Advanced Courses in Mathematics CRM Barcelona Centre de Recerca Matemàtica
Managing Editor: Manuel Castellet
Spiros A. Argyros Stevo Todorcevic
Ramsey Methods in Analysis
Birkhäuser Verlag Basel • Boston • Berlin
Authors: Spiros A. Argyros National Technical University Department of Mathematics Zografou Campus 157 80 Athens Greece
[email protected] Stevo Todorcevic Université Paris 7 – C.N.R.S. U.M.R. 7057 2, Place Jussieu – Case 7012 75251 Paris Cedex 05 France and Department of Mathematics University of Toronto Toronto, Ontario M5S 3G3, Canada
[email protected] 2000 Mathematical Subject Classification 46B20, 05D10, 03E75
A CIP catalogue record for this book is available from the Library of Congress, Washington D.C., USA
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Contents
Foreword
A
Saturated and Conditional Structures in Banach Spaces Spiros A. Argyros
vii
1
Introduction
3
I. Tsirelson and Mixed Tsirelson Spaces
7
II. Tree Complete Extensions of a Ground Norm II.1 Mixed Tsirelson Extension of a Ground Norm . . . . . . . . . . . . II.2 R.I.S. Sequences and the Basic Inequality . . . . . . . . . . . . . .
21 21 26
III. Hereditarily Indecomposable Extensions with a Schauder Basis III.1 The HI Property in X[G, σ] . . . . . . . . . . . . . . . . . . . . . . III.2 The HI Property in X[G, σ]∗ . . . . . . . . . . . . . . . . . . . . . .
39 39 43
IV. The Space of the Operators for HI Banach Spaces IV.1 Some General Properties of HI Spaces . . . . . . . . . . . . . . . . IV.2 The Space of Operators L(X[G, σ]), L(X[G, σ]∗ ) . . . . . . . . . . .
47 47 52
V. Examples of Hereditarily Indecomposable Extensions V.1 A Quasi-reflexive HI Space . . . . . . . . . . . . . . . . . . . . . . V.2 The Spaces p , 1 < p < ∞, are Quotients of HI Spaces . . . . . . . V.3 A Non Separable HI Space . . . . . . . . . . . . . . . . . . . . . . .
57 57 58 62
VI. The Space Xω1
71
VII. Finite Representability of JT0 and the Diagonal Space D(Xγ )
81
VIII. The Spaces of Operators L(Xγ ), L(X, Xω1 )
87
Appendix A. Transfinite Schauder Basic Sequences
99
Appendix B. The Proof of the Finite Representability of JT0
105
Bibliography
117
vi
B
Contents
High-Dimensional Ramsey Theory and Banach Space Geometry Stevo Todorcevic
121
Introduction
123
I. Finite-Dimensional Ramsey Theory 127 I.1 Finite-Dimensional Ramsey Theorem . . . . . . . . . . . . . . . . . 127 I.2 Spreading Models of Banach Spaces . . . . . . . . . . . . . . . . . 130 I.3 Finite Representability of Banach Spaces . . . . . . . . . . . . . . . 135 II. Ramsey Theory of Finite and Infinite Sequences II.1 The Theory of Well-Quasi-Ordered Sets . . . . . . . . . . . . . . II.2 Nash–Williams’ Theory of Fronts and Barriers . . . . . . . . . . II.3 Uniform Fronts and Barriers . . . . . . . . . . . . . . . . . . . . . II.4 Canonical Equivalence Relations on Uniform Fronts and Barriers II.5 Unconditional Subsequences of Weakly Null Sequences . . . . . . II.6 Topological Ramsey Theory . . . . . . . . . . . . . . . . . . . . . II.7 The Theory of Better-Quasi-Orderings . . . . . . . . . . . . . . . II.8 Ellentuck’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . II.9 Summability in Banach Spaces . . . . . . . . . . . . . . . . . . . II.10 Summability in Topological Abelian Groups . . . . . . . . . . . .
. . . . . . . . . .
143 143 147 153 165 169 177 180 185 188 192
III. Ramsey Theory of Finite and Infinite Block Sequences III.1 Hindman’s Theorem . . . . . . . . . . . . . . . . . III.2 Canonical Equivalence Relations on FIN . . . . . . III.3 Fronts and Barriers on FIN[ ω there exists some ξ < ω1 such that for the spaces T (M, θ) and T (Sξ , θ) the corresponding conclusion (iv) holds. For results related to this problem we refer the reader to [42]. A proof of the above theorem can be found in [20].
Chapter I. Tsirelson and Mixed Tsirelson Spaces
15
Mixed Tsirelson norms Schlumprecht space S is the completion of c00 endowed with the following norm. For x ∈ c00 , " xS = max x0 , sup sup n
# 1 Ei xS , log2 (n + 1) i=1 n
where the inside sup is taken over all choices E1 < E2 < · · · < En of subsets of N. The motivation for the construction of such a space was to provide an example of an arbitrarily distortable Banach space. The space S appeared as an ad hoc construction. However the next definition makes the relation of S with the Tsirelson type spaces more transparent. Let (Mn )n , (θn )n be two sequences with each Mn a compact family of finite subsets of N, 0 < θn < 1 and limn θn = 0. The mixed Tsirelson space T [(Mn , θn )n ] is the completion of c00 endowed with the norm k # " x∗ = max x0 , sup sup θn Ei x∗ , n
i=1
where the inside sup is taken over all choices E1 < E2 < · · · < Ek of Mn admissible families. Remark I.9. (a) In the above notation the Schlumprecht space S is the mixed 1 )n ], where An = {F ⊆ N : #F ≤ n}. Tsirelson space T [(An , log (n+1) 2
(b) The mixed Tsirelson space T [(Mk , θk )nk=1 ] defined by a finite family (Mk , θk )nk=1 is isomorphic to T (Mk0 , θk0 ) for some 1 ≤ k0 ≤ n. Thus in order to obtain really new spaces we need to use infinite sequences (Mn , θn )n . As in the case of Tsirelson spaces there is an alternative definition of the norm of mixed Tsirelson spaces resulting from a norming set W [(Mn , θn )n ]. This set is defined as the minimal subset of c00 satisfying the following properties. (i) W [(Mn , θn )n ] contains all ±e∗k , k ∈ N. (ii) It is closed under the operations (Mn , θn )n . It follows that for an f ∈ W [(Mn , θn )n ] the tree analysis (ft )t∈T is also defined, taking into account the necessary modifications. We shall discuss this topic more extensively in the next chapter. The main property of mixed Tsirelson spaces is that for appropriate choices (Mn , θn )n , they provide countable many equivalent norms ( · n )n such that for each (xk )k block sequence in T [(Mn , θn )n ] and every n ∈ N there exists a vector yn ∈ (xk )k such that yn = yn n and for m = n yn m ≤ εm where εm → 0. In other words, the sequence (·n )n has the property that in every block subspace none of them dominates the rest but also it is not dominated by the rest.
16
Chapter I. Tsirelson and Mixed Tsirelson Spaces
Theorem I.10. Let X = T [(Mn , θn )n ]. If for some n ∈ N it holds that i(Mn ) ≥ ω or i(Mn ) = r < ω and θn > 1r , then X is reflexive. Moreover if the first alternative holds, then X does not contain isomorphically any of the spaces p , 1 ≤ p < ∞, c0 . Proof. The proof of the reflexivity is similar to the original proof of Tsirelson [T]. According to a classical result of R.C. James [35], it is enough to show that the basis of X is boundedly complete and shrinking. (i) The basis (en )n of T [(Mn , θn )n ] is boundedly complete. On the contrary, assume n that there exist ε > 0 and a block sequence (xk )k of (en )n such that supn k=1 xk ≤ 1 and xk ≥ ε for all k = 1, 2, . . .. Since the basis is unconditional it is enough to find a finite subset A of N such that k∈A xk > 1. According to the assumption there exist n, r ∈ N such that i(Mn ) ≥ r (r−1) and θn > 1r . Therefore there exists n0 ∈ Mn . It follows that if (yk )k is a block sequence of (en )n there exist a sequence (nt )t∈N of positive integers and a subsequence (ykt )t∈N of (yk )k such that nt ≤ supp ykt+1 < nt+1 and for every l = 0, 1, . . . it holds {n0 , nlr+1 , . . . , nlr+(r−1) } ∈ Mn . It follows that the sequence (ylr+1 , . . . , y(l+1)r ) is Mn -admissible. Let s ∈ N be such that (θn r)s > 1ε . From the previous observation, we can choose a subsequence (xt )t of (xk )k such that for every l = 0, 1, . . . the rsequence (xlr+1 , . . . , x(l+1)r ) is Mn -admissible. For l = 0, 1, . . . , we set x(1,l) = i=1 xlr+i . It follows that r x(1,l) ≥ θn xlr+i ≥ θn rε . i=1
We repeat the same procedure for the sequence (x(1,l) )l∈N and we get a block sequence (x(2,l) )l∈N of (xn )n such that x(2,l) ≥ (θn r)2 ε. Repeating this procedure s times we get a block sequence (x(s,l) )l , where xs,l is sum of terms of the sequence (xn )n , such that x(s,l) ≥ (θn r)s ε > 1, a contradiction. (ii) The basis {en }∞ n=1 is a shrinking. Let θ = maxk θk < 1. For f ∈ X ∗ and m ∈ N, denote by Qm (f ) the restriction of f to the space generated by {ek }k≥m . It suffices to prove the following: For every f ∈ BX ∗ there is m ∈ N such that Qm (f ) ∈ θBX ∗ . Recall that BX ∗ = co(W ), where the closure is in the topology of pointwise convergence. We shall first prove the following: Claim. For every f ∈ W there is m such that Qm (f ) ∈ θ co(W ). To prove this, let f ∈ W and let {f n }∞ n=1 be a sequence in W converging pointwise to f . If f n = e∗kn for an infinite number of n, we have nothing to prove. So suppose that for every n there are kn ∈ N, a set {mn1 , . . . , mndn } ∈ Mkn and vectors fin ∈ W , i = 1, . . . , dn such that mn1 ≤ supp f1n < mn2 ≤ supp f2n < · · · < mndn ≤ supp fdnn and f n = θkn (f1n + · · · + fdnn ). If there is a subsequence of {θkn } converging to 0,
Chapter I. Tsirelson and Mixed Tsirelson Spaces
17
then f = 0. So we may suppose that there is a k such that kn = k for all n, i.e. θkn = θk and {mn1 , . . . , mndn } ∈ Mk . Since Mk is compact, if we substitute {f n } with a subsequence we get that there is a set {m1 , . . . , md } ∈ Mk such that the sequence of indicator functions of the sets {mn1 , . . . , mndn } converges to the indicator function of {m1 , . . . , md }. So, for large n, mni = mi , i = 1, . . . , d, and mnd+1 → ∞ as n → ∞. Passing to a further subsequence of (f n )∞ n=1 , we get that there exist fi ∈ W , i = 1, . . . , d, with supp fi ⊂ [mi , mi+1 ), i = 1, . . . , d − 1, and supp fd ⊂ [md , ∞) such that fjn → fj pointwise for j = 1, . . . , d. We conclude that f = θk (f1 + · · · + fd ), so Qmd (f ) = θk fd ∈ θ co(W ). The proof of the claim is complete. In particular we get that W is a weakly compact subset of c0 . Consider now an f ∈ BX ∗ = co(W ). In the set BX ∗ = co(W ) the pointwise topology coincides with the restriction of the w∗ topology of ∞ onto co(W ). According to Choquet’s theorem there exists a measure µ1 ∈ M1 (W ) such that Gdµ = G(f0 ) W ∗
for every linear w -continuous function G. Consider the sets Am = {f ∈ W , Qm (f ) ∈ θ 2 co(W )}, m ∈ N . The sequence (Am )m is an increasing sequence of closed sets converging to W , and hence there exists m0 such that |µ|(Am0 ) > 1 − θ(1 − θ). We claim that 1 θ Qm0 (f0 ) ∈ co(W ). Indeed if not, then by the Hahn–Banach theorem there exists a linear w∗ -continuous function G such that sup{G(f ) : f ∈ co(W )} = 1
0 such that n an xn ≤ 1 1 C( n |an |p ) p for all coefficients (an )n . Choose m ∈ N such that m1− p > θCn . Since i(Mn ) ≥ ω, there exist n1 , . . . , nm ∈ N such that the sequence (xn1 , . . . , xnm ) is Mn -admissible. It follows θn m ≤
m
1
xi ≤ Cm p ,
i=1
a contradiction.
