An Asymptotic Property of Schachermayer s Space under Renorming

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1 Journal of Mathematical Analysis and Applications 50, ) doi: /jmaa , available online at on An Asymptotic Property of Schachermayer s Space under Renorming Denka Kutzarova 1 Institute of Mathematics, Bulgarian Academy of Sciences, Sofia, Bulgaria denka@math.sc.edu and Denny H. Leung Department of Mathematics, National University of Singapore, Singapore matlhh@nus.edu.sg Submitted by Jean Mawhin Received February 8, 000 Let X be a Banach space with closed unit ball B. Given k, X is said to be k-β, respectively, k + 1 -nearly uniformly convex k + 1 -NUC), if for every ε>0there exists δ, 0<δ<1, so that for every x B and every ε-separated sequence x n B there are indices n i k, respectively, n i k+1, such that 1/ k + 1 x + k x n i 1 δ, respectively, 1/ k + 1 k+1 x n i 1 δ. It is shown that a Banach space constructed by Schachermayer is -β, but is not isomorphic to any -NUC Banach space. Modifying this example, we also show that there is a -NUC Banach space which cannot be equivalently renormed to be 1-β. 000 Academic Press Key Words: nearly uniform convexity; renorming; Schachermayer s space. 1. INTRODUCTION In [4], Huff introduced the notion of nearly uniform convexity NUC). A Banach space X with closed unit ball B is said to be NUC if for any ε>0 there exists δ<1 such that for every ε-separated sequence in B, co x n δb. Here co A denotes the convex hull of a set A; a sequence x n 1 Current address: Department of Mathematics, University of South Carolina, SC X/00 $35.00 Copyright 000 by Academic Press All rights of reproduction in any form reserved. 670

2 schachermayer s space 671 is ε-separated if inf x n x m m n ε. Huff showed that a Banach space is NUC if and only if it is reflexive and has the uniform Kadec Klee property UKK). Recall that a Banach space X with closed unit ball B is said to be UKK if for any ε>0 there exists δ<1 such that for every ε-separated sequence x n in B which converges weakly to some x X we have x δ. A recent result of Knaust et al. [5] gives an isomorphic characterization of spaces having NUC. They showed that a separable reflexive Banach space X is isomorphic to a UKK space if and only if X has a finite Szlenk index. More recent results concerning Szlenk indices and renormings are to be found in [, 3]. Another property related to NUC is the property β) introduced by Rolewicz [11]. In [6], building on the work of Prus [9, 10], the first author showed that a separable Banach space X is isomorphic to a space with β) if and only if both X and X are isomorphic to NUC spaces. In [7], a sequence of properties lying in between β) and NUC is defined. Let X be a Banach space with closed unit ball B. Given k, X is said to be k-β, respectively, k + 1 -nearly uniformly convex k + 1 -NUC), if for every ε>0 there exists δ, 0<δ<1, so that for every x B and every ε-separated sequence x n B there are indices n i k, respectively n i k+1, such that 1 k k + 1 x + x ni 1 δ respectively 1 k+1 k + 1 x ni 1 δ It follows readily from the definitions that every k-β space is k + 1 -NUC, every k + 1 -NUC space is k + 1 -β, and that every k-β space or k NUC space) is NUC. It is proved in [7] that property 1-β is equivalent to the property β) of Rolewicz. It is worth noting that the non-uniform version of property k-nuc has been well-studied. For k, a Banach space X is said to have property kr) if every sequence x n in X which satisfies lim n1 lim nk x n1 + +x nk =k lim n x n is convergent [1]. It is clear that the property kr) implies property k + 1 R). It follows from James characterization of reflexivity that every kr) space is reflexive. A recent result of Odell and Schlumprecht [8] shows that a separable Banach space is reflexive if and only if it can be equivalently renormed to have property R). Thus, all the properites kr) are isomorphically equivalent. Similarly, non-asymptotic properties known as k-uniform rotundity have been studied [13]. These properites are also isomorphically equivalent to each other as they are all equivalent to superreflexivity. In this paper, we find that

