Integrable Solutions of a Nonlinear Integral Equation via Noncompactness Measure and Krasnoselskii’s Fixed Point Theorem

Mahmoud Bousselsal and Sidi Hamidou Jah 1 Laboratoire d’Analyse Nonlineaire et HM, Department of Mathematics, ENS-Kouba, Vieux-Kouba 16250, Algiers, Algeria 2 Department of Mathematics, College of Science, Qassim University, P.O. Box 6644, Buraydah 51452, Saudi Arabia Correspondence should be addressed to Mahmoud Bousselsal; bousselsal55@gmail.com Received 20 September 2013; Revised 2 January 2014; Accepted 12 January 2014; Published 16 March 2014 Academic Editor: Seenith Sivasundaram Copyright © 2014 M. Bousselsal and S. H. Jah. 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. We study the existence of solutions of a nonlinear Volterra integral equation in the space L1[0, +∞).With the help of Krasnoselskii’s fixed point theorem and the theory of measure of weak noncompactness, we prove an existence result for a functional integral equation which includes several classes on nonlinear integral equations. Our results extend and generalize some previous works. An example is given to support our results.

Let us mention that the theory of functional integral equations has many useful applications in describing numerous events and problems of the real world. For example, integral equations are often applicable in engineering, mathematical physics, economics, and biology (cf. [3,4,[9][10][11][12]).
The paper is organized in five sections, including the introduction. Some preliminaries, notations, and auxiliary facts are presented in Section 2; in Section 3, we will introduce the main tools: measure of weak noncompactness and Krasnoselskii's fixed point theorem. The main theorem in our paper will be established in Section 4. In Section 5, we give an example to illustrate our results.

Preliminaries
Throughout this paper, we let R be the set of all real numbers, R + = [0, ∞), and 1 ( ) denotes the space of the Lebesgue integrable functions on a measurable subset of R with the standard norm The space 1 (R + ) and ‖ ‖ 1 (R + ) will be briefly denoted by 1 and ‖ ⋅ ‖, respectively. Let be an interval of R bounded or not.
Definition 1. Consider a function ( , , ) := : × R × R → R. We say that satisfies Carathéodory conditions if it is measurable in for any ( , ) ∈ R 2 and continuous in ( , ) for almost all ∈ . Now, we make a short note about the so-called superposition operator, which is one of the simplest and more important operators that are investigated in nonlinear functional analysis, see [13]. Consider a function ( , ) := : × R. Then for every function ( ) being measurable on , we may assign the function ( )( ) = ( , ( )), ∈ . The operator defined in such a way is called the superposition operator generated by the function , for more details about this theory the reader can see [5,7,14]. Krasnoselskiȋ [15] and Appell and Zabrejko [12] have proved the following assertion when is a bounded and an unbounded domain, respectively. Theorem 2 (see [16]). The superposition operator generated by the function ( , ) maps the space 1 ( ) continuously into itself if and only if | ( , )| ≤ ( ) + | | for all ∈ and ∈ R, where is a function from the spaces 1 ( ) and ∈ R + .
Theorem 3 (see [17]). Let be a bounded interval and let : × R → R be a function satisfying Carathéodory condition. Then it possesses the Scorza-Dragoni property; that is, for each > 0, there exists a closed subset of such that and | ×R is continuous.
Now, we are going to review a theorem from [14] about the continuity of the linear Volterra integral operator on the space 1 = 1 (R + ). Let Δ = {( , ) : 0 ≤ ≤ } and : Δ → R be measurable functions with respect to both variables. Consider We notice that is a linear Volterra integral operator generated by .
Theorem 4 (see [14]). Let be measurable on Δ such that Then the Volterra integral operator generated by maps continuously the space 1 (R + ) into itself and the norm ‖ ‖ of this operator is majorized by the number sup ≥0 ∫ ∞ | ( , )| .

