Journal of Applied Mathematics

VolumeΒ 2012Β (2012), Article IDΒ 756453, 10 pages

http://dx.doi.org/10.1155/2012/756453

## Fixed Point Theorems for -Contractive Mappings in Ordered Metric Spaces

Department of Applied Mathematics, National Hsinchu University of Education, Hsinchu 30014, Taiwan

Received 19 September 2011; Accepted 25 November 2011

Academic Editor: YanshengΒ Liu

Copyright Β© 2012 Chi-Ming Chen. 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 obtain some new fixed point theorems for -contractive mappings in ordered metric spaces. Our results generalize or improve many recent fixed point theorems in the literature (e.g., Harjani et al., 2011 and 2010).

#### 1. Introduction and Preliminaries

Throughout this paper, by , we denote the set of all real nonnegative 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 fixed point theorem asserts that if , is contractive and is complete, then has a unique fixed point in . It is well known that the Banach contraction principle [1] is a very useful and classical tool in nonlinear analysis. Also, this principle has many generalizations. For instance, a mapping is called a quasicontraction if there exists such that

for any . In 1974, ΔiriΔ [2] introduced these maps and proved an existence and uniqueness fixed point theorem.

In 1972, Chatterjea [3] introduced the following definition.

*Definition 1.1. *Let be a metric space. A mapping is said to be a -contraction if there exists such that for all , the following inequality holds:

Choudhury [4] introduced a generalization of -contraction as follows.

*Definition 1.2. *Let be a metric space. A mapping is said to be a weakly -contraction if for all ,
where is a continuous function such that if and only if .

In [3, 4], the authors proved some fixed point results for the -contractions. In [5], Harjani et al. proved some fixed point results for weakly -contractive mappings in a complete metric space endowed with a partial order.

In the following, we assume that the function satisfies the following conditions:(C1) is a strictly increasing and continuous function in each coordinate, and(C2)for all , , , , and .

*Example 1.3. *Let denote
Then, satisfies the above conditions (C1) and (C2).

Now, we define the following notion of a -contractive mapping in metric spaces.

*Definition 1.4. *Let be a partially ordered set and suppose that there exists a metric in such that is a metric space. The mapping is said to be a -contractive mapping, if
for .

Using Example 1.3, it is easy to get the following examples of -contractive mappings.

*Example 1.5. *Let endowed with usual ordering and with the metric given by
Let denote
where , , , and , for all . Let denote
Then, is a -contractive mapping.

*Example 1.6. *Let endowed with the coordinate ordering (i.e., and ) and with the metric given by
Let denote
where , , , , and , for all . Let denote
Then, is a -contractive mapping.

In this paper, we obtain some new fixed point theorems for -contractive mappings in ordered metric spaces. Our results generalize or improve many recent fixed point theorems in the literature (e.g., [5, 6]).

#### 2. Main Results

We start with the following definition.

*Definition 2.1. *Let be a partially ordered set and . Then one says that is monotone nondecreasing if, for ,

We now state the main fixed point theorem for -contractive mappings in ordered metric spaces when the operator is nondecreasing, as follows.

Theorem 2.2. *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 continuous and nondecreasing -contractive mapping. If there exists with , then has a fixed point in .*

*Proof. *If , then the proof is finished. Suppose that . Since is nondecreasing mapping, by induction, we obtain that
Put for . Then, for each , from (), and, as the elements and are comparable, we get
and so we can deduce that, for each ,
Let we denote . Then, is a nonincreasing sequence and bounded below. Thus, it must converge to some . If , then by the above inequalities, we have
Passing to the limit, as , we have
which is a contradiction. So .

We next claim that that the following result holds.

For each , there is such that for all ,
We will prove () by contradiction. Suppose that () is false. Then, there exists some such that for all , there exist and with such that
Using the triangular inequality:
and letting , we have

Since is a -contractive mapping, we also have
Letting . Then, we get
a contradiction. It follows from () that the sequence must be a Cauchy sequence.

Similary, we also conclude that for each ,
and so we have that for each ,

Let us denote . Then, is a nonincreasing sequence and bounded below. Thus, it must converge to some . If , then by the above inequalities, we have
Passing to the limit, as , we have
which is a contradiction. So . By the above argument, we also conclude that is a Cauchy sequence.

Since is complete, there exists such that . Moreover, the continuity of implies that

So we complete the proof.

In what follows, we prove that Theorem 2.2 is still valid for not necessarily continuous, assuming the following hypothesis in (which appears in Theorem 1 of [7]).If is a nondecreasing sequence in , such that

Theorem 2.3. *Let be a partially ordered set and suppose that there exists a metric in such that is a complete metric space. Assume that satisfies (), and let be a nondecreasing -contractive mapping. If there exists with , then has a fixed point in .*

*Proof. *Following the proof of Theorem 2.2, we only have to check that . As is a nondecreasing sequence in and , then the condition () gives us that for every . Since is a nondecreasing -contractive mapping, we have
Letting and using the continuity of , we have
and this is a contraction unless , or equivalently, .

In what follows, we give a sufficient condition for the uniqueness of the fixed point in Theorems 2.2 and 2.3. This condition is the following and it appears in [8]: In [7], it is proved that the above-mentioned condition is equivalent to the following:

Theorem 2.4. *Adding condition () to the hypothesis of Theorem 2.2 (or Theorem 2.3) and the condition for all , (or, ) to the function , one obtains the uniqueness of the fixed point of .*

*Proof. *Suppose that there exist which are fixed points of . We distinguish two cases.*Case 1. *If and are comparable and , then is comparable to for all , and
and this is a contradiction unless , that is, .*Case 2. *If and are not comparable, then there exists comparable to and . Monotonicity of implies that is comparable to and for all . We also distinguish two cases.*Subcase 2.1. * If there exists with , then we have
and this is a contradiction unless , that is, .*Subcase 2.2. * For all with , since is a nondecreasing -contractive mapping, we have
Using the above inequality, we claim that for each ,
If not, we assume that , then by the definition of and , we have
which implies a contradiction. Therefore, our claim is proved.

This proves that the nonnegative decreasing sequence is convergent. Put , . We now claim that . If , then making , we getthis is a contradiction. So , that is, .

Analogously, it can be proved that .

Finally, the uniqueness of the limit gives us .

This finishes the proof.

In the following, we present a fixed point theorem for a -contractive mapping when the operator is nonincreasing. We start with the following definition.

*Definition 2.5. *Let be a partially ordered set and . Then one says that is monotone nonincreasing if, for ,

Using a similar argument to that in the proof of Theorem 3.1 of [5], we get the following point results.

Theorem 2.6. *Let be a partially ordered set satisfying condition () and suppose that there exists a metric in such that is a complete metric space, and let be a nonincreasing -contractive mapping. If there exists with or , then . Moreover, if in addition, is compact and is continuous, then has a unique fixed point in .*

*Proof. *If , then it is obvious that . Suppose that (the same argument serves for ). Since is nonincreasing the consecutive terms of the sequence are comparable, we have
and so we conclude that for each ,

Thus, is a decreasing sequence and bounded below, and it must converge to . We claim that . If , then by the above inequalities and the continuity of , letting , we have
which is a contradiction. So , that is, . Consequently, .

Further, since is continuous and is compact, we can find such that

and, therefore, is a fixed point of .

The uniqueness of the fixed point is proved as in Theorem 2.4.

#### Acknowledgment

The authors would like to thank the referees whose helpful comments and suggestions led to much improvement in this paper.

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