2014
DOI: 10.1016/j.jmaa.2013.10.085
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Almost limited sets in Banach lattices

Abstract: Abstract. We introduce and study the class of almost limited sets in Banach lattices, that is, sets on which every disjoint weak * null sequence of functionals converges uniformly to zero. It is established that a Banach lattice has order continuous norm if and only if almost limited sets and L -weakly compact sets coincide. In particular, in terms of almost Dunford-Pettis operators into c0, we give an operator characterization of those σ-Dedekind complete Banach lattices whose relatively weakly compact sets a… Show more

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Cited by 51 publications
(39 citation statements)
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“…Recall from [4] that a subset A of a Banach lattice E is said to be almost limited if sup x∈A | f n (x)| → 0 holds for any disjoint weak* null sequence ( f n ) in E . In a standard style, we say that an operator T : E → F is almost limited if it maps the closed ball B E to an almost limited set of F. It is clear that an operator T : E → F is almost limited if and only if it maps norm bounded sequences in E to almost limited sets in F.…”
Section: Relations With Other Classes Of Operatorsmentioning
confidence: 99%
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“…Recall from [4] that a subset A of a Banach lattice E is said to be almost limited if sup x∈A | f n (x)| → 0 holds for any disjoint weak* null sequence ( f n ) in E . In a standard style, we say that an operator T : E → F is almost limited if it maps the closed ball B E to an almost limited set of F. It is clear that an operator T : E → F is almost limited if and only if it maps norm bounded sequences in E to almost limited sets in F.…”
Section: Relations With Other Classes Of Operatorsmentioning
confidence: 99%
“…(ii) ⇒ (i) We apply [4,Theorem 2.5]. Let (x n ) be a disjoint sequence in B + E and ( f n ) a disjoint weak* null sequence in (E ) + .…”
Section: Lemma 32 For a Dedekind σ -Complete Banach Lattice E The Fmentioning
confidence: 99%
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“…In fact, if we consider E = ℓ ∞ and F = c, the Banach lattice of all convergent sequences, it is clear that F = c is not σ-Dedekind complete, and it follows from the proof of Proposition 1 of [7] and Theorem 5.99 of [1] Proof: The result follows from Theorem 2.7 by noting that if E has the positive Schur property then the norm of E is order continuous. Now, from Corollary 2.8 and Theorem 4.9 (Nakano) of [1], we obtain the following result, which is just Proposition 3.3 of [2].…”
Section: ) Each Positive Almost Weak ⋆ Dunford-pettis Operator T : Ementioning
confidence: 71%
“…we see that h n (T (z n+1 )) ≥ ǫ 2 must hold for all n sufficiently large (because 2 −n h(T (z n+1 )) → 0), which contradicts with our hypothesis (2).…”
mentioning
confidence: 72%