Notes and Remarks. Tsirelson space is an important discovery in Banach space theory. It is the “first truly non-classical space” according to E. Odell and Th. Schlumprecht [53]. It is the first space with its norm inductively defined, and, more important, it introduces a fundamental method for saturating the structure of a Banach space with a property (P). To illustrate this we recall that J. Schreier [56] provided the first example of a weakly null sequence (xn )n with no norm Cesaro summable subsequence. Tsirelson space retains this property for all seminormalized weakly null sequences and it remains reflexive. Let us mention that Tsirelson’s initial construction used the forcing method and he actually defined the dual of what we call Tsirelson space. The implicit form is due to T. Figiel and W.B. Johnson [28]. We recommend the interested reader Tsirelson’s web page where it is explained how he discovered his space. A tentative study followed Tsirelson’s discovery. Thus T. Fiegel and W.B. Johnson [28], introduced the p-convexification of T and W.B. Johnson [39], defined the modified version of T spaces with remarkable properties. We refer the reader to [21] for a comprehensive presentation of the results concerning Tsirelson space. Tsirelson-type spaces of the form T (M, θ) were introduced in an unpublished paper [5], where also the proof of Theorem I.4 was presented. The later result was initially proved in [18] with a different proof. It is worth noticing that Lemmas I.6, I.7 are the simplest approach of the basic inequality, which will be discussed in the following chapters. The first mixed Tsirelson space is Schlumprecht’s space [55]. This space was the decisive ingredient for Gowers–Maurey example [33]. The concept of Mixed Tsirelson space was introduced in [6]. The modified versions of mixed Tsirelson spaces are discussed in [8] and [9]. It is interesting that while Tsirelson space is isomorphic to its modified version, genuine mixed Tsirelson spaces are totally incomparable to their modified version. A. Manoussakis in [43] introduced and studied a class of mixed Tsirelson defined as p-spaces. These are spaces of the form T [(Anj , θj )j ] satisfying the following property: For j ∈ N we denote by Tj = T [(Ani , θi )ji=1 ]. Then Tj is isomorphic to pj with 1 < pj ≤ ∞, [20]. A p-space is a space of the above such that the sequence (pj )j strictly decreases to p. Under this consideration Schlumprecht space S is an 1-space.
Chapter I. Tsirelson and Mixed Tsirelson Spaces
19
Theorem I.10 is proved in [6] and Theorem I.8 which is stated with no proof can be found in [20]. We refer the reader to survey papers [11] and [51] for related results to the content of the present section. The hierarchy of Schreier families (Sξ )ξ 2 . The Ramsey theorem yields that there exists L1 ∈ [N] such that either [L1 ]k ⊂ B1 or [L1 ]k ∩ B1 = ∅. From our assumption on the failure of the lemma the second alternative can not hold thus [L1 ]k ⊂ B1 . We may assume that L1 = N. We lk set f1,l = fl for l = 1, 2, . . .. As above we may assume, passing to a i=(l−1)k+1
subsequence, that f1,l1 + f1,l2 + · · · + f1,lk > 22 for every l1 < · · · < lk . After s steps, using the same argument, we arrive at a functional f ∈ Z which is of the ks fli with f > 2s . Since ks ≤ n2j we get that the functional m12j f is form f = i=1
the result of a (An2j , m12j ) on norm 1 functionals and hence f ≤ m2j . Therefore m2j < 2s , a contradiction which completes the proof of the lemma. Definition III.13. Let G be a ground set. The space YG has uniformly bounded averages if 1 → YG and for every ε > 0 and n0 ∈ N there exists k ∈ N such that for every weakly null block sequence (zn )n∈N in YG with zn G ≤ 1 there exist i1 < i2 < · · · < ik such that for every g ∈ G with min supp g < n0 we have that z +z +···+zik |g( i1 i2k )| < ε. Proposition III.14. Assume that YG has uniformly bounded averages and X[G, σ] is a strictly singular extension of YG . Let Z be an infinite dimensional subspace of X[G, σ]∗ and j ∈ N. Then there exists (1, 12, 2j) exact pair (y, φ) with dist(φ, Z) < 5m2j 5m2j n2j and yG < n2j . Proof. We may assume that Z is a block subspace of X[G, σ]∗ . We select 0 < ε < 1 n2j .
III.2. The HI Property in X[G, σ]∗
45
∗ ∗ Let k1 ∈ N. We choose a 2 − ck01 vector f1 = z1,1 + · · · + z1,k in Z and 1 k1 1 x1 = k1 (z1,1 + · · · + z1,k1 ) a 21 with ran f1 = ran x1 and f1 (x1 ) > 1. We set t1 = max supp f1 . Since YG has uniformly bounded averages there exists k2 = k( 4ε , t1 ) satisfying the property of Definition III.13; we may also as∗ ∗ )l in Z with min supp z2,1 > sume that k2 > k1 . We choose a block sequence (z2,l k2 ∗ ∗ max supp f1 such that z2,l > 12 and z2,i ≤ 2 for every i1 < · · · < ik2 in l l=1
∗ N. For each l we select z2,l ∈ X[G, σ] such that ran z2,l = ran z2,l , z2,l < 2 and ∗ z2,l (z2,l ) > 1. Since 1 → YG and (z2,l )l is a bounded sequence (in X[G, σ] and thus) in YG , we may assume that (z2,l )l is a weakly Cauchy sequence in YG . Thus the sequence (w2,l )l defined by w2,l = z2,2l−1 −z2,2l is a weakly null sequence in YG with w2,l ≤ 4. From our choice of k2 we may assume, passing to a subsequence k2 of (w2,l )l that |g( k12 w2,l )| < ε for every g ∈ G with min supp g ≤ t1 .
We set x2 =
1 k2
l=1 k2
w2,l and f2 =
k2
∗ z2l−1 . Observe that f2 ∈ Z with f2 ≤ 1
l=1
l=1
x2 ≥ f2 (x2 ) =
k2 1 ∗ z2l−1 (z2l−1 ) > 1 k2
while
l=1
which, in particular yields that x2 is a 4 − k12 average. We select k3 = k(ε, t2 ) with k3 > k2 satisfying the property of Definition III.13 where t2 = max supp f2 and we define x3 ∈ X[G, σ] and f3 ∈ Z similarly to the second step. Following this procedure we may construct a block sequence (xr )r∈N in X[G, σ] and a sequence (fr )r∈N in Z such that the following conditions are satisfied. (i) Each xr is a 4 − k1r average and each fr is a 4 − ck0r vector with fr (xr ) > 1 and ran ff = ran xr . (ii) For every g ∈ G with min supp g ≤ max supp xr−1 we have that |g(xr )| < ε. Thus for every g ∈ G we have that |g(xr )| > ε for at most one r. Passing to a subsequence we may additionally assume (Lemma II.24) that (xr )r∈N is a (6, ε) R.I.S. We set φ=
n2j n2j m2j 1 fr and x = xr . m2j r=1 n2j r=1
We have that f ∈ Z with w(f ) = m2j . Proposition II.19 yields that x ≤ 2·6 1 = 12. Thus 1 ≤ φ(x) ≤ x ≤ 12 hence we may select θ with 12 ≤θ≤1 m2j · m 2j such that φ(θx) = 1. We set y = θx. Using Proposition II.19 we easily get that (y, φ) is a (12, 2j) exact pair.
46 Chapter III. Hereditarily Indecomposable Extensions with a Schauder Basis It only remains to show that yG ≤ at most one r we get that |g(y)| ≤
5m2j n2j .
Let g ∈ G. Since |g(xr )| > ε for
m2j m2j (|g(x1 )| + · · · + |g(xn2j )|) ≤ (max xr + (n2j − 1)ε) n2j n2j r 5m2j 4m2j + 4ε < . < n2j n2j
Theorem III.15. If YG has uniformly bounded averages and X[G, σ] is a strictly singular extension of YG , then the space X[G, σ]∗ is HI. Proof. It follows from Proposition III.10 and III.14.
Definition III.16. Let G be a ground set. The space YG has the uniform weak Banach–Saks property if 1 → YG and for every ε > 0 there exists k ∈ N such that for every normalized weakly null block sequence (zn )n∈N in YG there exist z +z +···+zik i1 < i2 < · · · < ik such that i1 i2k G < ε. Corollary III.17. If YG has the uniform weak Banach–Saks property, then X[G, σ] is a strictly singular HI extension of YG and the space X[G, σ]∗ is HI. Notes and Remarks. There exists the unconditional counterpart of the reflexive HI space X[G0 , σ] (Theorem III.8). This space is a variation of the space presented by W.T. Gowers in [31]. To define this space we modify (v) of Definition III.2 by adding that for each spacial functional f and each E subset (not necessarily interval) of N, Ef ∈ DG . Then the resulting space denoted as Xu [G0 , σ] has an unconditional basis and it has a remarkable space of operators. Namely as it is shown in [34] every T ∈ L(Xu [G0 , σ]) is of the form D + S with D a diagonal operator and S strictly singular. This property yields that Xu [G0 , σ] is not isomorphic to any of its proper subspaces and every bounded linear projection is a strictly singular perturbation of a diagonal one. Other variants of the special sequences lead to indecomposable and unconditionally saturated Banach spaces (e.g. [14], [15]). The dual of a HI space X, even if X is reflexive, does not need to be HI. For example there exists a reflexive HI space X such that 2 is isomorphic to a subspace of X ∗ (e.g. [10]). More extreme is a recent result of a reflexive HI space X with X ∗ unconditionally saturated [17]. It is not known if such a divergent could be preserved in the subspaces of X. The following is open. Assume that X is a reflexive HI space. Does there exist a subspace Y of X such that Y ∗ is also HI? The HI constructions with the use of higher complexity saturation methods which appeared in [6] and [16] yield asymptotic 1 HI spaces. Variants of them could derive asymptotic p HI spaces for 1 < p < +∞ (e.g. [22]).
