Interpolating Hereditarily Indecomposable Banach Spaces

JOURNAL OF THE AMERICAN MATHEMATICAL SOCIETY Volume 13, Number 2, Pages 243–294 S 0894-0347(00)00325-8 Article electronically published on January 31, 2000

INTERPOLATING HEREDITARILY INDECOMPOSABLE BANACH SPACES

S. A. ARGYROS AND V. FELOUZIS

1. Introduction A Banach space X is said to be Hereditarily Indecomposable (H.I.) if for any pair of closed subspaces Y , Z of X with Y ∩ Z = {0}, Y + Z is not a closed subspace. (Throughout this section by the term “subspace” we mean a closed infinite-dimensional subspace of X.) The H.I. spaces form a new and, as we believe, fundamental class of Banach spaces. The celebrated example of a Banach space with no unconditional basic sequence, due to W. Gowers and B. Maurey ([GM]), is the first construction of a H.I. space. It is easily seen that every H.I. space does not contain any unconditional basic sequence. Actually, the concept of H.I. spaces came after W. Johnson’s observation that this was a property of the Gowers-Maurey example. To describe even further the peculiar structure of a H.I. space, we recall an alternative definition of such a space. A Banach space X is a H.I. space if and only if for every pair of subspaces Y , Z and ε>0thereexisty ∈ Y , z ∈ Z with ||y|| = ||z|| =1and||y − z|| <ε. Thus, H.I. spaces are structurally irrelevant to classical Banach spaces, in particular to Hilbert spaces. Other constructions of H.I. spaces already exist. We mention Argyros and Deliyanni’s construction of H.I. spaces which are asymptotic `1 spaces ([AD2]), V. Ferenczi’s example of a uniformly convex H.I. space ([F2]) and H.I. modified asymptotic `1 spaces contained in [ADKM]. Other examples of Banach spaces which are H.I. or which have a H.I. subspace are given in [G1], [H], [OS1]. The construction of such a space requires several steps and it uses two fundamental ideas. The first is Tsirelson’s recursive definition of saturated norms ([Ts]) and the second is Maurey-Rosenthal’s construction of weakly null sequences without unconditional basic subsequences ([MR]). An important ingredient in the Gowers- Maurey construction is the Schlumprecht space. This is a Tsirelson type Banach space which is arbitrarily distortable and has been used in the solution of important problems. Thus beyond its use in the constructions of H.I. spaces it plays a central role in the solution of the distortion problem for Hilbert spaces ([OS]). The essential difference between Schlumprecht and Tsirelson spaces became more transparent in [AD2] where the mixed Tsirelson spaces were introduced. It is natural to expect that H.I. spaces share special and interesting properties not found in the previously known Banach spaces. Indeed, the following theorem is proven in

Received by the editors April 14, 1998 and, in revised form, June 8, 1999. 2000 Mathematics Subject Classiﬁcation. Primary 46B20, 46B70; Secondary 46B03, 52A07, 03E05. Key words and phrases. Interpolation methods, hereditarily indecomposable spaces, thin convex sets, Schreier families, summability methods.

c 2000 American Mathematical Society 243

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[GM]: Every bounded linear operator from a complex H.I. space X to itself is of the form λI + S,whereI denotes the identity operator and S is a strictly singular operator. The same property of L (X, X) remains valid for certain real H.I. spaces like Gowers-Maurey space and Argyros-Deliyanni space. We recall that an operator is strictly singular if its restriction to any subspace is not an isomorphism. As a consequence of this theorem, every H.I. space (real or complex) is not isomorphic to any of its proper subspaces. Thus any H.I. space settles in the negative the famous Hyperplane Problem. Later on, Gowers proved a new dichotomy for Ba- nach spaces ([G]). His theorem states that every Banach space either contains an unconditional basic sequence or it contains a H.I. subspace. We will use this result to prove our dichotomy related to quotients of H.I. spaces. Let us mention that Gowers’ dichotomy leads to a solution of the homogeneous space problem. Also N. Tomczak-Jaegermann solved the distortion problem for H.I. spaces by showing that every H.I. space is arbitrarily distortable ([To]). In the present paper we demonstrate that in spite of the fact that the structure of H.I. spaces is irrelevant to that of spaces with an unconditional basis, still, there are ways to connect these two classes. Thus we show that the class of Banach spaces which are quotients of H.I. spaces is extensive and, further on, large classes of operators between Banach spaces are factorized through H.I. spaces. The structure of the quotients of H.I. spaces has also been studied by Ferenczi in [F1], who showed that every quotient of the Gowers-Maurey space is a H.I. space and, at the same time, there exists a quotient of a H.I. space which is not a H.I. space. The answer to the general question of whether every separable Banach space is a quotient of a H.I. space is negative, due to the lifting property of `1 (N). Indeed, as is well known, every Banach space that has `1 (N)asaquotienthasa complemented subspace isomorphic to `1 (N) and hence is not a H.I. space. The following result shows that `1 (N) is somehow the only exception. To be more precise we prove the following dichotomy: Theorem 1.1. Every Banach space X either contains a subspace isomorphic to `1(N) or it has a subspace which is a quotient of a H.I. Banach space. Further on, every member in the family of the so-called, classical Banach spaces not containing `1(N) is actually a quotient of a H.I. space. For example, we show p that separable Hilbert spaces, or even L (λ), 1

For a null sequence a =(an)n∈N of positive real numbers and a bounded convex symmetric subset W of a Banach space X we say that W is an a-thin set if the || || sequence ( n)n of equivalent norms on X, deﬁned by the Minkowski gauges of n the sets 2 W + anBX , is not uniformly bounded on the unit ball of every subspace of X. The second result of this paper concerns factorization of operators and it is the following:

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Theorem 1.2. Let T : X → Y be a bounded linear operator between Banach spaces such that T [BX ] is an a-thin set. Then there exist a H.I. space Z and bounded linear operators F1 : X → Z, F2 : Z → Y such that T = F2 ◦ F1 (i.e. T is factorized throughaH.I.space). As a consequence of this theorem we show that every T ∈L(`p,`q), where p =6 q, p, q ∈ [1, ∞), is factorized through a H.I. space. Additionally, the identity map I : L∞ (λ) → L1 (λ) is also factorized through a H.I. space and so is every strictly singular operator T ∈L(`p,`p) . The next result of the paper refers to the structure of the H.I. spaces. As we mentioned in the beginning, Gowers and Maurey have shown that for X acomplex H.I. space, every T in L (X, X)isoftheformλI + S where S is strictly singular. Further the same holds for some of the existing examples like [GM], [AD2]. It is not known whether a Banach space X such that every T in L (X, X)isoftheform λI + K,whereK is a compact operator, does exist. In general, the construction of strictly singular but not compact operators on a H.I. space does not seem easy. What is already known is a construction, due to T. Gowers ([G2]), of an operator T from a subspace Y of Gowers-Maurey space X to the whole space which is strictly singular and not compact. It is shown here that H.I. spaces with many strictly singular non-compact operators do exist. More precisely we have the following result: Theorem 1.3. There exists a H.I. space X with the property that for every subspace Y of X there exists a strictly singular non-compact operator T ∈L(X, X) with the range of T contained in Y . The last result of the paper that we would like to mention concerns `p-saturated Banach spaces. We recall that a Banach space X is said to be `p-saturated, for p ∈ [1, ∞), if every subspace of X contains a further subspace which is isomorphic p to ` .(Theclassofc0-saturated Banach spaces is similarly deﬁned.) Theorem 1.4. Every reﬂexive Banach space X with an unconditional basis contains a subspace Y with the following property: for every p ∈ [1, ∞) (resp. p =0) there exists an `p-saturated (resp. c0-saturated) space Z which has Y as a quotient. We also show that when X is some Lq(λ), 1 norm is not necessarily unconditional. Thus in the second section we introduce the d- Q∞ product norm of the space Ω00 = Xn ,where(Xn)n is a sequence of Banach n=1 00 spaces. Then for a sequence (|| ||n)n of equivalent norms on a given Banach space X Q∞ and a d-product norm on Ω00 = (X, || ||n) we consider the diagonal space n=1 00

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∆X˜,inthecompletionX˜ of Ω00, consisting of the vectors of the form x =(x, x, ...) such that x ∈ X and x belongs to X˜. The spaces ∆X˜ are the interpolation spaces that we are going to use. In the third section we define and study the key notions of thin and a-thin sets. The concept of thin sets was introduced in Neidinger’s Ph.D. thesis ([N]) and it is defined as follows. A convex bounded symmetric set W in a space X is called a thin set if for every subspace Z of X there exists ε>0 such that for every real λ the set BZ is not contained in the set λW + εBX . In a previous version of the paper we exclusively used the notion of a thin set. Then B. Maurey made the important observation that everything works if we use the concept of an a-thin set, introduced above, instead of a thin set. To explain the difference between these two notions we notice that if W is a thin set, then it is an a-thin set for every null sequence a =(an)n of positive real numbers. Let us point out that in most of the results, concerning thinness of sets, we are only able to establish a-thinness. The free choice of the sequence a is an essential advantage when we are dealing with such problems. Further on, 1 B. Maurey showed us how to prove that BL∞(λ) is an a-thin subset of L (λ)for an appropriate sequence a. Notice that Rosenthal had observed that BL∞(λ) is not 1 a thin subset of L (λ) ([N], [N1]). Therefore BL∞(λ) separates these two classes. We use the a-thin sets in the following manner: Start with a set W which for a sequence a =(an)n is an a-thin set. Denote also by || ||n the equivalent norm n defined by the set 2 W + anBX and suppose that there exists a d-product norm Q∞ on Ω00 = (X, || ||n) which is block H.I. (Definition 3.3). Then the diagonal n=1 00 space ∆X˜ is a H.I. space. Therefore the use of thin or a-thinsetsistheessential tool to construct H.I. interpolation spaces. The next three sections are devoted to the construction of a-thin (or thin) norming sets. We recall that a bounded set W in a Banach space norms a subspace Y of X∗ if there exists a constant C>0 such that for every x∗ in Y , ||x∗|| ≤ C sup {|x∗ (w) | : w ∈ W }. Our goal is, given a member A of a certain class of separable Banach spaces, to construct a corresponding space XA and W ∗ ∗ ∗ an a-thin subset of XA which norms a subspace Y of XA w -isometric to A .It does not seem obvious how to construct such a W even for concrete Banach spaces. However there were indications in the literature for the existence of such sets. We mention a result due to J. Bourgain in Radon-Nikodym theory that describes a similar phenomenon. It is proven in [Bo2] that if K is a bounded convex set in a Banach space and ε>0 such that the convex combinations of slices of K have diameter greater than ε, then the set K norms a subspace of X∗ isomorphic to `1 (N). It was somehow expected that one can construct such sets K, with the additional property that they be thin sets. Our construction concerns spaces A with an unconditional basis.P For such a space A we consider an auxiliary space XA which ⊕ 1 is of the form ` (kn) A and we “spread” the positive part of the unit ball of A on the branches of an appropriate tree in the ball of XA. Thus after defining such a set K,thesetW is the set co (K ∪−K) . As a result of this construction it ∗ ∗ easily follows that W norms a subspace of XA isometric to A . The difficult part is to show that W is an a-thin set. This is not always true. For example if A contains `1 (N), then W is not an a-thin set.

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InthefourthsectionofthepaperweprovethatW is an a-thin set for some special classes of reﬂexive Banach spaces with an unconditional basis, which include the classical spaces Lp (λ), 1

is a quotient of ∆XÃ. The sixth section contains the proof of the fact that the set W , defined for the space c0 (N), is a thin set. The seventh section is devoted to the general construction of block-H.I. d-product norms. In particular we prove that for every sequence (Xn)n∈N of separable Banach Q∞ spaces there exists a d-product norm on the space Ω00 = Xn which is n=1 00 block-H.I. This means that for every block normalized sequence (˜xn)n∈N in Ω00 the Banach space span[(˜xn)n∈N] is a H.I. space. The construction of such a norm is very similar to the corresponding constructions in the already existing examples. For somebody experienced in such constructions it will be clear that it is possible to use either the Gowers-Maurey scheme ([GM]) or the asymptotic `1 scheme ([AD2]) and obtain variations of block-H.I. d-product norms. We follow the scheme presented in the construction of H.I. spaces by Gowers and Maurey with few modifications in the definition of the norm. In the eight section the final results, including the results mentioned above, are presented.

2. Direct products-diagonal spaces In this section we present a generalization of Davis-Figiel-Johnson-Pelczynski’s interpolation method, by extending this to cases where the connecting external norm is not necessarily unconditional. We will use this in the next sections to construct interpolation spaces which are Hereditarily Indecomposable. Q∞ Notation. Let (Xn, || ||n)n∈N be a sequence of Banach spaces, and let Ω = Xn be n=1 their Cartesian product. The support of a vectorx ˜ ∈ Ω, denoted by supp(˜x), is the set of all n ∈ N such thatx ˜(n) =0.WesetΩ6 00 = {x˜ ∈ Ω:|supp(˜x)| < ∞}.The range of a vectorx ˜ ∈ Ω, denoted by range (˜x), is the interval of integers [minsuppx,˜ maxsuppx˜]. For( any A ⊆ N we deﬁne a linear transformation PA :Ω→ Ωby x˜(n)ifn ∈ A PA(˜x)(n)= .WealsosetPn = P{ },Pnc = P{ }, 0ifn/∈ A 1,...,n n+1,...

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∈ N → πn = P{n}, Ex˜ = PE (˜x) . For anyk we denote by ik : Xk Ω00 the natural x if n = k embedding deﬁned by i (x)(n)= . k 0ifn =6 k

Deﬁnition 2.1. A Banach space (X,˜ || ||) is called a d-product of the sequence (Xn, || ||n)n∈N if the following conditions are satisﬁed. 1. Ω00 ⊆ X˜ ⊆ ΩandΩ00 is a dense subset of X˜. 2. ||x||n = ||in(x)||, for every n ∈ N and x ∈ Xn. 3. For any n ∈ N andx ˜ ∈ X,˜ Pn : X˜ → X˜ is a bounded transformation andx ˜ = lim Pnx˜. n→∞

If (Xn, || ||n)n∈N is a sequence of Banach spaces and A is a Banach space with 1-unconditional basis (en)n∈N,then ! ( ) ∞ ∞ ∞ Y Y X ∈ || || ∞ Xn = (xn)n∈N Xn : xn nen < n=1 n=1 n=1 A A ∞ P with norm k(xn)n∈Nk = ||xn||nen is an example of a d-product of the n=1 A sequence (Xn)n∈N. ˜ || || If X is a d-product of (Xn, n)n∈N, then by the uniform boundedness principle there exists C>0 such that ||Pnx˜|| ≤ C||x˜||, for any n ∈ N andx ˜ ∈ X˜. Therefore the sequence (Xn)n∈N is a Schauder decomposition of the space X˜ (for a detailed study of this notion see [LT], pp. 47-52). ∞ The family (Pn)n=1satisfies the following properties: (P1) Pn ◦ Pm = Pmin{m,n},(P2)sup||Pn|| < ∞,(P3) For every x ∈ X, x = n∈N ∞ 0 S lim P x,orequivalently,(P3) X = P (X). →∞ n n n n=1 → ∞ Conversely, if a Banach space X admits a family (Pn : X X)n=1 of linear operators satisfying (P1), (P2), (P3), then the space X is the d-product of the − ∞ || || ˜ family ((Pn Pn−1)(X))n=1 where P0 = 0. The norm of the space X is called: (a) C-bimonotone if there is a C>0 such that ||PA(˜x)|| ≤ C||x˜|| for any interval A ⊆ N and anyx ˜ ∈ X˜. It follows from the definition that every d-product norm is C-bimonotone for some C; (b) unconditional if there is a C>0 such that ||PA(˜x)|| ≤ C||x˜|| for any A ⊆ N and anyx ˜ ∈ X˜; ∈ || || ∞ ∈ ˜ (c) boundedly complete if for anyx ˜ Ω, supn∈N Pnx˜ < implies thatx ˜ X; Pn (d) shrinking if for anyx ˜∗ ∈X˜ ∗,˜x∗ = lim P ∗x˜∗,where P ∗x˜∗(x)= x˜∗(π (˜x)) →∞ n n i n i=1 ∗ =˜x (Pnx˜). ∗ ∞ ˜ Since (Pn )n=1 satisfies (P1)and(P2) the norm of X is shrinking if and only ˜ ∗ ∗ ∞ if X is a d-product of the sequence (Xn)n=1. In the sequel we shall assume that || || is always 1-bimonotone and therefore ||x˜||∞ ≤||x˜|| ≤ ||x˜||1,where||x˜||∞ = P∞ || || || || || || ˜ supn∈N x˜(n) n, x˜ 1 = x˜(n) n. WenoticethatthespaceX is reflexive if n=1 and only if each Xn is reflexive and the norm of X˜ is boundedly complete and shrinking. (The proof is similar to that of the corresponding James’ theorem for spaces with Schauder basis [LT].)

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Deﬁnition 2.2. Let (Xn, || ||n)n∈N be a sequence of Banach spaces such that Xn = X1 and || ||n is equivalent to kk1 for any n ∈ N.LetX˜ be any d-product of the sequence (Xn, || ||n)n∈N.Thediagonal space ∆X˜ of X˜ is the (closed) subspace of X˜ consisting of allx ˜ ∈ X˜ such thatx ˜(n)=˜x(1), for any n ∈ N.

Define J :∆X˜ → X1,I:∆X˜ → X,˜ π1 : X˜ → X1 by J(˜x)=˜x(1),I(˜x)= x,˜ π1(˜x)=˜x(1). Then J = π1 ◦ I and, moreover, it is a 1-1, linear, continuous transformation. If the norm of X˜ is boundedly complete and shrinking, then J has the interesting property that J ∗∗ is 1-1 and it is a Tauberian operator. The latter means that ∗∗ −1 (J ) (X1) ⊆ ∆X˜. We can easily prove that if T : X → Y is a Tauberian operator, then T (W ) is a relatively weakly compact subset of Y iff W is a relatively weakly compact subset of X ([N]). The notion of a Tauberian operator is introduced in [KW] and for a study and characterizations of Tauberian operators see [N], [N1], [NR]. Notation. a) For A, B finite non-empty subsets of N we write Alinear span of a block sequence (˜xk)k∈N. Definition 2.3. a) A Banach space X is called a hereditarily indecomposable space or H.I. space if there is no subspace of X that can be decomposed as a linear topological direct sum of two infinite-dimensional subspaces, or equivalently for any pair of infinite-dimensional subspaces Y ,Z of X and any ε>0thereexist y ∈ Y , z ∈ Z such that ||y − z|| <ε||y + z||. b) A d-product X˜ is called a block-H.I. space if any block subspace of X˜ is an H.I. space.

