banach-Saks Property...... 28

3.2 Alternating Banach-Saks Property...... 31

3.3 Weak Banach-Saks property...... 35

4 Complementability of some ideals in L(X), Part II. 40

1 CONTENTS

4.1 Banach-Saks Operators...... 40

5 Geometry of Banach spaces, Part II. 43 5.1 Schur Property...... 43 5.2 Dunford-Pettis Property...... 45 5.3 B-convex Banach Spaces...... 46 5.4 Daugavet Property...... 47 5.5 Complete Continuous Property...... 49 5.6 Radon-Nikodym property...... 53 5.7 Analytic Radon-Nikodym property...... 58 5.8 The Equivalence Relation of Isomorphisms...... 59

6 Local Structure of Banach Spaces. 61 6.1 Finite Representability...... 61 6.2 Super Reﬂexibility...... 63 6.3 Local Unconditional Structure...... 65

7 On the Borel complexity of CP . 68

7.1 Pure classes not cointaining some `p...... 69 7.2 Spaces containing a minimal subspace...... 73

8 Non-Universality Results. 79 8.1 Non-Universality Results...... 79

9 Hausdorﬀ Distance and Openings Between Banach Spaces. 83 9.1 The Hausdorﬀ Distance on F(X)...... 85 9.2 Openings Between Banach Spaces...... 87

2 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 Ba- nach 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. Speciﬁ- cally, 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 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 c0.

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

1 least non Borel (it is also shown that the weak Banach-Saks property is at most Σ2). 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 `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 Ba- nach 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

1 Schur property is at least Σ2. After that, we show that both the set of B-convex Banach

4 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 CP = {Y ∈ SB|∃Z ∈ P,Z,→ Y }, for some specific classes of Banach spaces P ⊂ SB. Specifically, we show that CP 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 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 SB2 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 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 1.1. BANACH SPACES.

1.1 Banach spaces.

1.1.1 Basis and Sequences.

Given a Banach space X we say that a sequence (xn)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 combi-

N nation of (xn)n∈N, i.e., for all x ∈ X there exist an unique (an)n∈N ∈ R such that x = P a x . If X has a Schauder basis (x ) we can deﬁne, for all n ∈ , natural n∈N n n n n∈N N projections P (P a x ) = Pn a x . It can be shown that the norm of those projec- n i∈N i i i=1 i i tions are uniformly bounded ([AK]). If the supremum of these norms is K we say the

Schauder basis (xn)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 kId − Pnk = 1, for all n ∈ N, where Id is the identity on X.

A sequence (xn)n∈N is called a basic sequence if it is a Schauder basis for its closed span.

Equivalently (see [AK]), (xn)n∈N is a basic sequence if its elements are not zero and there exists K > 0 such that

k n X X aixi ≤ K aixi , i=1 i=1 for all k, n ∈ N (k ≤ n), and for all a1, ..., an ∈ 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 (xn)n∈N is unconditional if its elements are not zero and there exists K > 0 such

8 CHAPTER 1. BACKGROUND. that

n n X X aixi ≤ K bixi , i=1 i=1 for all n ∈ N, and all a1, ..., an, b1, ..., bn ∈ R such that |ai| ≤ |bi|, for all i ∈ {1, ..., n}. The inﬁmum of this constants is called the unconditional constant of the sequence. A sequence (x ) is called unconditionally converging if the limit P x exists for all n n∈N n∈N π(n) permutation (reordering) π : N → N. The sequence (xn)n∈N is called weakly unconditionally Cauchy (or sometimes weakly unconditionally convergent) if P |f(x )| exists n∈N π(n) for all permutation π and all linear functional f : X → R.

Given two sequences (xn)n∈N and (yn)n∈N we say that they are equivalent if there exists C ≥ 1 such that

k k k 1 X X X aixi ≤ aiyi ≤ C aixi , C i=1 i=1 i=1 for all k ∈ N, and all a1, ..., ak ∈ R.

1.1.2 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 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.

1.1.3 Some Classes of Banach Spaces.

A Banach space X is said to have the Banach-Saks property if every bounded sequence

−1 Pn (xn)n∈N in X has a subsequence (xnk )k∈N such that its Cesaro mean n k=1 xnk is norm convergent. A Banach space X is said to have the alternating Banach-Saks property if every bounded sequence (xn)n∈N in X has a subsequence (xnk )k∈N such that its alternating- −1 Pn k signs Cesaro mean n k=1(−1) xnk 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 kT kkT −1k, where T varies among all the isomorphisms between X and Y . A Banach space X is called B-convex if

10 CHAPTER 1. BACKGROUND.

n lim inf{D(Xn, `1 )|Xn is a n-dimensional subspace of X} = ∞, n→∞

n n n Pn where `1 is the n-dimensional space R endowed with the `1-norm k(xk)k=1k1 = k=1 |xk|. 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 L1[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 kId + T k = 1 + kT k.

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 `1 (see [AK]).

A Banach space X is said to have the Radon-Nikodym property if for any totally ﬁnite

11 1.1. BANACH SPACES. positive measure space (Y, Σ, µ) and any X-valued µ-continuous measure m on Σ, with

1 R |m|(X) < ∞, there exists f ∈ LX (Y, Σ, µ) such that m(E) = E fdµ, for all E ∈ Σ. 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.

1.1.4 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 kukku−1k < 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 kx−yk ≥ ε implies kx+yk/2 ≤ 1−δ, for all x, y ∈ SX .

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 )kukkwk ≤ 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 CHAPTER 1. BACKGROUND.

1.2 Descriptive set theory.

1.2.1 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

1 1 1 1 Σ1(X) (resp. Π1(X)). Hence, the terminology Σ1-set (resp. Π1-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 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.

1 1 Let X be a standard Borel space. We call a subset A ⊂ X Σ1-hard (resp. Π1-hard) if for all standard Borel space Y and all B ⊂ Y analytic (resp. coanalytic) there exists a Borel

1 1 reduction from B to A. Therefore, to say that a set A ⊂ X is Σ1-hard (resp. Π1-hard)

1 1 means that A is at least as complex as Σ1-sets (resp. Π1-sets) in the projective hierarchy. With this terminology we have that A ⊂ X is complete analytic (resp. complete coana-

1 1 lytic) if, and only if, A is Σ1-hard (resp. Π1-hard) and analytic (resp. coanalytic).

1 As there exist analytic non Borel (resp. coanalytic non Borel) sets we have that Σ1-hard

1 (resp. Π1-hard) sets are non Borel. Also, if X is a standard Borel space, A ⊂ X, and there

1 1 exists a Borel reduction from a Σ1-hard (resp. Π1-hard) subset B of a standard Borel

1 1 space Y to A, then A is Σ1-hard (resp. Π1-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

1 1 coanalytic sets) are also called Σ1-complete sets (resp. Π1-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 6= ∅}, 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 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 (Sn)n∈N :

F(X) → X such that {Sn(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 BX and SX , 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 {Sn}n∈N the sequence of Borel functions Sn : SB → C(∆) given by lemma 1 (more precisely, the restriction of those functions to SB).

Hence, for all X ∈ SB, {Sn(X)}n∈N is dense in X. Also, letting

  S (X), if S (X) 6= 0 0  n n Sn(X) = ,   Smin{m∈N|Sm(X)6=0}(X), if Sn(X) = 0

˜ 0 0 we obtain that Sn(X) = S (X)/kS (X)k is dense in SX for each X ∈ SB. Clearly, n n n∈N ˜ each Sn : SB → SC(∆) is also Borel (see [S]).

15 1.2. DESCRIPTIVE SET THEORY.

1.2.2 Trees.

Denote by N

< N N topology. If we think about Tr as a subset of 2 it is easy to see that Tr is a Gδ set in

< < < 2N N ([Ke]). Thus, it is also Borel in 2N N . As Tr is Borel in the Polish space 2N N we know

N that Tr is a standard Borel space ([Ke]). A β ∈ 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 = {T ∈ Tr|s∈ / T }. < n s∈N N |s|>n

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 CHAPTER 1. BACKGROUND.

T 0 = {s ∈ T |∃t ∈ T, s < t}.

By transﬁnite induction we deﬁne (Tξ)ξ∈ON, where ON denotes the ordinal numbers, as follows

T α = (T β)0, if α = β + 1, for some β ∈ ON, \ T α = T β, if α is a limit ordinal. β<α

We now deﬁne the order index on Tr. If there exists an ordinal number α < ω1, where ω1 denotes the smallest uncountable ordinal, such that T α = ∅ we say the order index of T

α is o(T ) = min{α < ω1| T = ∅}. If there is no such countable ordinal we set o(T ) = ω1. 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 ) < ω1.

Proposition 3. Let T ∈ WF with o(T ) > 1, then o(T (k)) < o(T ), for all k ∈ N.

θ Let θ ∈ Tr. We deﬁne c00(θ) = {(as)s∈θ ∈ R |as 6= 0 for ﬁnitely many s ∈ θ}. We denote by (es)s∈θ the standard unit basis of c00(θ).

Now we introduce a very useful way to construct a Borel ϕ : Tr → SB. Let (xn)n∈N be a basic sequence in a separable Banach space X. Pick θ ∈ Tr and p ∈ [1, ∞). We deﬁne a norm on c00(θ) as follows: For each x ∈ c00(θ), let

17 1.2. DESCRIPTIVE SET THEORY.

n 1 n X X p p o kxkθ = sup x(s)x|s| X | n ∈ N,I1, ..., In incomparable segments of θ , i=1 s∈Ii

where k.kX is the X norm. Deﬁne ϕ(θ) to be the completion of c00(θ) under the norm

Proposition 4. ϕ : Tr → SB is a Borel function. The same is true if we deﬁne k.kθ as

n X o kxkθ = sup x(s)x|s| X | I segment of θ . s∈I

For now on, every time we deﬁne a function on Tr as the completion of c00(θ) under one of those norms we will be considering an isometry of ϕ(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 Chapter 2

Complementability of some ideals in L(X).

“” (M. Gromov’s answer to my email)

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 ,−→ X if Y is isomorphic to a complemented 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.

Proof. In order to show this we only need to use that U(X) is complemented in L(X) if,

and only if, c0 does not embed in X (see [BBG], pag. 452). Therefore, U = NCc0 (where

19 2.1. UNCONDITIONALLY CONVERGING OPERATORS.

NCX = {Y ∈ SB|X 6,→ 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.

−1 Let’s deﬁne a Borel function ϕ : Tr → SB such that ϕ (U) = WF. Let (en)n∈N be the standart basis of c0. For each θ ∈ Tr and x ∈ c00(θ) we deﬁne

n 1 n X X 2 2 o kxk = sup x(s)e|s| | n ∈ ,I1, ..., In incomparable segments of θ , θ c0 N i=1 s∈Ii

where k.kc0 is the norm of c0. Now we deﬁne ϕ(θ) as the completion of c00(θ) under the norm k.kθ. Let’s now show that ϕ works as we want.

∼ If θ ∈ IF it’s clear that c0 ,→ ϕ(θ). Indeed, let β be a branch of θ, then c0 = ϕ(β) ,→ ϕ(θ), where by ϕ(β) we mean ϕ applied to the tree {s ∈ N

The only thing left to show is that if θ ∈ WF, then c0 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 c0 6,→ ϕ(θ) for all θ ∈ WF with o(θ) < α, for some α < ω1. Fix θ ∈ WF with o(θ) = α.

Let Λ = {λ ∈ N|(λ) ∈ θ} and enumerate Λ, say Λ = {λi|i ∈ N}. For each λ ∈ Λ, let

θλ = {s ∈ θ|(λ) ≤ s}. As θ ∈ WF Proposition 3 gives us

n [ o θ(λj) < o(θ) = α, ∀n ∈ N. j=1

20 CHAPTER 2. COMPLEMENTABILITY OF SOME IDEALS IN L(X).

Sn Our induction hypothesis implies that c0 6,→ ϕ( j=1 θ(λj)), for all n ∈ N. Therefore, Sn Sn as ϕ( j=1 θλj ) is a direct sum of ϕ( j=1 θ(λj)) with a ﬁnite dimensional space, c0 6,→ Sn ϕ( j=1 θλj ), for all n ∈ N. Consider now the projections

n [ Pλn : ϕ(θ) → ϕ θλj j=1

(as)s∈θ → (as) Sn . s∈ j=1 θλj

∼ Say E is a subspace of ϕ(θ). Assume E = c0 and let’s ﬁnd a contradiction.

Sj Case 1: ∃j ∈ N such that Pλj : E → ϕ( i=1 θλi ) is not strictly singular.

˜ Then there exists an inﬁnite dimensional subspace E ⊂ E such that Pλj |E˜ is an isomor- Sj phism. As E is minimal, this implies c0 ,→ ϕ( i=1 θλi ), absurd.

Sj Case 2: Pλj : E → ϕ( i=1 θλi ) is strictly singular, for all j ∈ N.

Claim: ∃(xn)n∈N a normalized sequence in E such that Pλj (xn) → 0, as n → ∞, ∀j ∈ N.

Indeed, by a well-known consequence of the deﬁnition of strictly singular operators, for

j j all j ∈ N, there exists a normalized sequence (xn)n∈N such that kPλj (xn)k < 1/n, for all

j n n ∈ N. Let (xn)n∈N be the diagonal sequence of the sequences (xn)n∈N, i.e., xn = xn, for all n ∈ N. As, i ≤ j implies kPλi (x)k ≤ kPλj (x)k, for all x ∈ E,(xn)n∈N has the required property.

Say (εi)i∈N is a sequence of positive numbers converging to zero. Using the claim above

21 2.1. UNCONDITIONALLY CONVERGING OPERATORS.

and the fact that Pλj (x) → x, as n → N, for all x ∈ ϕ(θ), we can pick increasing sequences of natural numbers (nk)k∈N and (lk)k∈N such that

i) kPλ (xn ) − xn kθ < εk, for all k ∈ , and lk k k N ii) kPλ (xn )kθ < εk, for all k ∈ . lk k+1 N

For all k ∈ , let yk = Pλ (xn ) − Pλ (xn ). Choosing εk small enough we can assume N lk k lk−1 k 2 1 kykkθ ∈ ( 2 , 2). It’s easy to see that (yk)k∈N is equivalent to (˜ek)k∈N, where (˜ek)k∈N is the standard `2-basis. Indeed, pick a1, ..., aN ∈ R, then

N N N N N 1 X X X X 2 X |a |2 ≤ ka y k2 = k a y k2 = ka y k ≤ 2 |a |2, 2 i i i θ i i θ i i θ i i=1 i=1 i=1 i=1 i=1

where the equalities above only hold because the supports of (yk)k∈N are completely incomparable. Therefore, by choosing (εk)k∈N converging to zero fast enough, the principle of small perturbations (see [AK]) gives us that (xnk )k∈N is equivalent to (yk)k∈N ∼ (˜ek)k∈N, ∼ absurd. Indeed, this would imply `2 ,→ E = c0.

