Weak Precompactness in the Space of Vector-Valued Measures of Bounded Variation

Ghenciu, Ioana

doi:https://doi.org/10.1155/2015/213679

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Research Article | Open Access

Volume 2015 | Article ID 213679 | https://doi.org/10.1155/2015/213679

Weak Precompactness in the Space of Vector-Valued Measures of Bounded Variation

Ioana Ghenciu¹

Academic Editor: José Rodríguez

Received16 Oct 2014

Revised15 Dec 2014

Accepted18 Dec 2014

Published06 Jan 2015

Abstract

For a Banach space and a measure space , let be the space of all -valued countably additive measures on of bounded variation, with the total variation norm. In this paper we give a characterization of weakly precompact subsets of .

1. Introduction

Let be a Banach space, a measure space, and the space of all -valued countably additive measures on of bounded variation (with the total variation norm). For we denote by its variation. For , a probability measure on , we denote .

Ulger [1] and Diestel et al. [2] gave a characterization of weakly compact subsets of , the Banach space of all -valued Bochner integrable functions on a probability space . In [3] we gave a characterization of weakly precompact subsets of . Randrianantoanina and Saab [4] gave a characterization of relatively weakly compact subsets of .

In this paper we use results of Talagrand [5], Ulger [6], and techniques of Randrianantoanina and Saab [4] to characterize weakly precompact subsets of . The characterization is obtained in two steps. In the first step we characterize the weakly precompact subsets of . We show that a subset of is weakly precompact if and only if for any sequence in and for any lifting of there exists a sequence with for each such that, for a.e. , the sequence is weakly Cauchy. In the second step we show that a subset of is weakly precompact if and only if there is a probability measure on such that, for any sequence in , there is a sequence with for each such that, for any , there is a positive integer and a weakly precompact subset of so that , where denotes the unit ball of .

This paper also contains several corollaries of these results. We show that if , then a subset of is weakly precompact if and only if is bounded and is uniformly countably additive.

2. Definitions and Notation

Throughout this paper, and will denote Banach spaces. The unit ball of will be denoted by . The unit basis of will be denoted by , and a continuous linear transformation will be referred to as an operator. The set of all compact operators from to will be denoted by . The set of all continuous compact operators from to will be denoted by .

A bounded subset of is said to be weakly precompact provided that every sequence from has a weakly Cauchy subsequence [5]. For a subset of , let denote the convex hull of . A series in is said to be weakly unconditionally convergent (wuc) if for every the series is convergent. An operator is weakly precompact if is weakly precompact and unconditionally converging if it maps weakly unconditionally convergent series to unconditionally convergent ones.

A bounded subset of (resp., of ) is called a -subset of (resp., a -subset of ) provided that for each wuc series in (resp., wuc series in ).

In his fundamental paper [7], Pełczyński introduced property and property . The Banach space has property (resp., ) if every -subset of (resp., -subset of ) is relatively weakly compact. The following results were also established in [7]: spaces have property ; -spaces have property ; the Banach space has property if and only if every unconditionally converging operator from to any Banach space is weakly compact; if has property , then has property ; every weakly Cauchy sequence in (resp., in ) is a -set (resp., a -set); consequently, every bounded weakly precompact set in (resp., in ) is a -set (resp., a -set). A Banach space has property weak if any -subset of is weakly precompact [8]. A Banach space has property weak if every -subset of is weakly precompact [9].

The Banach-Mazur distance between two isomorphic Banach spaces and is defined by , where the infimum is taken over all isomorphisms from onto . A Banach space is called an -space (resp., -space) [10] if there is a so that every finite dimensional subspace of is contained in another subspace with (resp., ) for some integer . Complemented subspaces of spaces (resp., spaces) are -spaces (resp., -spaces) (see [10, Proposition 1.26]). The dual of an -space (resp., -space) is an -space (resp., -space) (see [10, Proposition 1.27]).

Suppose is a compact Hausdorff space, and are Banach spaces, is the Banach space of all continuous -valued functions (with the supremum norm), and is the -algebra of Borel subsets of . It is known from [11] that .

If is a finitely additive vector measure and , then , , . For each , is a finitely additive vector measure.

