L’INSTITUTFOURIER DE ANNALES

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A L

E S

E D L ’IN IT ST T U

F O U R

ANNALES

DE

L’INSTITUT FOURIER

Spiros A. ARGYROS & Theocharis RAIKOFTSALIS

The cofinal property of the reflexive indecomposable Banach spaces Tome 62, n^o1 (2012), p. 1-45.

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THE COFINAL PROPERTY OF THE REFLEXIVE INDECOMPOSABLE BANACH SPACES

by Spiros A. ARGYROS & Theocharis RAIKOFTSALIS

Abstract. — It is shown that every separable reflexive Banach space is a quotient of a reflexive hereditarily indecomposable space, which yields that every separable reflexive Banach is isomorphic to a subspace of a reflexive indecomposable space. Furthermore, every separable reflexive Banach space is a quotient of a reflexive complementably`p-saturated space with 1< p <∞and of ac0saturated space.

Résumé. — On démontre que tout espace de Banach séparable réflexif est quotient d’un espace réflexif héréditairement indécomposable, ce qui implique que tout espace de Banach séparable réflexif est isomorphe à un sous-espace d’un espace ré- flexif indécomposable. De plus, tout espace de Banach séparable réflexif est quotient d’un espace réflexif complémentablement`p-saturé, où 1< p <+∞, et d’un espace c0-saturé.

1. Introduction

An infinite dimensional Banach space X is said to be indecomposable if it is not the topological direct sum of two infinite dimensional subspaces. In the 70^s J. Lindenstrauss [17] had asked if every infinite dimensional space is decomposable. Note that it was already known that the aforementioned problem has a positive answer for the members of a variety of classes of Banach spaces. For example, Banach spaces with an unconditional basis, nonseparable reflexive spaces [16] (or more generally nonseparable WCG spaces [2]), separable Banach spaces containingc0[22] are all decomposable spaces.

On the other hand since 1991 it is known that Lindenstrauss’ problem has an emphatically negative answer. Indeed W.T. Gowers and B. Maurey’s

Keywords:Banach space theory,`psaturated, indecomposable spaces, hereditarily indecomposable spaces, interpolation methods, saturated norms.

Math. classification:46B03,46B06,46B70.

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discovery of Hereditarily Indecomposable (HI) spaces ([13]) has provided examples of Banach spaces with no decomposable infinite dimensional subspace. Since the seminal work of Gowers and Maurey the classes of HI and Indecomposable spaces have been extensively studied leading to some remarkable results. In particular, new techniques have been developed concerning the existence of HI spaces having as a quotient a desired Banach space. These techniques follow two distinct directions.

The first one, which appeared in [4], is closely related to the DFJP interpolation method ([10]) and makes heavy use of the geometric aspect of thin sets, which can be traced back to A. Grothendieck’s work ([14]) and was explicitly defined in R. Neidinger’s PhD thesis ([20]). This method yielded that every reflexive space with an unconditional basis has a subspace which is a quotient of a reflexive HI space. In particular, separable Hilbert spaces and more generally any `p for 1 < p < ∞ are quotients of reflexive HI spaces. Using duality arguments, one may also conclude that reflexive `_p spaces can be embedded into a reflexive indecomposable.

The second method is based on saturated and HI extensions of a ground norming set and led to the most general result concerning quotients of HI spaces. Namely, as is shown in [9], every separable Banach space not containing`1 is a quotient of a separable HI space. Comparing the aforementioned techniques, one should point out that the second leads to more general results but by its own nature the dual of the resulting HI space is decomposable. Thus, for a given separable reflexive space X the corresponding HI spaceY which hasX as a quotient is never reflexive. On the contrary, whenever the first method is applicable it leads to HI spaces with structure similar to the starting one (i.e., starting with a reflexive space the obtained HI space remains reflexive).

The aim of the present work is to prove the following:

Theorem 1.1. — LetX be a separable reflexive space then, (1) X is a quotient of a reflexive HI space.

(2) X is isomorphic to a subspace of a reflexive indecomposable space.

Since the dual of a reflexive HI is indecomposable (2) is a direct con- sequence of (1). The proof of the theorem is based on a combination of the aforementioned methods and uses certain auxiliary spaces which are constructed either by interpolation or extension. More precisely, starting with a reflexive spaceX with a Schauder basis first we define a spaceX₀ with a Schauder tree basis (et)_t∈T, a weakly compact symmetric subsetW ofX₀ and a map Φ :X₀ →X such that Φ(W) is ¹₈ dense in the unit ball ofX, which implies thatX is a quotient ofX0. The definition ofX0 shares

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common features with the corresponding one in [4], however it requires certain modifications as in the present case the basis ofX is not necessarily unconditional. It is worth mentioning that if we were able to show thatW is a thin or (an)n-thin set (cf. [4]) ofX0 then applying a HI interpolation on the pair (X₀, W) we would arrive to the reflexive HI spaceXwhich has X as a quotient. This remains unclear and we proceed as follows.