Chapter IV
The Space of the Operators for Hereditarily Indecomposable Banach Spaces IV.1
Some General Properties of HI Spaces
An important property of the HI spaces concerns the structure of the spaces of their operators. As we will see these spaces have few operators and also remarkable properties of the spaces itself are obtained as consequence of the structure of their operators. We start with the following fundamental results due to Gowers and Maurey [33]. Theorem IV.1. Every bounded linear operator T : X → X with X a complex HI space, is of the form T = λI + S with S a strictly singular operator. This result is not extendable in real HI spaces. However, as we will see, in the case of the HI extensions X[G, σ] the above property remains valid. For an arbitrary HI space the following holds. Theorem IV.2. Let X be a HI space (real or complex). Then every Fredholm operator is of index zero. For complex HI this is a consequence of the above Theorem and Fredholm’s theory. For real HI spaces X it uses the complexification of X. Corollary IV.3. Let X be a HI space. Then X is not isomorphic to any proper subspace of it. In particular X is not isomorphic to its hyperplanes. For the proof of the above results we refer the reader to the Gowers–Maurey paper and also to Maurey’s survey in the Handbook of the Geometry of Banach spaces [46]
48
Chapter IV. The Space of the Operators for HI Banach Spaces
This rest of the section contains some results obtained from the definition of HI Banach spaces and equivalent reformulations on it. Corollary IV.8 describes the frame on which we shall build the non separable HI space which will be presented in the next chapters. Also, Theorem IV.6 shows that every HI space is a subspace of ∞ (N). The next proposition summarizes the equivalent reformulations of the definition of HI Banach spaces. Proposition IV.4. Let X be a Banach space. The following assertions are equivalent: (1) The space X is HI. (2) For every pair of infinite dimensional closed subspaces Y ,Z of X, dist(SY , SZ ) = 0. (3) For every pair of infinite dimensional closed subspaces Y ,Z of X and δ > 0 there exist y ∈ Y and z ∈ Z such that y − z ≤ δy + z. (4) [V.D. Milman] For every infinite dimensional closed subspace Y of X, ε > 0 and W ⊂ BX ∗ such that W ε-norms Y , the space W⊥ = {x ∈ X : f (x) = 0 for every f ∈ W } is a finite dimensional subspace of X. Proof. The assertions (1), (2) and (3) are trivially equivalent by the open mapping theorem and the triangle inequality. We first show that (1) implies (4). If Y is an infinite dimensional closed subspace of X, ε > 0 and W ⊂ BX ∗ are such that W ε-norms Y then Y ∩ W⊥ = {0} and Y ⊕ W⊥ is a closed subspace of X. The HI property of X yields that W⊥ is necessarily finite dimensional. It remains to show that (4) implies (1). Suppose that the space X is not HI Then X has two infinite dimensional closed subspaces Y, Z such that Y ∩ Z = {0} and Y + Z is a closed subspace of X. Then the projection P : Y + Z −→ Y is continuous; set ε = P1 . For every y ∗ ∈ εBY ∗ we may select, by the Hahn– Banach theorem, a functional y ∗ ∈ BX ∗ such that y ∗ extends y ∗ ◦ P and y ∗ X ∗ = y ∗ : y ∗ ∈ εBY ∗ }. Then W ε-norms Y thus by y ∗ ◦ P (Y +Z)∗ . We set W = { (4) the space W⊥ should be finite dimensional. This leads to a contradiction since clearly W⊥ contains the infinite dimensional space Z. Therefore X is HI. Proposition IV.5. Let X be a HI Banach space, and let T : X −→ Y be a bounded linear operator where Y is any Banach space. Then exactly one of the following two holds: (i) The operator T is strictly singular. (ii) Ker T is a finite dimensional subspace of X and X can be written as the direct sum of Ker T and a subspace Z of X such that T|Z is an isomorphism. If Y = X and T is one-to-one, then either T is strictly singular or T is an onto isomorphism.
IV.1. Some General Properties of HI Spaces
49
Proof. Assume that T = 1. Suppose that (i) does not hold. Then there exists an infinite dimensional closed subspace W of X and an ε > 0 such that T w ≥ ε for every w ∈ SW . Suppose also that (ii) does not hold. Then the restriction of T to any subspace of X of finite codimension is not an isomorphism, thus, by Proposition II.3 there ε exists an infinite dimensional subspace Z of X such that T|Z ≤ . 2 Then for every z ∈ SZ and w ∈ SW we have that z − w ≥ T z − T w ≥ T w − T z ≥ ε −
ε ε = , 2 2
ε so we get that dist(SZ , SW ) ≥ which contradicts the HI property of X. Hence 2 Ker T is finite dimensional and has a complement Z such that T|Z is an isomorphism. If Y = X and T is one-to-one and not strictly singular, then T is an isomorphism. As we will see later no HI space is isomorphic to any proper subspace of it. Therefore T is onto. Theorem IV.6. Every HI Banach space X embeds into ∞ . Proof. Let Y be any separable infinite dimensional closed subspace of X and select a countable subset D of the unit ball of X ∗ such that D 12 norms Y . Since X is HI Proposition IV.4 yields that D⊥ is finite dimensional. We enlarge D by a finite set F such that (D ∪ F )⊥ = {0}. Let D ∪ F = {x∗n : n ∈ N}. We define the operator T : X −→ ∞ by the rule T (x) = (x∗n (x))n∈N . Observe that T is one-toone and T restricted to Y is an isomorphism. Proposition IV.5 yields that T is an isomorphism. Theorem IV.7. Let X be a Banach space and let Z be an infinite dimensional closed subspace of X such that Z is HI and the quotient map Q : X −→ X/Z is strictly singular. Then the space X is HI. Proof. We begin the proof with the next two claims. Claim 1. If Z0 is a finite dimensional subspace of Z, Y is an infinite dimensional closed subspace of X and δ > 0, then there exist y ∈ SY and z ∈ SZ with y − z < δ such that z0 < (1 + δ)z0 + λz for every z0 ∈ Z0 and every scalar λ. Proof of Claim 1. We may assume that δ < 1. Let {x1 , x2 , . . . , xk } be a SZ0 and pick for each i = 1, 2, . . . , k an fi ∈ SX ∗ with fi (xi ) = 1. Then finite codimensional in X, thus the subspace Y ∩(
k i=1
k
δ net in 4 Ker fi is
i=1
Ker fi ) is infinite dimensional.
From the fact that the operator Q : X −→ X/Z is strictly singular, we may choose
50
Chapter IV. The Space of the Operators for HI Banach Spaces k
δ δ . If z ∈ Z with y − z < 16 16 i=1 z δ then setting z = we get that z ∈ SZ and y − z < . z 8 1 To finish the proof of the claim it is enough to show that z0 +λz > for 1+δ every z0 ∈ SZ0 and every scalar λ. It is also enough to consider only λ with |λ| < 2. δ Let z0 ∈ SZ0 and |λ| < 2. We choose i ∈ {1, 2, . . . , k} such that z0 − xi < . 4 Then
a y ∈ Y ∩(
Ker fi ) with y = 1 and Qy
|fi (xi + λz)| −
δ 4
δ δ > 1 − |λ| · (|fi (y)| + y − z) − 4 4 δ δ 1 δ . > 1 − 2(0 + ) − = 1 − > 8 4 2 1+δ ≥ 1 − |λ| · |fi (z)| −
Claim 2. For every infinite dimensional closed subspace Y of X and every ε > 0 there exists an infinite dimensional closed subspace W of Z such that dist(w, SY ) < ε for every w ∈ SW . Proof of Claim 2. Let (εn )n∈N be a sequence of positive reals with and SY
∞
∞ &
(1 + εn ) ≤ 2
n=1
ε . From Claim 1 we may inductively select a sequence (yn )n∈N in 8 n=1 n and a sequence (zn )n∈N in SZ such that yn − zn < εn and ai zi < εn
δ for all l ∈ L. The Hahn–Banach theorem yields that there exists φl ∈ BX[G,σ]∗ such that a) φl (yl ) = 0, φl (T yl ) > δ and b) ran φl ⊂ ran(supp yl ∪ supp T yl ). For simplicity we may assume that φl ∈ DG (the precise argument asserts that we can choose φl ∈ DG such that φl (yl ) ≥ δ and |φl (yl )| being as small as we wish). First we observe that for every ε > 0 and every j ∈ N there exist l1 < . . . < lnj ∈ L such that setting x=
yl1 +...+yln
j
nj
and φ =
φl1 +...+φln
j
m2j
gives (m2j x, φ) is a (0, 3C, 2j) exact pair such that m2j xG < ε (Lemmas II.22– n2j−1 II.24, Proposition II.25). Then for a given j ∈ N we may define {(xk , x∗k )}k=1 to 2 ∗ be (0, 18C, 2j − 1)-dependent sequence with xk G < 1/m2j−1 and xk (T xk ) ≥ δ for k = 1, . . . , n2j−1 . Proposition III.6 yields that
1 n2j−1
n2j−1
k=1
xk ≤
104C . m22j−1
(IV.1)
IV.2. The Space of Operators L(X[G, σ]), L(X[G, σ]∗ ) On the other hand setting x∗ =
1 n2j−1
1 m2j−1
n2j−1 k=1
n2j−1
k=1
1
∗
T xk ≥ x (
n2j−1
53
x∗k we get,
n2j−1
k=1
xk ) ≥
δ m2j−1
For sufficiently large j, (IV.1) and (IV.2) derive a contradiction.
.
(IV.2)
Theorem IV.12. Let Y be an infinite dimensional closed subspace of X[G, σ]. Every bounded linear operator T : Y → X[G, σ] takes the form T = λIY + S with λ ∈ R and S a strictly singular operator (IY denotes the inclusion map from Y to X[G, σ]). Proof. Assume that T is not strictly singular. We shall determine a λ = 0 such that T − λIY is strictly singular. Let Y be an infinite dimensional closed subspace of Y such that T : Y → T (Y ) is an isomorphism. By standard perturbation arguments and the fact that X[G, σ] is a strictly singular extension of YG , we may assume, passing to a subspace, that Y is a block subspace of X[G, σ] spanned by a normalized block sequence ∞ yn G < 1. From (yn )n∈N such that (T yn )n∈N is also a block sequence and n=1
Lemma II.22 we may choose a block sequence (yn )n∈N of 2 − n1 i averages of increasing lengths in span{yn : n ∈ N} with yn G → 0. Lemma IV.11 yields that lim dist(T yn , Ryn ) = 0. Thus there exists a λ = 0 such that lim T yn − λyn = 0. n
n
Since the restriction of T − λIY to any finite codimensional subspace of span{yn : n ∈ N} is clearly not an isomorphism and since also Y is a HI space, it follows from Proposition IV.5 that the operator T − λIY is strictly singular. Theorem IV.13. Let YG have uniformly bounded averages and X[G, σ] be the strictly singular extension of YG . Then every T ∈ L(X[G, σ]∗ ) is of the form T = λI + S with S strictly singular operator. We start with the following lemma. Lemma IV.14. Let X be an HI space with a Schauder basis (en )n . Assume that T : X → X is a bounded linear operator not of the form T = λI + S with S strictly singular. Then there exists n0 and δ > 0 such that for every z ∈ Xn0 = span{en : n ≥ n0 }, dist(T z, Rz) ≥ δz. Proof. If not, then there exists a normalized block sequence (zn )n such that dist(T zn , Rzn ) ≤ n1 . Choose λ ∈ R such that T zn − λzn n∈L → 0 for a subsequence (zn )n∈L . Then for a further subsequence (zn )n∈M , M ∈ [L], T − λI|span{zn : n∈N} is a compact operator. The HI property of X easily yields that T − λI is a strictly singular operator, contradicting our assumption. In the following lemma we assume that YG and X[G, σ] are as in the statement of the theorem.
