Banach spaces without minimal subspaces

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BANACH SPACES WITHOUT MINIMAL SUBSPACES

VALENTIN FERENCZI AND CHRISTIAN ROSENDAL

Abstract. We prove three new dichotomies for Banach spaces a la W. T.Gowers’ dichotomies. The three dichotomies characterise respectively thespaces having no minimal subspaces, having no subsequentially minimal basicsequences, and having no subspaces crudely finitely representable in all of theirsubspaces. We subsequently use these results to make progress on the programof Gowers of classifying Banach spaces by finding characteristic spaces presentin every space. Also, the results are used to embed any partial order of sizeℵ1 into the subspaces of any space without a minimal subspace ordered byisomorphic embeddability. Finally, we analyse several examples of spaces andclassify them according to which side of the dichotomies they fall.

Contents

1. Introduction 22. Preliminaries 72.1. Notation, terminology, and conventions 72.2. Gowers’ block sequence game 82.3. A trick and a lemma 93. Tightness 103.1. Tight bases 103.2. A generalised asymptotic game 123.3. A game for minimality 173.4. A dichotomy for minimality 184. Tightness with constants and crude stabilisation of local structure 205. Tightness by range and subsequential minimality 246. Refining Gowers’ dichotomies 317. Chains and strong antichains 358. Tight spaces of the type of Gowers and Maurey 398.1. A locally minimal space tight by support 418.2. Two HI spaces tight by range 449. Tight spaces of the type of Argyros and Deliyanni 509.1. A strongly asymptotically ℓ1 space tight by support 509.2. A strongly asymptotically ℓ∞ space tight by support 5710. Open problems 59References 60

Date: May 2007.2000 Mathematics Subject Classification. Primary: 46B03, Secondary 03E15.Key words and phrases. Minimal Banach spaces, Dichotomies, Classification of Banach spaces.The second author was partially supported by NSF grant DMS 0556368 and by FAPESP.

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http://arXiv.org/abs/0711.1350v1

2 VALENTIN FERENCZI AND CHRISTIAN ROSENDAL

1. Introduction

In the paper [17], W.T. Gowers initiated a celebrated classification theory forBanach spaces. Since the task of classifying all (even separable) Banach spaces upto isomorphism is extremely complicated (just how complicated is made precise in[10]), one may settle for a loose classification of Banach spaces up to subspaces, thatis look for a list of classes of Banach spaces such that:

(a) each class is pure, in the sense that if a space belongs to a class, then everysubspace belongs to the same class, or maybe, in the case when the property dependson a basis of the space, every block subspace belongs to the same class,

(b) the classes are inevitable, i.e., every Banach space contains a subspace in oneof the classes,

(c) any two classes in the list are disjoint,(d) belonging to one class gives a lot of information about operators that may

be defined on the space or on its subspaces.

We shall refer to this list as the list of inevitable classes of Gowers. Many classicalproblems are related to this classification program, as for example the questionwhether every Banach space contains a copy of c0 or ℓp, solved in the negative byB.S. Tsirelson in 1974 [38], or the unconditional basic sequence problem, also solvednegatively by Gowers and B. Maurey in 1993 [18]. Ultimately one would hope toestablish such a list so that any classical space appears in one of the classes, and sothat belonging that class would yield most of the properties which are known forthat space. For example any property which is known for Tsirelson’s space is alsotrue for any of its block subspaces, so Tsirelson’s space is a pure space, and as such,should appear in one of the classes with a reasonable amount of its properties. Also,presumably the nicest among the classes would consist of the spaces isomorphic toc0 or ℓp, 1 6 p <∞.

After the discovery by Gowers and Maurey of the existence of hereditarily in-decomposable (or HI) spaces, i.e., spaces such that no subspace may be written asthe direct sum of infinite dimensional subspaces [18], Gowers proved that everyBanach space contains either a HI subspace or a subspace with an unconditionalbasis [16]. These were the first two examples of inevitable classes. We shall callthis dichotomy the first dichotomy of Gowers. He then used his famous Ramseyor determinacy theorem [17] to refine the list by proving that any Banach spacecontains a subspace with a basis such that either no two disjointly supported blocksubspaces are isomorphic (which, for reasons that will become apparent later on,we shall call tight by support), or such that any two subspaces have further sub-spaces which are isomorphic. He called the second property quasi minimality. Thissecond dichotomy divides the class of spaces with an unconditional basis into twosubclasses (up to passing to a subspace). Finally, recall that a space is minimal ifit embeds into any of its subspaces. A quasi minimal space which does not containa minimal subspace is called strictly quasi minimal, so Gowers again divided theclass of quasi minimal spaces into the class of strictly quasi minimal spaces and theclass of minimal spaces.

Obviously the division between minimal and strictly quasi-minimal spaces is nota real dichotomy, since it does not provide any additional information. The main

BANACH SPACES WITHOUT MINIMAL SUBSPACES 3

result of this paper is to provide the missing dichotomy for minimality, which weshall call the third dichotomy.

A first step in that direction was obtained by A. Pe lczar, who showed that anystrictly quasi minimal space contains a further subspace with the additional prop-erty of not containing any subsymmetric sequence [30]. The first author proved thatthe same holds if one replaces subsymmetric sequences by embedding-homogeneoussequences (any subspace spanned by a subsequence contains an isomorphic copy ofthe whole space) [9].

A crucial step in the proofs of [30] and [9] is the notion of asymptoticity. Anasymptotic game of length k in a space E with a basis is a game where I playsintegers ni and II plays block vectors xi supported after ni, and where the outcomeis the length k sequence (xi). Asymptotic games have been studied extensivelyand the gap between finite dimensional and infinite dimensional phenomena wasusually bridged by fixing a constant and letting the length of the game tend toinfinity. For example, a basis is asymptotic ℓp if there exists C such that for any k,I has a winning strategy in the length k asymptotic game so that the outcome isC-equivalent to the unit vector basis of ℓkp.

In [30] it is necessary to consider asymptotic games of infinite length, which aredefined in an obvious manner. The outcome is then an infinite block-sequence. Theproof of the theorem in [30] is based on the obvious fact that if a basic sequence(ei) is subsymmetric, then II has a strategy in the infinite asymptotic game inE = [ei] to ensure that the outcome is equivalent to (ei). In [9] a similar fact forembedding homogeneous basic sequences is obtained, but the proof is more involvedand a more general notion of asymptoticity must be used. Namely, a generalisedasymptotic game in a space E with a basis (ei) is a game where I plays integers ni

and II plays integers mi and vectors xi such that supp(xi) ⊆ [n1,m1]∪ . . .∪ [ni,mi],and the outcome is the sequence (xi), which may no longer be a block basis.

Observe that in the first round of such a game, II may pick some integer m1

so large, depending on arbitrary large n, that he has enough room to choose anyvectors x1, . . . , xn supported in [n1,m1] as the n first vectors of the outcome. Thismeans that everything is decided at the first round in a finite generalised asymptoticgame, and so only infinite generalised asymptotic games are of interest.

The second author analysed infinite asymptotic games in [33], showing that themost obvious necessary conditions are, in fact, also sufficient for II to have a strategyto play inside a given set. This was done through relating the existence of winningstrategies to a property of subspaces spanned by vectors of the basis with indicesin some intervals of integers. His methods extend to the setting of generalisedasymptotic games and motivate the following definition. A space Y is tight in abasic sequence (ei) if there is a sequence of successive intervals I0 < I1 < I2 < . . .of N such that for all infinite subsets A ⊆ N, we have

Y 6⊑ [en

∣

∣ n /∈⋃

i∈A

Ii].

In other words, any embedding of Y into [ei] has a “large” image with respect tosubsequences of the basis (ei).

We then define a tight basis as a basis such that every subspace is tight in it,and a tight space as a space with a tight basis.

As we shall prove in Lemma 3.7, using the techniques of [33], essentially a blocksubspace Y = [yi] is not tight in (ei), when II has a winning strategy in the


generalised asymptotic game in [ei] for producing a sequence equivalent to (yi).This relates the notion of tight bases to the methods of [9], and by extending thesemethods we prove the main result of this paper:

Theorem 1.1 (3rd dichotomy). Let E be a Banach space without minimal sub-spaces. Then E has a tight subspace.

Theorem 1.1 extends the theorems of [30] and of [9], since it is clear that atight space cannot contain a subsymmetric or even embedding homogeneous block-sequence. This dichotomy also provides an improvement to the list of Gowers: astrictly quasi minimal space must contain a tight quasi minimal subspace. Example3.6 shows that this is a non-trivial refinement of the unconditional and strictly quasiminimal class, and Corollary 4.3 states that Tsirelson’s space is tight. Theorem 1.1also refines the class of HI spaces in the list, i.e., every HI space contains a tightsubspace, although it is unknown whether the HI property for a space with a basisdoes not already imply that the basis is tight.

Our actual examples of tight spaces turn out to satisfy one of two stronger formsof tightness. The first is called tightness with constants. A basis (en) is tight withconstants when for for every infinite dimensional space Y , the sequence of successiveintervals I0 < I1 < . . . of N witnessing the tightness of Y in (en) may be chosen sothat Y 6⊑K [en

∣

∣ n /∈ IK ] for each K. We show that this is the case of Tsirelson’sspace.

The second kind of tightness is called tightness by range. Here the range, range x,of a vector x is the smallest interval of integers containing its support, and the rangeof a block subspace [xn] is

⋃

n range xn. A basis (en) is tight by range when for forevery block subspace Y = [yn], the sequence of successive intervals I0 < I1 < . . . ofN witnessing the tightness of Y in (en) may be defined by Ik = range yk for each k.This is equivalent to no two block subspaces with disjoint ranges being comparable.We show that tightness by range is satisfied by some HI spaces and also by a spacewith unconditional basis constructed by Gowers.

It turns out that there are natural dichotomies between each of these strongforms of tightness and respective weak forms of minimality. For the first notion,we define a space X to be locally minimal if for some constant K, X is K-crudelyfinitely representable in any of its subspaces. Notice that local minimality is easilyincompatible with tightness with constants. Using an equivalent form of Gowers’game as defined by J. Bagaria and J. Lopez-Abad [3] we prove:

Theorem 1.2 (5th dichotomy). Any Banach space E contains a subspace with abasis that is either tight with constants or is locally minimal.

There is also a dichotomy concerning tightness by range. This direction forrefining the list of inevitable classes of spaces was actually suggested by Gowers[17]. P. Casazza proved that if a space X has a shrinking basis such that no blocksequence is even-odd ( the odd subsequence is equivalent to the even subsequence),then X is not isomorphic to a proper subspace, see [14]. So any Banach spacecontains either a subspace, which is not isomorphic to a proper subspace, or issaturated with even-odd block sequences, and, in the second case, we may find afurther subspace in which Player II has a winning strategy to produce even-oddsequences in the game of Gowers associated to his Ramsey theorem. This fact was


observed by Gowers, but it was unclear to him what to deduce from the propertyin the second case.

We answer this question by using Gowers’ theorem to obtain a dichotomy whichon one side contains tightness by range, which is a slightly stronger property thanthe Casazza property. On the other side, we define a space X with a basis (xn) tobe subsequentially minimal if every subspace of X contains an isomorphic copy ofa subsequence of (xn). This last property is satisfied by Tsirelson’s space and willalso be shown to be incompatible with tightness by range.

Theorem 1.3 (4th dichotomy). Any Banach space E contains a subspace with abasis that is either tight by range or is subsequentially minimal.

It is easy to check that the second case in Theorem 1.3 may be improved to thefollowing hereditary property of a basis (xn), that we call sequential minimality:every block sequence of [xn] has a further block sequence (yn) such that everysubspace of [xn] contains a copy of a subsequence of (yn).

Theorem 1.3 divides any class of strictly quasi minimal spaces into two subclasses.Combining Theorem 1.1 and Theorem 1.3 we therefore refine Gowers’ list to sixclasses which are enumerated in Theorem 6.3. We obtain two subclasses of strictlyquasi minimal unconditional spaces: no example of the first one is known, andthe other is illustrated by Tsirelson’s space. We also obtain two subclasses of HIspaces; we prove that a space constructed by Gowers [15] belongs to the first one,Proposition 8.8, but know of no example belonging to the second. It is an interestingproblem to look for examples for the two classes that haven’t yet been proved tobe non empty.

It has been proved in [6] that every minimal space with a strongly asymptoticallyℓp basis, 1 6 p < +∞, contains a subspace isomorphic to ℓp. A basis is stronglyasymptotically ℓp if finite families of disjointly supported (but not necessarily suc-cessive) normalised blocks supported “far enough” are uniformly equivalent to thebasis of ℓnp . Leaving aside the case of strongly asymptotically ℓ∞ spaces, this meansthat a space which does not contain a copy of c0 or ℓp must contain a further sub-space which either does not contain a minimal subspace, or does not contain astrongly asymptotically ℓp subspace. In the first case Theorem 1.1 implies thatsome further subspace is tight. In the second case, another dichotomy theorem dueto A. Tcaciuc [37] implies that some subspace Y is what we shall call uniformlyinhomogeneous, i.e. ∀ǫ > 0∃n ∈ N, ∀Y1, . . . , Yn ⊆ Y, ∃y1, z1 ∈ Y1, . . . ∃yn, zn ∈ Yn :‖∑n

i=1 yi‖ < ǫ‖∑ni=1 zi‖, where yn, zn are of norm 1. New subclasses for spaces

which do not contain a copy of c0 or ℓp may therefore be distinguished. We indi-cate how we may use this observation and the 5th dichotomy to refine the classesin the list of Theorem 6.3, to subclasses so that “spaces isomorphic to c0 or ℓp”appears as one subclass, Proposition 6.4. Interestingly, classical examples such asTsirelson’s space, its dual, and Schlumprecht’s space [34] each appear in a differentsubclass of this new list. So this seems a reasonable direction for an “ultimate” clas-sification in which each “pure” classical space, including c0 and ℓp, would appearas a representative of some subclass which would list most of its properties.

The six dichotomies and the interdependence of the properties involved can bevisualised in the following diagram.


Strongly asymptotic ℓp ∗ ∗ Tcaciuc’s dichotomy ∗ ∗ Uniformly inhomogeneous⇓ ⇑

Unconditional basis ∗ ∗ 1st dichotomy ∗ ∗ Hereditarily indecomposable⇑ ⇓

Tight by support ∗ ∗ 2nd dichotomy ∗ ∗ Quasi minimal⇓ ⇑

Tight by range ∗ ∗ 4th dichotomy ∗ ∗ Sequentially minimal⇓ ⇑

Tight ∗ ∗ 3rd dichotomy ∗ ∗ Minimal⇑ ⇓

Tight with constants ∗ ∗ 5th dichotomy ∗ ∗ Locally minimal

From a different point of view, coming from combinatorics and descriptive settheory, Theorem 1.1 also has important consequences for the isomorphic classifi-cation of separable Banach spaces. To explain this, suppose that X is a Banachspace and SB∞(X) is the class of all infinite-dimensional subspaces of X orderedby the relation ⊑ of isomorphic embeddability. Then ⊑ induces a partial orderon the set of biembeddability classes of SB∞(X) and we denote this partial orderby P(X). Many questions about the isomorphic structure of X translate directlyinto questions about the structure of P(X), e.g., X has a minimal subspace if andonly if P(X) has a minimal element and X is quasi minimal if and only if P(X)is downwards directed. In some sense, a space can be said to be pure in case thecomplexity of P(X) does not change by passing to subspaces and Gowers, Problem7.9 [17], motivated by this, asked for a classification of, or at least strong struc-tural information about, the partial orders P for which there is a Banach space Xsaturated with subspaces Y ⊆ X such that P ∼= P(Y ). A simple diagonalisationeasily shows that such P either consist of a single point (corresponding to a min-imal space) or are uncountable, and, using methods of descriptive set theory andmetamathematics, this was successively improved in [11] and [32] to either |P | = 1or P having a continuum size antichain. Using a strengthening of Theorem 1.1, weare now able to show that such P , for which |P | > 1, have an extremely complexstructure by embedding any partial order of size at most ℵ1 into them.

For A,B ⊆ N, we write A ⊆∗ B to mean that A \B is finite.

Theorem 1.4. Given a Banach space X, let P(X) be the set of all biembeddabilityclasses of infinite-dimensional subspaces of X, partially ordered under isomorphicembeddability. Let P be a poset for which there exist a Banach space X such thatX is saturated with subspaces Y such that P(Y ) ∼= P . Then either |P | = 1, or ⊆∗

embeds into P . In the second case it follows that

(a) any partial order of size at most ℵ1 embeds into P , and(b) any closed partial order on a Polish space embeds into P .

From the point of view of descriptive set theory, it is more natural to studyanother problem, part of which was originally suggested to us by G. Godefroysome time ago. Namely, the space SB∞(X), for X separable, can easily be madeinto a standard Borel space using the Effros–Borel structure. In this way, therelations of isomorphism, ∼=, and isomorphic embeddability, ⊑, become analyticrelations on SB∞(X) whose complexities can be measured through the notion of


Borel reducibility. We obtain Theorem 1.4 as a consequence of some finer resultsformulated in this language and that are of independent interest.

In Sections 8 and 9 various (and for some of them new) examples of “pure”spaces are analysed combining some of the properties of tightness or minimalityassociated to each dichotomy. We provide several examples of tight spaces fromthe two main families of exotic Banach spaces: spaces of the type of Gowers andMaurey [18] and spaces of the type of Argyros and Deliyanni [1]. Recall that bothtypes of spaces are defined using a coding procedure to “conditionalise” the normof some ground space defined by induction. In spaces of the type of Gowers andMaurey, the ground space is the space S of Schlumprecht, and in spaces of thetype of Argyros and Deliyanni, it is a mixed (in further versions modified or partlymodified) Tsirelson space associated to the sequence of Schreier families. The spaceS is far from being asymptotic ℓp and is actually uniformly inhomogeneous, andthis is the case for our examples of the type of Gowers-Maurey as well. On the otherhand, we use a space in the second family, inspired from an example of Argyros,Deliyanni, Kutzarova and Manoussakis [2], to produce strongly asymptotically ℓ1and ℓ∞ examples with strong tightness properties.

In the last section, a final list of 19 inevitable classes of “pure” Banach spacesassociated to the six dichotomies is given, with examples for 8 of these classes,Theorem 10.1. Open problems are also stated.

2. Preliminaries

2.1. Notation, terminology, and conventions. We shall in the following almostexclusively deal with infinite-dimensional Banach spaces, so to avoid repeating this,we will always assume our spaces to be infinite-dimensional. The spaces can alsosafely be assumed to be separable, but this will play no role and is not assumed.Moreover, all spaces will be assumed to be over the field of real numbers R, thoughthe results hold without modification for complex spaces too.

Suppose E is a Banach space with a normalised Schauder basis (en). Then, by astandard Skolem hull construction, there is a countable subfield F of R containingthe rational numbers Q such that for any finite linear combination

λ0e0 + λ1e1 + . . .+ λnen

with λi ∈ F, we have ‖λ0e0 +λ1e1 + . . .+λnen‖ ∈ F. This means that any F-linearcombination of (en) can be normalised, while remaining a F-linear combination.Thus, as the set of Q and hence also F-linear combinations of (en) are dense inE, also the set of F-linear normalised combinations of (en) are dense in the unitsphere SE .

A block vector is a normalised finite linear combination x = λ0e0 + λ1e1 + . . .+λnen where λi ∈ F. We insist on blocks being normalised and F-linear and willbe explicit on the few occasions that we deal with non-normalised blocks. Therestriction to F-linear combinations is no real loss of generality, but instead hasthe effect that there are only countably many blocks. We denote by Q the set ofblocks. The support, supp x, of a block x = λ0e0 + λ1e1 + . . . + λnen is the setof i ∈ N such that λi 6= 0 and the range, range x, is the smallest interval I ⊆ N

containing supp x.


A block (sub)sequence, block basis, or blocking of (en) is an infinite sequence (xn)of blocks such that supp xn < supp xn+1 for all n and a block subspace is theclosed linear span of a block sequence. Notice that if X is a block subspace, thenthe associated block sequence (xn) such that X = [xn] is uniquely defined up tothe choice of signs ±xn. So we shall sometimes confuse block sequences and blocksubspaces. For two block subspaces X = [xn] and Y = [yn], write Y 6 X if Y ⊆ X ,or, equivalently, yn ∈ span(xi) for all n. Also, let Y 6∗ X if there is some N suchthat yn ∈ span(xi) for all n > N .

When we work with block subspaces of some basis (en), we will assume that wehave chosen the same countable subfield F of R for all block sequences (xn) of (en),and hence a vector in [xn] is a block of (xn) if and only if it is a block of (en), so noambiguity occurs. We consider the set bb(en) of block sequences of (en) as a closedsubset of QN, where Q is equipped with the discrete topology. In this way, bb(en)is a Polish, i.e., separable, completely metrisable space. If ∆ = (δn) is a sequenceof positive real numbers, which we denote by ∆ > 0, and A ⊆ bb(en), we designateby A∆ the set

A∆ = {(yn) ∈ bb(en)∣

∣ ∃(xn) ∈ bb(en) ∀n ‖xn − yn‖ < δn}.If A is an infinite subset of N, we denote by [A] the space of infinite subsets of

A with the topology inherited from 2A. Also, if a ⊆ N is finite,

[a,A] = {B ∈ [N]∣

∣ a ⊆ B ⊆ a ∪ (A ∩ [maxa+ 1,∞[)}.Given two Banach spaces X and Y , we say that X is crudely finitely representable

in Y if there is a constant K such that for any finite-dimensional subspace F ⊆ Xthere is an embedding T : F → Y with constant K, i.e., ‖T ‖ · ‖T−1‖ 6 K.

Also, if X = [xn] and Y = [yn] are spaces with bases, we say that X crudely blockfinitely representable in Y if for some constantK and all k, there are (not necessarilynormalised) blocks z0 < . . . < zk of (yn) such that (x0, . . . , xk) ∼K (z0, . . . , zk).

Two Banach spaces are said to be incomparable if neither one embeds into theother, and totally incomparable if no subspace of one is isomorphic to a subspace ofthe other.

We shall at several occasions use non-trivial facts about the Tsirelson space, forwhich our reference is [5], and also facts from descriptive set theory that can all befound in [21]. For classical facts in Banach space theory we refer to [23].

2.2. Gowers’ block sequence game. A major ingredient in several of our proofswill be the following equivalent version of Gowers’ game due to J. Bagaria and J.Lopez-Abad [3].

Suppose E = [en] is given. Player I and II alternate in choosing blocks x0 <x1 < x2 < . . . and y0 < y1 < y2 < . . . as follows: Player I plays in the k’th round ofthe game a block xk such that xk−1 < xk. In response to this, II either chooses topass, and thus play nothing in the k’th round, or plays a block yi ∈ [xl+1, . . . , xk],where l was the last round in which II played a block.

I x0 . . . xk0 xk0+1 . . . xk1

II y0 ∈ [x0, . . . , xk0 ] y1 ∈ [xk0+1, . . . , xk1 ]

We thus see I as constructing a block sequence (xi), while II chooses a block sub-sequence (yi). This block subsequence (yi) is then called the outcome of the game.(Potentially the blocking could be finite, but the winning condition can be madesuch that II loses unless it is infinite.)