In the second step using DFJP `2 interpolation on the pair (X0, W) we obtain a reflexive spaceX₁ with a Schauder tree basis (˜e_t)t∈TK, whereTK

is a complete subtree ofT and a bounded closed convex set ˜W such that W˜ = J₁⁻¹(W). Here, J₁ : X₁ → X₀ is the usual operator mapping the diagonal space to the original one. Note that the composition operator Φ◦J₁ maps ˜W onto a ¹₈ dense subset ofBX and thus X is a quotient of X1. As in the case of X0 it remains unclear whether the set ˜W is a thin subset ofX₁. SinceX₁is a separable reflexive space there exists a countable ordinalξsuch that every weakly null sequence in X1 does not admit a `¹_ξ spreading model (cf.[5]).

The next step is the most critical. Here, using a ξ-saturated extension method ([9]) we pass to a new space denoted as Xξ and Iξ : Xξ → X₁ a bounded linear injection such that the set Wξ = I_ξ⁻¹( ˜W) is a weakly compact and also thin set. Let us note that the structure ofXξresembles the generalized Tsirelson spaceT_ξ (cf. [9]). In that senseX_ξ has a much richer local`1structure thanX1. Thus the thinness is established in a space with a strong presence of local`₁ structure which a-priori seems contradictory or at least peculiar. The final step is the expected one. Namely, we apply a HI interpolation on the pair (Xξ, Wξ) to obtain a diagonal reflexive spaceX and a bounded convex set ˜Wξ such that the natural operatorJξ :X→Xξ

satisfiesJξ( ˜Wξ) =Wξ. As the setWξ is thin the space Xis HI and is the desired one. Indeed the operatorQ= Φ◦J₁◦I_ξ◦J_ξ maps the set ˜W_ξ to a

1

8 dense subset ofX which yields thatQ:X→X is a quotient map.

The paper is organized as follows. Section 2 concerns preliminaries. Sec- tion 3 is devoted to the definition of the spaceX0 which as was mentioned has a Schauder tree basis (et)_T equipped with a partial form of uncondi- tionality defined as “segment-complete unconditional” tree basis. (Def. 3.1).

The main result of this section is that whenXis reflexive although the space X₀ is not necessarily reflexive the setW is weakly compact. In section 4 we prove the following:

Theorem 1.2. — Let (X,k · k) be a reflexive space with a segment complete unconditional tree basis(e_t)_t∈TandKbe a bounded subset ofX such that forx∈K, suppxis a segment of T. Let alsoξ < ω1 such that

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X does not admit a`¹_ξ spreading model. Then, there exists a norm k · kξ

onc00(T)such that setting Xξ to be the completion of(c00(T),k · kξ)the following hold:

(1) For everyx∈c00(T)kxk6kxkξ. (2) For everyx∈K,kxk=kxkξ.

(3) Denoting byWξ the setco(K∪−K)inXξwe obtain that it is weakly compact and thin.

This theorem provides a tool for constructing thin sets in spaces with a Schauder tree basis. The norm of the spaceX_ξ is defined via a norming set Gξ which contains as ground set a norming subset of the dual of the space Xand which isS_ξsaturated for finite sequences of functionals with pairwise incomparable, segment complete supports. In section 5 we show that the diagonal space in the DFJP `2 interpolation applied on the pair (X0, W) has a segment complete unconditional tree basis. In section 7 combining the results of sections 3, 4 and 5 we prove the following:

Theorem 1.3. — Let Y be a separable reflexive space. Then for every p∈ (1,∞) there exists a reflexive space Xp such that every subspace contains an isomorphic copy of`_p, complemented in the whole space and Xp has Y as a quotient. Additionally, Y is a quotient of a separable c0

saturated space.

It is worth mentioning that the first result in this direction was done by D.H. Leung in [15] proving that every separable Hilbert space is a quotient of ac0 saturated space and it was followed by the results in [4] mentioned earlier. More recently, following different techniques, I. Gasparis in [11]

and [12] has shown that certain members of the class of separable reflexive spaces are quotients of c0 saturated spaces. Let us also point out that Theorem 1.3 provides examples of reflexive Banach spaces with divergent structure between the spaces and their duals. For example, there exist spacesX_p as above such that their dual contains HI subspaces.

In Section 7 we present a variant of the HI interpolation method appear- ing in [4] which is traced to [3]. The necessity for modifying the initial HI interpolation method is the requirement that the diagonal space admits a Schauder basis. Similarly to [4] applying the new HI interpolation to a pair (X, W) withW being convex, symmetric, weakly compact and thin subset ofXwe obtain that the diagonal is reflexive HI and this proves Theorem 1.1 in the case whereX has a Schauder basis. The general case of a separable reflexive space mentioned in Theorem 1.1 follows by the classical result of

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M. Zippin that every separable reflexive space embeds into a reflexive space with a Schauder basis ([23]).

