Journal of Function Spaces and Applications

Volume 2013, Article ID 715789, 5 pages

http://dx.doi.org/10.1155/2013/715789

## The Use of an Isometric Isomorphism on the Completion of the Space of Henstock-Kurzweil Integrable Functions

Facultad de Ciencias Físico Matemáticas, Benemérita Universidad, Autónoma de Puebla, Avenida San Claudio y 18 Sur, Colonia San Manuel, 72570, Puebla, Mexico

Received 19 April 2013; Accepted 5 June 2013

Academic Editor: Nelson Merentes

Copyright © 2013 Luis Ángel Gutiérrez Méndez et al. 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

Employing an isometrically isomorphic space, we determine new properties for the completion of the space of the Henstock-Kurzweil integrable functions with the Alexiewicz norm.

#### 1. Introduction

Let be a compact interval in . In the vector space of Henstock-Kurzweil integrable functions on with values in , the Alexiewicz seminorm is defined as

The corresponding normed space is built using the quotient space determined by the relation if and only if , *except in a set of Lebesgue measure zero or, equivalently, if they have the same indefinite integral.* This normed space will be denoted by .

It is known that is neither complete nor of the second category [1]. However, is a separable space [1] and, consequently, its completion also has the same property. In addition, has “nice” properties, from the point of view of functional analysis, since it is an ultrabornological space [2]. As is not complete, it is natural to study its completion, which will be denoted by .

Talvila in [3] makes an analysis to determine some properties of the Henstock-Kurzweil integral on , such as integration by parts, Hölder inequality, change of variables, convergence theorems, the Banach lattice structure, the Hake theorem, the Taylor theorem, and second mean value theorem. Talvila makes this analysis by means of the space of the distributions that are derivatives of the continuous functions on , which is an isometrically isomorphic space to . Making use of this same isometrically isomorphic space, Bongiorno and Panchapagesan in [4] establish characterizations for the relatively weakly compact subsets of and .

In this paper, we make an analysis on by means of another isometrically isomorphic space to prove that *has the Dunford-Pettis property, it has a complemented subspace isomorphic to **, it does not have the Radon-Riesz property, it is not weakly sequentially complete, and it is not isometrically isomorphic to the dual of any normed space;* hence, we will also prove that *is neither reflexive nor has the Schur property*. Then, as an application of the above results, we prove that is not isomorphic to the dual of any normed space and that the space of all bounded, linear, weakly compact operators from into itself is not a complemented subspace in the space of all bounded, linear operators from into itself.

#### 2. Preliminaries

In this section, we restate the conventions, notations, and concepts that will be used throughout this paper.

All the vector spaces are considered over the field of the real numbers or complex numbers.

Let be a normed space. By , we denote the dual space of . A topological property that holds with respect to the weak topology of is said to be a *weak* property or to hold *weakly*. On the other hand, if a topological property holds without specifying the topology, the norm topology is implied.

Let , be two Banach spaces. We denote by (, resp.) we denote the Banach space of all bounded, linear (bounded, linear, weakly compact resp.), operators from into . If , then we write (resp. ) instead of (resp. ).

The symbols , , and represent, as usual, the vector spaces of all sequences of scalars convergent to 0, all sequences of scalars absolutely convergent, and all bounded sequences of scalars, respectively, neither one with nor usual norm.

Let be a compact metric space. We denote by the vector space of all continuous functions of scalar-values on together with the norm defined by .

By we denote the following collection: which is a closed subspace of and is therefore a Banach space.

*Definition 1. *Let be normed spaces and let be a lineal operator. We have the following. (i)is an *isomorphism* if it is one-to-one and continuous and its inverse mapping is continuous on the range of . Moreover, if , for all , it is said that is an *isometric isomorphism*. (ii) are *isomorphic*, which is denoted by , if there exists an isomorphism from onto . (iii) are *isometrically isomorphic* if there exists an isometric isomorphism from onto .

