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Journal of Function Spaces
Volume 2014 (2014), Article ID 324082, 9 pages
http://dx.doi.org/10.1155/2014/324082
Research Article

Existence and Asymptotic Stability of Solutions of a Functional Integral Equation via a Consequence of Sadovskii’s Theorem

1Department of Mathematics, Rzeszów University of Technology, al. Powstańców Warszawy 8, 35-959 Rzeszów, Poland
2Department of Mathematics, Sciences Faculty for Girls, King Abdulaziz University, Jeddah 21589, Saudi Arabia
3Department of Mathematics, Faculty of Science, Damanhour University, Damanhour 22511, Egypt
4Departamento de Matemáticas, Universidad de Las Palmas de Gran Canaria, Campus deTafira Baja, 35017 Las Palmas de Gran Canaria, Spain

Received 3 May 2014; Accepted 12 June 2014; Published 16 July 2014

Academic Editor: Józef Banaś

Copyright © 2014 Agnieszka Chlebowicz 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

Using the technique of measures of noncompactness and, in particular, a consequence of Sadovskii’s fixed point theorem, we prove a theorem about the existence and asymptotic stability of solutions of a functional integral equation. Moreover, in order to illustrate our results, we include one example and compare our results with those obtained in other papers appearing in the literature.

1. Introduction

Measures of noncompactness play very important role in nonlinear analysis. They are often applied to the theories of differential and integral equations as well as to the operator theory and geometry of Banach spaces. The concept of a measure of noncompactness was initiated by Kuratowski [1] and Darbo [2]. In [2] Darbo, by using the concept of a measure of noncompactness, proved a fixed point theorem. In [3] Sadovskii improved the results obtained in [2].

The purpose of this paper is to present a theorem on the existence and asymptotic stability of solutions of a functional integral equation. Our study will be placed in the Banach space of real functions which are defined, continuous, and bounded on the real half-axis . The functional integral equation studied in the paper contains as particular cases a lot of functional and integral equations appearing in the literature. The main tool used in our investigations is a consequence of Sadovskii’s fixed point theorem [3].

2. Notations, Definitions, and Auxiliary Facts

Let be a given real Banach space with a norm . For a nonempty subset of denote by the closure of and by the closed convex hull of . For and being subsets of , by and , , we denote the usual algebraic operations on and . Further, let denote the family of all nonempty and bounded subsets of and its subfamily consisting of all relatively compact subsets. If is a mapping defined on with real values, then by we denote the following family: This family will be called the kernel of the mapping . Following [4], we consider the following definition of the concept of a measure of noncompactness.

Definition 1. A mapping will be called a measure of noncompactness in if it satisfies the following conditions. (1)The family is nonempty and .(2).(3).(4).(5) for .(6)If is a sequence of closed subsets of such that and , then .

In [2] Darbo proved the following fixed point theorem.

Theorem 2. Let be a nonempty, bounded, closed, and convex subset of and let be a continuous mapping such that there exists a constant satisfying for any nonempty subset of , where is a measure of noncompactness.
Then has a fixed point in .

In [3], Sadovskii proved the following generalization of Theorem 2.

Theorem 3. Let be a nonempty, bounded, closed, and convex subset of and let be a continuous mapping such that for any nonempty and noncompact subset of , where is a measure of noncompactness in . Then has a fixed point in .

Notice that in [3] Theorem 3 is proved for a particular measure of noncompactness in , but the same argument serves for an arbitrary measure of noncompactness in [5, 6].

In our study, we will work in the Banach space consisting of all real, bounded, and continuous functions on . This space is furnished with the norm given by the formula In , we will use the measure of noncompactness which appears in [7, 8]. In order to present this measure of noncompactness, let us fix a nonempty, bounded subset of and a number . For and , we denote by the modulus of continuity of the function on the interval ; that is, Now, we consider the quantities Further, for a fixed number , we denote .

Finally, the measure of noncompactness which will be used in our study is defined as where . In [4], the authors proved that the function is a measure of noncompactness in .

In order to introduce the concept of asymptotic stability which will be used later, we assume that is a nonempty subset of and let be an operator. Also, consider the equation

Definition 4. One will say that solutions of (8) are locally attractive if there exists a ball in such that, for arbitrary solutions and of (8) belonging to , one has that
In the case when the limit in (9) is uniform with respect to the set , that is, when for each there exists such that for all being solutions of (8) and for any , one will say that solutions of (8) are asymptotically stable.

We will finish this section with the following generalization of Banach contraction mapping principle due to Geraghty [9] and where the class of functions is used satisfying By we denote the class of functions .

Theorem 5. Let be a complete metric space and let be an operator. Suppose that there exists such that for any Then has a unique fixed point in .

3. Main Result

We start this section with the following result which is a version of Theorem 5 in the context of measure of noncompactness.

Proposition 6. Let be a nonempty, bounded, closed, and convex subset of a Banach space and let be a continuous mapping such that for any nonempty and noncompact subset of , where and is an arbitrary measure of noncompactness in . Then has at least one fixed point.

