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International Journal of Analysis
Volume 2013 (2013), Article ID 307234, 7 pages
http://dx.doi.org/10.1155/2013/307234
Research Article

Some Common Fixed Point Results in Rectangular Metric Spaces

1Department of Mathematics, International Islamic University, H-10, Islamabad 44000, Pakistan
2Department of Mathematics, COMSATS Institute of Information Technology, Chak Shahzad, Islamabad 44000, Pakistan
3Department of Mathematics, Atilim University, İncek, 06836 Ankara, Turkey

Received 19 November 2012; Accepted 29 January 2013

Academic Editor: Ahmed Zayed

Copyright © 2013 Muhammad Arshad 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 obtain sufficient conditions for the existence of unique common fixed point of -weakly contractive mappings on complete rectangular metric spaces. In the process, we generalize several fixed point results from the literature. We also give an example to illustrate our work.

1. Introduction

Fixed point theorems are very important tools in nonlinear functional analysis. Banach Contraction Mapping Principle is the most frequently cited fixed point theorem in the literature. It asserts that if is a complete metric space and is a contraction, that is, there exists such that for all , then has a unique fixed point. The contraction definition (1) implies that is uniformly continuous, which is a very strong condition. It is quite natural to ask whether the inequality (1) can be replaced with another inequality which does not force to be continuous. This question was answered affirmatively by Kannan [1]. A self-mapping has a unique fixed point in a complete metric space if there are nonnegative real numbers with such that the following inequality is satisfied for all :

In 2000, Branciari [2] introduced the notion of a generalized (rectangular) metric space where the triangle inequality of a metric space was replaced by another inequality, the so-called rectangular inequality. In this paper [2], the author also extended the celebrated Banach Contraction Mapping Principle in the context of generalized metric spaces. Later, Azam and Arshad [3] obtained sufficient conditions for existence of a unique fixed point of Kannan type mappings in the framework of generalized/rectangular metric spaces. Subsequently, Azam et al. [4] proved an analog of Banach Contraction Principle in the setting of rectangular cone metric spaces. Following this trend, a number of authors focused on rectangular metric spaces and proved the existence and uniqueness of a fixed point for certain type of mappings (see e.g., [514]).

Recently, Di Bari and Vetro [15] obtained some common fixed point theorems for mappings satisfying a -weakly contractive condition in rectangular metric spaces. In this paper, we prove several fixed point results in rectangular metric spaces that can be considered as a continuation of [15].

We recall some basic definitions and necessary results on the topic in the literature.

Definition 1. Let be a nonempty set and let be a mapping such that for all and for all distinct points , each of them different from and , one has
Then, the map is called rectangular (generalized) metric. The pair is called a rectangular (generalized) metric space.

To avoid confusion, we prefer to use the term “rectangular metric space” for the spaces under consideration in this paper, because there are some other spaces that are also called generalized metric spaces. We abbreviate a rectangular metric spaces with RMS.

Definition 2 (see [2]). Let be a RMS.(i)A sequence is called RMS convergent to if and only if as . In this case, we use the notation .(ii)A sequence in is called a RMS Cauchy if and only if for each , there exists a natural number such that for all .(iii)A RMS is called a RMS complete if every RMS Cauchy sequence is RMS convergent in .

Definition 3. Let and be self-mappings of a nonempty set .(i)A point is said to be a common fixed point of and if .(ii)A point is called a coincidence point of and if . And if , then is said to be a point of coincidence of and .(iii)The mappings are said to be weakly compatible if they commute at their coincidence point that is, whenever .

Lemma 4. Let be a nonempty set. Suppose that the mappings have a unique coincidence point in . If and are weakly compatible, then and have a unique common fixed point.

Proof. Let be the coincidence point of , that is, Since and are weakly compatible, we observe that have a unique coincidence point, then . Hence, we have by (3).

Let denote all functions such that(a) is continuous,(b) if and only if .

2. Main Result

We start this section with the following theorem.

Theorem 5. Let be a Hausdorff RMS and let be self-mappings such that . Assume that is a complete RMS. Suppose that the following condition holds: for all and , where is nondecreasing and Then and have a unique coincidence point in . Moreover, if and are weakly compatible, then and have a unique common fixed point.