The following result is key to our principal results.

Theorem 2 (see [4]). *The space is isometrically isomorphic to . *

According to Theorem 2, we will prove that has the Dunford-Pettis property.

*Definition 3. *Let be a Banach space. It is said that has the *Dunford-Pettis property* if for every sequence in converging weakly to 0 and every sequence in converging weakly to 0, the sequence converges to 0.

If a Banach space has the Dunford-Pettis property, then not necessarily every closed subspace of inherits such property, except when the subspace is complemented in [5].

*Definition 4. *Let be a subspace of a normed space . It is said that is complemented in if it is closed in and there exists a closed subspace in such that .

Theorem 5 (see [5]). * Let be a Banach space with the Dunford-Pettis property. If is a complemented subspace in , then has the Dunford-Pettis property. *

To prove that has a complemented subspace isomorphic to the following result is essential.

Theorem 6 (see [6]). *Let be a compact metric space. If is an infinite-dimensional complemented subspace of , then contains a complemented subspace isomorphic to . *

On the other hand, making use again of Theorem 2 we will prove that is neither weakly sequentially complete nor has the Radon-Riesz property.

*Definition 7. *Let be a normed space, and let be a sequence in and . (i)If weakly converges whenever is weakly Cauchy, then it is said that is *weakly sequentially complete*. (ii)If converges to whenever weakly converges to and , then it is said that has the *Radon-Riesz property* or the *Kadets-Klee property*. (iii)If converges to whenever weakly converges to , then it is said that has the *Schur property*.

The following result establishes a characterization of weakly Cauchy sequences and weakly convergent sequences of the space .

Theorem 8 (see [7]). * Let and be a sequence and an element, respectively, in the space . Then we have the following. *(1)*The sequence is weakly convergent to if and only if (i) , for all , (ii)there exists such that , for all . *(2)

*The sequence is weakly Cauchy if and only if (i)*

*exists, for all ,*(ii)*there exists such that , for all .*We will also prove that is not isometrically isomorphic to the dual of any normed space and, as a consequence, we will prove that is not reflexive, for which we will use again Theorem 2 and the concept of extremal point.

*Definition 9. *Let be a vector space, , and . It is said that is an *extremal point* of if for all such that it holds that .

If is a normed space, then its closed unit ball will be denoted by and the collection of all extremal points of as .

The following theorem establishes that the extremal points are preserved under isometric isomorphisms.

Theorem 10 (see [8]). * Let be Banach spaces, and let be an isometric isomorphism. Then is an extremal point of if and only if is an extremal point of . *

Corollary 11 (see [9]). * An infinite-dimensional normed space whose closed unit ball has only finitely many extreme points is not isometrically isomorphic to the dual of any normed space. *

#### 3. Principal Results

Lemma 12. *The space has the Dunford-Pettis property. *

*Proof. *Let be the functional defined by . It is clear that is bounded and therefore is a hyperplane of the space , that is,
where and is a one-dimension, subspace in . Then, as has the Dunford-Pettis property [10] and according to Theorem 5, we obtain the desired conclusion.

Lemma 13. *The space has a complemented subspace isomorphic to . *

*Proof. *Using the proof of Lemma 12, we can see that is a complemented subspace in . Then, according to Theorem 6, we obtain the desired conclusion.

Lemma 14. *The space is not weakly sequentially complete. *

*Proof. *Without loss of generality, suppose that . Let be the function defined by
for all . Thus, (i) exists, for all , (ii), for all .

Therefore, according to Theorem 8 item , it follows that the sequence is weakly Cauchy in .

Now, suppose that there exists a function such that the sequence converges weakly to . Then according to Theorem 8 item , it holds that , for all , that is,
However, since is not continuous, it follows that the sequence does not converge weakly in .

Lemma 15. *The space does not have the Radon-Riesz property. *

*Proof. *Without loss of generality, suppose that . Let be the function defined by
for all . Thus, (i)the sequence converges pointwise to , where for all , (ii), for all .