3 67 kutzarova and leung the situation is different for the properites k-nuc and k-β. To be precise, we use the space constructed by Schachermayer in [1] and a variant to distinguish the properties 1-β, -NUC, and -β isomorphically. Let T = n=0 0 1 n be the dyadic tree. If ϕ = ε i m and ψ = δ i n are nodes in T, we say that ϕ ψ if m n and ε i = δ i for 1 i m. Also, ϕ for all ϕ T. Two nodes ϕ and ψ are said to be comparable if either ϕ ψ or ψ ϕ; they are incomparable otherwise. Let ϕ T, denote by T ϕ or T ϕ the subtree rooted at ϕ, i.e., the subtree consisting of all nodes ψ such that ϕ ψ. A node ϕ T has length n if ϕ 0 1 n. The length of ϕ is denoted by ϕ. Given ϕ = ε i n T, let S ϕ be the set consisting of all nodes ψ = δ i m such that m n, δ i = ε i if 1 i n, and δ i = 0 otherwise. Say that a subset A of T is admissible, respectively, acceptable, if there exists n 0 such that a) A ϕ =n T ϕ and b) A T ϕ 1 for all ϕ with ϕ =n, respectively, a ) A ϕ =n S ϕ, and b ) A S ϕ 1 for all ϕ with ϕ =n. For subsets A and B of T, say that A B if max ϕ ϕ A < min ϕ ϕ B. Letc 00 T be the space of all finitely supported real-valued functions defined on T.Forx c 00 T, let x X = sup k ϕ A i x ϕ ) ) 1/ where the sup is taken over all k and all sequences of admissible subsets A 1 A A k of T. The norm Y is defined similarly except that the sup is taken over all sequences of acceptable subsets A 1 A A k of T. Schachermayer s space X is the completion of c 00 T with respect to the norm X. The completion of c 00 T with respect to Y is denoted by Y. Remark. The space X defined here differs from Schachermayer s original definition and is only isomorphic to the space defined in [1]. In [7], it was shown that X with the norm given in [1]) is 8-NUC but is not isomorphic to any 1-β space. We first show that X X and Y Y are -β and -NUC respectively. We begin with a trivial lemma concerning the l -norm. Lemma 1. If α, β, and γ are vectors in the unit ball of l, and α + β + γ /3 1 δ, then max α β α γ β γ 18δ. Proposition. X X is -β. Proof. Let x and x n n 1, be elements in the unit ball of X such that x n is ε-separated. Choose δ>0 such that [ 1 1 3δ + 4δ 1/ 1 ε /9 ) ] 1/ > 1 1)