Measure of Weak Noncompactness
In this section, we collect a few auxiliary facts concerning mainly measures of noncompactness, see [18]. Let ( , ‖ ⋅ ‖) be a real Banach space with the zero element . Denote by ( , ) the closed ball centered at with radius . We will write to denote the ball ( , ).
The family of all nonempty and bounded subsets of will be denoted by ( ), while the subfamily consisting of all relatively weak compact sets is denoted by ( ).
Definition 5 (see [19]). A function : ( ) → R + is said to be a measure of weak noncompactness, if it satisfies the following conditions: International Journal of Analysis 3 The family ker described in (1) is said to be the kernel of the measure of weak noncompactness . Observe that the intersection set ∞ from (5) is a member of ker . Indeed, since ( ∞ ) ≤ ( ) for any natural, it follows that we get ( ∞ ) = 0. This simple observation will play an important role further on.
We mention that the first important example of measure of weak noncompactness was given by De Blasi [20] with the help of the following formula: The measure of the weak noncompactness has a lot of interesting properties and it is also applied in nonlinear analysis [20]. We observe that it is rather difficult to express the De Blasi measure of weak noncompactness with the help of a convenient and useful formula in a concrete Banach space. Such a formula is only known in the space 1 ( ), where is a bounded interval in R (cf. [21]).
The following nice example of measure of weak noncompactness is a typical measure of weak noncompactness, which was given by Banaś and Knap in [14]. For every nonempty and bounded subset of the space 1 , set where then : ( ) → R + is a measure of weak noncompactness.
The following results will be used in the sequel. 1 if and only if the following two conditions are satisfied:

Let be a closed convex and nonempty subset of a Banach space . Let and be two operators such that
is a contraction mapping,

) is relatively compact and is continuous.
Then there exists ∈ such that + = .
Lemma 8 (see [23]). Let be a Lebesgue measurable subset of R and 1 ≤ ≤ ∞. If { } is a convergent sequence to ∈ ( ) in the -norm, then there is a subsequence { } which converges to a.e., and there is ∈ ( ), ≥ 0, such that

Main Result
In this section, we consider (1) and we will study the existence of solution under the following assumptions.
We need to the following theorem in the sequel. Proof. Let { } be an arbitrary sequence in 1 which converges to in 1 -norm. By using Lemma 8, there is a subsequence { } which converges to a.e. and there is ℎ ∈ 1 such that Since → almost everywhere in R + and 3 is an absolutely continuous function; then from the continuity of with respect to the third variable, we get and we have ( , , ( 3 ( ))) ≤ 1 ( , ) + 2 ( , ) ℎ ( 3 ( )) . (21) Hence, by the Lebesgue dominated convergence theorem, we have Inequality (21) implies that for almost all ∈ R + . Regarding the assumptions on 1 and 2 , we obtain Then from (22), inequalities (23) and (24) and the Lebesgue dominated convergence theorem imply Since any sequence { } converging to in 1 has a convergent subsequence { } such that → in 1 , we can conclude that : 1 → 1 is a continuous operator.

Theorem 10. Under assumptions (i)-(vi), the problem
Then (1) may be written in the following form: The proof will be given as follows.
Now, by considering (55) and the last inequality, we obtain which shows that the sequence { } is a Cauchy sequence in the Banach space 1 . Then { } has a convergent subsequence, which implies that ( ∞ ) is a relatively compact subset of 1 .
Thus, we conclude by using Krasnoselskii's fixed point theorem that the problem (1) has at least one solution ∈ 1 .

Example
In this section, we give an example, which can be treated by Theorem 10, but not by the related theorem in [25], since it satisfies the assumptions of Theorem 10, but not fulfill the assumptions of results in [25]. To illustrate the new existence result, we consider the following nonlinear integral equation: ( ) = 1 2 + 1 + 1 6 ( ( ) + (2 )) + 1 + 2 cosh ( ) + ∫ for , ∈ R + and ∈ R. Therefore, we have All conditions of Theorem 10 are satisfied. Hence, the nonlinear equation (60) has at least one solution in 1 .