Publication: The Weak Banach-Saks Property On L(P)(Mu,E)
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1994
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Cambridge Univ Press
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Abstract
A Banach space E is said to have the Banach-Saks property (BS) if every bounded sequence {xn} in E has a subsequence with norm convergent Ces`aro means, i.e., 1 in E. If this occurs for every weakly convergent sequence in E, it is said that E has the weak Banach-Saks property (WBS).
It is known that uniformly convex spaces are BS,and E is BS iff E is WBS and reflexive. The spaces c0, `1, and L1 are WBS, whereas 1 and C[0, 1] are not. The BS and WBS properties do not pass from E to Lp(μ;E); in fact,
L2(c0) is not WBS [D. J. Aldous, Math. Proc. Camb. Philos. Soc. 85, 117-123 (1979;Zbl 0389.46027] and Bourgain constructed a Banach-Saks space E for which L2(E) is
not BS. Bourgain showed when the Banach-Saks property holds for Lp(μ,E) by using a property of L1 due to Koml´os; For every bounded sequence {fn} in L1(μ), there exists a
subsequence {f0 n } of {fn} and a f in L1(μ) such that 1
k Pk n=1 f0 n ! f almost everwhere for each subsequence {f0
n } of {f0 n }. When this holds in L1(μ,E), we say that L1(μ,E) has the Koml´os property. Bourgain showed that L1(μ,E) has the Koml´os property iff Lp(μ, e) is BS for some p 2 (1,1) iff Lp(μ,E) is BS for all p 2 (1,1). The author
uses ideas inspired by Bourgain’s work to similarly characterize WBS. She says that L1(μ,E) has the weak Koml´os property if every weakly null sequence {'n} in L1(μ,E)has the subsequence {'0 n} such that | 1 k Pk n=1 '0
n(·)| ! 0 almost everywhere for each subsequence {'0 n 0}of{'0 n}. She proves that L1(μ,E) is weak Banach-Saks iff Lp(μ,E)is WBS for some p 2 [1,1) iff Lp(μ,E) is WBS for all p 2 [1,1) iff L1(μ,E) is weak Koml´os. For example, if E is a B-convex Banach space, then L1(μ,E) is weak Koml´os and the above properties hold.