Journal of Function Spaces

Volume 2015 (2015), Article ID 796753, 13 pages

http://dx.doi.org/10.1155/2015/796753

## Some Classes of Continuous Operators on Spaces of Bounded Vector-Valued Continuous Functions with the Strict Topology

Faculty of Mathematics, Computer Science and Econometrics, University of Zielona Góra, Ulica Szafrana 4A, 65–516 Zielona Góra, Poland

Received 28 October 2014; Accepted 2 January 2015

Academic Editor: Luisa Di Piazza

Copyright © 2015 Marian Nowak. 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.

#### Abstract

Let be a completely regular Hausdorff space and let and be Banach spaces. Let be the space of all -valued bounded, continuous functions on , equipped with the strict topology . We study the relationship between important classes of -continuous linear operators (strongly bounded, unconditionally converging, weakly completely continuous, completely continuous, weakly compact, nuclear, and strictly singular) and the corresponding operator measures given by Riesz representing theorems. Some applications concerning the coincidence among these classes of operators are derived.

#### 1. Introduction and Terminology

Throughout the paper let and be real Banach spaces and let and denote the Banach duals of and , respectively. By and we denote the closed unit ball in and , respectively. By we denote the space of all bounded linear operators from to . Given a locally convex space by or we will denote its topological dual. We denote by the weak topology on with respect to a dual pair . Let stand for the collection of all finite subsets of the set of all natural numbers.

Assume that is a completely regular Hausdorff space. By (resp., ) we will denote the family of all zero sets (resp., of cozero sets) in , respectively. Let stand for the Banach space of all bounded continuous functions , equipped with the uniform norm . We write instead of . By we denote the Banach dual of . For let for .

Let (resp., ) stand for the algebra (resp., -algebra) of Baire sets in , respectively. Let (resp., ) stand for the Banach space of all totally -measurable (resp., totally -measurable) functions (see [1, 2]).

The strict topology (called also a superstrict topology and denoted by ) on and is of importance in the topological measure theory (see [3–9] for definitions and more details). is a closed subspace of the Banach space and -bounded sets in are -bounded. It is known that is -dense in if one of the following conditions holds (see [6, Theorems 5.1 and 5.2]):(i)has a -compact dense subset (e.g., separable).(ii) is a -space (see [10]).(iii) is a -space.

*Remark 1. *Throughout the paper we will assume that is -dense in .

For being a locally compact Hausdorff space, by we denote the Banach space of all continuous functions tending to zero at infinity, equipped with the uniform norm. If is a compact Hausdorff space, then coincides with the uniform norm topology on . In this case we write simply instead of .

Let stand for the Banach lattice of all Baire measures on , provided with the norm (= the total variation of ). Due to the Alexandrov representation theorem can be identified with through the lattice isomorphism , where for , and (see [4, Theorem 5.1]).

By we denote the set of all finitely additive measures with the following properties:(i)for each , the function defined by belongs to ;(ii), where stands for the variation of on .

Let denote the Banach space of all continuous functions such that is a relatively compact set in , equipped with the uniform norm . Then . In view of [11, Theorem 2.5] can be identified with through the linear mapping , where for and . Then one can embed into by the mapping , where, for ,

Assume that is a bounded linear operator. Then we can define the corresponding operator measure (called the* representing measure* of ) by setting
Here stand for the biconjugate of . Then , where the semivariation of on is defined by , where the supremum is taken over all finite -partitions of and for each . For let us put
Let stand for the variation of on . Then (see [1, §4, Proposition 5])
By we denote the set of all operator measures such that and for each .

Let denote the canonical embedding; that is, for , . Moreover, let stand for the left inverse of ; that is, .

For define

The following Bartle-Dunfor-Schwartz type theorem will be useful (see [12, Theorem 2], [13, Theorem 5, pages 153-154]).

Theorem 2. *Let be a bounded linear operator and be its representing measure. Then for each the following statements are equivalent:*(i)* is weakly compact.*(ii)* for each and is a relatively weakly compact set in .*(iii)* is strongly bounded.*

*Following [14–16] we have the following definition.*

*Definition 3. *A bounded linear operator is said to be* strongly bounded* if its representing measure is strongly bounded; that is, whenever is a pairwise disjoint sequence in .

