We study the existence of solutions of a nonlinear Volterra integral equation in the space . 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.

1. Introduction

In this paper, we present an existence result for the functional integral equation , where , , , and are given measurable functions while is an unknown function.

Equation (1) is a general form of many integral equations, such as the mixed Volterra-Fredholm integral equation which has been considered by many authors, see for example, [13] and references cited therein. Moreover, (1) contains the nonlinear Volterra and Fredholm integral equation on such as which is studied in [4, 5]. The existence of solution of Urysohn’s equation, was studied in [6] where he proved that (4) has a solution in space. The problem was studied in [1] where they obtained the existence of solution by using the classical Schauder fixed point principle. The nonlinear integral equation has been considered very recently by Liang et al. [7].

The main tool used in our research is a measure of weak noncompactness given by Banaś and Knab [3] to find a special subset of and also by applying the Krasnoselskii's fixed point theorem on this set. The existence results generalizing several previous works [1, 8] will be proved.

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, 912]).

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.

2. Preliminaries

Throughout this paper, we let be the set of all real numbers, , and denotes the space of the Lebesgue integrable functions on a measurable subset of with the standard norm The space and will be briefly denoted by and , respectively. Let be an interval of bounded or not.

Definition 1. Consider a function . We say that satisfies Carathéodory conditions if it is measurable in for any 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 . 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]. Krasnoselskiĭ [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 continuously into itself if and only if for all and , where is a function from the spaces and .

Theorem 3 (see [17]). Let be a bounded interval and let be a function satisfying Carathéodory condition. Then it possesses the Scorza-Dragoni property; that is, for each , there exists a closed subset of such that and is continuous.

Now, we are going to review a theorem from [14] about the continuity of the linear Volterra integral operator on the space . Let and 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 such that Then the Volterra integral operator generated by maps continuously the space into itself and the norm of this operator is majorized by the number .

3. 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 is said to be a measure of weak noncompactness, if it satisfies the following conditions:(1)the family is nonempty and .(2); (3), where denotes the convex hull of ;(4);(5)if is a sequence of nonempty, weak closed subsets of with bounded and with , then is nonempty.

The family 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 . Indeed, since for any natural, it follows that we get . 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 , where is a bounded interval in (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 , set where then is a measure of weak noncompactness.

The following results will be used in the sequel.

Theorem 6. A bounded set is relatively weakly compact in if and only if the following two conditions are satisfied:(a)for any there exists such that if meas then for all ;(b)for any there exists such that for all .

Theorem 7 (see [Krasnoselskii’s fixed point theorem [22]]). Let be a closed convex and nonempty subset of a Banach space . Let and be two operators such that(1),(2) is a contraction mapping,(3) is relatively compact and is continuous.

Then there exists such that .

Lemma 8 (see [23]). Let be a Lebesgue measurable subset of and . If is a convergent sequence to in the -norm, then there is a subsequence which converges to a.e., and there is , , such that

4. Main Result

In this section, we consider (1) and we will study the existence of solution under the following assumptions.(i)The functions satisfies Carathéodory conditions and there exist and functions , such that Additionally, the function satisfies the following Lipschitz condition for almost all : (ii)The function , , is measurable on , for all . Moreover, the function is continuous on the set for each .(iii) for , where , satisfies Carathéodory conditions. Moreover, the integral operator generated by , that is, is a continuous map from into itself and .(iv)The function , , is measurable on , for all . Moreover, the function is continuous on for each and for , where .(v) are absolutely continuous functions and satisfy , where is an arbitrary subset of and for all , .(vi).

We need to the following theorem in the sequel.

Theorem 9. Under the assumptions (ii), (iii), (iv), and (v), the operators map continuously into itself.