Proposition 2.1. Let X˜ be any d-product of a sequence (Xn, || ||n)n∈N of Banach spaces such that Xn = X1 and || ||n is equivalent to kk1 for any n ∈ N .IfX˜ is a block-H.I. space and the operator J :∆X˜ → X1 is strictly singular, then ∆X˜ is a H.I. space. Proof. We recall that the space X˜ is endowed with 1-bimonotone norm and let us suppose that ∆X˜ is not a H.I. space. Then there exist Y1,Y2 inﬁnite-dimensional closed subspaces of ∆X˜ such that Y1 ∩ Y2 = {0} and Y1 + Y2 is also closed. This is equivalent to the existence of some δ>0 such that for everyy ˜1 ∈ Y1,˜y2 ∈ Y2, 1 || || ≤ 3 || − || 2 < y˜i 2 , i =1, 2, we have that y˜1 y˜2 >δ. We may also assume that 1 δ<4 . ∞ Claim. There exists a block sequence (˜z ) in X˜ such that setting h i i i=1 h i ∞ ∞ Z1 = span (˜z2j)j=1 ,Z2 = span (˜z2j−1)j=1 ∈ ∈ ∈ ∈ || − || then for everyw ˜1 SZ1 ,˜w2 SZ2 there areu ˜1 Y1,˜u2 Y2 satisfying w˜i u˜i < δ || − || δ 4 for i =1, 2. Hence we get w˜1 w˜2 > 4 .

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∞ ∞ If such a sequence (˜zi)i=1 existed, then for the block subspace Z = span [(˜zi)i=1] we would have that Z = Z1 ⊕ Z2, contradicting the assumption that X˜ is a block H.I. space. To prove the claim we begin with the following two observations: 0 (1) Since || ||1, || ||n are equivalent, for every ε>0andn ∈ N there exists ε > 0 Pn ˜ || || 0 || || such that for everyy ˜ =(y, y, ....)in∆X if y 1 <ε,then y k <εand hence k=1 ||Pn (˜y)|| <ε. (2) Since J is a strictly singular operator it follows that for every Y˜ infinite- dimensional subspace of ∆X˜ and every ε0 > 0thereexists˜y ∈ Y˜ , ||y˜|| =1and || || 0 J (˜y) 1 <ε. These two observations permit us to apply the standard sliding hump argument ∞ ∞ and obtain sequences (˜yi)i=1,(˜zi)i=1 so that the following properties are satisfied: • y˜2j−1 ∈ Y1,˜y2j ∈ Y2 for j =1, 2, ... and ||y˜i|| = 1, for all i ∈ N. ∞ • (˜z ) is a block sequence of X˜ and ||y˜ − z˜ || < δ , for all i ∈ N. i i=1 h i i h i 2ii+4 ∞ ∞ ∈ || || We set Z1 = span (˜z2j−1)j=1 , Z2 = span (˜z2j)j=1 .Let˜w1 Z1, w˜1 =1. P∞ P∞ Thenw ˜1 = αjz˜2j−1 and we setu ˜1 = αjy˜2j−1.Weobservethat||u˜1 − w˜1|| ≤ j=1 j=1 P∞ | | δ δ αj 22j+3 < 4 . The last inequality holds from the monotonicity of the norm of j=1 ˜ ∈ − δ || || δ ∈ X. Thereforeu ˜1 Y1 and 1 4 < u˜1 < 1+ 4 . For the same reason forw ˜2 SZ2 ∈ − δ || || δ || − || δ there existsu ˜2 Y2 with 1 4 < u˜2 < 1+ 4 and u˜2 w˜2 < 4 .Thisproves the claim and as we have already mentioned it leads to a contradiction. Definition 2.4. Let W be a bounded convex symmetric non-empty subset of a ||| ||| ∞ Banach space (X, ), and a =(an)n=1 a null sequence of positive numbers. We n set Wn =2W + anBX and denote by || ||n the gauge of Wn. Then for every d-product norm of the sequence (X, || ||n)n∈N we shall call the diagonal space ∆X˜ an (a,W,X˜)-diagonal space or simply an (a,W,X˜)-space. P∞ −n In case an =2 for every n and X˜ = ⊕ Xn , then the (a,W,X˜)- n=1 `2 diagonal space is the interpolation space introduced in [DFJP]. The following proposition can be readily proven. Proposition 2.2. Let ∆X˜ be an (a,W,X˜)-diagonal space of the sequence (X, || ||n)n∈N. Then the following hold: ⊆ (a) W JB∆X˜ . ⊆ (b) For every ε>0 there exists λ>0 such that JB∆X˜ λW + εBX . Definition 2.5. Let (X, || ||) be a Banach space, A, B ⊆ X and ε>0. We say that (i) the set Aε-absorbs B if there exists a λ>0 such that B ⊆ λA + εBX ; (ii) the set A almost absorbs B if Aε-absorbs B for any ε>0. The following lemma is due to A. Grothendieck ([Gr]) and for a proof we refer to [D]. We are going to use this lemma in the proof of the last proposition of this section. Although we will not use these results we present them to illustrate that the factorization results in [DFJP] can be extended in the case of not necessarily unconditional d-products.

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Lemma 2.3. Let K be a weakly closed subset of the Banach space X. Suppose that for each ε>0 there exists a weakly compact set Kε in X such that K ⊂ Kε + εBX. Then K is weakly compact. Proposition 2.4. Let ∆X˜ be an (a,W, X˜)-diagonal space such that the norm of X˜ is boundedly complete and shrinking. Then ∆X˜ is reﬂexive if and only if W is a relatively weakly compact subset of X. Proof. If W is a relatively weakly compact subset of X, then by Proposition 2.2 (b)

W almost absorbs the set JB∆X˜ and hence from Lemma 2.3 JB∆X˜ is a relatively weakly compact set. Since J is a Tauberian operator, we get that B∆X˜ is a relatively weakly compact subset of ∆X˜.

3. Thin sets In this section we initiate the study of thin and a-thin sets that we will continue in the next two sections. The notion of a thin set was introduced in [N], [N1], and that of an a-thin set, which is weaker, was suggested to us by B. Maurey who also showed us how to prove that the unit ball of L∞ (λ)isana-thin subset of L1 (λ). It was observed by Rosenthal ([N], [N1]) that this set is not a thin subset of L1 (λ) . Thin or a-thin sets are important for our considerations since the interpolation diagonal space defined by an a-thin set (and hence by a thin set) and an appropriate external norm is, as we will show in Proposition 3.1, an H.I. space. Definition 3.1. a) A bounded convex symmetric non-empty subset W of a Banach space X is said to be a thin set if W does not almost-absorb the unit ball BY of any infinite-dimensional closed subspace Y of X (i.e. for any Y , infinite-dimensional closed subspace of X,thereexistsε>0 such that for all λ>0, BY * λW + εBX ). b) An operator T : Z → X is called thin if TBZ is a thin subset of X. Remark 3.1. (a) There are several examples of thin sets. The closed absolutely ∞ p N ∞ N convex hull of the basis (en)n=1 in any ` ( ), for 1

nach space X and a =(an)n∈N is a null sequence of positive numbers, then the || || sequence of equivalent norms ( n)n∈N on X defined by the Minkowski gauges n (2 W + anBX )n∈N is not uniformly bounded on BZ , for every infinite-dimensional closed subspace Z of X. ∞ Definition 3.2. a) Let a =(an)n=1 be a null sequence of positive numbers. The set W is said to be an a-thin subset of X if for any Y , infinite-dimensional closed || || ∞ || || subspace of X,supn∈N supy∈BY y n = ,where n is the gauge of Wn = n 2 W + anBX . Equivalently, a set W is an a-thin subset of X if for any Y , infinite- dimensional closed subspace of X, and every λ>0thereexistsann ∈ N such that n BY * λ (2 W + anBX ). b) An operator T : Z → X is called a-thin if TBZ is an a-thin subset of X. Remark 3.2. Every thin subset of X is a-thin for every null positive sequence a and every thin operator is a-thin for every null positive sequence a.Ana-thin operator is strictly singular. In particular if W is an a-thin set, then the operator

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J :∆X˜ → X is strictly singular. Combining these observations with Proposition 2.1 we get Proposition 3.1. Let ∆X˜ be the (a,W, X˜)-diagonal space deﬁned by an a-thin set W of a Banach space X and such that X˜ is a block-H.I. space. Then ∆X˜ is a H.I. space.

The next two results, due to B. Maurey, will enable us to show that BL∞(µ) is an a-thin subset of L1 (µ) for every probability measure µ. Lemma 3.2. Let (Ω,µ) be a probability space. Let Z be an inﬁnite-dimensional 2 ⊂ p ∞ subspacen of L (µ) such that Zo L (µ) for every p< and let Cp(Z)= Cp(Z) − 1 sup kfk : f ∈ Z and kfk ≤ 1 .Thenlim inf √ ≥ e 2 . p 2 p→+∞ p

Proof. Let us choose an orthonormal basis f =(f1, ..., fn, ...) for our subspace Z of 2 ⊂ ∞ ⊂ p L (µ)andletusﬁxp>2. By assumption,we have Z span[fj]j=1 L (µ)and Cp(Z) is equal to the smallest constant K such that

  1 2 Xn Xn Xn  2 cjfj ≤ K cjfj = K |cj| ,

j=1 j=1 j=1 p 2

for all n ≥ 1 and for all real scalars (cj) (if no such constant exists, we let Cp (Z)= ∞). Let g =(g1, ..., gn, ...) be a sequence of independent Gaussian random variables 2 − 1 − x 2 on (Ω,µ) with common distribution (2π) 2 e 2 dx and let G ⊂ LP(µ) be the closed subspace spanned by g. We know that all linear combinations c g such that P j j j | |2 j cj = 1 have the same distribution, namely that of g1; therefore

  1 2 Xn Xn  2 (1) cjgj = Cp(G) |cj|

j=1 j=1 p for all real scalars and   1 ∞ p Z+ p 1 r p  p − x2 dx  2 2 p +1 p (2) Cp(G)=kg1k = |x| e 2 √ = √ Γ ∼ p 2π π 2 e −∞ as p →∞, using Stirling’s formula. Pn Assume that Cp (Z) < ∞. Let us consider fj(s)gj(t). Then for p>2by(1) j=1 we get that

  p ZZ p Z 2 Xn Xn p  2 fj(s)gj(t) dsdt =(Cp(G)) |fj(s)| ds

j=1 j=1     p Z Xn 2 p   2  p p ≥ (Cp(G)) |fj(s)| ds = n 2 (Cp(G)) . j=1

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On the other hand,

p   p ZZ Z 2 Xn Xn p  2 fj(s)gj(t) dsdt ≤ (Cp(Z)) |gj(t)| dt

j=1 j=1 and   p ∞ Z Xn 2 Z Z+ 2 p − r2 − dr   p 2 n 1 ϕ(p, n)= |gj(t)| dt = kxk dγn(x)=υn−1 r e r n Rn (2π) 2 j=1 0 n where υn−1 is the surface area of the unit sphere in R . By change of variable, Z+∞ p + n −1 p −1 2 2 2 p n − − 2 2 n + p 2 + 2 1 u ϕ(p, n)=υn−1 n u e du = υn−1 n Γ . (2π) 2 π 2 2 0

p p n υn−1 ϕ(p,n) ϕ(p,n) − Γ( 2 + 2 ) n 2 1 Taking p =2givesn = ϕ(2,n)= n Γ 2 +1 , n = ϕ(2,n) =2 n π 2 Γ( 2 +1) p p n Γ( 2 + 2 ) 2 so that ϕ(p, n)=2 n . Γ( 2 ) 1 1 Summing up our information, n 2 Cp(G) ≤ Cp(Z)(ϕ(p, n)) p or Cp(G) ≤ q 1 n p p 2 Γ( 2 + 2 ) n n Cp(Z). Γ( 2 ) We know by Stirling’s formula that for any ﬁxed x>0wehavethatΓ(t + x) ∼ 1 q p Γ n + p x ∞ p 2 ( 2 2 ) t Γ(t)whent tends to + . Applying this for x = 2 we get lim n n n→+∞ Γ( 2 ) = 1 and it follows that Cp(G) ≤ Cp(Z). This says that the Gaussian case is the best possible inequality betweenp the L2 p p and L norms. By (2) we already know that Cp(G)isequivalentto e and this concludes the proof.

Proposition 3.3. Let (Ω,µ) be a probability space. The unit ball BL∞(µ) is an 1 − n a-thin subset of L (µ) for a =(an)n with an =exp( 16 ). Proof. Assume the contrary; then there exists an inﬁnite-dimensional closed sub- 1 1 ⊂ n ∞ space Z of L (µ)andM>0 such that M BZ 2 BL + anBL1 , for every ∈ N ∈ k k ≤ 1 n . So, every z Z with z 1 M has a decomposition z = z0 + z1 k k ≤ n k k ≤ ∈ | | n+1 ≤ with z0 ∞ 2 and z1 1 an. Thisimpliesthatµ s Ω: z(s) > 2 n −n µ (s ∈ Ω:|z1(s)| > 2 ) ≤ 2 an. Thus for every λ>2choosingn ≥ 0 such that 2n+1 <λ≤ 2n+2 the last inequality implies that 4 −λ4 µ (s ∈ Ω:|z(s)| >λ) ≤ 2−na ≤ exp . n λ 44 Therefore, using the deﬁnition of the Gamma function we get Z+∞ 4 −λ4 p − 1 kzkp ≤ 2p + pλp−1 exp dµ ≤ 2p + p4p−1Γ . p λ 44 4 2

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 254 S. A. ARGYROS AND V. FELOUZIS 1 k k ≤ 4 → ∞ Applying Stirling’s formula we have that z p O p ,whenp + , uniformly for z in the unit ball of Z.SoZ is a closed subspace of Lp (µ), for every 1 1 ≤ p<∞,andCp (Z) ≤ O p 4 , contradicting Lemma 3.2. Deﬁnition 3.3. A bounded subset W of a Banach space X is called a C-norming set for a subspace Z of X∗, C>0, if ||z|| ≤ C sup |z(w)|, for any z ∈ Z.Theset w∈W W is called a norming set for the subspace Z if it is C-norming for Z for some C>0. Proposition 3.4. (a) Let ∆X˜ be an a,W,X˜ -diagonal space, and let W be a C-norming set for a subspace Z of X∗.ThenZ is isomorphic to a closed subspace of ∆X˜ ∗. (b) Let X, W be as before, and let A be a Banach space such that A∗ is w∗- isomorphic to Z.ThenA is a quotient of ∆X˜. Proof. (a) Denote by J ∗ the adjoint of the operator J. By our assumption for ∈ ∈ || || ≤ | | ⊆ any z Z there exists wz W such that z 2C z(wz) .SinceW B∆X˜ , || ∗ || ≥ | ∗ −1 | | |≥ 1 || || ∗| J z J z(J (wz)) = z(wz) 2C z . Therefore J Z is an isomorphism onto a subspace of ∆X˜ ∗. (b) Let S : A∗ → Z be a w∗-continuous isomorphism. Then J ∗ ◦ S is also a w∗-continuous isomorphism; hence (J ∗ ◦ S)∗ maps the space ∆X˜ onto the space A.

4. Thin norming sets I (special reflexive cases) This is the first section devoted to the a-thin (thin) norming sets. For a reflexive Banach space A with an unconditional basis we shall define a space XA which is P∞ 1 of the form ⊕` (kn) and a symmetric convex closed subset W of the unit n=1 A ∗ ∗ ball of XA which norms a subspace of XA isometric to A , and for certain spaces A we shall show that the set W is an a-thin subset of XA.ThesetW is of the form co (K ∪−K)whereK forms a tree in the branches of which we have “spread” in a regular way a dense subset of the positive part of the unit ball of A.The fact that such a W norms A∗ is quite easy and it is true for all reflexive Banach spaces A with an unconditional basis. To show that W is an a-thin set is more difficult. For this we use two combinatorial results that we prove in this section. The first, Proposition 4.4, “the finite version”, will be applied in the results of the next section, and the second, Proposition 4.6, “the infinite version”, will be used in the present section. Property (P ) given in Definition 4.1, suggested by B. Maurey, is the main tool to show that for certain classes of reflexive spaces the corresponding set W is an a-thin set. Thus for example we are able to prove that for spaces like Lp (λ) ,`p (N) , 1

The space XA. We begin with the deﬁnition of the space XA and the subspace ∗ Y of XA. Consider a tree D of height ω with a least element ρ such that every δ in D with height equal to n has 34(n+1) + 1 immediate successors. We identify D with the set 4i of all ﬁnite sequences (k1, ..., kn), n ∈ N, of natural numbers such that 0 ≤ ki ≤ 3

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and with least element the empty sequence. Consider the usual partial order ≺ in ≺ 0 0 0 0 D, i.e. (k1, ..., kn) (k1, ..., k`)iff(k1, ..., kn) is an initial segment of (k1, ..., k`). A segment is a subset of D of the form d =[δ1,δ2]={δ ∈ D : δ1 ≺ δ ≺ δ2}. Asegmentd of the form d =[ρ, δ] is called an initial segment of D. For every segment d =[δ1,δ2] we define ext(d)=[ρ, δ2], that is, the minimal initial segment containing the segment d. A branch is a maximal linearly ordered subset of D. A branch is identified to an 4i infinite sequence (ki)i∈N of natural numbers, where 0 ≤ ki ≤ 3 . The height of δ =(k1, ..., kn) is denoted |δ| and it is equal to n.Weset|ρ| =0. We denote by c00(D) the linear space of all functions f : D → R such that supp(f)={δ ∈ D : f(δ) =06 } is a finite set. Also, denote by eδ the characteristic function of {δ}, δ ∈ D. The vectors (eδ )δ∈D form a Hamel basis for c00(D). If A, B are finite subsets of D we write A

P P∞ P ∈ || || | | For f = λδ eδ c00(D)weset f = λδ en . δ∈D n=0 |δ|=n A The space XA is the completion of c00(D) with the above defined norm. It is P∞ Qn 1 4i clear that XA is isometric to ⊕ ` (kn) for kn = 3 +1 and if A has n=0 A i=1 P∞ ∗ ⊕ ∞ a shrinking basis, then XA = ` (kn) .IfA is a reflexive Banach space, n=0 A∗ then XA is alsoP reflexive. ∗ ∗ ∗ || ∗ || Set yn = eδ and Z = span [(yn)n∈N]. Then yn = 1 and it is easily verified |δ|=n ∗ ∗ ∗ that the map T (yn)=en extends to an isometry between the spaces Z and A . Throughout this section we will denote by A a reflexive space with a 1-unconditional basis (en)n∈N.

kn The set W. For every δ =(k1, ..., kn)wesetaδ = 34n . P

For any inﬁnite branch γ of D we denote by xγ the formal series xγ = aδn eδn , n∈N where δn =(k1, ..., kn) ∈ D is the initial segment of γ with height equal to n.Also P∞ ||| ||| kn ||| ||| ∞ we set xγ = 34n en . It follows from the deﬁnitions that if xγ < , n=0 A then xγ ∈ XA and ||xγ || = |||xγ |||.WesetK = {xγ : ||xγ || ≤ 1},W0 = co(K ∪−K)

and W = W0.ThesetW is a closed bounded symmetric convex subset of BXA .