Every continuous linear function may be represented by a vector measure of finite semivariation (see [11, 12], and [13, page 182]) such that and , , where denotes the semivariation of . We denote this correspondence . We note that, for , even if is not -valued. A representing measure is called strongly bounded if for every decreasing sequence in , and an operator is called strongly bounded if is strongly bounded [11].

By Theorem 4.4 of [11], a strongly bounded representing measure takes its values in . It is well known that if is unconditionally converging, then is strongly bounded [14].

Let be a probability measure on , with , and let be a lifting of (see [12, 15]). For each , the scalar measure has a density (see [4, 5]). We define to be the element of defined by

It is well known (see [12, sect. , Theorem 5, page 269]) that,(i)for every , the map is -integrable;(ii)for every and all , (iii)the map is -integrable and for every ,

If is a dual space, then we define to be the element of defined by

3. Weak Precompactness in

Let be a Banach space, a measure space, and the space of all -valued countably additive measures on of bounded variation (with the total variation norm). Following [4], for , a probability measure on , we denote

Let be a lifting of [12, 15]. For a subset of and , let .

The following results will be useful in our study.

Lemma 1 (see [1, 6]). Let be a bounded subset of . Then is weakly precompact (resp., relatively weakly compact) if and only if for each sequence in there is a sequence so that for each and is weakly Cauchy (resp., weakly convergent).

Lemma 2 (see [5, Theorem 14]). Let be a sequence in . Suppose there is a probability measure with for each . Let be a lifting of . Then there is a sequence with for each and two measurable sets and such that and,(a)for , is weakly Cauchy;(b)for , there exists such that is equivalent to .

The most general result of this paper is the following theorem.

Theorem 3. Let be a subset of . Then is weakly precompact if and only if for any sequence in and for any/some lifting of there exists a sequence with for each such that, for a.e. , the sequence is weakly Cauchy in .

Proof. Suppose that is a weakly precompact subset of and let be a sequence in . By passing to a subsequence, we can assume that is weakly Cauchy. Let be a lifting of . By Lemma 2, there is a sequence with for each and two measurable sets and such that and,(a)for , is weakly Cauchy in ;(b)for , there exists such that is equivalent to .
If , then, by [5, Theorem 15], there exists such that is equivalent to . Since lies in the set , which is weakly precompact (see [16, page 377], [17, page 27]), one obtains a contradiction. Hence , and for a.e. , is weakly Cauchy in .
Conversely, let be a sequence in , and let be a lifting of . Let be a sequence with for each such that, for a.e. , the sequence is weakly Cauchy in . By [5, Theorem 15], is weakly Cauchy. By Lemma 1, is weakly precompact.

Theorem 4. Let be a subset of . Then is weakly precompact if and only if for any sequence in and for any/some lifting of there exists a sequence with for each such that, for a.e. , the sequence is weakly Cauchy in .

Proof. The proof is similar to that of Theorem 3.

Corollary 5. Let be a subset of .(i)Suppose that, for a.e. , the set is weakly precompact. Then is weakly precompact.(ii)Suppose that, for a.e. , the set is relatively weakly compact. Then is relatively weakly compact.

Proof. Let be a sequence in . By Lemma 2, there exist a sequence with for each and two sets and in with , such that conditions (a) and (b) of Lemma 2 are satisfied.(i)Since, for a.e. , the set is weakly precompact (see [16, page 377], [17, page 27]) and the sequence lies in this set, we have . Then for a.e. , the sequence is weakly Cauchy in . By Theorem 3, is weakly precompact.(ii)By (i), for a.e. , the sequence is weakly Cauchy in . Since, for a.e. , the set is relatively weakly compact (Krein-Smulian’s theorem) and the sequence lies in this set, we have that is relatively weakly compact. Then for a.e. , is weakly convergent. By [4, Theorem 1], is relatively weakly compact.

Corollary 6. Suppose that is a subset of .(i)If , then is weakly precompact.(ii)If is reflexive, then is relatively weakly compact.

Proof. Apply Corollary 5.

Corollary 7. Let be a subset of . If the set is weakly precompact for a.e. , then is weakly precompact.

Proof. The proof is similar to that of Corollary 5.

Corollary 8. Suppose that is a subset of . If , then is weakly precompact.

Proof. Apply Corollary 7.