54
Chapter IV. The Space of the Operators for HI Banach Spaces
Lemma IV.15. Let T : X[G, σ]∗ → X[G, σ]∗ , Z be a block subspace of X[G, σ]∗ such ∗ ∗ that Z = < (zn∗ )n > and for each n < m ran zn∗ ∪ ran T zn∗ < ran zm ∪ ran T zm . ∗ ∗ ∗ ∗ ∗ Assume further that for each z ∈ Z , with z = 1, dist(T z , Rz ) > δ. Then the following holds: (i) For each k ∈ N there exists a normalized block sequence (wl∗ )l∈N in Z satisfying the following properties: k ∗ ∗ For every l1 < . . . < lk setting w∗ = q=1 wlq we have that w ≤ 2. Further there exists an 2−k1 -average w ∈ X[G, σ] such that w∗ (w) = 0, T w∗ (w) > 1 and ran(w) ⊂ ran(supp w∗ ∪ supp T w∗ ). (ii) For every j ∈ N there exists (0, 18 δ , 2j) exact pair (x, φ) with φ ∈ Z, 18m x ∈ X[G, σ] and T φ(x) > 1, xG < δn2j2j . Proof. (i) The proof follows the arguments of Lemma III.12 using the following observation: Assume that w1∗ , . . . , wk∗ is a 2 − ck0 -vector. Our assumptions for the operator T yield that for q = 1, . . . , k there exists wq ∈ X[G, σ] such that wq∗ (wq ) = 0, T wq∗ (wq ) > 1 and wq < 1δ. Now using this and the arguments of Lemma III.12 we obtain the desired sequence. (ii) It only requires the adaptation described before the proof of Proposition III.14. Proof of Theorem IV.13. On the contrary assume that there exists T ∈ L(X[G, σ]∗) not of the desired form. Assume further that T = 1 and T e∗n is finitely supported with lim inf min supp T e∗n = ∞. (We may assume the later conditions from the fact that the basis (e∗n )n of (XG )∗ is weakly null.) In particular for every (zn∗ )n block sequence in X[G, σ]∗ there exists a subsequence (zn∗ )n∈L such that (ran zn∗ ∪ ran T zn∗ )n is a sequence of successive subsets of N. Let δ > 0 and n0 ∈ N as in Lemma IV.14. n2j−1 Then Lemma IV.15 yields that for every j ∈ N there exists (zk , zk∗ )k=1 ,a ∗ ∗ , 2j − 1) dependent sequence such that z (z ) = 0, T z (z ) > 1, ran z ⊂ (0, 18 k k k k k δ n2j−1 successive subsets of N and zk G ≤ m21 . ran zk∗ ∪ran T zk∗ , (ran zk∗ ∪ran T zk )k=1 2j−1
Proposition III.6 yields that
1 Finally m2j−1
1≥
1 m2j−1
n2j−1 k=1
1 n2j−1
k=1
zk ≤
144 . m22j−1 δ
T zk∗ ≤ 1 (since T ≤ 1 ) and also
n2j−1
k=1
n2j−1
T zk∗
n2j−1 m22j−1 δ 1 m2j−1 δ . ≥ T zk∗ (zk ) ≥ 144m2j−1 n2j−1 144 k=1
This yields a contradiction for sufficiently large j ∈ N.
IV.2. The Space of Operators L(X[G, σ]), L(X[G, σ]∗ )
55
Notes and Remarks. Our approach showing that the spaces X[G, σ] have few operators follows the lines of Gowers and Maurey [34]. In a recent paper ([4]) is shown that there exists a HI space with no reflexive subspace such that every operator T is of the form λI + W with W a weakly compact operator. The central remaining open problem is the existence of a Banach space with very few operators (i.e. T = λI + K with K compact). The recent results [14], [15] show that a Banach space X could admit few operators and also could have rich unconditional structure. It is not known if there exists a subspace Y of a space X with an unconditional basis, such that Y admits few operators. Actually it is unknown if there exists such a Y which is indecomposable.
Chapter V
Examples of Hereditarily Indecomposable Extensions In the present chapter we shall present HI extensions of YG when G is of specific form. Thus we shall show how we can obtain a quasi-reflexive HI space Xqr , a HI space Xp which has p as a quotient (1 < p < ∞) and also a non-separable HI space.
V.1
A Quasi-reflexive HI Space
We start with the quasi-reflexive HI space. We recall that a Banach space X is said to be quasi-reflexive if dim(X ∗∗ /X) < ∞. The set Gqr We consider the set Gqr = {±I ∗ : I is a finite interval of N}. Here we denote by I ∗ the function χI ∈ c00 . It follows readily that Gqr is a ground set. Moreover the basis of YGqr is equivalent to the summing basis of c0 , hence YGqr is c0 saturated and X[Gqr , σ] is a HI reflexive extension of YGqr . Theorem V.1. The space X[Gqr , σ] is a quasi-reflexive and HI space. Proof. Since X[Gqr , σ] is a reflexive extension of YGqr it follows that X[Gqr , σ]∗ = p span(Gqr ) (Proposition II.26). This yields that X[Gqr , σ]∗ = span({e∗n }n ∪ {N∗ }) from which we obtain that X[Gqr , σ]∗ /X[Gqr , σ]∗ ∼ = R. Since the basis of X[Gqr , σ] is boundedly complete we obtain that X[Gqr , σ]∗∗ /X[Gqr , σ] ∼ = R.
58
Chapter V. Examples of Hereditarily Indecomposable Extensions
Remark. Since every normalized weakly null sequence in YGqr contains a subsequence equivalent to the c0 basis, we conclude that YGqr has the uniform weak Banach–Saks property which yields that the predual of X[Gqr , σ] is also HI.
V.2
The Spaces p , 1 < p < ∞, are Quotients of HI Spaces
Next we define the HI space Xp which has p (1 < p < +∞) as a quotient. Let {Mi }i∈N be a disjoint partition of N into infinite sets, Mi = {mi1 < mi2 < · · · < min < · · · }. We consider the partially ordered set (L, ≺) where L = N and l ≺ d iff l < d and there exists i ∈ N such that l, d ∈ Mi . Clearly (L, ≺) is a tree. A segment s of L has the property that it is a segment of Mi for some i ∈ N. The set Gp is defined as follows. Gp =
n ) i=1
ai s∗i :
n
* |ai |q ≤ 1 and each si is a finite segment of L .
i=1
Here q is the conjugate of p. Clearly Gp is a ground set and the space YGp has the following properties. Proposition V.2. Let Z be a closed infinite dimensional subspace of YGp . Then either c0 → Z or p → Z. Moreover the space YGp has the uniform weak Banach– Saks property. Proof. It follows easily from the definition of YGp that YGp = ( ⊕Yi )p where Yi = < en >n∈Mi and (en )n∈Mi is equivalent to the summing basis of c0 . In particular n each Yi is isomorphic to c0 . Clearly for each n ∈ N, ( ⊕Yi )p remains isomorphic i=1
to c0 hence it is c0 saturated. Let Z be a subspace of YGp . Then one of the following two alternatives holds: 0 (a) There exists n0 ∈ N such that Pn0 |Z : Z → ni=1 ⊕Yi is not strictly singular. (b) For every n ∈ N, Pn |Z is a strictly singular operator. The preceding remarks immediately yield that in the first case c0 is isomorphic to a subspace of Z. If (b) occurs then a “sliding hump” argument derives a normalized sequence (z ) in Z of finite supported vectors and a sequence (wn )n∈N in YGp such that n n z n − wn < 1, and if we denote In = {i ∈ N : supp wn ∩ Mi = ∅}, then (In )n n consists of successive finite subsets of N. The later property yield that (wn )n is equivalent to the usual p basis and hence, (zn )n is equivalent to the p basis. This completes the proof of the first part. Next we show that YGp satisfies the uniform weak Banach–Saks property (Definition III.16). First we prove the following:
V.2. The Spaces p , 1 < p < ∞, are Quotients of HI Spaces
59
Claim. Let (yn )n be a normalized weakly null block sequence in YGp . Then for every ε > 0 and every M ∈ [N] there exists L ∈ [M ] such that the following holds: For every s segment of the tree L, #{ ∈ L : |s∗ (y )| ≥ ε} ≤ 2 . Proof of the claim. Assume on the contrary that for some M no such L exists. Then applying the classical Ramsey theorem for triples we obtain an L ∈ [M ] such that for 1 < 2 < 3 in L there exists a segment s of L with |s∗ (yi )| ≥ ε for i = 1, 2, 3. Let L = {1 < 2 < . . .}. For each n ∈ N and 1 < k < n choose s∗k,n with |s∗k,n (i )| ≥ ε for i = 1, . . . , n. Observe that there exists n0 ∈ N such that for each n ∈ N and each k < n, sk,n ⊂ Mi for some i < n0 . Indeed, sk,n ∩ supp y1 = ∅; now choose n0 such that for every i > n0 , max supp x1 < min Mi . To finish the proof, passing if required to a subsequence, we may assume that for every n ∈ N, w∗ − limn s∗k,n = s∗k with min sk < max supp x1 . Clearly sk is an infinite segment and sk ⊂ Min for some i ≤ n0 . Hence there exists Q ∈ [N] and i < n0 such that for each k ∈ Q, |Mi∗ (yk )| > ε, a contradiction since (y ) is weakly null. The uniform weak Banach–Saks property of YGp is obtained by the above claim in a similar manner as this property is established in p . As consequence we obtain the following. Proposition V.3. (i) The space X[Gp , σ] is a HI reflexive extension of YGp . (ii) The predual of X[Gp , σ]∗ is also HI. The next theorem describes the basic properties of X[Gp , σ]. Theorem V.4. There exists a surjective bounded linear operator Q : X[Gp , σ] → p . Additionally Q∗ [q ] is a complemented subspace of X[Gp , σ]∗ . In particular the quotient map Q is described by the rule Q(emik ) = ei for all i, k ∈ N. Also Q∗ (e∗i ) = Mi∗ and X[Gp , σ]∗ ∼ = X[Gp , σ]∗ ⊕ q . Proof. The proof follows easily from the next equation. For all {ai }ni=1 the following holds: n n + + ' (1/q + + (∗) ai Ii∗ + = |ai |q , + i=1
{Ii }ni=1
where next lemma.
i=1
are infinite subsegments of {Mi }ni=1 . This is a consequence of the
n Lemma V.5. For all {ai }ni=1 , ε > 0, n0 ∈ N there exist {Ji }i=1 finite segments of 1 (L, ≺) such that n0 < Ji ⊂ Mi , and setting xi = #Ji m∈Ji em , we have that d + + + + a i xi + + i=1
[Gp ,σ]
≤ (1 + ε)
d ' i=1
|ai |p
(1/p .
60
Chapter V. Examples of Hereditarily Indecomposable Extensions
Let us see how we finish the proof of the theorem. First we establish the equality (∗). Indeed the definition of Gp yields that d d + + ' (1/q + + ai Ii∗ + ≤ |ai |q . + i=1
i=1
The inverse inequality is obtained by the lemma and a simple duality argument. As consequence we have that span{Mi∗ : i ∈ N} is isometric to q and further X[Gp , σ]∗ /X[Gp , σ]∗ = q . This yields the proof of the theorem. d Proof of Lemma V.5. Assume that i=1 api = 1 and ε, n0 ∈ N are given. Choose j ∈ N such that (i)
1 mj
ε. n∈L
Using this fact we may inductively construct a sequence (sj )j∈N of pairwise disjoint segments of D and a decreasing sequence (Lj )j∈N of infinite subsets of the natural numbers such that |s∗j (xn )| > ε
∀j ∈ N
∀n ∈ Lj .
r . Since the ε segments s1 , s2 , s3 , . . . are pairwise disjoint we may choose an i0 ∈ N such that min si ≥ k for every i > i0 . Let n ∈ Li0 +k . Then n ∈ Li0 +t for each t ∈ {1, 2, . . . , k} i.e. |s∗i0 +t (xn )| > ε for each t ∈ {1, 2, . . . , k}. Setting εt = sgn(s∗i0 +t (xn )) we have that We set r = sup{xn : n ∈ N} and choose k ∈ N with k >
f=
k
εt s∗i0 +t ∈ Gns
w∗
t=1
thus f ∈ BYG∗ns . It follows that r ≥ xn ≥ f (xn ) =
k
εt s∗i0 +t (xn ) =
t=1
k
|s∗io +t (xn )| > kε,
t=1
a contradiction of the choice of k.
