#### 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. * 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.