Journal of Applied Mathematics

Volume 2013 (2013), Article ID 302438, 7 pages

http://dx.doi.org/10.1155/2013/302438

## Fixed Point Theory of Weak Contractions in Partially Ordered Metric Spaces

^{1}Department of Applied Mathematics, National Hsinchu University of Education, No. 521, Nandah Road, Hsinchu City, Taiwan^{2}Department of Applied Mathematics, Chung Yuan Christian University, Taiwan

Received 12 March 2013; Accepted 20 April 2013

Academic Editor: Erdal Karapınar

Copyright © 2013 Chi-Ming Chen et al. 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 prove two new fixed point theorems in the framework of partially ordered metric spaces. Our results generalize and improve many recent fixed point theorems in the literature.

#### 1. Introduction and Preliminaries

Throughout this paper, by , we denote the set of all nonnegative real numbers, while is the set of all natural numbers. Let be a metric space, a subset of , and a map. We say is contractive if there exists such that, for all , The well-known Banach's fixed point theorem asserts that if , is contractive and is complete, then has a unique fixed point in . In nonlinear analysis, the study of fixed points of given mappings satisfying certain contractive conditions in various abstract spaces has been investigated deeply. The Banach contraction principle [1] is one of the initial and crucial results in this direction. Also, this principle has many generalizations. For instance, Alber and Guerre-Delabriere in [2] suggested a generalization of the Banach contraction mapping principle by introducing the concept of weak contraction in Hilbert spaces. In [2], the authors also proved that the result of Eslamian and Abkar [3] is equivalent to the result of Dutta and Choudhury [4]. Later, weakly contractive mappings and mappings satisfying other weak contractive inequalities have been discussed in several works, some of which are noted in [4–16].

In 2008, Dutta and Choudhury proved the following theorem.

Theorem 1 (see [4]). *Let be a complete metric space, and let be such that
**
where are continuous and nondecreasing, and if and only if . Then has a fixed point in . *

Recently, Eslamian and Abkar [3] proved the following theorem.

Theorem 2 (see [3]). *Let be a complete metric space, and let be such that
**
where are such that is continuous and nondecreasing, is continuous, is lower semicontinuous, and
**
Then has a fixed point in . *

In the recent, fixed point theory has developed rapidly in partially ordered metric spaces (e.g., [17–22]).

In 2012, Choudhury and Kundu [23] proved the following fixed point theorem as a generalization of Theorem 2.

Theorem 3 (see [23]). *Let be a partially ordered set and suppose that there exists a metric in such that is a complete metric space and let be a nondecreasing mapping such that
**
where are such that is continuous and nondecreasing, is continuous, is lower semicontinuous, and
**
Also, if any nondecreasing sequence in converges to , then one assumes that
**
If there exists with , then and have a coincidence point in . *

In this paper, we prove two new fixed point theorems in the framework of partially ordered metric spaces. Our results generalize and improve many recent fixed point theorems in the literature.

#### 2. Fixed Point Results (I)

We start with the following definition.

*Definition 4. *Let be a partially ordered set and . Then is said to be monotone nondecreasing if, for ,

Let be a partially ordered set. are said to be comparable if either or holds.

In the section, we denote by the class of functions satisfying the following conditions: is an increasing, continuous function in each coordinate; for all , , and .

Next, we denote by the class of functions satisfying the following conditions: is a continuous, nondecreasing function; for and ; is subadditive; that is, for all .

And we denote the following sets of functions:

Let be a nonempty set, and let be a partially ordered set endowed with a metric . Then, the triple is called a partially ordered complete metric space.

We now state the main fixed point theorem for -contractions in partially ordered metric spaces, as follows.