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 `p, p ∈ [1, ∞). Say Y ⊂ X is a subspace with basis (en)n∈N. For each θ ∈ Tr, and each x ∈ c00(θ), deﬁne

n 1 n X X p p o kxkθ = sup x(s)e|s| Y | n ∈ N,I1, ..., In incomparable segments of θ , i=1 s∈Ii

22 CHAPTER 2. COMPLEMENTABILITY OF SOME IDEALS IN L(X).

where k.kY stands for the norm of Y . If we deﬁne a function ϕ : Tr → SB by letting ϕ(θ) be the completion of c00(θ) under the norm k.kθ we have that X,→ ϕ(θ) if, and only if,

θ ∈ IF. An analogous result holds if k.kθ is deﬁned as

n X o kxkθ = sup x(s)e|s| Y | I segment of θ , s∈I and we assume c0 6,→ X. If we forget the condition Y ⊂ X we still have that X 6,→ 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

1 compact operators on X. Then W is Π1-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 c0 ,→ X, then W(X) is not complemented in L(X). Let ϕ : Tr → SB be as in the proof of theorem 5 for X = c0 and (en)n∈N the standard c0-basis. Let’s ob-

−1 serve that ϕ (W) = WF, and we will be done. If θ ∈ IF we saw that c0 ,→ 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 2.2. WEAKLY COMPACT OPERATORS.

Radon-Nikodym property and the complete continuous property, we believe it is worth showing its complete proof here.

Let (si)i∈N be a compatible enumeration of θ, i.e., si ≤ sj implies i ≤ j. Notice that

(esi )i∈N is a basis for ϕ(θ), where each esi denotes the vector in ϕ(θ) with coordinate si being 1 and 0 elsewhere. This base is clearly bimonotone, which implies the biorthogonal functionals (e∗ ) are also a bimonotone basis for their closed span (see [AK]). si i∈N

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 (esi )si∈θ of ϕ(θ) is both boundedly complete and shrinking. We start by showing that (esi )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

(esi )si∈θ is boundedly complete for all θ ∈ WF of order smaller than α < ω1. Fix θ ∈ WF with o(θ) = α. Consider (as)s∈θ ∈ c00(θ) with a∅ = 0, then

Now say (asi )i∈N is a sequence of real numbers with the property that as1 = a∅ = 0 and PN supN k i=1 asi esi kθ < ∞. 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 CHAPTER 2. COMPLEMENTABILITY OF SOME IDEALS IN L(X). implies that

X asi esi exists for all k ∈ N. i∈N si∈θk

The equation above also implies that, for all ε > 0, there exists n0 ∈ N such that, for n > n0,

∞ ∞ 2 X X 2 ε asi esi < , θ 2 k=n i=1 si∈θk

There exists n1 ∈ N such that m ≥ n ≥ n1 implies

m 2 X 2 ε asi esi < , for all k ≤ n0. θ 2n i=n 0 si∈θk

Putting all of this together we have that, for all m ≥ n ≥ n1,

m m X 2 X X 2 asi esi θ = asi esi θ i=n k∈N i=n si∈θk n0 m ∞ ∞ X X 2 X X 2 ≤ asi esi θ + asi esi θ k=1 i=n k=n0+1 i=1 si∈θk si∈θk ε2 ε2 ≤ + = ε2. 2 2

Notice that here we used the fact that our basis is bimonotone. We just showed that, Pm for all ε > 0, there exists n1 ∈ N such that m ≥ n ≥ n1 implies k i=n asi esi kθ < ε, thus (esi )si∈θ is boundedly complete. We now show that (esi )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 (esi )si∈θ is shrinking for all θ ∈ WF of order smaller than α < ω1. Fix θ ∈ WF

25 2.2. WEAKLY COMPACT OPERATORS.

with o(θ) = α. Consider (as)s∈θ ∈ c00(θ) with a∅ = 0.

X X ∗ = sup ck ases θ (cn)∈B` 2 k∈N s∈θk 1 X X ∗ 2 2 = k aseskθ

k s∈θk

It’s well known that (e ) is shrinking if, and only if, (e∗ ) is boundedly complete si si∈θ si si∈θ (see [AK]). We show (e∗ ) is boundedly complete. Consider (a ) a sequence of real si si∈θ si i∈N numbers such that a = a = 0 and supk PN a e∗ k < ∞. The equation above implies s1 ∅ i=1 si si θ N

X X ∗ 2 ases θ < ∞. k∈N i∈N si∈θk

So, for all ε > 0, there exists n0 ∈ N such that

∞ 2 X X ∗ 2 ε asi e < , for all n ≥ n0. si θ 2 k=n i∈Nsi∈θk

By our inductive hypothesis, there exists n1 ∈ N such that m ≥ n ≥ n1 implies

m 2 X ∗ 2 ε asi e < , for all k ≤ n0. si θ 2n i=n 0 si∈θk

Therefore, if m ≥ n ≥ n1, we have

26 CHAPTER 2. COMPLEMENTABILITY OF SOME IDEALS IN L(X).

m m X 2 X X 2 a e∗ = a e∗ si si θ s s θ i=n k i=n s∈θk n0 m ∞ ∞ X X ∗ 2 X X ∗ 2 ≤ ases θ + ases θ k=1 i=n k=n0+1 i=1 s∈θk s∈θk ε2 ε2 ≤ + = ε2, 2 2

∗ here we use the fact that (es)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 (en)n∈N, and norm k.kX . Let p ∈ (1, ∞).

For each θ ∈ Tr and x ∈ c00(θ) we deﬁne

n 1 n X X p p o kxkθ = sup x(s)e|s| X | n ∈ N,I1, ..., In incomparable segments of θ , i=1 s∈Ii

and let ϕ(θ) be the completion of c00(θ) under the norm k.kθ. 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 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

−1 Pn (xn)n∈N in X has a subsequence (xnk )k∈N such that its Cesaro mean n k=1 xnk 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 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 (xn)n∈N in BX such that, for all subsequences (xnk )k∈N, ∀m ∈ N, and ∀` ∈ {1, ..., m}, the following holds

` m 1 X X ( xn − xn ) ≥ ε. m k k k=1 k=`+1 Theorem 11. BS is coanalytic in SB.

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.

0 Let’s associate a tree TBS(X, ε) on X to the pair (X, ε) in the following manner, for all

n ﬁnite sequences (xk)k=1 in BX ,

n 0 (xk)k=1 ∈ TBS(X, ε) ⇔∀n1 < ... < nm ≤ n, ∀` ∈ {1, ..., m}, ` m 1 X X ( xn − xn ) ≥ ε m k k k=1 k=`+1

0 It is clear, by theorem 10, that X does not have the BS property if, and only if, TBS(X, ε)

(Sn)n∈N be the sequence of Borel maps given by lemma 1 for X = C(∆). Given X ∈ SB

0 we deﬁne a tree TBS(X, ε) on N as, for all ﬁnite sequences (nk)k=1 ∈ N ,

n n 0 (nk)k=1 ∈ TBS(X, ε) ⇔ (Snk (BX ))k=1 ∈ TBS(X, ε).

As {Sn(BX )}n∈N is dense in BX , for all X ∈ SB, standard approximation arguments imply that X does not have the BS property if, and only if, TBS(X, ε) is ill-founded for some ε.

29 3.1. BANACH-SAKS PROPERTY.

a n

1 (m)a(n )n ∈ T (X) ⇔ (n )n ∈ T X, . k k=1 BS k k=1 BS m

It is clear that X/∈ BS if, and only if, TBS(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

` m \ n 1 X X o {X ∈ SB|s ∈ T (X)} = X ∈ SB| ( Sn (BX ) − Sn (BX )) ≥ ε , m k k n1<...

1 P` and we are done. Indeed, as Sn is Borel, ∀n ∈ N, the set {X ∈ SB|k m ( k=1 Snk (BX ) − Pm k=`+1 Snk (BX ))k ≥ 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.

1 Theorem 12. BS is Π1-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

1 WF is Π1-hard this will suﬃce. For each θ ∈ Tr and x ∈ c00(θ) deﬁne

n 1 n X X 2 2 o kxkθ = sup x(s)e|s| | n ∈ ,I1, ..., In incomparable segments of θ , `1 N i=1 s∈Ii

where (en)n∈N is the standard `1-basis and k.k`1 is its norm. We deﬁne ϕ(θ) as the com-

30 CHAPTER 3. GEOMETRY OF BANACH SPACES.

pletion of c00(θ) under the norm k.kθ. If θ ∈ IF we clearly have `1 ,→ ϕ(θ). Indeed, if β ∼ is a branch of θ we have ϕ(β) = `1. As `1 ,→ ϕ(θ) and `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 < ω1. 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

oθ(λ) < o(θ) = α, ∀λ ∈ Λ.

Our induction hypothesis implies that ϕ(θ(λ)) ∈ BS, for all λ ∈ Λ. Now, notice that

∼ M ϕ(θ) = R ⊕ ϕ(θ(λ)) , `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 `2-sum of spaces in BS is also L in BS. Hence, ( λ∈Λ ϕ(θ(λ)))`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 (xn)n∈N in X has a subsequence (xnk )k∈N such that its alternating-signs Cesaro

31 3.2. ALTERNATING BANACH-SAKS PROPERTY.

−1 Pn k mean n k=1(−1) xnk 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 (xn)n∈N in BX such that for all l ∈ N, if ` ≤ n(1) < ... < n(2`), where n(i) ∈ N, ∀i ∈ {1, ..., 2`}, then

2` 2` X X cixn(i) ≥ ε |ci|, i=1 i=1 for all c1, ..., c2` ∈ R.

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

0 n TABS(X, ε) on X to the pair (X, ε) in the following manner, for all ﬁnite sequences (xk)k=1 in BX ,

32 CHAPTER 3. GEOMETRY OF BANACH SPACES.

n 0 ` (xk)k=1 ∈ TABS(X, ε) ⇔∀` ∈ N, ∀n(1) < ... < n(2 )

` ` (n(k) ∈ N, ∀k ∈ {1, ..., 2 }, & n(1) ≥ l & n(2 ) ≤ n),

∀c1, ..., c2` ∈ R, 2` 2` X X ckxn(k) ≥ ε |ck| k=1 k=1

0 It is clear, by theorem 14, that X does not have the ABS property if, and only if, TABS(X, ε)

0 is ill-founded for some ε > 0. Now we use lemma 1 to make TABS(X, ε) into a tree on

0 we deﬁne a tree TABS(X, ε) on N as, for all ﬁnite sequences

n n 0 (nk)k=1 ∈ TABS(X, ε) ⇔ (Snk (BX ))k=1 ∈ TABS(X, ε).

By standard approximation arguments we see that X does not have the ABS property if, and only if, TABS(X, ε) is ill-founded for some ε. We now glue the trees (TABS(X, ε))ε

a n

1 (m)a(n )n ∈ T (X) ⇔ (n )n ∈ T (X, ). k k=1 ABS k k=1 ABS m

It is clear that X/∈ ABS if, and only if, TABS(X) is ill-founded. Therefore, we have just deﬁned the ϕ : SB → Tr we wanted. We only need to observe ϕ is Borel and

33 3.2. ALTERNATING BANACH-SAKS PROPERTY.

\ \ {X ∈ SB|s ∈ T (X)} = + `∈N c1,...,c l ∈Q ` 2 n(1)<...

P2` and we are done. Indeed, as Sn is Borel, ∀n ∈ N, the set {X ∈ SB|k k=1 ckSn(k)(BX )k ≥ 1 P2` m k=1 |ck|} is Borel.

Now we show that coanalyticity is the most we can get of ABS in relation to its Borel complexity.

1 Theorem 16. ABS is Π1-hard. Moreover, ABS is complete coanalytic.

Proof. Let ϕ : Tr → SB be deﬁne exactly as in the proof of theorem 12. We will show

−1 that ϕ (ABS) = WF. If θ ∈ IF, we have `1 ,→ ϕ(θ). As `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 < ω1. Pick θ ∈ WF with o(θ) = α.

Using the same notation as in the proof of theorem 12, we have

∼ M ϕ(θ) = R ⊕ ϕ(θ(λ)) . `2 λ∈Λ

By lemma 6, `1 6,→ ϕ(θ). B. Beauzamy showed in [Be] (pag. 386) that a Banach space not

34 CHAPTER 3. GEOMETRY OF BANACH SPACES.

containing `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), L ϕ(θ(λ)) is also in WBS. Thus, we conclude that ϕ(θ) ∈ WBS, λ∈Λ `2 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.

1 Theorem 17. WBS is Π1-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 `1 has the Schur property, `1 is clearly in WBS. We construct a similar function, for each θ ∈ Tr and x ∈ c00(θ) deﬁne

n 1 n X X 2 2 o kxkθ = sup x(s)e|s| C(∆) | I1, ..., In incomparable segments of θ , i=1 s∈Ii where (en)n∈N is a basis of C(∆) and k.kC(∆) is its norm. In the same fashion, we de-

ﬁne ϕ(θ) as the completion of c00(θ) under the norm k.kθ. 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 < ω1. Pick θ ∈ WF with o(θ) = α, let’s show that ϕ(θ) ∈ WBS.

35 3.3. WEAK BANACH-SAKS PROPERTY.

Using the same notation as in the proof of theorem 12, we have

∼ M ϕ(θ) = R ⊕ ϕ(θ(λ)) . `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), L ϕ(θ(λ)) is also in WBS, and we are λ∈Λ `2 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 (xn)n∈N be a bounded sequence in a Banach space X. (xn)n∈N is weakly null if, and only if, every subsequence of (xn)n∈N has a convex block subsequence converging to zero in norm. In particular, if (xn)n∈N is a weakly null sequence in a Banach space X, and if X embeds into another Banach space Y , then (xn)n∈N is weaklly null in Y .