The research included in the present paper was carried out in 2006. In April 2007 Richard Haydon visited us in Athens and with his collaboration we were able to prove that there exists an indecomposable space X containing`1. After the solution of the "scalar plus compact" problem ([6]) the aforementioned result was adapted to theL_∞ frame as follows:

Theorem 1.4. — There exists aL_∞spaceX with the scalar plus compact property containing`₁.

As it is also mentioned in [6] the ultimate problem concerning the cofinal properties of Indecomposable Banach spaces is the following:

Problem. — Does every separable Banach space not containingc₀em- bed into a separable indecomposable space?

We made an effort to make the present paper as self contained as possible.

Thus, except for a few technical or well known results all the other proofs are included.

2. Preliminaries

Let us recall some standard notation and definitions for trees.

Notation 2.1.

1. Let Λ be a countable set. By [Λ]^<ωand [Λ] we denote its finite and infinite sequences respectively. We consider [Λ]^<ω to be equipped with the partial ordering of the initial segment denoted byv.

2. By a tree on Λ we mean a subset T of [Λ]^<ω which is backwards closed underv.

3. Let T be a tree on Λ. A segment of T is a subset of T of the form {t ∈ T : t1 v t v t2} with t1, t2 ∈ T. We will usually denote segments of this form by [t₁, t₂] or more generally bys. Fort₁∈T we denote by ˆt1the set{t∈T :tvt1}. For a segments⊂T, ˆshas a similar meaning, namely ˆs={t∈T :∃ t⁰ ∈ssuch that tvt⁰}.

Fort ∈T we set |t| the cardinality of the set {t⁰ ∈ T : t⁰ vt} to be the height of t. For every n ∈ N the n^th-level of T is the set {t∈T :|t|=n}.

4. We identify the branches of a tree T with the elements of the set {(a_i)^∞_i=1: (a_i)^k_i=1∈T,∀k∈N} and we denote this set by [T]. For everyb∈[T] withb= (ai)^∞_i=1 we setb|n= (a1, ..., an).

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5. Two nodest₁, t₂∈T are called comparable if eithert₁vt₂ ort₂v t1. More generally, ifA1, A2⊆T thenA1, A2are called comparable if there existt₁∈A₁andt₂∈A₂which are comparable. Otherwise A1, A2 are called incomparable. We will writeA1 ⊥A2 to denote the fact thatA₁, A₂ are incomparable.

6. Fort∈T byOtwe denote the setOt= {b∈[T] :∃n∈Nsuch that b|n=t}. The sets {Ot:t∈T} form the usual basis of the topology of [T].

7. For everyt∈T andb∈[T] we will writet∈b if∃n∈Nsuch that b|n=t. For every segment sof T andb∈T we will writes⊆b if

∀t∈sit holds t∈b.

In the following sections all trees are considered countable, finitely splitting and with nonempty bodies. For the next definition we assume that Λ is a totally ordered set which later is going to be a subset of the real numbers with the natural order.

Definition 2.2. — For every such treeTwe fix a bijectionh_T: T7→N such that the following hold:

i. h_T(t₁)< h_T(t₂), whenever|t₁|<|t₂|.

ii. Ift1, t2∈Tand|t1|=|t2|i.e.,t1= (a1, ..., an)andt2= (b1, ..., bn) thenh_T(t₁)< h_T(t₂), wheneveran < bn.

When the tree T is understood we will refer to hT simply as h. We denote by c₀₀(T) the linear space of all functions f : T 7→ R such that supp(f) ={t∈T :f(t)6= 0}is a finite set. We also denote by (et)t∈Tthe standard Hamel basis ofc00(T) consisting of the characteristic functions of all singletons{t} ⊆T.

Definition 2.3. — Let(Ai)^∞_i=1 be a sequence of finite subsets ofT. We will say that(Ai)^∞_i=1 is

i. a block sequence ifmax{h(t) :t∈Ai}<min{h(t) :t ∈Ai+1} and we will writeA1< A2< ... < An < ...

ii. a level-block sequence if max{|t| : t ∈ Ai}< min{|(t| : t ∈ Ai+1} and we will writeA₁≺^lA₂≺^l...≺^lA_n≺^l....

iii. For a sequence(fi)^∞_i=1of elements ofc00(T)we will say that(fi)^∞_i=1 is block (level-block) if(suppf_i)^∞_i=1 is a block (level-block, respectively) sequence of subsets ofT.

For the sake of simplicity of notation if Ais a subset of T we will write minAfor min{h(t) :t∈A} and ifAis finite maxAfor max{h(t) :t∈A}.

We will also write min^lA for = min{|t| :t ∈ A} and max^lA for max{t : t∈A} .

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Definition 2.4. — LetA, I be subsets ofT.

i. We will callAsegment complete if for everyt1, t2in A, witht1vt2

[t₁, t₂]⊆A.

ii. Iwill be called an interval ofTifh(I) ={h(t) :t∈T} is an interval ofN.