4 schachermayer s space 673 Without loss of generality, we may assume that x n converges pointwise as a sequence of functions on T ) to some y 0 T. It is clear that if y z X and supp y supp z, then y + z X y X + z X. It follows easily that y 0 X. Lety n = x n y 0. It may be assumed that y n X converges. As x n is ε-separated, so is y n. We may thus further assume that y n X > ε/3 for all n. By going to a subsequence and perturbing the vectors x, y 0, and y n n 1, by as little as we please, it may be further assumed that a) they all belong to c 00 T, b) supp x supp y 0 supp y 1 supp y, and c) y 1 χ Tϕ = y χ Tϕ for all ϕ such that ϕ M, where is the sup norm and M = max ψ ψ supp x supp y 0. Claim. Let A be an admissible set such that min ϕ ϕ A M. If ϕ A y 1 ϕ = c, and ϕ A y ϕ = d, then there exists an admissble set B such that min ϕ ϕ A min ϕ ϕ B max ϕ ϕ B max ϕ ϕ A A supp y 0 B, and ϕ B y 1 ϕ c + d. To prove the claim, let N be such that A ϕ =N T ϕ and let A T ϕ 1 for all ϕ with ϕ =N. Then N M. Now, for each ψ A supp y, ψ T ϕ for some ϕ with ϕ =N M. It follows that y 1 χ Tϕ = y χ Tϕ y ψ Hence, there exists a ψ T ϕ such that y 1 ψ y ψ. Now let B = A supp y 0 supp y 1 ψ ψ A supp y It is easy to see that the set B satisfies the claim. Suppose that x + x 1 + x X /3 1 δ. Letx + x 1 + x = x + y 0 + y 1 + y be normed by a sequence of admissible sets A 1 A A k. Denote by α = a i k, β = b i k, γ = c i k, and η = d i k respectively the sequences ϕ A i x ϕ k, ϕ A i y 0 ϕ k, ϕ A i y 1 ϕ k, and ϕ A i y ϕ k. Now α + β + γ + β + η /3 x + x 1 + x X /3 1 δ But α x X 1. Similarly, β + γ, β + η 1. By Lemma 1, we obtain that α β γ, α β η, and γ η are all 18δ. Let j be the largest integer such that a j 0. Note that this implies supp x A j ; hence supp y 1 supp y ) A i = for all i<j. Thus, c i = d i = 0 for all i<j. Now b j+1 + d j+1 b k + d k α β η 18δ )

5 674 kutzarova and leung Moreover, Hence Therefore, 1 x = y X 0 + y y X 0 + y X β + y X X β 1 ε /9 3) 3 1 δ α + β + γ + β + η + β + η 1 3δ β + η = ) b 1 b j 1 b j + d j + ) b j+1 + d j+1 b k + d k ) ) b 1 b j 1 b j + d j + 18δ by ) β + d j ) + 18δ 1 ε /9 ) 1/ + dj ) + 18δ by 3) d j 1 4δ 1/ 1 ε /9 ) 1/ 4) Note that by the first part of the argument above we also obtain that β + γ 1 3δ 5) Since A j supp x, we may apply the claim to obtain an admissible set B. Using the sequence of admissible sets A 1 A j 1 B A j+1 A k to norm x 1 = y 0 + y 1 yields 1 y 0 + y 1 ) b X 1 b j 1 b j + c j + d j b j+1 + c j+1 b k + c k ) b 1 b j 1 b j + c j b j+1 + c j+1 b k + c k + d j = β + γ + d j [ 1 1 3δ ) 1/ + 4δ 1 ε /9 ) ] 1/ by 5) and 4). As the last expression is >1 by 1), we have reached a contradiction. Remark. The same method can be used to show that X is -β with the norm given in [1]. Proposition 3. Y Y is -NUC.

6 schachermayer s space 675 Proof. Let x n be an ε-separated sequence in the unit ball of Y. Choose δ>0 so that δ = 1δ + 8δ ε /18 6) and 1 δ + 8 δ> 1 ε/3 7) As in the proof of the previous proposition, it may be assumed that there exists a sequence y n n=0 in Y such that x n = y 0 + y n, supp y n 1 supp y n for all n, and y j χ Sϕ = y k χ Sϕ whenever ϕ M i and j k > i, where M i = max ψ ψ supp y i. We may also assume that y n Y converges. Since y n n=1 is ε-separated, η = lim y n Y ε/. The choice of δ in 6) guarantees that 4 η δ 1/ > 7η/ 3η + δ. Hence there exist η + >η>η >ε/3 such that 4θ 3η + + η + η + δ 8) where θ = η δ. We may now further assume that η + y n Y η for all n. Now suppose that x m + x n Y / > 1 δ for all m n. Claim. For all m<nin, there exists an acceptable set A such that ϕ A y i ϕ >θfor i = m n. First observe that there are acceptable sets A 1 A A k such that k ϕ A i y 0 + y m + y n ϕ > 4 1 δ.letα = a i k, β = b i k, and γ = c i k be the sequences ϕ A i y j ϕ k for j = 0 m n, respectively. Then α + β + γ > 1 δ and α + β y 0 + y m Y = x m Y 1. Similarly, α + γ 1. It follows from the parallelogram law that β γ < δ 8δ. Note also that α + β α + β + γ α + γ > 1 δ. Similarly, α + γ > 1 δ. Letj 1, respectively j, be the largest j such that a j 0, respectively b j 0. Since supp y 0 A j1, b 1 = = b j1 1 = 0. Similarly, c 1 = = c j 1 = 0. Moreover, j 1 j. Let us show that j 1 <j. For otherwise, j 1 = j = j. Then b j c j β γ < 8δ 9) Consider the set A j. Choose p 0 such that A j ϕ =p S ϕ and A j S ϕ 1 for all ϕ with ϕ =p. Note that p M 0.LetG = ϕ ϕ =p A j S ϕ supp y m.ifϕ G, y n χ Sϕ = y m χ Sϕ. Hence there exists ψ ϕ S ϕ supp y n such that y n ψ ϕ = y m χ Sϕ. It is easy to see that the set B = ψ ϕ ϕ G A j supp y 0 A j supp y n