Gowers [17] proved that if A ⊆ bb(ei) is an analytic set such that any (xi) ∈ bb(ei)has a block subsequence (yi) ∈ bb(ei) belonging to A, then for all ∆ > 0 and(zi) ∈ bb(ei), there is a block subsequence (vi) ∈ bb(ei) of (zi) such that II has astrategy to play in A∆ if I is restricted to play blockings of (vi).

2.3. A trick and a lemma. We gather here a couple of facts that will be usedrepeatedly later on.

We shall at several occasions use coding with inevitable subsets of the unit sphereof a Banach space, as was first done by Lopez-Abad in [24]. So let us recall herethe relevant facts and set up a framework for such codings.

Suppose E is an infinite-dimensional Banach space with a basis not contain-ing a copy of c0. Then by the solution to the distortion problem by Odell andSchlumprecht [27] there is a block subspace [xn] of E and two closed subsets F0

and F1 of the unit sphere of [xn] such that dist(F0, F1) = δ > 0 and such that forall block bases (yn) of (xn) there are block vectors v and u of (yn) such that v ∈ F0

and u ∈ F1. In this case we say that F0 and F1 are positively separated, inevitable,closed subsets of S[xn].

We can now use the sets F0 and F1 to code infinite binary sequences, i.e., α ∈ 2N

in the following manner. If (zn) is a block sequence of (xn) such that for all n,zn ∈ F0 ∪ F1, we let ϕ((zn)) = α ∈ 2N be defined by

αn =

{

0, if zn ∈ F0;1, if zn ∈ F1.

Since the sets F0 and F1 are positive separated, this coding is fairly rigid and canbe extended to block sequences (vn) such that dist(vn, F0 ∪ F1) < δ

2 by letting

ϕ((vn)) = β ∈ 2N be defined by

βn =

{

0, if dist(vn, F0) < δ2 ;

1, if dist(vn, F1) < δ2 .

In this way we have that if (zn) and (vn) are block sequences with zn ∈ F0 ∪ F1

and ‖vn − zn‖ < δ2 for all n, then ϕ((zn)) = ϕ((vn)).

One can now use elements of Cantor space 2N to code other objects in variousways. For example, let H denote the set of finite non-empty sequences (q0, q1, . . . , qn)

of rationals with qn 6= 0. Then as H is countable, we can enumerate it as ~h0,~h1, . . ..If now (yn) and (vn) are block sequences with ϕ((vn)) = 0n010n110n21 . . ., then(vn) codes an infinite sequence Ψ((vn), (yn)) = (un) of (not necessarily normalised)non-zero blocks of (yn) by the following rule:

uk = q0y0 + q1y1 + . . .+ qmym,

where ~hnk= (q0, . . . , qm).

We should then notice three things about this type of coding:

- It is inevitable, i.e., for all block sequences (yn) of (xn) and α ∈ 2N, thereis a block sequence (vn) of (yn) with ϕ((vn)) = α.

- It is continuous, i.e., to know an initial segment of (un) = Ψ((vn), (yn)), weonly need to know initial segments of (vn) and of (yn).

- It is stable under small perturbations. I.e., we can find some ∆ = (δn)only depending on the basis constant of (xn) with the following property.Assume that (vn) and (yn) are block bases of (xn) with vn ∈ F0 ∪ F1 forall n and such that Ψ((vn), (yn)) = (un) is in fact a block sequence of (yn)


with 12 < un < 2. Then whenever (v′n) and (y′n) are other block sequences

of (xn) with ‖vn − v′n‖ < δ2 and ‖yn − y′n‖ < δn for all n, the sequence

Ψ((v′n), (y′n)) = (u′n) will be a block sequence of (y′n) that is 2-equivalentto (un).

One can of course consider codings of other objects than sequences of vectorsand, depending on the coding, obtain similar continuity and stability properties.

The inevitability of the coding is often best used in the following form.

- Suppose B is a set of pairs ((yn), α), where (yn) is a block sequence of (xn)and α ∈ 2N, such that for all block sequences (zn) of (xn) there is a furtherblock sequence (yn) and an α such that ((yn), α) ∈ B. Then for all blocksequences (zn) of (xn) there is a further block sequence (yn) such that forall n, y2n+1 ∈ F0 ∪ F1 and ((y2n), ϕ((y2n+1))) ∈ B.

To see this, let (zn) be given and notice that by the inevitability of the coding thereis a block sequence (wn) of (zn) such that w3n+1 ∈ F0 and w3n+2 ∈ F1. Pick nowa block sequence (vn) of (w3n) and an α such that ((vn), α) ∈ B. Notice now thatbetween vn and vn+1 there are block vectors w3in+1 and w3in+2 of (zn) belongingto F0, respectively F1. Thus, if we let y2n = vn and set

y2n+1 =

{

w3in+1, if αn = 0;w3in+2, if αn = 1.

then ((y2n), ϕ((y2n+1))) ∈ B.

Lemma 2.1. Let (x0n) > (x1

n) > (x2n) > . . . be a decreasing sequence of block bases

of a basic sequence (x0n). Then there exists a block basis (yn) of (x0

n) such that (yn)

is√K-equivalent with a block basis of (xK

n ) for every K.

Proof. Let c(L) be a constant depending on the basis constant of (x0n) such that

if two block bases differ in at most L terms, then they are c(L)-equivalent. Findnow a sequence L0 6 L1 6 L2 6 . . . of non-negative integers tending to +∞ suchthat c(LK) <

√K. We can now easily construct an infinite block basis (yn) such

that for all K at most the first LK terms of (yn) are not blocks of (xKn )∞n=LK+1.

Then (yn) differs from a block basis of (xKn ) in at most LK terms and hence is√

K-equivalent with a block basis of (xKn ). �

3. Tightness

3.1. Tight bases. The following definition is central to the rest of the paper.

Definition 3.1. Consider a Banach space E with a basis (en) and let Y be anarbitrary Banach space. We say that Y is tight in the basis (en) if there is asequence of successive non-empty intervals I0 < I1 < I2 < . . . of N such that for allinfinite subsets A ⊆ N, we have

Y 6⊑ [en

∣

∣ n /∈⋃

i∈A

Ii].

In other words, if Y embeds into [en]n∈B, then B ⊆ N intersects all but finitelymany intervals Ii.

We say that (en) is tight if every infinite-dimensional Banach space Y is tightin (en).

Finally, an infinite-dimensional Banach space X is tight if it has a tight basis.


Also, the following more analytical criterion will prove to be useful. For simplic-ity, denote by PI the canonical projection onto [en]n∈I .

Lemma 3.2. Let X be a Banach space, (en) a basis for a space E, and (In) finiteintervals such that min In −→

n→∞∞ and for all infinite A ⊆ N,

X 6⊑ [en]n/∈S

k∈A Ik.

Then whenever T : X → [en] is an embedding, we have lim infk‖PIkT ‖ > 0.

Proof. Suppose towards a contradiction that T : X → E is an embedding such thatfor some infinite A ⊆ N, limk→∞

k∈A‖PIk

T ‖ = 0. Then, by passing to an infinite subset

ofA, we can suppose that∑

k∈A‖PIkT ‖ < 1

2‖T−1‖−1 and that the intervals (In)n∈A

are disjoint. Thus, the sequence of operators (PIkT )k∈A is absolutely summable and

therefore the operator∑

k∈A PIkT : X → E exists and has norm < 1

2‖T−1‖−1.But then for x ∈ X we have

‖∑

k∈A

PIkTx‖ 6 ‖

∑

k∈A

PIkT ‖ · ‖x‖ < 1

2‖T−1‖‖x‖ 61

2‖T−1‖‖T−1‖ · ‖Tx‖ =

1

2‖Tx‖,

and hence also

‖(

T −∑

k∈A

PIkT

)

x‖ > ‖Tx‖ − ‖∑

k∈A

PIkTx‖ > ‖Tx‖ − 1

2‖Tx‖ =

1

2‖Tx‖.

So T − ∑

k∈A PIkT is still an embedding of X into E. But this is impossible as

T − ∑

k∈A PIkT takes values in [en]n/∈

S

k∈A Ik. �

Proposition 3.3. A tight Banach space contains no minimal subspaces.

Proof. Suppose (en) is a tight basis for a space E and let Y be any subspace of E.Pick a block subspace X = [xn] of E that embeds into Y . Since Y is tight in (en),we can find a sequence of intervals (Ii) such that Y does not embed into [en]n∈B

whenever B ⊆ N is disjoint from an infinite number of intervals Ii. By passingto a subsequence (zn) of (xn), we obtain a space Z = [zn] that is a subspace ofsome [en]n∈B where B ⊆ N is disjoint from an infinite number of intervals Ii, andhence Y does not embed into Z. Since Z embeds into Y , this shows that Y is notminimal. �

The classical example of space without minimal subspaces is Tsirelson’s space Tand it is not too difficult to show that T is tight. This will be proved later on as aconsequence of a more general result.

Any block sequence of a tight basis is easily seen to be tight. And also:

Proposition 3.4. If E is a tight Banach space, then every shrinking basic sequencein E is tight.

Proof. Suppose (en) is a tight basis for E and (fn) is a shrinking basic sequence inE. Let Y be an arbitrary space and find intervals I0 < I1 < . . . associated to Y for(en), i.e., for all infinite subsets A ⊆ N, we have Y 6⊑ [en

∣

∣ n /∈ ⋃

i∈A Ii].We notice that, since (en) is a basis, we have for all m

‖PIk|[fi

∣

∣ i6m]‖ −→

k→∞0, (1)

and, since (fn) is shrinking and the PIkhave finite rank, we have for all k

‖PIk|[fi

∣

∣ i>m]‖ −→

m→∞0. (2)


Using alternately (1) and (2), we can construct integers k0 < k1 < . . . and intervalsJ0 < J1 < . . . such that

‖PIkn|[fi

∣

∣ /∈Jn]‖ < 2

n+ 1.

To see this, suppose kn−1 and Jn−1 have been defined and find some large kn > kn−1

such that

‖PIkn|[fi

∣

∣ i6max Jn−1]‖ 6

1

n+ 1.

Now, choose m large enough that

‖PIkn|[fi

∣

∣ i>m]‖ 6

1

n+ 1,

and set Jn = [max Jn−1+1,m]. Then ‖PIkn|[fi

∣

∣ i/∈Jn]‖ < 2

n+1 . It follows that if A ⊆N is infinite and T : Y → [fi]i/∈

S

n∈A Jnis an embedding, then limn∈A‖PIkn

T ‖ = 0,

which contradicts Lemma 3.2. So (Jn) witnesses that Y is tight in (fn). �

Corollary 3.5. If a tight Banach space X is reflexive, then every basic sequencein X is tight.

Notice that, since c0 and ℓ1 are minimal, we have by the classical theorem ofJames, that if X is a tight Banach space with an unconditional basis, then X isreflexive and so every basic sequence in X is tight.

Example 3.6. The symmetrisation S(T (p)) of the p-convexification of Tsirelson’sspace, 1 < p < +∞, does not contain a minimal subspace, yet it is not tight.

Proof. Since S(T (p)) is saturated with subspaces of T (p) and T (p) does not containa minimal subspace, it follows that S(T (p)) does not have a minimal subspace. Thecanonical basis (en) of S(T (p)) is symmetric, therefore S(T (p)) is not tight in (en)and so (en) is not tight. By reflexivity, no basis of S(T (p)) is tight. �

3.2. A generalised asymptotic game. Suppose X = [xn] and Y = [yn] are twoBanach spaces with bases. We define the game HY,X with constant C > 1 betweentwo players I and II as follows: I will in each turn play a natural number ni, whileII will play a not necessarily normalised block vector ui ∈ X and a natural numbermi such that

ui ∈ X [n0,m0] + . . .+X [ni,mi],

where, for ease of notation, we write X [k,m] to denote [xn]k6n6m. Diagramatically,

I n0 n1 n2 n3 . . .II u0,m0 u1,m1 u2,m2 u3,m3 . . .

We say that the sequence (ui)i∈N is the outcome of the game and say that II winsthe game if (ui) ∼C (yi).

For simplicity of notation, if X = [xn] is space with a basis, Y a Banach space,I0 < I1 < I2 < . . . a sequence of non-empty intervals of N and K is a constant, wewrite

Y ⊑K (X, Ii)

if there is an infinite set A ⊆ N containing 0 such that

Y ⊑K [xn

∣

∣ n /∈⋃

i∈A

Ii],


i.e., Y embeds with constant K into the subspace of X spanned by (xn)n/∈S

i∈A Ii.

Also, write

Y ⊑ (X, Ii)

if there is an infinite set A ⊆ N such that Y ⊑ [xn

∣

∣ n /∈ ⋃

i∈A Ii]. Notice that inthe latter case we can always demand that 0 ∈ A by perturbating the embeddingwith a finite rank operator.

It is clear that if Y = [yn] and II has a winning strategy in the game HY,X withconstant K, then for any sequence of intervals (Ii), Y ⊑K (X, Ii).

Modulo the determinacy of open games, the next lemma shows that the converseholds up to a perturbation.

Lemma 3.7. Suppose X = [xn] is space with a basis and K is a constant such thatfor all block bases Y of X there is a winning strategy for I in the game HY,X withconstant 2K. Then there is a Borel function f : bb(X) → [N] such that for all Y ifIj = [f(Y )2j , f(Y )2j+1], then

Y 6⊑K (X, Ij).

Proof. Notice that the game HY,X is open for player I and, in fact, if Q2K denotesthe set of blocks u with 1

2K 6 ‖u‖ 6 2K, then the set

A = {(Y, ~p) ∈ bb(X) × (N × Q2K × N)N∣

∣ either ~p is a legal run of the game HY,X

with constant 2K in which I wins or ~p is not a legal run of the game HY,X}is Borel and has open sections AY = {~p ∈ (N×Q2K ×N)N

∣

∣ (Y, ~p) ∈ A}. Also, sincethere are no rules for the play of I in HY,X , AY really corresponds to the winningplays for I in HY,X with constant 2K. By assumption, I has a winning strategy toplay in AY for all Y , and so by the theorem on strategic uniformisation (see (35.32)in [21]), there is a Borel function σ : Y 7→ σY that to each Y associates a winningstrategy for I in the game HY,X with constant 2K.

Now let ∆ = (δn) be a sequence of positive reals such that for all sequencesof blocks (wn) of X with 1

K 6 ‖wn‖ 6 K and sequences of vectors (un), if forall n, ‖wn − un‖ < δn, then (wn) ∼2 (un). We also choose sets Dn of finite (notnecessarily normalised) blocks with the following properties:

- for each finite d ⊆ N, the number of vectors u ∈ Dn such that supp u = dis finite,

- for all blocks vectors w with 1K 6 ‖w‖ 6 K, there is some u ∈ Dn with

supp w = supp u such that ‖w − u‖ < δn.

This is possible since the K-ball in [xi]i∈d is totally bounded for all finite d ⊆ N.So for all sequences (wn) of blocks with 1 6 ‖wn‖ 6 K there is some (un) ∈ ∏

n Dn

such that supp wn = supp un and ‖wn − un‖ < δn for all n, whence (wn) ∼2 (un).Suppose now that Y = [yn] is given. For each p = (n0, u0,m0, . . . , ni, ui,mi),

where uj ∈ Dj for all j and

I n0 n1 . . . ni

II u0,m0 u1,m1 . . . ui,mi

is a legal position in the game HY,X in which I has played according to σY , we writep < k if nj , uj,mj < k for all j 6 i. Notice that for all k there are only finitelymany such p with p < k, so we can define

α(k) = max(k,max{σY (p)∣

∣ p < k})


and set Ik = [k, α(k)]. Clearly, the sequence (Ik) can be computed in a Borelfashion from Y . The Ik are not necessarily successive, but their minimal elementstend to ∞, so to prove the lemma it is enough to show that Y does not K-embedinto [xn] avoiding an infinite number of Ik.

Suppose now for a contradiction that A ⊆ N is infinite, 0 ∈ A and yi 7→ wi isa K-embedding into [xn

∣

∣ n /∈ ⋃

k/∈A Ik]. Using the defining properties of Di, wefind ui ∈ Di such that ‖wi − ui‖ < δi and supp wi = supp ui for all i, whereby(ui) ∼2 (wi) ∼K (yi).

We now proceed to define natural numbers ni, mi, and ai ∈ A such that forpi = (n0, u0,m0, . . . , ni, ui,mi), we have

(i) a0 = 0 and [0, n0[⊆ Ia0 ,(ii) mi = ai+1 − 1,

(iii) pi is a legal position in HY,X in which I has played according to σY ,(iv) ]mi, ni+1[⊆ Iai+1 .

Let a0 = 0 and n0 = σY (∅) = α(0), whence Ia0 = [0, α(0)] = [0, n0]. Find a1 suchthat n0, u0, a0 < a1 and set m0 = a1 − 1. Then p0 = (n0, u0,m0) is a legal positionin HY,X in which I has played according to σY , p0 < a1, so n1 = σY (n0, u0,m0) 6

α(a1), and therefore ]m0, n1[⊆ Ia1 = [a1, α(a1)].Now suppose by induction that n0, . . . , ni and a0, . . . , ai have been defined. Since

[0, n0[⊆ Ia0 and ]mj , nj+1[⊆ Iaj+1 for all j < i, we have

ui ∈ X [n0,m0] + . . .+X [ni−1,mi−1] +X [ni,∞[.

Find some ai+1 greater than all of n0, . . . , ni, u0, . . . , ui, a0, . . . , ai and let mi =ai+1 − 1. Then

ui ∈ X [n0,m0] + . . .+X [ni−1,mi−1] +X [ni,mi]

and pi = (n0, u0,m0, . . . , ni, ui,mi) is a legal position played according to σY . Sincepi < ai+1 also

ni+1 = σY (n0, u0,m0, . . . , ni, ui,mi) 6 α(ai+1).

Thus ]mi, ni+1[⊆ Iai+1 = [ai+1, α(ai+1)].Now since p0 ⊆ p1 ⊆ p2 ⊆ . . ., we can let ~p =

⋃

i pi and see that ~p is a run ofthe game in which I followed the strategy σY and II has played (ui). Since σY iswinning for I, this implies that (ui) 6∼2K (yi) contradicting our assumption. �

Lemma 3.8. Suppose X = [xn] is a space with a basis and Y is a space such that

for all constants K there are intervals I(K)0 < I

(K)1 < I

(K)2 < . . . such that Y 6⊑K

(X, I(K)j ). Then there are intervals J0 < J1 < J2 < . . . such that Y 6⊑ (X, Jj).

Moreover, the intervals (Jj) can be computed in a Borel manner from (I(K)i )i,K .

Proof. By induction we can now construct intervals J0 < J1 < J2 < . . . such that

Jn contains one interval from each of (I(1)i ), . . . , (I

(n)i ) and if M = min Jn − 1 and

K = ⌈n · c(M)⌉, then maxJn > max I(K)0 + M , where c(M) is a constant such

that if two subsequences of (xn) differ in at most M terms then they are c(M)equivalent. It then follows that if A ⊆ N is infinite, then

Y 6⊑ [xn]n/∈S

i∈A Ji.


To see this, suppose towards a contradiction that A ⊆ N is infinite and that forsome integer N ,

Y ⊑N [xn]n/∈S

i∈A Ji.

Pick then a ∈ A such that a > N and set M = min Ja − 1 and K = ⌈a · c(M)⌉.Define an isomorphic embedding T from

[xn

∣

∣ n /∈⋃

i∈A

Ji]

into[xn

∣

∣ n /∈⋃

i∈A

Ji & n > maxJa] + [xn

∣

∣ max I(K)0 < n 6 max Ja]

by setting

T (xn) =

{

xn, if n > maxJa;x

max I(K)0 +n+1

, if n 6 M .

This is possible since maxJa > max I(K)0 +M . Also, since T only changes at most

M vectors from (xn), it is a c(M) embedding. Therefore, by composing with T andusing that N · c(M) 6 a · c(M) 6 K, we see that

Y ⊑K [xn

∣

∣ n /∈⋃

i∈A

Ji & n > maxJa] + [xn

∣

∣ max I(K)0 < n 6 maxJa].

In particular, as almost all Ji contain an interval I(K)l , we can find and infinite set

B ⊆ N containing 0 such that

Y ⊑K [xn

∣

∣ n /∈⋃

i∈B

I(K)i ],

which is a contradiction. �

Lemma 3.9. Let E = [en] be given and suppose that for all block subspaces Z 6 Eand constants C there is a block subspace X 6 Z such that for all block subspacesY 6 X, I has a winning strategy in the game HY,X with constant C. Then thereis a block subspace X 6 E and a Borel function f : bb(X) → [N] such that for allnormalised block bases Y 6 X, if we set Ij = [f(Y )2j , f(Y )2j+1], then

Y 6⊑ (X, Ij).

Proof. Using the hypothesis inductively together with Lemma 3.7, we can constructa sequence X0 > X1 > X2 > . . . of block subspaces XK and corresponding Borelfunctions fK : bb(XK) → [N] such that for all V 6 XK if Ij = [fK(V )2j , fK(V )2j+1],then V 6⊑K2 (XK , Ij).

Pick by Lemma 2.1 some block X∞ of X0 that is√K-equivalent with a block

sequence ZK of XK for every K. Then for any block sequence Y of X∞ and anyK there is some block sequence V 6 ZK 6 XK such that Y is

√K-equivalent with

V . Let (Ij) be the intervals given by fK(V ) so that V 6⊑K2 (XK , Ij). We can thenin a Borel way from (Ij) construct intervals (Jj) such that V 6⊑K2 (ZK , Jj) andtherefore also Y 6⊑K (X∞, Jj).

This means that there are Borel functions gK : bb(X∞) → [N] such that for allY 6 X∞ if JK

j (Y ) = [gK(Y )2j , gK(Y )2j+1], then Y 6⊑K (X∞, JKj (Y )). Using

Lemma 3.8 we can now in a Borel manner in Y define intervals LY0 < LY

1 < . . .such that

Y 6⊑ (X∞, LYj ).


Letting f : bb(X∞) → [N] be the Borel function corresponding to Y 7→ (LYj ), we

have our result. �

As will be clear in Section 7 it can be useful to have a version of tightnessthat not only assures us that certain intervals exist, but also tells us how to ob-tain these. Thus, we call a basis (en) continuously tight if there is a continuousfunction f : bb(en) → [N] such that for all normalised block bases X , if we setIj = [f(X)2j, f(X)2j+1], then

X 6⊑ ([en], Ij),

i.e., X does not embed into [en] avoiding an infinite number of the intervals Ij .We shall now improve Lemma 3.9 to conclude continuous tightness from its

hypothesis.

Lemma 3.10. Let E = [en] be given and suppose that for all block subspaces Z 6 Eand constants C there is a block subspace X 6 Z such that for all block subspacesY 6 X, I has a winning strategy in the game HY,X with constant C. Then there isa continuously tight block subspace X 6 E.

Proof. We observe that E does not contain a copy of c0. Indeed if Z is a blocksubspace of E spanned by a block sequence which is C-equivalent to the unit vectorbasis of c0, then for any Y 6 X 6 Z, II has a winning strategy in the game HY,X

with constant C2. We shall then use codings with inevitable subsets. So find firsta block subspace Z of E such that there are inevitable, positively separated, closedsubsets F0 and F1 of SZ . By Lemma 3.9, we can find a further block subspaceV of Z and and a Borel function g : bb(V ) → [N] such that for all Y 6 V , ifIj = [g(Y )2j , g(Y )2j+1], then Y 6⊑ (V, Ij). Define the set

A ={

(yn) ∈ bb(V )∣

∣ y2n ∈ F0 ⇔ n /∈ g((y2n+1)) and y2n ∈ F1 ⇔ n ∈ g((y2n+1)}

.