We note that every intervalI of T is segment complete.

Definition 2.5. — For every f ∈ c00(T) we define ranf to be the minimal intervalIofTsuch thatsuppf ⊆I. Similarly, we setran^lf to be the minimal intervalI^l ofT of the formI^l ={t∈T :m 6|t|6M} such thatsuppf ⊆I^l.

Remark 2.6. — It is clear that a sequence (f_i)^∞_i=1 inc₀₀(T) is i. block if ranfi<ranfi+1 ∀i∈N

ii. level-block if ran^lf_i<ran^lf_i+1 ∀i∈N.

3. Tree representation of the ball of a Banach space Let X be an arbitrary Banach space with a bimonotone, normalized Schauder basis (xi)_i∈Nand (x^∗_i)_i∈Nthe biorthogonal functionals of (xi)_i∈N in X^∗. In this section we define a tree T and, for a given X, a norm on c00(T) that will help us “spread” along the branches ofT a set K which is isometric (via a map Φ to be defined later) to a ¹₈-net in the unit ball ofX. We will denote byX0the completion of c00(T) with respect to this norm.

This technique gives X as a quotient of X0. In addition we show that if X is reflexive then the set K is weakly compact inX₀. We start with the following general definition.

3.1. The spaceX0

Definition 3.1. — LetT be a tree and a normk · kdefined onc00(T) such that the sequence(e_t)_t∈T is a Schauder basis, with the order induced by the functionh, for the completion ofc00(T)denoted byXT. Then

1. The normk · kand the basis(e_t)_t∈T will be calledSC-unconditional if for everyA⊆T segment complete andx=P

t∈Tλtet∈XT we have:

X

t∈A

λ_te_t 6

X

t∈T

λ_te_t .

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2. Let ψ : T → [−1,1] be a function assigning to each node of T a scalar ψ(t) ∈[−1,1]. Let alsoC > 0. We consider for each t ∈ T the vectory_t=P

t⁰vtψ(t⁰)e_t⁰ and set K¹_ψ ={y_t∈X_T :ky_tk6C}, K²_ψ={y_t|I:ky_tk6C andIis an interval ofT} andK_ψ= K²_ψ^k·k. For the rest of this section we assume thatXis a fixed Banach space with a Schauder basis (x_i)_iand (x^∗_i)_ithe biorthogonal functionals inX^∗. We pass on to define aSC-unconditional norm onc00(T) whereT is an appropriately defined tree such that the completion of this space with respect to this norm has a quotient isomorphic toX. We start with the definition ofT.

Definition 3.2. — Let(Fn)^∞_n=1 be the following sequence in([−1,1]∩ Q)^<ω

Fn=n

± i

8ⁿ : 16i68ⁿo

∀n∈N. We set

T ={(a1, a2, ..., ak) :ai∈Fi, i6k, k∈N}.

It can be readily seen that T is a countable, finitely splitting tree such that everyt∈ T with|t|=n, has 2·8ⁿ⁺¹immediate successors.

The norming set G0(X) ofX0 is defined as follows:

Definition 3.3. — LetG¹₀(X)be the following subset ofc₀₀(T) G¹₀(X) =nXⁿ

i=1

a_iX

|t|=i

e^∗_t :k

n

X

i=1

a_ix^∗_ikX^∗61o .

Set

G0(X) ={f|A:f ∈G¹₀(X)and Ais a segment complete subset ofT } wheref|_A denotes the restriction off onA.

We consider the norm onc00(T) induced by the set G0(X). Namely,

∀x∈c₀₀(T) we setkxk= sup{f(x) :f ∈G₀(X)}.

The spaceX0is the completion ofc00(T) under the norm defined above. It can be readily verified that the sequence (e_t)_t∈T (enumerated throughh) becomes a bimonotone, normalized Schauder basis ofX0. We also have the following easy observation:

Remark 3.4. — For every A ⊆ T segment complete the natural pro- jectionP_A : X₀ 7→ X₀ defined by P_A(P

t∈Tλ_te_t) = P

t∈Aλ_te_t has norm one.

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We will need the following Lemma that gives a description of the point- wise closure of the norming set G0(X).

Lemma 3.5. — Let G=n

f|A:f =

∞

X

i=1

ai

X

|t|=i

e^∗_t,

∞

X

i=1

aix^∗_i _X_∗61

andA is a segment complete subset ofTo where all limits are taken with respect to thew^∗-topology. ThenG0(X)^p= G.

Proof. — It is easy to see thatG⊆G₀(X)^p. Letf ∈G₀(X)^p. Then there exists a sequence (fn)n in G0(X) such thatfn

→p f. Eachfn is of the form fn=XA_n·(Pk_n

i=1aⁿ_i P

|t|=ie^∗_t), whereAn are finite segment complete subsets ofT, XA_n is the characteristic function ofAn andkPkn

i=1aⁿ_ix^∗_ik61.