The statements of Corollaries 6 and 8 can be considered just for (resp., ).

For the proofs of the following two theorems we will need the following lemmas. The first lemma contains a well-known result due to Grothendieck about relatively weakly compact sets (see [18, page 227]).

Lemma 9 (see [3, Lemma 2.15]; [18, page 227]; [9, Corollary 1.7]). Let be a bounded subset of . If for any there exists a weakly precompact (relatively weakly compact, resp., ) subset of such that , then is weakly precompact (relatively weakly compact, resp., a -set).

Lemma 10 (see [3, Lemma 2.12]). Let be a bounded subset of . Then is a -set if and only if for any sequence in there is a sequence so that for each and is a -set.

Lemma 11. (i) (see [19, Lemma ]) Let be a bounded subset of . If for any there exists a -subset of such that , then is a -set.
(ii) (see [19, Proposition ]) Let be a bounded subset of . Then is a -set if and only if for any sequence in there is a sequence so that for each and is a -set.

Lemma 12. (i) If is a -set in , then is uniformly countably additive.
(ii) If is a weakly precompact subset of , then is uniformly countably additive.

Proof. (i) Suppose is a -set in . Since each member of is a countably additive measure on the -algebra , the set is uniformly countably additive if and only if uniformly for whenever is a pairwise disjoint sequence in .
Let be a sequence in . Without loss of generality suppose that for all . Let be a pairwise disjoint sequence in and such that for all . For each , let be a partition of and let be points in such that Define the -valued simple functions by . Note that for all . Define by Note that is a well-defined operator, is wuc, and for each . Then is not a -set. This contradiction concludes the proof.
(ii) If is a weakly precompact set in , then is a -set in [7]. Apply (i).

Let denote the unit ball of .

Theorem 13. (i) Suppose is a bounded subset of such that is uniformly countably additive in . Then there is a probability measure on such that, for any sequence in , there is a sequence with for each such that, for any , there is a positive integer and a subset of so that .
(ii) Suppose is a bounded subset of . Then is weakly precompact (resp., a -set) if and only if there is a probability measure on such that, for any sequence in , there is a sequence with for each such that, for any , there is a positive integer and a weakly precompact (resp., a -set) subset of so that .

Proof. (i) Let be a bounded subset of such that is uniformly countably additive in . Then is relatively weakly compact in [18]. Hence there is a probability measure on so that is uniformly -continuous [18]; that is, for any , there is such that if , then for all .
Let be a sequence in . For each , let be the -density of . Since is relatively weakly compact in , is relatively weakly compact in . Choose a subsequence of so that converges weakly to some function . By Mazur’s theorem, there is a sequence with for each so that . By taking a subsequence, if necessary, , -a.e. Therefore -a.e., and thus , where is a set of measure zero.
Let . Choose from the definition of uniform -continuity. Choose a positive integer so that and let .
For each , . Suppose , with and , where the sums are finite. Let . Note that for each . Define and let .
For each , For , Then . For each , and thus for each . Therefore .
(ii) Suppose is a weakly precompact subset (resp., a -subset) of . By Lemma 12, is uniformly countably additive in . By (i), there is a probability measure on such that, for any sequence in , there is a sequence with for each such that, for any , there is a positive integer and a subset of so that . Since , which is weakly precompact (resp., a -set) (see [16, page 377], [17, page 27], resp., [9]), is weakly precompact (resp., a -set).
Conversely, let be a bounded subset of . Choose a probability measure as in the statement. Let be a sequence in . Let be a sequence with for each such that, for any , there is a positive integer and a weakly precompact subset (resp., a -subset) of so that . By Lemma 9, is weakly precompact (resp., a -set). By Lemma 1 (resp., 10), is weakly precompact (resp., a -subset).

Let denote the unit ball of .

Theorem 14. (i) Suppose is a bounded subset of such that is uniformly countably additive in . Then there is a probability measure on such that, for any sequence in , there is a sequence with for each such that, for any , there is a positive integer and a subset of so that .
(ii) Suppose is a bounded subset of . Then is weakly precompact (resp., a -subset) if and only if there is a probability measure on such that, for any sequence in , there is a sequence with for each such that, for any , there is a positive integer and a weakly precompact subset (resp., a -subset) of so that .

