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Journal of Function Spaces
Volume 2016 (2016), Article ID 9896502, 11 pages
http://dx.doi.org/10.1155/2016/9896502
Research Article

A Cone Measure of Noncompactness and Some Generalizations of Darbo’s Theorem with Applications to Functional Integral Equations

1Department of Mathematics, College of Science, King Saud University, P.O. Box 2455, Riyadh 11451, Saudi Arabia
2Department of Mathematics, Aligarh Muslim University, Aligarh 202002, India
3Departamento de Matemàticas, Universidad de Las Palmas de Gran Canaria, Campus de Tafira Baja, 35017 Las Palmas de Gran Canaria, Spain

Received 27 June 2016; Accepted 21 September 2016

Academic Editor: Jamshaid Ahmad

Copyright © 2016 Mohamed Jleli 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

We introduce the concept of cone measure of noncompactness and obtain some generalizations of Darbo’s theorem via this new concept. As an application, we establish an existence theorem for a system of integral equations. An example is also provided to illustrate the obtained result.

1. Introduction and Preliminaries

The measure of noncompactness concept is a very useful tool in nonlinear analysis, in particular when we deal with existence problems for functional operator equations. The measure of noncompactness concept was defined by many authors in different manners. See, for examples, Kuratowski [1], Akhmerov et al. [2], Appell [3], Deimling [4], Vath [5], Zeidler [6], Banaś and Goebel [7], and Dhage [8]. For the applications of the measure of noncompactness argument, we refer to [914] and the references therein.

In this paper, we introduce the concept of the cone measure of noncompactness and we establish some generalizations/extensions of Darbo’s fixed point theorem with respect to such a measure. The obtained results generalize several fixed point theorems obtained recently by many authors. Next, we present an application to functional integral equations.

At first, let us fix some notations and recall some basic concepts on cones in Banach spaces. For more details, we refer to books [4, 15, 16].

Let be a Banach space with respect to a certain norm . We denote by the zero vector of .

Definition 1. A subset of the Banach space is said to be a cone if it satisfies the following conditions: (K1) is a nonempty and closed subset of .(K2)For every and , one has .(K3)For every , one hasGiven a cone , one can define a partial order in by For , the notation means that and , while will stand for (interior of ).

Definition 2. Let be a cone of the Banach space . Then is called normal if there exists a number such thatThe least positive number satisfying (3) is called the normal constant of . It is clear that .

Definition 3. Let be a cone of the Banach space . One says that is nonnormal if is not a normal cone.

Example 4 (see [4]). Let be the set of functions such that is in . We endow the set with the norm where is the uniform norm. Then is a Banach space with respect to the norm . Let Then is a nonnormal cone of .

Definition 5. A cone is solid if it contains interior points; .

Definition 6. Let be a cone in . One says that the cone is regular if every decreasing sequence which is bounded from below is convergent; that is, if is a sequence such that for some , then there is some such that as .

Definition 7. Let be a solid cone of the Banach space . Let be a sequence in . One says that is -convergent if it satisfies the following condition: We denote to indicate that is -convergent.

Lemma 8 (see [17]). Let be a solid cone of the Banach space . Let be a sequence in such that . Then

Lemma 9 (see [17]). (i) If and , then .
(ii) If for every , then .

Lemma 10. Let be a solid cone of the Banach space . Let and be two sequences in such that Then

Proof. It follows from Lemma 8 and (i) in Lemma 9.

Remark 11. If is a nonnormal cone of the Banach space , then the sandwich theorem does not hold. In particular, if and are two sequences in such that this does not imply that as (see [17]).

Remark 12. Obviously, the sandwich theorem is satisfied when we deal with a normal cone.

We denote by the set of linear and bounded operators on . In the sequel, is supposed to be a solid cone of (not necessarily normal).

Lemma 13. Let be such that . Then

Proof. Let be a pair of points in such that . By the definition of the partial order , this means that . Since is linear and , we obtain ; that is, .

Let be a Banach space with respect to a certain norm with zero vector . For any subsets and of , we consider the following notations: denotes the closure of . denotes the convex hull of . denotes the set of nonempty subsets of . and () stand for algebraic operations on sets and .

We denote by the family of all nonempty bounded subsets of .

We introduce the concept of the cone measure of noncompactness as follows.

Definition 14. Let be a given mapping. One says that is a cone measure of noncompactness on if the following conditions are satisfied: (i)For every , is precompact.(ii)For every pair , one has (iii)For every , one has (iv)If is a decreasing sequence (with respect to ) of closed sets such that , then is nonempty.

Remark 15. Observe that if is a normal cone with normal constant and is a cone measure of noncompactness on , then the mapping defined by is a measure of noncompactness in the sense of Dhage [8].