*Note that is strongly bounded if and only if the family is uniformly strongly additive.*

*For each , for . It follows that if is strongly bounded, then is weakly compact, and hence for (see Theorem 2).*

*For being a compact Hausdorff space (resp., a locally compact Hausdorff space) different classes of bounded linera operators (resp., ) have been studied intensively; see [14–33]. The study of the relationship between operators (resp., and their representing operator-valued measures is a central problem in the theory. The main aim of the present paper is to extend to “the completely regular setting” some classical results concerning various classes of bounded operators (resp., ), where is a compact Hausdorff space (resp., a locally compact Hausdorff space). In [12] using the device of embedding the space into we establish general Riesz representation theorems for -continuous linear operators with respect to the representing measures (see Theorems 6 and 8 below). In Section 3 we show that if is -continuous and strongly bounded, then its representing measure has its values in and possesses a unique extension that is variationally semiregular; that is, the set is uniformly countably additive (see Theorem 11 below). In Sections 4–9 we study the folowing classes of -continuous linear operators unconditionally converging, weakly completely continuous, completely continuous, weakly compact, nuclear, and strongly singular. We show that if a -continuous linear operator belongs to any of these classes of operators, then is strongly bounded and, for each , the operator shares the property of (see Theorems 17, 23, 26, 29, 34, and 36 below). We derive some applications concerning to the coincidence among these classes of -continuous operators (see Corollary 13, Theorems 18 and 19, Corollary 27, Theorem 29).*

*2. Integral Representation of Continuous Operators on *

*The space of all -aditive members of will be denoted by (see [3, 4]). Then . Let
Then if (see [5, Proposition 3.9]).*

*For the integration theory of functions with respect to we refer the reader to [6, page 197], [5]. The following result will be of importance (see [6, Theorem 5.3]).*

*Theorem 4. The following statements hold:(i)for the following conditions are equivalent:(a) is -continuous;(b)there exists a unique such that
and ;(ii)for , for .*

*In view of [9, Corollary 5] we have the following characterization of convergence in .*

*Theorem 5. For a sequence in the following statements are equivalent:(i) for ;(ii) and in for each .*

*The following theorem gives a characterization of -continuous operators in terms of the corresponding operator measures (see [12, Theorem 9 and Corollary 7]).*

*Theorem 6. Let be a -continuous linear operator and be the representing measure of . Then the following statements hold.(i).(ii)For each , for .(iii)For each and there exists a unique vector in , denoted by , such that for each .(iv)For each , the mapping is a -continuous linear operator.(v)For , and .(vi).(vii)For and ; we have
*

*Following [34] by (= ), we denote the space of all bounded countably additive, real-valued, regular (with respect to zero sets) measures on .*

*We define to be the set of all measures such that the following two conditions are satisfied.(i)For each , the function defined by for , belongs to .(ii), where, for each , we define , where the supremum is taken over all finite -partitions of and all finite collections .It is known that if , then (see [34, Lemma 2.1]).*

*The following result will be of importance (see [34, Theorem 2.5]).*

*Theorem 7. Let . Then possesses a unique extension and .*

*From Theorem 7 and [13, Corollary 10, page 4] it follows that if , then for .*

*By we will denote the space of all operator measures such that and for each . By we will denote the space of all operator measures with such that for each .*

*The following theorem characterizes -continuous linear operators such that are weakly compact for each (see [12, Theorem 14 and Lemma 11]).*

*Theorem 8. Let be a -continuous linear operator such that is weakly compact for each , and let be the representing measure of . Then the following statements hold.(i) for each , and the measure defined by for , , belongs to and possesses a unique extension with which is countably additive both in the strong operator topology and the weak star operator topology. Moreover, for .(ii)For every and there exists a unique vector in , denoted by , such that for each , and
(iii)For each , the mapping defined by is a -continuous linear operator.(iv) for .*

*Remark 9. *As a consequence of Theorem 8 (for ) we have
where for , for and .

