`Abstract and Applied AnalysisVolume 2012 (2012), Article ID 267531, 23 pageshttp://dx.doi.org/10.1155/2012/267531`
Research Article

Krasnosel’skii Type Fixed Point Theorems for Mappings on Nonconvex Sets

1Department of Mathematics, Faculty of Science for Girls, King Abdulaziz University, P.O. Box 4087, Jeddah 21491, Saudi Arabia
2Department of Mathematics, School of Mathematics, Statistics and Applied Mathematics, National University of Ireland, Galway, Ireland
3Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah 21859, Saudi Arabia

Received 17 May 2012; Revised 28 August 2012; Accepted 29 August 2012

Copyright © 2012 Maryam A. Alghamdi 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 prove Krasnosel'skii type fixed point theorems in situations where the domain is not necessarily convex. As an application, the existence of solutions for perturbed integral equation is considered in p-normed spaces.

1. Introduction

Let be a linear space over with the origin . A functional with is called a norm on if the following conditions hold(a) if and only if ;(b), for all ; (c), for all .

The pair is called a normed space. If , then is a usual normed space. A normed space is a metric linear space with a translation invariant metric given by for all .

Let be a nonempty set, let be a algebra in , and let be a positive measure. The space based on the complete measure space is an example of a normed space with the norm defined by Another example of a normed space is , the space of all continuous functions defined on the unit interval with the sup norm given by The class of normed spaces is a significant generalization of the class of usual normed spaces. For more details about normed spaces, we refer the reader to [1, 2].

It is noted that most fixed point theorems are concerned with convex sets. As we know, there exists nonconvex sets also, for example, the unit ball with center in a normed space is not a convex set. It is a natural question whether the well-known fixed point theorems could be extended to nonconvex sets. Xiao and Zhu [3] established the existence of fixed points of mappings on convex sets in normed spaces, where .

Theorem 1.1 (see [3] (Krasnosel’skii-type)). Let be a complete normed space and a bounded closed convex subset of , where . Let be a contraction mapping and a completely continuous mapping. If for all , then there exists such that .

In this paper, we investigate the fixed point problem of the sum of an expansive mapping and a compact mapping. Our results extend and complement the classical Krasnosel’skii fixed point theorem. We also prove the Sadovskii theorem for convex sets in normed spaces, where , and from it we obtain some fixed point theorems for the sum of two mappings. In the last section, as an application of a Krasnosel’skii-type theorem, the existence of solutions for perturbed integral equation is considered in normed spaces.

2. Preliminaries

Throughout this paper, we denote the closure and the boundary of a subset of by and , respectively. will be the open ball of with center and radius .

Definition 2.1. Let and be two metric spaces and . The mapping is said to be Lipschitz, where is a positive constant, if is said to be nonexpansive if , and to be a contraction if .

It is clear that every Lipschitz mapping is continuous. Moreover, the Banach contraction principle holds for a closed subset in a complete normed space.

Definition 2.2 (see [3]). Let be a normed space and . A set is said to be convex if the following condition is satisfied Let . The convex hull of denoted by is the smallest convex set containing and the closed convex hull of denoted by is the smallest closed convex set containing .

In other words, the convexity of the set is equivalent to that For , we obtain the usual definition of convex sets. For a subset of , the convex hull of is given by It is easy to see that if is a closed convex set, then .

Lemma 2.3 (see [3]). Let be a normed space and .(a)The ball is convex, where .(b)If is convex and , then is convex. (c)If are convex, then is convex.(d)If are all convex, then is convex.(e) If and , then , where is the convex hull of .(f)If is a closed convex set and , then is a closed convex set.(g) If is complete and is a totally bounded subset of , then is compact.

Theorem 2.4 (see [3] (Schauder-type)). Let be a complete normed space and a compact convex subset of , where . If is continuous, then has a fixed point (i.e., there exists such that ).

Theorem 2.5. Let be a complete normed space and a closed convex subset of , where . If is a continuous compact map (i.e., the image of under is compact), then has a fixed point.

Proof. Let . Note is a closed compact -convex subset of and . The result follows from Theorem 2.4.

We will need the following definition.