7 676 kutzarova and leung is acceptable and that min ϕ ϕ A j min ϕ ϕ B. Hence A 1 A j 1 B. Thus by 9), 1 x n Y = y 0 + y n j 1 Y a i + y 0 + y n ϕ ϕ B j 1 a i + y 0 ϕ + y n ϕ + y n ψ ϕ ϕ A j ϕ A j ϕ G j 1 a i + a j + c j + ) y m χ Sϕ ϕ G j 1 a i + a j + c j + y m ϕ ϕ A j ) a 1 a j 1 a j + b j + c j a 1 a j 1 a j + b j ) α + β + b j 8δ ) > 1 δ + β 8δ) + c j ) ) ) Therefore, β < + 8 δ. It follows that α α + β β > 1 δ + 8) δ 10) However, α y 0 Y x m Y y m Y 1 η < 1 ε/3 11) Combining 10) and 11) with the choice of δ in 7) yield a contradiction. This shows that j 1 <j. Applying the facts that α + β > 1 δ and b j1 b j 1 β γ < 8δ, we obtain that b j > 1 δ α + ) 8δ 1 δ 1 η + 8δ) θ

8 schachermayer s space 677 Similarly, 1 δ < α + γ = α + c j + cj +1 c k ) α + c j + β γ 1 η ) + c j + 8δ Hence c j >θ. Thus the set A = A j satisfies the requirements of the claim. Taking m = 1, n =, and m =, n = 3, respectively, we obtain acceptable sets A and A from the claim. Since A supp y 1,ifϕ A supp y, ϕ S ϕ for some ϕ such that ϕ M 1. This implies that there exists ψ ϕ S ϕ such that y 3 ψ ϕ = y 3 χ Sϕ = y χ Sϕ y ϕ. Let q = min ϕ ϕ supp y 3 and = σ T σ =q. Forσ, define s σ = y 3 ψ ϕ if there exists ϕ A supp y such that ψ ϕ S σ ; otherwise, let s σ =0. Also, let t σ = y 3 ϕ if there exists ϕ A supp y 3 S σ ; otherwise, let t σ =0. Finally, let r σ = y 3 χ Sσ for all σ. Then r σ s σ 0 for all σ, σ r σ y 3 Y <η +, and σ s σ >θ. Hence σ r σ s σ <η + θ. Similarly, σ r σ t σ <η + θ. Therefore, σ t σ s σ < η + θ. LetB be the set of all nodes in A supp y that are comparable with some node in A supp y 3. Then y ϕ y 3 ψ ϕ t σ s σ < η + θ σ ϕ A\B ϕ A\B Hence ϕ B y ϕ >θ η + θ =3θ η +. Now let l = min ϕ ϕ A supp y. Divide B into B 1 = ϕ B ϕ <l and B = ϕ B ϕ l. Since B 1 and A supp y are acceptable sets such that B 1 A supp y, ) η+ > y y Y ϕ ) + y ϕ ) ϕ B 1 ϕ A > y ϕ ) + θ ϕ B 1 Thus ϕ B y ϕ > 3θ η + η + θ

