Chinese Journal of Mathematics

Volume 2014, Article ID 379049, 9 pages

http://dx.doi.org/10.1155/2014/379049

## Common Fixed Point of (, )-Generalized Contractive Mappings in Partially Ordered Metric Spaces

^{1}Department of Mathematics and Applied Mathematics, University of Pretoria, Lynnwood Road, Pretoria 0002, South Africa^{2}Department of Mathematics, COMSATS Institute of Information Technology, Lahore, Pakistan^{3}Faculty of Mechanical Engineering, University of Belgrade, Kraljice Marije 16, 11120 Beograd, Serbia

Received 13 September 2013; Accepted 5 December 2013; Published 19 February 2014

Academic Editors: C. Bai, F. Bobillo, Y. Fu, J. Huang, and D.-B. Wang

Copyright © 2014 M. Abbas 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

Gordji et al. (2012) gave a generalization of Geraghty’s theorem. The aim of this paper is to study the necessary conditions for the existence of coincidence and common fixed point of four mappings satisfying (, )-generalized contractive condition in the setup of partial ordered metric spaces. Some examples are given to validate the definitions and results presented herein.

#### 1. Introduction and Preliminaries

Abbas et al. [1] proved that a weakly contractive mapping defined on a Hilbert space is a Picard operator. Rhoades [2] established the same result considering the domain of mapping a complete metric space instead of Hilbert space. The study of common fixed points of mappings satisfying certain contractive conditions can be employed to establish existence of solutions of many types of operator equations such as differential and integral equations. Beg and Abbas [3] obtained common fixed points extending a weak contractive condition to two maps. In 2009, Kadelburg et al. [4] proved common fixed point theorems for generalized -weakly contractive mappings. Doric [5] obtained a common fixed point theorem for four maps. For more work in this direction, we refer to [2, 3, 5–13] and references mentioned therein.

In complete metric spaces equipped with a partial ordering, was first investigated in 2004 by Ran and Reurings [14], and then by Nieto and lpez [15] who subsequently extended the result of Ran and Reurings [14] for nondecreasing mappings and applied it to obtain a unique solution for a first order ordinary differential equation with periodic boundary conditions (see also [16]).

Abbas et al. [8] initiated the study of common fixed points for four mappings satisfying generalized weak contractive condition in complete partially ordered metric space. Recently, Esmaily et al. [17] coincidence point result for four mappings in partially ordered metric space and employed their result to find the common solution of two integral equations.

The aim of this paper is to obtain coincidence and common fixed points for four mappings under generalized contractive condition in complete partially ordered metric space. Our results extend, unify, and generalize the comparable results in [8, 17–19]. For some details also see [20, 21].

In the sequel, , , and denote the set of real numbers, the set of nonnegative real numbers, and the set of positive integers, respectively. The usual order on (resp., on ) will be indistinctly denoted by or by .

The following definitions and results will be needed in the sequel.

Let be the class of all mappings satisfying the condition whenever . Note that as if we take , , then .

Define is continuous, nondecreasing and if and only if . If we define by , then it is easy to check that .

Let be self-mapping on a set . If , for some in , then is called a fixed point of . The set of all fixed points of is denoted by . If and, for each in a complete metric space , the sequence , converges to , then is called a Picard operator.

In the following [22], Geraghty obtained the generalization of Banach’s contraction principle.

Theorem 1. *Let be a complete metric space and a self-map on . If there exists such that for all , then is a Picard operator.**A nonempty set equipped with a partial order is called partially ordered metric space if there exists a metric on . We will denote it by .*

*Definition 2 (see [23]). *Let be an ordered metric space. We say that is regular if is a nondecreasing sequence in with respect to such that as , then for all .

Harandi and Emami [24] proved Theorem 1 in partially ordered metric spaces.

Theorem 3. *Let be a partially ordered set, and suppose that there exists a complete metric on . Let be an increasing mapping such that there exists an element with . If there exists such that for each with , then has a fixed point provided that either is continuous or is regular. Moreover, if for each there exists which is comparable to and , then has a unique fixed point.*

Recently, [18] Gordji et al. proved the following result.