*3. Strongly Bounded Operators on *

*3. Strongly Bounded Operators on**Making use of [35, Theorem 8] we can state the following analogue (for Baire measures on a completely regular Hausdorff space) of the celebrated Dieudonné-Grothendieck’s criterion on weak compactness in the space of Borel measures on a compact Hausdorff space (see [36, Theorem 2], [37, Theorem 14, pages 98–103]), which will play a crucial role in the study of different classes of operators on .*

*By we denote the topology of simple convergence in . Then is generated by the family of seminorms, where for .*

*A completely regular Hausdorff space is said to be an -space if a subset which meets every zero-set in a zero-set must be a zero-set. One can note that every metrizable space is a -space.*

*From now on we will assume that is a -space.*

*Theorem 10. Assume that is a subset of such that . Then the following statements are equivalent.(i) is relatively -compact subset of .(ii) is uniformly countably additive, that is, whenever , .(iii) is uniformly strongly additive, that is, whenever is pairwise disjoint in .(iv) for every pairwise disjoint sequence in .*

*Proof. *(i)(ii) See [38, Theorem 7].

(ii)(iii) See [37, Theorem 10, pages 88-89].

(iv)(i) See [35, Theorem 8].

*Now we can state a characterization of -continuous strongly bounded operators .*

*Theorem 11. Let be a -continuous linear operator and let be its representing measure. Then the following statements are equivalent.(i)For each , is weakly compact and is variationally semiregular; that is, whenever , .(ii) is strongly bounded.(iii) whenever is a uniformly bounded sequence in such that in for each .(iv) whenever is a uniformly bounded sequence in such that for .*

*Proof. *(i)(ii) It follows from Theorem 8 and [12, Theorem 16].

(ii)(iii) It follows from [12, Theorem 17].

(iii)(iv) It is obvious.

(iv)(i) Assume that (iv) holds. First we shall show that for each , is weakly compact. Assume on the contrary that is not weakly compact for some . This means that is not strongly bounded. Since for , , we obtain that the family is not uniformly strongly additive. Hence the family is not uniformly countably additive. It follows that the family is not uniformly countably additive. In view of Theorem 10 there exist , a sequence in and a pairwise disjoint sequence in such that for , . Note that
Hence there exists a sequence in such that , and
Let for . Then for and by (iv), , which contradics (12). This means that is weakly compact for each , as desired.

In view of Theorem 8 can be uniquely extended to a measure . Assume that is not variationally semiregular. Then by Theorem 10 there exist , a pairwise disjoint sequence in and a sequence in such that . Hence by Theorem 7 there exists a sequence in and with for such that
Then, for ,
On the other hand, since for , by (iv), . This contradiction establishes that (i) holds.

*Corollary 12. Let be a -continuous and strongly bounded linear operator and let be its representing measure. Then the set is uniformly regular on ; that is, for each and , there exist with and with such that
*

*Proof. *In view of Theorem 11 the family is uniformly countably additive. Let be a control measure for and let and be given. Then there is such that whenever and . By the regularity of there exists with and with such that . Hence we get , .

*Corollary 13. Assume that is a -continuous linear operator and contains no isomorphic copy of . Then is strongly bounded.*

*Proof. *Let stand for the representing measure of . We shall first show that is weakly compact for each . Assume on the contrary that is not weakly compact for some . Then by the proof of implication (iv)(i) of Theorem 11 there exist , a sequence in , and a pairwise disjoint sequence in such that for . By the Rosenthall lemma (see [13, Lemma 1, page 18]) the sequence in and in can be chosen such that for ,
Since, for ,
there exists a sequence in such that with and
Let . We see that is an isomorphic copy of . Assume that for some sequence in . Then for we have
But , so
This means that is an isomorphism on , so contains an isomorphic copy of , which contradicts our assumption on . This means that is weakly compact for each . Hence in view of Theorem 8 is countably additive in the weak star operator topology and by [19, Remark 7, page 923] and Theorem 11 we derive that is strongly bounded, as desired.

*Remark 14. *If is a compact Hausdorff space, the equivalence (ii)(iii) of Theorem 11 was obtained by Brooks and Lewis (see [16, Theorem 2.1]).

*Let stand for the Banach space of all bounded strongly -measurable functions , equipped with the uniform norm . Assume that with is variationally semiregular. Then every is -integrable (see [39, Definition 2, page 523 and Theorem 5, page 524]) and whenever is a uniformly bounded sequence in converging pointwise to (see [40, Proposition 2.2]).*

*Note that if then is -measurable. Hence if is assumed to be separable then is strongly -measurable; that is, (see [2, Proposition 21, page 9]).*

*Recall that a function is -measurable if for each the function is -measurable. For by we denote the vector space of all weak ^{*}-measurable functions for which there exists such that -a.e. on (see [41, page 26]).*

*Following [40] we can distinguish an important class of operators on .*

*Definition 15. *A bounded linear operator is said to be *-smooth* if whenever is a uniformly bounded sequence in such that for each .

