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 [1–6]). 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 8–10 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 This implies Since is nonincreasing in 3rd variable, using triangle inequality, we get Therefore, from (), there exists , for such that If for some then Hence, and for , that is, and , so that (46) holds.
Suppose that for all ; then, from (49), we get Therefore is a decreasing sequence of nonnegative real numbers. Thus, there exists such that .
If , then, on letting in (49), we get .
Then, from (), we have , a contradiction. Hence . Since is also a coupled fixed point of , we have Hence, by the uniqueness of limit, we have and .

Remark 19. In Example 17, does not belong to (Example 15) and condition (40) does not hold and has two coupled fixed points.

Theorem 20. In addition to the hypotheses of Theorem 16 with suppose that and are comparable. Further, assume that (40) holds. Then has a unique coupled fixed point and .

Proof. From the proof of Theorem 16, we have that has a coupled fixed point say . That is, and . Suppose that . We show that using principle of mathematical induction on , where and , are the sequences defined in the proof of Theorem 16. Suppose that (55) is true for some . Then, by mixed monotone property of , Thus (55) is true for each . Now, from (9), we haveThis impliesOn letting , we get a contradiction to (), if . Hence, ; that is, .

Remark 21. Theorem 16 is an extension of Theorem 5 to coupled fixed points with an implicit relation of .

Corollary 22. 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 a continuous map , for , if and only if such that for each with and .
Suppose that either (a) or (b) of Theorem 16 holds.
If there exist such that and then has a coupled fixed point.

Proof. By defining in (9) as , , we obtain (60). Hence the conclusion follows from Theorem 16.

Corollary 23. 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 a continuous map , for , if and only if such that for each with and .
Suppose that either (a) or (b) of Theorem 16 holds.
If there exist such that and then has a coupled fixed point.

Proof. From (61), we obtain Adding (61) and (62), we obtain inequality (60). Hence the conclusion follows from Corollary 22.

Remark 24. Corollary 23 is another version of Theorem 4 with different .

Corollary 25. 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 a continuous map , for , such that for each with and .
Suppose that either (a) or (b) of Theorem 16 holds.
If there exist such that and then has a coupled fixed point.

Proof. By defining in (9) as , , we obtain (63). Hence the conclusion follows from Theorem 16.

Corollary 26. 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 a continuous map , for , such that for each with and .
Suppose that either (a) or (b) of Theorem 16 holds.
If there exist such that and then has a coupled fixed point.

Proof. From (64), we obtain Adding (64) and (65), we obtain inequality (63). Hence the conclusion follows from Corollary 25.

Remark 27. In addition to the hypotheses of Corollary 26, if satisfies (40), then has a unique coupled fixed point. Further, if and are comparable, then has a unique coupled fixed point and . The same is true in respect of Corollaries 22, 23, and 25.

Remark 28. In view of Examples 11, 13, and 14, we obtain new coupled fixed point results for mappings satisfying (7), the implicit relations in Examples 13 and 14, respectively, by using Theorem 16.

Remark 29. We obtain Theorem 3 as a corollary to Theorem 16 by choosing , .

Example 30. Let with the usual metric and usual ordering. We define by Now we define by Then satisfies mixed monotone property. There exists a point such that and . We now verify inequality (9) for with and in the following cases.
Case i. and .
In this case, , , and .
Hence Case ii. and .
In this case, , , and .
The verification is similar as in Case i.
Case iii. and .
This case does not arise, for we have and , and hence , which does not hold.
Case iv. and .
In this case, , , and .
The verification is similar as in Case i.
From all the above cases, we get
Thus, satisfies inequality (9). Hence all the hypotheses of Theorem 16 hold and (0,0) is the unique coupled fixed point of .
Here we observe that (2) fails to hold, for, at , , we have and . Hence, Theorem 3 is not applicable.

Example 31. Let with the usual metric and usual ordering. Suppose that be defined by We now define by Define by Then satisfies mixed monotone property. There exists a point such that and . We now verify inequality (9) for with and in the following cases.
Case i. and .
In this case, Hence
Case ii. and .
In this case, , , and .
The verification is similar as in Case i.
Case iii. and .
This case does not arise, for we have and , and hence , which does not hold.
Case iv. and .
In this case, The verification is similar as in Case i.
From all the above cases, we get
Hence satisfies inequality (9). Hence all the hypotheses of Theorem 16 hold and is the unique coupled fixed point of . Here we observe that satisfies the hypotheses of Theorem 18 and Theorem 20 too.

Remark 32. There is another viewpoint of coupled fixed points; that is, they can be viewed as fixed points of an operator on product spaces like what follows.
Let have a coupled fixed point , that is, and . We construct an operator as Then is a coupled fixed point of if and only if is a fixed point of . But this viewpoint may not be helpful always, a case being the present one. It is not immediately possible to find the characterization of the operator from the implicit inequality satisfied by . This justifies the study of coupled fixed points as it is traditionally done without an appeal to the product space.

4. An Application

In this section, we establish the existence of solution for an integral equation as an application to the above results in Section 3.

Consider the integral equation

Assumption A. Suppose that , and satisfy the following assumptions: (i), for all ;(ii)there exist and a nondecreasing mapping such that, for all , , (iii)there exists , for such that for all ;(iv).
As an example for the existence of mappings satisfying (iii), we can consider the following.
For , there exists , such that , for .

Definition 33. An element is called a coupled lower and upper solution of the integral equation (78), if for each and for each .

Theorem 34. Consider the integral equation (78) with , and and suppose that Assumption A holds. If the integral equation (78) has a coupled lower and upper solution, then it has a unique solution in .

Proof. Let . Define an ordering on as follows: With this relation, is a partially ordered set. Define a metric on by Then is a complete metric space.
Suppose that is a nondecreasing sequence in which converges to in . Then we have Therefore is a nondecreasing sequence of real numbers and converges to . Hence for all and so that for all .
Similarly, it is easy to see that if is a nonincreasing sequence in that converges to then for all .
Define by Now we show that satisfies mixed monotone property.
Let , with . That is, for all . Then we have Hence for all . That is, .
Now let . That is, for all . Then we have Hence for all . That is, . Thus satisfies mixed monotone property.
Now, we show that inequality (60) is satisfied. Let and ; that is, and . Then, we haveTherefore, for and , we have
Let be lower and upper solution of the integral equation (78); then we have and and for all . Hence, , and .
Thus satisfies all the hypotheses of Corollary 22 and hence has a coupled fixed point.
Also, with the partial ordering on defined by we have comparable with and for each .
Thus also satisfies condition (40). Hence has a unique coupled fixed point (say).
Since and are comparable, by Theorem 18, it follows that . Hence Hence is the unique solution of the integral equation (78).

The following is an example as an illustration of Theorem 34.

Example 35. Let . Let us now consider the integral equation where   and .
We transform this equation into Fredholm integral equation of the form (78). That is, where , , and for all , , and , , and , . Here and for all . It is easy to see that .
Write . We define by and . We choose and so that hold. Also, .
Hence the conditions (i) to (iv) of Assumption A hold. Then with and is a coupled lower and upper solution of the integral equation (78) and is the unique solution.

Conflict of Interests

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

Acknowledgment

Professor B. S. Choudhury suggested some remarks which are hereby acknowledged.