Obviously, A is Borel and, using inevitability, one can check that any block basisof V contains a further block basis in A. Thus, by Gowers’ Determinacy Theorem,we have that for all ∆ > 0 there is a block sequence X of V such that II has astrategy to play into A∆ when I plays block subspaces of X . Choosing ∆ > 0sufficiently small, this easily implies that for some block basis X of E, there is acontinuous function h : bb(X) → bb(X)× [N] that to each W 6 X associates a pair(

Y, (In))

consisting of a block sequence Y of W and a sequence of intervals (In)such that Y 6⊑ (V, Ij). Notice now that continuously in the sequence (Ij), we canconstruct intervals (Jj) such that Y 6⊑ (X, Jj) and hence also W 6⊑ (X, Ij). Sothe continuous function f : bb(X) → [N] corresponding to W 7→ (Jj) witnesses thecontinuous tightness of X . �

We shall need the following consequence of continuous tightness in Section 7.

Lemma 3.11. Suppose (en) is continuously tight. Then there is a continuousfunction f : [N] → [N] such that for all A,B ∈ [N], if B is disjoint from an infinitenumber of intervals [f(A)2i, f(A)2i+1], then [en]n∈A does not embed into [en]n∈B.

Proof. It is enough to notice that the function h : [N] → bb(en) given by h(A) =(en)n∈A is continuous. So when composed with the function witnessing continuoustightness we have the result. �


3.3. A game for minimality. For L and M two block subspaces of E, definethe infinite game GL,M with constant C > 1 between two players as follows. Ineach round I chooses a subspace Ei ⊆ L spanned by a finite block sequence ofL, a normalised block vector ui ∈ E0 + . . . + Ei, and an integer mi. In the firstround II plays an integer n0, and in all subsequent rounds II plays a subspace Fi

spanned by a finite block sequence of M , a (not necessarily normalised) block vectorvi ∈ F0 + . . . + Fi and an integer ni+1. Moreover, we demand that ni 6 Ei andmi 6 Fi.

Diagramatically,

I n0 6 E0 ⊆ L n1 6 E1 ⊆ L . . .u0 ∈ E0,m0 u1 ∈ E0 + E1,m1

II n0 m0 6 F0 ⊆M m1 6 F1 ⊆M . . .v0 ∈ F0, n1 v1 ∈ F0 + F1, n2

The outcome of the game is the pair of infinite sequences (ui) and (vi) and we saythat II wins the game if (ui) ∼C (vi).

Lemma 3.12. Suppose that X and Y are block subspaces of E and that player IIhas a winning strategy in the game HY,X with constant C. Then II has a winningstrategy in the game GY,X with constant C.

Proof. We shall in fact prove that II has a winning strategy in a game that isobviously harder for her to win. Namely, we shall suppose that II always playsni = 0, which obviously puts less restrictions on the play of I. Moreover, we do notrequire I to play the finite-dimensional spaces Ei, which therefore also puts fewerrestrictions on I in subsequent rounds. Therefore, we shall suppress all mention ofEi and ni and only require that the ui are block vectors in Y .

While playing the game GY,X , II will keep track of an auxiliary play of the gameHY,X in the following way. In the game GY,X we have the following play

I u0 ∈ Y,m0 u1 ∈ Y,m1 . . .

II m0 6 F0 ⊆ X m1 6 F1 ⊆ X . . .v0 ∈ F0 v1 ∈ F0 + F1

We write each vector ui =∑ki

j=0 λijyj and may for simplicity of notation assume

that ki < ki+1. The auxiliary run of HY,X that II will keep track of is as follows,where II plays according to her winning strategy for HY,X .

I m0 . . . m0 m1 . . . m1 . . .II w0, p0 . . . wk0 , pk0 wk0+1, pk0+1 . . . wk1 , pk1 . . .

To compute the vi and Fi in the game GY,X , II will refer to the play of HY,X andset

vi =

ki∑

j=0

λijwj ,

and letFi = X [mi,max{pki−1+1, . . . , pki}].

It is not difficult to see that mi 6 Fi ⊆ X , vi ∈ F0 + . . .+ Fi, and that the Fi andvi only depends on u0, . . . , ui and m0, . . . ,mi. Thus this describes a strategy for IIin GY,X and it suffices to verify that it is a winning strategy.


But since II follows her strategy in HY,X , we know that (wi) ∼C (yi) and there-fore, since ui and vi are defined by the same coefficients over respectively (yi) and(wi), we have that (vi) ∼C (ui). �

3.4. A dichotomy for minimality. We are now in condition to prove the centralresult of this paper.

Theorem 3.13 (3rd dichotomy). Let E be a Banach space with a basis (ei). Theneither E contains a minimal block subspace or a continuously tight block subspace.

Proof. Suppose that E has no continuously tight block basic sequence. By Lemma3.10, we can, modulo passing to a block subspace, suppose that for some constantC and for all block subspaces X 6 E there is a further block subspace Y 6 Xsuch that I has no winning strategy in the game HY,X with constant C. By thedeterminacy of open games, this implies that for all block subspaces X 6 E thereis a further block subspace Y 6 X such that II has a winning strategy in the gameHY,X with constant C.

A state is a pair (a, b) with a, b ∈ (Q′ × F)<ω, where F is the set of subspacesspanned by finite block sequences and Q′ the set of not necessarily normalisedblocks, such that |a| = |b| or |a| = |b| + 1. The set S of states is countable, andcorresponds to the possible positions of a game GL,M after a finite number of moveswere made, restricted to elements which do affect the outcome of the game fromthat position (i.e., mi’s and ni’s are forgotten).

For each state s = (a, b) we will define the game GL,M (s) in a manner similar tothe game GL,M depending on whether |a| = |b| or |a| = |b| + 1. To avoid excessivenotation we do this via two examples:

If a = (a0, A0, a1, A1), b = (b0, B0, b1, B1), the game GL,M (s) will start withII playing some integer n2, then I playing (u2, E2,m2) with n2 6 E2 ⊆ L andu2 ∈ A0+A1+E2, II playing (v2, F2, n3) with m2 6 F2 ⊆M and v2 ∈ B0+B1+F2,etc, and the outcome of the game will be the pair of infinite sequences (a0, a1, u2, . . .)and (b0, b1, v2, . . .).

If a = (a0, A0, a1, A1), b = (b0, B0), the game GL,M (s) will start with I playingsome integer m1, then II playing (v1, F1, n2) with m1 6 F1 ⊆M and v1 ∈ B0+F1, Iplaying (u2, E2,m2) with n2 6 E2 ⊆ L and u2 ∈ A0+A1+E2, etc, and the outcomeof the game will be the pair of infinite sequences (a0, a1, u2, . . .) and (b0, v1, v2, . . .).

The following lemma is well-known and easily proved by a simple diagonalisation.

Lemma 3.14. Let N be a countable set and let µ : bb(E) → P(N) satisfy either

V 6∗ W ⇒ µ(V ) ⊆ µ(W )

orV 6∗ W ⇒ µ(V ) ⊇ µ(W ).

Then there exists a stabilising block subspace V0 6 E, i.e., such that µ(V ) = µ(V0)for any V 6∗ V0.

Let now τ : bb(E) → P(S) be defined by

s ∈ τ(M) ⇔ ∃L 6 M such that player II has a winning strategy in GL,M (s).

By the asymptotic nature of the game we see that M ′ 6∗ M ⇒ τ(M ′) ⊆ τ(M),and therefore there exists M0 6 E which is stabilising for τ . We then define a mapρ : bb(E) → P(S) by setting

s ∈ ρ(L) ⇔ player II has a winning strategy in GL,M0(s).


Again L′ 6∗ L ⇒ ρ(L′) ⊇ ρ(L) and therefore there exists L0 6 M0 which isstabilising for ρ. Finally, the reader will easily check that ρ(L0) = τ(L0) = τ(M0).

Lemma 3.15. For every M 6 L0, II has a winning strategy for the game GL0,M .

Proof. Fix M a block subspace of L0. We begin by showing that (∅, ∅) ∈ τ(L0).To see this, we notice that as L0 6 E, there is a Y 6 L0 such that II has a winningstrategy for HY,L0 and thus, by Lemma 3.12, also a winning strategy in GY,L0 withconstant C. So (∅, ∅) ∈ τ(L0).

We will show that for all states

((u0, E0, . . . , ui, Ei), (v0, F0, . . . , vi, Fi)) ∈ τ(L0),

there is an n such that for all n 6 E ⊆ L0 and u ∈ E0 + . . .+ Ei + E, we have

((u0, E0, . . . , ui, Ei, u, E), (v0, F0, . . . , vi, Fi)) ∈ τ(L0).

Similarly, we show that for all states

((u0, E0, . . . , ui+1, Ei+1), (v0, F0, . . . , vi, Fi)) ∈ τ(L0)

and for all m there are m 6 F ⊆M and v ∈ F0 + . . .+ Fi + F such that

((u0, E0, . . . , ui+1, Ei+1), (v0, F0, . . . , vi, Fi, v, F )) ∈ τ(L0).

Since the winning condition of GL0,M is closed, this clearly shows that II has awinning strategy in GL0,M (except for the integers m and n, τ(L0) is a winningquasi strategy for II).

So suppose that

s = ((u0, E0, . . . , ui, Ei), (v0, F0, . . . , vi, Fi)) ∈ τ(L0) = ρ(L0),

then II has a winning strategy in GL0,M0(s) and hence there is an n such that forall n 6 E ⊆ L0 and u ∈ E0 + . . .+Ei +E, II has a winning strategy in GL0,M0(s′),where

s′ = ((u0, E0, . . . , ui, Ei, u, E), (v0, F0, . . . , vi, Fi)).

So s′ ∈ ρ(L0) = τ(L0).Similarly, if

s = ((u0, E0, . . . , ui+1, Ei+1), (v0, F0, . . . , vi, Fi)) ∈ τ(L0) = τ(M)

and m is given, then as II has a winning strategy for GL,M (s) for some L 6 M ,there are m 6 F ⊆M and v ∈ F0 + . . .+Fi +F such that II has a winning strategyin GL,M (s′), where

s′ = ((u0, E0, . . . , ui+1, Ei+1), (v0, F0, . . . , vi, Fi, v, F )).

So s′ ∈ τ(M) = τ(L0). �

Choose now Y = [yi] 6 L0 such that II has a winning strategy in HY,L0. Weshall show that any block subspace M of L0 contains a C2-isomorphic copy of Y ,which implies that Y is C2 + ǫ-minimal for any ǫ > 0.

To see this, notice that, since II has a winning strategy in HY,L0 , player I has astrategy in the game GL0,M to produce a sequence (ui) that is C-equivalent withthe basis (yi). Moreover, we can ask that I plays mi = 0. Using her winningstrategy for GL0,M , II can then respond by producing a sequence (vi) in M suchthat (vi) ∼C (ui). So (vi) ∼C2 (yi) and Y ⊑C2 M . �


Finally we observe that by modifying the notion of embedding in the definitionof a tight basis, we obtain variations of our dichotomy theorem with a weaker formof tightness on one side and a stronger form of minimality on the other.

Theorem 3.16. Every Banach space with a basis contains a block subspace E =[en] which satisfies one of the two following properties:

(1) For any [yi] 6 E, there exists a sequence (Ii) of successive intervals suchthat for any infinite subset A of N, the basis (yi) does not embed into[en]n/∈∪i∈AIi

as a sequence of disjointly supported blocks , resp. as a permu-tation of a block-sequence, resp. as a block-sequence.

(2) For any [yi] 6 E, (en) is equivalent to a sequence of disjointly supportedblocks of [yi], resp. (en) is permutatively equivalent to a block-sequence of[yi], resp. (en) is equivalent to a block-sequence of [yi].

The case of block sequences immediately implies the theorem of Pe lczar [30].The fact that the canonical basis of T ∗ is strongly asymptotically ℓ∞ implies

easily that it is tight for “embedding as a sequence of disjointly supported blocks”although T ∗ is minimal in the usual sense. We do not know of other examplesof spaces combining one form of minimality with another form of tightness in theabove list.

4. Tightness with constants and crude stabilisation of local

structure

We shall now consider a stronger notion of tightness, which is essentially local innature. Let E be a space with a basis (en). There is a particularly simple case whena sequence (Ii) of intervals associated to a subspace Y characterises the tightness ofY in (en). This is when for all integer constants K, Y 6⊑K [en]n/∈IK

. This propertyhas the following useful reformulations.

Proposition 4.1. Let E be a space with a basis (en). The following are equivalent:

(1) For any block sequence (yn) there are intervals I0 < I1 < I2 < . . . such thatfor all K,

[yn]n∈IK 6⊑K [en]n/∈IK.

(2) For any space Y , there are intervals I0 < I1 < I2 < · · · such that for allK,

Y 6⊑K [en]n/∈IK.

(3) No space embeds uniformly into the tail subspaces of E.(4) There is no K and no subspace of E which is K-crudely finitely repre-

sentable in any tail subspace of E.

A basis satisfying properties (1), (2), (3), (4), as well as the space it generates,will be said to be tight with constants.

Proof. The implications (1)⇒(2)⇒(3) are clear.To prove (3)⇒(4) assume some subspace Y of E is K-crudely finitely repre-

sentable in any tail subspace of E. Without loss of generality, we may assume thatY = [yn] is a block subspace of E. We pick a subsequence (zn) of (yn) in thefollowing manner. Let z0 = y0, and if z0, . . . , zk−1 have been chosen, we choosezk far enough on the basis (en), so that [z0, . . . , zk−1] has a 2K-isomorphic copy


in [en

∣

∣ k 6 n < min(supp zk)]. It follows that for any k, Z = [zn] has an M -

isomorphic copy in the tail subspace [en

∣

∣ n > k] for some M depending only on Kand the constant of the basis (en).

To prove (4)⇒(1), let c(L) be a constant such that if two block sequences differ inat most L terms, then they are c(L)-equivalent. Now assume (4) holds and let (yn)be a block sequence of (en). Suppose also that I0 < . . . < IK−1 have been chosen.By (4) applied to Y = [yn]∞n=max IK−1+1, we can then find m and l > max IK−1

such that [yn]ln=max IK−1+1 does not K · c(max IK−1 + 1)-embed into [en]∞n=m. Letnow

IK = [max IK−1 + 1, l+m]

and notice that, as [yn]ln=max IK−1+1 ⊆ [yn]n∈IK , we have that [yn]n∈IK does not

K · c(max IK−1 + 1)-embed into [en]∞n=m. Also, since (en)∞n=m and

(en)max IK−1

n=0⌢(en)∞n=max IK−1+1+m

only differ in max IK−1 + 1 many terms, [yn]n∈IK does not K-embed into

[en]max IK−1

n=0 + [en]∞n=max IK−1+1+m,

and thus not into the subspace [en]n/∈IKeither. �

It is worth noticing that a basis (en), tight with constants, is necessarily contin-uously tight. For a simple argument shows that in order to find the intervals IKsatisfying (1) above, one only needs to know a beginning of the block sequence (yn)and hence the intervals can be found continuously in (yn). From Proposition 4.1we also deduce that any block basis or shrinking basic sequence in the span of atight with constants basis is again tight with constants.

There is a huge difference between the fact that no subspace of E is K-crudelyfinitely representable in all tails of E and that no space is K-crudely finitely repre-sentable in all tails of E. For example, we shall see that while the former holds forTsirelson’s space, of course the latter fails, since Tsirelson’s space is asymptoticallyℓ1.

Recall that a basis (en) is said to be strongly asymptotically ℓp, 1 6 p 6 +∞,[6], if there exists a constant C and a function f : N → N such that for any n,any family of n unit vectors which are disjointly supported in [ek

∣

∣ k > f(n)] isC-equivalent to the canonical basis of ℓnp .

Proposition 4.2. Let E be a Banach space with a strongly asymptotically ℓp basis(en), 1 6 p < +∞, and not containing a copy of ℓp. Then (en) is tight withconstants.

Proof. Assume that some Banach space Y embeds with constant K in any tail sub-space of E. We may assume that Y is generated by a block-sequence (yn) of E and,since any strongly asymptotically ℓp basis is unconditional, (yn) is unconditional.By renorming E we may assume it is 1-unconditional. By a result of W.B. Johnson[20] for any n there is a constant d(n) such that (y0, . . . , yn) is 2K-equivalent to asequence of vectors in the linear span of d(n) disjointly supported unit vectors inany tail subspace of E, in particular in [ek

∣

∣ k > f(d(n))]. Therefore [y0, . . . , yn]2KC-embeds into ℓp. This means that Y is crudely finitely representable in ℓp andtherefore embeds into Lp, and since (yn) is unconditional asymptotically ℓp, that Ycontains a copy of ℓp (details of the last part of this proof may be found in [6]). �


Corollary 4.3. Tsirelson’s space T and its convexifications T (p), 1 < p < +∞, aretight with constants.

Observe that on the contrary, the dual T ∗ of T , which is strongly asymptoticallyℓ∞ and does not contain a copy of c0, is minimal and therefore does not containany tight subspace.

Suppose a space X is crudely finitely representable in all of its subspaces. Thenthere is some constant K and a subspace Y such that X is K-crudely finitelyrepresentable in all of the subspaces of Y . For if not, we would be able to construct asequence of basic sequences (xK

n ) in X such that (xK+1n ) is a block sequence of (xK

n )and such that X is not K2-crudely finitely representable in [xK

n ]. By Lemma 2.1,

we can then find a block sequence (yn) of (x0n) that is

√K-equivalent with a block

sequence of (xKn ) for any K and hence if X were K-crudely finitely representable in

[yn] for some K, then it would also be K3/2-crudely finitely representable in [xKn ],

which is a contradiction.When a space X is K-crudely finitely representable in any of its subspaces for

some K, we say that X is locally minimal. For example, by the universality prop-erties of c0, any space with an asymptotically ℓ∞ basis is locally minimal.

Theorem 4.4 (5th dichotomy). Let E be an infinite-dimensional Banach spacewith basis (en). Then there is a block sequence (xn) satisfying one of the followingtwo properties, which are mutually exclusive and both possible.

(1) (xn) is tight with constants,(2) [xn] is locally minimal.

Proof. If E contains c0, the result is trivial. So suppose not and find by the solutionto the distortion problem a block sequence (yn) and inevitable, positively separated,closed subsets F0 and F1 of the unit sphere of [yn]. Define for each integer K > 1the set

AK = {(zn) 6 (yn)∣

∣ z2n ∈ F0 ∪ F1 and (z2n) codes by 0’s and 1’s a block

sequence (vn) of (z2n+1) such that for all N, [vn] ⊑K [z2n+1]n>N

and moreover 1/2 < ‖vn‖ < 2}.Clearly AK is analytic, so we can apply Gowers’ Determinacy Theorem to get oneof two cases

(i) either there is a block sequence (xn) and a K such that player II has astrategy to play inside (AK)∆ whenever I plays a block sequence of (xn),where ∆ will be determined later,

(ii) or we can choose inductively a sequence of block sequences (xKn ) such that

(xK+1n ) 6 (xK

n ) and such that no block sequence of (xKn ) belongs to AK .

Consider first case (ii). Set wn = xn2n and choose now further block sequences

(xn) and (hn) of (wn) such that

x0 < h0 < h1 < x1 < h2 < h3 < x4 < . . .

and h2n ∈ F0, h2n+1 ∈ F1.We claim that (xn) is tight with constants. If not, we can find some block

sequence (un) of (xn) and a K such that [un] embeds with constant K into any tailsubspace of [xn]. By passing to tails of (xn) and of (un), we can suppose that (xn)is a block sequence of (xK

n ), (un) is a block sequence of (xn) and [un] K-embeds into


all tails of [xn]. By filling in with appropriate hi between xn and xn+1, we can nowproduce a block sequence (zn) of (xK

n ) such that (z2n) codes by 0’s and 1’s the blocksequence (un) of (z2n+1) with the property that for all N , [un] ⊑K [z2n+1]n>N . Inother words, we have produced a block sequence of (xK

n ) belonging to AK , whichis impossible. Thus, (xn) is tight with constants.

Consider now case (i) instead and let II play according to his strategy. Wesuppose that ∆ is chosen sufficiently small so that δi < dist(F0, F1)/3 and if twoblock sequences are ∆-close then they are 2-equivalent. Let (yn) ∈ (AK)∆ be theresponse by II to the sequence (xn) played by I and let (zn) ∈ AK be such that‖zn − yn‖ < δn for all n. Then (z2n) codes by 0’s and 1’s a block sequence (vn)of (z2n+1). Let (un) be the block sequence of (y2n+1) constructed in the sameway as (vn) is constructed over (z2n+1). We claim that (un) is 2K-crudely finitelyrepresentable in any block subspace of [xn].

For this, let [u0, . . . , um] be given and suppose that (fn) is any block subspace of(xn). Find a large k such that (z0, z2, . . . , z2k) codes the block sequence (v0, . . . , vm)of (z1, . . . , z2k+1) and let l be large enough so that when I has played x0, . . . , xl thenII has played y0, . . . , y2k+1. Consider now the game in which player I plays

x0, x1, . . . , xl, fl+1, fl+2, . . . .

Then, following the strategy, II will play a block sequence

y0, . . . , y2k+1, g2k+2, g2k+3, . . . ∈ (AK)∆.

So let (hn) ∈ AK be such that ‖hn − yn‖ < δn for all n. Then, as ‖h2n − z2n‖ <2δn <

23dist(F0, F1), we have that h2n ∈ Fi ⇔ z2n ∈ Fi. Also, (h2n+1) and (y2n+1)

are 2-equivalent, so (h2n) will code a block sequence (wn) such that (w0, . . . , wm)is 2-equivalent to (u0, . . . , um). Moreover, since (hn) ∈ AK , [wn] will K-embed intoevery tail subspace of [h2n+1], and thus, since (h2n+1) is block of (fn), [u0, . . . , um]will 2K-embed into [fn]. �

Local minimality can be reformulated in a way that makes the relation to localtheory clearer. For this, let Fn be the metric space of all n-dimensional Banachspaces up to isometry equipped with the Banach-Mazur metric

d(X,Y ) = inf(

log(‖T ‖ · ‖T−1‖)∣

∣ T : X → Y is an isomorphism)

.

Then for every Banach space X , the set of n-dimensional Y that are almost iso-metrically embeddable into X form a closed subset (X)n of Fn. It is well-knownthat this set (X)n does not always stabilise, i.e., there is not necessarily a subspaceY ⊆ X such that for all further subspaces Z ⊆ Y , (Z)n = (Y )n. However, ifinstead X comes equipped with a basis and for all block subspaces Y we let {Y }n

be the set of all n-dimensional spaces that are almost isometrically embeddableinto all tail subspaces of Y , then one can easily stabilise {Y }n on subspaces. Suchconsiderations are for example the basis for [26].

Theorem 4.4 gives a dichotomy for when one can stabilise the set (X)n in acertain way, which we could call crude. Namely, X is locally minimal if andonly if there is some constant K such that for all subspaces Y of X and all n,dH

(

(X)n, (Y )n

)

6 K, where dH is the Hausdorff distance. So by Theorem 4.4,the local structure stabilises crudely on a subspace if and only if a space is notsaturated by basic sequences tight with constants.


