Abstract

We prove new fixed point theorems in the framework of partially ordered metric spaces. The main result is an extension and a generalization of many existing results in the literature. An example is also considered to illustrate the main result.

1. Introduction and Preliminaries

Fixed point theory is one of fundamental tools of nonlinear functional analysis. Since the fixed point theory has a wide application area in almost all quantitative sciences, many authors have been working on this field. One of the impressive initial results in this direction was given by Banach [1], known as Banach Contraction Mapping Principle. It states that each contraction in a complete metric space has a unique fixed point. Since then, a number of papers have been reported on various generalization of celebrated Banach Contraction Mapping Principle.

In 2008, Dutta and Choudhury proved the following theorem.

Theorem 1.1 (see [2]). Let be a complete metric space and be such that where are continuous, nondecreasing and if and only if . Then, has a unique fixed point .

Remark 1.2. Notice that Theorem 1.1 remains true if the hypothesis on is replaced by is lower semicontinuous and if and only if (see, e.g., [3, 4]).

Eslamian and Abkar [5] stated the following theorem as a generalization of Theorem 1.1.

Theorem 1.3. Let be a complete metric space and be such that where are such that is continuous and nondecreasing, is continuous, is lower semicontinuous, continuous, Then, has a unique fixed point .

Aydi et al. [6] proved that Theorem 1.3 is a consequence of Theorem 1.1

Harjani and Sadarangani [7] extended Theorem 1.1 in the framework of partially ordered metric spaces in the following way.

Theorem 1.4. Let be a partially ordered complete metric space. Let be a continuous nondecreasing mapping such that where are continuous, nondecreasing and if and only if . If there exists such that , then has a fixed point .

Choudhury and Kundu [8] proved the following theorem as a generalization of Theorems 1.3 and 1.4.

Theorem 1.5. Let be a partially ordered complete metric space. Let be a nondecreasing mapping such that where are such that is continuous and nondecreasing, is continuous, is lower semicontinuous, continuous, If there exists such that , then has a unique fixed point .

Aydi et al. [6] proved that Theorem 1.5 is a consequence of Theorem 1.4.

2. Main Results

We define the following set of functions:

Let be ordered set. The pair is said to be comparable if either or holds.

Theorem 2.1. Let be an ordered metric space such that is complete and be a nondecreasing self-mappings. Assume that there exist , , and such that for all comparable . Suppose that either(a)is continuous, or(b)if a nondecreasing sequence is such that , then for all . If there exists such that , then has a fixed point.

Proof. Let . We define an iterative sequence in the following way: Since is nondecreasing and , we have and hence is a nondecreasing sequence. If for some , then the point is the desired fixed point of which completes the proof.

Hence we suppose that , that is, for all . Hence, (2.5) turns in to

We want to show that the sequence in nonincreasing. Suppose, to the contrary, that there exists some such that Since is nondecreasing, we obtain that By taking and , the condition (2.3) together with (2.8) we derive that which contradicts (2.2). Therefore, we conclude that hold for all . Hence is a nonincreasing sequence of positive real numbers. Thus, there exists such that . We will show that by method of reductio ad absurdum. For this purpose, we assume that . By (2.9) together with the properties of , , we have which is a contradiction. Hence

We will show that the sequence is Cauchy. Suppose, to the contrary, that that is, there is and sequences and such that for all positive integers with Additionally, corresponding to , we may choose such that it is the smallest integer satisfying (2.15) and . Thus, Now for all we have So Again, from Taking the limit as , by (2.13) and (2.18), we deduce Now from (2.3) we have Taking the as in the inequality above, we have So we have which contradicts the fact that for all . Hence that is, the sequence is Cauchy. Since is complete, there exists such that . Suppose that holds. Then, Hene, is a fixed point of .

Suppose that holds, that is, for all . We claim that is a fixed point of , that is, . Suppose, to the contrary, that . Due to condition (2.3), we have Taking the as in the inequality above, we obtain that which is a contradiction. Hence and hence, .

In the following theorem we investigate the uniqueness of fixed points in the theorem above. In order to assure the uniqueness of fixed points we need the following notion on the partially ordered metric space which is called the comparability condition:(C)  For every there exists such that either and or and .

Theorem 2.2. In addition to hypotheses of Theorem 2.1, suppose that is the comparability condition (C). Then has a unique fixed point.

Proof. Due to Theorem 2.1, we guarantee that has a fixed point. Suppose and are fixed points of with , that is, .
We need to consider two different cases. First case: If and are comparable, then which contradicts (2.2). Hence .
Let us examine the second and last case: if and are not comparable, then by (C) there exists such that and . Notice that and hence for each . Analogously we have for each . Then we have which is true only if . On the other hand, we have which yields that . Hence, .

Example 2.3. Let . We define a partial order by if and only if . Let and be defined by . Also define three functions , and by Clearly, is lower semicontinuous, also and for all . Let . Then we have where . And Then hold for all . Hence condition of Theorem 2.1 is hold. That is, has a fixed point. Theorem 1.5 cannot be applied to in this example. Since is lower semicontinuous and , , and . Further, the approach of Aydi et al. [6] cannot be modified for it.

Now, we use the following lemma, that is a consequence of the axiom of choice, to obtain common fixed point results for two self-mappings defined on a metric space.

Lemma 2.4 (see [9, Lemma 2.1]). Let be a nonempty set and a mapping. Then there exists a subset such that and is one-to-one.

Theorem 2.5. Let be an ordered metric space such that is complete and be two self-mappings. Assume that there exist , , and such that for all with . If the following conditions hold: (a) is -nondecreasing, g(X) is closed and ,(b)there exists such that ,(c)if a nondecreasing sequence is such that , then for all .
Then and have a coincidence point.

Proof. By Lemma 2.4, there exists such that and is one-to-one. Define Since is one-to-one on and , is well defined. By condition (2.35), for all , we have Since is -nondecreasing then is nondecreasing. Indeed, implies and then . Also, since is complete, by Theorem 2.1, we have that has a fixed point on , say . That is, . Then and have a coincidence point.

Remark 2.6. One can easily conclude that Theorem 2.5 is still valid if we replace the condition (c) of Theorem 2.5 with () if a nonincreasing sequence is such that , then for all .
We know state a new condition as follows:(D) The set of points of coincidence of and , say, is totally ordered.

Theorem 2.7. In addition to hypotheses of Theorem 2.5, suppose that is the comparability condition (D). Then and have a unique coincidence point. Moreover, if and are weakly compatible then and have a unique common fixed point.

Proof. Assume that and are two coincidence points of and with . By condition (2.35) we have which is a contradiction. Hence . Let . Since and are weakly compatible then . Now since and have a unique coincidence point then . That is, is a unique common fixed point of and .