Definition 2.6 (see [4]). Let be a metric space and a subset of . The mapping is said to be expansive, if there exists a constant such that

Theorem 2.7 (see [4]). Let be a closed subset of a complete metric space . Assume that the mapping is expansive and , then there exists a unique point such that .

Recently, Xiang and Yuan [4] established a Krasnosel’skii type fixed point theorem when the mapping is expansive. For other related results, see also [5, 6].

3. Main Results

Theorem 3.1. Let be a complete normed space and a closed convex subset of , where . Suppose that(i) is a continuous compact mapping (i.e., the image of under is compact);(ii) is an expansive mapping;(iii) implies where .
Then there exists a point such that .

Proof. Let . Then the mapping satisfies the assumptions of Theorem 2.7 by virtue of (ii) and (iii), which guarantees that the equation has a unique solution . For any , we have and so Since is expansive, there exists a constant such that As a result This implies that is continuous. Since is continuous on , it follows that is also continuous. Since is compact, so is . By Theorem 2.5, there exists , such that . From (3.1), we have that is, This completes the proof.

Corollary 3.2. Let be a complete normed space and a closed convex subset of , where . Suppose that(i) is a continuous compact mapping;(ii) is an expansive and onto mapping.
Then there exists a point such that .

The following example shows that there are mappings which are expansive and satisfy .

Example 3.3. Let with the usual metric and consider for . Then for all , we have Thus is expansive with .

Theorem 3.4. Let be a complete normed space and a closed convex subset of , where . Suppose that(i) is a continuous compact mapping;(ii) is an expansive mapping;(iii) implies where .
Then there exists a point such that .

Proof. Since is expansive, it follows that the inverse of exists, is a contraction and hence continuous. Thus is a closed set. Then, for each fixed , the equation has a unique solution . For any , we have so that Thus, This shows that is continuous. Since is continuous on , it follows that is also continuous and since is compact, so is . Then, by Theorem 2.5, there exists , such that . From (3.9) we have that is, that is, This completes the proof.

Lemma 3.5. Let be a normed space and . Suppose that the mapping is expansive with constant . Then the inverse of exists and

Proof. Let , we have From (3.17) we see that is one-to-one. Therefore, the inverse of exists. Thus, for , we have . Now, using and substituting for in (3.17), respectively, we obtain

Theorem 3.6. Let be a complete normed space and a closed convex subset of , where . Suppose that(i) is a continuous compact mapping;(ii) (or ) is an expansive mapping with constant ;(iii) and [ implies ] or .
Then there exists a point such that .

Proof. From (iii), for each since or there is an such that If then whereas if then Lemma 3.5 and (iii) imply . Now is continuous, and so is a continuous mapping of into . Since is compact, so is . By Theorem 2.5, has a fixed point with , that is, . This completes the proof.

Theorem 3.7 (Petryshyn-type). Let be a complete normed space and an open convex subset of with , where . Let be a continuous compact mapping and Then there exists such that .

Proof. The proof is exactly the same as the proof of Theorem  2.20 (b) [3]. Here we use Theorem 2.5 instead of Theorem  2.14 [3].

Theorem 3.8. Let be a complete normed space and an open convex subset of , where . Suppose that(i) is a continuous compact mapping; (ii)is an expansive map with constant ;(iii); (iv) for each .
Then there exists a point such that .

Proof. From (iii), for each , there is a such that Thus by Lemma 3.5, we have . Again by Lemma 3.5 and (i), we see that is compact. We now prove that (3.20) holds with replaced by . In fact, for each , from (3.21), we have so Since is expansive, we have It follows from (3.23) and (3.24) that Thus, by (3.25) and (iv), for each , we have which implies (3.20). This completes the proof.

Corollary 3.9. Let be a complete normed space and an open convex subset of , where . Suppose that(i) is a continuous compact mapping;(ii) is an expansive map with constant ; (iii) implies where ;(iv) for each .
Then there exists a point such that .

Theorem 3.10. Let be a complete normed space and an open convex subset of , where . Suppose that(i) is a continuous compact mapping;(ii) is a contraction with contractive constant ;(iii) for each .
Then there exists a point such that .