Proof. Let be an arbitrary sequence in which converges to in -norm. By using Lemma 8, there is a subsequence which converges to a.e. and there is such that Since almost everywhere in and is an absolutely continuous function; then from the continuity of with respect to the third variable, we get and we have Hence, by the Lebesgue dominated convergence theorem, we have Inequality (21) implies that for almost all . Regarding the assumptions on and , we obtain Then from (22), inequalities (23) and (24) and the Lebesgue dominated convergence theorem imply Since any sequence converging to in has a convergent subsequence such that in , we can conclude that is a continuous operator.
Similarly, we can prove that is a continuous operator.

Theorem 10. Under assumptions (i)–(vi), the problem (1) has at least one solution .

Proof. Denote by the nonlinear Volterra integral operator defined by the formula Then (1) may be written in the following form: The proof will be given as follows.
We show that is a contraction. From (i), it follows that From Theorem 9, the operators map continuously into itself; then by using Theorem 2, the Nemytskii operator generated by is a continuous operator from into itself. Therefore, is a continuous operator from into .
Now, our aim is to see whether there exists such that . For this, let us consider and be arbitrary functions in with to be determined later. In view of our assumptions we get where and is the norm of the bounded linear operator . Hence, from assumption , one can deduce that for , the operator takes into itself; that is, Now, we prove that for all bounded subset . Take arbitrary numbers and such that meas , for any ; we have Since the set having only one element is also weakly compact, then we have Therefore, We conclude that For and any . By similar calculation, we have Since the set having only one element is also weakly compact, then we have Therefore, Hence, we conclude that Therefore, from (35) and (39), we obtain for all bounded set contained in . Let Then and are nonempty, closed, and convex; so they are weakly closed. Moreover, which yields to Define Then . So, by the definition of measure of weak noncompactness, is nonempty, closed convex and relatively weakly compact. Moreover, Now, we prove that is relatively compact in . Let be an arbitrary sequence. Since , then which implies that and are relatively weakly compact.
Fix an arbitrary . Applying Theorem 9, for relatively weakly compact set , there exists such that Considering the functions , , , , and , by using Lusin theorem and the generalized version of Scorza-Dragoni theorem, see [24], there exists a closed subset of interval satisfying such that , , , , and , are continuous. Let us consider an arbitrary . Then for , we have where Let us take arbitrary , and assume that without loss of generality. By the monotone increasing property of , we have . Therefore, from uniform continuity of , we infer that where denotes the modulus of continuity of on the set .
Note that the sequence is weakly compact; then take Theorem 6 into account; we infer that the terms of the numerical sequence are arbitrary small provided that the numbers are small enough. Remark that if is small enough, then is small enough. Then the previous inequality tends to zero independent of as tends to zero. From (48), Therefore, the sequence is equibounded and equicontinuous on the set . Obviously from assumption (49) and similarly to the sequence , we can easily infer that the sequence is equibounded and equicontinuous on the set . Hence, uniform continuity of implies that the sequence is equibounded and equicontinuous on . Then, by Arzela-Ascoli theorem, the sequence has a convergent subsequence in the norm . Therefore, is a relatively compact subset of .
Furthermore, observe that the above reasoning does not depend on the choice of . Thus we can construct a sequence of a closed subsets of the interval such that as and such that the sequence is relatively compact in every space . Passing to subsequences if necessary we can assume that is a Cauchy sequence in each space , for .
Now, by using (46), we have From the relatively weakly compactness of , we can find such that for each closed subset of the interval with , we obtain Considering the fact that is a Cauchy sequence in for each . One can find such that meas and, for , Consequently, (56) and (57) imply that Now, by considering (55) and the last inequality, we obtain which shows that the sequence is a Cauchy sequence in the Banach space . Then has a convergent subsequence, which implies that is a relatively compact subset of .
Thus, we conclude by using Krasnoselskii’s fixed point theorem that the problem (1) has at least one solution .

5. 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: Put for and . Therefore, we have All conditions of Theorem 10 are satisfied. Hence, the nonlinear equation (60) has at least one solution in .

Conflict of Interests

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