1 Lemma 4.1. The set K (and hence the set W )isa 4 -norming set for the space ⊆ ∗ Z XA.

P∞ P∞ ∗ ∗ ∈ || ∗|| ∗ ∗ Proof. Let y = λkyk SZ ;then y = λkek = 1. Deﬁne z = k=1 k=1 A∗ P∞ P∞ ∗ ∗ ∗ ∗ ∈ || || ∗ || || ∈ λkek A and notice that z A = y = 1. Consider a vector z = µkek k=1 k=1

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P∞ ∗ SA such that |z (z)| =1= λkµk .Since(en)n∈N is 1-unconditional and bi- k=1

P∞ P P ≤ monotone, 1 = λkµk λkµk + λkµk . k =1 µk >0 µk <0

P Assume that λ µ ≥ 1 . k k 2 µk >0 For any k ∈ N we define nk =0ifµk ≤ 0, while if µk > 0, then nk is defined to n n +1 k ≤ k ∈N be the unique natural number with the property 34k µk < 34k .Since(en)n 4k is bimonotone, µk ≤ 1 for any k ∈ N and therefore nk ≤ 3 for any k ∈ N. P∞ || || nk ≤ Let γ =(n1,n2, ...) be an infinite branch of D.Then xγ = 34k ek k=1 A P∞ P∞ P ∗ µ e ≤ 1, i.e. x ∈ K.Also,|y (x )| = λ nk = λ nk ≥ k k γ γ k 34k k 34k k=1 A k=1 µk>0 P P P∞ ∗ λ µ − λ µ − nk ≥ 1 − 1 > 1 = 1 ||y ||,andtheproofis k k k k 34k 2 34k 4 4 µk >0 µk>0 k=1 complete.

In the following lemma φ will denote a function on XA which is either (i) φ(x)=y∗(x) for some non-negative functional y∗ or (ii) φ(x)=||x||. In fact we may consider any real valued bounded sublinear function φ on XA ⊂ ≤ ≤ suchthatforanytwosegmentsd1, d2 of D with d1 d2 we have that 0P φ(xd1 )

φ(xd2 ), where d is a segment of D. Here, by xd we denote the vector aδ eδ . δ∈d Lemma 4.2. Let E be a subset of D of the form E = {δ ∈ D : m ≤|δ|≤M}, φ afunctiononXA as before and ε>0. There exists a decomposition of E into two disjoint subsets E0, E00 such that 0 (1) |φ(E w)| <εfor every w ∈ W0,and (2) if d =[δ1,δ2] is a segment of the tree such that |δ1|≤m, |δ2|≥M and 00 d ∩ E =6 ∅,then|φ(Exd)|≥ε. E E Proof. For every δ ∈ E let extE(δ)=ext(δ) ∩ E =[δ ,δ], where δ is the unique element of E such that δE ≺ δ and |δE| = m.Weset 00 0 00 E = {δ ∈ E : ||xext(δ)|| ≤ 1andφ(xext (δ)) ≥ ε},E= E\E . E P Let w ∈ W0;thenEw can be represented as Ew = λdxd,whereL is a subset d∈L P of segments of E such that ||xext(d)|| ≤ 1 for every d ∈ L and |λd|≤1. For d∈L 0 0 ∈ ∩ 0 every d L theP set d = d E is either empty or a segment of E and φ(xd ) <ε. 0 00 So |φ(E w)|≤ |λd||φ(xd0 )| <ε, which proves property (1). Property (2) of E d∈L is obvious.

The following two propositions are of combinatorial nature and we will use them to show that the set W is an a-thin set. Both are related to the existence of incomparably supported elements of the set W . The first (Proposition 4.4) concerns finite families and the second (Proposition 4.6) is the infinite analog of Proposition 4.4. We begin with the following lemma.

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Lemma 4.3. Let (Ω,µ) be a measure space with µ(Ω) ≤ 1 and let B1, ..., BN be measurable subsets of Ω satisfying µ(Bi) ≥ ε for i =1, ..., N. Then for k

PN ≤ Proof. If not, then the function χBi is bounded by k on Ω and therefore Nε i=1 R PN ≤ χBi dµ k, a contradiction. Ω i=1 Next we prove the first incomparability result related to finite families of elements of the set W . n Proposition 4.4. Let (Ei)i=1 be a finite sequence of subsets of D of the form Ei = {δ ∈ D : mi ≤|δ|≤Mi},mi,Mi ∈ N,Mi 0, x ∈ B ∗ and (wi) a finite sequence of vectors of W0 there XA i=1 exists a partition D1, ..., DN of {1, 2, ..., n} and Fi ⊂ Ei for i =1, ..., n such that (1) If r, s belong to the same set Di and r

Proof. Applying Lemma 4.2 we find for every i =1, ..., n a decomposition of Ei 0 00 into two disjoint subsets Ei, Ei such that | ∗ 0 | ε (a) x (Eiwi) < 2 . | |≤ | |≥ ∩ 00 6 ∅ (b) If d =[δ1,δ2] is a segment such that δ1 mi, δ2 Mi and d Ei = , ∗ ε then x (Eixd) > 2 . P For every i =1, 2, ..., n the vector Eiwi has a representation Eiwi = λdxd, d∈Li || || ≤ ∈ whereP Li is a set of segments of Ei with xext(d) 1 for every d Li and |λd|≤1. d∈Li This representation defines a positive measure µi on the powerset of seg(Ei)= { || || ≤ } || || ≤ dP: d is a segment of Ei and xext(d) 1 ,with µi 1, as follows: µi(A)= |λd|, for every A ⊂ seg(Ei). d∈A∩Li For every S ⊂{1, ..., n} and every i ∈{1, ..., n} we set      [  i ∈ ∩  00 6 ∅ AS = d seg(Ei):ext(d) Es =  . s∈S\{i} Given any non-empty subset J of {1, ..., n} we inductively define a non-empty subset D(J)ofJ in the following manner: Suppose that J = {i1 < ... < im}.Weleti1 =minJ ∈ D(J). If 2 ≤ k ≤ m and we have selected all the elements among {i1, ..., ik−1} that belong to D (J), and denote them by {j1, ..., jr} (so that in particular j1 = i1), then ik ∈ D(J)ifand ik ε only if µi A{ } < . k j1,...,jr 2 S ∈ 00 ∩ { 00 ∈ } ∅ We also set Fj = δ Ej : ext(δ) Es : s

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d0 = E0 ∩d, d00 = F ∩d and d00 =(E00\F )∩d. It is clear that d00 ∈ Aj for every j a j b j j b D(J) d ∈ L , and since µ Aj < ε we get that |x∗((E \F )w )|≤|x∗(E0 w )| + j ! j D(J) 2 j j j j j P P ∗ ε x λdxd00 ≤ + |λd| <ε. We define the partition D1, ..., DN of b 2 d∈Lj ∈ j d AD(J) {1, 2, ..., n} as follows: J1 = {1, ..., n}, D1 = D(J1), J2 = J1\D1, D2 = D(J2)and we continue this process until the set JN+1 equals the empty set. 5 It remains to show that N<ε2 . Indeed, choose r ∈ J .Thenµ Ar ≥ ε for j =1, ..., N − 1. N r Dj 2 5 − ε 2 If N>ε2 ,then(N 1) 2 > ε . By Lemma 4.3 we can select Dj1 , ..., Djk with Tk Tk j < ... < j and Ar =6 ∅ for some k such that ε k>1. Let d ∈ Ar .Then 1 k Dj 2 Dj s=1 s s=1 s for every s =1, ..., k there exists r ∈ D such that d∩E00 =6 ∅ so x∗(E x ) ≥ ε s js rs rs ext(d) 2 ∗ ≥ ε || ∗|| ≤ and x (xext(d)) 2 k>1 which leads to a contradiction since x 1. Before passing to the proof of the infinite analog of Proposition 4.4 we give the statement and the proof of a known auxiliary result. ∞ Lemma 4.5. Let (Ω,µ) be a finite measure space, ε>0 and (Bn)n=1 asequence ≥ ∈ N of measurable subsets of Ω such thatT µ(Bn) ε for every n . There exists an infinite subset M of N such that Bn =6 ∅. n∈M Proof. Since µ(Ω) < ∞, it follows that ! \∞ [∞ µ(lim sup Bn)=µ Bn ≥ lim inf µ(Bn) ≥ ε. n=1m=n 6 ∅ ∈ N Hence lim sup BTn = and for ω lim sup Bn there exists an infinite subset M of such that ω ∈ Bn. n∈M ∞ Proposition 4.6. Let (En)n=1 be a sequence of subsets of D of the form En = {δ ∈ D : mn ≤|δ|≤Mn},withmn+1 >Mn, for every n ∈ N. Then for every ∞ N sequence (wn)n=1 of vectors in W0 and ε>0 there exist an infinite subset I of ⊂ ∈ and sets (Fn)n∈I with Fn En, for all n I, so that: (1) The sets (Fn)n∈I are pairwise incomparable. (2) For every n ∈ I, ||(En\Fn)wn|| <ε.

Proof. We apply Lemma 4.2 to find a decomposition of each En into two disjoint 0 00 sets En,En such that || 0 || ε (1) Enwn < 2 . | |≤ | |≥ ∩ 00 6 ∅ (2) If d =[δ1,δ2]isasegmentwith δ1 mn, δ2 Mn and d En = , then || || ≥ ε Enxd 2 . P For every n ∈ N the vector Enwn has a representation as Enwn = λdxd, P d∈Ln where Ln ⊆ seg(En)={d : d ⊂ En and ||xext(d)|| ≤ 1} and |λd|≤1. d∈Ln This representation definesP a positive measure µn on seg(En) as in the previous proposition: µn(A)= |λd|, for every A ⊂ seg(En). d∈A∩Ln Consider any probability measure ν on the compact metrisable space Γ of all infinite branches of D such that ν(Wd) =6 ∅ for every d segment of the tree, where

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Wd is the basic clopen subset of Γ consisting of all branches that contain d (for instance ν is the natural measure defined by ν(W )= 1 for every δ ∈ D, ext(δ) k|δ| { ∈ | | } where kn =# δ D : δ = n ). WeP define a measure µ on Γ as follows: For every ν(Wd∩B) clopen subset B of Γ, µ(B) = lim |λd| , where the limit is taken with n→U ν(Wd) d∈Ln respect to some non-trivial ultrafilter U on N. Using a diagonal argument we may actually assume that this limit is an ordinary limit, by passing to an appropriate subsequence. We shall also assume, without loss of generality, that this subsequence is the whole sequence of natural numbers. ∈ N { ∈ ∩ 00 6 ∅} s { ∈ For every n, s with n

Claim. For every inﬁnite subset I of N and θ>0thesetIθ = {n ∈ I : µ(Bn) <θ} is inﬁnite.

Indeed, otherwise there exists an infinite subset J of I such that µ(Bn) ≥ θ 0 for everyT n ∈ J and by Lemma 4.5 there exists an infinite subset J of J and ∈ ∈ 0 ∩ 00 6 ∅ || 00 || ≥ ε γ Bn. For every n J , γ En = and so Enxγ 2 for infinitely many n∈J0 ∈ 00 ∩ || || ≤ || || ≤ values of n. Select δn En γ.Then xext(δn ) 1andso xγ 1which contradicts the fact that the basis of XA is boundedly complete. We define I = N and inductively for k ≥ 0, 0 n o ε I˜ = n ∈ I : µ(B ) < ,n=minI˜ , k+1 k n 2k+2 k+1 k+1 n o s ε Ik+1 = s ∈ I˜k+1 : µn (A ) < k+1 nk+1 2k+2 and      [k  00  00  Fn = δ ∈ E : ext(δ) ∩ E = ∅ . k+1  nk+1 ni  i=1

By the previous claim the sets I˜k+1, Ik+1 are inﬁnite subsets of Ik and since ∈ nk+1 ε nk+1 Ik, µnk+1 (Ank ) < 2k+1 . { } ∞ We set I = n1,n2, ... and we observe that the sets (Fnj )j=1 are pairwise || \ || incomparable by their own deﬁnition. We show that for every k, (Enk Fnk )wnk < 00 ε.Ifk = 1 this follows from the fact that Fn1 = En1 . Let k ∈ N and set r = n . Consider the set: A = Ar ∪ ... ∪ Ar . Then, k+1 r n1 nk ≤ ε ε ε ∈ ∈ µr(Ar) 22 + ... + 2k+1 < 2 .Letd/Ar; then for every δ d and i =1, ..., k, ∩ 00 ∅ ⊂ ∈ ∩ ∅ ext(δ) Eni = Pso d Fr.Converselyifd Ar,thend Fr = .So k 00\ k≤ | |k 00\ k≤ ε (Er Fr)wr λd (Er Fr)xd µr(Ar) < 2 . d∈Ls∩Ar k \ k≤k 0 k k 00\ k ε ε Finally, we have (Er Fr)wr (Er)wr + (Er Fr)wr < 2 + 2 = ε. The present form of Propositions 4.4 and 4.6 is a variation of our original results suggested by B. Maurey.

Definition 4.1. Let (A, || ||A) be a reflexive Banach space with an unconditional ∞ ∞ basis (en) ,andlet(Ck) be a sequence of positive real numbers. We say that n=1 k=1 ∞ A satisfies property (P ) for the sequence (Ck)k=1 if for every block sequence ∞ ∞ ∈ N (xn)n=1 of (en)n=1 and every k there exists a normalized block sequence

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∞ ∞ (yn)n=1 of (xn)n=1 satisfying the following condition: For every inﬁnite subset M ∗ N ⊂ ⊆ R+ ⊆ ∗ of there exist E M and (λn)n∈E ,(yn)n∈E BA such that P P 1 (i) λn =1, λnyn < k , n∈E n∈E A ∗ ⊆ ∗ 1 ∈ (ii) suppyn suppyn and yn(yn) > 2 for every n E and P ∈ ∗ 6 (iii) for every x BA, yn(x)yn Ck. n∈E A

Remark 4.1. Condition (iii) in the above deﬁnition is equivalent to the factP that the → h i ∗ norm of the operator P : X (yn)n∈E deﬁned by the relation P (x)= yn(x)yn n∈E is dominated by Ck. There are other conditions which imply condition (iii) and in some cases they are more convenient to be checked. We give two of them for later use: P ¯ ∗ ≤ ¯ | |≤ (P1): yn Ck, (P2): E Ck. n∈E In the sequel by property (Pi),i=1, 2, we will mean property (P ) where condition (iii) has been replaced by (Pi),i=1, 2. To see that (Pi),i =1, 2, imply (P )weobservethat(P2) ⇒ (P1)isobvious, while the implication (P1) ⇒ (P ) follows from the unconditionality of the basis (en)n.

Lemma 4.7. If the Banach space (A, || ||A) satisfies property (P ) for the sequence ∞ (Ck) , then the same remains valid for the space (XA, || ||) and the same sequence ∞k=1 (Ck)k=1. ! P P∞ P Proof. For every x = λδ eδ we set x = |λδ | ei. It follows from the δ∈D i=1 |δ|=i definition of the space XA that!||x|| = ||x¯||. P ∞ Let (xn)n=1 = λδ eδ be a block sequence in XA. By passing to |δ|∈E n n∈N ∞ a subsequence of (xn)n=1, if necessary, we may assume that E1 0 there exists a normalized block (yn)n=1 of (¯xn) such that conditions (i), (ii), (iii)P of Definition 4.1 are fulfilled. P Eachy ¯n = λix¯i with F1 2 . P ∈ ∗ 6 (c) For every x BA, y¯n(x)¯yn Ck. n∈E A

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use INTERPOLATING HEREDITARILY INDECOMPOSABLE BANACH SPACES 261 ! P P ∗ ⊂ ∗ ∗ ∗ Since suppy¯n suppy¯n it follows thaty ¯n = αjej . We deﬁne yn = ∈ ∈ P P P i Fn j Ei ∗ | | (sgnλδ ) αjeδ ,wheresgnλδ =0ifλδ =0andλδ (sgnλδ )= λδ i∈Fn j∈Ei |δ|=j ∗ otherwise. It is easily checked that (yn)n∈E,(yn)n∈E,(θn)n∈E satisfy the conditions corresponding to (a), (b), (c) in the space XA and this completes the proof.

Remark 4.2. It can be shown that the analogous results for properties (Pi), i =1, 2, also hold. The proof of them follows similar steps and is left to the reader. The following result is well known [N] and says that if a closed set almost absorbs the unit ball BX of the whole space, then it actually absorbs BX . We will state it in the following form:

Lemma 4.8. Let W be a convex subset of a Banach space X such that BX ⊆ ⊂ λ λW + εBX ,forsomeλ>0 and ε<1.ThenBX 1−ε W.

