Abstract

Let be a completely regular Hausdorff space, and let and be Banach spaces. Let be the space of all -valued bounded, continuous functions defined on , equipped with the strict topologies , where . General integral representation theorems of -continuous linear operators with respect to the corresponding operator-valued measures are established. Strongly bounded and -continuous operators are studied. We extend to “the completely regular setting” some classical results concerning operators on the spaces and , where is a compact or a locally compact space.

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 . 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 .

Assume that is a completely regular Hausdorff space. Let stand for the Banach space of all bounded continuous, -valued functions on provided with the uniform norm . We write instead of . By we denote the Banach dual of . For let for .

Let (resp., ) be the algebra (resp., -algebra) of Baire sets in , which is the algebra (resp., -algebra) generated by the class of all zero sets of functions of . By we denote the family of all cozero sets in . Let stand for the Banach space of all totally -measurable functions (the uniform limits of sequences of -valued -simple functions) provided with the uniform norm (see [1, 2]). We will write instead of .

Strict topologies on and (for ) play an important role in the topological measure theory (see [312] for definitions and more details). Recall that a subset of is said to be solid if and with for imply that . Then are locally convex-solid topologies on ; that is, they have a local base at consisting of convex and solid sets (see [6, Theorem 8.1], [10, Theorem 5]). We have and . For a net in , for if and only if for in (see [6, 10]).

Let stand for the algebraic tensor product of and ; that is, is the space of all functions , where , for , and for . Then is dense in for (see [6, 8]). Moreover, is dense in if or is a -space (see [6, Theorem 5.2], [13]) and in if is real-compact (see [10, Theorem 7]).

Let denote the Banach space of all continuous functions such that is a relatively compact set in , provided with the uniform norm . Then .

Linear operators from the spaces and , equipped with the strict topologies to a locally convex space , were studied by Katsaras and Liu [14], Aguayo-Garrido, Nova-Yanéz and Sanchez [15, 16], and Khurana [17]. In particular, Katsaras and Liu found an integral representation of weakly compact operators and characterizations of -continuous and weakly compact operators for (see [14, Theorems 3, 4, 5]). Aguayo-Arrido and Nova-Yanéz derived a Riesz representation theorem for -continuous and weakly compact operators for in terms of their representing operator measures (see [15, Theorems 5 and 6]). If is a locally compact space, continuous operators on were studied by Dobrakov (see [18]) and Mitter and Young (see [19]).

In this paper we develop the theory of continuous linear operators from , equipped with the strict topologies to a Banach space . In particular, we extend to “the completely regular setting” some classical results of Brooks and Lewis (see [20, Theorem 5], [21, Theorem 5.2], [22, Theorem 2.1]) concerning operators on the spaces and , where is a compact or a locally compact space, respectively. In Section 2, using the device of embedding the space into (the Banach bidual of ), we state the integral representation of bounded linear operators from to . In Section 3 we derive general Riesz representation theorems for -continuous linear operators with respect to the corresponding measures (see Theorems 9 and 14 below). Section 4 is devoted to the study of -continuous and strongly bounded operators .

2. Integral Representation of Bounded Linear Operators on

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 .

In view of [23, Theorem 2.5] can be identified with through the linear mapping , where for and . Then one can embed into by the mapping , where for ,

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

Assume that is a bounded linear operator. Let where and denote the conjugate and biconjugate operators of , respectively. Then we can define a measure (called a representing measure of ) byThen , 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, Section 4, Proposition 5])

The following general properties of the operator are well known (see [1, Section 6], [2, Section 1], [13, 24]):and for each ,

For let From the general properties of it follows that Hence for each we get and hence . Moreover, we haveand using (5) we get

By we will denote the space of all measures such that and for each . Thus the representing measure of belongs to .

For any defineThen is a bounded linear operator. Let stand for the canonical embedding; that is, for , Let Then

The following lemma will be useful.

Lemma 1. Let be a bounded linear operator. Then for any and .

Proof. Let . Then for each , Hence we have On the other hand, for each ,  , and hence It follows that , as desired.

From Lemma 1 for and we get that is,

Now we are ready to prove the following Bartle-Dunford-Schwartz type theorem (see [25, Theorem 5, pages 153-154]).

Theorem 2. Let be a bounded linear operator and let 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.

