**Advances in Pure Mathematics** Vol.3 No.5(2013), Article ID:35428,4 pages DOI:10.4236/apm.2013.35068

Weak Integrals and Bounded Operators in Topological Vector Spaces

Mathematics Department, King Abdulaziz University, Jeddah, KSA

Email: mezianilakhdar@hotmail.com, alsulami@kau.edu.sa

Copyright © 2013 Lakhdar Meziani, Saud M. Alsulami. 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.

Received May 4^{th}, 2013; revised June 8^{th}, 2013; accepted July 11^{th}, 2013

**Keywords:** Bounded Operators; Integral Representation; Pettis Integral

ABSTRACT

Let X be a topological vector space and let S be a locally compact space. Let us consider the function space of all continuous functions, vanishing outside a compact set of S, equipped with an appropriate topology. In this work we will be concerned with the relationship between bounded operators, and X-valued integrals on. When X is a Banach space, such relation has been completely achieved via Bochner integral in [1]. In this paper we investigate the context of locally convex spaces and we will focus attention on weak integrals, namely the Pettis integrals. Some results in this direction have been obtained, under some special conditions on the structure of X and its topological dual. In this work we consider the case of a semi reflexive locally convex space and prove that each Pettis integral with respect to a signed measure, on S QUOTE gives rise to a unique bounded operator, which has the given Pettis integral form.

1. Topological Preliminaries

Suppose that S is a locally compact space and let X be a locally convex TVS. We denote by the set of all continuous functions vanishing outside a compact set of S, put if X = R. We are interested in representing linear bounded operators, by means of weak integrals against scalar measures on the Borel -field B_{S} of S. Before handling more closely this problem, we need some topological facts about the space.

If K is a compact set in S, let be the set of all continuous functions, vanishing outside K. It is clear that is a linear subspace of. We equip with the topology generated by the family of seminorms:

where is the family of seminorms generating the locally convex topology of X. The topology is the topology of uniform convergence on K.

Next let us observe that, the union being performed over all the compact subsets K of S. On the other hand if K_{1} is a subset of K_{2}, then the natural embedding is continuous. This allows one to provide the space with the inductive topology induced by the subspaces,. The facts we need about the space, is well known:

1.1. Proposition

1) The space, is locally convex Hausdorff and for each compact K, the relative topology of on is, this means that the canonical embedding is continuous.

2) Let be a linear operator of into the locally convex Hausdorff space V, then T is continuous if and only if the restriction of T to the subspace is continuous for each compact K.

1.2. Definition

For each in the topological dual of X and for each function, deﬁne the function on S by. Then sends into. Recall that is equipped with the uniform norm.

1.3. Lemma

The operator is linear and bounded. Moreover for each, is onto.

Proof: First it is clear that. Now by Proposition 1.1(b), we have to show that for each compact set K of S the operator is bounded. Since is bounded, there is a seminorm on X and a constant M such that for all. So we have if, and,; it follows that

.

Since by Formula (*), the right side of this inequality is, we deduce that is continuous. Now suppose. Then there exists such that and. It is clear that we can assume. Now let and define by, then and we have, because. It follows that is onto. ■

Now we consider the relationship between bounded operators, and weak integrals in the sense of the following definition. Such relationship is reminiscent to the classical Riesz theorem [2].

1.4. Definition

We say that a bounded operator has a Pettis integral form if there exists a scalar measure of bounded variation on B_{S} such that, for every continuous functional in, we have:

See Reference [3] for details on Pettis integral.

2. Integral Representation by Pettis Integral

In what follows, we introduce a class of bounded operators, which is, in this context, similar to the class used in [1].

2.1. Definition

Let P be the class of all bounded operators satisfying the following condition:

(I) For and, if then.

It is easy to check that P is a subspace of the space of all bounded operators from to X. Also one can prove that P is closed in the weak operator topology of. Note also that for a given bounded, Definition 1.4 implies condition (I) i.e.. The crucial point is that condition (I) implies the Pettis integral form of Definition 1.4, for some bounded scalar measure on B_{S}. This is the content of the following theorem proved in [4].