Theorem 4. *Let be a partially ordered set, and suppose that there exists a complete metric on . Let be a nondecreasing mapping such that there exists with . Suppose that there exist and such that
**
for all with . Assume that either is continuous or is regular. Then has a fixed point.*

It is worth to noticing that Condition (b) of subadditivity in [18] for the function is superfluous. Namely, all results in ([18], Theorems 2.2, 2.3, 3.3) are true only if is a continuous nondecreasing function with if and only if . However, it is easy to see that with Assumption (b) for the function [18] is not a generalization of [24]. For details see [25].

Let and be two self-mappings on a nonempty set . If , for some in , then is called a common fixed point of and .

Let be given self-mappings on a metric space . The pair is said to be compatible if , whenever is a sequence in such that, , for some .

Let be an ordered metric space and . If there exist and such that, for every two comparable elements , we have where then is said to be -order contractive pair with respect to and .

*Example 5. *Let be a partially ordered set and a usual metric on . Define four self-maps , , , and on as follows:
Take and . Then is -order contractive pair with respect to and .

Let be a partially ordered set. Two mappings are said to be weakly increasing if and hold for all [26].

*Example 6. *Let be endowed with usual order and usual topology. Let be defined by
Then, the pair is weakly increasing where is a discontinuous mapping on .

Let be a partially ordered set. Two mappings are said to be partially weakly increasing if for all [8].

*Definition 7 (see [23]). *Let be a partially ordered set and given mappings such that and . We say that and are weakly increasing with respect to if and only if, for all , we have: , for all , and , for all , where , for in .

*Definition 8 (see [17]). *Let be a partially ordered set and given mappings such that . We say that and are partially weakly increasing with respect to if and only if, for all , we have , for all .

If , we say that is weakly increasing with respect to .

Note that if (identity map on , then pair is weakly increasing.

Let be a partially ordered set. A self-mapping on is called (a) dominating if for each in [8] and (b) dominated if for each in .

*Example 9. *Let be endowed with usual ordering and let be defined by . Since for all , Therefore is a dominating map.

*Example 10. *Let be endowed with usual ordering and let be defined by . Since for all , therefore is a dominated map.

Assertion similar to the following lemma was used (and proved) in the course of proofs of several fixed point results in various papers [18, 27].

Lemma 11. *Let be a metric space and let be a sequence in such that is nonincreasing and that
**
If is not a Cauchy sequence, then there exist an and two sequences and of positive integers such that the following four sequences tend to when :
*

*2. Common Fixed Point Result*

*Now we start with the following result.*

*Theorem 12. Let be a partially ordered set such that there exists a complete metric on . Suppose that , , , and are self-mappings on with and such that a pair of dominating maps is -order contractive with respect to dominated maps and . If for a nondecreasing sequence with for all and implies that and either(a) are compatible, or is continuous, and are weakly compatible or(b) are compatible, or is continuous, and are weakly compatible,*

then , , , and have a common fixed point. Moreover, the set of common fixed points of , , , and is well ordered if and only if , , , and have one and only one common fixed point.

*Proof. *Let be an arbitrary point in . Construct sequences and in such that and . This can be done as and . By given assumptions, and . Thus, for all , we have . We claim that , for every . If not, then , for some . From inequality (2), we obtain
a contradiction. Hence, . Following the similar arguments, we obtain and so on. Thus becomes a constant sequence, and is the common fixed point of , , , and . Take for each . As and are comparable, so by inequality (2) we have
Now, if , then (9) gives a contradiction. Hence,
Similarly, we obtain
Thus the sequence is a nonincreasing sequence and bounded below. So exists. We claim that . If not, assume that ; then we have
and so
which, on taking limit as , implies that . By the property of , we have and so , a contradiction. Therefore . Now, we show that is a Cauchy sequence. It is sufficient to show that is a Cauchy sequence in . Suppose that this is not the case. Applying Lemma 11 to the sequence , we obtain that there exist and two sequences of positive integers and such that the sequences
all tend to when . Putting in (2) we have
where
Since
we obtain that . By the property of , we have which is a contradiction with . Therefore is a Cauchy sequence in , and hence is a Cauchy sequence. Since is complete, there exists a point in , such that converges to . Therefore,
Assume that is continuous. Since are compatible, we have . Now we show that . If not, that is . As , so from inequality (2), we have
where
On taking limit as , we obtain
a contradiction. Hence, . Now, since and as , so we have . We are to show that . If not, then from inequality (2), we obtain
which on taking limit as , implies that . Thus and so , a contradiction. Thus .

