On the Dunford-Pettis property in spaces of vector-valued bounded functions

Abstract

We show that £°°(ji, X) has the Dunford-Pettis property for some classical Banach spaces including L l (/i), C(K), the disc algebra A and H*".A Banach space X is said to have the Dunford-Pettis property if every weakly compact operator from X into an arbitrary Banach space is completely continuous, or equivalently, if given sequences (x n ) in X and (x^) in X*, both weakly convergent to zero, then (x^,x n ) tends to zero. A detailed exposition about this property can be found in [6]. In this reference, the following problem is posed [6, p.55]: assume that L°°(IJ,, X) denotes the Banach space of (equivalence classes of) essentially bounded, measurable and X-valued functions denned over a finite measure space (fl, E, fi). When does L°°(ii,X) have the Dunford-Pettis property? In general, this property does not lift from X to L°°(fi,X) [6, p.56]. On the other hand, the only non-trivial positive result, as far as we know, is that L°°(JJ,, £ 1 (i/)) has the property when fi is purely atomic [3, Theorem 1], The aim of this note is to provide some new positive examples. Namely, we show that L°° (IJ,,X) has the Dunford-Pettis property for every arbitrary finite measure \i, whenever X is either any £ 1 -space or any £°°-space or the disc algebra or the space H°° of bounded analytic functions on the disc.To avoid trivial situations, we always assume that there exists a pairwise disjoint sequence (C m ) in S such that fi{C m ) > 0. The notation is standard except, perhaps, the following one: if (A m ) is a sequence of pairwise disjoint S-measurable subsets of non-zero measure, we write =l J It is well-known that [A m ] is a complemented subspace of L°°((J.,X) isometrically isomorphic to l°°(X). In particular, if L°°(IJ.,X) has the Dunford-Pettis property,

The Dunford-Pettis Property for Symmetric Spaces

“…The sufficiency follows by [13] and [10]. In fact, in [13] it is shown that for finite measure space (Ω, µ), L ∞ (X) has the Dunford-Pettis property if and only if ∞ (X) has it.…”

“…The sufficiency follows by [13] and [10]. In fact, in [13] it is shown that for finite measure space (Ω, µ), L ∞ (X) has the Dunford-Pettis property if and only if ∞ (X) has it. It is also proved in [13] that if either X is any L 1 -space or any L ∞ -space, then L ∞ (X) has the Dunford-Pettis property.…”

“…In fact, in [13] it is shown that for finite measure space (Ω, µ), L ∞ (X) has the Dunford-Pettis property if and only if ∞ (X) has it. It is also proved in [13] that if either X is any L 1 -space or any L ∞ -space, then L ∞ (X) has the Dunford-Pettis property. On the other hand in [10] it is proved that if X is any L 1 -space or any L ∞ -space, then L 1 (X) has the Dunford-Pettis property.…”