Proof. Say every subsequence of (xn)n∈N has a convex block subsequence converging to zero in norm. First we show that (xn)n∈N has a weakly null subsequence. As (xn)n∈N is bounded, Rosenthal’s `1-theorem (see [R2]) says that we can ﬁnd a subsequence that is either weak-Cauchy or equivalent to the usual `1-basis. As `1’s usual basis has no subsequence with a convex block sequence converging to zero in norm, we conclude that

(xn)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 = Plk+1 a x ) , for k i=lk+1 i ni k∈N some subsequence (nk) of natural numbers.

36 CHAPTER 3. GEOMETRY OF BANACH SPACES.

∗ Say (xnk )k∈N is not weakly null. Then pick f ∈ X such that f(xnk ) 6→ 0. As (xnk )k∈N is weak-Cauchy, there exists δ 6= 0 such that f(xnk ) → δ. Hence, f(yk) → δ, absurd, because (yk)k∈N is norm convergent to zero.

∗ Now assume (xn)n∈N is not weakly null. Then we can pick f ∈ X , a subsequence (nk)k∈N, and δ 6= 0, such that f(xnk ) → δ. As the subsequence (xnk )k∈N has the same property as

(xn)n∈ , we can pick a weakly null subsequence, say (xn )l∈ . Hence f(xn ) → 0, absurd. N kl N kl

For the converse we only need to apply Mazur’s theorem.

Let’s denote the closed ball of radius r of a Banach space X by BX (r). For every X ∈ SB, we let

n N + Er(X) = (xk)k∈N, (nk)k∈N ∈BX (r) × [N]| ∀ε ∈ Q , ∀n ∈ N, n+l n+l + X X o ∃an, ..., an+l ∈ Q ai = 1 , aixni < ε , i=n i=n where [N] stands for the subset of NN consisting of the increasing sequences of natural

N numbers. As [N] is easily seen to be Borel, we have that Er(X) is Borel in ∪r∈NBX (r) × [N], for all r ∈ N. Deﬁne F (X) by

c c F (X) = π ∪r∈N Er(X) , where π denotes the projection into the ﬁrst coordinate. Notice that F (X) is coanalytic and that F (X) consists of all the bounded sequences in XN with the property that all of its subsequences have a convex block subsequence converging to zero in norm. By lemma 18, F (X) is the set of all weakly null sequences of X.

37 3.3. WEAK BANACH-SAKS PROPERTY.

Theorem 19. The set of weakly null subsequences F (X) ⊂ XN of X is coanalytic, for all X ∈ SB.

Say X = C(∆), and F = F (C(∆)). Let A = {(X, (xn)n∈N) ∈ SB × F |∀n ∈ N, xn ∈ X}, and

n ∗ G = π X, (xn)n∈N ∈ A| ∃ε ∈ Q , ∀n ∈ N, ∀n1 <... < nm ≤ n, ∀` ∈ {1, ..., m}, ` m 1 X X o ( xn − xn ) ≥ ε , m k k k=1 k=`+1 where π denotes the projection into SB. B. Beauzamy’s paper implies that WBS = Gc. We had just shown that WBS is the complement of a Borel image of a coanalytic set. If

1 a subset of a standard Borel space X has this property we say that it belongs to Σ2(X), see [Ke] or [S] for more details.

1 Theorem 20. WBS ∈ Σ2(SB).

Problem 21. Is WBS coanalytic? If yes, we had shown that WBS is complete coanalytic.

Remark: We had just seen that the set of weakly null subsequences F (X) ⊂ XN of a separable Banach space X is coanalytic in XN. It is easy to see that F (X) is actually

∗ ∗ Borel if X is separable. Indeed, if {fn}n∈N is dense in X , we have

\ \ [ \ N F (X) = (xn)n∈N ∈ X | |fl(xm)| < ε . n∈N ε∈Q+ n∈N m>n

On the other hand, as `1 is a Schur space, F (`1) consists of the set of norm null sequences

∗ in `1, and it is easily seen to be Borel. Which means, X does not need to be separable in order to F (X) to be Borel.

38 CHAPTER 3. GEOMETRY OF BANACH SPACES.

Problem 22. Is F (X) Borel, for all X ∈ SB? If not, under what conditions is F (X) (coanalytic) non Borel? Notice that, if F (C(∆)) is Borel, then we had shown that WBS is coanalytic.

39 Chapter 4

Complementability of some ideals in L(X), Part II.

“If they grow, why to stop them?” (Grigori Perelman about not cutting his ﬁngernails)

4.1 Banach-Saks Operators.

In the same spirit as Chapter 2, we now take a look at operator ideals of L(X). Let X be a Banach space, we say T ∈ L(X) is a Banach-Saks operator if for each bounded sequence

−1 Pn (xn)n∈N there is a subsequence (xnk )k∈N such that its Cesaro mean n k=1 T (xnk ) is norm convergent. We denote the space of Banach-Saks operators from X to itself by BS(X). We can now try to compute the Borel complexity of the set of Banach spaces such that BS(X) is a complemented subspace of L(X).

⊥ 1 Theorem 23. The set BS = {X ∈ SB|BS(X) ,−→ L(X)} is Π1-hard. In particular, BS is non Borel.

40 CHAPTER 4. COMPLEMENTABILITY OF SOME IDEALS IN L(X), PART II.

Proof. Let ϕ : Tr → SB be deﬁned as in the proof of theorem 17. The remark after this theorem says that if θ ∈ WF, then ϕ(θ) ∈ BS. Hence, BS(ϕ(θ)) = L(ϕ(θ)), and we have ϕ(θ) ∈ BS, for all θ ∈ WF. Let’s show that the same cannot be true if θ ∈ IF.

∼ Say θ ∈ IF. Then ϕ(θ) = C(∆) ⊕ Y , for some Y ∈ SB. Let P1 : C(∆) ⊕ Y → C(∆) be the standard projection. Suppose there exists a bounded projection P : L(C(∆) ⊕ Y ) →

BS(C(∆) ⊕ Y ). Deﬁne P0 : L(C(∆)) → BS(C(∆)) as, for all T ∈ L(C(∆)),

˜ P0(T ) = P1(P (T ))|C(∆), where T˜ : C(∆) ⊕ Y → C(∆) ⊕ Y is the natural extension, i.e., T˜(x, y) = (T (x), 0), for all (x, y) ∈ C(∆) ⊕ Y . Notice that P0(T ) ∈ BS(C(∆)), so P0 is well deﬁned. Also, if T ∈ BS(C(∆)), then T˜ ∈ BS(C(∆) ⊕ Y ), which implies P (T˜) = T˜ (because P is a projection). Therefore, P0 is a projection from L(C(∆)) onto BS(C(∆)). Let’s observe this gives us a contradiction.

It’s known that T : C(∆) → C(∆) has the Banach-Saks property if, and only if, T is weakly compact (see [DiSe], pag. 112). Hence, BS(C(∆)) = W(C(∆)) and, as c0 ,→ C(∆), we have that BS(C(∆)) is not complemented in L(C(∆)) ([BBG]). Absurd.

Problem 24. Is BS coanalytic? If yes, our previous proof would show that BS is complete coanalytic.

We had studied three classes of ideals of L(X)(U(X), W(X), and BS(X)) and whether those ideals are complemented in L(X) or not. Another natural question would be to study the Borel complexity of pairs (X,Y ) ∈ SB2 such that their respective ideals (U(X,Y ), W(X,Y ), and BS(X,Y )) are complemented in L(X,Y ). As mentioned in the introduction, this problem had been solved for the ideal of compact operators K(X,Y )

41 4.1. BANACH-SAKS OPERATORS. by D. Puglisi in [Pu].

2 Let ϕ : Tr → SB be as deﬁned above and deﬁne ϕ0(θ) = (ϕ(θ), ϕ(θ)) ∈ SB , for all θ ∈ Tr.

−1 2 ⊥ Clearly, we have that ϕ0 ({(X,Y ) ∈ SB |BS(X,Y ) ,−→ L(X,Y )}) = WF. Conclusion:

1 2 Theorem 25. The following sets are Π1-hard (hence, non Borel) in the product SB : ⊥ ⊥ {(X,Y ) ∈ SB2|BS(X,Y ) ,−→ L(X,Y )}, {(X,Y ) ∈ SB2|U(X,Y ) ,−→ L(X,Y )}, and ⊥ {(X,Y ) ∈ SB2|W(X,Y ) ,−→ L(X,Y )}.

42 Chapter 5

Geometry of Banach spaces, Part II.

“Luke, I am your father.” (Darth Vader)

5.1 Schur Property.

We say that a Banach space X has the Schur property if every weakly convergent sequence of X is norm convergent.

1 Theorem 26. Let S = {X ∈ SB|X has the Schur property}. S is Π1-hard. In particular, S is non Borel.

Proof. In the same fashion as our previous proofs, we start by deﬁning a norm on c00(θ) as, for all x ∈ c00(θ), let

n n X X o kxkθ = sup x(s)e|s| | n ∈ ,I1, ..., In incomparable segments of θ , c0 N i=1 s∈Ii

43 5.1. SCHUR PROPERTY.

where here (en)n∈N stands for the standard c0 basis. We deﬁne a function ϕ : Tr → SB by letting ϕ(θ) be the completion of c00(θ) under the norm k.kθ. As c0 ,→ ϕ(θ) if θ ∈ IF, we have ϕ(θ) 6∈ S, for all θ ∈ IF. Mimicking the proof of theorem 12 we have that

∼ M ϕ(θ) = R ⊕ ϕ(θ(λ)) , `1 λ∈Λ where Λ = {λ ∈ N|(λ) ∈ θ}. Proceeding by transﬁnite induction and using B. Tanbay’s result about the stability of the Schur property under `1-sums (see [T], pag. 350), we conclude that ϕ(θ) ∈ S, for all θ ∈ WF.

Let’s try to obtain more information about the Borel complexity of S. For this, notice that a Banach space X does not have the Schur property if, and only if, it has a weakly null sequence (xn)n∈N in SX .

Let F = F (C(∆)) be deﬁned as in Section 3.3, i.e., F is the set of all weakly null

N N subsequences of C(∆). Let E = F ∩ SC(∆), so E is coanalytic in SC(∆), and deﬁne

G = π {(X, (xn)n∈N) ∈ SB × E| ∀n ∈ N, xn ∈ X} , where π denotes the projection into SB. We can easily see that S = Gc. We had just shown that S is the complement of a Borel image of a coanalytic set.

1 Theorem 27. S ∈ Σ2(SB).

Remark: Notice that, if F = F (C(∆)) is Borel, then we had actually shown that S is coanalytic.

Problem 28. Is S coanalytic? If yes, our previous proof would show that S is complete coanalytic.

44 CHAPTER 5. GEOMETRY OF BANACH SPACES, PART II.

5.2 Dunford-Pettis Property.

A Banach space X is said to have the Dunford-Pettis property if every weakly compact operator T : X → Y from X into another Banach space Y takes weakly compact sets into norm-compact sets. In other words, X has the Dunford-Pettis property if every weakly compact operator from X into another Banach space Y is completely continuous. We have the following (see [FLR], and [Fa]):

Theorem 29. X∗ has the Schur property if, and only if, X has the Dunford-Pettis property and X does not contain `1.

1 Theorem 30. Let DP = {X ∈ SB|X has the Dunford-Pettis property}. DP is Π1-hard. In particular, DP is non Borel.

Proof. Deﬁne ϕ : Tr → SB by letting ϕ(θ) be the completion of c00(θ) under the norm

n X o kxkθ = sup x(s)e|s| | I segment of θ , `2 s∈I ∼ where (en)n∈N is the standard `2 basis and k.k`2 is its norm. If θ ∈ IF we have ϕ(θ) = `2⊕Y , for some Banach space Y . Hence, as `2 is reﬂexive, if is clear that T (x, y) = (x, 0) is a weakly compact operator from `2 ⊕Y to itself which is not completely continuous. There- fore, ϕ(θ) 6∈ DP, for all θ ∈ IF.

On the other hand, if θ ∈ WF, then lemma 6 implies that `1 6,→ ϕ(θ). Therefore, in order to show that ϕ(θ) ∈ DP it is enough to show that ϕ(θ)∗ has the Schur property. With the same notation as in the proofs of the previous theorems, we have

∼ M ϕ(θ) = R ⊕ ϕ(θ(λ)) , c0 λ∈Λ where Λ = {λ ∈ N|(λ) ∈ θ}. Hence, we have

45 5.3. B-CONVEX BANACH SPACES.

∗ ∼ M ∗ ϕ(θ) = R ⊕ ϕ(θ(λ)) . `1 λ∈Λ Therefore, if we proceed by transﬁnite induction and use the stability of the Schur property under `1-sums (exactly as we did in the proof of theorem 26), we will be done.

Problem 31. Is DP coanalytic? If yes, our previous proof would show that DP is complete coanalytic.

An operator T : X → Y is said to be completely continuous if T maps weakly compact sets into norm-compact sets. For a given X ∈ SB, let CC(X) be the set of completely continuous operators from X to itself.

⊥ Problem 32. Let CC = {X ∈ SB|CC(X) ,−→ L(X)}. Is CC non Borel? If yes, is it coanalytic?

5.3 B-convex Banach Spaces.

In this section we show that the set of B-convex Banach spaces in SB is Borel. In order to show that, we look at Borel functions ϕm : SB → Tr closely related to the `1-Bourgain index ϕ . NC`1

Theorem 33. BC is Borel in SB.