Example 16. Let be standard measures of noncompactness (real valued measures of noncompactness) on . Define the mapping by For , is a standard measure of noncompactness. However, for , is a cone measure of noncompactness on with respect to and the cone , but it is not a standard measure of noncompactness.

Now, we are ready to state and prove our main results. This is the aim of the next section.

2. Main Results

We continue to use the same notations fixed in the previous section.

Our first result is a Darbo-type fixed point theorem with respect to a cone measure of noncompactness.

We denote by the set of elements satisfying the following conditions:(A1) .(A2) For all , as .

Theorem 17. Let be a nonempty, bounded, closed, and convex subset of the Banach space . Let be a mapping satisfying the following conditions: (i) is continuous.(ii)There exist and a cone measure of noncompactness such that Then has at least one fixed point. Moreover, the set of fixed points of is precompact.

Proof. Consider the sequence of subsets of defined byBy induction, we obtain easily that Then is a decreasing sequence of closed and convex sets. On the other hand, we have Take in the above inequality; we obtain For , we have which yields from Lemma 13 Continuing this process, by induction, we obtain Since as , by Lemma 10, we obtainOn the base of axiom (iv) of Definition 14, we infer that the set is nonempty, closed, and convex. Since from axiom (ii) of Definition 14, we have Let be fixed. From (25), there exists , a positive integer, such that Using property (i) in Lemma 9, we obtain Then by property (ii) in Lemma 9, we deduce that which gives us from axiom (i) of Definition 14 that is precompact; then it is compact since it is closed. Observe that . Then the continuity of the mapping and Schauder’s fixed point theorem give us that has at least one fixed point in . Finally, since the set of fixed points of is a nonempty subset of and , on the base of axioms (i) and (ii) of Definition 14, we deduce that the set of fixed points of is precompact.

We denote by the set of functions satisfying the following condition: for every sequence in , we have

We have the following result.

Theorem 18. Let be a nonempty, bounded, closed, and convex subset of the Banach space . Let be a mapping satisfying the following conditions: (i) is continuous.(ii)There exist , , and a cone measure of noncompactness such that Then has at least one fixed point.

Proof. Let us consider the sequence of subsets of defined by (18). Then is a decreasing sequence of closed and convex sets. If for some we have , then by axiom (i) of Definition 14, will be compact. Since , Schauder’s fixed point theorem applied to the self-mapping gives the desired result. So we may suppose that for every . For , since and (from axiom (iii) of Definition 14) and then by assumption (ii), we haveAgain, for , we haveFrom (33) and (34), we obtain Continuing this process, by induction, we get Passing to the limit as , we obtain which yields By axiom (iv) of Definition 14, we infer that the set is nonempty, closed, and convex. The rest of the proof is similar to the proof of Theorem 17.

Let be the set of functions satisfying the following conditions: () is a nondecreasing function with respect to the partial order ; that is,  ()For all , the sequence converges to as .

Theorem 19. Let be a nonempty, bounded, closed, and convex subset of the Banach space . Let be a mapping satisfying the following conditions: (i) is continuous.(ii)There exist and a cone measure of noncompactness such that Then has at least one fixed point.

Proof. As previously mentioned, we consider the sequence of subsets of defined by (18). Then is a decreasing sequence of closed and convex sets. In the same manner as before, we may assume that for every Taking into account our assumptions, for all , we have that is, Since as , by Lemma 10, we obtain By axiom (iv) of Definition 14, we infer that the set is nonempty, closed, and convex. The rest of the proof is similar to the proof of Theorem 17.

Theorem 20. Let be a nonempty, bounded, closed, and convex subset of the Banach space and be a cone measure of noncompactness on . Let be a mapping satisfying the following conditions: (i) is continuous.(ii)For any with , there exists such that Moreover, we suppose that (iii) is a regular cone. Then has at least one fixed point.

Proof. We consider the sequence of subsets of defined by (18). Then is a decreasing sequence of closed and convex sets. From axiom (ii) of Definition 14, we have Since is a regular cone, there is some such that In the same manner as before, we may assume that for every Suppose now that . Take and ; we have Then there exists such that Passing to the limit as , we obtain On the other hand, since , we have Therefore, which is a contradiction with . As a consequence, we have which implies from Lemma 8 that By axiom (iv) of Definition 14, we infer that the set is nonempty, closed, and convex. The rest of the proof is similar to the proof of Theorem 17.

The following result is a Sadovskii’s fixed point theorem with respect to a cone measure of noncompactness.

Theorem 21. Let be a nonempty, bounded, closed, and convex subset of the Banach space and be a cone measure of noncompactness on satisfying the following condition: (i)There exists such that Let be a mapping satisfying the following conditions: (ii) is continuous.(iii)For every , we have Then has at least one fixed point.

