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BANACH SPACES OF THE TYPE OF TSIRELSON

arXiv:math/9207206v1 [math.FA] 6 Jul 1992BANACH SPACES OF THE TYPE OF TSIRELSONS. A. Argyros and I. DeliyanniAbstract .To any pair (M , θ ) , where M is a family of finite subsets of N , compactin the pointwise topology, and 0 < θ < 1 , we associate a Tsirelson-type Banach space T θM .It is shown that if the Cantor-Bendixson index of M is greater than n and θ > 1n then T θMis reflexive.

Moreover, if the Cantor-Bendixson index of M is greater than ω then T θM doesnot contain any ℓp , while if the Cantor-Bendixson index of M is finite then T θM containssome ℓp or co . In particular, if M = { A ⊂N : |A| ≤n } and1n < θ < 1 then T θM isisomorphic to some ℓp .Notation .

We denote by c00 the space of finitely supported sequences. {en}∞n=1 is the canonical basis in c00.If E ⊂N and x ∈c00 , we denote by Ex the restriction of x to E, that is Ex = x.XE .For E , F nonvoid subsets of N we write E < F for max E < min F .

We write n < Efor {n} < E .If Ei ⊂N, i = 1, . .

., n and E1 < E2 < . .

. < En we say that E1, .

. ., En are succes-sive sets.For x ∈c00 , supp(x) is the set {i : xi ̸= 0 } .If xi ∈c00 , xi ̸= 0 , i = 1, .

. ., n and supp(x1) < .

. .

< supp(xn) we say that x1, . .

. , xnare successive vectors.Finally, ∥.

∥p denotes the ℓp norm, 1 ≤p ≤∞.We recall that the norm ∥. ∥T on Tsirelson’s space T is defined by the following implicitequation.For all x ∈T∥x∥T = max{ ∥x∥∞, 12 supnXi=1∥Eix∥T } ,where the ”sup” is taken over all n and all sequences (E1, .

. ., En) of successive subsets ofN with n ≤min E1.It is well known that Tsirelson’s space is a reflexive space not containing any ℓp ([T]).ΛA basic notion in the definition of Tsirelson’s space is that of a ”Schreier set”.

A finiteset A is called a Schreier set if |A| ≤min A. The above mentioned properties of Tsirelson’sspace are consequences of the following two properties of the class S of Schreier sets:1) S does not contain infinite increasing sequences of sets.2) S contains arbitrarily large (finite) sets.This leads us to the following generalization:Let M be a set of finite subsets of N. We say that M is compact if the setX (M) = {XA : A ∈M } is a compact subset of the Cantor space 2N.– 1 –

Definition 1 . Let M be as above.

A family (E1, . .

., En) of successive finite subsetsof N is said to be M-admissible if there exists a set A = {m1, . .

., mn} ∈M such thatm1 ≤E1 < m2 ≤E2 < . .

. < mn ≤En .Definition 2 .

Let M be a compact family. Let 0 < θ < 1 .

We define the (M, θ)-Tsirelson-type space T θM as follows:T θM is the completion of c00 under the norm ∥. ∥satisfying the implicit equationFor all x ∈c00∥x∥= max{ ∥x∥∞, θ supnXi=1∥Eix∥} ,where the ”sup” is taken over all n and all M-admissible families (E1, .

. .

, En) .Remark . It is easy to see that {en}∞n=1 is a 1-unconditional basis for T θM .Let us now recall the notion of the Cantor-Bendixson index of a countable compact spaceW .Given a countable compact space W , one defines the sequence W (λ) , λ < ω1 as follows:W (0) = WW (λ+1) = { x ∈W : x is a limit point of W (λ)}and for a limit ordinal λ ,W (λ) = ∩µ<λW (µ)The Cantor-Bendixson index of Wis then defined as the least λ < ω1 for whichW (λ) = ∅.We state without proofs the following results:Theorem 1 .

Let M , θ be as in the definition of T θM . Suppose that there is n ∈Nsuch that the Cantor-Bendixson index of X (M) is at least n+1 and1n < θ < 1 .

Then T θMis reflexive.Theorem 2 . If the Cantor-Bendixson index of X (M) is greater than ω then T θMdoes not contain isomorphically any ℓp .In the opposite direction we have the following result:Theorem 3 .

If the Cantor-Bendixson index of X (M) is finite, then the space T θMcontains an isomorphic copy of ℓp , for some 1 < p < +∞, or c0 .We shall present here a special case of Theorem 3, namely the case whereM = { A ⊂N : |A| ≤n } for some n ∈N .Theorem 3a . Let n ∈N , n ≥2 be fixed.