Proof. (i)(ii) Assume that is weakly compact. Then by the Gantmacher theorem and is weakly compact (see [26, Theorem 17.2]). Hence and is weakly compact. In view of (21) for each , for and is strongly bounded (see [25, Theorem 1, page 148]). It follows that is a relatively weakly compact subset of (see [24, Theorem 7]).
(ii)(iii) It follows from [24, Theorem 7].
(iii)(i) Assume that is strongly bounded. Then by (21) is weakly compact and in view of (16) we derive that is weakly compact.

3. Integral Representation of Continuous Linear Operators on

The spaces of all -additive, -additive, perfect, -additive, and tight members of will be denoted by , , , , and , respectively (see [3, 4]). Then for .

For the integration theory of functions with respect to we refer the reader to [6, page 197], [5, Definition 3.10], [27, page 375]. For let Then if (see [5, Proposition 3.9], [6, Theorem 3.1], [10, Theorem 1]). For let us put, for , It is known that is additive and positively homogeneous and can be extended to a linear functional on (denoted by again) by for .

Theorem 3. Assume that and is dense in (resp., ; and is dense in ; ; ). Then the following statements hold.(i)For a linear functional on the following conditions are equivalent.(a) is -continuous.(b)There exists a unique such that (ii)For , for .

Proof. (i) See [6, Theorems 5.3 and 4.2, Corollary 3.9], [5, Theorem 3.13], and [10, Theorem 8].
(ii) See [6, Theorem 2.1].

Assume that is a subset of and , where . Then we say that satisfies the condition if we have the following:(1)for : whenever , ;(2)for : for every partition of unity for and every there exists a finite set in such that ;(3)for : for every continuous function from onto a separable metric space and every , there is a compact subset of such that ;(4)for : whenever , ;(5)for : for every there exists a compact subset of such that for each .

The following lemmas will be useful.

Lemma 4. Assume that is a subset of and , where and is -dense in (resp., ; and is -dense in ; ; ). Then the following statements are equivalent.(i) is -equicontinuous.(ii) is -equicontinuous.(iii) is -equicontinuous.(iv)The condition holds.

Proof. (i)(ii) See [9, Lemma 2].
(ii)(iii) It follows from Theorem 3.
(iii)(iv) See [4, Theorem 11.14] for ; [28, Proposition 3.6] for ; [28, Proposition 2.6] for ; [4, Theorem 11.24] for ; and [28, Proposition 1.1] for .

Lemma 5. Assume that and is -dense in (resp., ; , and is -dense in ;  ; ). Let . Then for the following statements hold.(i)A functional defined by is -continuous and can by uniquely extended to a -continuous linear functional , and one will write the following: (ii) for .

Proof. (i) Assume that is a net in such that for . Then Since for in and , we obtain that ; that is, is -continuous. Since is dense in , can be uniquely extended to a -continuous linear functional (see [29, Theorem 2.6]).
(ii) Assume that . Choose a net in such that for . Then for in . Then and hence . Since , we get

For let us put

Lemma 6. Assume that and is -dense in (resp., ;  , and is -dense in ; ; ). Assume that and the set satisfies the condition . Then for the following statements hold.(i)An operator defined by is -continuous and can be uniquely extended to a -continuous linear operator , and one will write the following. (ii)For each , for .

Proof. (i) In view of Lemma 5 the set is -equicontinuous in . Assume that is a net in such that for . Let be given. Then there exists a neighborhood of for in such that for . Since for in , choose such that for . Hence for . It follows that, for and each , and hence, This means that is -continuous. Since is -dense in possesses a unique -continuous extension (see [29, Theorem 2.6]). Let (ii) Let . Choose a net in such that for . By Lemma 5 and (7) for we have

Corollary 7. Assume that and is -dense in (resp., ; and is -dense in ; ; ). Assume that and the set satisfies the condition . Then for the following statements hold: In particular, if , then where the supremum is taken over all finite disjoint supported collections with and and . One has

Proof. Let and . Then by Lemma 5 for with we have On the other hand, let be given. Then there exist a finite -partition of and , , such that By the regularity of for , we can choose , such that for . Choose pairwise disjoint with for such that . Then for we can choose with , , and (see [4, page 115]). Define . Then and . Hence we get and hence . Thus the proof of (a) is complete.
In view of (5), (a), and Lemma 6 we get that is, (b) holds.
Assume now that . Let for . Then for . For choose with , , and . Let . Then and ; and hence by , . Note that , where supp are pairwise disjoint and supp for . Thus (c) holds.
Using (c) we easily show that (d) holds. Thus the proof is complete.