Often it is useful to have a bit more than local minimality. So we say that a basis(en) is locally block minimal if it is K-crudely finitely block representable in all of itsblock bases for some K. As with crude finite representability we see that there thenmust be a constant K and a block (yn) such that (en) is K-crudely finitely blockrepresentable in all block subspaces of (yn). We now have the following version ofTheorem 4.4 for finite block representability.

Theorem 4.5. Let (en) be a Schauder basis. Then (en) has a block basis (xn) withone of the following two properties, which are mutually exclusive and both possible.

(1) For all block bases (yn) of (xn) there are intervals I1 < I2 < I3 < . . . suchthat (yn)n∈IK is not K-equivalent to a block sequence of (xn)n/∈IK

,(2) (xn) is locally block minimal.

When (xn) is asymptotically ℓp and some block subspace [yn] of [xn] is uniformlyblock finitely representable in all tail subsequences of (xn), then it is clear that (yn)must actually be equivalent with the unit vector basis of ℓp, or c0 for p = ∞. Sothis shows that for asymptotically ℓp bases (xn), either [xn] contains an isomorphiccopy of ℓp or c0 or (xn) itself satisfies condition (1) of Theorem 4.5. This is thecounterpart of Proposition 4.2 for block sequences. In particular, since T ∗ does notcontain c0, but has a strongly asymptotically ℓ∞ basis, it thus satisfies (1).

Finally we note that there exist tight spaces which do not admit subspaces whichare tight with constants:

Example 4.6. There exists a reflexive, tight, locally block minimal Banach space.

Proof. E. Odell and T. Schlumprecht [29] have built a reflexive space OS with abasis such that every monotone basis is block finitely representable in any blocksubspace of OS. It is in particular locally block minimal and therefore contains nobasic sequence which is tight with constants. We do not know whether the spaceOS is tight. Instead, we notice that since the summing basis of c0 is block finitelyrepresentable in any block subspace of OS, OS cannot contain an unconditionalblock sequence. By Gowers’ dichotomy theorem it follows that some block subspaceof OS is HI, and by Theorem 3.13 and the fact that HI spaces do not containminimal subspaces, that some further block subspace is tight, which completes theproof. �

It is unknown whether there is an unconditional example with the above prop-erty. There exists an unconditional version of OS [28] but it is unclear whether ithas no minimal subspaces. We do have however:

Example 4.7. There exists a reflexive space with an unconditional basis which istight and locally minimal.

The example will be the dual of a space constructed by Gowers in [14]. Theproof requires some knowledge of the techniques used in that article and will begiven in Section 8.

5. Tightness by range and subsequential minimality

Theorem 1.1 shows that if one allows oneself to pass to a basis for a subspace,one can find a basis in which there is a close connection between subspaces spannedby block bases and subspaces spanned by subsequences. Thus, for example, if the


basis is tight there can be no space embedding into all the subspaces spanned bysubsequences of the basis. On the other hand, any block basis in Tsirelson’s spaceT is equivalent to a subsequence of the basis, and actually every subspace of a blocksubspace [xn] in T contains an isomorphic copy of a subsequence of (xn). In fact,this phenomenon has a deeper explanation and we shall now proceed to show thatthe connection between block sequences and subsequences can be made even closer.

Lemma 5.1. If (en) is a basis for a space not containing c0, then for all finite in-tervals (In) such that min In −→

n→∞∞ and all subspaces Y , there is a further subspace

Z such that

‖PIk|Z‖ −→

k→∞0.

Proof. By a standard perturbation argument, we can suppose that Y is gener-ated by a normalised block basis (yn). Let K be the basis constant of (en). Asmin In −→

n→∞∞ and each In is finite, we can choose a subsequence (vn) of (yn) such

that for all k the interval Ik intersects the range of at most one vector vm from(vn). Now, since c0 does not embed into [en], no tail sequence of (vn) can satisfyan upper c0 estimate. This implies that for all N and δ > 0 there is a normalisedvector

z =N ′

∑

i=N

ηivi,

where |ηi| < δ. Using this, we now construct a normalised block sequence (zn) of(vn) such that there are m(0) < m(1) < . . . and αi with

zn =

m(n+1)−1∑

i=m(n)

αivi

and |αi| < 1n whenever m(n) 6 i < m(n+ 1).

Now suppose u =∑

j λjzj and k are given. Then there is at most one vector zn

whose range intersect the interval Ik. Also, there is at most one vector vp from thesupport of zn whose range intersect Ik. Therefore,

‖PIk(u)‖ = ‖PIk

(λnzn)‖ = ‖PIk(λnαpvp)‖

6 2K‖λnαpvp‖ 6 |λn| ·2K

n6

4K2

n‖u‖.

It follows that ‖PIk|[zl]‖ 6 4K2

nk, where nk is such that Ik intersects the range of

znk(or nk = k if Ik intersects the range of no zn). Since min Ik −→

k→∞∞ and (zn) is

a block basis, nk → ∞ when k → ∞, and hence ‖PIk|[zl]‖ −→

k→∞0. �

Our next result should be contrasted with the construction by Pe lczynski [31]of a basis (fi) such that every basis is equivalent with a subsequence of it, andhence such that every space contains an isomorphic copy of a subsequence. Weshall see that for certain spaces E such constructions cannot be done relative to thesubspaces of E provided that we demand that (fn) lies in E too. Recall that twoBanach spaces are said to be incomparable if neither one embeds into the other.

Proposition 5.2. Suppose that (en) is a basis such that any two block subspaceswith disjoint ranges are incomparable. Suppose also that (fn) is either a block basis


or a shrinking basic sequence in [en]. Then [en] is saturated with subspaces Z suchthat no subsequence of (fn) embeds into Z.

Proof. Suppose first that (fn) is shrinking. Then, by taking a perturbation of (fn),we can suppose that each fn has finite support with respect to the basis (ei) and,moreover, that min range(fn) → ∞. Let In = range(fn).

Fix an infinite set N ⊆ N. Then for all infinite subsets A ⊆ N there is aninfinite subset B ⊆ A such that (fn)n∈B is a (not necessarily normalised) blocksequence and hence, since (en) is tight by range, [fn]n∈B 6⊑ [en]n/∈

S

i∈B Ii, and so

also [fn]n∈N 6⊑ [en]n/∈S

i∈A Ii. Applying Lemma 3.2 to X = [fn]n∈N , this implies

that for all embeddings T : [fn]n∈N → [en]n∈N, we have lim infn∈N‖PIkT ‖ > 0.

So find by Lemma 5.1 a subspace Z ⊆ Y such that ‖PIk|Z‖ −→

k→∞0. Then no

subsequence of (fn)∈N embeds into Z.The argument in the case (fn) is a block basis is similar. We set In = range fn

and repeat the argument above. �

We notice that in the above proof we actually have a measure for how “flat” asubspace Z of [en] needs to be in order that the subsequences of (fn) cannot embed.Namely, it suffices that ‖PIk

|Z‖ −→k→∞

0.

We should also mention that, by similar but simpler arguments, one can showthat if (en) is a basis such that any two disjoint subsequences span incompara-ble spaces, then some subspace of [en] fails to contain any isomorphic copy of asubsequence of (en).

The assumption in Proposition 5.2 that block subspaces with disjoint ranges areincomparable is easily seen to be equivalent to the following property of a basis(en), that we call tight by range. If (yn) is a block sequence of (en) and A ⊆ N isinfinite, then

[yn]n∈N 6⊑ [en

∣

∣ n /∈⋃

i∈A

range yi].

Thus, (en) is tight by range if it is tight and for all block sequences (yn) of (en) thecorresponding sequence of intervals Ii is given by Ii = range yi. This property isalso weaker than disjointly supported subspaces being incomparable, which we shallcall tight by support. It is trivial to see that a basis, tight by range, is continuouslytight.

We say that a basic sequence (en) is subsequentially minimal if any subspace of[en] contains an isomorphic copy of a subsequence of (en). It is clearly a weak formof minimality.

In [22] the authors study another notion in the context of certain partly modifiedmixed Tsirelson spaces that they also call subsequential minimality. Accordingto their definition, a basis (en) is subsequentially minimal if any block basis hasa further block basis equivalent to a subsequence of (en). However, in all theirexamples the basis (en) is weakly null and it is easily seen that whenever thisis the case the two definitions agree. They also define (en) to be strongly non-subsequentially minimal if any block basis contains a further block basis that hasno further block basis equivalent to a subsequence of (en). By Proposition 5.2, thisis seen to be weaker that tightness by range.

We shall now proceed to show a dichotomy between tightness by range andsubsequential minimality.


Theorem 5.3 (4th dichotomy). Let E be a Banach space with a basis (en). Thenthere exists a block sequence (xn) of (en) with one of the following properties, whichare mutually exclusive and both possible:

(1) Any two block subspaces of [xn] with disjoint ranges are incomparable.(2) The basic sequence (xn) is subsequentially minimal.

Arguably Theorem 5.3 is not a dichotomy in Gowers’ sense, since property (2)is not hereditary: for example the universal basis of Pe lczynski [31] satisfies (2)while admitting subsequences with property (1). However, it follows obviouslyfrom Theorem 5.3 that any basis (en) either has a block basis such that any twoblock subspaces with disjoint ranges are incomparable or has a block basis (xn)that is hereditarily subsequentially minimal, i.e., such that any block has a furtherblock that is subsequentially minimal. Furthermore, by an easy improvement ofour proof or directly by Gowers’ second dichotomy, if the first case of Theorem 5.3fails, then one can also suppose that [xn] is quasi minimal.

We shall call a basis (xn) sequentially minimal if it is both hereditarily subse-quentially minimal and quasi minimal. This is equivalent to any block basis of(xn) having a further block basis (yn) such that every subspace of [xn] contains anequivalent copy of a subsequence of (yn). We may therefore see Theorem 5.3 asproviding a dichotomy between tightness by range and sequential minimality.

Before giving the proof of Theorem 5.3, we first need to state an easy consequenceof the definition of Gowers’ game.

Lemma 5.4. Let E be a space with a basis and assume II has a winning strategyin Gowers’ game in E to play in some set B. Then there is a non-empty tree T offinite block sequences such that [T ] ⊆ B and for all (y0, . . . , ym) ∈ T and all blocksequences (zn) there is a block ym+1 of (zn) such that (y0, . . . , ym, ym+1) ∈ T .

Proof. Suppose σ is the strategy for II. We define a pruned tree T of finite blockbases (y0, . . . , ym) and a function ψ associating to each (y0, . . . , ym) ∈ T a sequence(z0, . . . , zk) such that for some k0 < . . . < km = k,

I z0 . . . zk0 zk0+1 . . . zk1 . . . zkm−1 . . . zkm

II y0 y1 . . . ym

has been played according to σ.

• The empty sequence ∅ is in T and ψ(∅) = ∅.• If (y0, . . . , ym) ∈ T and

ψ(y0, . . . , ym) = (z0, . . . , zk),

then we let (y0, . . . , ym, ym+1) ∈ T if there are some zk < zk+1 < . . . < zl

and k0 < . . . < km = k such that

I z0 . . . zk0 zk0+1 . . . zk1 . . . zkm+1 . . . zkl

II y0 y1 . . . ym+1

has been played according to σ and in this case we let

ψ(y0, . . . , yk, ym) = (z0, . . . , zk, zk+1, . . . , zl)

be some such sequence.


Now, if (y0, y1, y2, . . .) is such that (y0, . . . , ym) ∈ T for all m, then ψ(∅) ⊆ ψ(y0) ⊆ψ(y0, y1) ⊆ . . . and (yi) is the play of II according to the strategy σ in response to(zi) being played by I. So [T ] ⊆ B. It also follows by the construction that for each(y0, . . . , ym) ∈ T and block sequence (zi) there is a block ym+1 of (zi) such that(y0, . . . , ym, ym+1) ∈ T . �

We now pass to the proof of Theorem 5.3.

Proof. If E contains c0 the theorem is trivial. So suppose not. By the solution tothe distortion problem by Odell and Schlumprecht [27] and passing to a subspace,we can suppose there are two positively separated inevitable closed subsets F0 andF1 of the unit sphere of E, i.e., such that dist(F0, F1) > 0 and every block basishas block vectors belonging to both F0 and F1.

Suppose that (en) has no block sequence satisfying (1). Then for all block se-quences (xn) there are further block sequences (yn) and (zn) with disjoint rangessuch that [yn] ⊑ [zn]. We claim that there is a block sequence (fn) and a constantK such that for all block sequences (xn) of (fn) there are further block sequences(yn) and (yn) with disjoint ranges such that [yn] ⊑K [zn]. If not, we can constructa sequence of block sequences (fK

n ) such that (fK+1n ) is a block of (fK

n ) and suchthat any two block sequences of (fK

n ) with disjoint ranges are K2-incomparable.

By Lemma 2.1, we then find a block sequence (gn) of (en) that is√K-equivalent

with a block sequence of (fKn ) for every K. Find now block subspaces (yn) and

(zn) of (gn) with disjoint ranges and a K such that [yn] ⊑ [zn]. Then (gn) is√K-

equivalent with a block sequence of (fKn ) and hence we can find K3/2-comparable

block subspaces of (fKn ) with disjoint ranges, contradicting our assumption.

So suppose (fn) and K are chosen as in the claim. Then for all block sequences(xn) of (fn) we can find an infinite set B ⊆ N and a block sequence (yn) of (xn)such that [yn]n∈B K-embeds into [yn]n/∈B.

We claim that any block basis of (fn) has a further block basis in the followingset of normalised block bases of (fn):

A = {(yn)∣

∣ ∀n y2n ∈ F0 ∪ F1 & ∃∞n y2n ∈ F0 & [y2n+1]y2n∈F0 ⊑K [y2n+1]y2n∈F1}.To see this, suppose that (xn) is a block sequence of (fn) and let (zn) be a blocksequence of (xn) such that z3n ∈ F0 and z3n+1 ∈ F1. We can now find an infiniteset B ⊆ N and a block sequence (vn) of (z3n+2) such that [vn]n∈B ⊑K [vn]n/∈B.Let now y2n+1 = vn and notice that we can choose y2n = zi ∈ F0 for n ∈ B andy2n = zi ∈ F1 for n /∈ B such that y0 < y1 < y2 < . . .. Then (yn) ∈ A.

Choose now a sequence ∆ = (δn) of positive reals, δn < dist(F0, F1)/3, such thatif (xn) and (yn) are block bases of (en) with ‖xn − yn‖ < δn, then (xn) ∼2 (yn).Since A is clearly analytic, it follows by Gowers’ determinacy theorem that for someblock basis (xn) of (fn), II has a winning strategy to play in A∆ whenever I playsa block basis of (xn). We now show that some block basis (vn) of (xn) is such thatany subspace of [vn] contains a sequence 2K-equivalent to a subsequence of (vn),which will give us case (2).

Pick first by Lemma 5.4 a non-empty tree T of finite block sequences of (xn) suchthat [T ] ⊆ A∆ and for all (u0, . . . , um) ∈ T and all block sequences (zn) there is ablock um+1 of (zn) such that (u0, . . . , um, um+1) ∈ T . Since T is countable, we canconstruct inductively a block sequence (vn) of (xn) such that for all (u0, . . . , um) ∈ Tthere is some vn with (u0, . . . , um, vn) ∈ T .


We claim that (vn) works. For if (zn) is any block sequence of (vn), we constructinductively a sequence (un) ∈ A∆ as follows. Using inductively the extensionproperty of T , we can construct an infinite block sequence (h0

n) of (zn) that belongsto [T ]. Since [T ] ⊆ A∆, there is a shortest initial segment (u0, . . . , u2k0) ∈ T of(h0

n) such that d(u2k0 , F0) < δ2k0 . Pick now a term u2k0+1 from (vn) such that(u0, . . . , u2k0 , u2k0+1) ∈ T .

Again, using the extension property of T , there is an infinite block sequence (h1n)

of (zn) such that

(u0, . . . , u2k0 , u2k0+1)⌢(h1n)n ∈ [T ].

Also, as [T ] ⊆ A∆, there is a shortest initial segment

(u0, . . . , u2k0 , u2k0+1, . . . , u2k1) ∈ T

of

(u0, . . . , u2k0 , u2k0+1)⌢(h(1)n )n

that properly extends (u0, . . . , u2k0 , u2k0+1) and such that d(u2k1 , F0) < δ2k1 . Wethen pick a term u2k1+1 of (vn) such that (u0, . . . , u2k1 , u2k1+1) ∈ T . We continuein the same fashion.

At infinity, we then have a block sequence (un) ∈ A∆ and integers k0 < k1 < . . .such that d(u2n, F0) < δ2n if and only if n = ki for some i and such that for everyi, u2ki+1 is a term of (vn). Let now (wn) ∈ A be such that ‖wn − un‖ < δn. Then,as δn < dist(F0, F1)/3, we have that w2n ∈ F0 if and only if n = ki for some i andw2n ∈ F1 otherwise. Moreover, as (wn) ∈ A,

[w2ki+1]i∈N = [w2n+1]w2n∈F0 ⊑K [w2n+1]w2n∈F1 = [w2n+1]n6=ki .

So by the choice of δn we have

[u2ki+1]i∈N ⊑2 [w2ki+1]i∈N ⊑K [w2n+1]n6=ki ⊑2 [u2n+1]n6=ki .

Since [u2n+1]n6=ki is a subspace of [zn] and (u2ki+1) a subsequence of (vn) thisfinishes the proof. �

If for some constant C, all subspaces of [en] contain a C-isomorphic copy ofa subsequence of (en), we say that (en) is subsequentially C-minimal. Our proofshows that condition (2) in Theorem 5.3 may be improved to “For some constantC the basic sequence (xn) is subsequentially C-minimal”.

It is interesting to notice that if (xn) is hereditarily subsequentially minimal, thenthere is some C and a block sequence (vn) of (xn) such that (vn) is hereditarilysubsequentially C-minimal with the obvious definition. To see this, we first noticethat by Proposition 5.2, (xn) can have no block bases (yn) such that further blocksubspaces with disjoint ranges are incomparable. So, by the proof of Theorem 5.3,for any block base (yn) there is a constant C and a further block basis (zn) which issubsequentially C-minimal. A simple diagonalisation using Lemma 2.1 now showsthat by passing to a block (vn) the C can be made uniform. We do not know if anysubsequentially minimal basis is necessarily subsequentially C-minimal for someC, nor do we know if any hereditarily subsequentially minimal basis is hereditarilysubsequentially C-minimal for some C. Recall that Gowers also proved that a quasiminimal space must contain a further subspace which is C-quasi minimal [17].

We also indicate a variation on Theorem 5.3, relating the Casazza property to aslightly stronger form of sequential minimality. This answers the original problem


of Gowers left open in [17]. This variation is probably of less interest than Theorem5.3 because the Casazza property does not seem to imply tightness and also becausethe stronger form of sequential minimality may look somewhat artificial (althoughit is satisfied by Tsirelson’s space and is reminiscent of Schlumprecht’s notion ofClass 1 space [36]).

We say that two block sequences (xn) and (yn) alternate if either x1 < y1 <x2 < y2 < · · · or y1 < x1 < y2 < x2 < · · · .

Proposition 5.5. Let E be a Banach space with a basis (en). Then there existsa block sequence (xn) with one of the following properties, which are exclusive andboth possible:

(1) [xn] has the Casazza property, i.e., no alternating block sequences in [xn]are equivalent.

(2) There exists a family B of block sequences saturating [xn] and such thatany two block sequences in B have subsequences which alternate and areequivalent.

In particular, in case (2), E contains a block subspace U = [un] such that for everyblock sequence of U , there is a further block sequence equivalent to, and alternatingwith, a subsequence of (un).

Proof. If (en) does not have a block sequence satisfying (1), then any block sequenceof (en) has a further block sequence in A = {(yn)

∣

∣ (y2n) ∼ (y2n+1)}. Let ∆ be smallenough so that A∆ = A. By Gowers’ theorem, let (xn) be some block sequence of(en) so that II has a winning strategy to play in A∆ whenever plays a block sequenceof (xn). Let T be the associated tree given by Lemma 5.4. By construction, for anyblock sequence (zn) of (xn), we may find a further block sequence (vn) such thatfor any (y0, . . . , ym) ∈ T , there exists some vn with (y0, . . . , ym, vn) ∈ T . We setf((zn)) = (vn) and F = {f((zn))

∣

∣ (zn) 6 (xn)}. Given (vn) and (wn) in F , it isthen clear that we may find subsequences (v′n) and (w′

n) so that (v′0, w′0, v

′1, w

′1, . . .) ∈

T , and therefore (v′n) and (w′n) are equivalent. �

To conclude we may observe that there is no apparent relation between tight-ness by range and tightness with constants. Indeed Tsirelson’s space is tight withconstants and sequentially minimal. Similarly, Example 4.7 will be proved to betight by support and therefore by range, but is locally minimal. There also existsa space combining the two forms of tightness:

Example 5.6. There exists a space with a basis which is tight with constants andtight by range.

The example has a basis that is tight by support and strongly asymptotically ℓ1.It is based on techniques of [2] and will be constructed in Section 9.

Finally, if a space X is locally minimal and equipped with a basis which is tightby support and therefore unconditional (such as Example 4.7), then the readerwill easily check the following. The canonical basis of X ⊕X is tight (for a blocksubspace Y = [yn] of X ⊕X use the sequence of intervals associated to the rangesof yn with respect to the canonical 2-dimensional decomposition of X ⊕ X), butneither tight by range nor with constants. However, a more interesting questionremains open: does there exist a tight space which does not contain a basic sequencewhich is tight by range or with constants?


6. Refining Gowers’ dichotomies

We recall the list of inevitable classes of subspaces contained in a Banach spacegiven by Gowers in [17]. Remember that a space is said to be quasi minimal ifany two subspaces have a common ⊑-minorant, and strictly quasi minimal if itis quasi minimal but does not contain a minimal subspace. Two spaces are saidto be incomparable in case neither of them embeds into the other, and totallyincomparable if no space embeds into both of them.

Theorem 6.1 (Gowers [17]). Let X be an infinite dimensional Banach space. ThenX contains a subspace Y with one of the following properties, which are all possibleand mutually exclusive.

(1) Y is hereditarily indecomposable,(2) Y has an unconditional basis such that no two disjointly supported block

subspaces are isomorphic,(3) Y has an unconditional basis and is strictly quasi minimal,(4) Y has an unconditional basis and is minimal.

In case (2) the condition that any two disjointly supported block subspaces arenon-isomorphic may be replaced by the condition that any two such subspaces areincomparable, or even totally incomparable. It is easy to see that any space in case(2) is tight. Indeed if any two disjointly supported block subspaces in E = [en]are totally incomparable, then in particular (en) is tight by range. For this reasoncondition (2) is also called tightness by support.

Theorem 1.1 improves the list of Gowers in case (3). Indeed any strictly quasiminimal space contains a tight subspace, but the space S(T (p)), 1 < p < +∞ isstrictly quasi minimal and not tight: it is saturated with subspaces of T (p), whichis strictly quasi minimal, and, as was already observed, it is not tight because itscanonical basis is symmetric.

Concerning case (1), properties of HI spaces imply that any such space containsa tight subspace, but it remains open whether every HI space with a basis is tight.It is worth remembering that, by the study of Example 4.6, there exists a tight HIspace X such that no basic sequence in X is tight with constants.