Proof. For each X ∈ SB and ε > 0 we deﬁne a tree T 0 (X, ε) on X by NC`1

n 0 (xk) ∈ T (X, ε) ⇔∀c1, ..., cn ∈ , k=1 NC`1 R n n n 1 X X X |ck| ≤ ckxk ≤ ε |ck| ε k=1 k=1 k=1

46 CHAPTER 5. GEOMETRY OF BANACH SPACES, PART II.

As we did before, we deﬁne another tree T (X, ε) (now a tree on

n n 0 (nk) ∈ TNC (X, ε) ⇔ (Sn (X)) ∈ T (X, ε). k=1 `1 k k=1 NC`1

For each m ∈ let ϕ : SB → Tr be deﬁne as ϕ (X) = T (X, 1 ). Let’s notice that N m m NC`1 m n ϕm is Borel, ∀m ∈ N. Pick s = (nk)k=1 ∈ N. Then

1 \ X ∈ SB|s ∈ TNC X, = `1 m c1,...,cn∈R n n n n 1 X X X o X ∈ SB| |ck| ≤ ckSn (X) ≤ m |ck| . m k k=1 k=1 k=1

Hence ϕm is Borel. Now notice that X/∈ BC if, and only if, there exists m ∈ N such that,

n for all n ∈ N, there exists (xk)k=1 such that

n n n 1 X X X |ck| ≤ ckxk ≤ m |ck|, m k=1 k=1 k=1 for all c1, ..., cn ∈ R. Hence, X ∈ BC if, and only if, for all m ∈ N, ϕm(X) ∈ FTr.

−1 Therefore, BC = ∩mϕm (FTr). As FTr is Borel we are done.

We recall that the family of all separable B-convex Banach spaces coincides with the family of all separable Banach spaces with non-trivial type (see [MP]). Hence, this is an alternative proof to Bossard’s proof that the set of Banach spaces with non-trivial type is Borel ([Bo]).

5.4 Daugavet Property.

A Banach space X has the Daugavet property if every rank-1 operator T : X → X satisﬁes

47 5.4. DAUGAVET PROPERTY.

kId + T k = 1 + kT k.

+ For a Banach space X, x ∈ SX , and ε > 0, we deﬁne l (x, ε) = {y ∈ SX |kx + yk ≥ 2 − ε}. We write convl+(x, ε) for the convex hull of l+(x, ε). The following theorem will help up us to study the Borel complexity of separable Banach spaces with the Daugavet property (see [KaW], and [KaSSiW], pag. 1766).

Theorem 34. A Banach space X has the Daugavet property if, and only if, for every

+ x ∈ SX , and ε > 0, the set conv l (x, ε) is dense in BX .

Theorem 35. Let D = {X ∈ SB|X has the Daugavet property}. D is Borel.

˜ Proof. For this let’s ﬁrst remember that there exist a sequence of Borel functions Sn : ˜ SB → SC(∆) such that {Sn(X)}n∈N is dense in SX , for all X ∈ SB. Also, as X 7→ BX is a Borel function from SB into F(C(∆)), we have that X 7→ Sn(BX ) is Borel and

{Sn(BX )}n∈N is dense in BX , for all X ∈ SB. Now, we only need to observe that theorem 34 implies that

\ \ \ [ [ D = + < + n∈N ε∈Q m∈N (n1,...,nl)∈N N a1,...,al∈Q P ai=1 l l n ˜ X ˜ X ˜ o X ∈ SB| Sn(X) + aiSni (X) ≥ 2 − ε, Sm(BX ) − aiSni (X) < ε . i=1 i=1

˜ Checking this equality is just a matter of using the fact that {Sn(X)}n∈N is dense in SX , for all X ∈ SB, and that {Sn(BX )}n∈N is dense in BX , for all X ∈ SB. The details are left to the interested reader.

48 CHAPTER 5. GEOMETRY OF BANACH SPACES, PART II.

5.5 Complete Continuous Property.

A Banach space X is said to have the complete continuous property (or just to have the

CCP) if every operator from L1[0, 1] to X is completely continuous (i.e., if it carries weakly compact sets into norm-compact sets). It is well known that L1[0, 1] does not have this property.

1 Theorem 36. Let CCP = {X ∈ SB|X has the CCP}. CCP is Π1-hard. In particular, CCP is non Borel.

Proof. Let (en)n∈N be a basis of L1[0, 1] and k.kL1 its norm. For each θ ∈ Tr and x ∈ c00(θ), we deﬁne

n 1 n X X 2 2 o kxk = sup x(s)e|s| | n ∈ ,I1, ..., In incomparable segments of θ θ L1 N i=1 s∈Ii

and let ϕ(θ) be the completion of c00(θ) under the norm k.kθ. By lemma 8, if θ ∈ WF, then ϕ(θ) is reﬂexive, which implies ϕ(θ) = ϕ(θ)∗∗ is a separable dual. As separable duals have the Radon-Nikodym property (Dunford-Pettis theorem, see [DiU]) and RNP implies CCP (see [G]), we conclude that ϕ(θ) ∈ CCP, for all θ ∈ WF.

On the other hand, if θ ∈ IF we have that L1[0, 1] ,→ ϕ(θ). As L1[0, 1] does not have CCP, this clearly implies ϕ(θ) 6∈ CCP, for all θ ∈ IF.

M. Girardi had shown (see [G], pag. 70) that a Banach space X has the CCP if, and only if, X has no δ-Rademacher bush on it (the original terminology used by M. Girardi was δ-Rademarcher tree instead of bush, but in order to be coherent, specially with the next section of these notes, we chose to call it a bush). A δ-Rademarcher bush on X is a

l n set of the form {xn ∈ X|n ∈ N, l ∈ {1, ..., 2 }} satisfying

49 5.5. COMPLETE CONTINUOUS PROPERTY.

x2l−1 + x2l (i) xl = n n , for all n ∈ , and l ∈ {1, ..., 2n−1}. n−1 2 N

P2n−1 2l−1 2l n (ii) ∃δ > 0, such that l=1 (xn − xn ) > 2 δ, for all n ∈ N.

Theorem 37. A Banach space X has the CCP if, and only if, there exists no δ- Rademacher bush on X.

In the previous theorems of these notes, every time we could characterize a property of Banach spaces by the existence of some sequence in this space satisfying some geometric inequalities we could also obtain some information about its Borel complexity. M. Gi- rardi’s result does not give us the CCP in terms of the non existence of a sequence but it gives us in terms of the non existence of a δ-Rademacher bush on X. Let’s try to mimic our previous proofs in order to show that the set of separable Banach spaces with the complete continuous property is coanalytic.

2 First we introduce some notation. Deﬁne (Mn)n∈N inductively by M1 = N , and Mn =

2n Mn−1 × N . Let M = ∪nMn ∪ {∅}. Given s ∈ M, we say |s| = n if s ∈ Mn. If |s| = n,

2i+1 we can clearly represent s in a standard way by s = (s0, ..., sn−1), where si ∈ N , for all i ≤ n − 1. For s ∈ M we let s|i = (s0, ..., si−1), for all i ≤ |s|. Given s, t ∈ M, we say s ≤ t if |s| ≤ |t| and s = t|s|. We deﬁne s < t analogously. We say TM ⊂ M is a M-tree

(on N) if s ∈ TM implies si ∈ TM , for all i ≤ |s|.

As M is countable we have that 2M (the power set of M) is a Polish space with its

M standard topology. Let TrM be the set of M-trees on N, then TrM is Borel in 2 . Indeed,

M |s| M TrM = ∩s∈M {TM ∈ 2 | s 6∈ TM } ∩i=1 {TM ∈ 2 | s|i ∈ TM } .

50 CHAPTER 5. GEOMETRY OF BANACH SPACES, PART II.

M-tree branch. We say TM ∈ TrM is an M-well-founded tree if there is no such sequence.

We denote the set of M-ill-founded trees and the set of M-well-founded trees by IFM and

WFM , respectively.

Lemma 38. WFM is coanalytic non Borel, and IFM is analytic non Borel.

2i+1 π(sM i), ∀i ≤ |s|}, where π(sM i) is the ﬁrst coordinate of sM i ∈ N . Clearly we have

ϕ(WFM ) = WF. As ϕ is easily seen to be Borel and WF is non analytic we had shown that both WFM and IFM are non Borel.

Let’s see that IFM is analytic. For this notice that

∞ ∞ 2i E = {(TM , (sn)n=0) ∈ TrM ××(N )| (s0, ..., sn) ∈ TM , ∀n ∈ N} n=1 is Borel. Hence, as IFM is the projection of E on TrM , we are done.

Let’s notice that a δ-Rademacher bush can be realized as a M-tree branch on X. Say

l n B = {xn ∈ X|n ∈ N, l ∈ {1, ..., 2 }} is a δ-Rademacher bush. Here we should think of elements with subindex n as being in the “line” n. For each line n of B we can pick s ∈ Mn that represent all the elements of B in all the lines smaller than or equal

1 2 to n. Precisely, deﬁne a M-tree branch on X inductively by letting s(1) = (x1, x1),

1 4 s(2) = (s(1), (x2, ..., x2)), and (in general)

1 2n s(n) = s(1), ..., s(n − 1), xn, ..., xn .

51 5.5. COMPLETE CONTINUOUS PROPERTY.

It is clear that a δ-Rademacher bush on X can be understood as a M-tree branch with the identiﬁcation above. A cleaner way of writing s(n) is

n 1 2k s(n) = xk, ..., xk . k=1

Now, for each X ∈ SB, and δ > 0, we deﬁne a M-tree on X in the following manner

n 1 2k 0 k−1 xk, ..., xk ∈ TCCP (X, δ) ⇔ ∀k ≤ n, ∀l ∈ {1, ..., 2 }, k=1 x2l−1 + x2l xl = k k , k−1 2 2k−1 X 2l−1 2l k (xk − xk ) > 2 δ. l=1

0 By theorem 37, X has the CCP if, and only if, TCCP (X, δ) is M-well-founded for all

δ > 0. Now notice that, by taking rational linear combinations of the functions {Sn}, we can assume that, for all X ∈ N, all n, k ∈ N, and all p, q ∈ Q, there exists m ∈ N such that qSn(X) + pSk(X) = Sm(X). With this in mind, for all X ∈ SB, all K > 0, and all

δ > 0, we deﬁne a M-tree TCCP (X, δ) on N as

n n 1 2k 0 nk, ..., nk ∈ TCCP (X, δ) ⇔ Sn1 (X), ..., Sn 2k (X) ∈ TCCP (X, δ). k=1 k k k=1

Hence, standard approximation arguments, and Girardi’s result, imply that X has the CCP if, and only if, T (X, δ) is M-well-founded for all δ > 0.

Lemma 39. The function ϕδ : SB → TrM deﬁned by ϕ(X) = T (X, δ) is Borel, for all δ > 0.

52 CHAPTER 5. GEOMETRY OF BANACH SPACES, PART II.

Proof. Although notation doesn’t help us here, this proof follows by the same arguments as the proofs of theorems 11, and 15. The details are left to the interested reader.

n n+1 For a given s ∈ N2 let’s denote by s0 ∈ N2 the element that has its ﬁrst 2n coordinates coinciding with the ones of s and its other coordinates being 1 (1 is completely arbitrary here). Deﬁne a M-tree TCCP (X) on N by

1 (n, 1)a(s0 , ..., s0 ) ∈ T (X) ⇔ (s , ..., s ) ∈ T X, , 0 k 0 k n where the concatenation sat of elements of M is deﬁned in the natural way. By lemma

39, we can easily see that ϕ : SB → TrM , deﬁned by ϕ(X) = TCCP (X), is Borel. Hence,

−1 as ϕ (WFM ) = CCP, and WFM is coanalytic, we had just shown

Theorem 40. CCP is coanalytic.

Hence, by theorem 36, we have

Theorem 41. CCP is complete coanalytic.

5.6 Radon-Nikodym property.

A Banach space X is said to have the Radon-Nikodym property (or just to have the RNP) if for any totally ﬁnite positive measure space (Y, Σ, µ) and any X-valued µ-

1 continuous measure m on Σ, with |m|(X) < ∞, there exists f ∈ LX (Y, Σ, µ) such that R m(E) = E fdµ, for all E ∈ Σ.

1 Theorem 42. Let RNP = {X ∈ SB|X has RNP}. RNP is Π1-hard. In particular, RNP is non Borel.

Proof. See the proof of theorem 36. The same function and proof works here.

53 5.6. RADON-NIKODYM PROPERTY.

Observe that the same ϕ shows that the set of separable Banach spaces with separable dual, the set of Banach spaces that are separable duals, and the reﬂexive separable Ba- nach spaces, are all non Borel. For a proof that the later is coanalytic we refer to [D] and [S].

A well known equivalent definition of the RNP is: X has the Radon-Nikodym property if, and only if, X does not contain a δ-bush. Here a δ-bush on a Banach space X is a bounded partially ordered subset B ⊂ X for which i) each member has at least two (but finitely many) successors and is a convex combination of its successors, ii) there is a positive separation constant δ such that kx − yk ≥ δ if x is a successor of y, and iii) B has a first member to which each member of B can be joined by a linearly ordered chain of successive members of B (see [J2], and [J3]).

With the same spirit as in the previous section we see that, as the concept of a δ-bush can be deﬁned in terms of the existence of a special “tree” on X satisfying some geo-

1 metric conditions, this may help us to show RNP is coanalytic. Let’s call x1 the ﬁrst

1 element mentioned in the deﬁnition of a δ-bush. Say x1 has l(1, 1) immediate succes-

1 l(1,1) j sors, we call them x2(1), ..., x2 (1). Each x2(1) has l(2, j) successors, let’s call them

1 l(2,j) x3(j), ..., x3 (j). Proceeding like this we can write a δ-bush B in the following manner: Say

n i o B = xn(j) ∈ X| n ∈ N, j ∈ {1, ..., l(n − 1)}, i ∈ {1, ..., l(n − 1, j)} , (5.6.1)

Pl(n−1) where l(n) is deﬁne inductively as l(1) = l(1, 1) = 1, and l(n) = k=1 l(n − 1, k). Here we should think of elements with subindex n being in the “line” n. With this in mind, l(n) is the number of elements of B in the line n, and l(n − 1, j) is the number of successors of the j-th element of the line n − 1. We assume l(n − 1, j) ≥ 2, for

54 CHAPTER 5. GEOMETRY OF BANACH SPACES, PART II. all n ∈ N, and all j ∈ {1, ..., l(n − 1)}. Say n ∈ N, and j ∈ {1, ..., l(n − 1)}, we denote the j-th element of line n−1 by x(n−1, j). The set B is a δ-bush if, and only if, it satisﬁes

i (i) ∃K > 0 such that kxn(j)k < K, for all n ∈ N, all j ∈ {1, ..., l(n − 1)}, and all i ∈ {1, ..., l(n − 1, j)}.

(ii) For all n ∈ N, and j ∈ {1, ..., l(n − 1)}, there exist a1, ..., al(n−1,j) > 0 such that Pl(n−1,j) Pl(n−1,j) i i=1 ai = 1 and x(n − 1, j) = i=1 aixn(j).

