Abstract and Applied Analysis

Volume 2014 (2014), Article ID 580508, 8 pages

http://dx.doi.org/10.1155/2014/580508

Research Article

Expansive Mappings and Their Applications in Modular Space

A. Azizi, R. Moradi, and A. Razani

Department of Mathematics, Faculty of Science, Imam Khomeini International University, Qazvin 34149-16818, Iran

Received 19 September 2013; Revised 30 January 2014; Accepted 31 January 2014; Published 14 April 2014

Academic Editor: Mohamed Amine Khamsi

Copyright © 2014 A. Azizi 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.

Some fixed point theorems for ρ-expansive mappings in modular spaces are presented. As an application, two nonlinear integral equations are considered and the existence of their solutions is proved.

1. Introduction

Let be a metric space and a subset of .  A mapping is said to be expansive with a constant such that

Xiang and Yuan [1] state a Krasnosel’skii-type fixed point theorem as follows.

Theorem 1 (see [1]). Let be a Banach space and a nonempty, closed, and convex subset. Suppose that and map into such that

(I)is continuous; resides in a compact subset of ;(II) is an expansive mapping;(III) implies that , where .

Then there exists a point with .

For other related results, see also [2, 3].

In this paper, we study some fixed point theorems for , where is -expansive and resides in a compact subset of , where is a closed, convex, and nonempty subset of and . Our results improve the classical version of Krasnosel’skii fixed point theorems in modular spaces.

Finally, as an application, we study the existence of a solution of some nonlinear integral equations in modular function spaces.

In order to do this, first, we recall the definition of modular space (see [4–6]).

Definition 2. Let be an arbitrary vector space over or . Then we have the following.

(a)A functional is called modular if

(i) if and only if ;(ii) for with , for all ;(iii) if , , for all . If (iii) is replaced by(iii)′ for , , , for all , then the modular is called a convex modular.

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

(c)If is convex modular, the modular can be equipped with a norm called the Luxemburg norm defined by

Remark 3. Note that is an increasing function. Suppose that ; then property (iii), with , shows that .

Definition 4. Let be a modular space. Then we have the following.

(a)A sequence in is said to be

(i)-convergent to if as ;(ii)-Cauchy if as .

(b) is -complete if every -Cauchy sequence is -convergent.(c)A subset is said to be -closed if for any sequence and then .(d)A subset is called -bounded if , for all , where is called the -diameter of .(e) has the Fatou property if

 whenever and as .(f) is said to satisfy the -condition if whenever as .

2. Expansive Mapping in Modular Space

In 2005, Hajji and Hanebaly [7] presented a modular version of Krasnosel’skii fixed point theorem, for a -contraction and a -completely continuous mapping.

Using the same argument as in [1], we state the modular version of Krasnosel’skii fixed point theorem for , where is a -expansive mapping and the image of under that is, resides in a compact subset of , where is a subset of .

Due to this, we recall the following definitions and theorems.

Definition 5. Let be a modular space and a nonempty subset of . The mapping is called -expansive mapping, if there exist constants such that , and

for all .

Example 6. Let and consider with for and . Then for all , we have

Therefore is an expansive mapping with constant .

Theorem 7 (Schauder’s fixed point theorem, page 825; see [1, 8]). Let be a Banach space and is a nonempty, closed, and convex subset. Suppose that the mapping is continuous and resides in a compact subset of . Then has at least one fixed point in .

We need the following theorem from [6, 9].

Theorem 8 (see [6, 9]). Let be a -complete modular space. Assume that is a convex modular satisfying the -condition and is a nonempty, -closed, and convex subset of . is a mapping such that there exist such that , and for all one has

Then there exists a unique fixed point such that .

Theorem 9. Let be a -complete modular space. Assume that is a convex modular satisfying the -condition and is a nonempty, -closed, and convex subset of . is a -expansive mapping satisfying inequality (5) and . Then there exists a unique fixed point such that .

Proof. We show that operator is a bijection from to . Let and be in such that ; by inequality (5), we have ; also since it follows that the inverse of exists. For all ,

where . We consider , where denotes the restriction of the mapping to the set . Since , then is a -contraction. Also since is a -closed subset of , then, by Theorem 8, there exists a such that . Also is a fixed point of .

