Abstract

We consider the existence of a coupled fixed point for mixed monotone mapping satisfying a new contractive inequality which involves an altering distance function in partially ordered metric spaces. We also establish some uniqueness results for coupled fixed points, as well as the existence of fixed points of mixed monotone operators. The presented results generalize and develop some existing results. In addition to an example as well as an application, we establish some uniqueness results for a system of integral equations.

1. Introduction and Preliminaries

In this paper we aim to establish coupled fixed point theorems for a mixed monotone mapping in a metric space endowed with partial order. The concept of the mixed monotone operator was introduced by Guo and Lakshmikantham in [1]. Existence of fixed points in a metric space has been studied for a long time (see [210]). The Banach contraction principle, which plays a very important role in nonlinear analysis, is the most famous tool in the study of a fixed point theorem. In the last decade, there has been a trend to weaken the requirement on the contraction by considering metric spaces endowed with partial order. It is of interest to determine if it is still possible to establish the existence of a unique fixed point assuming that the operator considered is monotone in such a setting.

Recently, there have been a lot of generalizations of the Banach contraction-mapping principle in the literature (see [1128]). Investigation of the existence of fixed points for single-valued mappings in partially ordered metric spaces was initially considered by Ran and Reurings in [29] who proved the existence of a unique fixed point.

Following basically the same approach as the one in [29], Bhaskar and Lakshmikantham in [30] proved a fixed point theorem for a mixed monotone mapping in a metric space endowed with partial order. Bhaskar and Lakshmikantham extended [29, Theorem 2.1] to mixed monotone operators so that they can enlarge, in a unified manner, the class of problems that can be investigated. The authors in [30] also established some uniqueness results for coupled fixed points, as well as the existence of fixed points of .

Based on the works in [30], Zhao in [31] obtained a more general coupled fixed point theorem for mixed monotone operators. Another generalization of the contraction principle was suggested by Alber and Guerre-Delabriere. In 1997, Alber and Guerre-Delabriere [32] introduced the concept of weak contractions in Hilbert spaces. This concept was extended to metric spaces by Rhoades in [33].

Definition 1. A mapping , where is a metric space, is said to be weakly contractive if where and is an altering distance function.
The notion of altering distance function was introduced by Khan et al. [34]. An altering distance function is a control function that alters distance between two points in a metric space.
It was shown in [32] that, for Hilbert spaces, weakly contractive maps possess a unique fixed point, without any additional assumptions, and it was noted that the same is true at least for uniformly smooth and uniformly convex Banach spaces. In [33], Rhoades proved that the theorem remains true in arbitrary complete metric spaces, which was improved and extended by Dutta and Choudhury in [35]. Recently, a new version of the context of ordered metric spaces has been proved by Harjani and Sadarangani in [12]. We refer the readers to [3641] for more related works.
Motivated by the papers mentioned above, we aim to establish coupled fixed point results for mixed monotone operator which is weakly contractive in partially ordered complete metric spaces. Our main results are Theorems 3, 6, and 8. To the best of our knowledge, there are no similar results in the literature on the existence of the coupled fixed point. Compared with the results obtained in the [32, 33, 35], in our results is not necessary to be an altering distance function, which means that satisfies a more general contractive condition. On the other hand, our result is still valid for not necessarily continuous if we require that the underlying metric space has an additional property. Our main results will generalize and develop the results given in [12].

In Section 2, we give the proof of our main results. In Section 3, as an application of our theorems, we consider the existence of a unique solution to a system of integral equations.

2. Coupled Fixed Point Theorems

Let be a partially ordered set and let be a metric on such that is a complete metric space. Further, we endow the product space with the following partial order:

Definition 2 (see [34]). A function is called an altering distance function if the following conditions are satisfied:
(i) is continuous and nondecreasing
(ii)
Next we introduce a set of functions

It follows from Definition 2 that if is an altering distance function, .

The first main result is the following coupled fixed point result.