Question 6.2. Is every HI space with a basis tight?

Combining Theorem 6.1 with Theorem 1.1 and Theorem 1.3, we refine the listof inevitable spaces of Gowers as follows:

Theorem 6.3. Let X be an infinite dimensional Banach space. Then X containsa subspace Y with one of the following properties, which are all mutually exclusive.

(1) Y is hereditarily indecomposable and has a basis such that no two blocksubspaces with disjoint ranges are comparable,

(2) Y is hereditarily indecomposable and has a basis which is tight and sequen-tially minimal,

(3) Y has an unconditional basis such that no two disjointly supported blocksubspaces are isomorphic,

(4) Y has an unconditional basis such that no two block subspaces with disjointranges are comparable, and is quasi minimal,

(5) Y has an unconditional basis which is tight and sequentially minimal,(6) Y has an unconditional basis and is minimal.


We conjecture that the space of Gowers and Maurey is of type (1), although wehave no proof of this fact. Instead we prove that an asymptotically unconditional HIspace constructed by Gowers [15] is of type (1). The proof requires some familiaritywith Gowers and Maurey’s techniques and is postponed until the end of article.

We do not know whether type (2) spaces exist. If they do, they may be thoughtof as HI versions of type (5) spaces, i.e., of Tsirelson like spaces, so one might lookfor an example in the family initiated by the HI asymptotically ℓ1 space of Argyrosand Deliyanni, whose “ground” space is a mixed Tsirelson’s space based on thesequence of Schreier families [1].

In relation with the λId + K problem, it is not difficult to show that if X isa complex hereditarily indecomposable Banach space with a basis such that someblock subspace Y embeds into a block subspace with disjoint range, then thereexists an operator S : Y → X which is strictly singular and non-compact. Socandidates for solving the λId + K problem by the positive should probably belooked for among type (1) spaces.

The first example of type (3) was built by Gowers [14] and further analysed in[19]. We provide various other examples of such spaces in the end of this article.

Type (4) means that for any two block subspaces Y and Z with disjoint ranges, Ydoes not embed into Z, but some further block subspace Y ′ of Y does (Y ′ thereforehas disjoint support but not disjoint range from Z). It is unknown whether thereexist spaces of type (4). Gowers sketched the proof of a weaker result, namely theexistence of a strictly quasi minimal space with an unconditional basis and with theCasazza property, i.e., such that for no block sequence the sequence of odd vectorsis equivalent to the sequence of even vectors, but his example was never actuallychecked. Alternatively, results of [22] Section 4 suggest that a mixed Tsirelsonspace example might be looked for.

The main example of a space of type (5) is Tsirelson’s space. Actually sincespaces of type (1) to (4) are either HI or satisfy the Casazza property, they arenever isomorphic to a proper subspace. Therefore for example spaces with a basissaturated with block subspaces isomorphic to their hyperplanes must contain asubspace of type (5) or (6). So our results may reinforce the idea that Tsirelson’sspace is the canonical example of classical space without a minimal subspace.

It is worth noting that as a consequence of the theorem of James, spaces of type(3), (4) and (5) are always reflexive.

Finally it is interesting to note that a new refinement can be made to this listthanks to the 5th dichotomy and to a stabilisation theorem of A. Tcaciuc [37]generalising a result of [13]. We shall only indicate here how this refines the casesof type (5) and (6) - the classical cases. A general picture of the complete listof inevitable spaces will be given at the end of the article after new examples ofspaces have been analysed. It is easy to see that Tcaciuc’s theorem may be statedas follows: any Banach space contains either a strongly asymptotically ℓp subspace,1 6 p 6 +∞, or a subspace Y such that

∀ǫ > 0∃n ∈ N, ∀Y1, . . . , Yn ⊆ Y, ∃y1, z1 ∈ Y1, . . . ∃yn, zn ∈ Yn : ‖n

∑

i=1

yi‖ < ǫ‖n

∑

i=1

zi‖,

where yn, zn may be supposed of norm 1 in the above statement.


The second property in this dichotomy will be called uniform inhomogeneity.As strongly asymptotically ℓp bases are unconditional, while the HI property isequivalent to uniform inhomogeneity with n = 2 for all ǫ > 0, Tcaciuc’s dichotomyis only relevant for spaces with an unconditional basis. We combine Tcaciuc’s resultwith Proposition 4.2 (see also [6]), the 5th dichotomy, the fact that asymptoticallyℓ∞ spaces are locally minimal, and the classical theorem of James to obtain:

Proposition 6.4. Any space of type (5) or (6) in Theorem 6.3 contains a subspaceY with one of the following properties, which are all mutually exclusive.

(5a) Y has a tight with constants and sequentially minimal unconditional basis,and is uniformly inhomogeneous,

(5b) Y has a tight, locally and sequentially minimal unconditional basis, and isuniformly inhomogeneous,

(5c) Y has a tight with constants, sequentially minimal and strongly asymptoti-cally ℓp basis, 1 6 p < +∞,

(5d) Y has a tight, sequentially minimal and strongly asymptotically ℓ∞ basis,(6a) Y has an unconditional basis, is minimal and uniformly inhomogeneous,(6b) Y has a strongly asymptotically ℓ∞ basis, is minimal, and is reflexive,(6c) Y is isomorphic to c0 or ℓp, 1 6 p < +∞.

We know of no space close to being of type (5a) or (5b). A candidate for (5d)could be the dual of some partly modified mixed Tsirelson’s space not satisfyingthe “blocking principle” due to Casazza, Johnson, and Tzafriri [4] (see [22]). For-tunately, the remaining cases may be illustrated with well-known examples. Thereader will already have identified Tsirelson’s space and its convexifications as thecanonical examples of spaces of type (5c). Schlumprecht’s space S [34] does notcontain an asymptotically ℓp subspace, therefore it contains a uniformly inhomoge-neous subspace, which implies by minimality that S itself is of type (6a) - a directproof will also be given later on. Finally, the dual of Tsirelson’s space is of type(6b).

The interest of Proposition 6.4 lies in the fact that it provides a first idea ofa natural list of inevitable spaces such that the “nicest” class is, as it should be,the class of spaces c0 and ℓp. It would of course be interesting to refine Tcaciuc’scriterium, that is to see whether uniform inhomogeneity implies some strongerproperty on some further subspace. Maybe some property related to operators,like Schlumprecht’s definition of “Class 2” spaces [35], could be deduced from it, sothat (6a) could be replaced by some more informative assumption. Proposition 6.4also gives lines of research for finding new examples of “pure” Banach spaces: type(5a), (5b), and (5d) properties are obvious directions to follow.

For completeness we should mention that R. Wagner has also proved a dichotomybetween asymptotic unconditionality and a strong form of the HI property [39]. Hisresult could be used to refine the cases of type (1) and (2).

Concerning the operators living on the spaces of different types, it is known thatif X is a complex HI space and Y is a subspace, then any operator from Y into Xis a strictly singular perturbation of the inclusion map [7]; similar results hold inthe real case [8]. Also, Gowers proves in [17] a structure theorem on operators onspaces that are tight by support. However, his proof is not quite correct and wetherefore take the opportunity to give a direct short proof of this result here basedon his ideas.


Lemma 6.5 (Lemma 7.6 in [17]). Let E = [en] be a space with an unconditionalbasis such that any two disjointly supported subspaces are incomparable. Suppose(yn) and (zn) are block bases and T : yn 7→ zn an isomorphism. Then there is aninvertible diagonal operator D : E → E such that T − D|[yn] and T−1 − D−1|[zn]

are strictly singular.

Proof. For simplicity, let (en) be 1-unconditional and let c, C be positive constantssuch that for all x ∈ Y = [yn] we have c‖x‖ 6 ‖Tx‖ 6 C‖x‖. Define a partitionJ ∪K ∪ L of N by

m ∈ J ⇔ ∃n 2C|e∗m(yn)| < |e∗m(zn)|,m ∈ K ⇔ ∃n 0 <

c

2|e∗m(yn)| < |e∗m(zn)| 6 2C|e∗m(yn)|,

m ∈ L⇔ ∃n |e∗m(zn)| 6c

2|e∗m(yn)|.

Now define D : E → E by

D(e∗m(yn)em) = e∗m(zn)em for m ∈ K,

D(em) = em otherwise.

Since the basis is unconditional, D is an isomorphism.Now, by the definition of L and since the basis (en) is 1-unconditional, we have

for all x ∈ [yn]

‖LTx‖ 6c

2‖Lx‖.

Also by 1-unconditionality

c‖Lx‖ 6 c‖x‖ 6 ‖Tx‖ 6 C‖x‖,whence

2‖LTx‖ 6 c‖Lx‖ 6 ‖Tx‖ 6 ‖LTx‖ + ‖(J + K)Tx‖,and thus

‖LTx‖ 6 ‖(J +K)Tx‖.So

c‖x‖ 6 ‖Tx‖ 6 2‖(J +K)Tx‖ 6 2‖Tx‖ 6 2C‖x‖,and therefore the map x 7→ (J+K)Tx is an isomorphism on Y . Similarly, using thethe definition of J and 1-unconditionality, one proves that the map x 7→ (K + L)xis an isomorphism on Y .

So Y is isomorphic with subspaces of E supported on respectively [en]n∈J+K

and [en]n∈K+L, and therefore, as disjointly supported subspaces are totally incom-parable, any operator from Y into either [en]n∈L or from Y into [en]n∈J is strictlysingular. In particular, this applies to J : Y → E, L : Y → E, JT : Y → E andLT : Y → E. Now, on Y we have

T −D = JT +KT + LT − JD −KD − LD

= JT +KT + LT − J −KT − L

= JT + LT − J − L,

which is strictly singular. A similar argument applies to T−1 −D−1. �


7. Chains and strong antichains

The results in this section are in response to a question of Gowers from hisfundamental study [17] and concern what types of quasi orders can be realisedas the set of (infinite-dimensional) subspaces of a fixed Banach space under therelation of isomorphic embeddability.

Problem 7.1 (Problem 7.9. in [17]). Given a Banach space X, let P (X) be theset of all equivalence classes of subspaces of X, partially ordered as above. Forwhich posets P does there exist a Banach space X such that every subspace Y of Xcontains a further subspace Z with P (Z) = P?

Gowers noticed himself that by a simple diagonalisation argument any posetP (X) must either have a minimal element, corresponding to a minimal space, orbe uncountable. We shall now use our notion of tightness to show how to attackthis problem in a uniform way and improve on several previous results.

Definition 7.2. Suppose that X is a separable Banach space and E is an analyticequivalence relation on a Polish space P . We say that X has an E-antichain, ifthere is a Borel function f : P → SB(X) such that for x, y ∈ P

(1) if xEy, then f(x) and f(y) are biembeddable,(2) if x 6E y, then f(x) and f(y) are incomparable.

We say that X has a strong E-antichain if there is a Borel function f : P → SB(X)such that for x, y ∈ P

(1) if xEy, then f(x) and f(y) are isomorphic,(2) if x 6E y, then f(x) and f(y) are incomparable.

For example, if =R is the equivalence relation of identity on R, then =R-antichainsand strong =R-antichains simply correspond to a perfect antichain in the usualsense, i.e., an uncountable Borel set of pairwise incomparable subspaces. Also,having a strong E-antichain implies, in particular, that E Borel reduces to theisomorphism relation between the subspaces of X .

The main result of [10] reformulated in this language states that if EΣ11

denotes

the complete analytic equivalence relation, then C[0, 1] has a strong EΣ11-antichain.

We will now prove a result that simultaneously improves on two results duerespectively to the first and the second author. In [11], the authors proved that aBanach space not containing a minimal space must contain a perfect set of non-isomorphic subspaces. This result was improved by Rosendal in [32], in which itwas shown that if a space does not contain a minimal subspace it must containa perfect set of pairwise incomparable spaces. And Ferenczi proved in [9] that ifX is a separable space without minimal subspaces, then E0 Borel reduces to theisomorphism relation between the subspaces of X . Recall that E0 is the equivalencerelation defined on 2N by xE0y if and only if ∃m ∀n > m xn = yn.

Theorem 7.3. Let X be a separable Banach space. Then X either contains aminimal subspace or has a strong E0-antichain.

Proof. Suppose X has no minimal subspace. By Theorem 3.13 and Lemma 3.11,we can find a basic sequence (en) in X and a continuous function f : [N] → [N]such that for all A,B ∈ [N], if B is disjoint from an infinite number of intervals[f(A)2i, f(A)2i+1], then [en]n∈A does not embed into [en]n∈B. We claim that thereis a continuous function h : 2N → [N] such that


(1) if xE0y, then |h(x) \ h(y)| = |h(y) \ h(x)| <∞,(2) if x 6E0 y, then [en]n∈h(x) and [en]n∈h(y) are incomparable spaces.

This will clearly finish the proof using the fact that subspaces of the same finitecodimension in a common superspace are isomorphic.

We will construct a partition of N into intervals

I00 < I1

0 < I20 < I0

1 < I11 < I2

1 < . . .

such that if we set J0n = I0

n ∪ I2n and J1

n = I1n, the following conditions hold:

(1) for all n, |J0n| = |J1

n|,(2) if s ∈ 2n, a = Js0

0 ∪ Js11 ∪ . . . ∪ Jsn−1

n−1 ∪ I0n, and A ∈ [a,N], then for some i,

[f(A)2i, f(A)2i+1] ⊆ I0n,

(3) if s ∈ 2n, a = Js00 ∪ Js1

1 ∪ . . . ∪ Jsn−1

n−1 ∪ I1n, and A ∈ [a,N], then for some i,

[f(A)2i, f(A)2i+1] ⊆ I1n.

Assuming this is done, for x ∈ 2N we set h(x) = Jx00 ∪ Jx1

1 ∪ . . .. Then for all nthere is an i such that

[f(h(x))2i, f(h(x))2i+1] ⊆ Jxnn .

Therefore, if x 6E0 y, then h(y) = Jy0

0 ∪Jy1

1 ∪. . . is disjoint from an infinite numberof Jxn

n and thus also from an infinite number of intervals [f(h(x))2i, f(h(x))2i+1],whence [en]n∈h(x) does not embed into [en]n∈h(y). Similarly, [en]n∈h(y) does notembed into [en]n∈h(x).

On the other hand, if xE0y, then clearly |h(x) \ h(y)| = |h(y) \ h(x)| <∞.It therefore only remains to construct the intervals Ii

n. So suppose by inductionthat I0

0 < I10 < I2

0 < . . . < I0n−1 < I1

n−1 < I2n−1 have been chosen (the initial

step being n = 0) such that the conditions are satisfied. Let m = maxJ1n−1 + 1 =

max I2n−1+1. For each s ∈ 2n and a = Js0

0 ∪Js11 ∪. . .∪Jsn−1

n−1 , there are by continuityof f some ks > m, some interval m 6 Ms 6 ks and an integer is such that for allA ∈

[

a ∪ [m, ks],N]

, we have

[f(A)2is , f(A)2is+1] = Ms.

Let now k = maxs∈2n ks and I0n = [m, k]. Then if s ∈ 2n and a = Js0

0 ∪ . . .∪ Jsn−1

n−1 ,

we have for all A ∈ [a ∪ I0n,N] some i such that

[f(A)2i, f(A)2i+1] ⊆ I0n.

Again for each s ∈ 2n and a = Js0

0 ∪ Js1

1 ∪ . . . ∪ Jsn−1

n−1 there are by continuity off some ls > k+ 1, some interval k+ 1 6 Ls 6 ls and an integer js such that for allA ∈

[

a ∪ [k + 1, ls],N]

, we have

[f(A)2js , f(A)2js+1] = Ls.

Let now l = maxs∈2n ls + k and I1n = [k + 1, l]. Then if s ∈ 2n and a = Js0

0 ∪ . . . ∪J

sn−1

n−1 , we have for all A ∈ [a ∪ I1n,N] some j such that

[f(A)2j , f(A)2j+1] ⊆ I1n.

Finally, we simply let I2n = [l+ 1, l+ |I1

n|− |I0n|]. This finishes the construction. �


Definition 7.4. We define a quasi order ⊆∗ and a partial order ⊆0 on the space[N] of infinite subsets of N by the following conditions:

A ⊆∗ B ⇔ A \B is finite

and

A ⊆0 B ⇔(

A = B or ∃n ∈ B \A : A ⊆ B ∪ [0, n[)

.

Also, if (an) and (bn) are infinite sequences of integers, we let

(an) 6∗ (bn) ⇔ ∀∞n an 6 bn.

We notice that ⊆0 is a partial order refining the quasi order ⊆∗, namely, wheneverA ⊆∗ B we let A ⊆0 B if B 6⊆∗ A or A = B or A△B admits a greatest elementwhich belongs to B.

Proposition 7.5. (1) Any closed partial order on a Polish space Borel embedsinto ⊆0.

(2) Any partial order on a set of size at most ℵ1 embeds into ⊆0.(3) The quasi order ⊆∗ embeds into ⊆0, but does not Borel embed.(4) And finally ⊆0 Borel embeds into ⊆∗.

Proof. (1) By an unpublished result of A. Louveau [25], any closed partial order ona Polish space Borel embeds into (P(N),⊆). And if we let (Jn) be a partition of N

into countable many infinite subsets, we see that (P(N),⊆) Borel embeds into ⊆∗

and ⊆0 by the mapping A 7→ ⋃

n∈A Jn.(2) & (3) It is well-known that any partial order of size at most ℵ1 embeds into

⊆∗ and if we let s : [N] → [N] be any function such that |A△B| <∞ ⇔ s(A) = s(B)and |A△s(A)| <∞, i.e., s is a selector for E0, then s embeds ⊆∗ into ⊆0. To see thatthere cannot be a Borel embedding of ⊆∗ into ⊆0, we notice that if h : [N] → [N]was a Borel function such that A ⊆∗ B ⇔ h(A) ⊆0 h(B), then, in particular,|A△B| is finite ⇔ h(A) = h(B), contradicting that E0 is a non-smooth equivalencerelation on [N].

(4) To see that ⊆0 Borel embeds into ⊆∗, we define for an infinite subset A of N

a sequence of integers g(A) = (an) by

an =∑

i∈A∩[0,n]

2i.

Suppose now that g(A) = (an) and g(B) = (bn). Then for each n,

an = bn ⇔ A ∩ [0, n] = B ∩ [0, n]

and

an < bn ⇔ ∃m ∈ B \A, m 6 n, A ∩ [0, n] ⊆ B ∪ [0,m[.

Thus, we have an = bn for infinitely many n if and only if A = B, and if an < bnfor infinitely many n, then either B \ A is infinite or for some m ∈ B \ A we haveA ⊆ B ∪ [0,m[. Moreover, if B \A is infinite, then for infinitely many n, an < bn.So

B 6⊆∗ A⇒ (bn) 66∗ (an) ⇒(

B 6⊆∗ A or A ⊆0 B)

,

and thus by contraposition

(bn) 6∗ (an) ⇒ B ⊆∗ A.


Also, if (bn) 66∗ (an), then B 6⊆∗ A or A ⊆0 B, so if moreover B ⊆0 A, we wouldhave A ⊆0 B and hence A = B, contradicting g(B) = (bn) 66∗ (an) = g(A). Thus,

B ⊆0 A⇒ (bn) 6∗ (an).

To see that also

(bn) 6∗ (an) ⇒ B ⊆0 A,

notice that if (bn) 6∗ (an) but B 6⊆0 A, then, as B ⊆∗ A, we must have A ⊆0 Band hence (an) 6∗ (bn). But then an = bn for almost all n and thus A = B,contradicting B 6⊆0 A. Therefore,

B ⊆0 A⇔ (bn) 6∗ (an),

and we thus have a Borel embedding of ⊆0 into the quasi order 6∗ on the spaceNN. It is well-known and easy to see that this latter Borel embeds into ⊆∗ andhence so does ⊆0. �

Proposition 7.6. Any Banach space without a minimal subspace contains a sub-space with an F.D.D. (Fn) satisfying one of the two following properties:

(a) if A,B ⊆ N are infinite, then∑

n∈A

Fn ⊑∑

n∈B

Fn ⇔ A ⊆∗ B,

(b) if A,B ⊆ N are infinite, then∑

n∈A

Fn ⊑∑

n∈B

Fn ⇔ A ⊆0 B.

Proof. Suppose X is a Banach space without a minimal subspace. Then by The-orem 3.13, we can find a continuously tight basic sequence (en) in X . Using theinfinite-dimensional Ramsey Theorem for analytic sets, we can also find an infiniteset D ⊆ N such that

(i) either for all infinite B ⊆ D, [ei]i∈B embeds into its hyperplanes,(ii) or for all B ⊆ D, [ei]i∈B is not isomorphic to a proper subspace.

And, by Lemma 3.11, we can after renumbering the sequence (en)n∈D as (en)n∈N

suppose that there is a continuous function f : [N] → [N] that for A,B ∈ [N], if B isdisjoint from an infinite number of intervals [f(A)2i, f(A)2i+1], then [en]n∈A doesnot embed into [en]n∈B.

We now construct a partition of N into intervals

I0 < I1 < I2 < . . .

such that the following conditions hold:

- for all n, |I0 ∪ . . . ∪ In−1| < |In|,- if A ∈ [N] and In ⊆ A, then for some i,

[f(A)2i, f(A)2i+1] ⊆ In.

Suppose by induction that I0 < I1 < . . . < In−1 have been chosen such that theconditions are satisfied. Let m = max In−1 + 1. For each a ⊆ [0,m[ there are bycontinuity of f some la > m, some interval m 6 Ma 6 la and an integer ia suchthat for all A ∈

[

a ∪ [m, la],N]

, we have

[f(A)2ia , f(A)2ia+1] = Ma.


Let now l > maxa⊆[0,m[ la be such that |I0 ∪ . . .∪ In−1| < l−m, and set In = [m, l[.Then if a ⊆ [0,m[, we have for all A ∈ [a ∪ In,N] some i such that

[f(A)2i, f(A)2i+1] ⊆ In,

which ends the construction.Let now Fn = [ei]i∈In . Clearly,

∑n−1i=0 dimFi < dimFn, and if A\B is infinite and

we let A′ =⋃

n∈A In and B′ =⋃

n∈B In, then B′ will be disjoint from an infinitenumber of the intervals defined by f(A′) and hence

∑

n∈A Fn = [en]n∈A′ does notembed into

∑

n∈B Fn = [en]n∈B′ .In case of (i) we have that for all infinite C ⊆ N,

(en)n∈C ⊑ (en)n∈C′ ⊑ (en)n∈C′′ ⊑ (en)n∈C′′′ ⊑ . . . ,

where D′ denotes D \ minD. So, in particular, for any infinite A ⊆ N,∑

n∈A Fn

embeds into all of its finite codimensional subspaces and thus if A\B is finite, then∑

n∈A Fn ⊑ ∑

n∈B Fn. This gives us (a).

In case (ii), if A ⊆0 B but B 6⊆0 A, we have, as dimFn >∑n−1

i=0 dimFi, that∑

n∈A Fn embeds as a proper subspace of∑

n∈B Fn. Conversely, if∑

n∈A Fn ⊑∑

n∈B Fn, then A \ B is finite and so either A ⊆0 B or B ⊆0 A. But if B ⊆0 Aand A 6⊆0 B, then

∑

n∈B Fn embeds as a proper subspace into∑

n∈A Fn and thusalso into itself, contradicting (ii). Thus, A ⊆0 B. So assuming (ii) we have theequivalence in (b). �

We may observe that Tsirelson’s space satisfies case (a) of Proposition 7.6, whilecase (b) is verified by Gowers–Maurey’s space, or more generally by any space oftype (1) to (4).