For uniqueness, let and be two arbitrary fixed points of ; then

hence and .

We need the following lemma for the main result.

Lemma 10. Suppose that all conditions of Theorem 9 are fulfilled. Then the inverse of exists and

for all , where and is conjugate of ; that is, and .

Proof. For all ,

then

Now, we show that is an injective operator. Let and ; then by inequality (12), and . Therefore is an injective operator from into , and the inverse of exists. Also for all , we have . Then for all , by inequality (12) we get

Theorem 11. Let be a -complete modular space. Assume that is a convex modular satisfying the -condition and is a nonempty, -closed, and convex subset of . Suppose that

(I) is a -continuous mapping and resides in a -compact subset of ;(II) is a -expansive mapping satisfying inequality (5) such that ;(III) implies that , where .

There exists a point such that .

Proof. Let and . Consider the mapping ; then by Theorem 9, the equation has a unique solution . Now, we show that is a -contraction. For , and . Applying the same technique in Lemma 10,

where . Then

Therefore, mapping is a -contraction and hence is a -continuous mapping. By condition (I), is also -continuous mapping and, by -condition, is -continuous mapping. Also resides in a -compact subset of . Then using Theorem 7, there exists a such that which implies that .

The following theorem is another version of Theorem 11.

Theorem 12. Let be a -complete modular space. Assume that is a convex modular satisfying the -condition and is a nonempty, -closed, and convex subset of . Suppose that

(I) is a -continuous mapping and resides in a -compact subset of ;(II) or is a -expansive mapping satisfying inequality (5) such that ;(III) and implies that or .

Then there exists a point such that .

Proof. By condition (III), for each , there exists such that . If , then ; if , then by Lemma 10 and condition (III), . Now is a -continuous and so is a -continuous mapping of into . Since resides in a -compact subset of , so resides in a -compact subset of the closed set . By using Theorem 7, there exists a fixed point such that .

Using the same argument as in [2], we can state a new version of Theorem 11, where is -sequentially continuous.

Definition 13. Let be a modular space and a subset of . A mapping is said to be

(1)-sequentially continuous on the set if for every sequence and such that , then ;(2)-closed if for every sequence such that and , then .

Definition 14. Let be a modular space and , two subsets of . Suppose that and are two mappings. Define

Theorem 15. Let be a -complete modular space. Assume that is a convex modular satisfying the -condition and is a nonempty, -closed, and convex subset of . Suppose that

(I) is -sequentially continuous;(II) is a -expansive mapping satisfying inequality (5) such that ;(III) implies that , where ;(IV) is -closed in and is relatively -compact.

Then there exists a point such that .

Proof. Let , and . One considers the mapping ; by Theorem 9, the equation

has a unique solution .

Now, we show that exists. For any and by the same technique of Lemma 10, we have

where . This implies that exists and for all , and .

We show that is -sequentially continuous in . Let be a sequence in and such that . Since and is relatively -compact, then there exists such that . On the other hand, by condition (I), . Thus by (17), we get

then

therefore when , condition (IV) implies that ; that is, and

then is -sequentially continuous in . By -condition, is -sequentially continuous. Let , where denotes the closure of the convex hull in the sense of . Then and is a compact set. Therefore is -sequentially continuous from into . Then using Theorem 7, has a fixed point such that . From (17), we have

that is, .

The following theorem is another version of Theorem 15.

Theorem 16. Let be a -complete modular space. Assume that is a convex modular satisfying the -condition and is a nonempty, -closed, and convex subset of . Suppose that

(I) is -sequentially continuous;(II) is a -expansive mapping satisfying inequality (5), such that ;(III) and implies that (or .(IV) is -closed in and is relatively -compact.

Then there exists a point such that .

Proof. By (III) for each , there exists such that and . By the same technique of Theorem 15, is -sequentially continuous and there exists a such that .

3. Integral Equation for -Expansive Mapping in Modular Function Spaces

In this section, we study the following integral equation:

where is the Musielak-Orlicz space and . denote the space of all -continuous functions from to with the modular . Also is a real vector space. If is a convex modular, then is a convex modular. Also, if satisfies the Fatou property and -condition, then satisfies the Fatou property and -condition (see [9]).