If M = { A ⊂N : |A| ≤n } and1n < θ < 1then the space T θM is isomorphic to ℓp , where1p + logn( 1θ) = 1 .– 2 –

In other words: The norm ∥, ∥defined on c00 by the implicit equation∥x∥= max{ ∥x∥∞,1n1q supdXi=1∥Eix∥} ,where the ”sup” is taken over all d ≤n and all sequences of successive sets (E1, . .

.Ed) , isequivalent to the ℓp norm, where1p + 1q = 1 .Proof : For the proof we first need to deal a little with the dual of T θM .We define inductively a sequence {Ks}∞s=0 of subsets of [−1 , 1](N) as follows:K0 = {+−en : n ∈N}and for s ∈NKs+1 = Ks ∪{ θ · (f1 + . .

. + fd ) : d ≤n, fi , i = 1, .

. ., n are successive and fi ∈Ks }It is not difficult to see that for every x ∈c00∥x∥=supf∈∪∞s=1Ks< f , x >The proof of the Theorem goes through four steps:Step 1For every x ∈c00∥x∥≤∥x∥p.Proof : It is enough to show that for every s and every f ∈Ks , |f(x)| ≤∥x∥p .This is done by induction on s .For f ∈K0 it is trivial.

Suppose that for some s we have that for all f ∈Ks andall y ∈c00|f(y)| ≤∥y∥p . Let x ∈c00 and f =1n1q (f1 + .

. .

+ fd) ∈Ks+1 , whered ≤n , f1, . .

., fn are successive and belong to Ks .Then, setting xi = (supp(fi))(x) we have|f(x)| ≤1n1qdXi=1|fi(x)| = 1n1qdXi=1|fi(xi)| ≤1n1qdXi=i∥xi∥p ≤ dnq(dXi=i∥xi∥p)1p ≤∥x∥pby the induction hypothesis and H¨older’s inequality.Step 2For all m ∈N1n1p m1p ≤∥Pmi=1 ei∥.Proof : Suppose first that m = ns for some s ∈N. The functional f =1nsq Pnsi=1 eiclearly belongs to Ks .

So∥nsXi=1ei∥≥f nsXi=1ei= 1nsq ns = ns(1−1q ) = nsp = m1p– 3 –

Now let m ∈N and find s such that ns ≤m ≤ns+1. Then∥mXi=1ei∥≥∥nsXi=1ei∥= nsp = 1n1p ns+1p≥1n1p m1p .Step 3For every normalized block sequence (xk)∞k=1 of the basis (en)∞n=1 we have∥Xakxk∥≤2θ∥Xakek∥for all coefficients (ak) .Proof : It is enough to show that for every φ ∈∪∞s=1Ks one getsφ(Xakxk) ≤2θ ∥Xakek∥For the proof we need the following technical notions:Definition A.

Let m ∈N and φ ∈Km \ Km−1 . We call analysis of φ any sequence{F s(φ)}ms=0 of subsets of ∪∞s=1Ks such that:1) For every s F s(φ) consists of successive elements of Ks and∪f∈F s(φ)supp(f) = supp(φ) .2) If f belongs to F s+1(φ) then either f ∈F s(φ) or there is a d ≤n and successivef1, .

. ., fd ∈F s(φ) with f = θ(f1 + .

. .

+ fd) .3) F m(φ) = {φ}.Remark. It is clear by the definition of the sets Ks that each φ ∈∪Ks has an analysis.Also one can check that if f1 ∈F s(φ) , f2 ∈F s+1(φ) then either supp(f1) ⊆supp(f2)or supp(f1) ∩supp(f2) = ∅.So let φ ∈Km \ Km−1 .

By the 1-unconditionallity of (ek) we may and will assumethat there is ℓ∈N such that supp(φ) = ∪ℓk=1supp(xk) and the xk s and φ have onlynon-negative coordinates.Definition B. Let φ , (xk)ℓk=1 be as above.

Let {F s(φ)}ms=0 be a fixed analysis of φ .For k = 1, . .

., ℓwe setsk =max{ s : 0 ≤s < m and there are at least two f1, f2 ∈F s(φ)such that fi(xk) > 0, i = 1, 2 },when this set is non −empty0 ,when supp(xk) is a singleton .So for each k = 1, . .

., ℓthere exists a family f k1 , . .

., f kdk , 1 ≤dk ≤n of successivefunctionals in F sk(φ) such that for each i = 1, . .

