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`Journal of OperatorsVolume 2014, Article ID 389646, 16 pageshttp://dx.doi.org/10.1155/2014/389646`
Research Article

## Coupled Fixed Point Theorems with New Implicit Relations and an Application

1Department of Mathematics, Andhra University, Visakhapatnam 530 003, India
2Department of Mathematics, Lendi Institute of Engineering and Technology, Vizianagaram 535 005, India

Received 28 June 2014; Accepted 1 October 2014; Published 11 November 2014

Copyright © 2014 G. V. R. Babu and P. D. Sailaja. 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 two new classes of implicit relations and where is a proper subset of , and these classes are more general than the class of implicit relations defined by Altun and Simsek (2010). We prove the existence of coupled fixed points for the maps satisfying an implicit relation in . These coupled fixed points need not be unique. In order to establish the uniqueness of coupled fixed points we use an implicit relation , where . Our results extend the fixed point theorems on ordered metric spaces of Altun and Simsek (2010) to coupled fixed point theorems and generalize the results of Gnana Bhaskar and Lakshimantham (2006). As an application of our results, we discuss the existence and uniqueness of solution of Fredholm integral equation.

#### 1. Introduction

Existence of fixed point theorems in partially ordered metric spaces with a contractive condition has been considered by several authors (see [16]). Guo and Lakshmikantham [7] introduced mixed monotone operators. Gnana Bhaskar and Lakshmikantham [8] established the existence of coupled fixed points of mappings satisfying mixed monotone property in partially ordered metric spaces. Later, Lakshmikantham and Ciric [9] extended this property to two maps by introducing mixed -monotone property and established the existence of coupled coincidence point and coupled common fixed points for a pair of commuting maps.

Choudhury and Kundu [10] extended the existence of coupled coincidence and coupled common fixed points for a pair of noncommuting maps, particularly for a pair of compatible maps.

Definition 1 (see [7]). Let be a nonempty set. An element in is called a coupled fixed point of the mapping if and . A point is called a fixed point of if .

Definition 2 (see [7]). Let be a partially ordered set and be a mapping. We say that satisfies mixed monotone property if is monotone nondecreasing in and monotone nonincreasing in ; that is, for any :

Theorem 3 (see [8]). Let be a partially ordered set and suppose that is a metric on such that is a complete metric space. Let be a mapping satisfying mixed monotone property. Assume that there exists a with Suppose that either is continuous or the following conditions hold in : (i)if a nondecreasing sequence with , then for all and(ii)if a nonincreasing sequence with , then for all .
If there exist such that and then has a coupled fixed point.

In 2011, Luong and Thuan [11] proved the following coupled fixed point theorem.

Theorem 4 (see [11]). Let be a partially ordered set and suppose that is a metric on such that is a complete metric space. Let be a mapping satisfying mixed monotone property on and there exist such that and . Suppose that there exists a with for all and such that for each , and . Suppose that either(a) is continuous or(b) has the following property:if a nondecreasing sequence with , then for all andif a nonincreasing sequence with , then for all .
Then has a coupled fixed point.

In Section 2 of this paper, we introduce two new classes of implicit relations and where is a proper subset of , and these classes are more general than the class of implicit relations defined by Altun and Simsek [12]. In Section 3, we prove the existence of coupled fixed points for the maps satisfying an implicit relation in . These coupled fixed points need not be unique (Example 17). In order to establish the uniqueness of coupled fixed points we use an implicit relation , where . Our results extend the fixed point theorems on ordered metric spaces of Altun and Simsek [12] to coupled fixed point theorems and generalize the results of Gnana Bhaskar and Lakshimantham [8]. In Section 4, as an application to results of Section 3, we discuss the existence and uniqueness of solution of Fredholm integral equation.

#### 2. Implicit Relations

In 1999, Popa [13] introduced the idea of the class of implicit functions to prove the existence of fixed points as follows.

Let denote the set of all nonnegative real numbers. Let be the family of real lower semicontinuous functions satisfying the following conditions:: is nonincreasing in fifth and sixth variables;:there exists such that for every with: or:we have ;:, .

Many authors, namely, Altun and Simsek [12], Altun and Turkoglu [14, 15], Imdad et al. [16], Popa [17], Popa and Mocanu [18], Sharma and Deshpande [19], and Turkoglu and Altun [20], continued the study of the existence of fixed points in this direction. The purpose of using an implicit relation is that the condition involving implicit relation generalizes many contraction conditions, which in turn generalizes many existing results. In 2010, Altun and Simsek [12] defined a new class of implicit relations as follows and established a fixed point theorem in ordered metric spaces.

Let be the set of all continuous functions satisfying the following conditions:(T1): is nonincreasing in variables ;(T2):there exists a right continuous function , , for , such that for implies ;(T3): for .

Theorem 5 (see [12]). Let be a partially ordered set and suppose that is a metric on such that is a complete metric space. Let be a nondecreasing mapping such that, for all with , where . Also, suppose that either is continuous or the following condition holds: “if a nondecreasing sequence , then for all ”.
If there exists an with , then has a fixed point.

We now define a more general class of implicit relations than that of [12] as follows.

Let be the set of all continuous functions satisfying the following conditions:(): is nonincreasing in variables ;():there exists a mapping , for , such that for implies ;(): for .

Here we observe that .

Example 6. , where and .
Clearly holds. Let . Thus . Let . Thus . Now define by , where , for . Hence . Also, for . Therefore .