By Proposition 7.5 and Proposition 7.6 we now have the following result.

Theorem 7.7. Let X be an infinite-dimensional separable Banach space with-out a minimal subspace and let SB∞(X) be the standard Borel space of infinite-dimensional subspaces of X ordered by the relation ⊑ of isomorphic embeddability.Then ⊆0 Borel embeds into SB∞(X) and by consequence

(a) any partial order of size at most ℵ1 embeds into SB∞(X),(b) any closed partial order on a Polish space Borel embeds into SB∞(X).

We notice that this proves a strong dichotomy for the partial orders of Problem7.1, namely, either they must be of size 1 or must contain any partial order of size atmost ℵ1 and any closed partial order on a Polish space. In particular, in the secondcase we have well-ordered chains of length ω1 and also R-chains. This completesthe picture of [12].

8. Tight spaces of the type of Gowers and Maurey

In this section we prove that the dual of the type (3) space Gu constructed byGowers in [14] is locally minimal of type (3), that Gowers’ hereditarily indecompos-able and asymptotically unconditional space G defined in [15] is of type (1), andthat its dual G∗ is locally minimal of type (1). Theses spaces are natural variationon Gowers and Maurey’s space GM , and so familiarity with that construction willbe assumed: we shall not redefine the now classical notation relative to GM , suchas R.I.S. sequences, special functionals, the sets Q, K, L, etc., instead we shall tryto give details on the parts in which Gu or G differ from GM .


The idea of the proofs is similar to [14]. The HI property for Gowers-Maurey’sspaces is obtained as follows. Some vector x is constructed such that ‖x‖ is large,but so that if x′ is obtained from x by changing signs of the components of x, thenx∗(x′) is small for any norming functional x∗, and so ‖x′‖ is small. The upper boundfor x∗(x′) is obtained by a combination of unconditional estimates (not dependingon the signs) and of conditional estimates (i.e., based on the fact that |∑n

i=1 ǫi| ismuch smaller than n if ǫi = (−1)i for all i).

For our examples we shall need to prove that some operator T is unbounded.Thus we shall construct a vector x such that say Tx has large norm, and such thatx∗(x) is small for any norming x∗. The upper bound for x∗(x) will be obtained bythe same unconditional estimates as in the HI case, while conditional estimates willbe trivial due to the disjointness of the supports of the corresponding componentvectors and functionals. The method will be similar for the dual spaces.

Recall that if X is a space with a bimonotone basis, an ℓn1+-average with constant

1+ǫ is a normalised vector of the form∑n

i=1 xi, where x1 < · · · < xn and ‖xi‖ 6 1+ǫn

for all i. An ℓn∞+-average with constant 1 + ǫ is a normalised vector of the form∑n

i=1 xi, where x1 < · · · < xn and ‖xi‖ > 11+ǫ for all i. Recall that the function f

is defined by f(n) = log2(n+ 1). The space X is said to satisfy a lower f -estimateif for any x1 < · · · < xn,

1

f(n)

n∑

i=1

‖xi‖ 6 ‖n

∑

i=1

xi‖.

Lemma 8.1. Let X be a reflexive space with a bimonotone basis and satisfying alower f -estimate. Let (y∗k) be a normalised block sequence of X∗, n ∈ N, ǫ, α >0. Then there exists a constant N(n, ǫ), successive subsets Fi of [1, N(n, ǫ)], andvectors x∗i of the form λ

∑

k∈Fiy∗k, such that x∗ =

∑ni=1 x

∗i is an ℓn∞+- average

with constant 1 + ǫ. Furthermore, if for each i, xi is such that range xi ⊆ range x∗iand x∗i (xi) > α‖x∗i ‖, then x =

∑ni=1 xi is an ℓn1+-vector with constant 1+ǫ

α suchthat x∗(x) > α

1+ǫ‖x‖.

Proof. Since X satisfies a lower f -estimate, it follows by duality that any sequenceof successive functionals x∗1 < · · · < x∗n in G∗

u satisfies the following upper estimate:

1 6 ‖n

∑

i=1

x∗i ‖ 6 f(n) max16i6n

‖x∗i ‖.

Let N = nk where k is such that (1+ ǫ)k > f(nk). Assume towards a contradictionthat the result is false for N(n, ǫ) = N , then

y∗ = (y∗1 + . . .+ y∗nk−1) + . . .+ (y∗(n−1)nk−1+1 + . . .+ y∗nk)

is not an ℓn∞+-vector with constant 1 + ǫ, and therefore, for some i,

‖y∗ink−1+1 + . . .+ y∗(i+1)nk−1‖ 61

1 + ǫ‖y∗‖.

Applying the same reasoning to the above sum instead of y∗, we obtain, for somej,

‖y∗jnk−2+1 + . . .+ y∗(j+1)nk−2‖ 61

(1 + ǫ)2‖y∗‖.


By induction we obtain that

1 61

(1 + ǫ)k‖y∗‖ 6

1

(1 + ǫ)kf(nk),

a contradiction.Let therefore x∗ be such an ℓn∞+-average with constant 1 + ǫ of the form

∑

i x∗i .

Let for each i, xi be normalised such that range(xi) ⊆ range(x∗i ) and x∗i (xi) >

α‖x∗i ‖. Then

‖∑

i

xi‖ > x∗(∑

i

xi) >αn

1 + ǫ>

α

1 + ǫ‖∑

i

xi‖,

and in particular for each i,

‖xi‖ 61 + ǫ

αn‖∑

i

xi‖.

�

The following lemma is fundamental and therefore worth stating explicitly. Itappears for example as Lemma 4 in [15]. Recall that an (M, g)-form is a functionalof the form g(M)−1(x∗1 + . . .+ x∗M ), with x∗1 < · · · < x∗M of norm at most 1.

Lemma 8.2 (Lemma 4 in [15]). Let f, g ∈ F with g >√f , let X be a space with

a bimonotone basis satisfy a lower f -estimate, let ǫ > 0, let x1, . . . , xN be a R.I.S.

in X for f with constant 1 + ǫ and let x =∑N

i=1 xi. Suppose that

‖Ex‖ 6 sup{

|x∗(Ex)| : M > 2, x∗ is an (M, g)-form}

for every interval E such that ‖Ex‖ ≥ 1/3. Then ‖x‖ 6 (1 + ǫ+ ǫ′)Ng(N)−1.

8.1. A locally minimal space tight by support. Let Gu be the space definedin [14]. This space has a suppression unconditional basis, is tight by support andtherefore reflexive, and its norm is given by the following implicit equation, for allx ∈ c00:

‖x‖ = ‖x‖c0 ∨ sup{

f(n)−1n

∑

i=1

‖Eix‖∣

∣

∣2 6 n,E1 < . . . < En

}

∨ sup{

|x∗(x)|∣

∣

∣k ∈ K,x∗ special of length k

}

where E1, . . . , En are successive subsets (not necessarily intervals) of N.

Proposition 8.3. The dual G∗u of Gu is tight by support and locally minimal.

Proof. Given n ∈ N and ǫ = 1/10 we may by Lemma 8.1 find some N such thatthere exists in the span of any x∗1 < . . . < x∗N an ℓn+

∞ -average with constant 1+ǫ. Byunconditionality we deduce that any block-subspace of G∗

u contains ℓn∞’s uniformly,and therefore G∗

u is locally minimal.Assume now towards a contradiction that (x∗n) and (y∗n) are disjointly supported

and equivalent block sequences in G∗u, and let T : [x∗n] → [y∗n] be defined by Tx∗n =

y∗n.We may assume that each x∗n is an ℓn∞+-average with constant 1+ǫ. Using Hahn-

Banach theorem, the 1-unconditionality of the basis, and Lemma 8.1, we may alsofind for each n an ℓn1+-average xn with constant 1 + ǫ such that supp xn ⊆ supp x∗nand x∗n(xn) > 1/2. By construction, for each n, Tx∗n is disjointly supported from


[xk], and up to modifying T , we may assume that Tx∗n is in Q and of norm at most1 for each n.

If z1, . . . , zm is a R.I.S. of these ℓn1+-averages xn with constant 1 + ǫ, with m ∈[logN, expN ], N ∈ L, and z∗1 , . . . , z

∗m are the functionals associated to z1, . . . , zm,

then by [14] Lemma 7, the (m, f)-form z∗ = f(m)−1(z∗1 + . . .+ z∗m) satisfies

z∗(z1 + . . .+ zm) >m

2f(m)>

1

4‖z1 + . . .+ zm‖,

and furthermore Tz∗ is also an (m, f)-form. Therefore we may build R.I.S. vectorsz with constant 1 + ǫ of arbitrary length m in [logN, expN ], N ∈ L, so that zis 4−1-normed by an (m, f)-form z∗ such that Tz∗ is also an (m, f)-form. Wemay then consider a sequence z1, . . . , zk of length k ∈ K of such R.I.S. vectorsof length mi, and some corresponding (mi, f)-forms z∗1 , . . . , z

∗k (i.e z∗i 4−1-norms

zi and Tz∗i is also an (mi, f)-form for all i), such that Tz∗1 , . . . , T z∗k is a special

sequence. Then we let z = z1 + · · · + zk and z∗ = f(k)−1/2(z∗1 + . . . + z∗k). Since

Tz∗ = f(k)−1/2(Tz∗1 + . . .+ Tz∗k) is a special function it follows that

‖Tz∗‖ 6 1.

Our aim is now to show that ‖z‖ 6 3kf(k)−1. It will then follow that

‖z∗‖ > z∗(z)/‖z‖ > f(k)1/2/12.

Since k was arbitrary in K this will imply that T−1 is unbounded and provide thedesired contradiction.

The proof is now almost exactly the same as in [14]. LetK0 = K\{k} and let g bethe corresponding function given by [14] Lemma 6. To prove that ‖z‖ 6 3kf(k)−1

it is enough by [14] Lemma 8 and Lemma 8.2 to prove that for any interval E suchthat ‖Ez‖ > 1/3, Ez is normed by some (M, g)-form with M > 2.

By the discussion in the proof of the main theorem in [14], the only possiblenorming functionals apart from (M, g)-forms are special functionals of length k. Solet w∗ = f(k)−1/2(w∗

1 + · · · + w∗k) be a special functional of length k, and E be an

interval such that ‖Ez‖ > 1/3. We need to show that w∗ does not norm Ez.Let t be minimal such that w∗

t 6= Tz∗t . If i 6= j or i = j > t then by definitionof special sequences there exist M 6= N ∈ L, min(M,N) > j2k, such that w∗

i is an(M, f)-form and zj is an R.I.S. vector of sizeN and constant 1+ǫ. By [14] Lemma 8,

zj is an ℓN1/10

1+ -average with constant 2. If M < N then 2M < log log logN so, by

[14] Corollary 3, |w∗i (Ezj)| 6 6f(M)−1. If M > N then log log logM > 2N so, by

[14] Lemma 4, |w∗i (Ezj)| 6 2f(N)/N . In both cases it follows that |w∗

i (Ezj)| 6 k−2.If i = j = t we have |w∗

i (Ezj)| 6 1. Finally if i = j < t then w∗i = Tz∗i . Since

Tz∗i is disjointly supported from [xk] and therefore from zj, it follows simply thatw∗

i (Ezj) = 0 in that case.Summing up we have obtained that

|w∗(Ez)| 6 f(k)−1/2(k2.k−2 + 1) = 2f(k)−1/2 < 1/3 6 ‖Ez‖.Therefore w∗ does not norm Ez and this finishes the proof. �

It may be observed that G∗u is uniformly inhomogeneous. We state this in a

general form which implies the result for Gu, S and S∗ as well.

Proposition 8.4. Let f ∈ F and let X be a space with a bimonotone basissatisfying a lower f -estimate. Let ǫ0 = 1/10, and assume that for every n ∈


[logN, expN ], N ∈ L, x1, . . . , xn a R.I.S. in X with constant 1 + ǫ0 and x =∑N

i=1 xi,

‖Ex‖ 6 sup{

|x∗(Ex)| : M > 2, x∗ is an (M, f)-form}

for every interval E such that ‖Ex‖ ≥ 1/3. Then X and X∗ are uniformly inho-mogeneous.

Proof. Given ǫ > 0, let m ∈ L be such that f(m) > 16ǫ−1. Let Y1, . . . , Ym bearbitrary block subspaces of X . By the classical method for spaces with a lower festimate, we may find a R.I.S. sequence y1 < · · · < ym with constant 1 + ǫ0 withyi ∈ Yi, ∀i. By Lemma 8.2,

‖m

∑

i=1

yi‖ 6 2mf(m)−1.

Let on the other hand n ∈ [m10, expm] and E1 < · · · < Em be sets such that⋃m

j=1 Ej = {1, . . . , n} and |Ej | is within 1 of nm for all j. We may construct a R.I.S.

sequence x1, . . . , xn with constant 1 + ǫ0 such that xi ∈ Yj whenever i ∈ Ej . ByLemma 8.2,

‖∑

i∈Ej

xi‖ 6 (1 + 2ǫ0)(n

m+ 1)f(

n

m− 1)−1 6 2nf(n)−1m−1.

Let zj = ‖∑i∈Ejxi‖−1

∑

i∈Ejxi. Then zj ∈ Yj for all j and

‖m

∑

j=1

zj‖ > f(n)−1m

∑

j=1

(

‖∑

i∈Ej

xi‖−1∑

i∈Ej

‖xi‖)

> m/2.

Therefore

‖m

∑

i=1

yi‖ 6 4f(m)−1‖m

∑

i=1

zi‖ 6 ǫ‖m

∑

i=1

zi‖.

The proof concerning the dual is quite similar and uses the same notation. LetY1∗, . . . , Ym∗ be arbitrary block subspaces of X∗. By Lemma 8.1 we may find aR.I.S. sequence y1 < · · · < ym with constant 1 + ǫ0 and functionals y∗i ∈ Yi∗ suchthat range y∗i ⊆ range yi and y∗i (yi) > 1/2 for all i. Since ‖∑m

i=1 yi‖ 6 2mf(m)−1,it follows that

‖m

∑

i=1

y∗i ‖ > ‖m

∑

i=1

yi‖−1m

∑

i=1

y∗i (yi) > f(m)/4.

On the other hand we may construct a R.I.S. sequence x1, . . . , xn with constant1 + ǫ0 and functionals x∗i such that range x∗i ⊆ range xi, x

∗i (xi) > 1/2 for all i, and

such that x∗i ∈ Yj∗ whenever i ∈ Ej . Since ‖∑i∈Ejxi‖ 6 2nf(n)−1m−1, it follows

that

‖∑

i∈Ej

x∗i ‖ >n

2m

mf(n)

2n= f(n)/4.

Let z∗j = ‖∑i∈Ejx∗i ‖−1

∑

i∈Ejx∗i . Then z∗j ∈ Yj∗ for all j and

‖m

∑

j=1

z∗j ‖ 64

f(n)f(n) = 4.


Therefore

‖m

∑

i=1

z∗i ‖ 6 16f(m)−1‖m

∑

i=1

y∗i ‖ 6 ǫ‖m

∑

i=1

y∗i ‖.

�

Corollary 8.5. The spaces S, S∗, Gu, and G∗u are uniformly inhomogeneous.

8.2. Two HI spaces tight by range. We show that Gowers’ space G constructedin [15] and its dual are of type (1). The proof is a refinement of the proof that Gu orG∗

u is of type (3), in which we observe that the hypothesis of unconditionality maybe replaced by asymptotic unconditionality. The idea is to produce constituentparts of vectors or functionals in Gowers’ construction with sufficient control ontheir supports (and not just on their ranges, as would be enough to obtain the HIproperty for example).

The space G has a norm defined by induction as in GM , with the addition ofa new term which guarantees that its basis (en) is 2-asymptotically unconditional,that is for any sequence of normalised vectors N < x1 < . . . < xN , any sequence ofscalars a1, . . . , aN and any sequence of signs ǫ1, . . . , ǫN ,

‖N

∑

n=1

ǫnanxn‖ 6 2‖N

∑

n=1

anxn‖.

The basis is bimonotone and, although this is not stated in [15], it may be provedas for GM that G is reflexive. It follows that the dual basis of (en) is also 2-asymptotically unconditional. For the definition, the norm on G is given by theimplicit equation, for all x ∈ c00:

‖x‖ = ‖x‖c0 ∨ sup{

f(n)−1n

∑

i=1

‖Eix‖∣

∣

∣2 6 n,E1 < . . . < En

}

∨ sup{

|x∗(Ex)|∣

∣

∣k ∈ K,x∗ special of length k,E ⊆ N

}

∨ sup{

‖Sx‖∣

∣

∣S is an admissible operator

}

,

where E, E1, . . . , En are intervals of integers, and S is an admissible operator

if Sx = 12

∑Nn=1 ǫnEnx for some sequence of signs ǫ1, . . . , ǫN and some sequence

E1, . . . , EN of intervals which is admissible, i.e. N < E1 and 1+maxEi = minEi+1

for every i < N .

R.I.S. pairs and special pairs are considered in [15]; first we shall need a moregeneral definition of these. Let x1, . . . , xm be a R.I.S. with constant C, m ∈[logN, expN ], N ∈ L, and let x∗1, . . . , x

∗m be successive normalised functionals.

Then we call generalised R.I.S. pair with constant C the pair (x, x∗) defined byx = ‖∑m

i=1 xi‖−1(∑m

i=1 xi) and x∗ = f(m)−1∑m

i=1 x∗i .

Let z1, . . . , zk be a sequence of successive normalised R.I.S. vectors with constantC, and let z∗1 , . . . , z

∗k be a special sequence such that (zi, z

∗i ) is an R.I.S. pair for

each i. Then we shall call generalised special pair with constant C the pair (z, z∗)

defined by z =∑k

i=1 zi and z∗ = f(k)−1/2(∑k

i=1 z∗i ). The pair (‖z‖−1z, z∗) will be

called normalised generalised special pair.


Lemma 8.6. Let (z, z∗) be a generalised special pair of length k ∈ K with constant2 in G such that supp z∗ ∩ supp z = ∅. Then

‖z‖ 65k

f(k).

Proof. The proof follows classically the methods of [18] or [14]. Let K0 = K \ {k}and let g be the corresponding function given by [15] Lemma 5. To prove that‖z‖ 6 5kf(k)−1 it is enough by Lemma 8.2 to prove that for any interval E suchthat ‖Ez‖ > 1/3, Ez is normed by some (M, g)-form with M > 2.

By the discussion in [15] after the definition of the norm, the only possiblenorming functionals apart from (M, g)-forms are of the form Sw∗ where w∗ is aspecial functional of length k and S is an “acceptable” operator. We shall not statethe definition of an acceptable operator S, we shall just need to know that sincesuch an operator is diagonal of norm at most 1, it preserves support and (M, g)-forms, [15] Lemma 6. So let w∗ = f(k)−1/2(w∗

1 + · · ·+w∗k) be a special functional of

length k, S be an acceptable operator, and E be an interval such that ‖Ez‖ > 1/3.We need to show that Sw∗ does not norm Ez.

Let t be minimal such that w∗t 6= z∗t . If i 6= j or i = j > t then by definition

of special sequences there exist M 6= N ∈ L, min(M,N) > j2k, such that w∗i and

therefore Sw∗i is an (M, f)-form and zj is an R.I.S. vector of size N and constant

2. By [15] Lemma 8, zj is an ℓN1/10

1+ -average with constant 4. If M < N then

2M < log log logN so, by [15] Lemma 2, |Sw∗i (Ezj)| 6 12f(M)−1. If M > N then

log log logM > 2N so, by [15] Lemma 3, |Sw∗i (Ezj)| 6 3f(N)/N . In both cases it

follows that |Sw∗i (Ezj)| 6 k−2.

If i = j = t we simply have |Sw∗i (Ezj)| 6 1. Finally if i = j < t then w∗

i = z∗i .and since supp Sz∗i ⊆ supp z∗i and supp Ezi ⊆ supp zi, it follows that Sw∗

i (Ezj) = 0in that case.

Summing up we have obtained that

|Sw∗(Ez)| 6 f(k)−1/2(k2.k−2 + 1) = 2f(k)−1/2 < 1/3 6 ‖Ez‖.

Therefore Sw∗ does not norm Ez and this finishes the proof. �

The next lemma is expressed in a version which may seem technical but this willmake the proof that G is of type (1) more pleasant to read. At a first reading, thereader may simply assume that T = Id in its hypothesis.

Lemma 8.7. Let (xi) be a normalised block basis in G and T : [xi] → G be anisomorphism such that (Txi) is also a normalised block basis. Then for any n ∈ N

and ǫ > 0, there exist a finite interval F and a multiple x of∑

i∈F xi such thatTx is an ℓn1+-average with constant 1 + ǫ, and a normalised functional x∗ such thatx∗(x) > 1/2 and supp x∗ ⊆ ⋃

i∈F range xi.

Proof. The proof from [15] that the block basis (Txi) contains an ℓn1+-average withconstant 1 + ǫ is the same as for GM , and gives that such a vector exists of theform Tx = λ

∑

i∈F Txi with |F | 6 nk, where k = min{i∣

∣ f(ni) < (1 + ǫ)i}. Wemay therefore assume that 2|F | − 1 < supp x. Let y∗ be a unit functional whichnorms x and such that range y∗ ⊆ range x. Let x∗ = Ey∗ where E is the unionof the |F | intervals range xi, i ∈ F . Then x∗(x) = y∗(x) = 1 and by unconditionalasymptoticity of G∗, ‖x∗‖ 6 3

2‖y∗‖ < 2. �


The proof that G is HI requires defining “extra-special sequences” after havingdefined special sequences in the usual GM way. However, to prove that G is tightby range, we shall not need to enter that level of complexity and shall just usespecial sequences.

Proposition 8.8. The space G is of type (1).

Proof. Assume some normalised block-sequence (xn) is such that [xn] embeds intoY = [ei, i /∈

⋃

n range xn] and look for a contradiction. Passing to a subsequenceand by reflexivity we may assume that there is some isomorphism T : [xn] → Ysatisfying the hypothesis of Lemma 8.7, that is, (Txn) is a normalised block basisin Y . Fixing ǫ = 1/10 we may construct by Lemma 8.7 some block-sequence ofvectors in [xn] which are 1/2-normed by functionals in Q of support included in⋃

n range xn, and whose images by T form a sequence of increasing length ℓn1+-averages with constant 1 + ǫ. If Tz1, . . . , T zm is a R.I.S. of these ℓn1+-averageswith constant 1 + ǫ, with m ∈ [logN, expN ], N ∈ L, and z∗1 , . . . , z

∗m are the

functionals associated to z1, . . . , zm, then by [15] Lemma 7, the (m, f)-form z∗ =f(m)−1(z∗1 + . . .+ z∗m) satisfies

z∗(z1 + . . .+ zm) >m

2f(m)>

1

4‖Tz1 + . . .+ Tzm‖ > (4‖T−1‖)−1‖z1 + · · · + zm‖.