To study the integral equation (23), we consider the following hypotheses.

(1) is a -expansive mapping; that is, there exist constants such that , and for all

 and is onto. Also for ,   is -continuous.(2) is a function from into such that is -continuous on for almost all and is measurable function on for each and for almost all . Also, there are nondecreasing continuous functions such that

 for all , and .(3)There exists measurable function such that

 for all and ; also .(4) for all and .

Remark 17 (see [7]). We consider , the Musielak-Orlicz space. Since is convex and satisfies the -condition, then

as on . This implies that the topologies generated by and are equivalent.

Theorem 18. Suppose that the conditions (1)–(4) are satisfied. Further assume that satisfies the -condition. Also and ; also . Then integral equation (23) has at least one solution .

Proof. Suppose that

Conditions (1) and (2) imply that and are well defined on . Define the set . Then is a nonempty, -bounded, -closed, and convex subset of . Equation (23) is equivalent to the fixed point problem . By Theorem 12, we find the fixed point for in . Due to this, we prove that satisfies the condition of Theorem 12. For , we show that . Indeed,

then . Since and is -bounded, is -bounded and by -condition -bounded.

We show that is -equicontinuous. For all and such that ,

then by condition (3),

since , then is -equicontinuous. By using the Arzela-Ascoli theorem, we obtain that is a -compact mapping. Next, we show that is -continuous. Suppose that is given; we find a such that , for some . Note that

also

then

therefore is -continuous.

Since is -continuous, it shows that transforms into itself. In view of supremum and condition (1), it is easy to see that is -expansive with constant . For ,

then

where is conjugate of . Let ; since , then

Now, assume that for some . Since , then , and

which shows that . Now, define a map as follows:

for each ; by

for all ,

therefore

then is -expansive with constant and is onto. By Theorem 9, there exists such that ; that is, . Hence and condition (III) of Theorem 12 holds. Therefore by Theorem 12, has a fixed point with ; that is, is a solution to (23).

Now, we consider another integral equation.

Let be the Musielak-Orlicz space and . Suppose that is convex and satisfies the -condition. Since topologies generated by and are equivalent, then we consider Banach space and denote the space of all -continuous functions from to with the modular ; also is a real vector space. Consider the nonlinear integral equation

where

(1) is a -expansive mapping; that is, there exists constant such that

 for all and is onto; also for , is -continuous;(2) is function from into such that is a -continuous and is measurable for every . Also, there exist functions and a nondecreasing continuous function such that

 for all and . Also for , is nondecreasing on ;(3) is function from into such that is -continuous and there exists a such that

 for all and ; also for , is nondecreasing on and for , is nondecreasing on ;(4) is function from into . For each , is measurable on . Also is bounded on and . The map is continuous from to . Also for , is nondecreasing on .

Theorem 19. Suppose that the conditions (1)–(4) are satisfied and there exists a constant such that for all ,

where and also . Then integral equation (43) has at least one solution .

Proof. Define

then is a nonempty, -bounded, -closed, and convex subset of . Consider

It is easy that by the hypothesis and are well defined on .

For , we show that . Consider

Let and assume that such that , for a given positive constant . We have

since

then is -equicontinuous. By using the Arzela-Ascoli Theorem, we obtain that is a -compact mapping.

We show that is -continuous. Suppose that is given. We find a such that . We have

Since is -continuous, it shows that transforms into itself. In view of supremum and condition (1), it is easy to see that is -expansive with constant .

For ,

then

since , then

Now, assume that for some . Then

which shows that . Now for each we define a map as follows:

by

for all ,

therefore

then is -expansive with constant and is onto. By Theorem 9, there exists such that ; that is, . Hence . Therefore by Theorem 12, has a fixed point with ; that is, is a solution of (43).

Finally, some examples are presented to guarantee Theorems 18 and 19.

Example 20. Consider the following integral equation:

where , .

For and , we have

Therefore by Theorem 18, the integral equation (62) has at least one solution.

Example 21. Consider the following integral equation:

where , , , and . Also , . Therefore by Theorem 19, the integral equation (64) has at least one solution.

Conflict of Interests

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

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