Theorem 5. *Let be a partially ordered complete metric space. Let be monotone nondecreasing, and
**
for all comparable , where , , , and , and
**
Suppose that either ** is continuous or ** if any nondecreasing sequence in converges to , then one assumes that
**If there exists with , then has a fixed point in . *

* Proof. *Since is nondecreasing, by induction, we construct the sequence recursively as
Thus, we also conclude that
If any two consecutive terms in (14) are equal, then the has a fixed point, and hence the proof is completed. So we may assume that
Now, we claim that for all . If not, we assume that for some ; substituting and in (10) and using the definition of the function , we have
and hence
Since for all , we have that , which contradicts (15). Therefore, we conclude that
From the previous argument, we also have that for each
It follows in (18) that the sequence is monotone decreasing; it must converge to some . Taking limit as in (19) and using the continuities of and and the lower semicontinuous of , we get
which implies that . So we conclude that

We next claim that is a Cauchy sequence; that is, for every , there exists such that if , then .

Suppose, on the contrary, that there exists such that, for any , there are with satisfying Further, corresponding to , we can choose in such a way that it the smallest integer with and . Therefore . Now we have that for all

By letting . we get that On the other hand, we have Letting , then we get By (14), we have that the elements and are comparable. Substituting and in (10), we have that, for all , By the previous argument and using inequality (10), we can conclude that which implies that , a contradiction. Therefore, the sequence is a Cauchy sequence.

Since is complete, there exists such that Suppose that holds. Then

Thus, is a fixed point in .

Suppose that holds; that is, for all . Substituting and in (10), we have that Taking limit as in equality (31), we have which implies that ; that is . So we complete the proof.

If we let it is easy to get the following theorem.

Theorem 6. *Let be a partially ordered complete metric space. Let be monotone nondecreasing, and
**
for all comparable , where , , and , and
**
Suppose that either* * is continuous or * *if any nondecreasing sequence in converges to , then one assumes that
** If there exists with , then has a fixed point in .*

#### 3. Fixed Point Results (II)

In the section, we denote by the class of functions satisfying the following conditions: is an increasing and continuous function in each coordinate; for , , , and .

Next, we denote by the class of functions satisfying the following conditions: is continuous and nondecreasing; for , and .

And we denote by the class of functions satisfying the following conditions. is continuous; for , and .

We now state the main fixed point theorem for the -contractions in partially ordered metric spaces, as follows.

Theorem 7. *Let be a partially ordered complete metric space, and let be monotone nondecreasing, and
**
for all comparable and , , , where and
**
Suppose that either* * is continuous or* *if any nondecreasing sequence in converges to , then one assumes that
**If there exists with , then has a fixed point in . *

* Proof. *If , then the proof is finished. Suppose that . Since is nondecreasing, by induction, we construct the sequence recursively as
Thus, we also conclude that
We now claim that
Put and in (37). Note that
So, we obtain that
where
We now claim that
If not, we assume that ; then , since is non-decreasing. Using inequality (44) and the conditions of the function , we have that, for each ,
which implies that , and hence . This contradicts our initial assumption.

From the previous argument, we have that, for each ,
And since the sequence is decreasing, it must converge to some . Taking limit as in (48) and by the continuity of and , we get
and so we conclude that and .

We next claim that is Cauchy; that is, for every , there exists such that if , then .

Suppose, on the contrary, that there exists such that, for any , there are with satisfying
Further, corresponding to , we can choose in such a way that it the smallest integer with and . Therefore . By the rectangular inequality, we have
Letting , then we get
On the other hand, we have
By letting , we get that
Using inequalities (37), (52), and (54) and putting and , we have that
where
Letting , then we obtain that
This implies that , and hence . So we get a contraction. Therefore is a Cauchy sequence.

Since is complete, there exists such that

Suppose that holds. Then
Thus, is a fixed point in

Suppose that holds; that is, for all . Substituting and in (37), we have that
where
Letting , then we obtain that
which implies that ; that is, . So we complete the proof.

If we let it is easy to get the following theorem.

Theorem 8. *Let be a partially ordered complete metric space, and let be monotone nondecreasing, and
**
for all comparable and , , where and
**
Suppose that either* * is continuous or * *if any nondecreasing sequence in converges to , then one assumes that
**If there exists with , then has a fixed point in .*

#### Acknowledgment

The authors would like to thank the referee(s) for the many useful comments and suggestions for the improvement of the paper.

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