Therefore we may build R.I.S. vectors Tz with constant 1 + ǫ of arbitrary lengthm in [logN, expN ], N ∈ L, so that z is (4‖T−1‖)−1-normed by an (m, f)-form z∗

of support included in⋃

n range xn. Therefore (Tz, z∗) is a generalised R.I.S. pair.We then consider a sequence Tz1, . . . , T zk of length k ∈ K of such R.I.S. vectors,some special sequence of corresponding functionals z∗1 , . . . , z

∗k, and finally the pair

(z, z∗) where z = z1 + · · · + zk and z∗ = f(k)−1/2(z∗1 + . . .+ z∗k): observe that thesupport of z∗ is still included in

⋃

n range xn. Since (Tz, z∗) is a generalised specialpair, it follows from Lemma 8.6 that

‖Tz‖ 6 5kf(k)−1.

On the other hand,

‖z‖ > z∗(z) > (4‖T−1‖)−1kf(k)−1/2.

Since k was arbitrary in K this implies that T−1 is unbounded and provides thedesired contradiction. �

As we shall now prove, the dual G∗ of G is of type (1) as well, but also locallyminimal.

Lemma 8.9. Let (x∗i ) be a normalised block basis in G∗. Then for any n ∈ N andǫ > 0, there exists N(n, ǫ), a finite interval F ⊆ [1, N(n, ǫ)] and a multiple x∗ of∑

i∈F x∗i which is an ℓn∞+-average with constant 1 + ǫ and an ℓn1+-average x with

constant 2 such that x∗(x) > 1/2 and supp x ⊆ ⋃

i∈F range x∗i .

Proof. We may assume that ǫ < 1/6. By Lemma 8.1 we may find for each i 6 nan interval Fi, with |Fi| 6 2 minFi, and a vector y∗i of the form λ

∑

k∈Fix∗k, such

that y∗ =∑n

i=1 y∗i is an ℓn∞+-average with constant 1 + ǫ. Let, for each i, xi be

normalised such that y∗i (xi) = ‖y∗i ‖ and range xi ⊆ range y∗i . Let yi = Eixi, whereEi denotes the canonical projection on [en, n ∈ ⋃

k∈Firange x∗k]. By the asymptotic


unconditionality of (en), we have that ‖yi‖ 6 3/2. Let y′i = ‖yi‖−1yi, then

y∗i (y′i) = ‖yi‖−1y∗i (yi) = ‖yi‖−1y∗i (xi) >2

3‖y∗i ‖.

By Lemma 8.1, the vector x =∑

i y′i is an ℓn1+-average with constant less than 2,

such that x∗(x) > ‖x‖/2, and clearly supp x ⊆ ⋃

i∈F range x∗i . �

Proposition 8.10. The space G∗ is locally minimal and tight by range.

Proof. By Lemma 8.9 we may find in any finite block subspace of length N(n, ǫ)and supported after N(n, ǫ) an ℓn∞+-average with constant 1 + ǫ. By asymptoticunconditionality we deduce that uniformly, any block-subspace of G∗ contains ℓn∞’s,and therefore G∗ is locally minimal.

We prove that G∗ is tight by range. Assume towards a contradiction thatsome normalised block-sequence (x∗n) is such that [x∗n] embeds into Y = [e∗i , i /∈⋃

n range x∗n] and look for a contradiction. If T is the associated isomorphism, wemay by passing to a subsequence and perturbating T assume that Tx∗n is successive.

Let ǫ = 1/10. By Lemma 8.9, we find in [x∗k] and for each n, an ℓn∞+-average y∗nwith constant 1 + ǫ and an ℓn1+-average yn with constant 2, such that y∗n(yn) > 1/2and supp yn ⊆ ⋃

k range x∗k. By construction, for each n, Ty∗n is disjointly supportedfrom [x∗k], and up to modifying T , we may assume that Ty∗n is in Q and of normat most 1 for each n.

If z1, . . . , zm is a R.I.S. of these ℓn1+-averages yn with constant 2, with m ∈[logN, expN ], N ∈ L, and z∗1 , . . . , z

∗m are the ℓn∞+-averages associated to z1, . . . , zm,

then by [14] Lemma 7, the (m, f)-form z∗ = f(m)−1(z∗1 + . . .+ z∗m) satisfies

z∗(z1 + . . .+ zm) >m

2f(m)>

1

6‖z1 + . . .+ zm‖,

and furthermore Tz∗ is also an (m, f)-form. Therefore we may build R.I.S. vectorsz with constant 2 of arbitrary length m in [logN, expN ], N ∈ L, so that z is 6−1-normed by an (m, f)-form z∗ such that Tz∗ is also an (m, f)-form. We may thenconsider a sequence z1, . . . , zk of length k ∈ K of such R.I.S. vectors of length mi,and some corresponding functionals z∗1 , . . . , z

∗k (i.e., z∗i 6−1-norms zi and Tz∗i is also

an (mi, f)-form for all i), such that Tz∗1 , . . . , T z∗k is a special sequence. Then we let

z = z1 + · · ·+ zk and z∗ = f(k)−1/2(z∗1 + . . .+ z∗k), so that (z, T z∗) is a generalised

special pair. Since Tz∗ = f(k)−1/2(Tz∗1 + . . .+ Tz∗k) is a special function it followsthat

‖Tz∗‖ 6 1.

But it follows from Lemma 8.6 that ‖z‖ 6 5kf(k)−1. Therefore

‖z∗‖ > z∗(z)/‖z‖ > f(k)1/2/30.

Since k was arbitrary in K this implies that T−1 is unbounded and provides thedesired contradiction. �

It remains to check that G∗ is HI. The proof is very similar to the one in [15]that G is HI, and we shall therefore not give all details. There are two maindifferences between the two proofs. In [15] some special vectors and functionals areconstructed, and the vectors are taken alternatively in arbitrary block subspacesY and Z of G. In our case we need to take the functionals in arbitrary subspacesY∗ and Z∗ of G∗ instead. This is possible because of Lemma 8.1. We also needto correct what seems to be a slight imprecision in the proof of [15] about the


value of some normalising factors, and therefore we also get worst constants for ourestimates.

Let ǫ = 1/10. Following Gowers we define an R.I.S. pair of size N to be ageneralised R.I.S. pair (x, x∗) with constant 1+ǫ of the form (‖x1+. . .+xN‖−1(x1+. . . , xN ), f(N)−1(x∗1 + · · · + x∗N )), where x∗n(xn) > (1/3) and range x∗n ⊂ range xn

for each n. A special pair is a normalised generalised special pair with constant 1+ǫof the form (x, x∗) where x = ‖x1+. . .+xk‖−1(x1+. . .+xk) and x∗ = f(k)−1/2(x∗1+· · · + x∗k) with range x∗n ⊆ range xn and for each n, x∗n ∈ Q, |x∗n(xn) − (1/2)| <10−min supp xn . By [15] Lemma 8, z is a R.I.S. vector with constant 2 whenever(z, z∗) is a special pair. We shall also require that k 6 min supp x1, which willimply by [15] Lemma 9 that for m < k1/10, z is a ℓm1+-average with constant 8 (seethe beginning of the proof of Proposition 8.11).

Going up a level of “specialness”, a special R.I.S.-pair is a generalised R.I.S.-pairwith constant 8 of the form (‖x1+. . .+xN‖−1(x1+. . . , xN ), f(N)−1(x∗1+· · ·+x∗N )),where range x∗n ⊂ range xn for each n, and with the additional condition that(xn, x

∗n) is a special pair of length at least min supp xn. Finally, an extra-special

pair of size k is a normalised generalised special pair (x, x∗) with constant 8 of theform x = ‖x1 + . . .+ xk‖−1(x1 + . . .+ xk) and x∗ = f(k)−1/2(x∗1 + · · · + x∗k) withrange x∗n ⊆ range xn, such that, for each n, (xn, x

∗n) is a special R.I.S.-pair of length

σ(x∗1, . . . , x∗n−1).

Given Y∗, Z∗ block subspaces of G∗ we shall show how to find an extra-specialpair (x, x∗) of size k, with x∗ built out of vectors in Y∗ or Z∗, such that the signs ofthese constituent parts of x∗ can be changed according to belonging to Y∗ or Z∗ toproduce a vector x′∗ with ‖x′∗‖ 6 12f(k)−1/2‖x∗‖. This will then prove the result.

Consider then an extra-special pair (x, x∗). Then x splits up as

ν−1k

∑

i=1

ν−1i

Ni∑

j=1

ν−1ij

kij∑

r=1

xijr

and x∗ as

f(k)−1/2k

∑

i=1

f(Ni)−1

Ni∑

j=1

f(kij)−1

kij∑

r=1

x∗ijr

where the numbers ν, νi and νij are the norms of what appears to the right. Thesespecial sequences are chosen far enough “to the right” so that kij 6 min supp xij1,

and also so that (max supp xi j−1)2k−1ij 6 4−(i+j). We shall also write xi for

ν−1i

∑Ni

j=1 ν−1ij

∑kij

r=1 xijr and xij for ν−1ij

∑kij

r=1 xijr .

We define a vector x′ by

k∑

i=1

ν′−1i

Ni∑

j=1

ν′−1ij

kij∑

r=1

(−1)rxijr ,

where the numbers ν′i and ν′ij are the norms of what appears to the right. We shall

write x′i for ν′−1i

∑Ni

j=1 ν′−1ij

∑kij

r=1(−1)rxijr and x′ij for ν′−1ij

∑kij

r=1(−1)rxijr .

Finally we define a functional x′∗ as

f(k)−1/2k

∑

i=1

f(Ni)−1

Ni∑

j=1

f(kij)−1

kij∑

r=1

(−1)kx∗ijr .


Proposition 8.11. The space G∗ is HI.

Proof. Fix Y∗ and Z∗ block subspaces of G∗. By Lemma 8.9 we may construct anextra-special pair (x, x∗) so that x∗ijr belongs to Y∗ when r is odd and to Z∗ whenr is even.

We first discuss the normalisation of the vectors involved in the definition of x′.By the increasing condition on kij and xijr and by asymptotic unconditionality, wehave that

‖kij∑

r=1

(−1)rxijr‖ 6 2‖kij∑

r=1

xijr‖,

which means that ν′ij 6 2νij . Furthermore it also follows that the functional

(1/2)f(kij)−1/2∑kij

r=1(−1)rx∗ijr is of norm at most 1, and therefore we have that

‖∑kij

r=1(−1)rxijr‖ > (1/2)kijf(kij)−1/2. Lemma 9 from [15] therefore tells us that,

for every i, j, x′ij is an ℓmij

1+ -average with constant 8, if mij < k1/10ij . But the kij

increase so fast that, for any i, this implies that the sequence x′i1, . . . , x′i Ni

is arapidly increasing sequence with constant 8. By [15] Lemma 7, it follows that

‖Ni∑

j=1

x′ij‖ 6 9Ni/f(Ni).

Therefore by the f -lower estimate in G we have that ν′i 6 9νi.We shall now prove that ‖x′‖ 6 12kf(k)−1. This will imply that

‖x′∗‖ >x′∗(x′)

‖x′‖ >f(k)

12k[f(k)−1/2

k∑

i=1

f(Ni)−1ν′−1

i

Ni∑

j=1

f(kij)−1ν′−1ij

kij∑

r=1

x∗ijr(xijr)]

> f(k)1/2(12k)−1.18−1[

k∑

i=1

f(Ni)−1ν−1

i

Ni∑

j=1

f(kij)−1ν−1ij

kij∑

r=1

x∗ijr(xijr)]

= f(k)1/2(216k)−1k

∑

i=1

x∗i (xi) > 648−1f(k)1/2.

By construction of x∗ and x′∗ this will imply that

‖y∗ − z∗‖ > 648−1f(k)1/2‖y∗ + z∗‖for some non zero y∗ ∈ Y∗ and z∗ ∈ Z∗, and since k ∈ K was arbitrary, as well asY∗ and Z∗, this will prove that G∗ is HI.

The proof that ‖x′‖ 6 12kf(k)−1 is given in three steps:

Step 1. The vector x′ is a R.I.S. vector with constant 11.

Proof. We already know the sequence x′i1, . . . , x′i Ni

is a rapidly increasing sequence

with constant 8. Then by [15] Lemma 8 we get that x′i is also an ℓMi1+ -average with

constant 11, if Mi < N1/10i . Finally, this implies that x′ is an R.I.S.-vector with

constant 11, as claimed. �


Step 2. Let K0 = K \ {k}, let g ∈ F be the corresponding function given by [15]Lemma 5. For every interval E such that ‖Ex′‖ > 1/3, Ex′ is normed by an(M, g)-form.

Proof. The proof is exactly the same as the one of Step 2 in the proof of Gowersconcerning G, apart from some constants which are modified due to the change ofconstant in Step 1 and to the normalising constants relating νi and νij respectivelyto ν′i and ν′ij . The reader is therefore referred to [15]. �

Step 3. The norm of x′ is at most 12kg(k)−1 = 12kf(k)−1

Proof. This is an immediate consequence of Step 1, Step 2 and of Lemma 8.2. �

We conclude that the space G∗ is HI, and thus locally minimal of type (1). �

Let us observe that the examples of locally minimal, non-minimal, spaces we haveproduced so far could be said to be so for trivial reasons: since they hereditarilycontain ℓn∞’s uniformly, any Banach space is crudely finitely representable in any oftheir subspaces. It remains open whether there exists a tight and locally minimalBanach space which does not contain ℓn∞’s uniformly, i.e., which has finite cotype.

9. Tight spaces of the type of Argyros and Deliyanni

By Proposition 8.4, spaces built on the model of Gowers-Maurey’s spaces areuniformly inhomogeneous. We shall now consider a space of Argyros-Deliyannitype, more specifically of the type of a space constructed by Argyros, Deliyanni,Kutzarova and Manoussakis [2], with the opposite property, i.e., with a basis whichis strongly asymptotically ℓ1. This space will also be tight by support. By Proposi-tion 4.2 this basis will therefore be tight with constants as well, making this examplethe “worst” known so far in terms of minimality.

9.1. A strongly asymptotically ℓ1 space tight by support. In [2] an exampleof HI space Xhi is constructed, based on a “boundedly modified” mixed Tsirelsonspace XM(1),u. We shall construct an unconditional version Xu of Xhi in a similarway as Gu is an unconditional version of GM . The proof that Xu is of type (3)will be based on the proof that Xhi is HI, conditional estimates in the proof of [2]becoming essentially trivial in our case due to disjointness of supports.

Fix a basis (en) and M a family of finite subsets of N. Recall that a familyx1, . . . , xn is M-admissible if x1 < · · · < xn and {min supp x1, . . . ,min supp xn} ∈M, and M-allowable if x1, . . . , xn are vectors with disjoint supports such that{min supp x1, . . . ,min supp xn} ∈ M. Let S denote the family of Schreier sets, i.e.,of sets F such that |F | 6 minF , Mj be the subsequence of the sequence (Fk) ofSchreier families associated to sequences of integers tj and kj defined in [2] p 70.

We need to define a new notion. For W a set of functionals which is stable underprojections onto subsets of N, we let convQW denote the set of rational convexcombinations of elements of W . By the stability property of W we may writeany c∗ ∈ convQW as a rational convex combination of the form

∑

i λix∗i where

x∗i ∈ W and supp x∗i ⊆ supp c∗ for each i. In this case the set {x∗i }i will be calleda W -compatible decomposition of c∗, and we let W (c∗) ⊆ W be the union of allW -compatible decompositions of c∗. Note that if M is a family of finite subsets ofN, (c∗1, . . . , c

∗d) is M-admissible, and x∗i ∈ Wi(c

∗i ) for all i, then (x∗1, . . . , x

∗d) is also

M-admissible.


Let B = {∑n λnen : (λn)n ∈ c00, λn ∈ Q ∩ [−1, 1]} and let Φ be a 1-1 functionfrom B<N into 2N such that if (c∗1, . . . , c

∗k) ∈ B<N, j1 is minimal such that c∗1 ∈

convQAj1 , and jl = Φ(c∗1, . . . , c∗l−1) for each l = 2, 3, . . ., then Φ(c∗1, . . . , c

∗k) >

max{j1, . . . , jk} (the set Aj is defined in [2] p 71 by Aj = ∪n(Knj \K0) where the

Knj ’s are the sets corresponding to the inductive definition of XM(1),u).

For j = 1, 2, . . ., we set L0j = {±en : n ∈ N}. Suppose that {Ln

j }∞j=1 have beendefined. We set Ln = ∪∞

j=1Lnj and

Ln+11 = ±Ln

1 ∪ {1

2(x∗1 + . . .+ x∗d) : d ∈ N, x∗i ∈ Ln,

(x∗1, . . . , x∗d) is S − allowable},

and for j > 1,

Ln+12j = ±Ln

2j ∪ { 1

m2j(x∗1 + . . .+ x∗d) : d ∈ N, x∗i ∈ Ln,

(x∗1, . . . , x∗d) is M2j − admissible},

L′ n+12j+1 = ±Ln

2j+1 ∪ { 1

m2j+1(x∗1 + . . .+ x∗d) : d ∈ N such that

∃(c,1 . . . , c∗d) M2j+1 − admissible and k > 2j + 1 with c∗1 ∈ convQL

n2k, x

∗1 ∈ Ln

2k(c∗1),

c∗i ∈ convQLnΦ(c∗1,...,c∗i−1)

, x∗i ∈ LnΦ(c∗1 ,...,c∗i−1)

(c∗i ) for 1 < i 6 d},

Ln+12j+1 = {Ex∗ : x∗ ∈ L′ n+1

2j+1 , s ∈ N, E subset of N}.We set Bj = ∪∞

n=1(Lnj \ L0) and we consider the norm on c00 defined by the set

L = L0 ∪ (∪∞j=1Bj). The space Xu is the completion of c00 under this norm.

In [2] the space Xhi is defined in the same way except that E is an interval ofintegers in the definition of Ln+1

2j+1, and the definition of L′ n+12j+1 is simpler, i.e., the

coding Φ is defined directly on M2j+1-admissible families x∗1, . . . , x∗d in L<N and in

the definition each x∗i belongs to LnΦ(x∗

1,...,x∗

i−1). To prove the desired properties for

Xu one could use the simpler definition of L′ n+12j+1 ; however this definition doesn’t

seem to provide enough special functionals to obtain interesting properties for thedual as well.

The ground space forXhi and forXu is the spaceXM(1),u associated to a normingset K defined by the same procedure as L, except that Kn

2j+1 is defined in the sameway as Kn

2j, i.e.

Kn+12j = ±Kn

2j ∪ { 1

m2j(x∗1 + . . .+ x∗d) : d ∈ N, x∗i ∈ Kn,

(x∗1, . . . , x∗d) is M2j+1 − admissible}.

For n = 0, 1, 2, . . . , we see that Lnj is a subset of Kn

j , and therefore L ⊆ K. Thenorming set L is closed under projections onto subsets of N, from which it followsthat its canonical basis is unconditional, and has the property that for every j andevery M2j–admissible family f1, f2, . . . fd contained in L, f = 1

m2j(f1 + · · · + fd)


belongs to L. The weight of such an f is defined by w(f) = 1/m2j. It follows thatfor every j = 1, 2, . . . and every M2j–admissible family x1 < x2 < . . . < xn in Xu,

‖n

∑

k=1

xk‖ >1

m2j

n∑

k=1

‖xk‖.

Likewise, for S–allowable families f1, . . . , fn in L, we have f = 12 (f1 + · · ·+fd) ∈ L,

and we define w(f) = 1/2. The weight is defined similarly in the case 2j + 1.

Lemma 9.1. The canonical basis of Xu is strongly asymptotically ℓ1.

Proof. Fix n 6 x1, . . . , xn where x1, . . . , xn are normalised and disjointly supported.Fix ǫ > 0 and let for each i, fi ∈ L be such that fi(xi) > (1 + ǫ)−1 and supp fi ⊆supp xi. The condition on the supports may be imposed because L is stable underprojections onto subsets of N. Then 1

2

∑ni=1 ±fi ∈ L and therefore

‖n

∑

i=1

λixi‖ >1

2

n∑

i=1

|λi|fi(xi) >1

2(1 + ǫ)

n∑

i=1

|λi|,

for any λi’s. Therefore x1, . . . , xn is 2-equivalent to the canonical basis of ℓn1 . �

It remains to prove that Xu has type (3). Recall that an analysis (Ks(f))s off ∈ K is a decomposition of f corresponding to the inductive definition of K,see the precise definition in Definition 2.3 [2] . We shall combine three types ofarguments. First L was constructed so that L ≺ K, which means essentially thateach f ∈ L has an analysis (Ks(f))s whose elements actually belong to L (seethe definition on page 74 of [2]); so all the results obtained in Section 2 of [2] for

arbitrary K ≺ K (and in particular the crucial Proposition 2.9) are valid in ourcase. Then we shall produce estimates similar to those valid for Xhi and which areof two forms: unconditional estimates, in which case the proofs from [2] may beapplied directly up to minor changes of notation, and thus we shall refer to [2] fordetails of the proofs; and conditional estimates, which are different from those ofXhi, but easier due to hypotheses of disjointness of supports.

Recall that if F is a family of finite subsets of N, then

F ′ = {A ∪B : A,B ∈ F , A ∩B = ∅}.Given ε > 0 and j = 2, 3, . . ., an (ε, j)-basic special convex combination ((ε, j)-basic s.c.c.) (relative to XM(1),u) is a vector of the form

∑

k∈F akek such that: F ∈Mj, ak > 0,

∑

k∈F ak = 1, {ak}k∈F is decreasing, and, for every G ∈ F ′tj(kj−1+1),

∑

k∈G ak < ε.

Given a block sequence (xk)k∈N inXu and j > 2, a convex combination∑n

i=1 aixki

is said to be an (ε, j)-special convex combination of (xk)k∈N ((ε, j)-s.c.c), if thereexist l1 < l2 < . . . < ln such that 2 < supp xk1 6 l1 < supp xk2 6 l2 < . . . <supp xkn 6 ln, and

∑ni=1 aieli is an (ε, j)-basic s.c.c. An (ε, j)-s.c.c.

∑ni=1 aixki is

called seminormalised if ‖xki‖ = 1, i = 1, . . . , n and

‖n

∑

i=1

aixki‖ >1

2.

Rapidly increasing sequences and (ε, j)–R.I. special convex combinations in Xu

are defined by [2] Definitions 2.8 and 2.16 respectively, with K = L.Using the lower estimate for M2j-admissible families in Xu we get as in [2]

Lemma 3.1.


Lemma 9.2. For ǫ > 0, j = 1, 2, . . . and every normalised block sequence {xk}∞k=1

in Xu, there exists a finite normalised block sequence (ys)ns=1 of (xk) and coefficients

(as)ns=1 such that

∑ns=1 asys is a seminormalised (ǫ, 2j)–s.c.c..

The following definition is inspired from some of the hypotheses of [2] Proposition3.3.

Definition 9.3. Let j > 100. Suppose that {jk}nk=1, {yk}n

k=1, {c∗k}nk=1 and {bk}n

k=1

are such that

(i) There exists a rapidly increasing sequence

{x(k,i) : k = 1, . . . , n, i = 1, . . . , nk}with x(k,i) < x(k,i+1) < x(k+1,l) for all k < n, i < nk, l 6 nk+1, such that:

(a) Each x(k,i) is a seminormalised ( 1m4

j(k,i)

, j(k,i))–s.c.c. where, for each k, 2jk+2 <

j(k,i), i = 1, . . . nk.