Proof. We first prove that the coincidence point of and is unique if it exists. Let and be coincidence points of and . Thus, there exists some such that and . By (5), we derive that where Thus, we conclude that by (7).
Remember that and are weakly compatible. Since is the unique coincidence point of and , the point is the unique common fixed point of and by Lemma 4.
Now, we will prove the existence of a coincidence point of and . Let be an arbitrary point. Since , we define two iterative sequences and in as follows: for all . If then clearly and have a coincidence point in . Indeed, and is the desired point. Thus, we assume that , that is, for all . Moreover, if , then we choose , for all .
We assert that Now from (5), we have where If , then we have which implies that , and hence . Then , which contradicts with the initial assumption. Thus, we have , and hence Since is nondecreasing, then for all , that is, the sequence is nonincreasing and bounded below. Hence, it converges to a positive number, say . Taking the limit as in (14), we get which leads to , and hence . Thus, In this step, we show that the second limit in (5) is also 0. To prove this claim, we set and in (5) and get that where We consider all possible cases for . If , then we have Letting in (19), the right hand side of (19) tends to 0. Hence, . Since, is continuous, we find that . For the case , we get analogously .
Let us consider the last case, that is, . The inequality (17) turns into Therefore, the sequence is non-increasing and bounded below. Hence, the sequence converges to a number, . Taking limit as in (20), we get which implies that , and hence . In other words, Suppose that for all and prove that is a Cauchy sequence. If possible, let be not a Cauchy sequence. Then there exists for which we can find subsequences and of with such that Furthermore, corresponding to , we can choose in such a way that it is the smallest integer with and satisfying (23). Then, Using (23), (24), and the rectangular inequality , we have Taking limit as in (23) and using (16), (22) we get Again, using the rectangular inequality , we obtain Letting in (27), and by using (16) and (22) we get Now, we substitute and in (5). Consider where Clearly, as , we have Then letting in (29), we have This implies that which contradicts the fact that . Thus, is a RMS Cauchy sequence. Since is RMS complete, there exists such that as . Let such that . Applying the inequality (5), with , we obtain where Now, if or , we have since is nondecreasing. In either case, letting , we get . Since is Hausdorff, we deduce that . If, on the other hand, , then taking limit as in we get , hence , that is, . Let . Then is a point of coincidence of and . Suppose that there exists such that , we can choose in such a way that it is the smallest positive integer satisfying . We aim to prove that , then and so is a point of coincidence of and . Assume that . This implies that . Now we have where If , then we have which is a contradiction. Thus .

Corollary 6. Let be a Hausdorff and complete RMS and let be self-mappings such that satisfying for all and . Then and have a unique common fixed point in .

Proof. Let and . Then by Theorem 5,   and have a unique common fixed point.

Corollary 7. Let be a Hausdorff and complete RMS and let be self-mappings such that satisfying for all and . Then and have a unique common fixed point in .

Proof. It is obvious that Let and . Then by Theorem 5, and have a unique common fixed point.

Corollary 8. Let be a Hausdorff and complete RMS and let be self-mappings such that satisfying for all where Then and have a unique common fixed point.

Proof. Let . Then by Theorem 5, and have a unique common fixed point.

Theorem 9. Let be a Hausdorff and complete RMS and let be self-mappings such that satisfying for all and , where is nondecreasing and Then and have a unique common fixed point.

Proof. Let be an arbitrary point. Since , we define the sequence as follows for all . Assume that for all . Now from (5),we get where The rest of the proof is the same as the proof of Theorem 5.

By , we denote the class of functions satisfying the following.(a) is Lebesgue integrable function on each compact subset of ,(b) for any .

Theorem 10. Let be a Hausdorff RMS and let be self-mappings such that . Assume that is a complete RMS and that the following condition holds: for all and where Then and have a unique common fixed point.

Proof. Let and . Then and are function in . By Theorem 5,   and have a unique common fixed point.

Theorem 11. Let be a Hausdorff RMS and let be self-mappings such that . Assume that is a complete RMS and that the following condition holds: for all and and some , where Then and have a unique common fixed point.

Proof. Let . Then by Theorem 10, and have a unique common fixed point.

Theorem 12. Let be a Hausdorff RMS and let be a self-mappings such that . Assume that is a complete RMS and that the following condition holds: for all and where Then and have a unique common fixed point.

Proof. Let and . Then and are function in . By Theorem 9,   and have a unique common fixed point.

Example 13. Let , where and . Define the generalized metric on as follows:
It is easy to show that does not satisfy the triangle inequality on . Indeed, Thus holds, so is a rectangular metric. Notice that is usual metric space, and hence it is Hausdorff. On the other hand, each singleton is closed and open in , and hence is Hausdorff rectangular metric space.
Let be defined as follows: Define and . Then and satisfy the condition of Theorem 5 and have a unique common fixed point of , that is, .

Acknowledgment

The authors thank the referees for their appreciation, valuable comments, and suggestions.

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