(iii) kx − yk > δ if x is a successor of y.

The equivalence between both deﬁnitions of a δ-bush should be clear.

Theorem 43. A Banach space X has the Radon-Nikodym property if, and only if, X does not contain a δ-bush, for some δ > 0.

As, for this new kind of bush each element can have more than two successors, we cannot use the same M as we used in the previous section to deﬁne M-trees. But we can do some-

Let M = ∪nMn ∪ {∅}. Given s ∈ M, we say |s| = n if s ∈ Mn. If |s| = n, we can clearly

M As M is countable we have that 2 is a Polish space. Let TrM be the set of trees on M,

M then TrM is Borel in 2 . Indeed,

55 5.6. RADON-NIKODYM PROPERTY.

M |s| M TrM = ∩s∈M {TM ∈ 2 | s 6∈ TM } ∩i=1 {TM ∈ 2 | s|i ∈ TM } .

M As TrM is a Borel set in 2 we have that TrM is actually a Polish space (see [Ke]). We say TM ∈ TrM is an M-ill-founded tree if there exists a sequence (s(n))n∈N in TM such that s(n)|s(m)| = s(m), for all m ≤ n. If this sequence is maximal, we called it a M-tree branch. We say TM ∈ TrM is an M-well-founded tree if there is no such sequence. We denote the set of M-ill-founded trees and the set of M-well-founded trees by IFM and

WFM , respectively.

Lemma 44. WFM is coanalytic non Borel, and IFM is analytic non Borel.

Proof. This is analogous to the proof of lemma 38.

Let’s notice that a δ-bush can be realized as a M-tree branch on X. Say B is as in (5.6.1).

For each line n of the bush B we can pick s ∈ Mn that represent all the elements of B in all the lines smaller than or equal to n. Precisely, deﬁne a M-tree branch on X inductively

1 1 l(1,1) by letting s(1) = x1, s(2) = (s(1), (x2(1), ..., x2 (1))), and (in general)

1 l(n−1,1) 1 l(n−1,l(n−1)) s(n) = s(1), ..., s(n − 1), xn(1), ..., xn (1), xn(2), ..., xn (l(n − 1)) .

It is clear that a δ-bush on X can be understood as a M-tree branch with the identiﬁcation above. A cleaner (but not clean!) way of writing s(n) is

n i l(k−1,1) i l(k−1,l(k−1)) s(n) = xk(1) , ..., xk(l(k − 1)) . i=1 i=1 k=1

Now, for each X ∈ SB, K > 0, and δ > 0, we deﬁne a M-tree on X in the following manner

56 CHAPTER 5. GEOMETRY OF BANACH SPACES, PART II.

n i l(k−1,1) i l(k−1,l(k−1)) 0 xk(1) , ..., xk(l(k − 1)) ∈ TRNP (X, K, δ) ⇔ i=1 i=1 k=1 ∀k ≤ n, ∀j ∈ {1, ..., l(k − 1)},

∀i ∈ {1, ..., l(k − 1, j)},

l(k−1,j) + X ∃a1, ..., al(k−1,j) ∈ Q ai = 1 , i=1

i such that kx(k − 1, j) − xk(j)k > δ l(k−1,j) i X i kxk(j)k < K & x(k − 1, j) = aixk(j). i=1

0 By theorem 43 X has the RNP if, and only if, TRNP (X, K, δ) is M-well-founded for all

K, δ > 0. Now notice that by taking rational linear combinations of the functions {Sn} we can assume that, for all X ∈ N, all n, k ∈ N, and all p, q ∈ Q, there exists m ∈ N such that qSn(X) + pSk(X) = Sm(X). With this in mind, for all X ∈ SB, all K > 0, and all

δ > 0, we deﬁne a M-tree TRNP (X, K, δ) on N as

n i l(k−1,1) i l(k−1,l(k−1)) nk(1) , ..., nk(l(k − 1)) ∈ TRNP (X, K, δ) ⇔ i=1 i=1 k=1 n l(k−1,1) l(k−1,l(k−1)) 0 Sni (1)(X) , ..., Sni (l(k−1))(X) ∈ TRNP (X, K, δ). k i=1 k i=1 k=1

Hence, standard approximation arguments, and theorem 43, imply that X has the RNP if, and only if, TRNP (X, K, δ) is M-well-founded for all K, δ > 0. As we had before, we have:

Lemma 45. The function ϕK,δ : SB → TrM deﬁned by ϕ(X) = TRNP (X, K, δ) is Borel, for all K, δ > 0.

2 Choose a bijection N → N , say n 7→ (n1, n2). Deﬁne a M-tree TRNP (X) on N by

57 5.7. ANALYTIC RADON-NIKODYM PROPERTY.

1 a (n) (s0, ..., sk) ∈ TRNP (X) ⇔ (s0, ..., sk) ∈ TRNP X, n1, . n2

By lemma 45, we can easily conclude that ϕ : SB → TrM , deﬁned by ϕ(X) = TRNP (X),

−1 is Borel. Hence, as ϕ (WFM ) = RNP, and WFM is coanalytic we had just shown

Theorem 46. RNP is coanalytic.

Hence, by theorem 42, we have

Theorem 47. RNP is complete coanalytic.

We would like to point out that after showing the results in this section we found out that B. Bossard had already shown that RNP is coanalytic non Borel in [Bo]. However, Bossard’s proof is diﬀerent and does not rely on the existence of bushes in the space.

5.7 Analytic Radon-Nikodym property.

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 derivative with respect to the Lebesgue measure on T.

So far, we had only being working with real Banach spaces. But, as CC(∆) (the space of the complex valued continuous functions endowed with the supremum norm) is universal for the class of separable complex Banach spaces, we can code the class of separable complex Banach spaces in an analogous way. Precisely, we let SBC = {X ⊂

CC(∆)|X is a closed linear subspace}. Analogously as we had before, SBC endowed with the Eﬀros-Borel structure is a Polish space and it makes sense to wonder whether classes of

58 CHAPTER 5. GEOMETRY OF BANACH SPACES, PART II. separable complex Banach spaces with speciﬁc properties are Borel or not in this coding. With this in mind we, have:

Theorem 48. Let a-RNP = {X ∈ SB|X has the analytic Radon-Nikodym property.}.

1 a-RNP is Π1-hard. In particular, a-RNP is non Borel.

For the proof of this result two well known theorems will do the work (see [HN]).

Theorem 49. If X has the Radon-Nikodym property, then X has the analytic Radon- Nikodym property.

Theorem 50. If X has the analytic Radon-Nikodym property, then X does not contain c0.

Proof. (of theorem 48). Let ϕ : Tr → SB be deﬁned as in the proof of theorem 5. Say θ ∈ WF. Then ϕ(θ) is reﬂexive, hence ϕ(θ) = ϕ(θ)∗∗ is a separable dual, therefore it has the RNP. By theorem 48, ϕ(θ) ∈ a-RNP, for all θ ∈ WF.

On the other hand, if θ ∈ IF, then c0 ,→ ϕ(θ), hence, by theorem 50, ϕ∈ / a-RNP.

5.8 The Equivalence Relation of Isomorphisms.

As we had already mentioned, the set of pairs (X,Y ) is analytic non Borel in SB2, i.e.,

ISO = {(X,Y ) ∈ SB2|X ∼= Y } ⊂ SB2 is analytic non Borel in the product of the Eﬀros- Borel structure of SB. In this section we observe that ISO is actually complete analytic. Hence, every analytic set is the inverse image of ISO through a Borel function. Essentially, this had already been done in [Bo], although Bossard had not stated this result.

Theorem 51. ISO is complete analytic.

59 5.8. THE EQUIVALENCE RELATION OF ISOMORPHISMS.

Proof. Let U be the unique, up to isomorphism, separable Banach space that contains all Banach spaces that have a Schauder basis as a complemented set, i.e., U is the Pelczynski space (see [Pe1]). Let (un)n∈N be the standard basis of U, and k.kU its norm. Let’s construct a Borel function ϕ : Tr → SB2 such that ϕ−1(ISO) = IF. For each θ ∈ SB let

ϕ0(θ) be the completion of c00(θ) under the norm

n 1 n X X 2 2 o kxkθ = sup x(s)u|s| U | n ∈ N,I1, ..., In incomparable segments of θ . i=1 s∈Ii

By the uniqueness of U and the properties of this completion we had already seen, we ∼ have that ϕ0(θ) = U if, and only if, θ ∈ IF. Deﬁne ϕ(θ) = (U, ϕ0(θ)) and we are done.

We refer to [FLR] for a proof that ISO is actually much more than just complete analytic as a set. Precisely, the authors of [FLR] show that ISO is a complete analytic equivalence relation.

60 Chapter 6

Local Structure of Banach Spaces.

“Finish him, Ultraman!” (Boy running away from a monster in Tokyo)

6.1 Finite Representability.

A Banach space Y is said to be finitely representable in a Banach space X if for each 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 kukku−1k < 1 + ε. In other words, Y is finitely representable in X if for each 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. Let’s study pairs of Banach spaces (X,Y ) such that Y is finitely representable in X.

Theorem 52. Let FR = {(X,Y ) ∈ SB2|Y is ﬁnitely representable in X}. FR is Borel.

Proof. In order to make the idea behind the notation below clear, let’s remember some simple facts about linear algebra. Let X be a Banach space and x1, ..., xl ∈ X \{0}. Then

61 6.1. FINITE REPRESENTABILITY.

+ Pk span{x1, ..., xl} has dimension l if, and only if, there exists K ∈ Q such that k i=1 aixik Pl ≤ Kk i=1 aixlk, for all k ≤ l, and all a1, ..., al ∈ Q.

Also, if x1, ..., xl ∈ X are linear independent, F is a n-dimensional Banach space, and ε >

−1 0, then a linear function u : span{x1, ..., xl} → F is an isomorphism such that kukku k < Pl 1 + ε, if, and only if, there exists A, B ∈ Q such that AB < 1 + ε, k i=1 aiu(xn) ≤ Pl Pl Pl Ak i=1 aixnk and k i=1 aixnk ≤ Bk i=1 aiu(xn)k, for all a1, ..., al ∈ Q.

Now only need to notice that

\ [ \ [ [ \ FR = + 0 0 + + ε∈ , l∈ K∈N k≤l n1,...,n ∈N A,B∈ a1,...,al∈Q Q N + l Q n1,...,nl∈N b1,...,bl∈Q AB<1+ε k l n 2 X X (X,Y ) ∈ SB | biSni (Y ) ≤ K biSni (Y ) i=1 i=1   l l P 0 P aiSn (X) ≤ A aiSni (Y )  i=1 i i=1    o ⇒  &  .    Pl Pl  a S (Y ) ≤ B a S 0 (X) i=1 i ni i=1 i ni

As the conditions inside of the set above are all Borel we are done.

Corollary 53. For every X ∈ SB both {Y ∈ SB|Y is ﬁnitely representable in X} and {Y ∈ SB|X is ﬁnitely representable in Y } are Borel.

Proof. Remember, injective images of Borel sets under Borel maps are Borel (see [Ke]).

Let P be a property of separable Banach spaces, i.e., P ⊂ SB. We say X ∈ SB has

62 CHAPTER 6. LOCAL STRUCTURE OF BANACH SPACES. property super-P if Y ∈ SB being finitely represented in X implies X ∈ P . Notice that, as X is always finitely represented in itself, X ∈ P if X has property super-P . Also, if Y is finitely represented in X, and X has property super-P , then not only Y ∈ P but Y has property super-P . For a given property P ⊂ SB, we denote the set of X ∈ SB with property super-P by sP .

Theorem 54. Let P ⊂ SB be a Borel property (a Borel set). Then sP is at least coanalytic.

Proof. As P is Borel we have that SB × P c is Borel. Let π : SB × P c → SB be the standard projection. Let E = FR ∩ SB × P c and notice that

sP c = πE).

We can draw some corollaries now.

Corollary 55. The sets sBC and sD, i.e., the super property of B-convex spaces and the super Daugavet property, respectively, are coanalytic.

6.2 Super Reﬂexibility.

A Banach space X is called super reflexive if every Banach space Y that is finitely representable in X is reflexive. We know that the set of reflexive separable Banach spaces is coanalytic non Borel (see [D]), so we cannot apply theorem 54 in order to obtain some information about the Borel complexity of this new class of separable Banach spaces.

On the other hand, it is well known that the set of uniformly convex separable Banach spaces is Borel in SB. P. Enﬂo had shown ([E]) that a Banach space X is super reﬂexive

63 6.2. SUPER REFLEXIBILITY. if, and only if, X admits an equivalent uniformly convex norm. In other words, let SR stand by the set of super reﬂexive separable Banach spaces and UC the set of uniformly convex separable Banach spaces, then SR = {X ∈ SB|∃Y ∈ UC,X ∼= Y }. As the class of isomorphic spaces ISO = {(X,Y ) ∈ SB2|X ∼= Y } is analytic (see [S]), and the set of uniformly convex separable Banach spaces is well known to be Borel ([D]), this trivially implies that SR is at least analytic in SB. The usual natural question then arises: Is SR Borel?

Theorem 56. SR is Borel.

Proof. For this, we remember an equivalent way to define super reflexive spaces ([J4]): X is not super reflexive if, and only if, there exists α, β > 0 such that, for all n ∈ N, there exists x1, ..., xn ∈ X with the property that, for all k < n, and all a1, ..., an ∈ R we have

n k X X aixi ≥ α max{|ai|} and xi < β. i i=1 i=1 With this deﬁnition, it is easy to see that

[ \ [ \ SRc =

n,m∈N l∈N n1,...,nl∈N a1,...,al∈Q k

A word on this equality may be useful. Say X ∈ SR, let α and β be as above. Pick l ∈ N and let x1, ..., xl ∈ X be as above. Pick n ∈ N such that α > 1/n. So,

l X 1 aixi − max{|ai|} > 0, n i i=1 for all a1, ..., al ∈ R. By compactness of the ﬁnite dimensional unit ball, there exists δ > 0

64 CHAPTER 6. LOCAL STRUCTURE OF BANACH SPACES. such that

l X 1 aixi − max{|ai|} > δ, n i i=1 for all a1, ..., al ∈ R. The end of the argument now follows from standard approximation arguments (the other inclusion is trivial).