Definition 8. Let be a bounded linear operator. Then the measure defined by will be called a representing measure of .

Now we state general Riesz representation theorems for continuous linear operators on , provided with the strict topologies , where .

Theorem 9. Assume that and is -dense in (resp., ; , and is -dense in ; ; ).(I)Let be a -continuous linear operator and let be its representing measure. Then the following statements hold.(i) and satisfies the condition .(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).(II)Let and let the set satisfy the condition . Then the statements (iii) and (iv) hold and for , and the mapping defined by is a -continuous linear operator. Moreover, coincides with the representing measure of and the statements (ii) and (vi) hold.

Proof. (I) In view of (10) for each , for . By Theorem 3 for each there exists a unique such that for . It follows that, for each , (see [23, Theorem 2.5]) and this means that . Hence Since is -equicontinuous in , by Lemma 4 the set satisfies the condition . Thus (i) and (ii) hold. In view of Lemma 6, (iii) and (iv) are satisfied.
According to (9) for each , and . Hence by Lemma 6, . Let . Choose a net in such that for . Hence Thus (v) holds. Using (v) and Corollary 7 we get .
(II) By Lemma 6 the statements (iii) and (iv) are satisfied.
Now let . Choose a net in such that for . Then by Lemma 6, because , and it follows that is -continuous.
Let stand for the representing measure of . Note that, for , , and we have that is, . By the first part of the proof (ii) and (vi) hold. Thus the proof is complete.

Following [14, 27] 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 [27, Lemma 2.1]).

The following result will be of importance (see [27, Theorem 2.5]).

Theorem 10. Let . Then possesses a unique extension and .

Arguing as in the proof of Lemma 6 we can obtain the following lemma.

Lemma 11. Assume that is -dense in and . Then for the following statements hold.(i)A functional defined by is -continuous and can be uniquely extended to a -continuous linear functional , and one will write the following: (ii)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 .

Remark 12. Note that in view of the Orlicz-Pettis theorem every is countably additive in the strong operator topology; that is, for each , the measure defined by for is countably additive. Moreover, in view of [30, Theorem 2] for each , is inner regular by zero sets and outer regular by cozero sets; that is, for each and there exist with and with such that and , ( denotes the semivariation of on ).

According to [14, Theorem 7] we have the following theorem.

Theorem 13. Assume that and is a relatively weakly compact subset of for each . Then possesses a unique extension such that .

For a linear operator and let for . For and let for .

Theorem 14. Assume that is -dense in .(I)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) and whenever , .(ii), 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 in the weak star operator topology. Moreover, for .(iii)For every and there exists a unique vector in , denoted by , such that, for each , .(iv)For each , the mapping defined by is a -continuous linear operator.(v) for .(II)Let be such that whenever and for each , let be strongly bounded. Then the operator defined by is -continuous and is weakly compact for each , and the statements (ii)(v) hold.

Proof. (I) (i) It follows from Theorem 9.
(ii) In view of Theorem 2   for , , and is a relatively weakly compact in for each . Since , by Theorem 13   possesses a unique extension with . By the Orlicz-Pettis theorem is countably additive in the strong operator topology. Moreover, since, for each , , we obtain that . This means that is countably additive in the weak star operator topology.
Let . Then for and we have , and by Theorem 10, .
(iii) For let for . Proceeding as in the proof of Lemma 6 we can show that is a -continuous linear operator, and hence possesses a unique -continuous linear extension (see [29, Theorem 2.6]). Let us write the following: Let . Choose a net in such that for . For each , (see (i)) and by Lemma 11 we have (iv) It follows from the proof of (iii).
(v) Let . In view of Theorem 9, for each , . On the other hand by (ii) for we have . It follows that .
(II) Since satisfies the condition , by Theorem 9 for , and the mapping defined by is a -continuous linear operator, and coincides with the representing measure of . Hence in view of Theorem 2   is a weakly compact operator. Thus by the first part of the proof the statements (ii)–(v) are satisfied.

4. Strongly Bounded Operators on

Definition 15. 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.

Now we are ready to state our main results that extend some classical results of Lewis (see [20, Theorem 5], [31, Lemma 1]) and Brooks and Lewis (see [22, Theorem 2.1], [21, Theorem 5.2]) concerning operators on the spaces and , where is a compact or a locally compact space, respectively.