*Proposition 16. Assume that is separable. Let be a ,-continuous and strongly bounded linear operator, and let be its representing measure. Then for each there exists such that
where is a control measure for .*

*Proof. *Since is supposed to be separable, . Moreover, since is variationally semiregular (see Theorem 11), the corresponding integration operator is -smooth and for we have (see [40, Proposition 2.2])
It follows that for each .

Let . Then is a -smooth functional on , and is -absolutely continuous; that is, . According to the Radon-Nikodym type theorem (see [41, Theorem ]) there exists a weak^{*}-measurable function which satisfies the following conditions.(1)The function is -measurable and -integrable; that is, .(2)For every and ,

It follows that . Note that for every the mapping is -measurable and using (2) we get
Now let . Then there exists a sequence in such that and for each and (see [2, Theorem 1.6, page 4]). Then the mapping is -measurable. Using the Lebesgue dominated convergence theorem we have
It follows that

*4. Unconditionally Converging Operators on *

*4. Unconditionally Converging Operators on*

*Recall that a series in a Banach space is called weakly unconditionally Cauchy (wuc) if, for each , . We say that a bounded linear operator is unconditionally converging if, for every weakly unconditionally Cauchy series in , the series converges unconditionally in a Banach space .*

*If is a compact Hausdorff space, Swartz [33] proved that every unconditionally converging operator is strongly bounded. Dobrakov (see [28, Theorem 3]) showed that if is a locally compact Hausdorff space, then every unconditionally converging operator is strongly bounded and for every Borel set in , the operator is unconditionally converging. Moreover, Brooks and Lewis [27, Theorem 5.2] showed that if contains no isomorphic copy of , then every strongly bounded operator is unconditionally converging. We will extend these results to the setting when is a -continuous linear operator and is a completely regular Hausdorff space.*

*Theorem 17. Let be a -continuous and unconditionally converging linear operator, and stand for the representing measure of . Then the following statements hold.(i) is strongly bounded.(ii)For each , is an unconditionally converging operator.*

*Proof. *(i) Assume that is a uniformly bounded sequence in such that for . Then is bounded in and, since is unconditionally converging, we obtain that . Hence by Theorem 11 is strongly bounded.

(ii) Let and assume that is in . Then . In view of Theorem 11 is uniformly countably additive and let stand for the control measure of (see Corollary 12). Let be given. Then there is such that whenever , . Then there exist with and with such that . Hence
Then one can choose with , , and . Define for . We shall show that is unconditionally converging. Indeed, for , . Hence the series is unconditionally convergent; that is, is unconditionally converging, as desired. Then for each , we have
Hence and since the class of all unconditionally converging operators from to is a closed linear subspace of (see [28, page 20]), we derive that is unconditionally converging.

*Theorem 18. Assume that is separable and contains no isomorphic copy of . Then for a -continuous linear operator the following statements are equivalent.(i) is unconditionally converging.(ii) is strongly bounded.*

*Proof. *(i)(ii) See Theorem 17.

(ii)(i) See [12, Corollary 18].

*Recall that a subset of a Banach space is said to be weakly precompact if every bounded sequence in contains a subsequence so that converges for each . An operator is said to be weakly precompact if is weakly precompact in a Banach space .*

*Abbott et al. [17, Theorem 2.8] discussed the relationship between strongly bounded and unconditionally converging operators whenever is a compact Hausdorff space. They showed that if contains no isomorphic copy of and has the RNP, then the classes of strongly bounded and unconditionally converging operators coincide. Now we state an analogue of Theorem 2.8 of [17] for -continuous linear operator , where is a completely regular Hausdorff space.*

*Theorem 19. Assume that contains no isomorphic copy of and has the RNP. Then for a -continuous linear operator the following statements are equivalent.(i) is weakly precompact.(ii) is unconditionally converging.(iii) is strongly bounded.*

*Proof. *(i)(ii) See [17, Theorem 2.7].

