2014
DOI: 10.1007/s10587-014-0157-y
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On the complexity of some classes of Banach spaces and non-universality

Abstract: Abstract. These notes are dedicated to the study of the complexity of several classes of separable Banach spaces. We compute the complexity of the Banach-Saks property, the alternating Banach-Saks property, the complete continuous property, and the LUST property. We also show that the weak Banach-Saks property, the Schur property, the Dunford-Pettis property, the analytic Radon-Nikodym property, the set of Banach spaces whose set of unconditionally converging operators is complemented in its bounded operators,… Show more

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Cited by 7 publications
(9 citation statements)
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“…In particular, these classes are not Σ 1 2 . This result answers two questions posed by B. M. Braga in [6]. We recall that a Banach space X is said to have the Schur property if every weakly convergent sequence in X is norm convergent.…”
Section: 3(vi)])supporting
confidence: 64%
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“…In particular, these classes are not Σ 1 2 . This result answers two questions posed by B. M. Braga in [6]. We recall that a Banach space X is said to have the Schur property if every weakly convergent sequence in X is norm convergent.…”
Section: 3(vi)])supporting
confidence: 64%
“…Therefore, it follows from the Tsirelson space method that the class of all separable Banach spaces with the Schur property is not coanalytic. Using a tree space method developed in [3], it can be shown that this class is not analytic (see [6,Theorem 27]). Our proof of the proper complexity result (Theorem 1.3) can be considered as a combination of these two methods.…”
Section: Lemma 38 There Exists a Borel Mappingmentioning
confidence: 99%
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“…(V) If a separable Banach space X is isomorphically universal for separable Schur spaces, then it is actually isomorphically universal for all separable Banach spaces. This follows from methods in [3] (see [6,Corollary 51]). We are able to prove the isometric version of this statement (see Remark 3.7).…”
Section: Introduction and The Main Resultsmentioning
confidence: 99%
“…For each T ∈ Tr, let X 2,C[0,1],T be the metric space define above. An easy transfinite induction on the order of T gives us that, for all well-founded trees T ∈ Tr, the space X 2,C[0,1],T has the Banach-Saks property (see [Bra14], Theorem 14). Also, if T ∈ IF, it is clear that C[0, 1] linearly isometrically embeds into X 2,C[0,1],T .…”
Section: Complexity Of Some Asymptotic Notions and Applicationsmentioning
confidence: 99%