#### Abstract

Two concepts—one of the coupled fixed point and the other of the generalized metric space—play a very active role in recent research on the fixed point theory. The definition of coupled fixed point was introduced by Bhaskar and Lakshmikantham (2006) while the generalized metric space was introduced by Mustafa and Sims (2006). In this work, we determine some coupled fixed point theorems for mixed monotone mapping satisfying nonlinear contraction in the framework of generalized metric space endowed with partial order. We also prove the uniqueness of the coupled fixed point for such mappings in this setup.

#### 1. Introduction

Fixed point theory is a very useful tool in solving variety of problems in the control theory, economic theory, nonlinear analysis and, global analysis. The Banach contraction principle [1] is the most famous, most simplest, and one of the most versatile elementary results in the fixed point theory. A huge amount of literature is witnessed on applications, generalizations, and extensions of this principle carried out by several authors in different directions, for example, by weakening the hypothesis, using different setups, and considering different mappings.

Recently, the idea of generalized metric spaces was introduced and studied by Mustafa and Sims [2] originated from the concept of metric spaces. Some fixed point theorem in this setup was first determined by Mustafa et al. [3]; particularly, the Banach contraction principle was established in this work. Since then several fixed point, coupled fixed point, and triple fixed point theorems in the framework of generalized metric spaces have been investigated in [411].

In the recent past, many authors obtained important fixed point theorems in partially ordered metric spaces, that is, metric spaces endowed with a partial ordering (see [1224]).

The aim of this paper is to determine some coupled fixed point theorems for nonlinear contractions in the framework of partially ordered generalized metric spaces.

#### 2. Definitions and Preliminary Results

We will assume throughout this paper that the symbol and will denote the set of real and natural numbers, respectively. In this section, we recall some definitions and preliminary results which we will use throughout the paper. Mustafa and Sims [2] have recently introduced the concept of generalized metric space as follows.

Let be a nonempty set and a mapping . Then is called a generalized metric (for short, -metric) on and a generalized metric space or simply -metric space if the following conditions are satisfied: (i) if , (ii), for all and , (iii), for all and , (iv) (symmetry in all three variables), (v), for all (rectangle inequality).

We remark that every -metric on defines a metric on by , for all .

Example 2.1 (see [2]). Let be a metric space. The function defined by or for all , is a -metric on .
The concepts of convergence and Cauchy sequences and continuous functions in -metric space are studied in [2].
Let be a -metric space. Then, a sequence is said to be convergent in or simply -convergent to if for every there exists such that , for all .
Let be a -metric space. Then, is said to be Cauchy in or simply -Cauchy if for every there exists such that , for all . A -metric space is said to be complete if every -Cauchy sequence is -convergent.
Let be a -metric space and a mapping. Then, is said to G-continuous at a point if and only if it is -sequentially continuous at ; that is, whenever is -convergent to , we have -convergent to .

Proposition 2.2 (see [2]). Let be a -metric space and a sequence in . Then, for all , the following statements are equivalent: (i) is -convergent to ,(ii) as ,(iii) as ,(iv) as .

Proposition 2.3 (see [2]). Let be a -metric space and a sequence in . Then, the following statements are equivalent: (i) is -Cauchy, (ii)For every there exists such that , for all .

Lemma 2.4 (see [2]). If is a -metric space then for all .

Let be a -metric space and a mapping. Then, a map is said to be continuous [10] in if for every -convergent sequences and , is -convergent to .

Quite recently, Bhaskar and Lakshmikantham [14] defined and studied the concepts of mixed monotone property and coupled fixed point in partially ordered metric space.

Let be a partially ordered set and a mapping. Then, a map is said to have mixed monotone property if is monotone nondecreasing in and is monotone nonincreasing in ; that is, for any ,

An element is said to be a coupled fixed point of the mapping if

The following class of functions are considered in [25]. Denote with the set of all functions which satisfy that(i) is continuous and nondecreasing, (ii) if and only if , (iii), for all .

By we denote the set of all functions which satisfy , for all and .

For example, functions , where , , , and , and the functions , where , and

#### 3. Main Results

In this section, we establish some coupled fixed point results by considering a map on generalized metric spaces endowed with a partial order.

Theorem 3.1. Let be a partially ordered set, and let be a -metric on such that is a complete -metric space. Suppose that there exist , , and a mapping such that for all with and where either or . Suppose has a mixed monotone property and also suppose that either (a) is continuous or(b) has the following property:(i)if a nondecreasing sequence is -convergent to , then , for all ,(ii)if a nonincreasing sequence is -convergent to , then , for all .If there exist such that and , then has a coupled point; that is, there exist such that and .