(ii)(iii) See Theorem 17.

(iii)(i) Assume that is strongly bounded. Since , we have to show that is a weakly precompact subset of the Banach space . By Theorem 11 is uniformly countably additive, and let be a control measure for . Since is supposed to have the RNP, for each there exists such that and for . It follows that is a uniformly integrable subset of and since contains no isomorphic copy , is a weakly precompact subset of (see [42]). Since (= the Banach space of all -continuous members of ) and the Radon-Nikodym theorem establishes the isometry between and , we obtain that is a weakly precompact subset of because for .

*5. Weakly Completely Continuous Operators on *

*5. Weakly Completely Continuous Operators on**Recall that a bounded linear operator from a Banach space to a Banach space is said to be a Dieudonné operator if maps -Cauchy sequences in into weakly convergent sequences in .*

*If is a compact Hausdorff space, then Dieudonné operators from the Banach space to were studied by Bombal and Cembranos [23] and Abbott et al. (see [17, Theorems 3.1, 3.5 and Theorem, page 334].*

*Definition 20. *A bounded linear operator is said to be* weakly completely continuous* if is -convergent in whenever is a uniformly boundd sequence in such that is a -Cauchy sequence in for each .

*Proposition 21. Let be a bounded linear operator. Then the following statements are equivalent.(i) is weakly completely continuous.(ii) maps -Cauchy sequences in onto -convergent sequences in .*

*Proof. *(i)(ii) Assume that is weakly completely continuous and is a -Cauchy sequence in . Then for each , is a -Cauchy sequence in because , where for . Since is -bounded, we get . It follows that is -convergent.

(ii)(i) Assume that (ii) holds and is a uniformly bounded sequence in such that is a -Cauchy sequence in for each . We shall show that is a -Cauchy sequence. Assume on the contrary that is not a -Cauchy sequence. Then there exist and and a subsequence of satisfying for . Since for each , by Theorem 5 for , . Hence . This contradiction establishes that is a , -Cauchy sequence, and it follows that a sequence is -convergent in .

*From Proposition 21 it follows that every weakly completely continuous operator is a Dieudonné operator. As a consequence, we get the following result (see [37, Problem 8, page 54]).*

*Corollary 22. Assume that is a weakly completely continuous operator. Then is unconditionally converging.*

*Theorem 23. Let be a -continuous and weakly completely continuous linear operator and stand for its representing measure. Then the following statements hold.(i) is strongly bounded.(ii)For each , is a Dieudonné operator.*

*Proof. *(i) It follows from Corollary 22 and Theorem 17.

(ii) Let and assume that is a -Cauchy sequence in . Hence . Since is strongly bounded, arguing as in the proof of Theorem 17 for a given there exist with and with such that
Then we can choose with , , and . Define for . We shall show that is a Dieudonné operator. Let for . Then and is a -Cauchy sequence in for each . Hence is -convergent in and this means that is a Dieudonné operator. Then arguing as in the proof of Theorem 17, we obtain that and since the class of all Dieudonné operators from to is a closed linear subspace of (see [17, Theorem 3.5]), we derive that is a Dieudonné operator.

*6. Completely Continuous Operators on *

*6. Completely Continuous Operators on**Recall that a bounded linear operator from a Banach space to a Banach space is said to be a Dunford-Pettis operator if in for implies (see [43, Section 19]).*

*Definition 24. *A bounded linear operator is said to be* completely continuous* if whenever is a uniformly bounded sequence in such that in for each .

*Using Theorem 5 one can get the following result.*

*Proposition 25. Let be a bounded linear operator. Then the following statements are equivalent.(i) is completely continuous.(ii) whenever in .*

*Theorem 26. Let be a -continuous and completely continuous operator and its representing measure. Then the following statements hold.(i) is strongly bounded.(ii)For each , is a Dunford-Pettis operator.*

*Proof. *(i) In view of [43, Theorem 19.1] and Proposition 25 maps , Cauchy sequences in onto norm convergent sequences in . It follows that is a Dieudonné operator and hence is unconditionally converging. Thus is strongly bounded (see Theorem 17).

(ii) Let and assume that in for . Then . Since is strongly bounded, arguing as in the proof of Theorem 17 for a given there exist with and with such that
Then we can choose with ,