Theorem 3. Assume(H1) ( is not necessary to be an altering distance function)(H2) being a mixed monotone mapping, there exist a constant such thatand, for each , is an altering distance function which satisfies(H3)there exist such that and (H4)(a) is continuous or(b) has the following properties:(i)If a nondecreasing sequence , then (ii)If a nonincreasing sequence , then Then there exist , such that

Proof. Sincelet , , , and ; we have Denote Note that , ; it follows from the mixed monotone property of thatwhich implies For , let i.e.,It is easy to see that i.e., By (4) (12), we haveand thusFrom , we haveSince is continuous and nondecreasing, Let . From , is a Cauchy sequence, and thus there exist such thatNow we claim . In fact, by (14) and (), one haswhich implies , so is .
Next we will show and are Cauchy sequences.
Arguing indirectly we suppose is not a Cauchy sequence. Thus, there exists a positive constant such that, for any , there exist such that For , let be the smallest integer satisfying and (21). Thus, one has which impliesand, by (21) and (), we have The triangular inequality gives usand thusBy (14), From (26) (27), one hasand, letting in the above inequality,From , note that , ; we have which is a contradiction. This shows that and are Cauchy sequences. Since is a complete metric space, there exist such thatCase (a). Assume is continuous. Thenand, thus, is a coupled fixed point of .
Case (b). Assume has the following properties:
(i) If a nondecreasing sequence , then
(ii) If a nonincreasing sequence , then
By (14) (31), we have Again, the triangular inequality gives usFrom , observing that is nondecreasing and using , , we getand we get . Then . Similarly, we get The proof is completed.

Corollary 4. Let be a partially ordered set and let be a metric on such that is a complete metric space. Assume is a fixed monotone operator and(f1)there exist such that and (f2)there exist , for with , ,(f3)(a) is continuous or(b) has the following properties:(i)If a nondecreasing sequence , then (ii)If a nonincreasing sequence , then Then there exist such that

Proof. Let . The proof is finished by Theorem 3.

Note that, in (36), is not necessary to be an altering distance function.

Corollary 5. Let be a partially ordered set and let be a metric on such that is a complete metric space. Assume is a fixed monotone operator and(f1)there exist such that and (f2)there exist , for with , ,(f3)(a) is continuous or(b) has the following property:(i)If a nondecreasing sequence , then (ii)If a nonincreasing sequence , then Then there exist such that

Proof. Let . The proof is finished by Corollary 4.

Next, we discuss the uniqueness of the coupled fixed point. Since the contractivity assumption is made only on comparable elements in , Theorem 3 cannot guarantee the uniqueness of the coupled fixed point. Before stating our uniqueness results, we require that the product space endowed with the partial order mentioned earlier have the following property:

For every , there exists which is comparable to and .

Note that is equivalent to (see [42]) the following:

Every pair of elements in has either a lower bound or an upper bound.

Next, we state our second main result.

Theorem 6. Assume and (or ) hold. Then the coupled fixed point of is unique.

Proof. By , applying Theorem 3, has a coupled fixed point. Let be the coupled fixed points of , i.e.,To show the uniqueness, we need to prove .
By , there exists which is comparable to and . Let Since is comparable to with respect to the ordering in , we suppose Then, for , In fact, (42) implies that For , by the mixed monotone property, we have For , we suppose . Therefore, for , we have which implies (43) holds.
From (41) ,which impliesSet . By (48), we have . Then, there exists such that If , by ,which is a contradiction. So . We haveand thenSimilarly, we also have It follows from (52) (53) that

Corollary 7. In addition to the hypothesis of Corollary 4, suppose holds. Then has a unique coupled fixed point.

Next, we state our third main result.