Theorem 4.9. Let (A, || ||A) be a reﬂexive Banach space with an unconditional ∞ ∞ basis which satisﬁes property (P ) for a sequence (Ck) .Weseta =(an) = ∞ k=1 n=1 1 . Then the set W is an a-thin subset of the space XA. nCn2n n=1

Proof. Suppose that W is not an a-thin subset of XA. Then there exists a normal- ∞ ized sequence (xn)n=1 of successive blocks of (eδ )δ∈D and λ>0 such that for every n ∈ N, ⊆ n (1) BX λ(2 W + anBXA ) ∞ ∈ N where X = span(xn)n=1.Fixk with k>16λ. The space XA also satisfies property (P ) for the same sequence (Ck)k, by Lemma ∞ ∞ 4.7. Hence there exists a normalized block sequence (yn)n=1 of (xn)n=1 such that + ∗ for every infinite subset M of N we can find E ⊂ M,(θ ) ∈ ⊆ R ,(y ) ∈ ⊆ B ∗ n n E n n E XA such that

X X 1 (2) θ =1, θ y < , n n n k2k n∈E n∈E 1 (3) suppy∗ ⊂ suppy ,y∗ (y ) > n n n n 2 and

(4) ||P || 6 C k P k2 ∗ where P (x)= yn(x)yn. n∈E k Let wn ∈ W be such that ||yn − λ2 wn|| <λak and set En = suppyn. By Proposition 4.6 there exists an inﬁnite subset M of N and a family (Fn)n∈M of pairwise incomparable subsets of D such that for every n ∈ N a F ⊂ E and ||E w − F w || < k . n n n n n n 2kλ ∗ ⊂ ⊂ R+ ⊂ ∗ Let E M,(θn)n∈E P ,(yn)n∈E BX satisfy the above conditions and ∗ ∗ | ∗ A set zn = yn Fn , R(x)= zn(x)yn. n∈E Since the basis of XA is 1-unconditional it follows that

(5) ||R|| ≤ ||P || ≤ Ck2k

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and 1 (6) |z∗(y )| > . n n 4 Set YE = span[(yn)n∈E]. Then (1) yields that ⊂ k (7) R(BYE ) λ2 R(W )+λakR(BXA ) ⊂ ⊂ λ ⊂ 1 and by (5) we get that λakR (BXA ) λakCk2k BYE κ BYE 16 BYE . 1 ⊂ Also (6) implies that 4 BYE R (BYE ). Substituting in (7) we obtain that ⊂ k 1 BYE 4λ2 R (W )+ 4 BYE and from Lemma 4.8 we deduce that ⊂ k (8) BYE 8λ2 R(W ). ∗ ⊂ Since the sets (suppzn)n∈E are pairwise incomparable we get that R(W ) co[(yi)i∈E ]. P ⊂ k k ∈ Therefore BYE 8λ2 co[( yi)i∈E] and since k2 θnyn BYE we con- n∈E clude that k ≤ 8λ, which derives a contradiction since we had selected k>16λ. ∞ Remark 4.3. Assuming that the sequence (Ck)k=1 in the above theorem is bounded or even if Ck ≤ Ck and (P1) holds, then our arguments can show that the set W is actually a thin subset of XA. Next we state the following well known lemma (cf. [LT], 1c8). Lemma 4.10. Let A be a reflexive Banach space with an unconditional basis and P : A → Y a bounded projection of A onto a block subspace Y of A spanned ∞ by a normalized block sequence (yn)n=1. Then there exists a bounded projection P∞ → ∗ ∗ ∞ R : A Y of the form R(x)= yn(x)yn,where(yn)n=1 is a block sequence in n=1 ∗ ∗ ⊂ ∈ N A with supp(yn) supp(yn) for every n . Remark 4.4. It is easily seen that if a reflexive Banach space A satisfies the complemented subspace property described in the statement of Lemma 4.10, then the same property holds for the space XA. The following result has been proved by N.Tomczak-Jaegermann who kindly permitted us to include it here. Theorem 4.11. Let A be a reflexive Banach space with an unconditional basis such that every block subspace of A has a block subspace complemented in A. Then the set W is a thin subset of XA.

Proof. We recall that the basis of A, and therefore the basis of XA,isaswell bimonotone. Suppose that the set W is not a thin subset of XA. Then there exists a normalized block sequence (xn)n of XA so that letting X = span [(xn)n] we have that for every ⊂ ε>0thereexistsaλ>0 such that BX λW + εBXA . Lemma 4.10 and Remark 4.4 allow us to assume (by passing to a normalized ∗ ⊂ ∗ block sequence of (xn)n, if necessary) that there exists a sequence (xn)n XA, ∈ N ∗ ⊂ ∗ such that for every n , supp(xn) range(xn), xn(xn) = 1 and the operator P∞ ∗ P (x)= xn(x)xn is bounded. Since the basis of XA is bimonotone we have that n=1 || ∗ || ≤ || || 1 ⊂ xn P for all n.Letε = 8kP k and λ>0 such that BX λW + εBXA .

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For every n ∈ N we select wn ∈ W0 such that kxn − λEnwnk <ε,whereEn = range(xn). Applying Proposition 4.6 we obtain an inﬁnite subset I of N and (Fn)n∈I pairwise incomparable subsets of D such that for every n ∈ I, Fn ⊂ En and kEnwn − Fnwnk P∞ 1 ∗ ∗ ∗ < 4λ .Wesetyn = Fnxn and R(x)= yn(x)xn. n=1 k k≤k k | ∗ | 1 We observe that R P and yn(xn) > 4 .LetY = span[(xn)n∈I ]. Since | ∗ | 1 yn(xn) > 4 we obtain that 1 (1) B ⊂ R (B ) . 4 Y Y 1 Also since ε = 8||P || we have that 1 (2) R (B ) ⊂ λR (W )+ B . Y 8 Y (1) and (2) and Lemma 4.8 yield that

BY ⊂ 8λR(W ). ∗ ∗ Since the operator R is deﬁned by the sequence (yn)n∈I and the sets (suppyn)n∈I are pairwise incomparable we have that R(W ) ⊂||P ||co[(xn)n∈I ], and ﬁnally

(3) B ⊂ (8λ||P ||) co[(x ) ∈ ]. Y n n I P To derive a contradiction we select a convex combination z = θnxn such that n∈E k k 1 || || ∈ z < 16λ||P || and observe that (16λ P ) z BY , contradicting (3). The proof of the theorem is complete. Remark 4.5. It is known that `p(N)where1 0 such that for every Pn ∞ ∞ 1 ≤ p normalized block sequence (xi)i=1 of (en)n=1 we have that xi Cn . i=1 Remark 4.6. If a space X has an upper p-estimate for some p>1, then it satisfies p − property (P2)withCk ≤ (Ck) p 1 ,whereC is the constant in Definition 4.2, and hence it also satisfies property (P ) (see Remark 4.1). Spaces with an upper p- estimate are Lp (µ)where1

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Proof. Choose any ε<1. Then there exists an inﬁnite-dimensional closed subspace p + Z of ` and λ ∈ R such that BZ ⊂ λT [BX ]+εB`p .Further,wemayassume that Z is a block subspace of `p and hence there exists a projection P : `p → Z with ||P || = 1. Therefore BZ ⊂ λP (T [BX]) + εBZ and from Lemma 4.8 BZ ⊂ λ ◦ ◦ 1−ε P T [BX] which implies that P T is a quotient map. The proof of the following lemma is similar to the proof of the lifting property of `1 (see [LT], p.107). Lemma 4.13. If T : `p → `p,where1 ≤ p<∞, is a bounded linear onto operator, p then there exists an inﬁnite-dimensional closed subspace Z of ` such that T |Z is an isomorphism. Corollary 4.14. (a) If 1 ≤ p =6 r<∞, then every T ∈L(`p,`r) is a thin operator. (b) If 1 ≤ p<∞ and T ∈L(`p,`p),thenT is a strictly singular operator if and only if it is a thin operator.

5. Thin norming sets II (general reflexive case) In this section we prove that every reflexive space A with an unconditional basis has a subspace B such that the set W in XB is an a-thin set for an appropriate null sequence a. The proof heavily depends on the uniform control on a subspace of A of the rate of convergence to zero in norm of certain convex combinations. Therefore the ordinal rank of the complexity of weakly null sequences seems to be necessary and we strongly use the results of [AMT]. The main parts of this section are the following: The first part contains the definitions of Schreier families, initially introduced in [AA], and RA-Hierarchy from [AMT]. We also present some results from [AMT] that we will use here. The most important of them are the dichotomy principle (Proposition 5.3) and the large families theorem (Theorem 5.4). In the second part, for a countable ordinal ξ, a natural number n and B a subspace of A we introduce the numerical quantity τξ,n(B) which is basic for our proof. We show that there exists a countable ordinal ξ0 such that τξ,n(B) = 0 for every ξ ≥ ξ0 and n ∈ N. Moreover, there exists a subspace B of A so that τξ,n is stabilized over all further subspaces. This subspace B is the desired subspace of A.To accomplish this, we need to show that for every ε>0thereexists(ξ,n) such that 0 <τξ,n(B) <ε. Finally, in the last part (Propositions 5.18, 5.20, Theorem 5.21), we prove that the set W in XB is an a-thin set. Schreier families. By [N](resp.[N]<ω) we will denote the set of all infinite (resp. finite) subsets of the set N of natural numbers. Also for M ∈ [N]wedenoteby[M] the set of all infinite subsets of M. For every countable ordinal ξ we define a family Sξ of finite subsets of N as follows: (1) S0 = {{n} : n ∈ N} ∪ {∅}. (2) If S has been defined, then ξ ( ) [n Sξ+1 = Fi : n ∈ N,n≤ F1

(3) If ξ is a limit ordinal and for every ζ<ξthe family Sζ has been defined, then we fix a strictly increasing sequence (ξn)n∈N of non-limit ordinals with supξn = ξ n∈N { ≤ ∈ ∈ N}∪{∅} and define Sξ = F : n min F and F Sξn for some n .

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Notation. Let M ∈ [N], and let ξ be a countable ordinal. We introduce two more M { ⊂ families related to Sξ and the set M.IfM =(mk)k∈N we set Sξ = G M : ∈ { ∈ }} { ⊂ there exists F Sξ such that G = m` : ` F .Wealsodenote Sξ [M]= F Sn ∈ } M n ∈ M M : F Sξ ,[Sξ ] = Fi : Fi Sξ ,F1 0, and ξM (m) ≤ 1 for all m ∈ M. n `1 n minM M M M ∈ (2) suppξn < suppξn+1 and suppξn Sξ . S∞ M (3) M = suppξn . n=1 S∞ ∞ 0 M (4) If (nk)k=1 is a strictly increasing sequence, then setting M = suppξnk we k=1 M 0 M ∈ N have that ξk = ξnk for all k .

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M We recall the deﬁnition of (ξn )n∈N from [AMT]. It is given by transﬁnite induction. { } M (1) For ξ =0,M = m1,m2, ... , m1

Proposition 5.3 (Dichotomy Principle). Let ξ<ω1, M ∈ [N] and n ∈ N.We n set I = {(ξN , ..., ξN ):k < ... < k ,N ∈ [M]}. Clearly I ⊂ S+ .Then M,n k1 kn 1 n M,n `1 for every real function φ : IM,n → R and τ ∈ R one of the following holds: (i) There exists L ∈ [M] such that φ(s) >τ for all s ∈ IL,n. (ii) There exists L ∈ [M] such that φ(s) ≤ τ for all s ∈ IL,n. Definition 5.2. An adequate family F of subsets of N is said to be ∈ N ∈ (a) (ξ,M,ε)-large,whereξ<ω1, M [ ]andP ε>0, if for every L [M]and ∈ N ∈ h L i L n there exists F F such that ξn ,F = ξn (k) >ε; k∈F ∈ {h N i}n (b) (n, ζ, M, ε)-large if for every N [M], sup inf ζi ,F i=1 >ε. F ∈F i The following result has been proved in [AMT] and it will be one of the basic tools for our proofs. Theorem 5.4. Let F be an adequate family of finite subsets of N which is (ξ,M,ε)- large ((n, ξ, M, ε)-large for some n ∈ N) for some countable ordinal ξ, M ∈ [N] and L ⊂ L n ⊂ ε>0. Then there exists an infinite subset L of M such that Sξ F ([Sξ ] F). Proof. For n = 1 the result follows from Proposition 2.32 and Theorem 2.26 of [AMT]. If n ≥ 2 the result follows from Lemma 2.3.5 in [AMT] and the Remark after Corollary 2.2.8 in [AMT]. From Lemma 5.1 and Theorem 5.4 we obtain the following. Corollary 5.5. Let F be a (ξ,M,ε)-large family. For any δ>0 there exists L ∈ [M] such that for every N ∈ [L] and n ∈ N there exists F ∈ F with the property N − ξn ,F > 1 δ.

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For a more detailed study of Schreier families and RA-Hierarchy we refer to [AG]. Notation. Let X be a Banach space with a basis. We denote by: (1) Σ(X) the set of all normalized block sequences of the basis, i.e. Σ(X)= { ∞ || || ∈ } (xi)i=1 : xi

Lemma 5.6. Suppose that for some s ∈ Σ(X), ξ<ω1 we have that τξ (s)=δ>0. Then the following properties hold: F ∞ ∈ N ∈ N ∈ (1) For every =(Fi)i=1 Σ( ) and M [ ] there exists L [M] such that || N F || ∈ ∈ N ξn (s, ) < 2δ for every N [L] and n . F ∞ ∈ N ∈ N ∈ (2) There exist =(Fi)i=1 Σ( ) and M [ ] such that for all N [M],we || N F || 1 have that ξ1 (s, ) > 2 δ. (3) There exist F∈Σ(N) and M ∈ [N] such that for all N ∈ [M] we have that 1 || N F || 2 δ< ξ1 (s, ) < 2δ. Proof. Property (1) follows from Proposition 5.3 (the Dichotomy Principle) and property (2) follows directly from the deﬁnition. Property (3) is a combination of (1) and (2).

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To explain the nature of the quantity τξ (s) let us assume that τξ (s)=δ>0. Then from the previous Lemma 5.6 (2) we have that there exist F∈Σ(N)and ∈ N || N F || δ ∈ M [ ] such that ξ1 (s, ) > 2 for every N [M]. It follows readily that for ∈ N ∈ F || N · || δ every N [M]thereexistst Σ(s, ) such that ξ1 t > 2 . The diﬃculty is that by changing from N to L it is possible that the sequence tN will also change to some other sequence tL. This forces us to be more careful in our proofs. The advantage of the above deﬁnition comes essentially from part (1) of Lemma 5.6. Indeed, if for a given ε>0 we knew that 0 <τξ (s)=δ<ε, then for every F∈ N ∈ N ∈ || N F || Σ( ) and every M [ ]thereexistsL [M] such that ξ1 (s, ) < 2δ<2ε for all N ∈ [L]. This permits us to have a uniform 2δ-control over all convex combinations of a certain form and at the same time we know by part two of Lemma 5.6 that there are convex combinations of the same form which remain δ greater than 2 . This key observation is crucial for our approach.

Lemma 5.7. Let X be a Banach space with a basis and ξ<ω1. (a) If s1,s2 ∈ Σ(X) and s2 ≺ s1, then τξ (s1) ≥ τξ (s2). (b) If Y ≺ Z ≺ X,thenτξ (Y ) ≥ τξ (Z). (c) If s1,s2 ∈ Σ(X) and s2 is a tail subsequence of s1,thenτξ (s1)=τξ (s2). (d) If Z is a tail subspace of Y ,thenτξ (Y )=τξ (Z).

Proof. (a) It is enough to show that if for θ ∈ (0, 1) we have τξ (s2) >θ,then also τξ (s1) >θ.Suppose that τξ (s2) >θfor some θ ∈ (0, 1). Then there exist F ∞ ∈ N ∈ N ∈ N ∈ =(Fn)n=1 Σ( )andM [ ] such that for every N [M] we can selectP t F || N · N || ∞ ∞ Σ(s2, )with ξ1 t >θ.Let s1 =(yn)n=1 and s2 =(zn)n=1, zn = λiyi, i∈En ∞ N where (En)n=1 isS a sequence of successive subsets of . 0 F 0 0 ∞ F 0 ⊂ F 0 We set Fn = Ei, =(Fn)n=1.ThenΣ(s2, ) Σ(s1, ) and therefore i∈Fn N 0 for every N ∈ [M] the sequence t also belongs to Σ(s1, F ). We thus obtain that || N · N || ξ1 t >θ. Properties (b), (c), (d) are obvious.

Lemma 5.8. For every ξ<ω1 there exists a block subspace Y of X such that τξ (s)=τξ (Y ) for every s ∈ Σ(Y ). ∞ ∞ Proof. We inductively define sequences (si)i=1, (Yi)i=1 such that s1 =(en)n∈N is ≺ ∈ − 1 the basis of X, sn+1 sn, Yi = span [si], si+1 Σ(Yi)andτξ (si+1) τξ (Yi) < i+1 . ∞ ∈ Let s be a diagonal block sequence of (si)i=1 and set Y = span[s]. If t Σ(Y ), then for every k there exists tk ≺ sk, a tail subsequence of t, which satisfies τξ (Yk) ≤ τξ (tk)=τξ (t)=τξ (tk+1) ≤ τξ (sk+1), by Lemma 5.7 (a) and (c). Therefore for 0 ∈ | − 0 | 1 ∈ N 0 t, t Σ(Y ), τξ (t) τξ (t ) < k for all k and hence τξ (t)=τξ (t ). The following definition is the natural extension of Definition 5.3 to n-averages N of (ξk )k∈N and it will be used at a certain stage of our proof.

Deﬁnition 5.4. (a) For every s ∈ Σ(X), ξ<ω1, n ∈ N we set: N N ξ1 + ... + ξn τξ,n(s)= sup sup inf || (s, F)||. F∈Σ(N) M∈[N] N∈[M] n

(b) We deﬁne τξ,n(X)= inf τξ,n(s). s∈Σ(X)

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Remark 5.2. (a) For n = 1 the above definition coincides with Definition 5.3. (b) The corresponding results of Lemmas 5.7 and 5.8 stated for τξ,n instead of τξ also hold and the proofs are completely analogous. In particular for a given ξ<ω1, n ∈ N, there exists a block subspace Y of X such that τξ,n(Y )=τξ,n(s) for every s ∈ Σ(Y ). ∞ ∈ F ∞ ∈ N Definition 5.5. For δ>0, s =(yi)i=1 Σ(X)and =(Fi)i=1 Σ( )we <∞ ∗ ∗ define: F (s, F)={G ∈ [N] : ∃x ∈ S ∗ such that ||x | || >δ ∀l ∈ G}, δ G X G YFl

where YFl = span [(yi)i∈Fl ]. Remark 5.3. If X is a reflexive Banach space with a 1-unconditional basis, then it is easy to check that the set Fδ (s, F) is an adequate and compact family of finite subsets of N. In the sequel X will denote a reflexive Banach space with a 1-unconditional basis. ∞ ∈ F Lemma 5.9. Suppose that for some ξ<ω1, s =(yi)i=1 Σ(X) and = ∞ ∈ N ∈ N || N F || (Fi)i=1 Σ( ) there exists M [ ] and δ>0 such that ξ1 (s, ) >δfor ∈ ∈ L ⊂ F all N [M]. Then there exists L [M] such that Sξ Fδ/2(s, ), i.e. for every L ∗ ∗ δ G ∈ S there exists x ∈ S ∗ satisfying ||x | || > for every l ∈ G. ξ G X G YFl 2 Proof. It follows from stability property (4) of the summability methods that for ∈ N ∈ N F ∈ F all n , N [ M ]wehavethat ξn (s, ) >δ. Further, given t Σ(s, ) ∗ ∗ N · ∈ ∗ N · suchP that ξn t >δwe choose x SX such that x (ξn t) >δ. It follows that {ξN (l):||x∗| || > δ } > δ . Therefore the compact and adequate family n YFl 2 2 F δ Fδ/2(s, )is(ξ,M, 2 )-large. The result follows from Theorem 5.4 and the proof is complete. ∈ F∈ N ∈ N Lemma 5.10. Let s Σ(X), Σ( ), M [ ] and δ> 0. Assume that for ζN +...+ζN 2 ∈ 1 l F k>n δ +1 and ζ<ω1 we have that for all N [M], k (s, ) >δ. ∈ L n ⊂ F Then there exists L [M] such that [Sζ ] Fδ/4(s, ).

