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Abstract and Applied Analysis
Volume 2014, Article ID 301675, 6 pages
http://dx.doi.org/10.1155/2014/301675
Research Article

Implicit Vector Integral Equations Associated with Discontinuous Operators

1Department of Mathematics and Computer Science, University of Messina, Viale F. Stagno d’Alcontres 31, 98166 Messina, Italy
2Center for Fundamental Science, Kaohsiung Medical University, Kaohsiung 807, Taiwan
3Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia

Received 19 February 2014; Accepted 25 March 2014; Published 14 April 2014

Academic Editor: Chong Li

Copyright © 2014 Paolo Cubiotti and Jen-Chih Yao. 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 . We consider the vector integral equation for a.e. where and are given functions and are suitable subsets of . We prove an existence result for solutions , where the continuity of with respect to the second variable is not assumed. More precisely, is assumed to be a.e. equal (with respect to second variable) to a function which is almost everywhere continuous, where the involved null-measure sets should have a suitable geometry. It is easily seen that such a function can be discontinuous at each point . Our result, based on a very recent selection theorem, extends a previous result, valid for scalar case .

1. Introduction

Let . Recently, in the paper [1], the following integral equation was studied: given , , , , and ; find such that In [1], the following existence result was proved, where, unlike other results in the field (see, for instance, the papers [25] and references therein, to which we also refer for motivations for studying (1)), the continuity of with respect to the second variable was not assumed.

Theorem 1 (Theorem 1 of [1]). Let , a closed interval, a continuous function, , and two given functions. Let , , with and (the conjugate exponent of ), , and . Assume that  (i)there exists a function , two negligible sets , with closed, and a countable dense subset of , such that for all the function is measurable, and for a.e. one has  (ii) for all (the interior of ), one has ; (iii)if one puts, for all , then for a.e. one has  (iv)one has  (v)for all , the function is measurable; (vi)for a.e. , the function is continuous in , differentiable in and Then, there exists a solution to (1).

Of course, the main peculiarity of Theorem 1 resides in the kind of discontinuity that is allowed for . Indeed, it is easy to construct examples of functions , , and satisfying the assumptions of Theorem 1 and such that for all the function is discontinuous at all points .

Theorem 1 extends a previous result (Theorem 1 of [6]), valid for the case where does not depend on explicitly. At this point, it is natural to ask if Theorem 1 above can be extended to the more general case where the function takes its values in the space . In this direction, we note that some results exist for the vector explicit equation (see [7, 8]), while for the implicit equation (1) the problem is still unsolved.

The aim of this note is exactly to provide such an extension. In the following, if and , we will denote by the projection over the th axis. Moreover, we will denote by the -dimensional Lebesgue measure over . If , with for all , we will put . Finally, if and are as above, we will denote by the family of all subsets such that there exist sets , with for all , such that . The following is our main result (where denotes the positive open orthant of , and is the interior of in ).

Theorem 2. Let , and let be a nonempty, closed, connected, and locally connected subset of , with for all . Let be a continuous function and and two given functions. Let , , with and , , and . Finally, let be a countable dense subset of .
Assume that there exists a function and two sets , with closed, such that (i)for all the function is measurable; (ii)for a.e. one has Moreover, assume that (iii), for all ; (iv)if one puts, for all ; then for a.e. and all one has (where denotes the th component of the function ); (v)one has  (vi)for all , the function is measurable; (vii)for a.e. , the function is continuous in , differentiable in and Then, there exists such that

Theorem 2 will be proved as an application of the following selection theorem, recently proved in [9], which we now state for the reader’s convenience (in the following, if is a topological space, we will denote by the Borel family of ).

Theorem 3 (Theorem 2.2 of [9]). Let and be complete separable metric spaces, with , and let (endowed with the product topology). Let be positive regular Borel measures over , respectively, with finite and -finite.
Let be a separable metric space, and let be a multifunction with nonempty complete values. Finally, let be a given set, and, for each , let be the projection over . Assume that(i)the multifunction is -measurable (where denotes the completion of the Borel -algebra of with respect to the measure );(ii)for a.e. , one has Then, there exist sets , with and for all , and a function such that(a) for all ;(b)for all , the function is -measurable over ;(c)for a.e. , one has

The proof of Theorem 2 will be given in Section 2. Further, we will point out some counterexamples to possible improvements of Theorem 2.