Therefore, according to Theorem 8 item , it follows that the sequence converges weakly to in ; in addition, as , it holds that . However, the sequence does not converge to in .

It is not difficult to prove that if a Banach space has the Dunford-Pettis property, or if it has a complemented subspace isomorphic to , or if it is not weakly sequentially complete, or if it has the Radon-Riesz property, then these properties are preserved under isometric isomorphisms. Therefore, according to Theorem 2 and Lemmas 12, 13, 14 and 15, we obtain the following result.

Proposition 16. *The space *(1)has the Dunford-Pettis property, (2)has a complemented subspace isomorphic to , (3)is not weakly sequentially complete, (4)does not have the Radon-Riesz property.

*Remark 17. *According to Definition 7 and Proposition 16 item , it follows that does not have the *Schur property*.

Lemma 18. *The collection of all extremal points of the closed unit ball of the space is empty. *

*Proof. *Let . Since is continuous on , it holds that for there exits such that
Now, define the following functions:
where the function can be any continuous function defined over the interval such that and .

Since the functions and are continuous, , and , it holds that cannot be an extremal point; therefore, .

It is a known fact that the collection is formed only by the constant functions . However, since in general there is not a relationship between the extremal points of the closed unit ball of a subspace with the extremal points of the closed unit ball of all space, we need Lemma 18 for the following result.

Proposition 19. *The space is not isometrically isomorphic to the dual of any normed space. *

*Proof. *By Lemma 18 and Theorems 2 and 10, it holds that . Then, as the space is dimensionality infinite and according to Corollary 11, we obtain the desired conclusion.

Corollary 20. *The space is not reflexive. *

*Proof. *Suppose that the space is reflexive. Then coincides under the canonical imbedding with its second dual. Therefore, the space is isometrically isomorphic to the dual of the space , which is a contradiction by Proposition 19.

In general, it is important to know when a Banach space enjoys certain functional analysis properties. However, in certain contexts, also it is useful to know when a Banach space does not have certain properties; Propositions 22 and 24 are examples of both facts.

Lemma 21 (see [11]). * Let be two Banach spaces. Assume that and contain a complemented copy of . Then is uncomplemented in . *

Proposition 22. *The space is uncomplemented in the space . *

*Proof. *According to Proposition 16 item , it holds that contains a complemented copy of . Therefore, on the basis of Lemma 21, we obtain the desired conclusion.

By Proposition 19, we can see that is not isometrically isomorphic to the dual of any normed space. However, we can ask ourselves the following. Is there a normed space X such that is isomorphic to the dual of ? To answer this question, we need of the following result.

Lemma 23 (see [7]). * Let be a normed space and let be a Banach space with the Dunford-Pettis property that does not have the Schur property. If contains a copy of , then contains a copy of . *

Proposition 24. *The space is not isomorphic to the dual of any normed space. *

*Proof. *Suppose that there exists a normed space such that
Since is separable [1] and consequently also is separable, it follows from (9) that is separable.

On the other hand, by Proposition 16 item and according to the isomorphism from (9), it holds that has a complemented subspace isomorphic to . Since has the Dunford-Pettis property [9] and does not have the Schur property [9], it holds that has a copy of , by Lemma 23.

As has a copy of , it holds that is isometrically isomorphic to , where denotes the annihilator of . Since is isometrically isomorphic to it holds that, in particular,

Therefore, since is separable and according to the isomorphism from (10), we obtain that is separable, which is a contradiction.

On this way, we can see that Proposition 19 and Corollary 20 are consequences of the above result. We did not do it this way because one of the principal objectives of this paper is to show the importance of knowing explicitly a closed subspace of which is isometrically isomorphic to and it is a known fact that every separable Banach space of infinite dimension is isometrically isomorphic to a closed subspace of ; however, this information is not sufficient to prove, in particular, the results that we have shown in this paper.

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