9 678kutzarova and leung Finally, since B A supp y is acceptable, η + > y Y y ϕ + y ϕ ϕ B ϕ A supp y > 3θ η + This contradicts inequality 8). η + θ + θ Before proceeding further, let us introduce some more notation. A branch in T is a maximal subset of T with respect to the partial order. If γ is a branch in T and n 0, let ϕ γ n be the node of length n in γ. A collection of pairwise distinct branches is said to have separated at level L if for any pair of distinct branches γ and γ in the collection the nodes of length L belonging to γ and γ respectively are distinct. Finally, if γ 1 γ k is a sequence of pairwise distinct branches which have separated at a certain level L, we say that a sequence of nodes ϕ 1 ϕ k S γ 1 γ k L if ϕ i T ϕ γ i L,1 i k. Let us note that in this situation χ ϕi 1 i k X = k. Suppose is an equivalent norm on X which is -NUC. It may be assumed that there exists ε>0 so that ε x X x x X for all x X. Letδ = δ ε > 0 be the number obtained from the definition of -NUC for the norm. Proposition 4. Let n 0. Then there are pairwise incomparable nodes ϕ 1 ϕ n such that whenever γ i γ i are distinct branches passing through ϕ i, 1 i n, and γ i γ i 1 i n have separated at level L, there is a sequence of nodes ψ 1 ψ n+1 S γ 1 γ 1 γ n γ n L) satisfying χ ψi 1 i n+1 1 δ n+1. Proof. Assume that n is the first non-negative integer where the proposition fails. Let ϕ 1 ϕ n 1 be the nodes obtained by applying the proposition for the case n 1. If n = 0, begin the argument with any node ϕ 1.) For each i, 1 i n 1, let ψ i 1 1 and ψ i 1 be a pair of incomparable nodes in T ϕi. If n = 0, let ψ 1 1 be any node in T ϕ1.) Since the proposition fails for the nodes ψ 1 1 ψ n 1, there are distinct branches γ i 1, γ i 1 passing through ψ i 1, 1 i n, and a number L 1 so that γ i 1 γ i 1 1 i n have separated at level L 1, but χ ξi 1 i n+1 > 1 δ n+1 for any sequence of nodes ξ 1 ξ n+1 S γ 1 1 γ 1 1 γ n 1 γ n 1 L 1. However, since the proposition holds for the nodes ϕ 1 ϕ n 1, we obtain a sequence of nodes ξ 1 1 ξ n 1 S γ 1 1 γ n 1 L 1 such that χ ξi 1 1 i n 1 δ n

