Banach-Saks properties and spreading models.

Beauzamy, Bernard

doi:10.7146/math.scand.a-11818

Cited by 80 publications

(59 citation statements)

References 7 publications

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“…Assume now that (x n ) has no Cauchy subsequence. We follow in part the line of the proof of Theorem II.2 of [2]. We extract a subsequence (x n ) of (x n ) that is the fundamental sequence of the spreading model F built on (x n ).…”

Section: Resultsmentioning

confidence: 99%

“…The case p = 1 was examined by Szlenk [14] who proved that every weakly convergent sequence in L 1 [0, 1] contains a subsequence with strongly convergent Cesàro means. This variant of the BS property is considered also for operators (see [2]). A bounded linear operator T between Banach spaces X and Y is said to have the weak Banach-Saks (WBS) property if every weakly null sequence (x n ) in X contains a subsequence (x n ) such that (T x n ) is Cesàro convergent in Y .…”

Section: Introduction a Banach Space X Is Said To Have The Banach-samentioning

confidence: 99%

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Deviation from weak Banach–Saks property for countable direct sums

Kryczka¹

2014

Annales UMCS, Mathematica

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Abstract. We introduce a seminorm for bounded linear operators between Banach spaces that shows the deviation from the weak Banach-Saks property. We prove that if (Xν ) is a sequence of Banach spaces and a Banach sequence lattice E has the Banach-Saks property, then the deviation from the weak Banach-Saks property of an operator of a certain class between direct sums E(Xν ) is equal to the supremum of such deviations attained on the coordinates Xν . This is a quantitative version for operators of the result for the Köthe-Bochner sequence spaces E(X) that if E has the Banach-Saks property, then E(X) has the weak Banach-Saks property if and only if so has X. Introduction.A Banach space X is said to have the Banach-Saks (BS) property if every bounded sequence in X contains a subsequence (x n ) whose Cesàro means

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Section: Resultsmentioning

confidence: 99%

Section: Introduction a Banach Space X Is Said To Have The Banach-samentioning

confidence: 99%

Deviation from weak Banach–Saks property for countable direct sums

Kryczka¹

2014

Annales UMCS, Mathematica

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“…The connection between Rosenthal's theorem and the notion of a spreading model was first suggested by L. Tzafriri (see [7]) and was studied in full detail by B. Beauzamy (see [1]). The following corollary is a consequence of Theorem 4.1 (see [1] Proof.…”

Section: Theorem 42mentioning

confidence: 99%

“…More precisely, we assume that (x s ) s∈S is a normalized sequence and ( i ) is a sequence of positive real numbers converging to 0. Then we find a subtree T of S such that for any chain β = {s 1 …”

Section: Theorem 14mentioning

confidence: 99%

Regular methods of summability on tree-sequences in Banach spaces

Poulios

2011

Proc. Amer. Math. Soc.

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Abstract. Suppose that X is a Banach space, a ij is a regular method of summability and (x s ) s∈S is a bounded sequence in X indexed by the dyadic tree S. We prove that there exists a subtree S ⊆ S such that: either (a) for any chain β of S the sequence (x s ) s∈β is summable with respect to a ij or (b) for any chain β of S the sequence (x s ) s∈β is not summable with respect to a ij . Moreover, in case (a) we prove the existence of a subtree T ⊆ S such that if β is any chain of T , then all the subsequences of (x s ) s∈β are summable to the same limit. In case (b), provided that a ij is the Cesàro method of summability and that for any chain β of S the sequence (x s ) s∈β is weakly null, we prove the existence of a subtree T ⊆ S such that for any chain β of T any spreading model for the sequence (x s ) s∈β has a basis equivalent to the usual l 1 -basis. Finally, we generalize the theory of spreading models to tree-sequences. This also allows us to improve the result of case (b).

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“…Fix 0 < < min{η, 1} and put γ = (1−c) /6. Then there exists a weakly null sequence (y n ) in X such that (1 − γ ) |α| + |β| ≤ αy m + βy n ≤ (1 + γ ) |α| + |β| , (2.56) for all m < n and α, β ∈ R (see [3]). Similarly, as in the proof of Theorem 2.5 we can choose m and y ∈ S E for which y m − y < 3γ .…”

Section: Corollary 29 Let X Be a Reflexive Space Then X Is Nuc If mentioning

confidence: 99%