∗ ∗ ζN +...+ζN ∈ ∗ ∈ F 1 k · Proof. We observe that if x BX , t Σ(s, )aresuchthatx ( k t) >δ, then there exist {k δ for all 1 2 n S ki 2 { ≤ i =1, ..., n. Hence we can deﬁne a partition of [M]= A(k1,...,kn) :1 k1 < ... < ∗ ≤ } ∈ ∈ ∗ kn k with the rule N A(k1,...,kn) if and only if there exists x BX with x∗(ζN · t) > δ for all i =1, ..., n. ki 2 From the dichotomy principle (Proposition 5.3) there exist L ∈ [M]and1≤ ≤ ⊂ k1 < ... < kn k such that [L] A(k1,...,kn). It follows from stability property (4) of the repeated averages hierarchy that for every N ∈ [L]thereexistsN 0 ⊂ N such 0 that N ⊂ A(1,...,n). From this we can conclude that actually [L] ⊂ A(1,...,n). Let ∗ ∗ ∈ ∈ ∗ N · δ N [L]andchoosex BX nsuch that x ( ζi t) > o2 for i =1, ..., n.Thenfor P every i =1, ..., n we get that ζN (d): x∗ > δ > δ . Therefore F (s, F) i YFd 4 4 δ/4 δ is (n, ζ, L, 4 )-large and the result follows from Theorem 5.4. Lemma 5.11. Let ζ<ω, s ∈ Σ(X), F =(F ) ∈ Σ(N), δ>0 and k ∈ N. 1 h ni n∈N k ∈ N L ⊂ F Assume that for some L [ ] we have that Sζ Fδ (s, ). Then there exists 0 k G∈Σ(N), G⊂F , such that for every G ⊂Gand M ∈ [N] we have that [Sζ ] ⊂ 0 ζM +...+ζM 0 G 1 k G Fδ (s, ) and k (s, ) >δ.

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G { } { } Proof. Let L =(li)i∈N.Weset = Gi i∈N = Fli i∈N. It follows from the L G k ⊂ G G0 ⊂G deﬁnition of Sξ and the choice of the family that [Sξ ] Fδ (s, ). If and G0 ⊂ N ∈ G0 0 Σ( ), then there exists M [L] such that =(Fli )i∈M .SetM =(li)i∈M . h ik h ik 0 ∈ M 0 ⊂ L ⊂ G 0 Then M [L] and it is easy to check that Sζ Sζ Fδ (s, ). If M = G0 G k ⊂ G0 (mi)i∈N,then =(Fmi )i∈N andasinthecaseof we get that [Sζ ] Fδ (s, ). ζM +...+ζM 0 ∈ N 1 k G Therefore for every M [ ], k (s, ) >δand the proof is complete. Remark 5.4 . The content of the above lemmas is that at the moment when we ζN +...+ζN 1 k F F ∈ know that k (s, ) >δfor some δ>0, the pair (s, )andallN [L], then by passing to a certain G⊂Fa similar inequality holds for the pair (s, G), δ N the number 4 instead of δ, and, what is most important, for all L subsets of .

Lemma 5.12. Let s ∈ Σ(X) and ξ<ω1 such that τξ (s) >δ>0. Then for every δ ζ<ξ, τζ (s) > 2 .

Proof. Lemmas 5.9 and 5.11 (see also Remark 5.4) yield G⊂Fsuch that Sξ ⊂ F δ (s, G). By property (5) of the Schreier families there exists n ∈ N such that 2 Sζ [N\{1, ..., n − 1}] ⊂ Sξ ⊂ F δ (s, G). Then for L ∈ [N\{1, ..., n − 1}] it follows 2 L G δ δ that ζ1 (s, ) > 2 and hence τζ (s) > 2 .

Remark 5.5. It follows from Lemma 5.12 that if τξ (s)=0forsomeξ<ω1 and 0 s ∈ Σ(X), then τξ0 (s) = 0 for every ξ ≥ ξ.

Before passing to the next result we make a brief introduction to c0-trees. A tree T is said to be well-founded if every linearly ordered subset of T is finite. For a well-founded tree we define the derivative tree T 0 to be the subtree of T consisting of all non-maximal elementsT of T ; then we inductively define the ξ-derivative as T (ζ+1) T (ζ) 0 T (ξ) T (ζ) T =( ) and = ζ<ξ if ξ is a limit ordinal. The order o( )ofa well-founded tree T is the smallest ordinal ξ such that the ξ-derivative of T is an empty set. c0-trees: Let X be a separable Banach space. For δ>0, we define T (c0,δ)= { || || 1 ∞} (x1, ..., xn): xi =1and(x1, ..., xn)is δ -equivalent to the usual basis of `n . Then T (c0,δ) with the natural order is a tree and further if c0 is not isomorphic to a subspace of X,thenT (c0,δ) is a well-founded tree. The c0-index of a space X not containing c0 is defined as o(X)=sup{o(T (c0,δ)) : δ>0}. The following result is due to J. Bourgain [Bo1]:

Theorem 5.13. Let X be a separable Banach space not containing c0. Then the c0-index o(X) is a countable ordinal.

In the next proposition we will show that the assumption τξ (s) > 0 implies that ∗ o (X ) ≥ ξ and from this we will conclude that there exists a countable ordinal ζ0 such that τξ (s) = 0 for all s ∈ Σ(X)andξ ≥ ζ0. We begin with some notation and preliminary remarks. (a) For s =(y ) ∈ Σ(X), F =(F ) ∈ Σ(N)andk ∈ N we denote Y˜ = h S l l∈N i k n Fok ∈ ˜ ∈ ˜ span ei : ei ∈ suppyl and we denote by suppYFk the set i : ei YFk . l Fmk (b) For s ∈ Σ(X), F∈Σ(N)andδ>0considerD ⊂ Fδ (s, F). We denote by ∗ ∗ C { ∈ }⊂ ∗ ∈ F (D)anyset xG : G D BX where each xG witnesses that G Fδ (s, ). F Observe that from the deﬁnition of Fδ (s, ) and the fact that YFk is a subspace

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˜ ∈ F { } ≤ ≤ of YFk we get that if G Fδ ( s, )andG = m1, ..., md , then for all 1 i d,

x∗ | ∈ X∗ and x∗ | >δ. Therefore if X has a 1-unconditional basis, G Y˜F G Y˜F nmi omi then x∗ | : i =1, ..., d is 1 equivalent to the uvb of c (d). Furthermore if G Y˜F δ 0 mi Pd ε< δ and {x∗}d are such that x∗ | − x∗ <ε,then{x∗ : i =1, ..., d} is 2 i i=1 G Y˜F i i i=1 k 1 δ2 equivalent to the uvb of c0 (d). ⊂ F C { ∗ ∈ } Lemma 5.14. Assume that for D Fδ (s, ) and (D)= xG : G D there exists M ∈ [N] such that Sξ [M] ⊂ D. Then for every k0 ∈ N there exists L ∈ [M] ∗ and x ∈ BX∗ such that the following hold: ∗ ⊂ ˜ (i) suppx suppYFm . k0 ⊂{ }∪ (ii) mk0 < min L and if G mk0 L is a maximal member of Sξ with

∗ | − ∗ min G = mk0 ,then x ˜ x <ε. G YFm k0 Proof. Since Y ∗ is a finite-dimensional subspace of X∗ there exists an ε-net Fm k0 { ∗ ∗} x1, ..., xd in BY˜ ∗ . We recall that property (4) of Schreier families implies that Fm k0 ∈ ∞ ∈ N d for L [M], L =(lj)j=1 there exists d so that (lj)j=1 is a maximal element of Sξ .For1≤ i ≤ d define the set  h i   N ∈ M\{m }k0 :ifN 0 = N ∪{m } and G is the unique maximal   i i=1 k0  element of S defined by an initial segment of N 0, Ai = ξ .    ∗ | − ∗  then xG Y˜ xi <ε Fm k0 Sd \{ }k0 It is immediate that [M mi i=1]= Ai and each Ai is open. Hence by the i=1 ∈ \{ }k0 infinite Ramsey Theorem (Theorem 5.2) we obtain L [M mi i=1] such that ⊂ ∈{ } ∗ ∗ [L] Ai0 for some i0 1, ..., d .Setx = xi and observe that for every maximal ⊂{ }∪ 0 ∈ { }∪ element G mk0 L of Sξ with min G = mk0 there exists N [ mk0 L]such that G is an initial segment of N. Therefore the pair (x∗,L) satisfies the desired conclusion. Proposition 5.15. Let X be a reflexive Banach space with an unconditional basis. ∗ Assume that for some ξ<ω1 and s ∈ Σ(X), τξ (s) > 0. Then the c0-index of X is greater than or equal to ξ. Proof. We prove by induction the following statement: The inductive hypothesis: If D ⊂ Fδ/2(s, F) is such that Sξ [M] ⊂ D and C { ∗ ∈ } T⊂{ ∗ ∗ (D)= xG : G D , then for every ε>0 there exists a tree (x1, ..., xd): ∗ ∈ N ∈ ∗ } d , xi BX satisfying the following properties: (i) o (T ) ≥ ξ. ∗ ∗ ∈T (ii) If (x1, ..., xd) ξ , then there exists a maximal element G of Sξ , contained Pd in M and G0 ⊂ G such that G0 = {m , ..., m }, ||x∗ | − x∗|| <εand 1 d G Y˜F i i=1 mi suppx∗ ⊂suppY˜ . i Fmi Proof of the inductive hypothesis: For ξ = 0 it is obvious. Suppose that

ξ is a limit ordinal, (ξn)n∈N is the increasing sequence that deﬁnes Sξ and D,

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C (D), ε>0 are given. Suppose that M ∈ [N] such that Sξ [M] ⊂ D.Welet { ∈ ≥ } ⊂ Mn = m M : m n . It follows from the deﬁnition of Sξ that Sξn [Mn] ∈ Sξ [M]. To each U Sξn [Mn] there corresponds a unique maximal element GU of ⊂ { ∈ } C Sξ [M]withU GU . Then for the families Dn = U : U Sξn [Mn] , (Dn)= { ∈ } GU : U Sξn [Mn] and the given ε>0 the inductive assumptions are fulﬁlled

and hence there existsS Tξn satisfying conditions (i)and(ii). It is easily checked T T that the tree ξ = ξn satisﬁes conditions (i)and(ii) for the families D and n C (D)andthegivenε>0. This completes the proof for the limit ordinal case. Suppose now that ξ = ζ +1 and D, C (D), ε>0aregivensuchthatSξ [M] ⊂ D. ∈ ≥ 0 ε Choose mk0 M with mk0 2. Then by Lemma 5.14 for ε = 2 there exists ∗ ∗ ∈ ∗ ⊂ ˜ ∈ ⊂ x BX with suppx suppYFm and L [M] such that for every G L, k0

∗ | − ∗ 0 G maximal in Sξ ,minG = mk0 we have that x ˜ x <ε.Observe G YFm k0 ∈ that the deﬁnition of Sζ+1 implies that for every G Sζ [L], mk0 0thereexist Σ( ) ∈ N ∈ N F τξ (s) and M [ ] such that, for every N [M], ξ1 (s, ) > 2 = δ. Therefore ∈ L ⊂ F F by Lemma 5.9 there exists L [M] such that Sξ F δ (s, ). If =(Fn)n∈N we G 2⊂ G set =(Gn)n∈N =(Fk)k∈L. Then it follows that Sξ Fδ/2(s, ) (Lemma 5.11). ∈ G ∗ ∈ G To each G Fδ/2(s, ) we assign xG witnessing that G Fδ/2(s, )andweset C ∗ ∈ G δ = xG : G Fδ/2(s, ) .Let0<ε< 4 . Then applying the inductive hypothesis for the families Fδ/2(s, G), C, and the number ε we get a tree T which satisﬁes properties (i)and(ii). Property (i) implies that o (T ) ≥ ξ. Also property (ii) ∗ ∗ ∈T 2 and the unconditionality of the basis of X show that every (x1, ..., xd) is δ2 ∗ equivalent to the uvb of c0 (d). Hence indeed o (X ) ≥ ξ.

Proposition 5.16. Every reﬂexive Banach space X with an unconditional basis contains a subspace Y with the following property: There exists a unique ξY <ω1 such that 6 (1) τξY (Y )=0and τξ (Y ) =0for every ξ<ξY ,and (2) for every ξ<ω1, s ∈ Σ(Y ) and n ∈ N, τξ,n(s)=τξ,n(Y ). Proof. As was shown in Lemma 5.8, for a countable ordinal ξ and Z ≺ X there exists Yξ ≺ Z such that τξ (s)=τξ (Yξ ). By similar arguments it is also shown that there exists Yξ,n such that the same holds for the quantity τξ,n (Remark 5.1). For such a subspace we say that τξ,n is stabilized in Yξ,n. Therefore for every Z ≺ X and ξ<ω1 we inductively choose Z Yξ,1 Yξ,2 ... Yξ,n ... such { } that τξ,n is stabilized in Yξ,n.IfYξ is any diagonal subspace of Yξ,n n∈N,then (τξ,n)n∈N are simultaneously stabilized on Y . From Theorem 5.13 and Proposition 5.15 we obtain a countable ordinal ζ0 such that for every ξ ≥ ζ0 and s ∈ Σ(X) we have that τξ (s) = 0. We deﬁne ξX =min{ξ : for every s ∈ Σ(X), τξ (s)=0}. Observe that whenever τξ (s) = 0, then also τξ,n (s) = 0 for all n ∈ N. Clearly ξX is a countable ordinal and ξX > 0. Enumerate the set {ξ : ξ<ξX } as a sequence

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(ξn)n∈N and inductively choose a sequence (Yn)n∈N of block subspaces of X such ≺ that Yn+1 Yn and (τξn ,k)k∈N is stabilized in Yn.LetY be any diagonal subspace of the sequence (Yn)n∈N. It is readily checked that Y is the desired subspace and ξY , which is smaller than or equal to ξX , satisfies the desired property (1). Definition 5.6. A Banach space Y which satisfies the conclusion of the previous proposition is called a stabilized space. F 1 F 2 F ≺F ∈ Notation. For 1 =(Fl S)and 2 =(Fl ),wewrite 2 1 if there exists (Gl)l∈N N 2 1 ∈ N Σ( ) such that Fl = Fd , for all l . d∈Gl Proposition 5.17. Let X be a stabilized reflexive space with an unconditional basis. Then the following hold: (a) If ξ = ξX is a limit ordinal and (ξn)n∈N is the increasing sequence that defines

Sξ ,thenlim τξn =0. (b) If ξ = ξX is of the form ξ = ζ +1,thenlimn τζ,n =0. Proof. We ﬁrst show that (a) holds. By Lemma 5.12 it is enough to prove that

limτξn = 0. Assume on the contrary that limτξn > 0. Since the space X is stabilized ∈ N ∈ there exists δ>0 such that τξn (s) >δfor all n and every s Σ(X). Let F ⊂ N s =(en)n be the basis of X. We inductively construct ( n)n∈ N Σ( )such

F F ∈ N M G δ that n n+1 for all n , and we have that (ξn)1 (s, n) > 2 for every Gn ∈ Σ(N), every Gn ⊂Fn, and every M ∈ [N]. We show how to construct F1, F2 and in the same manner we get the general inductive step. G ∈ N ∈ N Since τξ1 (s) >δ,Lemma 5.6 (2) and Lemma 5.9 yield 1 Σ( )andL1 [ ] L1 such that S ⊂ F δ (s, G1). Hence from Lemma 5.11 there exists F1 ⊂G1 satisfying ξ1 2 F 1 ∈ 1 the inductive hypothesis for n =1.Let 1 = Fl l∈N and choose nl Fl .Weset

s1 =(enl )l∈N.SinceX is stabilized, τξ2 (s1) >δ. Therefore as in the case n =1 L2 there exists G2 ∈ Σ(N)andL2 ∈ [N] such that S ⊂ F δ (s1, G2). We show that ξ2 2 G0 ≺F L ⊂ G0 G there exists 2 1 such that S F δ (s, 2). Indeed, if 2 =(Gd) ∈N we set S ξ2 2 d G0 = F 1 . Then clearly G0 =(G0 ) ≺F .Further,ifY = he : l ∈ G i, d nl 2 d d∈N 1 Gd nl d l∈Gd 0 0 0 h ∈ i δ G ⊂ δ G then YGd is a subspace of YG = en : n G . Therefore F (s1, 2) F (s, ). d d 2 2 2 L 0 0 Hence S ⊂ F δ (s, G ). Again from Lemma 5.11 there exists F2 ∈ Σ(N), F2 ⊂G , ξ2 2 2 2 which satisfies the inductive hypothesis for n =2.Togofromthenth to the (n +1)th step we use exactly the same arguments as in the case from the first F step to the second. Therefore the sequence ( n)n∈N has been constructed. Let F n F n n =(Fk )k∈N and set ∞ =(Fn )n∈N. ∈ N || M · F || δ Claim. For every M [ ], we have that ξ1 (s, ∞) > 2 . It follows immediately that the proof of the claim derives a contradiction since we have assumed that τξ (s)=τξ (X) = 0. To prove the claim choose any M = M M M (mn)n∈N. Then from the definition of ξ1 we have that ξ1 =(ξm1 )1 .Sincefor m all n ≥ m , F ≺F it follows that there exists (F 1 ) ≥ such that l = m 1 n m1 ln n m1 m1 1 and F m1 ⊂ F n. We define G = {F m1 , ..., F m1 }∪{F m1 } . Clearly G⊂F , ln n 1 m1 ln n>m1 m1 therefore by the inductive assumption for L ∈ [N], we get that ||(ξ )L ·(s, G)|| > δ . m1 1 2 In particular for the set M, we conclude that ||(ξ )M ·(s, G)|| > δ .IfG =(G ) , m1 1 2 n n∈N

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≥ ⊂ n then for n m1 we have that Gn Fn . Hence for the given M it follows readily δ M M that < ||(ξ ) · (s, G)|| ≤ ||(ξ ) · (s, F∞)||. 2 m1 1 m1 1 This ﬁnishes the proof for the limit ordinal case. The case of successor ordinal is proven by similar arguments.