Proof. For each , the mapping is a contraction. Thus, the equation has a unique solution . For any , from it follows that Since is a contraction with contractive constant , we have Thus, we have It follows from (3.31) and (i) that is compact. From (3.27), we have For every , from (3.32) and (iii), we deduce that which implies (3.20). This completes the proof.

4. Condensing Mappings

Now, we extend the above results to a class of condensing mappings. For convenience, we recall some definitions, see [4, 7].

Definition 4.1. Let be a bounded subset of a metric space . The Kuratowski measure of noncompactness of is defined as follows: where denotes the diameter of set .

It is easy to prove the following fundamental properties of , see [7](i). (ii) if and only if is compact. (iii). (iv). (v) If are bounded, then . (vi) If is bounded and , then .

For (ii) one should remember that a set is compact if and only if it is closed and totally bounded.

Proposition 4.2. Let be a complete normed space and let , be two bounded subsets of . Then where .

Proof. Let and the open ball with center and radius . Note holds for any bounded set in a metric space; here . Then, if is a bounded set, we have that
Assume first that and are bounded convex sets. Let . We now prove that To do it, suppose that and . Then there exist and , , such that . Let for and for . Then, we have Consequently and then (4.7) leads to Assume that for each , there is a positive integer such that . For each , let Note if , then for some . This implies that if and , then for some . Thus, By (iv) and (4.4), we have For each , by (v) and (vi), we deduce that Since we have Hence, Since is arbitrary we obtain (4.5).
Consequently, if , where each is bounded convex, we have that
Now, to prove (4.2), let . Then , where . We now claim that . In fact, since , we have for each . Also, let , such that and . Then that is, Hence, .
Now we may assume that each is convex for each . By (4.16), we have Since this is true for all then so (4.2) is proved.

Definition 4.3. Let , be two metric spaces and a subset of . A bounded continuous map is set contractive if for any bounded set , we have is strictly set contractive if is set contractive and for all bounded sets with . We say is a condensing map if is a bounded continuous 1-set contractive map and for all bounded sets with .

Notice that is a compact map if and only if is a 0-set contractive map.

Now, we extend Sadovskii theorem in [7] to a map, that is, defined on a normed space.

Theorem 4.4 (Sadovskii type). Let be a complete normed space and a closed convex subset of , where . If is condensing and is bounded, then has a fixed point.

Proof. For each fixed , let be the family of all closed convex subsets of , such that and . Suppose that Since and , it follows that . This implies that . On the other hand, since , we have and . Therefore, .
As a result, we have and Since is condensing, it follows that , that is, is compact. Therefore is a compact mapping of convex set into itself. By Theorem 2.5, has a fixed point.

We next make use of the main ideas established in [5, 6, 8] to obtain a Krasnosel’skii fixed point theorem in a normed space.

Theorem 4.5. Let be a complete normed space and a closed convex subset of , where . Suppose that and(i) is such that the inverse of exists;(ii); (iii) is condensing and is bounded.
Then there exists a point such that .

Proof. From (ii), we have . Thus, is a condensing map of into itself. By Theorem 4.4, has a fixed point. This completes the proof.

The following lemma is easy to prove.

Lemma 4.6. Let be a complete normed space and . Assume that is a Lipschitizian map, that is, Then for each bounded subset of , we have .

Theorem 4.7. Let be a complete normed space and a closed convex subset of , where . Suppose that(i) is a 1-set contractive map (condensing) and is bounded;(ii) is an expansive map with constant ; (iii) implies where .
Then there exists a point such that .

Proof. Let be the function defined as in Theorem 3.1. We will show that is a condensing map. Let be bounded in . From (3.5) and Lemma 4.6, it follows that Suppose first that is 1-set contractive. Then which implies that is a condensing map. The other case when is condensing and also guarantees that is a condensing map. The result follows from Theorem 4.4. This completes the proof.

Theorem 4.8. Let be a complete normed space and a closed convex subset of , where . Suppose that(i) is a 1-set contractive map (condensing) and is bounded;(ii) (or ) is an expansive map with constant ;(iii) and [ implies ] or .
Then there exists a point such that .

