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2008
DOI: 10.1007/s00209-008-0389-3
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Relatively weakly open sets in closed balls of Banach spaces, and the centralizer

Abstract: We prove that, if the centralizer of a Banach space X is infinite-dimensional, then every nonempty relatively weakly open subset of the closed unit ball of X has diameter equal to 2. This result, together with a suitable refinement also proven in the paper, contains (and improves in some cases) previously known facts for C * -algebras, J B * -triples, spaces of vector valued continuous functions, and spaces of operators.

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Cited by 7 publications
(5 citation statements)
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“…Indeed, it is known that the image of Z (X) under this embedding is contained in Z (X (∞ ) (see [10,Proposition 4.3]). …”
Section: Lemma 25 Let X and Y Be Banach Spaces And Assume That Z (Xmentioning
confidence: 97%
See 1 more Smart Citation
“…Indeed, it is known that the image of Z (X) under this embedding is contained in Z (X (∞ ) (see [10,Proposition 4.3]). …”
Section: Lemma 25 Let X and Y Be Banach Spaces And Assume That Z (Xmentioning
confidence: 97%
“…In the case of the injective tensor product, it will be enough to assume one restriction only to one of the spaces in order to obtain a positive result, as we will see later. [10,Theorem 4.4].) Let X be a Banach space failing the diameter two property.…”
Section: Injective Tensor Productmentioning
confidence: 99%
“…Afterwards, we give the necessary definitions and background in order to prove Theorem 3.1. We follow the notation from [9]; given a Banach space E, we denote by C(E) the Cunningham algebra of E (see the paragraph preceding Theorem 3.3 for a formal definition) and by E (∞ the completion of the normed space ∞ n=0 E (2n , where (E (2n ) ∞ n=0 is the sequence of even duals such that E ⊆ E * * ⊆ E (4 ⊂ . .…”
Section: Octahedral Norms In Free Banach Lattices In Terms Of the Cun...mentioning
confidence: 99%
“…Let p be an extreme point of BX. Then p(t)=1,tK, by [, Lemma 2.1]. Consider the analytic mapping given by (λi)1nCnLpx+12T(λi)1npX. We claim that Lp((λi)1n<1 if (λi)1n1. Indeed, if tK, then either there exists a unique kS such that tOk or tKiSOi. If tOk, we have that Lp((λi)1n)(t)x(t)+12|λk|mk(t)p(t)13+12<1.While if tKiSOi, then Lp((λi)...…”
Section: Banach Function Modulesmentioning
confidence: 99%