Lemma V.10. Let (xn )n∈N be a bounded sequence in YGns . There exists a subsequence (xn )n∈M of (xn )n∈N such that for every segment s the sequence (s∗ (xn ))n∈M is convergent. Proof. From Lemma V.9 we may inductively construct a decreasing sequence (Lk )k∈N of infinite subsets of the natural numbers and a sequence (Fk )k∈N of finite sets of pairwise disjoint segments, Fk = {sk1 , sk2 , . . . , skmk }, such that for each m k k ∈ N and segment s with s ∩ ( ski ) = ∅ we have that i=1
lim sup |s∗ (xn )| < n∈Lk
1 . k
V.3. A Non Separable HI Space
65
Let F be the countable set consisting of all the finite segments of D and all ∞ Fk . We choose a diagonal set L of the subsegments of segments contained in k=1
decreasing sequence (Lk )k∈N . Since {s∗ : s ∈ F } is a countable subset of BYG∗ns and the sequence (xn )n∈L is bounded we may choose, by a diagonal argument, an M ∈ [L] such that the sequence (s∗ (xn ))n∈M is convergent for every s ∈ F . It remains to show that the sequence (s∗ (xn ))n∈M is convergent for every segment s. Let s be a segment. We show that (s∗ (xn ))n∈M is a Cauchy sequence. Let 1 ε ε > 0 and choose k ∈ N with < . We have k 4 s=
m k
(ski ∩ s) ∪ (s \
i=1
and s \
m k i=1
m k
ski )
i=1
ski can be written as the finite union of pairwise disjoint segments with
at most one of them being infinite. We assume that one of them is infinite and let s\
m k i=1
ski
=(
nk
tj ) ∪ t
j=1
where t1 , t2 , . . . , tnk , t are pairwise disjoint segments, t is infinite while t1 , t2 , . . . , tnk are finite. m k 1 It is clear that t ∩ ( ski ) = ∅ thus lim sup |t∗ (xn )| < and by the construck n∈Lk i=1 1 tion of M as a (subset of a) diagonal set we have that lim sup |t∗ (xn )| < . Thus k n∈M we may choose an ma ∈ M such that |t∗ (xn )|
n ≥ max{ma , mb } we have that |s∗ (xm ) − s∗ (xn )|
ε y . for each i = 1, 2, . . . , t then t ≤ ε y y Proof. Suppose that t > and choose A ⊂ {1, 2, . . . , t} with #A = [ ] + 1. ε ε p Set εj = sgn(s∗j (y)) for each j ∈ A and f = εj s∗j . We have f ∈ Gns ⊂ BYG∗ns .
Lemma V.11. Let ε > 0 and y ∈ YGns such that min supp y ≥ [
j∈A
Indeed, the segments sj , j ∈ A are pairwise disjoint and min sj ≥ min supp y ≥ y ] + 1 = #A. Thus [ ε y ] + 1) > y, y ≥ f (y) = εj s∗j (y) = |s∗j (y)| > ε(#A) = ε([ ε j∈A
j∈A
a contradiction.
y . ε The following also holds. Therefore t ≤
Lemma V.12. Let (xn )n be a bounded block sequence such that b∗ (xn ) → 0 for every b branch of D. Then for every ε > 0 there exists L ∈ [N] such that for every segment s of D the following holds: #{n ∈ L : |s∗ (xn )| ≥ ε} ≤ 2. Proof. We set A
) =
{k1 , k2 , k3 } ∈ [N]3 : there exists a segment s * such that |s∗ (ykl )| > ε for l = 1, 2, 3
and B = [N]3 \ A. It follows from Ramsey’s theorem that there exists an L ∈ [N] such that [L]3 ⊂ A or [L]3 ⊂ B. The conclusion of the lemma is exactly that there exists an L ∈ [N] such that [L]3 ⊂ B. Therefore it is enough to exclude the possibility of existing a L ∈ [N] with [L]3 ⊂ A. Suppose that there exists an L ∈ [N], L = {l1 < l2 < l3 < · · · } such that 2r 3 [L] ⊂ A. We set r = sup yn . We may assume that min L > . ε n For each k ≥ 3 we apply the following procedure. We consider the set ) Tk = ran(yl2 + ylk−1 ) ∩ s : s is a segment , such that |s∗ (yl1 )| > ε, * |s∗ (ylk )| > ε, and there exists a t ∈ {2, . . . , k−1} such that |s∗ (ylt )| > ε .
V.3. A Non Separable HI Space
67
r . Indeed, let t1 , t2 , . . . , tm be pairwise different eleε ments of Tk and for each i = 1, 2, . . . , m let si be a segment such that ran(yl2 + ylk−1 ) ∩ si = ti and |si (ylk )| > ε. The segments ri = (ran ylk ) ∩ si i = 1, 2, . . . , m are pairwise disjoint. We also have that |ri∗ (ylk )| > ε and ri ⊂ ran ylk for each i = 1, 2, . . . , m. From the fact that ylk ≥ |ri∗ (ylk )| > ε we get that We claim that #Tk ≤
min supp ylk ≥ lk > l1 > We easily get that m ≤ r m0 = [ ]. Then ε
2ylk yl 2r ≥ > [ k ] + 1. ε ε ε
ylk r r , thus m ≤ . We conclude that #Tk ≤ . We set ε ε ε Tk = {sk1 , sk2 , . . . , sktk }
for some tk ≤ m0 . For k ≥ 3 and i = 1, 2, . . . , tk we set ) * Aki = j ∈ {2, . . . , k − 1} : |(ski )∗ (ylj )| > ε while for tk < i ≤ m0 we set Aki = ∅. Hence {2, . . . , k − 1} =
m0
Aki .
(V.4)
i=1
We consider the following partition of [N \ {1}]2 . For j = 1, 2, . . . , m0 we set ) * Bj = {p, q} ∈ [N \ {1}]2 : p < q and p ∈ Aqj . Observe that (V.4) yields that [N \ {1}]2 =
m 0
Bj ; thus by Ramsey’s theorem
j=1
there exist an M ∈ [N \ {1}] and a j0 ∈ {1, 2, . . . , m0 } such that [M ]2 ⊂ Bj0 . Let M = {m1 < m2 < m3 < · · · }. Then for each t ∈ N and i ≤ t we have m that {mi , mt+1 } ∈ Bj0 , thus mi ∈ Aj0 t+1 . So there exists a segment st such that |s∗t (ymi )| > ε
∀i = 1, 2, . . . , t.
The sequence (s∗t )t∈N has a w∗ convergent subsequence; its w∗ limit is of the form s∗ for some infinite segment s. We deduce that |s∗ (ymi )| > ε for all i which contradicts to the assumption that the sequence (yn )n∈N is weakly null. The next result uses the above lemmas. Lemma V.13. For every Z block subspace of YGns and every ε > 0 there exists z ∈ Z with z = 1 and |s∗ (z)| < ε for every s segment of (D, ≺).
68
Chapter V. Examples of Hereditarily Indecomposable Extensions
Proof. Assume that the conclusion fails. Then there exists a Z block subspace of YGns and ε0 > 0 such that for every z ∈ Z there exists a segment s of D with s∗ (z) ≥ ε0 z. Choose n0 > 4/e0 and let (yn )n∈N be a normalized weakly null block sequence in Z. Lemma V.12 yields that we may assume that for every segment s of D ε0 (V.5) #{n ∈ N : s∗ (xn ) ≥ } ≤ 2 . n Select ym , . . . , ym+n0 with n0 < ym . Further for every m ≤ k ≤ m+n0 choose a segs∗k } > ment sk of D with supp sk ⊂ ran(yk ) and s∗k (yk ) ≥ ε0 . Since min{min n0 supp ∗ 0 is Schreier admissible, hence s ∈ Gns . n0 the family {min supp s∗k }nk=1 k=1 k Therefore n n0 n0 0 ∗ ym+k ≥ ( sk ) ym+k ≥ n0 ε0 > 4 . (V.6) k=1
k=1
k=1
Also from (V.5) for a segment s of D |s∗ (
n0
ym+n )| ≤ 3ε0 .
(V.7)
k=1
(V.6) and (V.7) derive a contradiction completing the proof.
Proof of Theorem V.6. (i) Let Z be a block subspace of YGns and let ε > 0. Using Lemma V.13 we may inductively select a normalized block sequence (yn )n∈N in Y such that, setting dn = max supp yn for each n and d0 = 1, |x∗ (yn )| < 2n dεn−1 for every σF special functional x∗ . We claim that (yn )n∈N is 1 + ε isomorphic to the standard basis of c0 . Indeed, let (βn )N n=1 be a sequence of scalars. We shall show that max |βn | ≤ 1≤n≤N
N n=1
βn yn Gns ≤ (1+ε) max |βn |. We may assume that max |βn | = 1. The left 1≤n≤N
1≤n≤N
inequality follows directly from the bimonotonicity of the Schauder basis (en )n∈N d of YGns . To see the right inequality we consider a g ∈ Gns , g = ai s∗i , where i=1
(si )di=1 are finite segments with min supp si ≥ d. Let n0 be the minimum integer n such that d ≤ dn . Since min supp g ≥ d > dn0 −1 we get that g(yn ) = 0 for n < n0 . Therefore g(
N
βn yn )
N
≤ |g(yn0 )| +
N d
|g(yn )| ≤ 1 +
n=n0 +1
n=1
< 1+
N n=n0 +1
d
|s∗i (yn )|
n=n0 +1 i=1 N
ε < 1+ 2n dn−1 n=n
0
ε < 1 + ε. n 2 +1
(ii) We pass now to show that YGns has uniformly bounded averages. Let 0 < ε < 1 and n ∈ N. We set k0 = k(n, ε) > 3n/ε and we claim that k0 satisfies the requirements of Definition III.13.
V.3. A Non Separable HI Space
69
Indeed, let (yl )l be a normalized weakly null block sequence in YGns . It follows from Lemma V.12 that we can assume the following: For every segment s of D, #{l : |s∗ (yl )| >
ε } ≤ 2. k02
Consider g ∈ Gns , g = dj=1 s∗j with min g < n. This yields that d < n. Consider 0 yli , l1 < . . . < lk0 . Then also x = k10 ki=1 |g(x)| ≤
d j=1
|s∗j (x)| ≤
d 3 3d 3n = < < ε, k k k0 0 i=1 0
and this completes the proof of (ii) and the entire proof of the theorem.
Proposition V.14. Let B denote the set of all branches of the binary tree D. Then we have the following: (i) X[Gns , σ]∗ = span({e∗n : n ∈ N} ∪ {b∗ : b ∈ B}). (ii) X[Gns , σ]∗ /X[Gns , σ]∗ = span{b∗ + X[Gns , σ]∗ : b ∈ B} = c0 (B). (iii) The space X[Gns , σ]∗∗ is isomorphic to X[Gns , σ] ⊕ 1 (B). Proof. (i) It is easy to see that the set Gns Gns
p
=
d "
w∗
is equal to the set
εi s∗i : εi ∈ {−1, 1}, i = 1, 2, . . . , d, (si )di=1 are pairwise
i=1
# disjoint segments and {min s1 , min s2 , . . . , min sd } ∈ S1 , d ∈ N . Therefore, from Proposition II.26 we get that X[Gns , σ]∗ = span({e∗n : n ∈ N} ∪ {b∗ : b ∈ B}). (ii) By Remark II.29 the basis (en )n∈N is boundedly complete and the space X[Gns , σ]∗ = span{e∗n : n ∈ N} is the predual of X[Gns , σ]. It follows from (i) that X[Gns , σ]∗ /X[Gns , σ]∗ = span{b∗ + X[Gns , σ]∗ : b ∈ B} so it remains to show that span{b∗ + (X[Gns , σ])∗ : b ∈ B} = c0 (B). It is enough to show that if b1 , b2 , . . . , bn are pairwise different elements of B and ε1 , ε2 , . . . , εn ∈ {−1, 1} then ε1 (b∗1 + X[Gns , σ]∗ ) + ε2 (b∗2 + X[Gns , σ]∗ ) + · · · + εn (b∗n + X[Gns , σ]∗ ) = 1.
