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International Journal of Analysis

Volume 2013 (2013), Article ID 852727, 6 pages

http://dx.doi.org/10.1155/2013/852727

## On a Theorem of Khan in a Generalized Metric Space

^{1}Department of Mathematics, COMSATS Institute of Information Technology, Chak Shahzad, Islamabad 44000, Pakistan^{2}Department of Mathematics, International Islamic University, H-10, Islamabad 44000, Pakistan^{3}Dipartimento di Matematica e Informatica, Università degli Studi di Palermo, Via Archirafi, 34, 90123 Palermo, Italy

Received 25 August 2012; Revised 20 January 2013; Accepted 23 January 2013

Academic Editor: Ying Hu

Copyright © 2013 Jamshaid Ahmad 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

Existence and uniqueness of fixed points are established for a mapping satisfying a contractive condition involving a rational expression on a generalized metric space. Several particular cases and applications as well as some illustrative examples are given.

#### 1. Introduction and Preliminaries

In the last decades, several attempts have been made in order to generalize the concept of metric space, and many important results have been reported. For instance, see quasi metric spaces [1], generalized quasi metric spaces [2], pseudometric spaces ([3], Chapter 2), approach spaces [4], -spaces [5], inframetric spaces [6], and -metric spaces [7]. Sometimes, as in [8, 9], even the very notion of generalized metric spaces (or even gms [10]) is used, but it has a different meaning. In 2000, Branciari [11] introduced the notion of generalized metric space where the triangle inequality of a metric space is replaced by a rectangular inequality involving four terms instead of three. He also extended the Banach’s contraction principle in such spaces. In 2008, Azam and Arshad [12] obtained sufficient conditions for existence of unique fixed point of Kannan type mappings defined on generalized metric spaces. Recently, Samet [13] and Sarma et al. [14] showed that some propositions in [11] are not true. Moreover, in [14], a rigorous and nice proof of the Banach’s contraction principle is presented, by assuming that the generalized metric space is Hausdorff. Afterwards, many authors studied various existence theorems of fixed points in such spaces. For more details about fixed point theory in generalized metric spaces, we refer the reader to [13, 15–24].

On the other hand, in [25] Khan proved the following fixed point theorem.

Theorem 1. * Let be a complete metric space and let be a self-mapping on that satisfies the following contractive condition:
**
for all and for some . Then has a unique fixed point in .*

*Remark 2. *In (1) if the denominator vanishes, then and and consequently also the numerator vanishes. Moreover, we have , and so the contractive condition is not well defined.

The aim of this paper is to give a version of Theorem 1 in the setting of generalized metric spaces.

The following definitions will be needed in the sequel.

*Definition 3 (see [11]). *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 (gm1) if and only if ,(gm2),
(gm3) (the rectangular inequality).

Then is called generalized metric and the pair is called generalized metric space (gms, for short). Note that the rectangular inequality and simple induction show that this inequality holds with terms, where .

The next example gives a method of construction a new generalized metric space from a family of given generalized metric spaces.

*Example 4. *Let be a family of disjoint generalized metric spaces and let . Define by with
Then, is a gms.

By referring to the Minkowski’s inequality for sums, we give an example of a gms that is not a metric space.

*Example 5. *Let , and define the generalized metric by
Clearly, (gm1) and (gm2) are obvious. Also (gm3) is immediate if at least one between and is not in . Then, consider , , , and so that we have and . Note that
and so (gm3) holds true. Then is a gms. On the other hand, since . Here, the triangle inequality is violated; that is, is not a metric space.

*Definition 6 (see [11]). * Let be a gms, and let be a sequence in and . We say that is gms convergent to if and only if as . We denote this by .

*Definition 7 (see [11]). *Let be a gms and let be a sequence in . We say that is a gms Cauchy sequence if and only if for each there exists a natural number such that for all .

*Definition 8 (see [11]). *Let be a gms. is called a complete gms if every gms Cauchy sequence is gms convergent in .

*Definition 9. *Let be a set and let be a topology on . is a Hausdorff topological space, if for every with there are open subsets of such that , , and are disjoint. Equivalently, a topological space is Hausdorff if and only if every convergent net in has a unique limit, and in particular the limit of a convergent sequence is unique.

Now, we recall an example of a complete gms that is not Hausdorff.

*Example 10 (see [14]). * Let and define by with
Then is a complete gms.

#### 2. Main Results

Our main theorem is essentially inspired by Khan [25]. More precisely, we state and prove the following result.