Theorem 8. In addition to the hypothesis of Theorem 6, suppose one of the followings holds:
(i) Elements of the coupled fixed point are comparable
(ii) and are comparable
(iii) Every pair of elements of X has an upper bound or a lower bound in
Then, we have ; that is, has a unique fixed point:

Proof. From Theorem 6, has a unique coupled fixed point .
Case (i). Since and are comparable, we have or . Suppose ; then and are comparable. By (4) in , one obtainsNoting , it follows from (56) that , i.e., .
Case (ii). Since and are comparable, we have or . Suppose we are in the first case, . LetFrom Theorem 3, the coupled fixed point satisfies By and the mixed monotone property of , one has and, hence, by induction one obtains On the one hand, it follows from the continuity of the distance thatOn the other hand, by (4) in , one getswhich implies and therefore , which finishes the proof.

Case (iii). If are comparable, see Case (i). If are not comparable, then there exists comparable to and . Without loss of generality, we suppose (the other case is similar). In view of the order of (or for short), one has that is, are comparable in .

Inspired by [31], we consider the functional defined by is a metric on and, moreover, if is complete, then is a complete metric space, too. Define another operator as follows:For , let ; one has

By contractive condition (4) in , we get a Banach type contraction condition:

and thus and, hence, by induction we have

Now, applying (70) to the comparable pairs , one obtains

Now, for , note that ; we get

So, using the triangle inequality and (4) (71)-(74), we havewhich implies , so .

Corollary 9. In addition to the hypothesis of Corollary 7, suppose one of the followings holds:
(i) Elements of the coupled fixed point are comparable
(ii) and are comparable
(iii) Every pair of elements of X has an upper bound or a lower bound in
Then, we have ; that is, has a unique fixed point:

Remark 10. Corollary 9 includes theorems in [31].

Example 11. Let and be defined by Then,
(i) has a coupled fixed point
(ii) has a unique coupled fixed point
(iii) has a unique fixed point

Proof. Let , , . Since one gets, for ,and Note thatand, for , and, hence, (81) (83) imply that On the other hand, is continuous and , satisfy(i) As mentioned above, by Theorem 3, has a coupled fixed point
(ii) Since has property () or (), by Theorem 6, the uniqueness of the coupled fixed point is obtained
(iii) Noting that and are comparable, by Theorem 8 (ii), has a unique fixed point

3. Applications to Integral Equations

As an application to the fixed point theorem proved in Section 2, we shall study a class of integral equation where . We will establish existence and uniqueness results. It is well known that some boundary value problems are equivalent to an integral equation or a system of integral equations.

Let be a partially ordered set such that, for , is endowed with the sup metric:so is a complete metric space. The corresponding metric on is defined by and then consider on the partial order relation: A pair is called a solution of (86) if

If , will be a solution of (86).

Let be a nondecreasing function such that where satisfies , .

We make the following assumptions:

(i) ,

(ii)

(iii) )

(iv) there exist constants such that, for

(v) is a solution of (86)

Theorem 12. Suppose hold. Then (86) has a unique solution.

Proof. In order to obtain the (unique) solution of (86), define for the operator byIf is a coupled fixed point of , then we have It is obvious that the fixed point of is the solution of (86). In what follows, we will show that has a unique fixed point.
(1) We will show that the operator has a unique coupled fixed point in
Firstly, by (v), we haveSecondly, we check is mixed monotone for such that . From (i) and (iv), we haveSimilarly, for such that , we havewhich yields that is mixed monotone.
Thirdly, we show that verifies the contraction condition. Let us consider with , ; we haveSimilarly, one obtains Sincewhich together with (iv) imply thatwhich proves that verifies contraction condition (36) in Corollary 4.
Next, we consider a monotone nondecreasing sequence converging to . Then, for every , the sequence of real numbers converges to . So, for all Hence, , for all . Similarly, we can verify that the limit of monotone nonincreasing sequence in is a lower bound for all the elements in the sequence. That is, for all .
Therefore, it follows from Corollary 4 that has a coupled fixed point , i.e.,On the other hand, has property since, for any , , , for each , are in and are the upper and lower bounds of , respectively. Then, by Corollary 7, has a unique coupled fixed point.
(2) Noting that is a coupled solution of (86), one obtains for all . So, and are comparable. By Corollary 9, has a unique fixed point

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Authors’ Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Acknowledgments

This work was supported by the NSFC (11471187, 11571197).