70
Chapter V. Examples of Hereditarily Indecomposable Extensions
For k = 1, 2, . . . we denote by Ek the interval Ek = {n ∈ N : n ≥ 2k }. Since b1 , b2 , . . . , bn are pairwise different we may select a k0 ≥ n such that b1 ∩ Ek , b2 ∩ Ek , . . . , bn ∩ Ek are incomparable segments for all k ≥ k0 . Thus ε1 (b∗1 + X[Gns , σ]∗ ) + · · · + εn (b∗n + X[Gns , σ]∗ ) = dist(ε1 b∗1 + · · · + εn b∗n , X[Gns , σ]∗ ) = dist(ε1 b∗1 + · · · + εn b∗n , span{e∗j : j ∈ N}) = dist(ε1 b∗1 + · · · + εn b∗n , span{e∗j : j ∈ N}) = lim dist(ε1 b∗1 + · · · + εn b∗n , span{e∗j : j ≤ 2k − 1}) k
= lim ε1 (b1 ∩ Ek )∗ + · · · + εn (bn ∩ Ek )∗ k
= 1. The last equality holds since for every k ≥ k0 the functional gk =
n
εi (bi ∩ Ek )∗
i=1
belongs to Gns , thus gk ≤ 1, while for n ∈ b1 ∩ Ek we have that |gk (en )| = 1. (iii) As is well known, for every Banach space Z the space Z ∗∗∗ is isomorphic to Z ∗ ⊕(Z ∗∗ /Z)∗ . For Z = X[Gns , σ]∗ we get that the space X[Gns , σ]∗∗ is isomorphic to the space X[Gns , σ] ⊕ (X[Gns , σ]∗ /X[Gns , σ]∗ )∗ = X[Gns , σ] ⊕ (c0 (B))∗ = X[Gns , σ] ⊕ 1 (B). Theorem V.15. The space X[Gns , σ]∗ is a non separable HI Banach space. Proof. The space X[Gns , σ], being HI, contains no isomorphic copy of 1 , the space X[Gns , σ]∗ = span{e∗n : n ∈ N} is HI and the quotient space X[Gns , σ]∗ /X[Gns , σ]∗ is isometric to c0 (B). From Corollary IV.8 we deduce that X[Gns , σ]∗ is HI while it is clear that X[Gns , σ]∗ is non separable. Remark. The three examples presented in this chapter are from [16] where they have been constructed with the use of higher complexity saturation methods and have the additional property that they are asymptotic 1 spaces. In particular for the space corresponding to X[Gns , σ]∗ , it is shown that every bounded linear operator is of the form λI +W with W weakly compact. Since the space X[Gns , σ]∗ is the dual of a separable space we conclude that the weakly non-compact operator W has separable range. We do not include this property here as it requires much more efforts.
Chapter VI
The Space Xω1 In this part we present a reflexive Banach space Xω1 with a transfinite basis (eα )α κΦ,Ψ , l = λΦ,Ψ . Suppose that l > λΦ,Ψ . Since w(ψl ) = w(φi ) for all i ≤ n2j+1 , we obtain that |ψl (xi )| ≤ n212 . Now suppose that κΦ,Ψ < l < λΦ,Ψ . By (DS.4) we have 2j+1
/ (κΦ,Ψ , λΦ,Ψ ), using that ψl (xi ) = 0 for every κΦ,Ψ < i < λΦ,Ψ . And for i ∈ the fact that w(ψ ) = w(φ ), we can conclude that |ψ (x )| ≤ 12/n22j+1 . Hence, l i l i ( l>κΦ,Ψ ,l=λΦ,Ψ ψl )(xi ) ≤ 12/n2j+1 for every 1 ≤ i ≤ n2j+1 , as desired. Combining (a), (b) and (c) we obtain that 1 | m2j+1
n2j+1 i=κΦ,Ψ
ψi (x)| ≤ 1 +
#E . n22j+1
(VI.12)
From (VI.9) and (VI.12) we conclude that |ψ(x)| ≤ 12(1 + #E/n22j+1 ), as desired. Proposition VI.11. Let (yn )n be a block sequence of vectors of Xω1 . Then the closed linear span of (yn )n is hereditarily indecomposable. Proof. Fix a block sequence (yn )n of Xω1 , two block subsequences (zn )n and (wn )n of (yn )n and ε > 0. Let j be large enough such that m2j+1 ε > 1. By Proposition VI.7 we can choose a (1, j)-dependent sequence (x1 , φ1 , . . . , xn2j+1 , φn2j+1 ) such n2j+1 that x2i−1 ∈ zn n , and x2i ∈ wn n . Set z = (1/n2j+1 ) i=1,i odd xi and w = n2j+1 (1/n2j+1 ) i=1,i even xi . Notice that z ∈ zn n and w ∈ wn n . By Proposition VI.9, we know that z + w ≥ 1/m2j+1 and z − w ≤ 1/m22j+1 . Hence z − w ≤ εz + w. Corollary VI.12. (a) The distance between the unit spheres of every two normalized block sequences (xn ) and (yn ) in Xω1 such that supn max supp xn = supn max supp yn is 0. (b) There is no unconditional basic sequence in Xω1 .
Chapter VI. The Space Xω1
79
(c) Every infinite dimensional closed subspace of Xω1 contains a hereditarily indecomposable subspace. (d) The distance between the unit spheres of two non separable subspaces of Xω1 is equal to 0. Proof. (b) follows from Proposition VI.11 and 4. of Proposition A.3. (c) This result follows from the previous corollary and Gowers’ dichotomy. Moreover, every subspace of Xω1 isomorphic to the closed linear span of a block sequence with respect to the basis (eα )α ε and from the Hahn–Banach and Goldstime theorems there exists a finitely supyα ≤ 1, α ≤ min supp y˜α , x∗ (˜ yα ) > ε and | β ε − ε/4 > ε/2 and β F1 (1/m2j2 ) k∈F2 fk ∈ Kω1 and z2 = (m2j2 /n2j2 ) k∈F2 yk . Notice that φ2 > φ1 , φ2 T z2 > δ and φ2 z2 = 0. Pick p2 ≥ max{p1 , p (supp z1 ∪ supp z2 ∪ supp T z1 ∪ supp T z2 ∪supp Φ1 ∪supp Φ2 ), # supp z2 ·n22j+1 }} and set 2j3 = σ (φ1 , m2j1 , p1 , φ2 , m2j2 , p2 ), and so on. Let us check that (z1 , φ1 , . . . , zn2j+1 , φn2j+1 ) is a (0, j)-dependent sequence: Conditions (DS0.1) and (DS0.2) are rather easy to check from the definition of this sequence. Let us check (DS0.3). There are two cases: (a) Suppose that max supp zk ≤ max supp φk for every 1 ≤ k ≤ n2j+1 . Then supp zk ⊆ pλ −1 supp φλΦ,H −1 Φ,H for every κΦ,H < k < λΦ,H . Then part 2 of (TP.3) gives the desired result. (b) Suppose that max supp φk ≤ max supp zk for every 1 ≤ k ≤ n2j+1 , then supp φk ⊆ supp zλΦ,H −1 pλΦ,H −1 for every κΦ,H < k < λΦ,H , and we are done by part 1 of (TP.3). Fix a (0, j)-dependent sequence (z1 , φ1 , . . . , zn , φn2j+1 ) as in the claim, and set z=
1 n2j+1
n2j+1 k=1
(−1)k+1 zk and φ =
1 m2j+1
n2j+1 k=1
φk .
n2j+1 Then φT z = 1/n2j+1 k=1 (−1)k+1 φT zk ≥ δ/m2j+1 and z ≤ 12/m22j+1 . So, T (z) ≥ δ/m2j+1 ≥ δm2j+1 z/12 > εz as desired. Corollary VIII.4. Let (yk )k be a (C, ε)-RIS, Y its closed linear span and T : Y → Xω1 be a bounded operator. Then limn→∞ d(T yk , Ryk ) = 0. Proof. If not, by previous Proposition VIII.3, we can find a vector z ∈ yk k such that z < (1/T )T z which is impossible if T is bounded. Lemma VIII.5. Let (xn )n be a (C, ε)-RIS, X its closed span and T : X → Xω1 be a bounded operator. Then λT : N → R defined by d(T xn , Rxn ) = T xn − λT (n)xn is a convergent sequence. Proof. Fix any two strictly increasing sequences (αn )n and (βn )n with supn αn = supn βn , and suppose that λT (αn ) →n λ1 , λT (βn ) →n λ2 . By going to a subsequences, we can assume that xαn < xβn for every n. Since the closed linear span of {xαn }n ∪ {xβn }n is an HI space, we can find for every ε two normalized vectors w1 ∈ xαn n and w2 ∈ xβn n such that T w1 −λ1 w1 ≤ ε/3, T w2 −λ2 w2 ≤ ε/3 and w1 − w2 ≤ ε/3T . Then we have that λ1 w1 − λ2 w2 ≤ T w1 − λ1 w1 + T w1 − T w2 + T w2 − λ2 w2 ≤ ε, (VIII.2) and hence, ε ≥ λ1 w1 − λ2 w2 ≥ |λ1 − λ2 |w1 − |λ2 |w1 − w2 ≥ |λ1 − λ2 | − |λ2 |ε. (VIII.3) So, |λ1 − λ2 | ≤ ε(1 + |λ2 |) for every ε. This implies that λ1 = λ2 .
90
Chapter VIII. The Spaces of Operators L(Xγ ), L(X, Xω1 )
Definition VIII.6. Recall that for a set A of ordinals, A(0) is the set of isolated points of A. Fix a transfinite block sequence (xα )α 0 and every finite-dimensional block subspace X0 of X there is a finitedimensional block subspace Y0 of Y and an isomorphism T : X0 → Y0 such that T T −1 < 1 + . Clearly, if (ei ) is a spreading model of a subsequence of some basic sequence (xn ) in some Banach space X, then the space generated by (ei ) is finitely block representable in (xn ). So the following result summarizes the results obtained so far. Theorem I.3.6. Suppose (xn ) is a normalized basic sequence in some Banach space X. Then there is a Banach space Y with a Schauder basis (yn ) which is finitely block representable in (xn ) and has the property
n i=0
ai yi =
n
σi ai yki
i=0
for every n, every sequence k0 < k1 < · · · < kn of integers, every sequence a0 < a1 < · · · < an of scalars, and every sequence σ0 , σ1 , . . . , σn of signs. Corollary I.3.7. For every positive integer n and > 0 there is an integer m such that every basic sequence (ei )m i=0 in some Banach space X contains a block subsequence (fi )ni=0 such that
n
σi ai fi ≤ (1 + )
i=0
n
ai fi
i=0
for every choice a0 , a1 , . . . , an of scalars and σ0 , σ1 , . . . , σn of signs.