2.2. Theorem

Let be in the class P. Then there is a unique bounded signed measure on B_{S} such that holds for all in and. Moreover for each seminorm on X we have, where is the total variation of and is the -norm of T deﬁned by

with

.

By this theorem we may denote each operator T in the class P by the conventional symbol

where the letter P stands for Pettis integral.

3. Operators Associated to Scalar Measures via Pettis Integrals

In this section we start with a bounded scalar measure on and we seek for a linear bounded such that the correspondence between and T would be given by formula (W). First let us make some observations.

3.1. Operators via Pettis Integrals

A little inspection of (W) suggests the following quite plausible observations: First the integral

, as a linear functional of on, should beat least continuous for some convenient topology on Also the existence of the corresponding Tf in (W) will require that such topology on X should be compatible for the dual pair. Finally, to get the continuity of the functional, one can seek conditions such that if in an appropriate manner, then goes to 0 uniformly for. Since is bounded this will give

.

Such a program has been realized in [4], for a locally convex space having the convex compactness property [5], according to the following theorems (see [4] for details).

3.2. Theorem

Let X be a locally convex space with the convex compactness property, and whose dual is equipped with the Mackey topology If is a bounded scalar measure on B_{S}, then there is a unique bounded operator in the class P satisfying (W), with for each seminorm on X.

3.3. Theorem

Let X be a locally convex Hausdorff space whose dual is a barrelled space. If is a bounded signed measure on B_{S}, then there is a unique bounded operator in the class P satisfying (W) with respect to and such that.

Most of these results have been obtained for a space whose dual is a Mackey space. It is natural to ask if similar representations can be established if the dual is endowed with another topology, e.g. the strong topology.

3.4. Definition

The strong topology of is the topology generated by the family of the seminorms:

where B is running over all the bounded sets of X.

It is the topology of uniform convergence on the bounded sets of X. When we restrict to the ﬁnite sets B of X we get the so called weak * topology, which is the topology of simple convergence on X. We shall denote by the space equipped with the -topology (the - topology). Then we have:

3.5. Proposition

1) For each there exists a unique

such that:.

2), that is, every weak * continuous functional on is strongly continuous .

3.6. Definition

We say that the space X is semireﬂexive if

.

Now we are in a position to state the main results of this paper.

3.7. Theorem

Let X be a locally convex Hausdorff semireﬂexive space. If is a bounded signed measure on, then there is a unique bounded operator in the class P satisfying:

.

where is the variation of.

Proof: Fix f in and define the functional

, by. It is clear that is linear. Moreover. Indeed it is enough to prove that. If, in, then for each bounded subset B of X, uniformly for. But since, the set is bounded, so uniformly in. Therefore, , because the measure μ is of bounded variation. Hence.

Since X is semireflexive,; by Proposition 3.5(a), there is a unique such that,. Now let us define the operator by,. It is easily checked that T is linear, and satisfies the condition of the theorem by construction. We have to show that T is bounded. Let be a seminorm on X, and let K be a compact subset of X. For, we have:

which proves the continuity of T.

Now to compute, observe from the integral form of that.

Taking the supremum in both sides over, the polar set of the unit ball of X, we get:

So we deduce that. To see the reverse inequality, let us consider a function of the form, with satisfying and x fixed in X such that. With this choice, the function f belongs to the unit ball. Then we have

and

so that

since. So we get

because.

Therefore

■

By appealing to theorem 2.3, we get the following rather precise theorem:

3.8. Theorem

Let X be a locally convex Hausdorff semireflexive space. Then there is a one to one correspondence between the bounded operators of the class P and the X-valued Pettis integrals with respect to some bounded signed measure on B_{S}. This correspondence is given by the relation:

4. Acknowledgements

This work has been done under the Project No. 121/130/ 1432. The authors are grateful to the Deanship of Scientific Research of the King Abdulaziz University, Jeddah, for their financial support.

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