2. Proof of Theorem 2

Before giving the proof of Theorem 2, we fix some notations. If , the space will be considered with its Euclidean norm . Moreover, if and , we put If , the space will be considered with the usual norm As usual, we put . For the basic definitions and facts about multifunctions, we refer to [10].

Proof of Theorem 2. Without loss of generality, we can assume that (8) and (10) hold for all . Moreover, we can suppose that . Firstly, we prove that the functions and are measurable. Observe that, by assumption (ii), for all , one has To see this, fix , and let . Since we get Now, assume that . Hence, there is such that . Since the function is continuous at , there exist such that Since , we get which is absurd. Therefore, the second equality in (18) is proved. The first one can be checked in analogous way.
Since is closed, it can be easily checked that the set is nonempty, countable, and dense in . Consequently, by Lemma at page 198 of [11], the function is -measurable (where denotes the family of all Lebesgue-measurable subsets of ). By (18) and Lemma of [12], it follows that the functions and are measurable over , as claimed.
By assumption (iii) and Theorem 2.4 di [13], there exists a set such that and the function is open (it carries open subsets of onto open subsets of ). Consequently, the multifunction defined by putting, for each , is lower semicontinuous in with nonempty values. Let be defined by putting, for all , Clearly, the function is -measurable and, by (18), one has Moreover, assumption (ii) and the closedness of imply that for all one has Let be the multifunction defined by setting, for each , Observe that is well-defined since for all one has Moreover, by the lower semicontinuity of and by (25), for all , we get Let (more precisely, ) be the multifunction defined by putting, for each , . By (28), for all , we get Moreover, the values of are closed (in ) subsets of .
Since is -measurable and is lower semicontinuous, by Proposition of [10] the multifunction is -weakly measurable. That is, for each set , with open in the relative topology of , the set belongs to . By Proposition 2.6 and Theorem 3.5 of [14], the multifunction is -measurable.
Since , there exist sets such that and for all . By Theorem 3, there exist sets , with and for all , and a function such that(a)for all one has ;(b)for all , the function is -measurable;(c)for a.e. , one has
Since is closed and is continuous, for all , the set is closed in . Consequently, for all , we get Now, let By (27), (32), and assumption (iv) we get Let be defined by putting By (34) and (35) we easily get Let and let be any countable dense subset of . Since , it is easily seen that is dense in . Let be any countable dense subset of . Then, the set is countable and dense in , and for all the function is measurable by the above construction. Moreover, taking into account (31), for all , one has Let be defined by putting, for all , where “” stands for “closed convex hull.” By Proposition 2 of [8], taking into account (36) and (38), we have that(a) has nonempty closed convex values;(b)for all , the multifunction is measurable;(c)for all , the multifunction has closed graph;(d)for all , one has Moreover, by (36), we have Now we want to apply Theorem 1 of [15], choosing , , , , , , , , , and To this aim, we can argue as in [8]. In particular, observe the following.(a). This follows easily from our assumptions (vi) and (vii) and the classical Lebesgue’s dominated convergence theorem.(b)If and is a sequence in , weakly convergent to in , then the sequence converges to strongly in . This follows by Theorem 2 at page 359 of [16], since is th power summable in (note that is measurable on by the classical Scorza-Dragoni theorem; see [17] or also [11]).(c)By (41), the function belongs to and .
Therefore, taking into account the above construction, all the assumptions of Theorem 1 of [15] are satisfied. Consequently, there exists a function and a set , with , such that That is, We now prove that the function satisfies our conclusion. To this aim, observe that, since , there exist sets , with for all , such that .
Fix . Let be the function By (45) we get hence for all we get the inequality hence . By (vi), (vii), and (45) we have that is strictly increasing. Moreover, by Lemma 2.2. at page 226 of [18] we get Consequently, by Theorem 2 of [19], the function is absolutely continuous. By Theorem 18.25 of [20], the set has null Lebesgue measure. Now, put Of course, . Choose any point . Since , by (45) we get For each , since , taking into account (48), we have Therefore, and for all we have Consequently, we get By (40) we get By (32) and (56) we get hence Since , we get
This completes the proof.

Remark 4. Of course, a function satisfying the assumptions of Theorem 2 can be discontinuous at each point . The Example at the end of [8] shows that, in the statement of Theorem 2, none of the sets and can be assumed to depend on . Moreover, the Example at the end of [6] shows that the second inequality in assumption (vii) cannot be weakened by assuming that

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

This research was partially supported by Grant NSC 103-2923-E-037-001-MY3.

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