Let gn = Pk_n

i=1aⁿ_ix^∗_i. Then there exists a M ∈ [N] such that the sequence (g_n)_n∈M converges w^∗ to a g ∈ B_X^∗. Let i ∈ N. Denote by a_i the limit limn∈Maⁿ_i and setA= lim infn∈MAn which can be readily seen to be a segment complete subset of T. Considering the functional ˆf = XAP∞

i=1aiP

|t|=ie^∗_t, we claim that the sequence (fn)n∈M convergesw^∗to f. Indeed, letˆ t∈ T with|t|=i. Ift∈A, then lim_n∈Mf_n(e_t) =a_i= ˆf(e_t).

Assuming thatt /∈A then lim_n∈Mfn(et) = ˆf(et) = 0. Hence, ˆf =f ∈G

and the proof is complete.

We pass on to define the map ψ : T → N that will give us the corresponding set Kψ

3.2. The set K

The set K and the map Φ are defined as follows:

Definition 3.6. — Let ψ : T :→ N be the following assignment. For everyt= (a1, a2, ..., an)∈ T we setψ(t) =at=an andyt=P

t⁰vtat⁰et⁰. We setK = K_ψ as in Definition 3.1 with constantC= 1 + ¹₇.

Definition 3.7. — We consider a map Φ : X0 7→ X defined as Φ(P

t∈T λtet) =P∞ i=1(P

|t|=iλt)xi.

Remark 3.8. — We can see that Φ is a bounded linear operator with kΦk 61. In addition, for b ∈[T] if we denote by Xb the subspace Xb =

< et:t∈b >^k·kthen we have that Φ restricted toXb is an isometry.

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We also have the following

Proposition 3.9. — The setΦ(K)is a ¹₈-net in the unit ballBX ofX. Moreover, the mapΦis onto.

Proof. — Let x=P

nbnxn ∈ BX. Since the basis of X is bimonotone we have that |b_n| 6 1, for all n ∈ N. For each n ∈ N we can choose a_n in the set Fn = {±₈ⁱn : 1 6 i 6 8ⁿ} such that |bn −an| 6 8¹ⁿ. If we setσ= (a_n)^∞_n=1 ∈[T] theny_σ =P

na_ne_σ|n ∈K and kΦ(y_σ)−xk_X 6 ¹8. Indeed, if we letXσ=< et:t∈σ >, then by Remark 3.8 Φ :Xσ7→Xis an isometry. Thus if we denote the restriction of Φ onXσ by Φσ we have that Φ⁻¹_σ (P

nb_nx_n) = P

nb_ne_σ|n and kP

nb_ne_σ|nkX0 = kΦ(P

nb_ne_σ|n)kX = kP

nbnxnkX61. Thus,

X

n

ane_σ|n _X

0

6

Xbne_σ|n _X

0

+X

n

|an−bn|61 +1 8. This gives us thatyσ is an element of K. Finally,

X

n

bne_σ|n−X

n

ane_σ|n _X

0

= Φ X

n

bne_σ|n

−Φ X

n

ane_σ|n _X

=kΦ(y_σ)−xk_X 61 8.

The following Lemma shows the behavior of incomparably supported sequences of vectors in K if we assume thatX has a shrinking basis.

Lemma 3.10. — Suppose(xi)iis a shrinking basis forX, then for every sequence(yn)n in Ksuch that (suppyn)n are finite and mutually incomparable subsets ofT we have thatyn

→w 0.

Proof. — Let (y_n)_n be as above. In order to prove that (y_n)_n is weakly null it is enough to show thatf(yn)→0,∀f ∈G0

p. Choosef ∈G0 p. By Lemma 3.5 there exist a g =P∞

i=1bix^∗_i ∈ BX^∗ and a segment complete A ⊂ T so that f = XAP∞

i=1b_iP

|t|=ie^∗_t. Let s^A_n = suppy_n∩A, z_n = y_n|_A=P

t∈s^A_na_te_tand observe the following, f(y_n) =

∞

X

i=1

b_iX

|t|=i

e^∗_t(z_n) =g(Φ(z_n)).

Hence, it is enough to show thatg(Φ(z_n))→0. As Φ(z_n) =P

t∈s^A_na_tx_|t|

it can be seen that for i ∈ N, x^∗_i(Φ(zn)) = at, if sÂ_n ∩Li 6= ∅ and zero otherwise. SinceT is finitely branching andsÂ_n ⊥sÂ_mfor alln6=m∈N, we deduce that for a fixedi∈Nthe set{n∈N:sÂ_n ∩Li6=∅}is finite. Thus,

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x^∗_i(Φ(z_n))→0. Finally, as (x_i)i is shrinking we obtain thatg(Φ(z_n))→0

and this completes the proof.

Proposition 3.11. — For every reflexive Banach space X with a bimonotone normalized Schauder basis(x_i)_i the setKis weakly compact.