Theorem 16. Assume that is -dense in . Let be a -continuous linear operator and let be its representing measure. Then and the following statements are equivalent.(i) is strongly bounded.(ii) whenever , (here denotes the unique extension of ).(iii)If is a sequence in such that , then there exists a nested sequence in such that for and , and .

Proof. In view of Theorem 9  .
(i)(ii) Assume that is strongly bounded. Since the family is uniformly strongly additive, according to [25, Lemma 1, page 26] the family is uniformly countably additive (see Theorem 16).
(ii)(i) It follows from [25, Lemma 1, page 26].
(ii)(iii) Assume that (ii) holds and is a sequence in such that . Then there exists such that is uniformly -continuous (see [25, Theorem 4, pages 11-12]). Let be given. Hence there exists such that whenever and . Since is zero-set regular, there exists a nested sequence in so that and for . Hence for . In view of (ii) there exists such that for . Hence for ; that is, .
Let , , and . Then by Theorem 9 we have It follows that , .
(iii)(ii) Assume that (iii) holds and , . Then there exists a nested sequence in such that for and Assume that (ii) does not hold. Then there exist and such that and whenever , , and . It follows that there exists such that . Hence there exist a finite -partition of and , , such that Since is zero-set regular (see [4, page 118]), we can choose , , such that for . Choose pairwise disjoint with for such that . Let for . Then and for . For choose such that , , and (see [4, page 115]). Let . Then , , and Hence we get Hence Thus we get a contradiction to .
Thus the proof is complete.

Theorem 17. Assume that is -dense in . Let be a -continuous and strongly bounded operator and let be its representing measure. Then the following statements hold.(i) and for , , and the measure , defined by for , , belongs to and possesses a unique extension with which is variationally semiregular; that is, whenever , .(ii)For every and there exists a unique vector in , denoted by , such that, for each , .(iii)For each , whenever is a uniformly bounded sequence in such that for .(iv) for .(v) whenever is a uniformly bounded sequence in such that for .

Proof. (i) Note that, for , for . Hence is strongly bounded, and by Theorems 2 and 14   and possesses a unique extension with . Since for , by Theorem 16 we have whenever .
(ii) It follows from Theorem 14 because for each , is weakly compact (see Theorem 2).
(iii) In view of (i) there exists such that is -continuous (see [25, Theorem 4, pages 11-12]). Let be a sequence in such that and for every . Let be given. Then there exists such that whenever , . Since for , by the Egoroff theorem there exists with and . Choose such that for .
Let . Note that for . Then by Lemma 11 and (ii), for and we get Hence for , as desired.
(iv) It follows from Theorem 14.
(v) It follows from (iii) and (iv).

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 [32, Definition 2, page 523 and Theorem 5, page 524]) and whenever is a uniformly bounded sequence in converging pointwise to (see [33, Proposition 2.2]).

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

As an application of Theorem 17 we have the following result.

Corollary 18. Assume that is -dense in , where is a separable Banach space which contains no isomorphic copy of . Let be a -continuous and strongly bounded operator. Then is unconditionally converging.

Proof. Assume that is a wuc series in the Banach space . Hence for each and because , where for . It follows that is an unconditionally convergent series in for each because contains no isomorphic copy of (see [34]). Let for , where for , . Then because is wuc (see [34]) and because is assumed to be separable (see [2, Theorem 21, page 9]). Hence (see [2, Theorem 10, page 6]).
Let be the representing measure of and let be a unique extension of (see Theorem 17). Since is variationally semiregular, in view of [33, Proposition 2.2] we have Hence . Finally, if is any permutation of , then for . Then , as desired.

Remark 19. A related result to Corollary 18 for strongly bounded operators on the space of -valued continuous functions vanishing at infinity defined on a locally compact space was obtained by Brooks and Lewis (see [21, Theorem 5.2]).

Recall that a Banach space is said to be a Schur space if every weakly convergent sequence in is norm convergent.

As a consequence of Theorem 17 we derive the following Dunford-Pettis type theorem for operators on .

Theorem 20. Assume that is -dense in , where is a Schur space. Let be a -continuous and strongly bounded operator. Then in whenever is a convergent to sequence in .

Proof. Assume that for . Then according to [11, Corollary 5], we obtain that and in for each . It follows that for because is supposed to be a Schur space. Using Theorem 17 we derive that in , as desired.

Conflict of Interests

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

Acknowledgment

The author wishes to thank the referee for useful remarks and suggestions that have improved the paper.