Proof. Let us denote by the set of subsets satisfying the following conditions: , is closed, is convex, and . Clearly is a nonempty set since . Set Then is a nonempty (), closed, and convex set. Moreover, we have . Set We claim that . In fact, we have and , which yields . On the other hand, the inclusion implies that . Note also that . Then and . This proves our claim. Next, from (i) and axiom (iii) of Definition 14, we obtain Suppose that ; then from (iii), we have which is a contradiction. As a consequence, , which implies from axiom (i) of Definition 14 that is precompact, so it is compact since it is closed. Finally, by Schauder’s fixed point theorem, the mapping has at least one fixed point.

Theorem 22. Let be a nonempty, bounded, closed, and convex subset of the Banach space and be a cone measure of noncompactness on . Let be a given mapping. Suppose that (i) is continuous.(ii)There exists such that, for all and , (iii) is closed, where is the identity mapping.(iv)One has Then has at least one fixed point.

Proof. Let be a sequence in such that as . Consider the sequence of operators defined by Note that is well-defined since is a closed set. Using the considered assumptions, for all , for all , we have Define the sequence of operators by Clearly, we have By Theorem 17, for all , the operator has a fixed point ; that is, This yields Passing to the limit as and using the fact that is a bounded sequence, we get Since is closed, we deduce that . As a consequence, there is some such that , which means that is a fixed point of .

Let be two cone measures of noncompactness on , where is a normal cone with normal constant . We define the mapping by We endow the product set with the norm defined by Let be the partial order on defined by Observe that We denote by the set of linear and bounded operators on . We denote by the set of elements satisfying the following conditions: (A1) .(A2) For all , as .

We endow also the product set with the norm defined by

We are interested to study the existence of solutions to the problem: find such thatwhere is a nonempty, bounded, closed, and convex subset of and , , are continuous mappings.

We have the following result.

Theorem 23. Suppose that there exists such thatThen Pb. (75) has at least one solution.

Proof. Let us define the mapping by Observe that is a solution to Pb. (75) if and only if is a fixed point of . Let us consider the two sequences and of subsets of defined by Then is a decreasing sequence of closed and convex sets. Similarly, is a decreasing sequence of closed and convex sets. On the other hand, we have Similarly, we have Then by the definition of , we have Using (76), we obtain Using the properties of the operator , by induction, we obtain which yields Passing to the limit as , we obtain which is equivalent to Since (for ) is a cone measure of noncompactness, we deduce that and are nonempty, convex, and compact sets of . Moreover, we have and . Then the operator is well-defined. Finally, Schauder’s fixed point theorem gives us the desired result.

3. An Application to a System of Functional Integral Equations

In this section, we provide an application to study the existence of solutions to the following system of integral equations:where , , and , .

At first, let us fix some notations and recall some results that will be used later.

We denote by the set of real continuous functions defined in . We endow this set with the norm defined by Then is a Banach space over .

Let , where is the set of nonempty and bounded subsets of . For and , set We define the mapping by It was proved in [7] that the mapping defined by is a measure of noncompactness (in the sense of Banaś and Gobel) on the Banach space . Then it is a cone measure of noncompactness on with respect to the normal cone of the Banach space . Let be the mapping defined by For , let

We consider the following assumption: (A1)The functions and , , are continuous.

The following result is immediate.

Lemma 24. Under assumption (A1), for all , the mapping maps into ; that is, is a well-defined mapping.

Now, we consider the following additional assumptions: (A2)For , where are constants. (A3)For , where are nondecreasing functions. (A4)There exists some such that where , , , , , and , .

We denote by the closed ball in with center and radius ; that is,

Lemma 25. Under assumptions (A1)–(A4), for all , the mapping maps into ; that is, is a well-defined mapping.

Proof. Let be fixed. Let . For all , we have Then for all , we have This proves our result.

Lemma 26. Under assumptions (A1)–(A4), for all , the mapping maps continuously into .

Proof. Let and such that ; that is, . Let be fixed. For all , we have where and Note that from the uniform continuity of the function , we have as . Then we have This gives us the desired result.

Next, we consider the following assumption: (A5)The parameters , , satisfy the following inequality:

Our main result in this section is the following existence theorem.

Theorem 27. Under assumptions (A1)–(A5), Pb. (87) has at least one solution with and .

Proof. Let . Let , , and be such that . Without restriction of the generality, we may assume that . For all , we have(i) Estimate of . Observe that Set We obtainNote that, by the uniform continuity of the function we have as .
(ii) Estimate of . We have Therefore,Using (106), (109), and (112), we obtain Passing to the limit as , we obtain Let be the bounded operator defined by where is the matrix given by As a consequence, we have Since for , then . Moreover, from (A5), we have where denotes the spectral radius of the matrix . Then . Finally, from Theorem 23, Pb. (87) has at least one solution .

We end the paper with the following illustrative example.

Example 28. Consider the system of integral equations:Pb. (119) can be written in the form (87) with