Proof. — Let (wn)n be a sequence in K. Up to an arbitrarily small perturbation we may assume that suppwn is finite for all n ∈ N. We set sn = suppwn and we observe that each sn is a finite segment of T. So, w_n = P

t∈s_na_te_t. We may also assume (by passing to a subsequence if needed) that for each t ∈ T, e^∗_t(wn) → w(t) ∈ R. We set S ={t∈ T :w(t)6= 0}. We know that S is a segment ofT (finite or infinite). Thus, we may assume that eachyn has a decompositionyn=un+yn

where

i. (suppu_n)_nis an increasing (with respect to⊆) sequence of segments ofS

ii. (suppy_n)_n is a sequence of incomparable segments ofT. We setws=P

t∈SatetifS 6=∅and 0 otherwise. We claim thatwn

→w ws. Lemma 3.10 yields that (yn)n is weakly null. To finish the proof we shall show that u_n ^k·k→ w_s. Indeed, for every n ∈ Nwe have that suppu_n v S and sincewn∈K we have thatun∈K for alln∈N. Thus, as the basis of X is boundedly complete,u_n→^p w_sandkΦ(un)kX =kunkX0 we have that ws∈K andun

k·k→ws.

Remark 3.12. — The connection between the set K andBX whenX is an arbitrary Banach space with a basis is not completely clear to us. For example, if one considersX to bec0with the summing basis then it turns out that K contains a sequence equivalent to the standard`¹basis. Indeed, we notice that the norming set in this case becomes G₀^p ={±XA :A ⊆ T segment complete} where by XA we denote the characteristic function ofA. We can construct a sequence (w_n)_nin K which has no weakly Cauchy subsequence. To see this choose a sequence (tn)n ⊂ T with the following properties,

i. Eacht_n is of the formt_n= (a₁, ..., a_k_n) anda_k_n =a_t_n= ¹₂; ii. tn⊥tm for alln6=m∈N.

For each n ∈ N set t⁰_n = tn a ⁻¹2 = (a1, ..., ak_n,⁻¹₂ ) and wn =at_net_n+ a_t⁰

ne_t⁰

n. To see that (w_n)_n satisfies the desired property, choose a subsequence (wm_i)_i∈N. Let t(i) = tm_i when i = 2k and t(i) = t⁰_m_i when i= 2k−1. It is clear that (t(i))_iare mutually incomparable. Therefore, the setA=∪^∞_i=1{t(i)} is segment complete. So the functionalf =XA ∈G0

p

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estimates|f(w_m_i)−f(w_m_i+1)|= 1 for alli. By Rosenthal’s`¹theorem [21]

we obtain that (wn)n is equivalent to the`¹ basis.

4. Thin subsets of Banach spaces

LetTbe a tree andk·kX_T be aSC-unconditional norm defined onc00(T).

Denote byXT the completion ofc00(T) with respect to k · kX_T. Fix also a functionψ :T →[−1,1] and Kψ (referred to as K for simplicity) as in Definition 3.1.

In this section we present a general method for extending the norm of XT to a new norm defined onc00(T) such that the completion of this space contains K as athin subset. Namely, the entire section is devoted to the proof of the following theorem:

Theorem 4.1. — Suppose thatXT is reflexive andKis a weakly compact subset of XT. Then there exists a space Xξ such that the following hold:

1. The identity operatorI:Xξ→XT is continuous.

2. K⊂I(Xξ) and the closed convex hull of(I⁻¹(K)∪I⁻¹(−K)) is a weakly compact and thin subset ofXξ.

The notion of thinness was introduced in [19] and was extensively used in [4] where several methods for proving that a set satisfies this property were developed. We give the corresponding definition in subsection 4.3 where we also prove the aforementioned Theorem. Before doing so though we need some preparatory work which is done in the following subsection.

4.1. Tsirelson type spaces and norms

We start with some preliminary results concerning families of finite subsets ofN and Tsirelson type norms. Most of these results are well known and have been extensively used in the relevant literature (see [5] and the references therein), with the exception of Lemmas 4.6, 4.9, 4.11 and Re- mark 4.7 which can be found in [18] and were brought to our attention by the authors. We include this subsection in order to make the text as self-contained as possible. We start by recalling the following notions concerning families of finite subsets ofN.

Definition 4.2. — LetMbe a family of finite subsets ofN.Mis called

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i. Compact if the set of characteristic functions{XA : A ∈ M}is a compact subset of{0,1}^N.

ii. Hereditary if for everyA∈ MandB⊆Awe haveB ∈ M.

iii. Spreading if for everyA={t1< t2< ... < tr} ∈ MandB ={t⁰₁<

t⁰₂< ... < t⁰_r}withti6t⁰_i ∀i= 1, ..., r we haveB ∈ M.