10 schachermayer s space 679 Note that the preceding statement holds trivially if n = 0.) For each i, choose a node ψ i in γ i 1 such that ψ i >L 1. Then ψ i 1 and ψ i are a pair of incomparable nodes in T ϕi, and the argument may be repeated. If n = 0, repeat the argument using the node ψ 1.) Inductively, we thus obtain sequences of branches γ 1 r γ 1 r γ n r γ n r r=1, a sequence of numbers L 1 <L <, and sequences of nodes ξ 1 r ξ n r r=1 such that 1. the branches γ i r γ i r 1 i n have separated at level L r, r 1,. χ ξi 1 i n+1 > 1 δ n+1 for any sequence of nodes ξ 1 ξ n+1 S γ 1 r γ 1 r γ n r γ n r L r) 3. ξ 1 r ξ n r S γ 1 r γ n r L r), and χ ξi r 1 i n 1 δ n r 1 4. ξ i r T ϕ λ ) i s L s whenever r>s, and 1 i n. It follows that if r>s, then ξ 1 r ξ 1 s ξ n r ξ n s S γ 1 s γ 1 s γ n s γ n s L s ) 1) Let x r = 1 δ n χ ξi r 1 i n, r 1. By Item 3, x r 1. Moreover, because of 1), if r>s, then x r x s ε x r x s X = n+1 ε/ 1 δ n ε Thus x r is ε-separated in the norm. By the choice of δ, there are r>ssuch that x r + x s / 1 δ. Therefore, χ ξ1 r ξ 1 s ξ n r ξ n s 1 δ n+1. But this contradicts Item and the condition 1). Theorem 5. There is no equivalent -NUC norm on X. Proof. In the notation of the statement of Proposition 4, we obtain, for each n, nodes ψ 1 ψ n+1 such that χ ψi 1 i n+1 1 δ n+1 and χ ψi 1 i n+1 X = n+1. Hence cannot be an equivalent norm on X. The proof that the space Y has no equivalent 1-β norm follows along similar lines. Suppose that is an equivalent 1-β norm on Y.We may assume that ε Y Y for some ε>0. Let δ = δ ε be the constant obtained from the definition of 1-β for the norm. Let n 0 and denote the set ϕ T ϕ =n by. Proposition 6. For any m, 0 m n, any subset of with = m, and any p, there exists an acceptable set A ϕ S ϕ such that A = m, min ϕ ϕ A p, and χ A m 1 δ m.

11 680 kutzarova and leung Proof. The case m = 0 is trivial. Suppose the proposition holds for some m, 0 m<n.let, = m+1, and let p. Divide into disjoint subsets 1 and such that 1 = = m.bythe inductive hypothesis, there exist acceptable sets B and C j, j, such that B ϕ 1 S ϕ, B = m, min ϕ ϕ B p, and χ B m 1 δ m ; and also C j ϕ S ϕ, C j = m, min ϕ ϕ C 1 p, C j C j+1, and χ Cj m 1 δ m for all j. It is easily verified that the sequence ) m 1 δ m χ Cj is ε-separated and has norm bounded by 1 with respect to. It follows that there exists j 0 such that m 1 δ m χ B + χ Cj0 1 δ. The induction is completed by taking A to be B C j0. Using the same argument as in Theorem 5, we obtain Theorem 7. There is no equivalent 1-β norm on Y. We close with the obvious problem. Problem. For k 3, can every k-nuc Banach space, respectively, k-β Banach space, be equivalently renormed to be k 1 -β, respectively, k-nuc? REFERENCES 1. K. Fan and I. Glicksberg, Fully convex normed linear spaces, Proc. Nat. Acad. Sci. USA ), G. Godefroy, N. J. Kalton, and G. Lancien, Subspaces of c 0 and Lipschitz isomorphisms, preprint. 3. G. Godefroy, N. J. Kalton, and G. Lancien, Szlenk indices and uniform homeomorphisms, preprint. 4. R. Huff, Banach spaces which are nearly uniformly convex, Rocky Mountain J. Math ), H. Knaust, E. Odell, and Th. Schlumprecht, On asymptotic structure, the Szlenk index and UKK properties in Banach spaces, Positivity 31999), D. Kutzarova, An isomorphic characterization of property β) of Rolewicz, Note Mat ), D. Kutzarova, k-β and k-nearly uniformly convex Banach spaces, J. Math. Anal. Appl ), E. Odell and Th. Schlumprecht, Asymptotic properties of Banach spaces under renorming, J. Amer. Math. Soc ), S. Prus, Finite dimensional decomposition s of Banach spaces with p q -estimates, Dissertationes Math ), S. Prus, Nearly uniformly smooth Banach spaces, Boll. Un. Mat. Ital. B 3) 71989), S. Rolewicz, On uniform convexity and drop property, Studia Math ), W. Schachermayer, The class of Banach spaces which do not have c 0 as a spreading model is not L -hereditary, Ann. Inst. H. Poincaré ), F. Sullivan, A generalization of uniformly rotund Banach spaces, Can. J. Math ),