6.3 Local Unconditional Structure.

As we are taking a look at local structures it makes sense to wonder what happens for spaces with local unconditional structure. 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 )kukkwk ≤ 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..

Theorem 57. Let LUST = {X ∈ SB|X has l.u.st.}. LUST is Borel.

Proof. We ﬁrst remember some well known facts about linear algebra other then the ones we had already remembered in the proof of theorem 52. Let X be a Banach space and x1, ..., xl ∈ X be linear independent. Then x1, ..., xl are M-unconditional if, and only if, Pl Pl k i=1 aixik ≤ Mk i=1 bixik, for all a1, ..., al, b1, ..., bl ∈ Q such that |ai| ≤ |bi|, for all i ∈ {1, ..., l}.

By taking rational linear combinations of the functions {Sn}, we can (and we do) assume that, for all X ∈ N, all n, k ∈ N, and all p, q ∈ Q, there exists m ∈ N such that

65 6.3. LOCAL UNCONDITIONAL STRUCTURE.

0 0 qSn(X) + pSk(X) = Sm(X). Say X,Y ∈ SB, n1, ..., nk ∈ N, and n1, ..., nk ∈ N. If

k 0 (Sni (X))i=1 is linearly independent, we denote by P (X,Y, (ni), (ni)) the linear function from span{S (X), ..., S (X)} to span{S 0 (Y ), ..., S 0 (Y )} such that S (X) 7→ S 0 (Y ), n1 nk n1 nk ni ni for all i ∈ {1, ..., k}. Now notice that

[ \ [ \ [ \ [ \ LUST = + K∈ 0 0 + e ,...,e ∈ + λ∈Q k∈N N m≤k n1,...,nk∈N a1,...,al∈Q 1 l Q w1,...,wl∈Q n1,...,nk∈N c ,...,c ∈Q+ + + + 1 k l≥k, M∈Q b1,...,bl∈Q A,B∈Q 00 00 (|a |≤|b |, ∀i) AB<λ n1 ,...,nl ∈N i i 000 000 d1,...,dk∈ n1 ,...,nl ∈N Q ( m k ! X X X ∈ SB| ciSni (X) ≤ K ciSni (X) i=1 i=1   Pk Pl diSn0 (C(∆)) = eiSn00 (C(∆))  i=1 i i=1 i     &     Pl Pl   aiSn00 (C(∆)) ≤ M biSn00 (C(∆))   i=1 i i=1 i       &    )  k k  ⇒ P 0 P .  wiSn (C(∆)) ≤ A wiSni (X)   i=1 i i=1       &     Pl Pl   i=1 wiSn000 (X) ≤ B i=1 wiSn00 (C(∆))   i i     &     00 000  P (C(∆),X, (n ), (n )) S 0 (C(∆)) = S (X) i i ni ni

There are a couple of comments about the equality above that should be made. First, no- Pk Pl Pl tice that the restrictions d S 0 (C(∆)) = e S 00 (C(∆)) and a S 00 (C(∆)) i=1 i ni i=1 i ni i=1 i ni Pl ≤ M b S 00 (C(∆)) do not depend on X, i.e., those restrictions should actually be i=1 i ni incorporated in the unions and intersections preceding the set. We believe this would only make the notation harder, so we take the liberty of writing it as above. Also, the only thing

00 000 in the equality above that is not clearly Borel is X 7→ P (C(∆),X, (n ), (n )) S 0 (C(∆)) . i i ni

66 CHAPTER 6. LOCAL STRUCTURE OF BANACH SPACES.

00 000 But P (C(∆),X, (ni ), (ni )) is nothing more than a matrix with coordinates depending on the Borel functions X 7→ S 000 (X). So we are done. ni

67 Chapter 7

On the Borel complexity of CP.

“Is Brazil in Africa?” “Wait, isn’t Hugo Chavez the president of South America?” (Common knowledge)

Given a separable Banach space X ∈ SB, a natural question is whether the isomorphism class hXi = {Y ∈ SB|Y ∼= X} is Borel or not in SB. This question had been solved for some speciﬁc Banach spaces. For example, S. Kwapien showed in [Kw] that `2 is the only Banach space which has both type and cotype equal 2, up to isomorphism. Using this result it is not hard to show that h`2i is Borel (see [Bo]). In fact, h`2i is the only example of a Borel isomorphism class and G. Godefroy had conjectured in [Go] that h`2i is indeed the only Borel isomorphism class. For some classical spaces, such as Lp[0, 1], Pelczynski’s universal space U, and C(∆), it had been shown that their isomorphism classes are analytic non Borel ([Bo], [P], and [Ke], respectively).

As the problem of classifying Banach spaces up to isomorphism is extremely complex (see [FLR]) it makes sense to study an easier problem, i.e., classifying Banach spaces up to its

68 CHAPTER 7. ON THE BOREL COMPLEXITY OF CP . subspaces. We have then another natural problem, given a Banach space X ∈ SB, what can we say about the Borel complexity of CX = {Y ∈ SB|X,→ Y }? This question had been solved in [Bo].

Theorem 58. Let X ∈ SB. If X is ﬁnite dimensional, then CX is Borel. If X is inﬁnite dimensional, then CX is complete analytic.

In this chapter we work with a similar problem. Instead of ﬁxing a separable Banach space X ∈ SB and looking at CX = {Y ∈ SB|X,→ Y }, we ﬁx classes of Banach spaces

P ⊂ SB and we look at CP = {Y ∈ SB|∃Z ∈ P,Z,→ Y }.

1 We give a suﬃcient condition for CP to be Σ1-hard and we give some concrete examples of classes satisfying this condition (tight Banach, HI spaces, B-convex spaces, super-reﬂexive spaces, etc).

Using other methods, we also show that CM, where M = {X ∈ SB|X is minimal}, is

1 Σ1-hard.

7.1 Pure classes not cointaining some `p.

Let’s deﬁne a Borel function ϕ : Tr → SB in the following manner. Let (en)n∈N be a basis of C(∆). For each θ ∈ Tr and x ∈ c00(θ) we deﬁne

n 1 n X X p p o kxkθ = sup x(s)e|s| C(∆) | n ∈ N,I1, ..., In incomparable segments of θ , i=1 s∈Ii

where k.kC(∆) is the norm of C(∆). Now we deﬁne ϕ(θ) as the completion of c00(θ) under the norm k.kθ.

69 7.1. PURE CLASSES NOT COINTAINING SOME `P .

Lemma 59. The Borel function ϕ : Tr → SB deﬁned above has the following properties:

(i) ϕ(θ) is universal, if θ ∈ IF.

(ii) ϕ(θ) is `p saturated, if θ ∈ WF, i.e., every inﬁnite dimensional subspace of ϕ(θ)

contains a copy of `p, if θ ∈ WF.

Proof. If θ ∈ IF, clearly ϕ(θ) contains C(∆). Say θ ∈ WF. Let’s proceed by transﬁnite induction on the order of θ. If o(θ) = 1 the result is clear. Assume ϕ(θ) is `p saturated, for all θ ∈ WF with o(θ) < α, for some α < ω1. Fix θ ∈ WF with o(θ) = α.

Let Λ = {λ ∈ N|(λ) ∈ θ} and enumerate Λ, say Λ = {λi|i ∈ N}. For each λ ∈ Λ, let

θλ = {s ∈ θ|(λ) ≤ s}. As θ ∈ WF Proposition 3 gives us

n [ o θ(λj) < o(θ) = α, ∀n ∈ N. j=1

Sn Our induction hypothesis implies that X 6,→ ϕ( j=1 θ(λj)), for all X ∈ P, and all n ∈ N. Sn Sn Therefore, as ϕ( j=1 θλj ) is a direct sum of ϕ( j=1 θ(λj)) with a ﬁnite dimensional space, Sn X 6,→ ϕ( j=1 θλj ), for all X ∈ P, and all n ∈ N. Consider now the projections

n [ Pλn : ϕ(θ) → ϕ θλj j=1

(as)s∈θ → (as) Sn . s∈ j=1 θλj

Say E ⊂ ϕ(θ).

70 CHAPTER 7. ON THE BOREL COMPLEXITY OF CP .

Sj Case 1: ∃j ∈ N such that Pλj : E → ϕ( i=1 θλi ) is not strictly singular.

˜ Then there exists an inﬁnite dimensional subspace E ⊂ E such that Pλj |E˜ is an isomor- Sj phism onto its image. By our inductive hypothesis, ϕ( i=1 θλi ) is `p saturated, so we are done.

Sj Case 2: Pλj : E → ϕ( i=1 θλi ) is strictly singular, for all j ∈ N.

In this case an easy sliding hump argument gives us a sequence (yn)n∈N in E equivalent to the `p basis, so we are done.

Let P ⊂ SB. We say P is a class of Banach spaces if X ∈ P and Y ∼= X implies Y ∈ P. We say a class of Banach spaces P ⊂ SB is pure if X ∈ P implies Y ∈ P, for all subspace Y ⊂ X, and P contains only inﬁnite dimensional spaces. We say a class P ⊂ SB is almost-pure if X ∈ P and Y ⊂ X imply that there exists a subspace Z ⊂ Y such that Z ∈ P, and P contains only inﬁnite dimensional spaces. The following theorem is a simple consequence of lemma 59.

Theorem 60. Say P ⊂ SB is almost-pure and that `p does not embed in any Y ∈ P, for

1 some p ∈ [1, ∞). Then CP = {Y ∈ SB|∃Z ∈ P,Z,→ Y } is Σ1-hard. In particular, the same is true if P is pure and does not contain `p, for some p ∈ [1, ∞).

We say an inﬁnite dimensional Banach space X is hereditatily indecomposable, or HI, if none of X subspaces can be decomposed as a direct sum of two inﬁnite dimensional subspaces. Clearly, the class HI = {X ∈ SB|X is hereditarily indecomposable} is a pure class.

1 Corollary 61. CHI is Σ1-hard.

71 7.1. PURE CLASSES NOT COINTAINING SOME `P .

Let I0,I1 ⊂ N be two ﬁnite intervals. We say I0 < I1 if max I0 < min I1. A Banach space

X is said to be tight if it has a basis (en)n∈N such that, for all subspace Y ⊂ X there exist a sequence of ﬁnite intervals I0 < I1 < I2 < ... such that, for all inﬁnite subset A ⊂ N,

Y 6,→ span{en|n 6∈ ∪i∈AIi}.

Proposition 3.3 and theorem 1.1 of [FR] gives us that T = {X ∈ SB|X is tight} is an almost-pure class. We have the following corollary.

1 Corollary 62. CT is Σ1-hard.

Proposition 63. Let P ⊂ SB be a Borel set. Then CP is analytic.

Proof. We only need to notice that

N X ∈ CP ⇔∃Y ∈ P, ∃(ni)i∈N, (mi)i∈N, (li)i∈N ∈ N ,

s.t. {Sn (Y )} = Y,T (Sn (Y )) := Sm (X) is well-defined, i i∈N i li kT k < ∞, where by “T (S (Y )) := S (X) is well-defined” we mean “the function T : {S (Y )} → ni mli ni i∈N {S (X)} defined by T (S (Y )) := S (X) is well-defined”. As we can easily write nli i∈N ni mli the relations “{Sn(Y )} = Y ”, “T : {Sn (Y )}i∈ → {Sn (X)}i∈ is well-defined” and n∈N i N li N “kT k < ∞” in a Borel fashion, we are done.

Theorem 64. Let P ⊂ SB be as in lemma 60. If P is Borel, then CP is complete analytic.

Let BC = {X ∈ SB|X is B-convex} and SR∞ = {X ∈ SB|X is inﬁnite dimensional and super reﬂexive}. It was shown in previous chapters that both BC and SR∞ are Borel sets. Applying the theorem above we have the following.

72 CHAPTER 7. ON THE BOREL COMPLEXITY OF CP .

Corollary 65. CBC and CSR∞ are complete analytic.

Proof. Both BC and SR∞ are clearly pure and they do not contain `1.

We can also apply theorem 60 to some of the other classes of Banach spaces studied in previous chapters. Let BS = {X ∈ SB|X has the Banach-Saks property}, ABS = {X ∈ SB|X has the alternating Banach-Saks property}, S = {X ∈ SB|X has the Schur property}, RNP = {X ∈ SB|X has the Radon-Nikodym property}, and CCP = {X ∈ SB|X has the complete continuous property} (for the deﬁnition of those properties see chapter 1).

1 Corollary 66. CP is Σ1-hard for all the classes of separable Banach spaces deﬁned above.

7.2 Spaces containing a minimal subspace.

We now turn our attetion to the following: Although M = {X ∈ SB|X is minimal} is clearly a pure class, M contains `p, for all p ∈ [1, ∞). Therefore, theorem 60 does not say anything about the Borel complexity of CM. We now use the classical construction of Tsirelson space by T. Figiel and W. B. Johnson ([FJ]) in order to construct a ϕ : Tr → SB that will solve our problem.

Given θ ∈ Tr, E ⊂ θ, and x = P a e ∈ c (θ) (for some (a ) ∈ N), we let n∈N sn sn 00 sn n∈N R Ex = P a e . sn∈E sn sn

73 7.2. SPACES CONTAINING A MINIMAL SUBSPACE.

We deﬁne a Borel function ϕ : Tr → SB as, for each θ ∈ Tr and each x ∈ c00(θ) let

(k.kθ,m)m∈N be inductively deﬁned as

k 1 X kxk = max{kxk , max kE xk }, θ,m+1 0 2 i θ,m i=1 for all m ∈ N, where the “inner” maximum above is taken over all k ∈ N and all completely

k incomparable ﬁnite sets (Ei)i=1 (Ei ⊂ θ, for all i ∈ {1, ..., k}) such that k ≤ E1 < ... < Ek.

Exactly as we have for the standard Tsirelson space, we can deﬁne a norm k.kθ as

kxkθ = lim kxkθ,m, n→∞ for all x ∈ c00(θ). We deﬁne ϕ(θ) to be the completion of c00(θ) under this norm. This norm can be implicitly deﬁned as

k 1 X kxk = max{kxk , max kE xk }, θ 0 2 i θ i=1 where the “inner” maximum above is taken over all k ∈ N and all completely incompara-

k ble ﬁnite sets (Ei)i=1 (Ei ⊂ θ, for all i ∈ {1, ..., k}) such that k ≤ E1 < ...Ek.