Definition 4.3. — LetM ⊆[N]^<ω.

i. A finite sequence(E₁, ..., E_n)of successive and finite subsets ofNis calledM-admissible if there existsF ∈ M withF ={m1< m2<

... < m_n} such thatm₁6E₁< m₂6E₃< ... < m_n6E_n.

ii. A finite sequence(f1, ..., fn)of vectors inc00(N)is calledM-admissible if(suppfi)ⁿ_i=1 isM-admissible.

Definition 4.4. — Let F,G be two families of finite subsets of N we define:

i. The block sumF ⊕ G ={M∪N:M < N, M ∈ G, N ∈ F }.

ii. The convolution F ⊗ G = {∪ⁿ_i=1Fi : F₁ < ... < Fn, Fi ∈ F, i = 1, ..., nand{minFi}ⁿ_i=1∈ G}.

Definition 4.5. — The Schreier hierarchy was first defined in [1]. It is a set of families (Sξ)ξ<ω₁ of finite subsets of N which can be defined recursively as follows:

S₁={F ⊂N: #F 6minF} ∪ {∅}

Letξ < ω₁and suppose thatSξ have been defined for allζ < ξThen i. Ifξ=ζ+ 1we setSξ =Sζ⊗ S1.

ii. Ifξ is a limit ordinal then we fix a strictly increasing sequence of non-limit ordinals(ξn)n withsupξn = ξand set Sξ =S∞

n=1{F ∈ S_ξ_n:F>n}.

It can be verified by transfinite induction that each Sξ for ξ < ω1 is compact hereditary and spreading. We will need the following two results found in [18] concerning the families (Sξ)ξ<ω₁ which can be proved by transfinite induction.

Lemma 4.6. — For every ordinalξ < ω1andM ∈[N]we have i. [M]⁶³⊗ Sξ ⊆ Sξ⊗[M]⁶².

ii. IfminM >3, then [M]⁶³⊗(Sξ⊕[M]⁶¹)⊆ Sξ⊗[M]⁶³. Remark 4.7. — By Lemma 4.6 we have that∀M ∈[N] and ξ < ω₁

[M]⁶⁸⊗(Sξ⊗[M]⁶²)⊆([M]⁶³⊗([M]⁶³⊗ Sξ))⊗[M]⁶²

⊆(Sξ⊗[M]⁶⁴)⊗[M]⁶²⊆ Sξ⊗[M]⁶⁸.

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The Spaces T(θ,F).Let 0 < θ < 1 and F be a compact hereditary family of finite subsets ofN.

Definition 4.8. — LetG_θ,Fbe the minimal subset ofc00(N)such that i. ±e_n∈G_θ,F ∀n∈N;

ii. G_θ,F is closed under the (θ,F)-operation. That is if (fi)^d_i=1 is an F-admissible family inc₀₀(N)thenθPd

i=1f_i ∈G_θ,F .

The spaceT(θ,F)is the completion ofc00(N)under the following norm

∀x∈c00(N) we set kxk_(θ,F)= sup{f(x) :f ∈Gθ,F}.

Detailed expositions of the T(θ,F) type spaces can be found in [8]. In the sequel we shall denote Tξ = T(¹₂,Sξ), T_ξ¹ = T(¹₂,Sξ ⊗[N]⁶²) and T_ξ² =T(¹₂,Sξ⊕[N]⁶¹). The following two Lemmas are results in [18] but for the sake of completion we include their proofs here.

Lemma 4.9. — Let ξ < ω1. Then for every finite sequence (bi)^k_i=1 of scalars we have

I. kPk

i=1bieik_T1

ξ 68kPk

i=1bieikT_ξ

II. kPk

i=1bieik_T2

ξ 63kPk

i=1bieikT_ξ

where by(ei)_i∈N we denote the standard Hamel basis ofc00(N).

Proof. — I. By Remark 4.7 we have that [N]⁶⁸⊗(Sξ⊗[N]⁶²)⊆ Sξ⊗[N]⁶⁸. LetG_T1

ξ be the norming set of T_ξ¹ and let f ∈ G_T1

ξ. We will define g₁ <

... < g_lwithl68 andg_i∈ G_T_ξ, i= 1, ..., lsuch thatf =Pl

i=1g_i. We use induction on the complexity off. Let f =±en for somen∈N,then there is nothing to prove. Letf = ¹₂Pd

i=1 such that i. f1< ... < fd.

ii. (fi)^d_i=1 is anSξ⊗[N]⁶²admissible sequence.

iii. For everyf_ithere exists a sequenceg₁ⁱ < ... < g_lⁱ

isuch thatgⁱ_j∈G_T_ξ forj = 1, ..., li andfi=Pl_i

j=1g_jⁱ.