Theorem 11. *Let be a complete gms and let be a self-mapping such that
**
for all and , and for some with . Then has a unique fixed point in .*

*Proof. *Let be an arbitrary point. Define , for . If for some , then and hence is a fixed point of . This completes the proof. Therefore, we suppose for all and distinguish two cases.*Case 1*. Assume that , for . Then
that is,
or equivalently
Thus in general, for , we have
and so, since , as . Now, we consider the following two subcases. *Subcase 1.1*. We assume that for some with . Letting , then , that is, where , . Now, if , from (6) we have
that implies , since . *Subcase 1.2*. We assume that for all with . Clearly we have
and, after routine calculations, it is also easy to obtain
Using the previous inequality, we get
and hence
Now, if is odd, then writing , and using the fact that for with , we can easily show that
On the other hand, if is even, then writing , and using the fact that for with , we can easily show that
Thus, combining all the cases, for all we have

Therefore, as . This implies that is a Cauchy sequence in . Since is complete, there exists a such that . If , for some , then , and, by (6), we get . It follows that ; that is, is a fixed point of . Then, without loss of generality we can assume that for all . Now, if , and , by the rectangular inequality we have
Letting , since , we obtain easily that leads to a contradiction since and so must be .

On the other hand, if for every and is infinite, then
that implies and so . If is finite, then (or generally ) only for a finite set of values of , and therefore (19) holds for all large enough. Now, we show that has a unique fixed point. For this, we assume that is another fixed point of in such that . By (6) we have
or, equivalently, that leads to a contradiction and hence .*Case 2*. Assume that , for some . By condition (6), it follows that and hence . This completes the proof of the existence of a fixed point of , say .

The uniqueness follows as in Case .

As consequence of Theorem 11, we give the following corollary.

Corollary 12. * Let be a complete gms and let be a self-mapping such that
**
for all and , for some and for some with . Then has a unique fixed point in .*

*Proof. *By Theorem 11, let be the unique fixed point of the mapping . Then, we have and so is a fixed point of . Therefore, by uniqueness of the fixed point it must be that .

By setting in Corollary 12, we draw the following result which can be viewed as an extension of Bryant’s theorem [26] to generalized metric spaces.

Corollary 13. *Let be a complete gms and let be a self-mapping such that
**
for all and , for some and for some . Then has a unique fixed point in .*

*Proof. * Note that if , then and and so from (23) we have
Since , this implies , but by hypothesis, condition (23) must hold true only for all with . Then, without loss of generality we can assume that . The rest of the proof is a direct consequence of Corollary 12 and Theorem 11.

Our first example illustrates Theorem 11.

*Example 14. *Let and define the generalized metric by with
By routine calculations, it is easy to check that holds and so is a generalized metric; obviously, is complete. On the other hand, does not satisfy the triangle inequality. Indeed, we get
Now, define by
The reader can easily show that the mapping satisfies the contractive condition of Theorem 11 with and , for example. Then, is the unique fixed point of .

The following example illustrates Corollary 12 but essentially shows the usefulness of Bryant’s theorem in respect of Banach’s contraction principle.

*Example 15. *Let and . Define the generalized metric by with

By routine calculations, it is easy to check that (gm3) holds and so is a generalized metric. On the other hand, does not satisfy the triangle inequality on . Indeed, we get
Next, we show that is complete. Indeed, if there exists such that for all , then the sequence is Cauchy. Now, we assume that for all there exists such that . Let be fixed and assume that there exists such that . This implies that is at most singleton (each two distinct points in have distance greater than , by definition). Also by definition of , it follows that is Cauchy in , endowed with the usual metric of real numbers, and so convergent in . Consequently, the same holds in the gms .

Now, define by
Then, for and , we get
which leads to a contradiction since . Therefore, the Banach’s contraction principle is not applicable in this case. On the other hand, since , we have
which shows that satisfies the contractive condition of Corollary 12 and is the unique fixed point of .

Finally, as an application of Theorem 11 we give the following result.

Corollary 16. *Let be a complete gms and let be a self-mapping satisfying (6). If is a self-mapping such that , then and have a unique common fixed point in ; that, is there exists such that . *

*Proof. *By Theorem 11, let be the unique fixed point of the mapping . Then, we have and so is a fixed point of . Therefore, by uniqueness of the fixed point it must be that .

#### Acknowledgments

The authors would like to thank the editor and reviewers for their helpful comments. The third author is supported by Università degli Studi di Palermo, Local University Project R.S. ex 60%.

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