Proposition 5.18. Let A be a stabilized reﬂexive Banach space with an unconditional basis such that ξA is a successor ordinal. Then A satisﬁes property (P2).

Proof. Let ξA = ζ +1 andτ = τζ (A) > 0. It is easy to see that also τζ,n (A) > 0 for all n ∈ N. According to Proposition 5.17 we have that lim τζ,n(A)=0andso for every k ∈ N there exists n ∈ N such that τ (A) < τ . k ζ,nk 2k ∈ N ∞ ∈ For every k we set Ck = nk. We will prove that for every (xi)i=1 Σ(A)and ∈ N ∞ ≺ ∞ { } ∈ N k there exists t =(zi)i=1 (xi)i=1 such that for every M = mn n∈N [ ], zm +...+zm 1 nk 1 < , which actually implies property (P2) . nk k Let s =(x )∞ ∈ Σ(A)andB = h(x )∞ i.Sinceτ (s)=τ (A) < τ ,for i i=1 i i=1 ζ,nk ζ,nk 2k every F∈Σ(N )andM ∈ [N]thereexistsN ∈ [M] such that if t ∈ Σ(s, F), then ζN +...+ζN 1 nk τ · t < . On the other hand τζ (s)=τ and so there exist F1 ∈ Σ(N) nk 2k ∈ N || N F || τ ∈ ∈ N and M1 [ ] such that ζn (s, 1) > 2 for every N [M1] and every n . (n) ∈ N ∈ F || M1 · || τ For every n we choose tn =(yi )i Σ(s, 1) such that ζn tn > 2 . M1 Let kn =maxsuppζn and deﬁne

s˜ =(˜y )∞ =(y(1), ..., y(1),y(2) , ...y(2),y(3) , ...). i i=1 1 k1 k1+1 k2 k2+1

ζM1 ·s˜ Then it is readily checked that ||ζM1 · s˜|| > τ for every n ∈ N.Wesetz = n , n 2 n M1 ζn s˜ ∞ t =(zn)n . =1 S∞ Let M ∈ [M ], M = {m ,m , ...}and set M = suppζM1 . 1 1 2 2 mk k=1 Observe that by property (4) of the repeated averages hierarchy we have that M1 · M1 · zm +...+zm ζm s˜+...+ζm s˜ || M1 · || || M2 · || 1 nk ≤ 2 1 nk ζm s˜ = ζ s˜ . Therefore, = k k nk τ nk ζM2 ·s˜+...+ζM2 ·s˜ 2 1 n 1 k < , and the proof is complete. τ nk k

Next we will deal with stabilized Banach spaces A for which ξA is a limit ordinal. We start with some notation and an auxiliary lemma.

Notation. (a) We recall that for A as above, XA denotes the space deﬁned in the fourth section, and for a ﬁnitely supported x in XA we denote by range (x)the set {δ ∈ D : ∃α, β ∈ supp (x)with|a|≤|δ|≤|β|}.Wewriterange(x)

Lemma 5.19. Let A be a stabilized reﬂexive Banach space with an unconditional basis. Then for every ξ<ω1 and s ∈ Σ(XA) we have that τξ (s)=τξ (A).

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Pmi ∈ ∞ Proof. Let s Σ(XA), s =(xi)i=1.Toeachxi = λα eα we assign a vector | | ! α =ni Pmi P | | ∈ ∞ ∈ x¯i = λα ek A. Clearlys ¯ =(¯xi)i=1 Σ(A). Furthermore if k=ni |α|=k F∈σ (N)andYk (F,s)=span [xi : i ∈ Fk], Y¯k (F,s)=span [¯xi : i ∈ Fk], then Yk (F,s) is isometric to Y¯k (F,s). It follows immediately from the deﬁnitions that τξ (s)=τξ (¯s) and since A is stabilized we get τξ (s)=τξ (A). The proof is complete.

We recall that W denotes the closed symmetric convex subset of XA deﬁned in Section 4. Proposition 5.20. Let A be a stabilized reﬂexive Banach space with an unconditional basis such that ξA is a limit ordinal. Then there exists a null sequence a =(an)n∈N of positive numbers such that the set W in XA is an a-thin set.

Proof. Since τξA (XA) = 0 it follows from Proposition 5.17 that for the sequence

(ξn)n∈N used in the definition of SξA we have that lim τξn (XA) = 0. Therefore for τ ∈ N 1 ξnk every k there exists ξ such that τ < k and we define a = .Our nk ξnk k2 k 32k goal is to show that for a =(ak)k∈N, W is an a-thin set. Assume on the contrary that this is not the case. Then there exists an infinite- dimensional closed subspace Z of XA and λ>0 such that for all n ∈ N, BZ ⊂ k λ(2 W + akBXA ). Choose k>1000λ and for the rest of the proof we will denote by ζ the ordinal ξ and by τ the positive real number τ (A). We may also assume that Z is a nk ξnk block subspace generated by a sequence (zl)l∈N with range (zl)

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There exists M ∈ [N] such that for all N ∈ [M], 1 (1) ||ζN (t, F)|| > τ. 1 2 M According to Lemma 5.9 there exists M ∈ [M] such that S 1 ⊂ F τ (t, F). 1 ζ 2 Furthermore if G =(F ) ,thenS ⊂ F τ (t, G) (Lemma 5.11). Therefore we m m∈M1 ζ 4 may assume that the family F and the set M satisfy the following stronger property: ∗ ∗ For every N ∈ [M]thereexistsx ∈ B ∗ with ||x |Z || >τ/4 ∀m ∈suppζN . XA Fm 1 We choose N ∈ [M] such that condition (*) is also fulfilled. Consider the following family of finite subsets of N:   ∗  G ⊂ N : ∃x ∈ BX∗ ,and(ym) ∈ such that   G A m G   ∈ || || ≤   (a) ym ZFm and ym 1  0 ∗ 0 τ F = (b) ∀m∃E ⊂ rangeym : x (E ym) ≥ .  m G m 16   0   and the sets (Em)m∈G are pairwise incomparable  || ∗ | || (c) xG XFm < 8τ Claim. There exists ε>0 such that F is (ζ,N,ε)-large. Suppose that the claim has been proved. Then we derive a contradiction in the following manner. By Theorem 5.4 and Lemma 5.1 (see also Corollary 5.5) there exists an L ∈ [N] ∈ Q N \{ N } such that if Q [L], then the set G =suppζ1 minsuppζ1 belongs to F. By property (1) of the repeated averages hierarchy we can find Q ∈ [L]such that X 1 (2) ζQ(m) > . 1 2 m∈GQ ∗ ∗ Q ∈ ∈ ∗ Since G F, there exist a functional x = x Q BX and a sequence G PA Q (ym)m∈GQ satisfying conditions (a), (b), (c). We sety ˜ = ζ1 (m)ym,y= P m∈GQ Q 0 ζ1 (m)Emym. m∈GQ ∈ || ||−1 − · k 0 PChoosew ˜ W0 such that y˜ y˜ λ 2 w˜ <λak and set wm = Emw,˜ w = −1 k wm,u= ||y˜|| y − λ · 2 w.Then ∈ Q m G −1 k (3) kuk = ||y˜|| y − λ · 2 w <λak. By (1) we get that (4) ||y˜|| < 2τ, and by (2) and condition (b) of the definition of F τ (5) x∗ (y) > . 32 0 Since the sets (Em)m∈GQ are pairwise incomparable we get that w is a convex 0 ∈ ∈ Q combination of vectors wm BXFm ,m G . Therefore by property (c) of the definition of the family F we have that (6) x∗ (w) ≤ 8τ. ∗ 1 τ − Combining relations (4), (5) and (6) above, we obtain that x (u) > 2τ 32 k 2 λ8τ>λak which contradicts (3). Hence it remains to prove the claim.

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∗ ∗ τ For any L ∈ [N]setG =suppζL and choose x ∈ B ∗ such that ||x | || > n XA ZFm 4 for all m ∈ G. ( ∗ ∗ ∗ ∗ + ∗ − ∗ + x (eδ)ifx (eδ ) > 0 We write x =(x ) − (x ) ,where(x ) (eδ )= and ( 0otherwise ∗ ∗ ∗ − −x (eδ)ifx (eδ ) < 0 (x ) (eδ )= . 0otherwise ∗ + τ ∗ − τ We set G = m ∈ G : ||(x ) | || > , G− = m ∈ G : ||(x ) | || > . + P ZPFm 8 ZFm 8 L 1 L 1 Then either ζ1 (m) > 2 or ζ1 (m) > 2 . m∈G+ m∈G− Suppose that X 1 (7) ζL(m) > 1 2 m∈G+ ∗ ∗ + 1 ∗ { ∈ || | || } ∈ and set z =(x ) , G = m G+ : z XFm < 8τ .Sinceforeacht =(tm)m G, ∈ || || || L · || tm XFm , tm =1wehave ζ1 t < 2τ, (7) yields that X 1 (8) ζL(m) > . 1 4 m∈G1 Further choose τ (9) y ∈ Z such that ||y || =1andx∗(y ) > . m Fm m m 4 1 For every m ∈ G choose wm ∈ W such that k (10) ||ym − λ2 wm|| <λak. 1 τ Set ε1 = 32 2k λ and Em = range (ym) . We apply Proposition 4.4 to ﬁnd a 0 ⊂ ∗ \ 0 partition D1, ..., Dp of the set G and Em Em such that (i) z ((Em Em) wm) <ε1, 0 (ii)thesets(E ) ∈ are pairwise incomparable for every i =1, ..., p and (iii) m m Di P ε2 5 ∈{ } N 1 1 p< ε2 .By(8)thereexistsi0 1, ..., p such that ζ1 (m) > 4p > 20 .By 1 ∈ m Di0 ∗ 0 τ − k − τ − 1 τ − τ τ (9), (10) and (i)wehavethatz (Emym) > 8 2 λε1 λak > 8 32 2kλ λ 32k > 16 . ε2 ∈ 1 Therefore Di0 F. Hence for ε = 20 the set F is (ζ,N,ε)-large as claimed. This completes the entire proof. Theorem 5.21. Let A be a reﬂexive Banach space with an unconditional basis. Then there exists a block subspace B and a null sequence a such that the set W in XB is an a-thin set. Proof. From Proposition 5.16 we choose a stabilized block subspace B of the space A.ThenifξB is a successor ordinal, then the result follows from Proposition 5.18, while if ξB is a limit ordinal, then it follows from Proposition 5.20.

6. Thin norming sets III (c0(N)case)

In this section! for A = c0 (N) we construct a Banach space XA isometric to P∞ ⊕ 1 n ∗ ∗ ` (2 ) and a set W which norms a subspace of XA w -isometric to n=1 ∗ 0 (c0 (N)) .ThesetW is in a way similar to the reﬂexive space case. The proof that ∗ W is a thin set uses the structure of the w -closure of W in the second dual of XA.

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Theorem 6.1. ∗ There exists a Banach space Xc0 with separable dual Xc0 and a

closed, bounded convex symmetric subset W of Xc0 such that W is a thin subset of X , c (N)∗ is w∗-isometric to a closed subspace Y of X∗ and W is a 1 -norming c0 0 c0 2 set for Y .

We define the spaces Xc0 and Y in a manner similar to the reflexive case. We denote by D thedyadictree,i.e.everyδ ∈ D has exactly two immediate successors. The tree D naturally coincides with the set of all finite sequences {(ε1, ..., εn):n ∈ N and εi ∈{0, 1},fori =1, 2, ..., n}. We will use the notation and the definitions about the tree D introduced in Section 4. In the! vector space P P

c (D) we deﬁne the following norm: λδ eδ =max |λδ | . 00 ∈ δ∈D n N |δ|=n

The space Xc0 is the completion of c00(D) with the above defined norm. It is clear P ⊕ 1 n ∗ that Xc0 is isometric to the space ` (2 ) .ThespaceXc0 is isometric to n∈N 0 P P ⊕ ∞ n ∗∗ ⊕ 1 n the space ` (2 ) and Xc0 is isometric to ` (2 ) .Thelast n∈N 1 n∈N ∞ 1 n space is not a d-product of (` (2 ))n∈N. P ∗ ∗ The space Y is defined as follows: We set yn = eδ ,whereδ = |δ|=n,δ(n)=1 ∗ (ε1, ..., εn), δ(n)=εn,andY = span [(yn)n∈N]. It is easy to see that the space Y ∗ ∗ N is w -isometricP to c0( ). Next we define the set W . For any initial segment s of D we let xs = eδ . Clearly, ||xs|| =1. δ∈s { } We set W = coP xs : s is an initial segment . For any infinite branch γ of D ∗ ∗ ∗∗ we set x = w − e = w − lim x | . Clearly each x ∈ X \X .Itremains γ δ →∞ γ n γ c0 c0 δ∈γ n to prove the following:

1 Proposition 6.2. The set W is 2 -norming for the space Y and it is a thin subset

of Xc0 .

1 Proof. The conclusion that the set W is a 2 -norming set can be proved by similar arguments as in Lemma 4.1. It remains to be shown that W is a thin subset of

Xc0 . Assume on the contrary that W is not a thin set. Then for every ε>0thereexists a normalized block sequence (z ) ∈N in X and λ>0 such that B ⊂ λW + εB , n n Z Xc0 ∗ ∗∗ h ∈Ni ˜ where Z = (zn)n . WedenotebyW the w -closure of W in Xc0 and by the ∗ compactness of W˜ in the w -topology we get that BZ∗∗ ⊂ λW˜ + εBX∗∗ .We c0 denote by Br(D), S(D) the sets of infinite branches and finite initial segments of D respectively. Define Kc = {xγ : γ ∈ Br(D)}, Kd = {xs : s ∈ S(D)} and K = w∗ Kc ∪ Kd. It is easy to check that W˜ = co (K ∪−K). Finally M(Kc) denotes the space of the regular finite Borel measures on Kc, M1(Kc)={µ ∈M(K):||µ|| ≤ 1} ∗∗ → ∗∗ and Q : Xc0 Xc0 /Xc0 is the natural quotient map. We will use the following three lemmas.

Lemma 6.3. If R : M1(Kc) → W˜ is the natural aﬃne map from the unit ball of the Borel measures on Kc onto W˜ ,thenQR is an isometry between M1(Kc) and Q(W˜ ).

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∗ Proof. Notice ﬁrst that (Kc,w ) is a compact metric space homeomorphic to the N Cantor set {0, 1} . Hence the family (Nδ )δ∈D,whereNδ = {xγ : δ ∈ γ},isabasis N ∈M for the topologyP of Kc and each δPis a clopen subset of Kc.Nowforµ 1(Kc), ||µ|| = lim |µ(N )| = lim |Rµ(e )| = d(Rµ, X )=||QRµ|| and this →∞ δ →∞ δ c0 n |δ|=n n |δ|=n completes the proof of the lemma. { } ⊂M { } Let µi i∈I 1(Kc)andδ>0. We say that µi i∈I is δ-singular if there { } ⊂M || || { − } exists νi i∈I 1(Kc) such that vi <δand µi νi i∈I is a pairwise singular family.

Lemma 6.4. Let (µξ )ξ∈Ξ be an uncountable family of measures in M1(Kc).Then for any δ>0 there exists an uncountable family {{ξi,ζi},i∈ I} of pairwise disjoint two-point sets such that {µξ − µζ ,i∈ I} is δ-singular. i i P 1 Proof. It is well known ([L]) that M1(Kc) is isometric to ⊕ L (µα ) where w α<2 1 {µα }α<2w is a maximal family of pairwisen o singular probability measures in M1(Kc). dµ ∈M The isometry assigns the vector dµ to each µ (Kc). Since Kc is a α α<2w 1 compact metric space, each L (µα ) is a separable Banach space. Given δ>0, for

P w dµξ δ each ξ ∈ Ξ we choose a ﬁnite subset Fξ of 2 such that µξ − < . Apply dµα 4 α∈Fξ the Erd¨os-Rado theorem ([KM]) to get an uncountable subset Ξ0 of Ξ and a subset ω ∈ 0 6 ∩ ∅ F of 2 such that for ξ1,ξ2 Ξ , ξ1 = ξ2,wehaveFξ1 Fξ2 = F .IfF = ,then δ the measures (µξ )ξ∈Ξ0 are pairwise -singular, hence for any family {{ξi,ζi},i∈ I} 4 0 of pairwise disjoint two-point subset of Ξ , the measures {µ − µ } ∈ are pairwise ξiP ζi i I δ -singular. Therefore we assume that F =6 ∅.Wesetτ = dµξ for any ξ ∈ Ξ0. 2 ξ dµα α∈F P 1 Since ⊕ L (µα ) is a separable Banach space there exists an uncountable α∈F 1 {{ }} 0 family ξi,ζi i∈I of pairwise disjoint two-point subsets of Ξ such that for each ∈ || − || δ { − i I, τξi τζi < 4 . It is easy to check that the members of the family µξi µζi : i ∈ I} are pairwise δ-singular.

In the sequel by a normalized block sequence in the space Xc0 we understand a

sequence (zn)n∈N of vectors of Xc0 with ﬁnite supports (En)n∈N andsuchthat

max{|δ| : δ ∈ En} < min{|δ| : δ ∈ En+1}.

Lemma 6.5. Let (zn)n∈N be a normalized block sequence in the space Xc0 and { } N ∈ Mξ ξ∈Ξ anP uncountable almost disjoint family of subsets of . For any ξ Ξ ∗∗ ∗ { ∗∗} set zξ = zn,inthew -sense. Then the family Qzξ ξ∈Ξ is isometrically n∈Mξ equivalent to the usual basis of c0(Ξ).