Now since{minf_i}^d_i=1∈ S_ξ⊗[N]⁶²by Remark 4.7 thatSd

i=1{ming_jⁱ :j6 l_i} ∈ Sξ⊗[N]⁶⁸. Thus there existB₁, ..., B_k withk68 such that

1. Bm∈ Sξ for allm= 1, ..., kand 2. B₁< ... < B_k

such thatSd

i=1{ming_jⁱ:j6li}=Sk

m=1Bm. By settingg^(m)=¹₂(P

ming_jⁱ=∈Bm

g^j_i) we get a. f =Pk

m=1g^(m)and b. g^(m)∈GT_ξ

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as desired.

II. By Lemma 4.6 we have [N]⁶³⊗(Sξ⊕[N]⁶¹)⊆ Sξ⊗[N]⁶³. Using the same arguments as in the proof of I. we conclude that for everyf ∈G_T2 ξ

there exist g1 < g2 < g3 with gi ∈ GT_ξ for i = 1,2,3 such that f = P3

i=1gi.

Definition 4.10. — LetM ∈[N]withM ={m₁< m₂< ...}.

i. For everym∈M we setm⁺ to be the immediate successor ofmin M, that ism⁺_i =m_i+1for alli∈N.

ii. IfA∈[M]⁶^ωthen we set A⁺={m⁺ :m∈A}.

Lemma 4.11. — Letξ < ω1 andM ∈[N] withM ={m1< m2< ...}.

Then for every finite sequence(bi)^k_i=1 of scalars we havekPk

i=1biem_ik6 kPk

i=1bie_m+

ik 6 3kPk

i=1biem_ik where all norms are considered in the spaceTξ.

Proof. — LetA∈[M]⁶^ω. Suppose thatA⁺={m⁺:m∈A} ∈ Sξ. Since Sξ is a spreading family it follows that A\minA ∈ Sξ. Thus, A ∈ Sξ⊕ [M]⁶¹. So, if we consider f ∈GT_ξ with the property suppf ⊆ {m⁺_i :i= 1, ..., k} there is an f⁰ ∈G_T2

ξ such that f(Pk

i=1b_ie_m+

i) 6f⁰(Pk

i=1b_ie_m_i) and this gives

i. kPk

i=1b_ie_m+

ikTξ6kPk

i=1b_ie_m_ik_T²

ξ.

On the other hand sinceSξ is spreading it is easily verified that ii. kPk

i=1biem_ikT_ξ 6kPk

i=1bie_m⁺

ikT_ξ

combining i. and ii. we have

k

X

i=1

biem_i

_T

ξ

6

k

X

i=1

bie_m⁺

i

_T

ξ

6

k

X

i=1

biem_i

_T₂

ξ

and by Lemma 4.9 we get the desired.

4.2. The norming setG_ξ

In this subsection starting withX_T, as in the introductory paragraph of the section, assuming thatXT does not contain an isomorphic copy of`¹we defineX_ξand prove that it satisfies the first two properties ofY mentioned in Theorem 4.1. Namely, we show that the identity mapI : Xξ →XT is continuous and that the setI⁻¹(K) is a weakly compact subset ofX_ξ. We start with some well known results concerning`¹-spreading models.

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Definition 4.12. — A bounded sequence (x_n)n in a Banach space Y is an`¹_ξ-spreading model, forξ < ω1, if there exists a constant C >0such that

X

i∈F

aixi

>CX

i∈F

|ai|

for everyF ∈ Sξ and all choices of scalars(ai)i∈F.

The following is a well known result and for its proof we refer the inter- ested reader to [7].

Lemma 4.13. — If a separable Banach space Y contains`¹_ξ-spreading model, for everyξ < ω1, thenY contains an isomorphic copy of`¹.

As we have supposed, the space XT does not contain `¹ therefore it follows that there isξ < ω1 such thatXT contains no `¹_ξ-spreading model.

We fix this countable ordinalξ < ω₁and we use the following norming set forXT:

G1=n X

t∈F

bte^∗_t :

X

t∈F

bte^∗_t

X_T^∗61 andF⊆T finite and segment completeo .

We also consider a bijectionh:T →Nas in Definition 2.2 and make use of the following piece of notation:

Notation 4.14. — For every sequence (fi)^d_i=1 inc₀₀(T) such that i. (fi)^d_i=1 is block

ii.

min{h(t) :t∈suppf_i} ^d_i=1∈ Sξ

iii. {suppfi}^d_i=1 are incomparable subsets ofT. We will call (f_i)^d_i=1 a (T, ξ)-admissible sequence.

The definition of the norming set is the following

Definition 4.15. — LetG_ξ be the minimal subset ofc₀₀(T)such that 1. G1⊆Gξ

2. G_ξ is closed under the (¹₂,Sξ)-operation on (T, ξ)-admissible sequences. That is, for every(T, ξ)-admissible sequence f1, ..., fd in Gξ we have that ¹₂Pd

i=1fi is an element ofGξ. We define a norm on c₀₀(T) as follows:

For every x∈c00(T) we let kxkX_ξ = sup{f(x) :f ∈Gξ} and set

X_ξ =< e_t:t∈T >^k.k^X^ξ.