Proof. Let be such that and . We can choose such that and . Write for all . Due to the mixed monotone property of , we can find and . By straightforward calculation, we obtain Using (3.1) and (3.2), we obtain and similarly Adding (3.4) and (3.5), we get Using the property for all , we get which implies that Since is nondecreasing, we have For all , set then a sequence is decreasing. Therefore, there exists some such that Now we have to show that . On the contrary, suppose that . Letting in (3.7) (equivalently, is -convergent to ) and using the property of and , we get which is a contradiction. Thus ; from (3.11), we have Again, we have to show that and are Cauchy sequences in the -metric space . On the contrary, suppose that at least one of or is not a Cauchy sequence in . Then there exists , for which we can find subsequences , and , of the sequences and , respectively, with , for all such that We may also assume that by choosing to be the smallest number exceeding , for which (3.14) holds. From (3.14) and (3.15) and using the rectangle inequality, we obtain Letting in the above inequality and using (3.13), we get Again, by using rectangle inequality, we obtain By using Lemma 2.4, the above inequality becomes this implies that Operating on both sides of the above inequality, Now we find the expressions and in terms of and by using (3.1) and (3.2); that is, Adding (3.22) and (3.23), we get From (3.21) and (3.24), we obtain Taking limit as on both sides of the above inequality, we get which is a contradiction, and hence and are Cauchy sequences in the -metric space . Since is complete -metric space, hence and are -convergent. Then, there exist such that and are -convergent to and , respectively. Suppose that condition (a) holds. Letting in (3.2), we get and . Lastly, suppose that assumption (b) holds. Since a sequence is nondecreasing and -convergent to and also is nonincreasing sequence and -convergent to , by assumption (b), we have and for all . Using the rectangle inequality, write Applying the function on both sides of the above equation and using (3.1), we have Letting , we get . Hence . Similarly we obtain . Thus, we conclude that has a coupled fixed point.

Corollary 3.2. Let be a partially ordered set, and let be a -metric on such that is a complete -metric space. Suppose that is a mapping having mixed monotone property. Assume that there exists such that for all , with and where either or . Suppose that either (a) is continuous or (b) has the following property: (i)if a nondecreasing sequence is -convergent to , then , for all ,(ii)if a nonincreasing sequence is -convergent to , then , for all .If there exist such that and , then there exist such that and ; that is, has a coupled point in .

Proof. Taking in Theorem 3.1 and proceeding the same lines as in this theorem, we get the desired result.

Corollary 3.3. Let be a partially ordered set, and let be a -metric on such that is a complete -metric space. Suppose is a mapping having mixed monotone property and assume that there exists such that for all with and where either or . Suppose that either (a) is continuous or (b) has the following property: (i)if a nondecreasing sequence is -convergent to , then , for all ,(ii)if a nonincreasing sequence is -convergent to , then , for all .If there exist such that and , then has a coupled point in .

Proof. Taking and in Theorem 3.1 and proceeding the same lines as in this theorem, we get the desired result.

Remark 3.4. To assure the uniqueness of a coupled fixed point, we shall consider the following condition. If is a partially ordered set, we endowed the product with for all .

Theorem 3.5. In addition to the hypothesis of Theorem 3.1, suppose that for all , there exists such that is comparable with and . Then, has a unique coupled fixed point.

Proof. It follows from Theorem 3.1 that the set of coupled fixed points is nonempty. Suppose and are coupled fixed points of the mappings ; that is, , , and , . By assumption there exists in such that is comparable to and . Put and and choose such that and . Thus, we can define two sequences and as Since is comparable to , we can assume that . Then it is easy to show that and are comparable; that is, , for all . Thus, from (3.1), we have Using the property of and adding (3.33) and (3.34), we get which implies that Therefore, We see that the sequence is decreasing; there exists some such that Now we have to show that . On the contrary, suppose that . Letting in (3.35), we get which is not possible. Hence . Therefore, (3.38) becomes which implies Similarly, we can show that . We conclude that and . Thus, has a unique coupled fixed point.

Theorem 3.6. In addition to the hypothesis of Theorem 3.1, suppose that and are comparable. Then, has a unique fixed point.

Proof. Proceeding the same lines as in the proof of Theorem 3.1, we know that a mapping has a coupled fixed point . Now we need to show that . Since and are comparable, we can assume that . It is easy to show that for all , where and . Suppose that , for all with . Using the rectangle inequality, write Operating on both sides of the above inequality, we get From (3.1), we have Similarly, we obtain Adding (3.44) and (3.45) and using the property of , we get Letting in the above inequality, we get which implies that which is not possible, and hence . Thus , whence the result.

Theorem 3.7. Let be a partially ordered set, and let be a -metric on such that is a complete -metric space. Let be a mapping such that has a mixed monotone property and whenever . Suppose that there exist and such that for all , (3.1) holds with , and where either or . Assume that either (a) is continuous or (b) has the following property: (i)if a nondecreasing sequence is -convergent to , then , for all ,(ii)if a nonincreasing sequence is -convergent to , then , for all .If there exist such that , and , then has a coupled point; that is, there exist such that and .

Proof. Let be such that and . We can choose such that and . Since , we have . Accordingly, Continuing this process, we can construct two sequences and in such that for all . Therefore, The rest of the proof can be done on the same lines as in Theorem 3.1.

#### 4. Example and the Concluding Remark

In the following, we construct an example of a -metric space involving the idea of coupled fixed point to see the applicability of our results.

Let . Define a mapping from to by Then is a complete -metric space [10]. Let us consider a partial order on be such that holds if , divides , and and hold, for all . Consider a mapping defined by Suppose that holds. Therefore, we have . It follows that .

Now, applying the function to this equality and then using the hypothesis of this function, we see that (3.1) is satisfied since the left-hand side of (3.1) becomes 0. For and , Theorem 3.7 is applicable. In this case, the coupled fixed point is not unique. Hence and are two coupled fixed points of .

We remark that inequality (3.1) is not satisfied when , , and , and hence Theorem 3.1 does not work for this example. We know that a -metric naturally induces a metric given by [2], but inequality (3.1) does not reduce to any metric inequality with the metric due to the condition that either or . Hence our theorems are more general, different from the classical results, and do not reduce to fixed point problems in the corresponding metric space .

#### Acknowledgment

The authors have benefited from the reports of the anonymous referees and they are thankful for their valuable comments on the first draft of this paper which improved the presentation and readability.