Proof. Since ( Mξ )ξ∈Ξ are almost disjoint and (zn)n∈N is a normalized block se- Pn Pn ∗∗ ∗∗ quence, then λiQz = lim λiz |{δ : |δ|≥n} =max{|λi| :1≤ i ≤ ξi →∞ ξi i=1 n i=1 n}.

1 Completion of the proof of Proposition 6.2. For ε = 16 choose a normalized block sequence (z ) ∈N and λ>0 such that B ⊂ λW +εB ,whereZ = span [(z ) ∈N]. n n Z Xc0 n n

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N Let (Nξ )ξ∈Ξ be an uncountableP almost disjoint family of subsets of and for any ∗∗ ∗∗ ∈ ∗∗ ξ Ξ deﬁne zξ = zn.Since(zξ )ξ∈Ξ is a subset of BZ there exists a family n∈Nξ { } M || − ∗∗|| || ∗∗|| µξ ξ∈Ξ in 1(Kc) such that λQRµξ Qzξ <ε.Since Qzξ = 1, it follows that 1 − ε<||λµξ || = ||λQRµξ || < 1+ε. 1 { − } ∈ By Lemma 6.4 we get an uncountable family λµξi λµζi i I which is 16 - singular. Since ||Q(z∗∗ − z∗∗)|| =1,wegetthat1− 2ε<||λQL(µ − µ )|| < ξi ζi ξi ζi 1+2ε and also we have that ||λQL(µ − µ ) − (z∗∗ − z∗∗)|| < 2ε. Hence, 1 = ξi ζi ξi ζi Pn Pn Q(z∗∗ − z∗∗) ≥ λQR(µ − µ ) −2ε ≥ (1−2ε−2δ)n−2ε ≥ 1 − 1 n− 2 ξi ζi ξi ζi 4 4 i=1 i=1 which leads to a contradiction for n ≥ 4. The proof of the proposition is complete. Theorem 6.1 follows immediately from Proposition 6.2.

7. Constructions of block-H.I. d-products In this section we proceed to the construction of block-H.I. d-product norms. The section is divided into two parts. The first is devoted to a brief presentation of mixed Tsirelson d-product spaces and the second to the construction of a Gowers- Maurey type block-H.I. norm. The norm defined here is an adaptation of the Gowers-Maurey definition. We follow closely their proof and there are only a few points where we proceed in a different manner. ∞ ∈ 7.1. Mixed Tsirelson d-products. We start by recalling that forx ˜ =(xn)n=1 Q∞ Ω00 = Xn we have denoted by supp (˜x)theset{n ∈ N : xn =06 },by n=1 00 range (˜x) the interval of N [min supp (˜x) , max supp (˜x)], and by Ex˜ the vector PE (˜x) for E ⊂ N. We also use the notationx ˜1 < x˜2 to denote blocksx ˜1, x˜2 with supp (˜x1) < supp (˜x2). Let M be an adequate family of finite subsets of N. A sequence (E1, ..., En)of finite subsets of N is called M-admissible if there exists a set {m1, ..., mn}∈M such that m1 ≤ E1

Proposition 7.1. Let (Xn, ||.||)n∈N be a sequence of Banach spaces. For every Q∞ ∈ N F ⊆ ∗ ⊂ ∗ ∗ n we choose n Xn Ω00 = Xn . Then for every sequence n=1 00 S∞ M ∞ K⊂ ∗ F (θk, k)k=1 there exists a smallest set Ω00 containing n and closed n=1 under the (θk, Mk)-operation for all k ∈ N. Furthermore if each Fn is countable so is K. S∞ Proof. To deﬁne the set K we proceed by induction. For l =0wesetK0 = Fn. n=1 If K has been deﬁned we set l ( ) Xd K d ⊂K M ∈ N ∪K l+1 = θk φi :(φi)i=1 l is k-admissible, k l. i=1

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S∞ Finally, we set K = Kl. It is easily verified that K is closed under the (θk, Mk)- l=0 operation for all k ∈ N and moreover any other K0 having this property and con- S∞ taining Fn contains K as well. Hence K is the smallest set with these properties. n=1 S∞ Also if Fn is countable, then inductively we show that every Kl is countable, n=1 hence K is countable. M ∞ M Let (θk, k)k=1 be such that each k is a compact and adequate family of finite subsets of N and 0 <θk < 1, limk θk =0. ∞ F ⊂ Definition 7.1. Let (Xn)n=1 be a sequence of Banach spaces and choose n ∗ K BXn a symmetric and norming set. We denote by the set resulting from Propo- {F }∞ { M }∞ M ∞ sition 7.1 applied on the families n n=1, (θk, k) k=1. Then the (θk, k)k=1- mixed Tsirelson d-product norm on Ω00 is defined by: ||x˜|| =sup{φ (˜x):φ ∈K}.

Remark 7.1. (i) If Xn = R and Fn = {en, −en}, then the above definition coincides with the mixed Tsirelson spaces introduced in [AD1]. M ∞ || || (ii) An alternative description of the (θk, k)k=1 norm is given by the implicit formula ( ( )) Xd ||x˜|| =max ||x˜||∞, sup sup θk ||Eix˜|| ∈N k i=1 || || {|| || ∈ N} M where x˜ ∞ =max xn n : n and the inside sup is taken over all k-admissible sequences (E1, ..., En). (iii) Most of the already existing examples use the following two collections of A A { ⊂ N | |≤ } compact adequate families. The first is ( k)k∈N where k = F : F k , and the second is (Sk) ∈N where Sk is the Schreier family defined in Section 5. k ∞ 1 Thus the Schlumprecht space S is defined by , Ak and the Argyros- log2(k+1) k=1 ∞ 1 Deliyanni spaces are built with the use of families of the form , Sn . mk k ∞ k=1 1 S (iv) The spaces obtained by families m , nk have the property that every k k=1 block subspace is an asymptotic `1-space. (v) All the mixed Tsirelson d-product norms are shrinking. Assuming further M −1 that there exists some k such that the Cantor-Bendixson index i ( k) >θk ,then the norm is boundedly complete. For example the later condition is always satisfied

in the AD-constructions and also in Schlumprecht space. In such a case if (Xn)n∈N are reflexive, then the d-product X˜ is also reflexive. k 1 A (vi) Let us point out that the finite collections of the form m , ni define P P i i=1 ⊕ ∞ ⊕ norms equivalent to ( Xn)p for some 1

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 282 S. A. ARGYROS AND V. FELOUZIS Q∞ ∗ ∗ K Consider the vector space Ω00 = Xn and denote by the smallest subset n=1 00 S∞ ∗ F of Ω00 containing n and closed under the following operations. n=1 PN K 1 (1) For every φ1 < ... < φN elements of the vectors log (N+1) φi and 2 i=1 PN √ 1 φi also belong to K. log2(N+1) i=1 (2) For every φ ∈Kand E segment of N, Eφ ∈K. (3) For every E ∈Kand r ∈ Q∩ [−1, 1], rφ ∈K.(Q denotes the set of rational numbers.) The set K is defined in a similar manner as in Proposition 7.1. Indeed, by K ∞ induction, we produce an increasing sequence ( l)l in the following manner. S∞ =0 K0 = Fn and if Kl has been defined we set n=1  P   r N  g(N) i=1pEφi : g (N)=log2 (N) K = K ∪ ∈ Q∩ − . l+1 l  or g (N)= log2 (N), r [ 1, 1] ,  E ⊂ N is an interval and φ1 < ... < φN belong to Kl S∞ It is easily checked that the set K = Kl satisfies the desired properties. Fur- l=0 ther the following additional properties can be established for the set K. (4) K is a countable set. ∈K ∈ N || || ≤ th (5) For every φ and n , φ (n) n 1. Here by φ (n)wedenotethen ∗ coordinate of φ which of course belongs to Xn. Next, as in the Gowers-Maurey proof, we will define an injection σ from the (N) set K = {(φ1, ..., φn):φ1 < ... < φn, φi ∈K} to a certain set L, a subset of N ∞ N . We begin by considering a very fast increasing sequence J =(jn)n=1 in . ∈ N More precisely we define J such that log2 (j1 +1) > 256 and for all n , 2 { }∞ log log log(jn+1) > 4jn. WedenotebyK the set j2i+1 i=0 and by L the set { }∞ K(N) → ∈K(N) j2i i=1. We define an injection σ : L such that for (φ1, ..., φN ) if N P 1 1 40 ≥| | S = σ (φ1, ..., φN ), φ = φi,then 20 log2 S +1 range (φ) . i=1 The H.I. norm we wish to define will be obtained by a symmetric subset D of the set K defined above. We start with the following two definitions. D ∗ ∞ → ∞ Definition 7.2. Let be a subset of Ω00 and let g :[1, ] [1, ] be a function. A vector φ in D is said to be an (M,g)-form with respect to D if there exists Pd ≤ d 1 d M and a sequence (φi)i=1 of successive blocks such that φ = g(M) φi and i=1 φi ∈Dfor i =1, ..., d. K ∈K An easy consequence of the inductive p definition of the set is that every φ K is either (M,log2 (x + 1)) or an M, log2 (x +1) -form with respect to . Definition 7.3. Let K and σ : K(N) → L be the set and the function defined above. (N) A finite sequence (φ1, ..., φN ) ∈K is said to be a special sequence provided { }⊂ that there exists M1, ..., MN L such that each φi is an (Mi, log2 (x + 1))-form with respect to K and for i =2, ..., N, Mi = σ (φ1, ..., φi−1).

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The norming set D. The set D is inductively deﬁned as follows. We set D0 = S∞ 0 0 φ : φ = rφ , r ∈ [−1, 1] ∩ Q, φ ∈ Fn . Assume that Dn has been constructed n=1 and we deﬁne Dn+1 as follows: We set ( ) r XN E = φ : φ < ... < φ , φ ∈D , r ∈ Q∩ [−1, 1] , n+1 log (N +1) i 1 N i n 2 i=1

n N r X Fn+1 = p φi :(φ1, ..., φN ) is a special sequence, log (N +1) 2 i=1 o N ∈ K, r ∈ Q∩ [−1, 1] and ﬁnally

Qn+1 = {Eφ : E is an interval of N, φ ∈En+1 ∪Fn+1} . D D ∪Q D {D }∞ We define n+1 = n n+1. The desired set is the union of all n n=0. Remark 7.2. The definitions of the function σ and the norming set D presented here are slightly different from the corresponding ones in the Gowers-Maurey space. The reason we follow this approach is to ensure that every φ belonging to D is D anp (M,g)-form with respect the set ,whereg is either equal to log2 (x +1)or ∈D log2 (x + 1). To see this, first we observe that each φ 0 is a (1,g)-form; next PM ∈D r D suppose that φ n+1 so φ = g(M) Eφi where φ1, ..., φM are in n.Butfrom i=1 the inductive definition for all i =1, ..., M the functional rEφi belongs to Dn. PM 1 0 0 ∈D Hence φ = g(M) φi with φi n. Later we will make use of this observation i=1 to define the analysis of the elements of D. It is also clear that D is a symmetric subset of the set K defined before. Therefore every (M,g)-form with respect to D is also an (M,g)-form with respect to K. In the sequel by the term (M,g)-form we will mean an (M,g)-form with respect to the set D. Q∞ ∈ || || { ∈D} Next forx ˜ Ω00 = Xn we set x˜ GM =sup φ (˜x):φ and denote n=1 00 ˜ || || by XGM the completion of Ω00 under GM . It follows from the properties of the D ˜ || || norming set that XGM is a d-product of (Xn, n)n∈N which is additionally 1-bimonotone. We shall prove that:

Theorem 7.2. The space X˜GM is a block-H.I. Banach space. The proof of this theorem is done in several steps. As we have mentioned we will follow the Gowers-Maurey approach. We begin by recalling Schlumbrecht’s class of functions. That is, the class H of all functions f :[1, ∞] → [1, ∞] satisfying the following properties: (i) f (1) = 1 and f (x) 1. (ii) f is strictly increasing and tends to inﬁnity. (iii) lim x−qf (x) = 0 for every q>0. x→∞ x (iv) The function f(x) is concave and non-decreasing. (v) f (xy) ≤ f (x) f (y) for every x, y ≥ 1.

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H ∈H √ It is easily verified that the function log2 (x + 1) belongs to and if f then f also belongs to H. Definition 7.4. Let X,˜ || || be a d-product of the sequence (X , || || ) ,and n n n∈N let f ∈H.Wesaythat X,˜ || || satisfies a lower f-estimate if for everyx ˜ ∈ X ( ) XN −1 ||x˜|| ≥ sup f (N) ||Eix˜|| : N ∈ N, E1 < ... < En, Ei is an interval of N . i=1

Remark 7.3. It follows from the properties of the norming set D that X˜GM satisﬁes alowerlog2 (x + 1)-estimate. ˜ || || || || We recall that for X, a d-product of (Xn, n)n∈N, a block subspace is

a subspace generated by a sequence (˜yk)k∈N such that suppy˜k < suppy˜k+1 (see the n notation at the beginning of Section 2). Let (˜xi)i=1 be a ﬁnite block sequence in Pn ˜ n ≥ X. Then the vectorx ˜ = x˜i is an `1+ average with constant C 1provided i=1 || || || || ≤ −1 n x˜ =1and x˜i Cn .An`1+ vector with constant C is any positive multiple n of an `1+ average with constant C. ∈H ˜ || || || || Lemma 7.3. Let f and let X, be a d-product of (Xn, n)n∈N satisfying a lower f-estimate. Then for every block subspace Y of X˜ and every C>1, Pn ∈ N n n ,thereexisty˜1, ..., y˜n successive blocks in Y so that y˜ = y˜i is an `1+ i=1 average with constant C.

Proof. Suppose that the result is false and let Y be a block subspace of X, n ≥ 2 n and C>1 such that Y does not contain any `1+ -average with constant C.Then Y satisﬁes the following condition: Ifx ˜1 < ... < x˜n, n ∈ N, is any block sequence in Y with ||x˜i|| ≤ M for each i, −1 then ||x˜1 + ... +˜xn|| ≤ nMC . ∞ Let (˜xi)i=1 be any normalized block sequence of vectors of Y .Usingtheprevious observation we prove by induction that for every s ∈ N we have

s Xn ≤ s −s (1) x˜i n C . i=1 Since X satisﬁes a lower f-estimate,

s Xn ≥ s −1 s (2) x˜i f (n ) n . i=1 s −s ≥ ∈ N s s −q By (1) and (2) f (n ) C 1 for every s .Setq =logn C;thenf (n )(n ) ≥ 1 for every s ∈ N which contradicts the fact that lim f (x) x−q =0forevery x→∞ q>0.

n The next more technical result also concerning `1+ averages will be used in the ﬁnal constructions.

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use INTERPOLATING HEREDITARILY INDECOMPOSABLE BANACH SPACES 285 Lemma 7.4. Let X,˜ || || satisfy a lower f-estimate, D⊂B ∗ a norming set and X 1 ∈ N ˜ || || 0 <ε< 4 . Then for every n and every block subspace Y of X, there exist: Pn ∈ n (i) x˜ = x˜i Y ,an`1+ average with constant 1+ε. i=1 (ii) y˜∗ ∈Dsuch that y˜∗ (˜x) > 1 − ε and range (˜x) ⊂ range (˜y∗). ∈ N 0 0 ε Proof. For a given n we select ε > 0 such that ε < n2 .Then (1 + ε0)(n − 1) 1+ 1 (n − 1) 1 (1) < n = 1 − < 1 − ε0. n +2 n n2

nP+2 From the previous lemma for every block subspace Y of X˜ there existsx ˜ = x˜i, i=1 n+2 0 ∗ ∈D ∗ − 0 an `1+ average with constant 1 + ε .Wechoose˜x such thatx ˜ (˜x) > 1 ε . nP+1 nP+1 Pn 0 x˜i − ε ≤|| || ≤ We setz ˜ = x˜i andy ˜ = ||z˜|| = y˜i.Evidently,1 2 n+2 z˜ 1. Using i=2 i=2 i=1 ∗ ∗ (1) we get thatx ˜ (˜xi) =0forall6 i =1, ..., n +2 andsorange (˜y) ⊂ range (˜x ). 0 0 1−ε0−2 ε 1+ε 0 0 ∗ ≥ n+2 − || || ≤ n+2 1+ε ≤ 1+ε 1+ε Also, x (˜y) − ε0 > 1 ε. Finally, y˜i − ε0 = n+2−2ε0 n < n 1 2 n+2 1 2 n+2 n and thereforey ˜ is an `1+ average with constant 1 + ε. Lemma 7.5. Let M,N ∈ N,letC ≥ 1,let X,˜ || || be a bimonotone d-product ∈ ˜ n and let x˜ X be an `1+ vector with constant C.AlsoletE1 < ... < EM be a sequence of intervals. Then XM 2M ||E x˜|| ≤ C 1+ ||x˜|| . j N j=1

PM In particular, if M ≤ N,then ||Ej x˜|| ≤ 3C ||x˜||. j=1 The proof of this lemma is quite easy and we refer the reader to Lemma 4 of [GM]. R → R Let f (x)=log2 (x +1)andletMf : be the function deﬁned by Mf (x)= f −1 36x2 .Forε>0wedenotebyε0 the number min {ε, 1}. We shall say that a sequencex ˜1 < ... < x˜N is a rapidly increasing sequence (or R.I.S.) of `1+ averages, nk for f of length N with constant 1+ε,ifxk is an `1+ average with constant 1 + ε 0 ≥ 2(1+ε)Mf (N/ε ) for each k, n1 ε0f 0(1) ,and ε0 p f (n ) ≥|range (˜x − )| 2 k k 1 for k =2, ..., N.Heref 0 (1) is the derivative of f at 1. Next we give the statements of two lemmas which correspond to Lemmas 10 and 12 in [GM]. Their proofs are exactly the same as those of Lemmas 10 and 12 in [GM]. Thus we refer the reader to Gowers-Maurey’s paper for the proof of these results. We should point out that Lemmas 10 and 12 in [GM] constitute the hard part of the proof that the space is H.I. In their proofs Gowers and Maurey make elegant use of the properties of Schlumprecht class H. Lemma 10 is of an

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unconditional nature and the important role of special sequences appears in Lemma 12.