Proof. Let be a nonempty and noncompact subset of . Then . We can distinguish two cases.
Case 1 . In this case, from (13) we get and therefore .
Case 2 . In this case, since the function has as range , from (13) we have .
Since is an arbitrary nonempty and noncompact subset of , the contractive condition appearing in Theorem 3 is satisfied. Finally, Theorem 3 says that has a fixed point in . This completes the proof.

Now, we present the following result which belongs to the classical metric fixed point theory.

Corollary 7. Let be a nonempty, bounded, closed, and convex subset of a Banach space and let be an operator satisfying for any , where is a nondecreasing function with . Then has a unique fixed point in .

Proof. Let be the function defined by , where It is easy to see that is a measure of noncompactness in [4].
Now, we take a nonempty subset of with . Using (14) and the fact that is nondecreasing, we have When , we infer that is a singleton; thus is also a singleton. Consequently, . Therefore, (16) is also satisfied when . Since , Proposition 6 gives us the existence of at least one fixed point in .
In order to prove the uniqueness of the fixed point, we take into account , since and, consequently, . Finally, since consists of singletons, is a singleton and this proves the uniqueness of the fixed point. The proof is complete.

An example of the function appearing in Corollary 7 is .

Now, we present the main result of the paper.

Theorem 8. Consider the following functional integral equation: under the following assumptions. (a)The function is continuous and the function is bounded.(b)There exists a continuous and nondecreasing function with , satisfying(b1) for any ,(b2),  such that (c)The function is continuous and there exist continuous functions such that(c1),(c2) for any and .(d)There exists a positive solution of the inequality , where Then (17) has at least one solution . Moreover, solutions of (17) are asymptotically stable.

Proof. Let us consider the operator defined on as follows: For convenience, we divide the proof into several steps.
Step 1 ( maps into itself). In fact, since and are continuous functions, for we infer that is continuous on . Now, we prove that for the function is bounded. In fact, for arbitrarily fixed we get This proves that Therefore, maps into itself.
Step 2 ( maps into itself). It follows from assumption that maps into itself.
Step 3 (an estimate of with respect to the quantity . For fixed and let us take with . Without loss of generality, we may assume that . Then for we obtain the following estimate: where we denoted From the uniform continuity of the functions and on the sets and , respectively, it follows that and when . Notice that, since and are continuous on , we have that . Therefore, we derive the following estimate: Since is nondecreasing, we obtain Hence Finally, we get
Step 4 (an estimate of with respect to the diameter). For and , we have Since is nondecreasing, from the last inequality it follows that Consequently, from assumption and the continuity of , we get
Step 5 ( satisfies the contractive condition of Proposition 6). In fact, from assumption , (28), (31), and the definition of the measure of noncompactness , we infer Now, considering in the measure of noncompactness defined by , the last estimate can be written in the form Therefore, if , then or equivalently where .
In the case we have that is a relatively compact subset of and, since is continuous, is also relatively compact and thus . This proves that (35) is also satisfied when . Summarizing, for any nonempty subset of , we have where (assumption ) and is a measure of noncompactness in .
In the sequel, let us consider the sequence of sets , where , , and so on. Notice that the sequence is decreasing; that is, for . Moreover, and the sets in this sequence are closed, convex, and nonempty.
On the other hand, in view of (32), we get and, by using induction, where we have used the nondecreasing character of and where denotes the th iteration. Taking into account , since , we have for and as is continuous, it follows that for [10].
Therefore, we deduce that Now, taking into account Definition 1, we deduce that the set is nonempty, bounded, closed, and convex. Moreover, since for any , is a member of the kernel of the measure of noncompactness . Let us also observe that the operator transforms the set into itself.
Next, we will prove that is continuous on the set . To do this let us fix a number and we take a sequence and such that . We have to prove that .
In fact, since , we have and, particularly, . Then, for we can find such that for any and . Particularly, since we have for any , and, thus, for , On the other hand, since is continuous on a compact set, it is uniformly continuous. This means that for we can find such that if for and , we have .
Taking into account that , we can find such that, for , .
For and , we have where we have used the fact that for and the nondecreasing character of .
From (40) and (41), for . This proves our claim.
Finally, taking into account that as and, consequently, is relatively compact, is a continuous operator, applying the classical Schauder fixed point theorem, we infer that the operator has at least one fixed point in .
In order to prove that solutions of (17) are asymptotically stable, we notice that any solution of (17) in is a fixed point of . Now, taking into account that transforms into itself, we have Since for any nonempty subset of , we have Further, we distinguish two cases.
Case 1 . In this case, by (43), .
Case 2 . In this case, by (43) and taking into account that the range of the function is , we infer which is a contradiction. Therefore .
Since for any nonempty subset , we deduce that . Taking into account the definition of the measure of noncompactness (see Section 2), we have But this means that for any we can find such that As all solutions of (17) in are in , by (46) we have that for there exists such that where and they are solutions of (17). This means that solutions of (17) are asymptotically stable. The proof is complete.

4. Example

In order to present an example which illustrates our results, we need to prove some properties about the inverse tangent function.