., dk , f ki (xk) > 0 and– 4 –

supp(xk) ⊆∪dki=1supp(fi) . We define the initial partx′k and the final part x′′k of xkwith respect to {F s(φ)}ms=0 as follows:x′k = xk|supp(f k1 )x′′k = xk| ∪dki=2 supp(f ki )( x′′k = 0 if supp(xk) is a singleton).Our aim is to show thatφ(ℓXk=1x′k) ≤1θ ∥ℓXk=1akek∥andφ(ℓXk=1x′′k) ≤1θ ∥ℓXk=1akek∥Since the proofs of these inequalities are similar we shall only prove the first.

In particularwe shall show by induction on s ≤m that for every J ⊆{1, . .

., ℓ} and every f ∈F s(φ)(∗)|f(Xk∈Jakxk)| ≤1θ ∥Xk∈Jakek∥. (∗) is clear for s = 0 .

Suppose that we know (∗) for s < m ; we shall prove it for s +1 .Let J ⊆{1, . .

., ℓ}. Let f ∈F s+1(φ) , f = θ(f1 + .

. .

+ fd) with d ≤n , (fi)di=1successive members of F s(φ) . Consider the setsK = { k ∈J : there exists i ∈{1, .

. ., d −1} such that fi(x′k) > 0 and fi+1(x′k) > 0 }I = { i : 1 ≤i ≤d and there exists K ∈J such that supp(x′k) ⊂supp(fi) }Claim .

|K| + |I| ≤n .Proof of the claim . Let k ∈K .

Then there exists i < d such that fi(x′k) > 0 andfi+1(x′k) > 0 . This means that s < sk , so there exists f ∈F sk(φ) with supp(fi+1) ⊆supp(f) .

Then f(x′k) > 0 . But if f ∈F sk(φ) and f(x′k) > 0 then by the definition ofx′k , max supp(x′k) = max supp(f) .

So supp(fi+1) ⊂supp(x′k) . Thus i+1 /∈I .

Therefore wecan define a one-to-one map G : K →{1, . .

., d} \ I ; hence |K| ≤d −|I| ≤n −|I| .We proceed with the proof of the inductive step. For i ∈I setEi = { k ∈J : supp(x′k) ⊆supp(fi) } .

Ofcourse Ei ∩K = ∅for every i ∈I . We havef(Xk∈Jakx′k) = θXk∈Jfi(akx′k) = θXi∈Ifi(Xk∈Eiakx′k) +Xk∈K(dXk=1fi)(akx′k).By the inductive hypothesis and the fact that for each k ∈J– 5 –

Pk∈K(Pdk=1 fi)(akx′k) ≤1θ∥akxk∥= 1θ∥akek∥, we getf(Xk∈Jakx′k) ≤θ1θXi∈I∥Ei(Xk∈Jakek)∥+ 1θXk∈K∥akek∥≤1θ ∥Xk∈Jakek∥using the fact that |K| + |I| ≤n and the implicit equation satisfied by the norm.The proof of the inductive step and thus the proof of Step 3 are complete.Step 4For all ℓand all rational non-negative (rj)ℓj=1∥ℓXj=1r1pj ej∥≥12n(ℓXj=1rj)1pProof . Write rj = kjk , kj, k ∈N .

Set s0 = 0 , sj = k1 + . .

. + kj and uj = Psji=sj−1+1 ei ,j = 1, .

. ., ℓ.

By Step 1, ∥uj∥≤k1pj . So∥ℓXj=1r1pj ej∥= 1k1p ∥ℓXj=1k1pj ej∥≥1k1p ∥ℓXj=1∥uj∥ej∥by unconditionallity.By Step 31k1p ∥ℓXj=1∥uj∥ej∥≥θ21k1p ∥ℓXj=1∥uj∥uj∥uj∥∥= θ21k1p ∥ℓXj=1sjXi=sj−1+1ei∥= θ21k1p ∥sℓXi=1ei∥By Step 2, ∥Psℓi=1 ei∥≥1n1p s1pℓso setting θ =1n1qwe getθ21k1p ∥sℓXi=1ei∥≥12nsℓ1pk1p = 12nPℓj=1 kjk 1p = 12nℓXj=1rj 1p .Step 4 and the unconditionallity of (en)n∈N imply that ∥P akek∥≥12n(P |ak|p)1p forall coefficients (ak) .

This fact combined with Step 1 completes the proof of the Theorem.Remark . The result of the Theorem can also be deduced by Steps 1, 2 and 3 using awell known Theorem of Zippin [Z].REFERENCES[T]B. S. Tsirelson, Not every Banach space contains ℓp or co , Funct.

Anal. App.8(1974), p. 138-141[Z]M. Zippin, On perfectly homogeneous bases in Banach spaces, Israel J. of Math.

4(1966), p. 265-272– 6 –

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출처: arXiv:9207.206 • 원문 보기