Lemma 7.6. Let N ∈ L, n ∈ [log N,exp N] and ε>0. Also let x˜1, ..., x˜n be a Pn 0 n R.I.S. with constant 1+ε.Then x˜i ≤ (1 + ε + ε ) . log2(n+1) i=1 GM ˜ Let us observe that since XGM satisﬁes a lower log2 (n + 1)-estimate we get that Pn n ≤ x˜i . log2(n+1) i=1 GM ∈ ∗ ∗ Lemma 7.7. Let k K and let x˜1, ..., x˜k be a special sequence of length k,where ∗ each x˜i is an (Mi, log2 (x +1))-form. Let x˜1, ..., x˜k be a sequence of successive vectors such that every x˜i is a normalized R.I.S. vector of lengthMi with constant Pk Pk ε 1 ⊂ ∗ ∗ ≤ 1+ 4 , ε = 10 . Assume that range (˜xi) range (˜xi ) and x˜i Ex˜i 2 i=1 i=1 for every interval E.Then

Xk k ≤ x˜i (1 + 2ε) . log2 (k +1) i=1 GM Let us notice that the statement of Lemma 7.7 is slightly diﬀerent from the corresponding statement in Lemma 12 in [GM]. In spite of this change the proof of this lemma goes exactly along the lines of the proof of Lemma 12 in [GM].

Completion of the proof of Theorem 7.2. Let Y and Z be block subspaces of ε ˜ || || 1 ∈ ∈ 4 ≥ XGM , GM .Forε = 10 , k K we choose M1 = j2k L such that M1 4M k ≥ f ( ε ) ∗ ∗ N1 εf 0(1) where f (x)=log2 (x + 1). We inductively deﬁnex ˜1, ..., x˜k ,˜x1, ..., x˜k so that the following conditions are fulﬁlled:

• x˜2j+1 ∈ Y ,˜x2j ∈ Z andx ˜1 < ... < x˜k. • ∗ D ⊂ ∗ x˜i is an (Mi,f)-form with respect to the set , range (˜xi) range (˜xi )and ∗ ∗ (˜x1, ..., x˜k) is a special sequence. • ∗ − 1 1 Eachx ˜i is a normalized R.I.S. of length Mi and x˜i (˜xi) 2 < k . To prove that such a choice is possible we follow closely the corresponding con- ∗ struction in [GM], p.868. We begin with the deﬁnitions ofx ˜1 andx ˜1.Weﬁrstchoose ε ∗ + x˜11, ..., x˜1M1 a R.I.S. of `1 averages with constant 1+ 4 .Foreach˜x1j we select x ˜1j D ∗ − ε ⊂ ∗ in such thatx ˜1j (˜x1j) > 1 4 and further assume that range (˜x1j) range x˜1j (Lemma 7.3). AccordingtoLemma7.6wehavethat

XM1 ε −1 x˜1j ≤ 1+ M1 log (M1 +1) 2 2 j=1 GM ! MP1 MP1 1 ∗ ε M1 and also x˜ x˜1j ≥ 1 − . log (M1+1) 1j 4 log (M1+1) 2 j=1 j=1 2

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MP1 Therefore settingx ˜1 to be the normalized multiple of x˜1j we get that j=1 MP1 1 x˜∗ (˜x ) > 1 and so there exists r ∈ [0, 1] ∩ Q such that log (M1+1) 1j 1 2 2 j=1

XM1 r ∗ 1 1 x˜ (˜x1) − < . log (M +1) 1j 2 k 2 1 j=1

MP1 We setx ˜∗ = r x˜∗ which clearly belongs to D and further range (˜x ) ⊂ 1 log (M1+1) 1j 1 2 j=1 ∗ ∗ ∗ range (˜x1). Thus the choice ofx ˜1,˜x1 is complete. To choosex ˜2,˜x2 we set M2 = ∗ σ (˜x1) and we repeat the above procedure for the number M2 instead of M1 and ∈ x˜2j Z, j =1, ..., M2, such thatx ˜1 < x˜21 < ... < x˜2M2 . In the general inductive step we proceed in a similar manner. It remains to prove that Y + Z does not form a topological direct sum. We deﬁney ˜ =˜x1 +˜ x3 + ... andz ˜ =˜x2 +˜x4 + .... Then Pk Pk √ 1 ∗ y˜ ∈ Y andz ˜ ∈ Z.Further||y˜ +˜z|| = x˜i ≥ x˜ (˜xi) ≥ GM log (k+1) i i=1 GM 2 i=1 k √ 1 k − − i−1 ∗ k 2 1 . Also since ( 1) x˜i ,(˜xi )i=1 satisfy the assumptions of log2(k+1) i=1 Pk i−1 1 Lemma 7.7 we have that ||y˜ − z˜|| = (−1) x˜i ≤ (1 + 2ε) k . GM log2(k+1) i=1 GM ∈ ∈ || || ≥ Hence for every δ>0thereexist˜y Y ,˜z Z such that y˜ +˜z GM −1 || − || δ y˜ z˜ GM which immediately shows that the sum Y + Z is not closed. The proof of the theorem is complete.

Proposition 7.8. For every sequence (Xn)n∈N of separable Banach spaces the d- product space X˜GM defined above is boundedly complete and shrinking. Therefore if each Xn is a reflexive space, then the space X˜GM is also reflexive. (For a proof we refer to p. 869 of [GM].) Remark 7.4. It is also possible to define block-H.I. spaces with an adaptation of the construction of asymptotic `1 H.I. spaces contained in [AD2]. Such a construction will produce block-H.I. spaces with the additional property that every block subspace is an asymptotic `1 space. The rest of this section is devoted to some results which will be used to show that there exist H.I. spaces X and T : X → X which is a strictly singular and non- compact operator. For the proof of these results we use the analysis of a functional φ ∈Dintroduced in [AD2] and defined below. ∈D D m Definition 7.5. Let φ .Ananalysis of φ is a sequence ( s(φ))s=0 such that D { } ∈D (1) s(Sφ)= f1, ..., fds where f1 < ... < fds and fi s for i =1, ..., ds, (2) suppf =suppφ, f∈Ds(φ)) s s−1 (3) if s>0andψ ∈ K (φ)\K (φ), then there exist M ∈ N and ψ1, ..., ψd, Pd ≤ D 1 d M,elementsof s−1(φ) such that ψ = g(M) ψi where g (M)=log2 (M +1) p i=1 or g (M)= log2 (M +1)and (4) Dm(φ)={φ}.

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∈D D m Let us observe that every φ has an analysis ( s)s=0 (see Remark 7.2). ∞ ˜ Definition 7.6. A sequence (˜xi)i=1 of successive blocks of XGM is called an infi- ni nite rapidly increasing sequence or a R.I.S. ofp constant C ifx ˜i is an `1+ av- ∈ N i | | erage of constant C and for each i it holds that log2 (ni+1) > 2 range (˜xi) . √ 1 In the sequel by S3 we denote the mixed Tsirelson space T , A3n log (n+1) 2 n∈N K and by S3 the norming set of S3 as it is defined in Definition 7.1 and Remark 7.1 (i).

∞ ˜ ni Proposition 7.9. Let (˜xi)i=1 be a R.I.S. in XGM of `1+ average with constant 4 ∈ 0 ∈ ∞ ⊂ R C< 3 . Then for every φ K there exists φ KS3 such that for every (λi)i=1 , P∞ P∞ P∞ 0 φ λix˜i ≤ 4φ |λi|ei +2max{|λi| : i ∈ N}. Therefore λix˜i ≤ i=1 i=1 i=1 GM P∞

6 λiei . i=1 S3 ∈D D m { ∈ N Proof. For φ , we choose an analysis ( s(φ))s=0 of φ.LetI = i : supp (˜xi) Sm ∩supp (φ) =6 ∅}.Foreachi ∈ I there exist a unique element φi ∈ Ds(φ)andan s=0 element si ∈{0, ..., m} satisfying the following two conditions: (1) suppφi∩suppx˜i = suppφ∩suppx˜i and si s (2) φi ∈K (φ)andfors 0letMi be the natural number such that φi = Pdi 1 ψj,wheredi ≤ Mi and ψ1, ..., ψd are successive elements of Ds −1 (φ). g(Mi) i i j=1 0 00 0 We set I = {i ∈ I : Mi ≥ ni+1} and I = I\I .

| | P ∈ 0 | |≤√range(˜xi) 1 Observe that for every i I , φ (˜xi) < i hence φ λix˜i < log (n +1) 2 2 i i∈I0 max{|λi| : i ∈ N}. P Sm

It remains to estimate φ λix˜i . For every ψ ∈ Ds(φ)wesetDψ = i∈I00 s=0 00 {i ∈ I :suppψ∩suppx˜i =suppφ∩suppx˜i}. Sm ∈ D 0 ∈ K For every ψ s(φ) we inductively deﬁne a functional ψ 2 S3 such that s=0 the following are fulﬁlled: 0 (a) suppψ ⊂ Dψ , 0 (b) |ψ(˜xi)|≤4ψ (ei)and (c) one of the following two alternatives holds: 0 ∈K (c1) ψ S3 or (1) ∈K ∈ 00 (1) 0 ∗ (1) (c2)thereexistsψ S3 and i I , i

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Case 1. For every i ∈ J, M/∈ (ni,ni+1]. In this case we deﬁne   Xd X 0 √ 1  0 ∗ ψ = ψj + ei . log2(M+1) j=1 i∈J

Evidently condition (a) is fulﬁlled. To show that condition (c1)alsoholdsweset { ∈{ } 0 ∈K } { }\ F1 = j 1, ..., d : ψ S3 and F2 = 1, ..., d F1. 0 (1) ∗ (1) 0 For every j ∈ F2, ψ = ψ + e where ij < min supp ψ .Soψ = " j j ij # j P P P P √ 1 ψ0 + ψ(1) + e∗ + e∗ . j j ij i log2(M+1) j∈F1 j∈F2 j∈F2 i∈J | | | | | |≤ | |≤ 0 ∈K Since F1 + F2 + J 2M and F2 M it follows that ψ S3 .Itremains to show that condition (b) holds. Indeed if i/∈ J, then there exists a j ∈{1, ..., d} such that suppψj ∩suppx˜i =suppφ∩suppx˜i. By the inductive hypothesis |ψ (˜xi)| = 1 | |≤√ 1 | |≤√ 4 0 0 ∈ g(M) ψj (˜xi) ψj (˜xi) ψj (ei)=4ψ (ei). If i J,then log2(M+1) log2(M+1) ψ = φi and so M ≤ ni+1.ButM/∈ (ni,ni+1] and therefore M ≤ ni. From Lemma 7.5 we get that |ψ (˜x )|≤ 1 C 1+ 2M < √ 4 =4ψ0 (e ). i g(M) ni i log2(M+1) ∈ ≤ Case 2. There exists (a unique) i0 J such that ni0

In the last part of this section we will show that the space S3 satisfies property (P2) (see Remark 4.1). Let us observe that S3 satisfies a lower g estimate where x ≥ g (x)=log2 3 +1 for x 3. Therefore from Lemma 7.3 every block subspace Y n ≥ of S3 contains `1+ averages with constant C for all n 3andC>1. Also it is obvious that S3 is a mixed Tsirelson space with an unconditional basis and hence ∈K n every φ S3 admits an analysis. The definitions of `1+ averages with constant C and the infinite R.I.S. in block subspaces of S are exactly the same as in the space 3 ∞ √ 1 X˜GM . Finally we denote by S9 the space T , A9N . log (N+1) 2 N=1 ∞ Lemma 7.11. Let Y be a block subspace of S3 and let (xi)i=1 be an infinite R.I.S. ∞ ⊂ R | |≤ in Y with constant 1+ε, 0 <ε<1. Then for every (λi)i=1 with λi 1, ∞ ∞ P P λixi ≤ 6 λixi . i=1 S3 i=1 S9

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The proof of this lemma is similar to that of Proposition 7.9. In fact we may P∞ ∈K 0 ∈K ≤ prove that for every φ S3 there exists a φ S9 such that φ xi i=1 P∞ 0 4φ |λi| ei +2max{|λi| : i ∈ N}. This can be accomplished by using the anal- i=1 ysis of φ. Next we notice that for every n ∈ N

X9n 9n p (1) ei = . log2 (n +1) i=1 S9 ∞ 4 Therefore for every inﬁnite R.I.S. (xi)i=1 in the space S3 with constant C<3 we obtain via Lemma 7.11 and (1), that for every k ∈ N there exists nk such that for i < ... < i , 1 nk

xi + ... + xi 1 nk 6 1 S3 < p < . nk log2 (nk +1) k Thus we have the following

Corollary 7.12. The space S3 satisﬁes property (P2). Hence the set W in the

space XS3 is an a-thin set.

8. Final results This final section contains the main results of this paper. These are theorems concerning quotients of H.I. spaces and factorization of a-thin operators through H.I. spaces. In the final part we prove similar results for quotients of `p-saturated Banach spaces. Theorem 8.1. Let A be a reflexive Banach space with an unconditional basis satisfying property (P). Then there exists a reflexive H.I. Banach space X such that A is a quotient of X. Proof. Recall that property (P ) is defined in Definition 4.1. By Theorem 4.9 it follows that the set W in XA is an a-thin set for an appropriate positive null ∗ ∗ sequence a =(an)n∈N and it also norms a subspace Y of XA isometric to A .If || || n n is the equivalent norm defined by Minkowski’s gauge 2 W + anBXA ,thenby ˜ || || Proposition 7.8 the d-product XGM of the sequence ((XA, n))n∈N is reflexive. Since W is an a-thin subset of XA it follows by Theorem 7.2 and Proposition 3.1 that the diagonal space ∆X˜GM is a H.I. space. The space ∆X˜GM as a subspace of a reflexive space is in itself reflexive and further, A∗ is isomorphic to a subspace of ˜ ∗ ˜ ∆XGM (Proposition 3.4). It follows that ∆XGM is the desired space and the proof is complete

Theorem 8.2. Suppose that A is either c0 or a reflexive space with an unconditional basis such that every block subspace B of A contains a further block subspace Z complemented in A.ThenA is a quotient of a H.I. space X. In the later case X can be chosen to be reflexive. Proof. The proof is completely analogous to that of Theorem 8.1. We consider the space XA and the set W defined in Section 6 (Theorem 6.1) if A = c0 (N)and Section 4 (Theorem 4.11) if the space A is a reflexive space. In either case W is a

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thin norming subset of XA and the argument given in the last part of Theorem 8.1 is still carried over yielding the assertion of Theorem 8.2. Next, we list some classical spaces that are quotients of a H.I. space. Corollary 8.3. Every Lp (λ), 1

a-thin, for a positive null sequence a =(an)n∈N,ifT [BX ]isana-thin subset of Y . Theorem 8.5. If X, Y are Banach spaces and T : X → Y is an a-thin operator, then there exists a H.I. space D such that T is factorized through D.

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Proof. Set W = T [BX], which by our assumption is an a-thin subset of Y .Denote by || ||n the equivalent norm on Y defined by the Minkowski gauge of the set n ˜ || || 2 W + anBY . Consider next the d-product XGM of the sequence ((Y, n))n∈N which is by Theorem 7.2 a block-H.I. space. The diagonal space ∆X˜GM of X˜GM is a H.I. space (Proposition 3.1) and further on, the unit ball of ∆X˜GM contains the set W . Therefore the operator Q : X → ∆X˜GM , defined by Q (x)=T (x), is a bounded linear operator and T = j ◦ Q,wherej :∆X˜GM → Y , is the natural inclusion map. Setting D =∆X˜GM the result follows. Theorem 8.6. (a) For every r =6 p, 1 ≤ r, p<∞,andT ∈L(`r,`p), T is factorized through a H.I. Banach space. The same remains valid if one of `r (N), p ` (N) is substituted by c0 (N). (b) Every strictly singular operator T ∈L(`r (N) ,`p (N)) is factorized through a H.I. space. (c) The identity map I : L∞ (λ) → L1 (λ) is factorized through a H.I. space. Proof. (a) and (b) follow from Theorem 8.5 and Corollary 4.14. (c) follows from the above Theorem and Proposition 3.3 which assert that BL∞ is an a-thin subset of L1 (λ). Next we present a result related to the structure of L (X, X)forX a H.I. space. In particular we show that there exist H.I. spaces X with “many” T : X → X which are strictly singular and not compact. Theorem 8.7. There exists a H.I. space X such that for every infinite-dimensional closed subspace Z of X there exists a strictly singular operator T : X → Z which is not compact. ∞ √ 1 Proof. We recall that S3 denotes the space T , A3n .Thenby log (n+1) 2 n=1 Corollary 7.12 the set W in XS3 is an a-thin set and hence the diagonal space ∆X˜GM defined by the set W is a H.I. space that has the space S3 as a quotient. Let Z be any closed subspace of ∆X˜GM ; then there exists a normalized basic ∞ ˜ sequence in Z equivalent to a block sequence (˜yi)i=1 in XGM . It follows (Corollary 7.10) that there exists a non-compact bounded operator U : S3 → Z.IfQ denotes the surjection of ∆X˜GM onto S3,thenU ◦ Q is a non-compact strictly singular operator from ∆X˜GM to Z and the proof is complete.

A by-product of our method is the following result concerning `p(N)(1

Definition 8.1. A Banach space X is said to be `p(N)-saturated (c0 (N)- saturated) if every subspace Y of X contains a further subspace isomorphic to `p(N)(c0 (N)). Theorem 8.8. Let A be a reflexive Banach space with an unconditional basis. Then there exists a subspace B of A such that for every p ∈ (1, ∞) there exists an `p(N)-saturated reflexive space Xp which has the space B as a quotient. Further there exists a c0 (N)-saturated space with the same property. Proof. According to Theorem 5.21 there exists B, a subspace of A, with an unconditional basis such that the closed bounded symmetric set W in XB is an ∗ ∗ || || a-thin set, norming a subspace Y of XA isometric to B . We denote by n the

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n equivalent norm deﬁned on XA by the gauge 2 W + anBXA and by Xn the space (X, || ||n) . Then each Xn is a reﬂexive space and the same holds for the space P∞ X˜p = ⊕Xn ,1

Acknowledgments We would like to thank Bernard Maurey for his valuable help which allowed us to make an essential step forward and arrive at the present form of the paper. We also thank Nicole Tomczak-Jaegermann for the discussions we had during her visit at Herakleion and Athens. Her suggestion to use quantities similar to Deﬁnition 5.3 was really important to us. Finally, we thank Antonis Manoussakis, Irene Deliyanni and Apostolos Giannopoulos for their help during the preparation of the paper.

References

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Department of Mathematics, University of Athens, Athens, Greece E-mail address: [email protected] Department of Mathematics, University of Athens, Athens, Greece

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