Lemma 9. The function defined as has the following properties. (a) is continuous, nondecreasing and satisfies .(b)For .

Proof. It is clear that is continuous, nondecreasing (since ) and .
Since for any , we infer that is concave and, therefore, for any .

Definition 10. A function is said to be subadditive if

Lemma 11. Suppose that is subadditive. Then, for , one has that

Proof. In fact, since the desired result follows.

Remark 12. From Lemma 11, we infer that if is subadditive, then

Lemma 13. Let be a concave function with . Then is subadditive.

Proof. Since is concave and , we have for Adding these inequalities side by side, we obtain Therefore is subadditive.

Remark 14. Since the function defined by is concave and , by Lemma 13, is subadditive. Taking into account Remark 12, we get or

Lemma 15. The function defined by for , where , belongs to the class .

Proof. From mathematical analysis, we know that , for . Therefore, the function maps into . This completes the proof.

Now, we are ready to present an example illustrating our results.

Example 16. Consider the following functional integral equation: Notice that (57) is a particular case of (17), provided we put It is clear that is continuous and, moreover, Thus, assumption of Theorem 8 is satisfied.
On the other hand, taking into account Remark 14, for any and , we obtain Thus, by Lemmas 9 and 15, the function satisfies assumption of Theorem 8.
Further, notice that is continuous and, for any and for , one has Putting and , it is clear that and are continuous functions on . Moreover, Since assumption of Theorem 8 is satisfied.
Now, we estimate the constant appearing in assumption of Theorem 8. Indeed, we have Notice that if , then and when , we have Thus, from the last inequality, we obtain Now, we consider the inequality An application of Bolzano’s theorem gives that this inequality is satisfied by a number . Therefore, assumption of Theorem 8 is satisfied.
Finally, by Theorem 8, we conclude that (57) has at least one solution in satisfying .

5. Final Remarks

In [10] the authors proved the following result.

Theorem 17 (Theorem 2.2 of [10]). Let be a nonempty, bounded, closed, and convex subset of a Banach space and let be a continuous operator satisfying for any nonempty subset of , where is an arbitrary measure of noncompactness and is a nondecreasing function such that for each , where denotes the -iteration of . Then has at least one fixed point in .

Notice that we can rewrite condition (69) in the form for any nonempty subset of with . When , this means that is a relatively compact subset of and, since is continuous, is also relatively compact subset of and, therefore, . Consequently, condition (70) is satisfied for any nonempty subset of . This tells us that Theorem of [10] can be reformulated in the following way.

Theorem 18. Let be a nonempty, bounded, closed, and convex subset of a Banach space and let be a continuous operator satisfying for any nonempty subset of , where is an arbitrary measure of noncompactness and belongs to the class of functions with , where is a nondecreasing function such that for each . Then has at least one fixed point in .

Now, we compare the classes of functions and appearing in Proposition 6 and Theorem 18, respectively. To do this, we need the following lemma which appears in [10] under weaker assumptions. For the paper to be self-contained, we present a proof.

Lemma 19. Let be a continuous and nondecreasing function. Then the following conditions are equivalent: (a) for any ,(b) for any .

Proof. (a)⇒(b) Suppose that the conclusion is not true. This means that we can find such that . Since is nondecreasing, we obtain for any and the sequence is nondecreasing. Therefore and this contradicts .
(b)⇒(a) Let be an arbitrary number but fixed with . Since and is nondecreasing we infer that the sequence of nonnegative real numbers is decreasing. Thus, for certain . Suppose that . Then, by , . On the other hand, since for any , the continuity of gives us which leads to a contradiction. Therefore, and this completes the proof. In virtue of Lemma 19, it is obvious that .

Now, we will prove that . To this end consider the function given by It is clear that maps into and therefore .

If , then , where is a nondecreasing and for . In view of the equality it is obvious that is not nondecreasing and, consequently, . This proves that .

In [7], the authors investigated the following functional integral equation: under the following assumptions. (i)The functions are continuous and as .(ii)The function is continuous and there exist positive constants , such that for and for . Moreover, we assume that .(iii)The function is bounded on with .(iv)The function is continuous and there exist functions such that for . Moreover, we assume that The main result of [7] is formulated as follows.

Theorem 20. Under the above assumptions, the functional integral equation (75) has at least one solution in the space .

Notice that (17) is a particular case of (75) with .

If we compare assumptions of Theorems 8 and 20, then we see that assumption of Theorem 8 and assumption of Theorem 20 are essentially distinct.

Next, we prove that assumption of Theorem 20 is a particular case of assumption of Theorem 8. Indeed, consider the function defined by with . Obviously, is continuous, , and is nondecreasing (since ). Since , is concave and by Lemma 13, is subadditive. Moreover, since we have that .

Therefore, we infer that assumption of Theorem 8 is more general than assumption of Theorem 20. Consequently, Theorem 8 generalizes and improves Theorem 20 (which is the main result of [7]) when for .

Conflict of Interests

The authors declare there is no conflict of interests in the submitted paper.

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