Starting from the basic sequence (yn ) of Theorem I.3.6 one can find its blocksubsequence whose spreading model is equivalent to the basis of one of the standard sequence space p (1 ≤ p < ∞) or c0 . This is the content of the following result. Theorem I.3.8 (Krivine). Suppose (ei ) is a basic sequence in some Banach space such that n n ai ei = σi ai eki i=0
i=0
for every n ≥ 0, sequence of integers 0 ≤ k0 < k1 < · · · < kn , sequence a0 , a1 , . . . , an of scalars and sequence σ0 , σ1 , . . . , σn of signs. Then the unit vector basis of either p for some 1 ≤ p < ∞ or of c0 is finitely block representable in (ei ).
140
Chapter I. Finite-Dimensional Ramsey Theory
Proof. We need a convenient way to “double” a given finitely supported vector of (ei ) as well as to “triple” it. In other words to have bounded operators x → D(x) and x → T (x) which commute and which perform these two operations, respectively. A natural way to formalize this is to re-enumerate (ei ) as (ei )i∈Q and define D and T on the span X1 of (ei )i∈Q∩[0,1) by letting D(eq ) = eq/2 + e(q+1)/2 and T (eq ) = eq/3 + e(q+1)/3 + e(q+2)/3 . Note that this indeed defines bounded linear operators on X1 which commute. In fact D ≤ 2 and T ≤ 3. Taking an ultrapower over a block subsequence (xn ) of (ei ) such that for some 1 ≤ λ ≤ 2, the sequence D(xn ) − λxn converges to 0 one finds a sequence (y˜n ) in the ultrapower and 1 ≤ µ ≤ 3 such that T (xn ) − µy˜n converges to 0. So, for some normalized block subsequence (zn ) of (xn ) we have that D(zn ) − λzn → 0 and D(zn ) − µzn → 0. For n, k ≥ 0 let zn k be the copy of the vector zn when Q ∩ [0, 1) is shifted to Q ∩ [k, k + 1) in the natural way. Using (fk ) to denote the standard basis of c00 we use (zn k ) to define a norm on c00 by
l
ak fk = lim n→U
k=0
l
ak zn k .
k=0
Let Y be the completion of c00 relative to this norm. Then for every pair x and y of finitely supported vectors of Y and index k such that x < k, k + 1, k + 2 < y, we have x + fk + fk+1 + y = x + λfk + y and x + fk + fk+1 + fk+2 + y = x + µfk + y. It follows that for every pair of integers m and n,
m n 2 3
fk = λm µn
(I.4)
k=1
More generally,
l k=1
ek =
j
λmi µni ei
(I.5)
i=1
m whenever l = ji=1 2mi 3ni . So if λ = 1 or µ = 1 we would get that 2k=1 fk = 1 3n or k=1 fk = 1, respectively, so the basis (fk ) of Y is 1-equivalent to the basis of c0 . So we may concentrate on the case when λ, µ > 1. Note that by (I.4) the function ϕ(2m 3−n ) = λm µ−n is a nondecreasing multiplicative function, and since {2n 3−n : m, n ∈ N} is dense in R+ , it extends to a nondecreasing multiplicative function ϕ : R+ → R+ . So there is 1 ≤ p < ∞ such that ϕ(v) = v 1/p . In particular, λ = 21/p and µ = 31/p . l Consider now the function ψ(l) = k=1 fk . Then taking mi = m and ni = n 1 (1 ≤ i ≤ j) in (I.5), we obtain that ψ(j · 2m · 3n ) = λm · 3n · ψ(j) = (2m · 3n ) p · ψ(j).
I.3. Finite Representability of Banach Spaces
141
Since ψ is nondecreasing and subadditive and since {2n 3−n : m, n ∈ N} is dense in R+ one easily concludes that, in fact, ψ(j) = j 1/p for all j ∈ N. Using this and (I.5) again we see that if for each i ≤ j we have scalars ai of the form λmi µni , then for l = ji=1 2mi 3ni we have that
j i=1
ai fi =
l k=1
1 p
fk = l = (
j
1
api ) p .
i=1
Since the sequence normalizations of sequences (a1 , . . . , aj ) of this form are dense in the unit sphere of pj we have this equality for every choice of scalars. Corollary I.3.9. For every basic sequence (xn ) in some Banach space X the standard basis of some p (1 ≤ p < ∞) or of c0 is finitely block representable in the closed linear span of (xn ). It is worth stating also a finite form of Corollary I.3.9 proved via the usual compactness argument. Corollary I.3.10. For every integer n and constants 0 < , C < ∞ there is an integer m = m(n, , C) such that if X is an m-dimensional Banach space with normalized basis (xj )m j=1 and constant C there is a p ∈ [1, ∞] and a block subsequence (fi )ni=1 of (xj )m j=1 which is (1 + )-equivalent to the standard basis of pn . We finish this section by mentioning another finite-dimensional result. Theorem I.3.11 (Odell–Rosenthal–Schlumprecht). Suppose h is a uniformly continuous real-valued function defined on the unit sphere of some Banach space X with Schauder basis (xn ). Then there is λ ∈ R such that for every > 0 and every integer n > 0 there is a block subsequence (fi )ni=0 such that |h(y) − λ| < for all y ∈ (fi )ni=0 of norm 1. It turns out that an infinite-dimensional analogue of this result is false even for Banach spaces of the form p (1 ≤ p < ∞), though it is true for the Banach space c0 . We return to this point in another section of these notes.
Chapter II
Ramsey Theory of Finite and Infinite Sequences II.1
The Theory of Well-Quasi-Ordered Sets
Infinite-dimensional Ramsey theory is a branch of Ramsey theory initiated in the early 1960s by Nash–Williams in his attempts to extend a well-known theorem of Kruskal that the finite trees are well-quasi-ordered (w.q.o.) under the inf-preserving embeddings (see Theorem II.1.9 below). Recall that a quasi-order (q.o.) is a set Q with a binary transitive and reflexive relation ≤. We write x < y if x ≤ y and y x, we write x|y if x y and y x, and we write x ≡ y if x ≤ y and y ≤ x. A quasi-ordered set Q is well-quasi-ordered (w.q.o.) if there are no infinite descending sequences x0 > x1 > · · · > xn > · · · and no infinite antichains x m | xn
(m = n).
Lemma II.1.1. A quasi-ordered set Q is w.q.o. iff for every infinite sequence (xn ) ⊆ Q there exists m < n such that xm ≤ xn iff for every infinite sequence (xn ) ⊆ Q there exists an infinite sequence n0 < n1 < · · · < nk < · · · of integers such that xni ≤ xnj whenever i < j. Proof. Use the 2-dimensional Ramsey theorem.
Corollary II.1.2. If Q0 and Q1 are w.q.o., then so is their cartesian product Q0 × Q1 . Lemma II.1.3. Suppose (xn ) is a sequence of elements of a q.o. set Q. Then there is an infinite subsequence (xnk ) such that either
144 (1) xnk | xn
Chapter II. Ramsey Theory of Finite and Infinite Sequences whenever k = , or
(2) xnk ≤ xn
whenever k < , or
(3) xnk > xn
whenever k < .
Proof. Apply the 2-dimensional Ramsey theorem to the 3-coloring given by (1), (2) and (3). A quasiordered set Q is well-founded if it contains no infinite sequence (xn ) such that xm > xn whenever m < n. Corollary II.1.4. If a well-founded q.o. set Q is not w.q.o., then it contains an infinite sequence (xn ) such that xm xn whenever m < n. An infinite sequence (xn ) of elements of some q.o. set Q is a bad array if xm xn holds for all m < n. Definition II.1.5 (Nash–Williams). A bad array (xn ) of a q.o. set Q is minimal bad if there is no bad array (yn ) of Q such that: (1) ∀m∃n ym ≤ xn , (2) ∃m∃n ym < xn . Lemma II.1.6. Suppose Q is a well-founded but not well-quasi-ordered set. Then there is a minimal bad array (xn ) in Q. Proof. By Corollary II.1.4 we can start with a bad array (x0n ) of Q and assume that its first term x00 is minimal first term of all possible bad arrays. Choose now a bad array (x1n ) such that x11 is minimal second term among all bad arrays (yn ) such that yo = x00 . Then choose a bad array (x2n ) ≤ (x1n ) such that x22 is minimal third term among all bad arrays (yn ) such that y0 = x00 and y1 = x11 , and so on. We claim that the diagonal sequence (xnn ) is a minimal bad array of Q. For suppose there is a bad array (yi ) such that ∀i∃ n yi ≤ xnn and ∃ i∃ n yi < xnn . Let m be the minimal n such that for some i, yi < xnn . Going to a subsequence of (yi ) we may assume that in fact y0 < xm m and that for all i there is n ≥ m such that yi < xnn . Then the sequence x00 , . . . , xn−1 n−1 , y0 , y1 , y2 , . . . is a bad array, contradicting the choice of xnn . Corollary II.1.7. If (xn ) is a minimal bad array of a quasi-ordered set Q then {q ∈ Q : (∃ i)q < xi } is w.q.o. Given a q.o. set Q we let the set Qk
Let Fk = Ek \ l>k Ek for k ∈ N. Then (Fk ) is an infinite sequence of disjoint sets from B such that |νnk (Fk )| > (2/3)ε and |νnk (Fl )| < ε/3 for all k < l. In particular, |νnk+1 (Fk+1 ) − νnk (Fk )| > ε/3 for all k. As in the previous proof applying Lemma II.10.1 we get a set E ∈ B such that |νnk+1 (E) − νnk (E)| > ε/3 for infinitely many k’s. This means that (νn (E))∞ n=0 is not a convergent sequence, a contradiction. Remark II.10.10. We have taken Lemma II.10.1 (but not its proof) from a paper of Matheron [53]. We refer the reader to papers [15] and [19] for some further uses of these ideas. Definition II.10.11. A topological group G is Mazur–Orlicz complete if every null ∞ (x ) such that the series sequence (xn ) in G contains a subsequence n k k=0 xnk ∞ is convergent in G. If in addition k=0 xnk is unconditionally convergent then we say that G is unconditionally Mazur–Orlicz complete . Clearly every completely metrizable abelian group is unconditionally Mazur– Orlicz complete. The following result is a partial converse to this implication. Theorem II.10.12 (Burzyk–Klis–Lipecki). Every metrizable Mazur–Orlicz complete abelian group G is a Baire space. Proof. Consider a decreasing sequence (Un )∞ n=0 of dense open subsets of G. We shall show that 0 is in the closure of the intersection from which it follows easily that ∞ n=0 Un is dense in G. Let d be a fixed translation invariant metric on G giving us its topology and let ε > 0. A simple recursive construction will give us ∞ a sequence (xn )∞ n=0 ⊆ G and a sequence (Vn )n=0 of open subsets of G such that for all n: (1) d(0, xn ) ≤ ε · 2−n−1 , Vn ⊆ Un , and (2) n n (3) k=0 σk xk ∈ k=0 Vkσk for all (σk )nk=0 ⊆ {0, 1}, where for a subset S of G we let S 0 = X and S 1 = S. Applying the Mazur–Orlicz completeness of G to (xn ) we get a subsequence (xni ) such that the series ∞ i=0 xni
196
Chapter II. Ramsey Theory of Finite and Infinite Sequences
converges in G. It follows that required.