Proof. (i) The proof is similar to the proof of Theorem 13(i).
(ii) If is a weakly precompact subset of , then is a -subset of [7]. Hence is uniformly countably additive in (see [9, Proposition 2.1], [20, Proposition 2]). The remainder of the proof is similar to that of Theorem 13(ii), using Lemma 9 (resp., 11) and Lemma 1 (resp., 11).

Corollary 15. (i) Assume . Then a subset of is weakly precompact if and only if is bounded and is uniformly countably additive.
(ii) Assume is reflexive. Then a subset of is relatively weakly compact if and only if is bounded and is uniformly countably additive.

Proof. If is a weakly precompact subset of , then is uniformly countably additive, by Lemma 12.
Now suppose is a bounded subset of and is uniformly countably additive. By Theorem 13(i), there is a probability measure on such that, for any sequence in , there is a sequence with for each such that, for any , there is a positive integer and a subset of so that .(i)By Corollary 6, is weakly precompact, since . By Theorem 13(ii), is weakly precompact.(ii)By Corollary 6, is relatively weakly compact, since is reflexive. By Lemma 9, is relatively weakly compact. By Lemma 1, is relatively weakly compact.

Corollary 16. (i) If , then has property .
(ii) If is reflexive, then has property .

Proof. Let be a -subset of . By Lemma 12, is uniformly countably additive.(i)By Corollary 15(i), is weakly precompact.(ii)By Corollary 15(ii), is relatively weakly compact.

Corollary 17. Suppose . Then a subset of is weakly precompact if and only if is bounded and is uniformly countably additive.

Proof. The proof is similar to that of Corollary 15, using Theorem 14 and Corollary 8.

Corollary 18 (see [21], Theorem 3.12, [8]). Suppose is a compact Hausdorff space and is a Banach space. If and is a strongly bounded operator, then is weakly precompact.

Proof. Suppose that and is a strongly bounded operator. We claim that is weakly precompact. Recall that takes values in and that . Let be a sequence in and let for each . Without loss of generality suppose that for each . Since is strongly bounded, is uniformly countably additive (see [11, Lemma 3.1]). By Corollary 17, is weakly precompact. Hence is weakly precompact.

Corollary 19. If is a compact Hausdorff space and , then has property .

Proof. Let be an unconditionally converging operator. Then is strongly bounded [14]. By Corollary 18, is weakly precompact. Then has property [8].

Corollary 20. Suppose . Then the following are equivalent: (i) is not a quotient of ;(ii)for any compact Hausdorff space and any Banach space , an operator has weakly precompact adjoint whenever is strongly bounded and is weakly precompact for every .

Proof. If is not a quotient of and , then , by [21, Proposition 3.8]. Apply Corollary 18.
Suppose is a surjection. By [21, Theorem 2.4], there is a compact space and a continuous linear surjection so that is strongly bounded and is compact for all . Since is a surjection onto , is an isomorphism on , and thus is not weakly precompact.

Corollary 21. (i) [6, 11] If is reflexive, then every strongly bounded operator is weakly compact.
(ii) [7] If is reflexive, then has property .

Proof. (i) Let be a strongly bounded operator. Let be a sequence in and , . By Corollary 18, is weakly precompact in . Since is weakly sequentially complete, is weakly sequentially complete [5]. Hence is relatively weakly compact. Then is weakly compact. Hence is weakly compact.
(ii) Every unconditionally converging operator on is strongly bounded [14] and thus weakly compact (by (i)). Then has property [7].

A Banach space is injective if it is complemented in any superspace. We recall that property (resp., property ) is stable under quotients.

Corollary 22. (i) Suppose that is injective and . Then has property .
(ii) Suppose that is injective and is reflexive. Then has property .

Proof. The space is isomorphic to . Since is injective, is complemented in . Now, is isomorphic to [22].(i)By Corollary 19, has property . Hence has property .(ii)By Corollary 21, has property . Hence has property .

Corollary 23. (i) Suppose that is an -space and . Then has property .
(ii) Suppose that is an -space and is reflexive. Then has property .

Proof. The space is isomorphic to [22]. Since is an -space, is an -space [10], and thus is injective [23]. Apply Corollary 22.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

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Copyright © 2015 Ioana Ghenciu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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