Example 7. , where is a continuous map with , for .
Clearly holds. Let and . If then , which is a contradiction. Hence and hence . Let and ; that is, . If then clearly . Thus is satisfied with . Also, for . Therefore .

Example 8. , where is a continuous map with if and only if , for .
Clearly holds. Let . Let ; that is, . Thus with by . Hence is satisfied. Also, for . Hence .

Example 9. , where .
Clearly holds. Let ; that is, . Let ; that is, . Thus, with . Hence, is satisfied. Also, for . Thus .

Example 10. , where is a continuous map with , for .
Clearly holds. Let ; that is, . Let ; that is, . Thus with . Hence is satisfied. Also, for . Hence .

Example 11.
is nonincreasing in and but not in any of , . Let , , a contradiction. Now let , a contradiction. If then clearly . Thus is satisfied. Also, for . Thus . But does not belong to .
In view of Example 11, the class of implicit functions is larger than that of introduced by Altun and Simsek [12] and we use the implicit relation involving implicit function of to prove our main results.
In this paper, we prove a coupled fixed point theorem (Theorem 16) using an implicit relation in . In order to obtain the uniqueness of coupled fixed point we define another class of implicit relations as follows.
Let be the set of all continuous functions satisfying the following conditions: : is nonincreasing in variables ;:there exists a mapping , for , such that, for each , implies ;: for .
Here we observe that .
All functions in Examples 810 are in . Some more examples in this direction are as follows.

Example 12. , where and and .
Clearly holds. Let . Thus . Let . Thus . Now let . This implies . Now define by , where , for . Hence . Also, for . Therefore .

Example 13. , where with for .
Clearly holds. Let and . If , then , which is a contradiction. Hence and hence . Let and ; that is, . Let and . If , then , which is a contradiction. Hence and hence . If , then clearly . Thus () is satisfied with . Also, for . Therefore, .

The following examples suggest that is a proper subset of .

Example 14. , where defined by . In view of Example 7, . But is not in for let , and suppose that . This implies . Therefore . Hence, there cannot exist a function with for such that . Hence .

Example 15. Let us consider as in Example 11. Then . But does not belong to , since is not nonincreasing in , so that is a proper subset of .

#### 3. Existence of Coupled Fixed Points Using a New Implicit Relation

Theorem 16. Let be a partially ordered set and suppose that is a metric on such that is a complete metric space. Let be a mapping satisfying mixed monotone property and suppose that there exists such that for each with and .
Suppose that either (a) is continuous or(b) has the following property:if a nondecreasing sequence with , then for all andif a nonincreasing sequence with , then for all .
If there exist such that and then has a coupled fixed point.

Proof. Suppose that there exists such that and . Choose such that and . Then and . Now choose such that and . Since has mixed monotone property, we get Continuing this process, we obtain sequences and in such that for and
Let and for each .
If and for some , then is a coupled fixed point of . So, without loss of generality, we assume that or for all .
Now using (9), we haveSince is nonincreasing in 6th variable, using triangle inequality, we get From , there exists , for such that Then for all . Hence, from (14), we have Therefore is strictly decreasing sequence of positive reals. Hence there exists such that We show that . Suppose that . On letting in (13), we get Then from (), we have , a contradiction. Hence . Now we show that and are Cauchy sequences in . Suppose that at least one of or is not Cauchy. Then there exist and sequences of positive integers and with such that Now choose the least positive integer such that (19) holds. Then Now On taking limit supremum as , we get Hence Similarly, we can show that Hence Now Since , we have and for each . Hence, from (9), we get Since is nonincreasing in 5th and 6th variable, using triangle inequality, we get On letting , we get From (), we get Now on letting , in (26), we get Hence Therefore and are Cauchy sequences in . Since is complete there exists with and .
Suppose that is continuous. Then Therefore is a coupled fixed point of .
Now suppose that conditions (b) hold. Hence, we have for all and for all .
Now from (9), we have On letting , we get If , then is a coupled fixed point of .
If , then (35) yields a contradiction to (). Hence and . Therefore is a coupled fixed point of .

Theorem 16 does not guarantee the uniqueness of coupled fixed point of . The following example supports this assertion.

Example 17. Let with the usual metric. We define a partial order on as follows: We define by , and . Then (Example 11).
Let We define by Clearly satisfies the mixed monotone property. We choose . Then and so that and . The elements , in such that and are the following: , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , . We now verify inequality (9) for the above pairs.
Case i. , such that , , , are either in or or or .
In this case so that inequality (9) holds trivially.
Case ii. , or , .
In this case Hence inequality (9) holds. Therefore all the hypotheses of Theorem 16 hold and and are two coupled fixed points of .

In the following theorem, we prove the uniqueness of coupled fixed points by using an implicit relation in .

We define a partial order on by if , .

Theorem 18. In addition to the hypotheses of Theorem 16 with suppose that the following condition holds: Then has a unique coupled fixed point.

Proof. With the hypotheses of Theorem 16, has a coupled fixed point , that is, and . Suppose that if possible there exists such that and . We show that and .
Case i. and are comparable.
Without loss of generality, we assume that . That is, and .
Suppose that either or . Then Then, from (), there exists , for such that a contradiction. Hence and .
Case ii. There exists which is comparable with and .
Without loss of generality, we assume that and , that is, , and , .
We denote for each .
Since has mixed monotone property, we have
Now suppose that
Hence by mathematical induction, We now show that Using (9), we get