∞ i=0
xni ∈
∞ n=0
Un and d(0,
∞ i=0
xni ) ≤ δ, as
Corollary II.10.13. Suppose that G is a metrizable abelian group which has the property of Baire in its completion. Then G is complete if and only if it is Mazur– Orlicz complete. Corollary II.10.14. Let Y be a subspace of a Banach space X. If Y is an analytic subset of X and if it is Mazur–Orlicz complete then it must in fact be closed. Remark II.10.15. This corollary has been first observed by A.R.D. Mathias who was using Silver’s theorem to prove it (see [39] and [88]). Note that there are interesting groups that are unconditionally Mazur–Orlicz complete but are not Baire spaces. One such example is the group G = (1 , w). They of course cannot be metrizable. It turns out however that non-metrizable group do share some of the properties of the metrizable ones as the following result shows. Theorem II.10.16. Let G be an unconditionally Mazur–Orlicz complete abelian group. Then every Borel subgroup H of G which is Mazur–Orlicz complete must in fact be closed in G. Proof. Suppose H is not closed in G and pick a sequence (xn ) ⊆ H converging to a point of G \ H. Applying the unconditional Mazur–Orlicz completeness of x G and going to a subsequence of (xn ) we may assume that the series ∞ n=0 n is unconditionally convergent in G. For n ∈ N let y = x − x . Note that n n+1 n ∞ [∞] y ∈ / H. For M ∈ N , let n n=0 yM =
∞ M (2k+1)−1
yn ,
k=0 n=M (2k)
where (M (k))∞ k=0 is the increasing enumeration of M . Let XM be the collection of all M ∈ N[∞] such that yM ∈ H. Then XH is a Borel subset of N[∞] , so Galvin–Prikry theorem applies to XH . Pick an M ∈ N[∞] such that M [∞] ⊆ XH or M [∞] ∩ XH = ∅. Note that for N ∈ N[∞] and n ∈ N at most one of the / H. It follows that elements yN and yN \{n} can belong to XH since ∞ n=0 yn ∈ M [∞] ∩ XH = ∅. For k ∈ N, set
M (2k+1)−1
zk =
n=M (2k)
yk .
The fact that the series yM = ∞ k=0 zk is convergent in G means in particular that the sequence (zk ) of elements of H converges to 0. Applying the Mazur–Orlicz ∞ completeness of H to (zk ) we get that a subsequence (zki ) such that i=0 zki is ∞ convergent in H. Note however that i=0 zki is equal to yN for some N ∈ M [∞] , so M [∞] ∩ XH = ∅, a contradiction.
Chapter III
Ramsey Theory of Finite and Infinite Block Sequences III.1 Hindman’s Theorem Hindman’s theorem in its finite unions form (rather than non-repeating sums form) can be considered as the first non-trivial result of so-called block Ramsey theory. To state this theorem, let FIN denote the collection of all finite nonempty subsets of N. A finite or infinite sequence X = (xi ) of elements of FIN is a block sequence if xi < xj 1 whenever i < j. For a block sequence X = (xi ) set [X] = {xi0 ∪ · · · ∪ xik : k ∈ N, i0 < · · · < ik < |X|}. We call [X] a partial subsemigroup of (FIN, ∪) generated by X = (xi ). Note that when X = (xi )∞ i=0 is infinite then ([X] , ∪) is isomorphic to (FIN, ∪) via isomorphism which sends {i} to xi . So the following well-known result gives us a basic pigeon-hole principle for the semigroup FIN. Theorem III.1.1 (Hindman). For every finite coloring of FIN there is an infinite block sequence X in FIN such that [X] is monochromatic. Proof. Let γFIN be the collection of all ultrafilters U on FIN such that for all n ∈ N, {x ∈ FIN : n < x} ∈ U. For U, V ∈ γFIN let U ∪ V = {A ⊆ FIN : {x : {y : x ∪ y ∈ A} ∈ V} ∈ U}. 1 For
x, y ∈ FIN by x < y we denote the fact that max(x) < min(y).
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Chapter III. Ramsey Theory of Finite and Infinite Block Sequences
Note that U ∪ V is an ultrafilter and that it belongs to γFIN. So we have defined an operation ∪ on γFIN which is easily seen to be associative, i.e., (U ∪ V) ∪ W = U ∪ (V ∪ W). ˇ One can view γFIN as a closed subset of the Stone–Cech compactification βFIN of FIN taken with its discrete topology. Thus a basic open set of γFIN is determined by a subset A of FIN as below: A∗ = {U ∈ γFIN : A ∈ U}.
Lemma III.1.2. For each V ∈ γFIN the function U → U ∪ V is a continuous function from γFIN into γFIN. Proof. It suffices to show that for every A ⊆ FIN, the set T = {U ∈ γFIN : A ∈ U ∪ V} is open in γFIN. Let B = {x : {y : x ∪ y ∈ A} ∈ V}. Then T = B ∗ , and so T is open in γFIN. Lemma III.1.3 (Ellis). Suppose (S, ∗) is a compact semigroup such that x → x ∗ y is a continuous map for each y ∈ S. Then (S, ∗) has an idempotent. Proof. By compactness, S contains a minimal closed nonempty subsemigroup T . Pick t ∈ T . By continuity of x → x ∗ t the subsemigroup T ∗ t is compact and therefore closed subsemigroup of T. It follows that T ∗ t = T so V = {x ∈ T : x ∗ t = t}. is nonempty. It is also closed being a preimage of the point t under x → x ∗ t. It follows that V = T and therefore t ∗ t = t, as required. The following lemma is the last step in the proof of Theorem III.1.1. Lemma III.1.4. Suppose U is an idempotent of (γFIN, ∪). Then every A ∈ U contains a subsemigroup of (FIN, ∪) generated by an infinite block sequence. Proof. Since A ∈ U = U ∪ U we can find x0 ∈ A such that A1 = {y ∈ A : x0 < y and x0 ∪ y ∈ A} ∈ U. Then again since A1 ∈ U ∪ U we can find x1 ∈ A1 such that A2 = {y ∈ A1 : x1 < y and x1 ∪ y ∈ A1 } ∈ U, and so on. This procedure gives us a decreasing sequence A = A0 ⊇ A1 ⊇ · · · ⊇ An ⊇ · · · of elements of U and an infinite block sequence X = (xn ) such that xn ∈ An and An+1 = {y ∈ An : xn < y and xn ∪ y ∈ An }.
III.1. Hindman’s Theorem
199
Inductively on k ∈ N one shows that for every sequence n0 < n1 < · · · < nk of elements of N, xn0 ∪ xn1 ∪ · · · ∪ xnk ∈ An0 . This is clearly so for k = 0. To see the recursive step, let y = xn1 ∪ · · · ∪ xnk . Then by the inductive hypothesis, y ∈ An1 . Since An1 ⊆ An0 +1 we get that y ∈ An0 +1 and therefore xn0 ∪ y ∈ An0 as required. Exercise III.1.5. Show that V → U ∪ V is not necessarily a continuous function on γFIN, when V is taken to be an arbitrary member of γFIN. For an integer k ≥ 1, let FIN[k] denote the collection of all block sequences of elements of FIN of length k. Similarly, let FIN[∞] denote the set of all infinite block sequences of elements of FIN. The following is an analogue of the Ramsey theorem in the context of FIN. Theorem III.1.6 (Taylor). For every integer k ≥ 1 and every finite coloring of [k] FIN[k] there exists an infinite block sequence X = (xi ) such that [X] is monochromatic. Before proving the theorem, we give some definitions: For two block sequences X = (xi ) and Y = (yj ) of finite subsets of N we put X ≤ Y iff xi ∈ [Y ] for all i. For a block sequence X = (xi ) ⊆ FIN and s ∈ FIN, let X/s = (xm+j )j≥1 where m = min{i : xi > s}. Thus X/s is the block sequence that enumerates the tail of X determined by s. It is clear that Theorem III.1.6 follows from the next lemma. Lemma III.1.7. For every P ⊆ FIN[k] and every infinite block sequence Y of elements of FIN there is an infinite block sequence X ≤ Y such that [X][k] ⊆ P or [k] [X] ∩ P = ∅. Proof. The proof is by induction on the integer k. The case k = 1 is given by the proof of Hindman’s theorem restricted to the semigroup [Y ] instead of FIN. So we assume k > 1 and that the Lemma is true for k − 1. Recursively on n, we build a block sequence Z = (zn ) in [Y ] and a decreasing sequence (Yn ) of members of FIN[∞] such that Y0 = Y and such that: (1) zn is the first term of Yn . k−1 (2) For every n and every (x0 , . . . , xk−2 ) ∈ [(zi )ni=0 ] either (x0 , . . . , xk−2 , x) ∈ P / P for all x ∈ [Yn+1 ]. for all x ∈ [Yn+1 ], or (x0 , . . . , xk−2 , x) ∈ There is no problem in constructing (zn ) and (Yn ) since (2) is provided by the case k = 1 of the lemma.
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Chapter III. Ramsey Theory of Finite and Infinite Block Sequences
Having constructed Z = (zn ) we note (see (2)) that for (x0 , . . . , xk−2 , xk−1 ) ∈ [Z] , the sentence (x0 , . . . , xk−2 , xk−1 ) ∈ P [k]
does not depend on xk−1 , or in other words, P ∩ [Z][k] is a cylinder over its [k−1] . So we are done using the inductive hypothesis. projection on [Z]
III.2 Canonical Equivalence Relations on FIN Consider the following five equivalence relations on FIN: xE0 y xE1 y xE2 y xE3 y
iff iff iff iff
x=y x=x min(x) = min(y) max(x) = max(y)
xE4 y iff min(x) = min(y) and max(x) = max(y). The purpose of this section is to show that these are all essential equivalence relations on FIN provided one is willing to shrink to a subsemigroup of FIN generated by an infinite block sequence. Theorem III.2.1 (Taylor). For every equivalence relation E on FIN there is i < 5 such that E [X] = Ei [X] for some infinite block sequence X = (xi ). Proof. An equation (in 4 variables v0 , v1 , v2 , v3 ) is a formula ϕ(a, b, v0 , v1 , v2 , v3 ) of the form ϕ(a, b, v0 , v1 , v2 , v3 ) ≡ a ∪ vi0 ∪ · · · ∪ vi3 E b ∪ vj0 ∪ · · · ∪ vj3 where i0 , j0 , ..., i3 , j3 ∈ {0, 1, 2, 3} and a, b ∈ FIN ∪ {∅}. Given a block sequence Z and an equation ϕ ≡ ϕ(a, b, v0 , ..., v3 ) we say that ϕ is true in Z iff ϕ(a0 , . . . , a3 ) holds for every (a0 , . . . , a3 ) ∈ [Z/(a ∪ b)]4 . We say that ϕ is false in Z iff ¬ϕ(a0 , . . . , a3 ) holds for every (a0 , . . . , a3 ) ∈ [Z/(a ∪ b)]4 . We say that ϕ is decided in Z if either ϕ is true or ϕ is false in Z. By Theorem III.1.6, and using a non difficult diagonal procedure, we can find a block sequence Z = (zn )n such that all equations ϕ(a, b, v0 , ..., v3 ) with a, b ∈ [Z] ∪ {∅} are decided in Z. Our aim is to show that there is some i < 5 such that E [Z] = Ei [Z]. Case 1. v0 E v1 is true in Z. Then E [Z] = E1 [Z]: Let s, t ∈ [Z] and pick u ∈ [Z] such that u > s, t. Then s E u and t E u, and hence s E t. Case 2. v0 E v1 is false in Z, v0 ∪ v1 E v0 is true in Z and v0 ∪ v1 E v1 is false in Z. Let us check that E is equal to E2 on [Z]. Fix s, t ∈ [Z]. Suppose first that s E2 t, and write s = zn ∪ s and t = zn ∪ t , with s , t ∈ [Z] and zn < s , t . Using that v0 ∪ v1 E v0 is true in Z, we obtain that s, t E zn , and hence s E t.
III.3. Fronts and Barriers on FIN[