(b) Each yk is a ( 1m4

2jk

, 2jk)– R.I.s.c.c. of {x(k,i)}nki=1 of the form yk =

∑nk

i=1 b(k,i)x(k,i).

(c) The sequence {bk}nk=1 is decreasing and

∑nk=1 bkyk is a ( 1

m42j+1

, 2j + 1)–s.c.c.

(ii) c∗k ∈ convQL2jk, and max(supp c∗k−1∪ supp yk−1) < min(supp c∗k ∪ supp yk),

∀k.(iii) j1 > 2j + 1 and 2jk = Φ(c∗1, . . . , c

∗k−1), k = 2, . . . , n.

Then (jk, yk, c∗k, bk)n

k=1 is said to be a j-quadruple.

The following proposition is essential. It is the counterpart of Lemma 8.6 for thespace G.

Proposition 9.4. Assume that (jk, yk, c∗k, bk)n

k=1 is a j-quadruple in Xu such thatsupp c∗k ∩ supp yk = ∅ for all k = 1, . . . , n. Then

‖n

∑

k=1

bkm2jkyk‖ 6

75

m22j+1

.

Proof. Our aim is to show that for every ϕ ∈ ∪∞i=1Bi,

ϕ(

n∑

k=1

bkm2jkyk) 6

75

m22j+1

.

The proof is given in several steps.1st Case: w(ϕ) = 1

m2j+1. Then ϕ has the form ϕ = 1

m2j+1(Ey∗1 + · · · + Ey∗k2

+

Ey∗k2+1 + · · ·Ey∗d) where E is a subset of N and where y∗k ∈ L2jk(c∗k) ∀k 6 k2 and

y∗k ∈ L2jk(d∗k) ∀k > k2 + 1, with d∗k2+1 6= c∗k2+1 (this is similar to the form of such

a functional in Xhi but with k1 = 1).If k 6 k2 then c∗s and therefore y∗s is disjointly supported from yk, so Ey∗s(yk) = 0

for all s, and therefore ϕ(yk) = 0. If k = k2 + 1 then we simply have |ϕ(yk)| 6

‖yk‖ 6 17m−12jk

, [2] Corollary 2.17. Finally if k > k2 + 1 then since Φ is 1-1, wehave that jk2+1 6= jk and for all s = k2 + 1, . . . , d, d∗s and therefore y∗s belong toB2ts with ts 6= jk. It is then easy to check that we may reproduce the proof of [2]Lemma 3.5, applied to Ey∗1 , . . . , Ey

∗d, to obtain the unconditional estimate

|ϕ(m2jkyk)| 6

1

m22j+1

.


In particular instead of [2] Proposition 3.2, which is a reformulation of [2] Corollary

2.17 for Xhi, we simply use [2] Corollary 2.17 with K = L.Summing up these estimates we obtain the desired result for the 1st Case.2nd Case: w(ϕ) 6 1

m2j+2. Then we get an unconditional estimate for the evalu-

ation of ϕ(∑n

k=1 bkm2jkyk) directly, reproducing the short proof of [2] Lemma 3.7,

using again [2] Corollary 2.17 instead of [2] Proposition 3.2. Therefore

|ϕ(

n∑

k=1

bkm2jkyk)| 6

35

m2j+26

35

m22j+1

.

3rd Case: w(ϕ) > 1m2j+1

. We have yk =∑nk

i=1 b(k,i)x(k,i) and the sequence

{x(k,i), k = 1, . . . n, i = 1, . . . nk} is a R.I.S. w.r.t. L. By [2] Proposition 2.9 thereexist a functional ψ ∈ K ′ (see the definition in [2] p 71) and blocks of the basisu(k,i), k = 1, . . . , n, i = 1, . . . , nk with supp u(k,i) ⊆ supp x(k,i), ‖uk‖ℓ1 6 16 andsuch that

|ϕ(

n∑

k=1

bkm2jk(

nk∑

i=1

b(k,i)x(k,i)))| 6 m2j1b1b(1,1)+ψ(

n∑

k=1

bkm2jk(

kn∑

i=1

b(k,i)u(k,i)))+1

m22j+2

6 ψ(

n∑

k=1

bkm2jk(

kn∑

i=1

b(k,i)u(k,i))) +1

m2j+2.

Therefore it suffices to estimate

ψ(

n∑

k=1

bkm2jk(

nk∑

i=1

b(k,i)u(k,i))).

In [2] ψ is decomposed as ψ1 + ψ2 and different estimates are applied to ψ1 andψ2. Our case is simpler as we may simply assume that ψ1 = 0 and ψ2 = ψ. Weshall therefore refer to some arguments of [2] concerning ψ2 keeping in mind thatψ2 = ψ.

Let Dk1 , . . . , D

k4 be defined as in [2] Lemma 3.11 (a). Then as in [2],

4⋃

p=1

Dkp =

nk⋃

i=1

supp u(k,i) ∩ supp ψ.

The proof that

(1) ψ|Sk Dk

2(∑

k

bkm2jk(∑

i

b(k,i)u(k,i))) 61

m2j+2,

(2) ψ|Sk Dk

3(∑

k

bkm2jk(∑

i

b(k,i)u(k,i))) 616

m2j+2,

and

(3) ψ|Sk Dk

1(∑

k

bkm2jk(∑

i

b(k,i)u(k,i))) 61

m2j+2.

may be easily reproduced from [2] Lemma 3.11. The case of Dk4 is slightly different

from [2] and therefore we give more details. We claim


Claim: Let D =⋃

k Dk4 . Then

(4) ψ|D(∑

k

bkm2jk(∑

i

b(k,i)u(k,i))) 664

m2j+2,

Once the claim is proved it follows by adding the estimates that the 3rd Case isproved, and this concludes the proof of the Proposition.

Proof of the claim: Recall that Dk4 is defined by

Dk4 = {m ∈

nk⋃

i=1

A(k,i) : for all f ∈⋃

s

Ks(ψ) with m ∈ suppf, w(f) >1

m2jk

and

there exists f ∈⋃

s

Ks(ψ) with m ∈ suppf, w(f) =1

m2jk

and

for every g ∈⋃

s

Ks(ψ) with supp f ⊂ supp g strictly, w(g) >1

m2j+1}.

For every k = 1, . . . , n, i = 1, . . . , nk and every m ∈ supp u(k,i) ∩ Dk4 , there

exists a unique functional f (k,i,m) ∈ ⋃

sKs(ψ) with m ∈ supp f , w(f) = 1

m2jkand

such that, for all g ∈ ⋃

sKs(ψ) with supp f ⊆ supp g strictly, w(g) > 1

m2j+1. By

definition, for k 6= p and i = 1, . . . , nk, m ∈ supp u(k,i), we have suppf (k,i,m)∩Dp4 =

∅. Also, if f (k,i,m) 6= f (k,r,n), then supp f (k,i,m) ∩ supp f (k,r,n) = ∅.For each k = 1, . . . , n, let {fk,t}rk

t=1 ⊆ ⋃

Ks(ϕ) be a selection of mutually disjointsuch functionals with Dk

4 =⋃rk

t=1 supp fk,t. For each such functional fk,t, we set

afk,t =

nk∑

i=1

b(k,i)

∑

m∈supp fk,t

am.

Then,

(5) fk,t(bkm2jk(∑

i

b(k,i)u(k,i))) 6 bkafk,t .

We define as in [2] a functional g ∈ K ′ with |g|∗2j 6 1 (see definition [2] p 71), andblocks uk of the basis so that ‖uk‖ℓ1 6 16, supp uk ⊆ ⋃

i supp u(k,i) and

ψ|D4(∑

k

bkm2jk(∑

i

b(k,i)u(k,i))) 6 g(2∑

k

bkuk),

hence by [2] Lemma 2.4(b) we shall have the result.

For f = 1mq

∑dp=1 fp ∈ ⋃

sKs(ψ|D4) we set

J = {1 6 p 6 d : fp = fk,t for some k = 1 . . . , n, t = 1, . . . , rk},T = {1 6 p 6 d : there exists fk,t with suppfk,t ⊆ suppfp strictly}.

For every f ∈ ⋃

sKs(ψ|D4) we shall define by induction a functional gf , by gf = 0

when J∪T = ∅, while if J∪T 6= ∅ we shall construct gf with the following properties.Let Df =

⋃

p∈J∪T suppfp and uk =∑

afk,tefk,t , where efk,t = emin suppfk,t , then:

(a) supp gf ⊆ supp f .(b) gf ∈ K ′ and w(gf ) > w(f),(c) f |Df

(∑

k bkm2jk(∑

i b(k,i)u(k,i))) 6 gf(2∑

k bkuk).


Let s > 0 and suppose that gf have been defined for all f ∈ ⋃s−1t=0 K

t(ψ|D4) and let

f = 1mq

(f1 + . . . + fd) ∈ Ks(ψ|D4)\Ks−1(ψ|D4) where the family (fp)dp=1 is Mq-

admissible if q > 1, or S-allowable if q = 1. The proofs of case (i) (1/mq = 1/m2jk

for some k 6 n) and case (ii) (1/mq > 1/m2j+1) are identical with [2] p 106.Assume therefore that case (iii) holds, i.e., 1/mq = 1/m2j+1. For the same reasonsas in [2] we have that T = ∅.

Summing up we assume that f ∈ Ks(ψ|D4)\Ks−1(ψ|D4) is of the form

f =1

m2j+1

d∑

p=1

fp =1

m2j+1(Ey∗1 + . . .+ Ey∗k2

+ Ey∗k2+1 + . . .+ Ey∗d),

where (y∗i )i is associated to (c∗1, . . . , c∗k2, d∗k2+1, . . .) with d∗k2+1 6= c∗k2+1, that T = ∅

and J 6= ∅, and it only remains to define gf satisfying (a)(b)(c).Now by the proof of [2] Proposition 2.9, ψ = ψϕ was defined through the analysis

of ϕ, in particular by [2] Remark 2.19 (a),

ψ =1

m2j+1

∑

k∈I

ψEy∗

k

for some subset I of {1, . . . , d}. Furthermore, for l ∈ I, l 6 k2 and 1 6 k 6 d,supp Ey∗l ∩ supp xk = ∅, therefore there is no functional in a family of type I andII w.r.t. xk of support included in supp Ey∗l (see [2] Definition 2.11 p 77). Thisimplies that DEy∗

l= ∅ ([2] Definition p. 85), and therefore that ψEy∗

l= 0 ([2]

bottom of p. 85).For l ∈ I, l > k2 + 1, then since Φ is 1 − 1, w(Ey∗l ) = w(Ed∗l ) 6= 1/m2jk

∀k.Therefore w(ψEy∗

l) 6= 1/m2jk

∀k, [2] Remark 2.19 (a). Then by the definition of Dk4 ,

supp ψEy∗

l∩Dk

4 = ∅ for all k.

Finally this means that ψ|D4= 1

m2j+1ψEy∗

k2+1|D4and J = {k2 + 1}, Df =

supp fk2+1. Write then fk2+1 = fk0,t and set gf = 12e

∗fk2+1

, therefore (a)(b) are

trivially verified. It only remains to check (c). But by (5),

f |Df(∑

k

bkm2jk(∑

i

b(k,i)u(k,i))) 6 bk0afk2+1

= bk0afk2+1e∗fk2+1

(efk2+1) = gf (2bk0afk2+1

efk2+1)

= gf(2∑

t

bk0afk,tefk,t) = gf(2

∑

k

bkuk).

So (c) is proved. Therefore gf is defined for each f by induction, and the Claim isverified. This concludes the proof of the Proposition. �

Proposition 9.5. The space Xu is of type (3).

Proof. Assume towards a contradiction that T is an isomorphism from some block-subspace [xn] of Xu into the subspace [ei, i /∈ ⋃

n supp xn]. We may assumethat max(supp xn, supp Txn) < min(supp xn+1, supp Txn+1) and min supp xn <min supp Txn for each n, and by Lemma 9.2, that each xn is a ( 1

m42n, 2n) R.I.s.c.c.

([2] Definition 2.16). We may write

xn =

pn∑

t=1

an,txn,t


where (xn,1, . . . , xn,pn) is M2n-admissible. Let for each n, t, x∗n,t ∈ L be such thatsupp x∗n,t ⊆ supp Txn,t and such that

x∗n,t(Txn,t) >1

2‖Txn,t‖ >

1

4‖T−1‖ ,

and let x∗n = 1m2n

(x∗n,1 + . . .+ x∗n,pn) ∈ L2n. Note that supp x∗n ∩ supp xn = ∅ and

that

x∗n(Txn) >1

m2n

pn∑

t=1

an,t

4‖T−1‖ = (4‖T−1‖m2n)−1.

We may therefore for any j > 100 construct a j-quadruple (jk, yk, c∗k, bk)n

k=1 sat-isfying the hypotheses of Proposition 9.4 and such that yk ∈ [xi]i and c∗k(Tyk) >

(4‖T−1‖m2jk)−1 for each k (note that we may assume that c∗k ∈ Lj2k

for each k).From Proposition 9.4 we deduce

‖n

∑

k=1

bkm2jkyk‖ 6

75

m22j+1

.

On the other hand ψ = 1m2j+1

∑nk=1 c

∗k belongs to L therefore

‖T (

n∑

k=1

bkm2jkyk)‖ > ψ(

n∑

k=1

bkm2jkTyk) >

1

4‖T−1‖m2j+1.

We deduce finally that

m2j+1 6 300‖T ‖‖T−1‖,which contradicts the boundedness of T . �

9.2. A strongly asymptotically ℓ∞ space tight by support. Since the canon-ical basis of Xu is tight and unconditional, it follows that Xu is reflexive. Inparticular this implies that the dual basis of the canonical basis of Xu is a stronglyasymptotically ℓ∞ basis of X∗

u. It remains to prove that this basis is tight withsupport.

It is easy to prove by duality that for any M2j-admissible sequence of functionalsf1, . . . , fn in X∗

u, we have the upper estimate

‖∑

i

fi‖ 6 m2j supi‖fi‖.

We use this observation to prove a lemma about the existence of s.c.c. normed byfunctionals belonging to an arbitrary subspace of X∗

u. The proof is standard exceptthat estimates have to be taken in X∗

u instead of Xu.

Lemma 9.6. For ǫ > 0, j = 1, 2, . . . and every normalised block sequence {fk}∞k=1

in X∗u, there exists a normalised functional f ∈ [fk] and a seminormalised (ǫ, 2j)–

s.c.c. x in Xu such that supp f ⊆ supp x and f(x) > 1/2.

Proof. For each k let yk be normalised such that supp yk = supp fk and fk(yk) =1. Recall that the integers kn and tn are defined by k1 = 1, 2tn > m2

n andkn = tn(kn−1 + 1) + 1, and that Mj = Fkj for all j.

Applying Lemma 9.2 we find a successive sequence of (ǫ, 2j)–s.c.c. of (yk) ofthe form (

∑

i∈Ikaiyi)k with {fi, i ∈ Ik} Fk2j−1+1-admissible. If ‖∑i∈Ik

fi‖ 6 2 for


some k, we are done, for then

(∑

i∈Ik

fi)(∑

i∈Ik

aiyi) >1

2‖∑

i∈Ik

fi‖.

So assume ‖∑i∈Ikfi‖ > 2 for all k, apply the same procedure to the sequence

f1k = ‖∑i∈Ik

fi‖−1∑

i∈Ikfi, and obtain a successive sequence of (ǫ, 2j)–s.c.c. of the

sequence (y1k)k associated to (f1

k )k, of the form (∑

i∈I1ka1

i y1i )k, with {fl : supp fl ⊆

∑

i∈I1kf1

i } a Fk2j−1+1[Fk2j−1+1]-admissible, and therefore M2j-admissible set. Then

we are done unless ‖∑i∈I1kf1

i ‖ > 2 for all k, in which case we set

f2k = ‖

∑

j∈I1k

f1j ‖−1

∑

j∈I1k

f1j

and observe by the upper estimate in X∗u that

1 = ‖f2k‖ = ‖

∑

j∈I1k

∑

i∈Ij

‖∑

j∈I1k

f1j ‖−1‖

∑

i∈Ij

fi‖−1fi‖ 6 m2j/4.

Repeating this procedure we claim that we are done in at most t2j steps. Otherwisewe obtain that the set

A = {fl : supp fl ⊆∑

i∈It2j−1k

ft2j−1

i }

is M2j-admissible. Since ft2j

k =∑

fl∈A αlfl, where the normalising factor αl is less

than (1/2)t2j for each l, we deduce from the upper estimate that

1 = ‖f t2j

k ‖ 6 2−t2jm2j ,

a contradiction by definition of the integers ti’s. �

To prove the last proposition of this section we need to make two observations.First if (f1, . . . , fn) ∈ convQL is M2j-admissible, then 1

m2j

∑nk=1 fk ∈ convQL2j.

Indeed we may easily find convex rational coefficients λi such that each fk is of theform

fk =∑

i

λifki , f

ki ∈ L, supp fk

i ⊆ supp fk ∀i.

Then 1m2j

∑nk=1 fk =

∑

i λi(1

m2j

∑nk=1 f

ki ) and each 1

m2j

∑nk=1 f

ki belongs to L2j .

Likewise if ψ = 1m2j+1

(c∗1 + . . . + c∗d), k > 2j + 1, c∗1 ∈ convQL2k and c∗l ∈convQLΦ(c∗1,...,c∗l−1)

∀l > 2, then ψ ∈ convQL. Indeed as above we may write

ψ =∑

i

λi(1

m2j+1

d∑

l=1

f li ), f1

i ∈ L2k, fli ∈ LΦ(c∗1,...,c∗i−1)

(c∗i ) ∀l > 2,

and each 1m2j+1

∑dl=1 f

li belongs to L′n+1

2j+1 ⊆ L.

Proposition 9.7. The space X∗u is of type (3).

Proof. Assume towards a contradiction that T is an isomorphism from some block-subspace [fn] of X∗

u into the subspace [e∗i , i /∈ ∪nsupp fn]. We may assume thatmax(supp fn, supp Tfn) < min(supp fn+1, supp Tfn+1) and min supp Tfn <min supp fn for each n. Since the closed unit ball of X∗

u is equal to convQL we


may also assume that fn ∈ convQL for each n. Applying Lemma 9.6, we may alsosuppose that each fn is associated to a ( 1

m42n, 2n) s.c.c. xn with Tfn(xn) > 1/3 and

supp xn ⊂ supp Tfn, and we shall also assume that ‖Tfn‖ = 1 for each n. Buildthen for each k a ( 1

m42k, 2k) R.I.s.c.c. yk =

∑

n∈Akanxn such that (Tfn)n∈Ak

and

therefore (fn)n∈Akis M2k-admissible. Then we note that by the first observation

before this proposition,

c∗k := m−12k

∑

n∈Ak

fn ∈ convQL2k,

and we observe that supp c∗k ∩ supp yk = ∅ and that Tc∗k(yk) > (3m2k)−1.We may therefore for any j > 100 construct a j-quadruple (jk, yk, c

∗k, bk)n

k=1

satisfying the hypotheses of Proposition 9.4 and such that c∗k ∈ [fi]i and Tc∗k(yk) >

(3m2jk)−1 for each k. From Proposition 9.4 we deduce

‖n

∑

k=1

bkm2jkyk‖ 6

75

m22j+1

.

Therefore

‖d

∑

k=1

Tc∗k‖ >

∑dk=1 bkm2jk

Tc∗k(yk)

‖∑nk=1 bkm2jk

yk‖>m2

2j+1

225,

but on the other hand

‖d

∑

k=1

c∗k‖ 6 m2j+1

since by the second observation the functional m−12j+1

∑dk=1 c

∗k belongs to convQL.

We deduce finally that

m2j+1 6 225‖T ‖,

which contradicts the boundedness of T . �

10. Open problems

In the following theorem, we present the final list of possible classes of spacescontained in a Banach space, obtained by combining the five dichotomies and thedichotomy of Tcaciuc. We obtain 19 inevitable classes of spaces, and examples for8 of them. The class (2) is divided into two subclasses and the class (4) into foursubclasses, which are not made explicit here for lack of an example of space of type(2) or (4) to begin with. Recall that the spaces contained in any of the 12 subclassesof type (1)-(4) are never isomorphic to their proper subspaces, and in this sensethese subclasses may be labeled “exotic”. On the contrary “classical” pure spacesmust belong to one of the 7 subclasses of type (5)-(6).

Theorem 10.1. Any infinite dimensional Banach space contains a subspace of oneof the types listed in the following diagram:


Type Properties Examples(1a) HI, tight by range and with constants ?(1b) HI, tight by range, locally minimal G∗

(2) HI, tight, sequentially minimal ?(3a) tight by support and with constants, uniformly inhomogeneous ?(3b) tight by support, locally minimal, uniformly inhomogeneous G∗

u

(3c) tight by support, strongly asymptotically ℓp, 1 6 p <∞ Xu

(3d) tight by support, strongly asymptotically ℓ∞ X∗u

(4) unconditional basis, quasi minimal, tight by range ?(5a) unconditional basis, tight with constants, sequentially minimal, ?

uniformly inhomogeneous(5b) unconditional basis, tight, sequentially and locally minimal, ?

uniformly inhomogeneous(5c) tight with constants, sequentially minimal, T , Tp

strongly asymptotically ℓp, 1 6 p <∞(5d) tight, sequentially minimal, strongly asymptotically ℓ∞ ?(6a) unconditional basis, minimal, uniformly inhomogeneous S(6b) minimal, reflexive, strongly asymptotically ℓ∞ T ∗

(6c) isomorphic to c0 or lp, 1 6 p <∞ c0, ℓp

We finally conclude with a list of open problems.

Problem 10.2. (1) Does there exist a tight Banach space admitting a basiswhich is not tight?

(2) Does there exist a tight, locally block minimal and unconditional basis?(3) Find a locally minimal and tight Banach space with finite cotype.(4) Does there exist a tight Banach space which does not contain a basic se-

quence that is either tight by range or tight with constants? In other words,does there exist a locally and sequentially minimal space without a minimalsubspace?

(5) Suppose [en] is sequentially minimal. Does there exist a block basis all ofwhose subsequences are subsequentially minimal?

(6) Is every HI space tight?(7) Is every tight basis continuously tight?(8) Do there exist spaces of type (2), (4), (5a), (5b), (5d)?(9) Suppose (en) is tight with constants. Does (en) have a block sequence that

is (strongly) asymptotically ℓp for some 1 6 p <∞?(10) Does there exist a separable HI space X such that ⊆∗ Borel embeds into

SB∞(X)?(11) If X is a separable Banach space without a minimal subspace, does ⊆∗

Borel embed into SB∞(X)? What about more complicated quasi orders, inparticular, the complete analytic quasi order 6Σ

11?

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Address of V. Ferenczi:Institut de Mathematiques de Jussieu,Projet Analyse Fonctionnelle,Universite Pierre et Marie Curie - Paris 6,Boıte 186, 4, Place Jussieu,75252, Paris Cedex 05,[email protected]

Current address:Departamento de Matematica,Instituto de Matematica e Estatıstica,Universidade de Sao Paulo.05311-970 Sao Paulo, SP,[email protected]

Address of C. Rosendal:Department of Mathematics,University of Illinois at Urbana-Champaign,273 Altgeld Hall, MC 382,1409 W. Green Street,Urbana, IL 61801, [email protected]

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Banach spaces without minimal subspaces