Proof. For each , by (iii), there exists a such that If then whereas if then it follows from Lemma 3.5 and (iii), that . Now, if is bounded, then by Lemma 4.6 and (3.16), we have Suppose first that is 1-set contractive. Then which implies since that is a condensing map. The other case when is condensing and also guarantees that is a condensing map.
The result follows from Theorem 4.4. This completes the proof.

Lemma 4.9. Let be a normed space and . Suppose that the mapping is a contraction with contractive constant . Then the inverse of exists and

Proof. Since, for each , we have Then is one-to-one and the inverse of exists. Suppose that From the identity we have that Thus, and so Therefore,

Remark 4.10. If is a contraction with contractive constant , by Lemma 4.9, it follows that exists and is continuous. If in addition holds, then . To see this first note, since , for each , the mapping is a contraction. Thus, has a unique solution . Hence, and we have . This shows that .
From Theorem 4.4, we generalize a Krasnosel’skii type theorem (Theorem 1.1) as follows.

Theorem 4.11. Let be a complete normed space and a closed convex subset of , where . Suppose that(i) is a strictly set contractive map (or a set contractive map with ) and is bounded; (ii) is a contraction with contractive constant ; (iii)any imply .
Then there exists a point such that .

Proof. Let be bounded. By Remark 4.10, Lemma 4.6 and (4.32), we have and so is a condensing map. Hence, from Theorem 4.4, there exists a point such that . This completes the proof.

Remark 4.12. If exists and is continuous on and if in addition holds, then . To see this first note if , then there exists such that . Let , so . This implies , and then . Hence, and we have . This shows that .

Theorem 4.13. Let be a complete normed space and a closed convex subset of , where . Suppose that(i) is a strictly set contractive map (or a set contractive map with ) and is bounded;(ii) is a contraction with contractive constant ; (iii).
Then there exists a point such that .

Proof. Let be bounded. By Remark 4.12, Lemma 4.6 and (4.32), we have and so is a condensing map. Hence, from Theorem 4.4, there exists a point such that . This completes the proof.

5. Application

In [9], Hajji and Hanebaly presented a modular version of Krasnosel’skii fixed point theorem and applied their result to the existence of solutions to perturbed integral equations in modular spaces. In this section, we use the same argument as in [9] to give an application of Krasnosel’skii fixed point theorem to a normed space. For more details about modular spaces, we refer the reader to [10, 11]. Now, we recall some definitions.

Definition 5.1 (see [12]). Let be an arbitrary linear space over , where or . A functional is called a modular if (i) if and only if ; (ii) if with , for all ; (iii) if , with , for all .

If in place of (iii), we have(iv) for , and with an ,then the modular is called an convex modular, and if , is called convex modular.

A modular defines a corresponding modular space, that is, the space given by

The subset of is bounded if The diameter of is defined by

A simple example of a modular space is a normed space .

Now, we discuss the existence of solutions for the following perturbed integral equation in the modular space (normed space) , where and is the space of all continuous functions from to under the modular (norm) Consider the following assumptions:

(i) is a bounded closed convex subset of with , where ;

(ii) is a mapping satisfying

for all where , and is a continuous compact mapping such that ;

(iii) is a fixed element of .

Theorem 5.2. Suppose that (i)–(iii) are satisfied then the integral equation (5.4) has a solution .

Proof. Consider with for with . First we show where is given by Suppose that and as . Since and are continuous, then is continuous at . In fact, Let so we have , that is, is continuous at . Since is arbitrary, is continuous from into . Next, in the complete space , we have where . Since , it follows that and since is closed -convex, we have . Therefore, for all . Thus .
Let Thus . We will show that the hypotheses of Theorem 4.11 are satisfied.
Now since we have for any and . Hence . For any , we have Fix . Using the same argument in the proof of Lemma  2.1 [12] we now show that is a contraction. Let be any subdivision of . We know that is convergent to in when as . Therefore On the other hand, since then we have that is, This implies that Note also Hence, and so for any . Therefore, is a contraction.
Now we show that is compact. Let . Then is equicontinuous. To see this let . Then Hence, Recall the functions and are uniformly continuous on . Therefore, for , there exists such that if then and there exists such that if then