Since , there exists a point such that . Now we show that . If not, that is . Since implies , therefore from inequality (2), we obtain
a contradiction. Hence . Since and are weakly compatible, . Thus is a coincidence point of and .

Next we show that . If not, then . As and as implies that , so from (2), we have
which, on taking limit as , gives , a contradiction, thus . Therefore . The proof is similar when is continuous. Similarly, the result follows when (b) holds.

*Now suppose that the set of common fixed points of , , , and is well ordered. We claim that common fixed point of , , , and is unique. Assume on contrary that and but . By given assumption, we can replace by and by in (2) to obtain
a contradiction. Hence . Conversely, if , , , and have only one common fixed point then the set of common fixed points of , , , and being singleton is well ordered.*

*Example 13. *Let be a partially ordered set defined as if and only if and a usual metric on . Define self-maps , , , and as
It is easy to verify that mappings and are dominated and and are dominating. Take , , and (see Table 1).

*The mappings , , , and satisfy all the conditions given in Theorem 12. Moreover, is a unique common fixed point of , , , and .*

*Remark 14. *If in above example, we take , then
for any value of . Also,
does not hold for any altering distance functions . So Theorem 2.1 in [8] is not applicable.

Now we prove existence of coincidence point of two pairs of compatible mappings on partially ordered metric spaces.

*Theorem 15. Let be a partially ordered set such that there exists a complete metric on and given mappings. Suppose that pairs and are compatible, , and are continuous, and and are partially weakly increasing with respect to and , respectively. If there exist and such that
holds for every for which and are comparable, where
then the pairs and have a coincidence point . Moreover, if and are comparable, then is a coincidence point of , , , and .*

*Proof. *Suppose be an arbitrary point in . Following similar arguments to those given in Theorem 12, sequences and in are given by
By construction, we have . Using the fact that is partially weakly increasing with respect to , we obtain
Also, . As is partially weakly increasing with respect to , so
Hence
That is,
We will prove the result in four steps.*Step **1*. First we show that
*First Case.* There exists an such that . If there exists an such that , then by (35) we have . Now we claim that . If not, then . As and are comparable so from inequality (29), we have
a contradiction. Thus for every , we have . This implies that . Similarly, (36) remains valid if .*Second Case.* For every , . Since and are comparable, so by (29) we obtain
which further implies that
Similarly, as and are comparable, so by inequality (29), we have
which implies that
Combining (39) and (41), we have for any . Hence is a nonincreasing sequence and bounded below. So exists. We claim that . If not, assume that . Note that
gives
Hence,
Which, on taking limit as , implies that , so , which further implies that , a contradiction. So . Thus .*Step* *2*. Now, we show that is a Cauchy sequence. It is sufficient to show that is a Cauchy sequence in . Suppose that this is not the case. Applying Lemma 11 to the sequence we obtain that there exist and two sequences of positive integers and such that the sequences
all tend to when . Putting in (29), we obtain
where
Since
we now have that . By the property of the function follows that a contradiction with . Hence, is a Cauchy sequence in , and hence is a Cauchy sequence.*Step* *3*. Now we show the existence of a coincidence point for and . From the completeness of , there is a such that . From (31), we obtain that . Since the pairs and are compatible, so . Now using the continuity of , , , , we have . The triangular inequality and (31) yield
On taking limit as , we obtain , and . Hence and .*Step* *4*. Existence of a coincidence point for , , , and .

Since and are comparable, now are to show that . If not, that is . So from inequality (29), we have
a contradiction. Hence , which means that is a coincidence point of , , , and .

*Conflict of Interests*

*Conflict of Interests*

*The authors declare that there is no conflict of interests regarding the publication of this paper.*

*Acknowledgment*

*Acknowledgment*

*Stojan Radenovi is thankful to the Ministry of Education, Science and Technological Development of Serbia.*

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