By the universality of C(∆) for separable Banach spaces, we can identify ϕ(N

Theorem 67. Let ϕ : Tr → SB be deﬁned as above. Then ϕ is a Borel function with the following properties

(i) co ,→ ϕ(θ), for all θ ∈ IF.

(ii) ϕ(θ) has the following property if θ ∈ WF: For every inﬁnite dimensional subspace

74 CHAPTER 7. ON THE BOREL COMPLEXITY OF CP .

E ⊂ ϕ(θ), there exists a further subspace F ⊂ E isomorphic to an inﬁnite dimensional subspace of Tsirelson’s space.

1 In particular, CM is Σ1-hard.

Proof. If θ ∈ IF it is clear that c0 ,→ ϕ(θ). Say θ ∈ WF, let’s show that every infinite dimensional subspace of ϕ(θ) contains a subspace isomorphic to an infinite dimensional subspace of Tsirelson’s space. We proceed by transfinite induction on the order of θ. If o(θ) = 1 the result is clear. Assume X 6,→ ϕ(θ), for all θ ∈ WF with o(θ) < α, for some

α < ω1. Fix θ ∈ WF with o(θ) = α.

For now on we use the same notation as in theorem 60. Say E is a subspace of ϕ(θ).

Sj Case 1: ∃j ∈ N such that Pλj : E → ϕ( i=1 θλi ) is not strictly singular.

˜ Then there exists an inﬁnite dimensional subspace E ⊂ E such that Pλj |E˜ is an isomor- Sj phism onto its image. By our inductive hypothesis ϕ( i=1 θλi ) has the desired property, so we are done.

Sj Case 2: Pλj : E → ϕ( i=1 θλi ) is strictly singular, for all j ∈ N.

In this case an easy sliding hump argument gives us that E contains a subspace isomorphic to a block subspace generated by a block sequence (yn)n∈N with completely incomparable supports.

Let k.kT,θ be the standard Tsirelson norm on c00(θ), i.e.,

75 7.2. SPACES CONTAINING A MINIMAL SUBSPACE.

k 1 X kxk = max{kxk , max kE xk }, T,θ 0 2 i T,θ i=1

k where the “inner” maximum above is taken over all k ∈ N and all ﬁnite sets (Ei)i=1

(Ei ⊂ θ, for all i ∈ {1, ..., k}) such that k ≤ E1 < ...Ek. Here we just ignore the

k restriction that (Ei)i=1 should be completely incomparable. Clearly, we have

k k X X k aiyikθ ≤ k aiyikT,θ, i=1 i=1 for all a1, ..., ak ∈ R. Mimicking the proof of lemma II.1 of [CS] we have the following lemma

Lemma 68. Let (pn)n∈N be an increasing sequence of natural numbers. Let (esn )n∈N be the standard basis of ϕ(θ). Let y = Ppn+1 b e (for all n ∈ ) be a normalized block n i=pn+1 i si N sequence of (esn )n∈N and assume (yn)n∈N has completely incomparable supports. Then

X X k anespn+1 kθ ≤ k anynkθ, n∈N n∈N for all sequences of scalars (an)n∈N.

Proof. It is enough to show that, for all sequences (an)n∈N, we have

X X k anespn+1 kθ,m ≤ k anynkθ, n∈N n∈N for all m ∈ N. Let’s proceed by induction on m ∈ N. For m = 0 the result is clear. Assume the equation above holds for a ﬁxed m ∈ . Let x = P a e and y = P a y . N n∈N n pn+1 n∈N n n k Fix k ∈ N, and completely incomparable ﬁnite sets (En)n=1 (En ⊂ θ, for all n ∈ {1, ..., k}) such that k ≤ E1 < ... < Ek. Consider the sum

76 CHAPTER 7. ON THE BOREL COMPLEXITY OF CP .

k 1 X kE xk . 2 i θ,m i=1

Since the support of x is contained in {pn + 1|n ∈ N}, we may assume that

Ei ⊂ {pn + 1|n ∈ N}, for all i ∈ N. Applying the inductive hypothesis we have

k k 1 X 1 X X kE xk ≤ k a y k 2 i θ,m 2 n n θ i=1 i=1 n∈N pn+1∈Ei

As pn + 1 ∈ Ei implies k ≤ pn + 1, and as (yn)n∈N has completely incomparable supports, the sum on the right of the equation above is allowed as an “inner” sum in the deﬁnition of the norm k.kθ. We had just shown that

k 1 X kE xk ≤ kyk , 2 i θ,m θ i=1

k for all k ∈ N, and all completely incomparable ﬁnite sets (En)n=1 (En ⊂ θ, for all n ∈

{1, ..., k}) such that k ≤ E1 < ... < Ek. Therefore,

X X k anespn+1 kθ,m+1 ≤ k anynkθ, n∈N n∈N and we are done.

Let (bn)n∈N be the sequence of scalars such that our block sequence (yn)n∈N can be written as y = Ppn+1 b e . Then lemma 68 gives us that n i=pn+1 i si

X X k anespn+1 kθ ≤ k anynkθ, n∈N n∈N

77 7.2. SPACES CONTAINING A MINIMAL SUBSPACE.

for all sequences of scalars (an)n∈N. As the supports of (yn)n∈N are completely incomparable we have that

X X k anespn+1 kθ = k anespn+1 kT,θ, n∈N n∈N for all sequences of scalars (an)n∈N. By proposition II.4 of [CS] we have

1 X X k a y k ≤ k a e k , 18 n n T,θ n spn+1 T,θ n∈N n∈N

N for all (an)n∈N ∈ R . We had just proved that the space generated by the block sequence

(yn)n∈N can be embedded into Tsirelson space. The last part of the theorem follows from the well known fact that Tsirelson’s space contains no minimal subspaces (see [CS]).

78 Chapter 8

Non-Universality Results.

“The old fart is in his oﬃce.” (Joe Diestel)

8.1 Non-Universality Results.

In this chapter we use ideas that can be found in [S] to show the non existence of universal spaces for some speciﬁc classes of Banach spaces. Precisely, we say P is a property (or a class) of separable Banach spaces, i.e., P ⊂ SB and P is closed under isomorphism. For a given property P, can we ﬁnd a Banach space X with property P such that all Banach spaces with property P can be isomorphically embedded in X? If yes, we say X is a P-universal element of P. Analogously, we say that X ∈ P is a complementedly P- universal element of P ⊂ SB if every element of P can be complementedly isomorphically embedded in X. We have the following easy lemma.

Lemma 69. Say P ⊂ SB is non analytic. Assume property P is pure, i.e., X ∈ P and Y,→ X (Y ∈ SB) implies Y ∈ P. Then P has no P-universal element. If P is assumed

79 8.1. NON-UNIVERSALITY RESULTS.

⊥ to be complementedly pure, i.e., X ∈ P and Y ,→ X implies Y ∈ P, then we have that P has no complementedly P-universal element.

Proof. Say X ∈ P is P-universal. Let A = {Y ∈ SB|Y,→ X}, so A is analytic (see [S]). Clearly P = A, absurd. For the complementedly universal case we let A = {Y ∈ ⊥ SB|Y ,−→ X} and as A is also well known to be analytic we are done.

This lemma together with our previous results easily give us some interesting corollaries.

Corollary 70. Let U and W be as in Chapter 2. There is no complementedly universal space X ∈ U for the class U. The same is true for W.

Proof. First notice that we had actually shown that both these classes are not only non ∼ Borel but non analytic. Now, we only need to notice that if X = X1 ⊕X2 and P : L(X) → ˜ U(X) is a projection then P (T ) = P1 ◦P (T )|X1 , where P1 : X1 ⊕X2 → X1 is the standard projection, is a projection from L(X1) to U(X1) (the same works for the class W).

Corollary 71. Let X ∈ SB, then there is no Y ∈ NCX universal for NCX .

Proof. It is well known (and we had actually remembered this fact in these notes) that

NCX is coanalytic non Borel.

Corollary 72. There is no X ∈ BS universal for the class BS. The some holds for ABS and WBS.

Proof. One way of noticing WBS is hereditary is lemma 18.

Corollary 73. There is no X ∈ BS complementedly universal for the class BS.

Corollary 74. There is no X ∈ S universal for the class S.

Corollary 75. There is no X ∈ DP complementedly universal for the class DP.

80 CHAPTER 8. NON-UNIVERSALITY RESULTS.

Corollary 76. There is no X ∈ RNP universal for the class RNP. The some holds for CCP and a-RNP.

After getting the corollary above we discovered that its ﬁrst claim had already been discovered by M. Talagrand by completely diﬀerent methods. Talagrand’s proof remains unpublished though.

Let’s take a look at another easy (but proﬁtable) lemma.

Lemma 77. Say P1, P2 ⊂ SB. Assume there exists a Borel ϕ : Tr → SB such that

ϕ(WF) ⊂ P1 and ϕ(IF) ⊂ P2. Let A ⊂ SB be an analytic subset containing P1. Then

A ∩ P2 6= ∅. In particular, if P2 ⊂ {X ∈ SB|X is universal for SB}, we have that if X is universal for P1, then X is universal for SB.

Proof. As WF ⊂ ϕ−1(A) and WF is non analytic we cannot have equality. Hence, there exists θ ∈ IF such that ϕ(θ) ∈ A. As ϕ(θ) ∈ P2 we are done. For the second claim, let X be universal for P1, deﬁne A = {Y ∈ SB|Y,→ X}, and apply the ﬁrst claim.

Notice that all the corollaries we obtained before could be obtained by this lemma, but the new corollaries we will derive could not be obtained with the ﬁrst lemma. The proofs of the following corollaries are either contained in the proofs of their respective sections or are just a slightly modiﬁcation of them.

Corollary 78. If X ∈ SB is universal for either U or W, then X is universal for SB. In particular, those classes admit no element universal for themselves.

Corollary 79. Let X ∈ SB. If Y is universal for NCX , then it is universal for SB.

Corollary 80. If X ∈ SB is universal for the class BS, then X is universal for SB. The some holds for ABS and WBS.

Corollary 81. If X ∈ SB is universal for the class S, then X is universal for SB.

81 8.1. NON-UNIVERSALITY RESULTS.

Corollary 82. If X ∈ SB is universal for the class RNP, then X is universal for SB.. The some holds for CCP and a-RNP.

82 Chapter 9

Hausdorﬀ Distance and Openings Between Banach Spaces.

“Quotes are stupid.” (Jakub Kolacz)

In these chapter we are interested in studying the Borel complexity of the Hausdorff distance between the unit spheres of two given separable Banach spaces and relating this with the Vietoris topology and the Effros-Borel structure of a Polish space X. For this let’s first discuss the motivation for this work.

Let X be a topological space. In this chapter we denote by K(X) the set of all nonempty compact subsets of X. We endow K(X) with the Vietoris topology, i.e., the topology generated by

{K ∈ K(X)|K ⊂ U0,K ∩ U1 6= ∅},

83 where U0 and U1 vary among all the open subsets of X. It is well known that if (X, d) is a metric space, then the Vietoris topology on K(X) coincides with the Hausdorff topology on K(X), i.e., the topology generated by the Hausdorff distance distance in K(X), which is defined as

δH (K,L) = max{max d(k, L), max d(l, K)}, k∈K l∈L for all K,L ∈ K(X). As in this chapter we are only concerned with metric spaces, we shall make no distinction between the Hausdorﬀ topology and the Vietoris topology on K(X). The Hausdorﬀ metric on K(X) is known to have all kind of nice properties. Precisely, if X is separable, completely metrizable, or compact metrizable, so is K(X) (see [Ke], or [S]). Hence, if X is a Polish space, so is K(X).

For a given topological space X we denote by F(X) the set of all nonempty closed subsets of X. We endow F(X) with the Eﬀros-Borel structure. Hence, the similarity between the Vietoris topology and the Eﬀros-Borel structure is evident. The following well known proposition makes this similarity to be even more interesting (see [S]).

Proposition 83. Let X be a compact metrizable space. Denote by BK(X) the Borel structure on K(X) generated by the Vietoris topology. Then BK(X) is generated by {K ∈ K(X)| K ∩ U 6= ∅}, where U varies among all the open subsets of X.

Therefore, the proposition above states that if X is compact metrizable (which clearly implies F(X) = K(X)) then there is no distinction between the Eﬀros-Borel structure and the Borel structure generated by the Vietoris topology.

As said before, our goal for this chapter is to study the Borel complexity of openings between Banach spaces that are deﬁned in a similar way to the Hausdorﬀ distance on

84 CHAPTER 9. HAUSDORFF DISTANCE AND OPENINGS BETWEEN BANACH SPACES.

2 2 SB . Say X is a metric space. Of course δH : K(X) → [0, ∞) is continuous, after all this is how the Hausdorff topology on K(X) was defined, therefore it is clearly Borel in relation to the Borel σ-algebra generated by the Hausdorff topology. If X is assumed to be compact proposition 83 implies that δH is also Borel in relation to the Effros-Borel structure. But what can we say if X is not compact? More precisely, what can we say if X = C(∆)?

Unless stated diﬀerently, d : C(∆)2 → [0, ∞) will denote the distance on C(∆) given by its standard norm.

9.1 The Hausdorﬀ Distance on F(X).

We can deﬁne the Hausdorﬀ distance for all pairs (K,L) of closed subsets of X if we allow

2 our function to attain inﬁnity, i.e., if we consider δH as a function F(X) → [0, ∞].

2 Theorem 84. Let (X, d) be a Polish space. The Hausdorﬀ distance δH : F(X) → [0, ∞]

2 is Borel. In particular, δH : K(X) → [0, ∞] is Borel if we consider K(X) endowed with the Eﬀros-Borel structure.

Proof. First we notice that, as {K ∈ F(X)|K is compact} is Borel in F(X) (see [S]), we have that K(X) endowed with the Eﬀros-Borel structure is a standard Borel space and if

2 δH : F(X) → [0, ∞] is Borel its restriction to K(X) will be Borel as well. Now we show

2 that δH : F(X) → [0, ∞] is Borel.

2 In order to do that it is enough to show that δ0 : F(X) → [0, ∞] is Borel, where

δ0(K,L) = sup d(k, L). k∈K

85 9.1. THE HAUSDORFF DISTANCE ON F(X).

Pick α ∈ R. Let’s show that δ−1 ((−∞, α]) is Borel in F(X). Let S be a countable dense subset of X. We will refer to elements of S as kn or ln, depending on what makes the notation more intuitive. In order to see that δ−1 ((−∞, α]) is Borel we show that

−1 \ \ 2 δ ((−∞, α]) = {(K,L) ∈ F(X) | B(kn, δ) ∩ K = ∅}∪ δ∈Q+ kn∈S ! \ [ 2 {(K,L) ∈ F(X) | B(ln, ε) ∩ L 6= ∅} , ε∈Q+ ln∈S d(ln,kn)<α+δ+ε where B(k, δ) = {k0 ∈ X|d(k, k0) < δ}, for all k ∈ X and δ > 0. As the right hand side is clearly Borel, once we show that the equality above holds we will be done. In order to simplify notation, let E denote the set on the left hand side of the equation above and F be the set on its right hand side.

Let’s show E ⊂ F . Pick (K,L) ∈ E. Then,

∀k ∈ K, ∀ε¯ > 0, ∃l ∈ L such that d(k, l) < α +ε. ¯ (9.1.1)

+ 2 Fix δ ∈ Q and kn ∈ S. If B(kn, δ) ∩ K = ∅, great, (K,L) ∈ {K,L) ∈ F(X) |B(kn, δ) ∩

0 0 K = ∅} ⊂ F . Otherwise, pick k ∈ B(kn, δ) ∩ K. Let d(k, kn) = δ , clearly δ < δ. Fix ε > 0. Then, by equation (9.2.1), ∃l ∈ L such that d(k, l) < α + ε. As S is dense in L we

0 can pick ln ∈ S such that d(ln, l) < min{ε, δ − δ }. Then we have

0 0 d(ln, kn) ≤ d(ln, l) + d(l, k) + d(k, kn) < δ − δ + α + ε + δ = α + δ + ε.

Hence, as d(l, ln) < ε, and l ∈ L,

86 CHAPTER 9. HAUSDORFF DISTANCE AND OPENINGS BETWEEN BANACH SPACES.

[ 2 (K,L) ∈ {(K,L) ∈ F(X) | B(ln, ε) ∩ L 6= ∅},

ln∈S d(ln,kn)<α+δ+ε and we are done.

Let’s now show that F ⊂ E. For this assume (K,L) ∈/ E. Then, there exists β > α and k ∈ K such that d(k, l) > β, for all l ∈ L. Suppose (K,L) ∈ F , let’s look for a

β−α contradiction. Let δ = 4 and pick kn ∈ S such that kn ∈ B(k, δ). As B(kn, δ) ∩ K 6= ∅, we must have

\ [ 2 (K,L) ∈ {(K,L) ∈ F(X) | B(ln, ε) ∩ L 6= ∅}. ε∈Q+ ln∈S d(ln,ln)<α+δ+ε

β−α Pick ε = 4 . Then, there exists ln ∈ S such that B(ln, ε)∩L 6= ∅, and d(ln, kn) < α+δ+ε.

Pick l ∈ B(ln, ε) ∩ L. Then we have

d(k, l) ≤ d(k, kn) + d(kn, ln) + d(ln, l) < α + 2δ + 2ε = α + β − α = β, absurd. Therefore, (K,L) ∈/ E implies (K,L) ∈/ F , and we are done.

9.2 Openings Between Banach Spaces.

We now turn our attention to some examples of openings, or gaps, between Banach spaces. There are many ways of defining a kind of distance between Banach spaces that will gen- erate a topology. Here we study some of them. Specifically, we study the Borel complexity of the geometric opening, spherical opening and the ball opening between separable Ba- nach spaces. The interested reader can find many properties of those openings in [O]. In this paper M. Ostrovskii not only discusses all those openings but also many applications

87 9.2. OPENINGS BETWEEN BANACH SPACES. to the geometry of Banach spaces.

The spherical opening between two separable Banach spaces X,Y ∈ SB is deﬁned as

Ω(X,Y ) = max{ sup d(x, SY ), sup d(y, SX )}. x∈SX y∈SY

Notice, Ω(X,Y ) is nothing more than the Hausdorﬀ distance between SX and SY . The geometric opening between X,Y ∈ SB is deﬁned as

Θ(X,Y ) = max{ sup d(x, Y ), sup d(y, X)}. x∈SX y∈SY

At last, we deﬁne the ball opening between X,Y ∈ SB by

Λ(X,Y ) = max{ sup d(x, BY ), sup d(y, BX )}. x∈SX y∈SY

Clearly, all those openings have range [0, 1].

Theorem 85. The spherical opening Ω: SB2 → [0, 1] is Borel.

The proof of this result is basically the same as the proof of theorem 84 but with some small technicalities. For the sake of completeness and because of those technicalities we give its proof here.

2 Proof. We show that Ω0 : SB → [0, 1] is Borel, where

Ω0(X,Y ) = sup d(x, SY ). x∈SX

Pick α ∈ R. Let S be a countable dense subset of SC(∆). We will refer to elements of S as xn or yn, depending on what makes the notation more intuitive. We will be done once we show that

88 CHAPTER 9. HAUSDORFF DISTANCE AND OPENINGS BETWEEN BANACH SPACES.

−1 \ \ 2 Ω0 ((−∞, α]) = {(X,Y ) ∈ SB | B(xn, δ) ∩ X = ∅}∪ δ∈Q+ xn∈S ! \ [ 2 {(X,Y ) ∈ SB | B(yn, ε) ∩ Y 6= ∅} . ε∈Q+ yn∈S d(yn,xn)<α+2δ+ε

To simplify notation, let E denote the set in the left hand side of the equation above and F be the set in its right hand side. Before we show that the equality holds let’s introduce a simple lemma that will help us to deal with some technicalities.

Lemma 86. Let X be a normed space. Say x0 ∈ SX , δ > 0, and x ∈ B(x0, δ). Assume

x x 6= 0. Then kxk ∈ B(x0, 2δ).

Proof. We only need to notice that kx − kxkx0k < 2kxkδ. By the triangle inequality we have

kx − kxkx0k ≤ kx − kxkxk + kkxkx − kxkx0k ≤ |1 − kxk|kxk + kxkkx − x0k < 2kxkδ.

Let’s show E ⊂ F . Pick (X,Y ) ∈ E. Then,

∀x ∈ SX , ∀ε¯ > 0, ∃y ∈ SY such that d(x, y) < α +ε. ¯ (9.2.1)

+ 2 Fix δ ∈ Q and xn ∈ S. If B(xn, δ)∩X = ∅, great, (X,Y ) ∈ {(X,Y ) ∈ SB |B(xn, δ)∩X =

0 ∅} ⊂ F . Otherwise, pick x ∈ B(xn, δ) ∩ X. As B(xn, δ) ∩ X is open in X we can assume

0 x 6= 0. Using lemma 86, and the fact that X is a linear space, we pick x ∈ B(xn, 2δ)∩SX .

0 0 Let d(x, xn) = δ , clearly δ < 2δ. Fix ε > 0. Then, by equation (9.2.1), ∃y ∈ SY such that

89 9.2. OPENINGS BETWEEN BANACH SPACES.

0 d(x, y) < α+ε. As S is dense in SY we can pick yn ∈ S such that d(yn, y) < min{ε, 2δ−δ }. Then we have

0 0 d(yn, xn) ≤ d(yn, y) + d(y, x) + d(x, xn) < 2δ − δ + α + ε + δ = α + 2δ + ε.

Hence, as d(y, yn) < ε, and y ∈ Y ,

[ 2 (X,Y ) ∈ {(X,Y ) ∈ SB | B(yn, ε) ∩ Y 6= ∅},

yn∈S d(yn,xn)<α+2δ+ε and we are done.

Let’s now show that F ⊂ E. For this assume (X,Y ) ∈/ E. Then, there exists β > α and x ∈ SX such that d(x, y) > β, for all y ∈ SY . Suppose (X,Y ) ∈ F , let’s look for a

β−α contradiction. Let δ = 6 and pick xn ∈ S such that xn ∈ B(x, δ). As B(xn, δ) ∩ X 6= ∅, we must have

\ [ 2 (X,Y ) ∈ {(X,Y ) ∈ SB | B(yn, ε) ∩ Y 6= ∅}. ε∈Q+ yn∈S d(yn,xn)<α+2δ+ε

β−α Pick ε = 6 . Then, there exists yn ∈ S such that B(yn, ε) ∩ Y 6= ∅, and d(yn, xn) <

α + 2δ + ε. Using Lemma 86 again, we can pick y ∈ B(yn, 2ε) ∩ SY . Then we have

d(x, y) ≤ d(x, xn) + d(xn, yn) + d(yn, y) < α + 3δ + 3ε = α + β − α = β, absurd. Therefore, (X,Y ) ∈/ E implies (X,Y ) ∈/ F , and we are done.

Theorem 87. The geometric opening Θ: SB2 → [0, 1] is Borel.

Proof. This proof is completely analogous to the proof of theorem 85. We only need to notice that

90 CHAPTER 9. HAUSDORFF DISTANCE AND OPENINGS BETWEEN BANACH SPACES.

−1 \ \ 2 Θ0 ((−∞, α]) = {(X,Y ) ∈ SB | B(xn, δ) ∩ X = ∅}∪ δ∈Q+ xn∈S ! \ [ 2 {(X,Y ) ∈ SB | B(yn, ε) ∩ Y 6= ∅} , + ε∈Q yn∈S¯ d(yn,xn)<α+2δ+ε

¯ where S is a countable dense subset of C(∆) and Θ0 is deﬁned analogously as Ω0 was deﬁned.

Theorem 88. The geometric opening Θ: SB2 → [0, 1] is Borel.

Proof. We can either reproduce the proofs above or notice that ϕ : SB → F(C(∆)) deﬁned as ϕ(X) = BX is Borel.

As we are discussing the Borel complexity of “distances” between Banach spaces it is natural to consider the Banach-Mazur distance between Banach spaces. Given two Banach spaces X,Y ∈ SB we define the Banach-Mazur distance D(X,Y ) as the infimum of the quantities kT kkT −1k, where T : X → Y is an isomorphism (D(X,Y ) is defined to be infinite if X and Y are not isomorphic). As kT kkT −1k is always bigger than 1 the Banach-Mazur distance has range [1, ∞].

Theorem 89. The Banach-Mazur distance D : SB2 → [1, ∞] is non Borel.

Proof. We only need to notice that there are Banach spaces such that its isomorphism equivalence class is non Borel, e.g., C(∆) (see [Ke]). So, D(C(∆),.)−1((−∞, ∞)) is non Borel.

Problem 90. It is well known that h`2i, the set of Banach spaces X ∈ SB such that X is isomorphic to `2, is Borel (see [Bo]). Therefore, h`2i is a standard Borel space and it

2 2 makes sense to restrict D to h`2i . Is D : h`2i → [1, ∞) Borel?

91 9.2. OPENINGS BETWEEN BANACH SPACES.

Theorem 91. Let FD = {X ∈ SB|X is ﬁnite dimensional}. The Banach-Mazur distance D : FD2 → [1, ∞] is Borel.

To prove this result we will make use of lemma 1. Once again, we denote by (Sn)n∈N the restrictions to SB of the functions given by the lemma 1 for X = C(∆).

Proof. First we observe that FD is Borel ([S]), so it makes sense to show D is Borel on it. Pick α ∈ R. Notice that, if X and Y are l-dimensional Banach spaces, T : X → Y is

−1 an isomorphism such that kT kkT k < α, if, and only if, there exist x1, ..., xl ∈ X, and Pl A, B ∈ Q, such that AB < α, x1, ..., xl are linearly independent and k i=1 ciT (xi)k < Pl Pl Pl Ak i=1 cixik, and k i=1 cixik < Bk i=1 ciT (xi)k, for all c1, ..., cl ∈ Q. With this in mind, we have

[ [ \ [ [ \ D−1((−∞, α)) = l l+1 + l∈N n1,...,nl∈N (a1,...,al)∈Q \{0} (b1,...,bl+1)∈Q \{0} A,B∈Q c1,...,cl∈Q m1,...,ml∈N AB<α l l n 2 X X (X,Y ) ∈ FD | aiSni (X) 6= 0 & aiSmi (Y ) 6= 0 & i=1 i=1 l+1 l+1 X X biSni (X) = 0 & biSmi (Y ) = 0 & i=1 i=1 l l X X ciSmi (Y ) < A ciSni (X) & i=1 i=1 l l X X o ciSni (X) < B ciSmi (Y ) . i=1 i=1

As all the functions considered above are Borel we are done.

Another natural distance to look at is the operator opening. Let GL(C(∆)) denote the group of all isomorphisms from C(∆) to itself. For X,Y ∈ SB, deﬁne

92 CHAPTER 9. HAUSDORFF DISTANCE AND OPENINGS BETWEEN BANACH SPACES.

r0(X,Y ) = inf {kT − Idk|T ∈ GL(C(∆)),T (Y ) = X}

(this inﬁmum is taken to be 1 in case there is no such T ∈ GL(C(∆))). We deﬁne the operator opening between two separable Banach spaces X and Y as

r(X,Y ) = max{r0(X,Y ), r0(Y,X)}.

As {T ∈ GL(C(∆))|T (Y ) = X} is closed under scaler multiplication it is clear that r takes values in [0, 1].

Problem 92. Is r : SB2 → [0, 1] Borel?

Notice that r(X,Y ) may be 1 even though X and Y are isomorphic, so we cannot show that r is non Borel in the same way as we did for the Banach-Mazur distance.

We don’t know the answer for the problem above, but if we simplify (a lot) this problem we can easily ﬁnd a positive solution. Precisely, let Subs(`2) = {X ⊂ `2|X is a closed subspace} and endow Subs(`2) with the Eﬀros-Borel structure. Analogously as we had before, we obtain that Subs(`2) is a standard Borel space.

2 Theorem 93. Let r : Subs(`2) → [0, 1] be the operator opening on the subspaces of `2. Then r is Borel.

Proof. As `2 is a Hilbert space we have that r(X,Y ) = Θ(X,Y ), for all X,Y ∈ Subs(`2) (see [O]). As Θ is Borel the result follows.

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98 “Excuse me, I can’t stand up.”

Groucho Marx’s epitaph.