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Journal of Operators
Volume 2013 (2013), Article ID 404573, 8 pages
http://dx.doi.org/10.1155/2013/404573
Research Article

A Fixed Point Theorem in Orbitally Complete Partially Ordered Metric Spaces

1Department of Mathematics, Andhra University, Visakhapatnam 530 003, India
2Department of Mathematics, Lendi Institute of Engineering and Technology, Vizianagaram 535 005, India
3Department of Mathematics, Jimma University, Jimma-378, Ethiopia

Received 14 June 2013; Revised 30 October 2013; Accepted 2 November 2013

Academic Editor: Ram U. Verma

Copyright © 2013 G. V. R. Babu 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

Let be a partially ordered set and be a mapping. We prove a fixed point theorem for the map satisfying a contractive condition in orbits, when is -orbitally complete. Our result extends and generalizes the results of Samet et al. (2013) to partially ordered sets. Also, we generalize the results of Ran and Reurings (2004).

1. Introduction

In 1922, Banach [1] established a very fundamental result in fixed point theory, namely, the Banach contraction principle. Large number of generalizations have been made on this principle (we refer to Berinde and Vetro [2], Chatterjea [3], Ćirić [4], and Kannan [5, 6]). The Banach contraction principle forces the map to be continuous. In 1968, Kannan [5] established a fixed point theorem for a map which need not be continuous. Later in 1971, Reich [7] proved a fixed point theorem through which the Banach contraction principle and Kannan's fixed point theorem are generalized.

Recently, Samet et al. [8] established new fixed point theorems in complete metric spaces that generalize the Banach contraction principle and Reich’s and Kannan’s fixed point theorems.

Theorem 1 (see [8]). Let be a complete metric space, a lower semicontinuous function, and a given mapping. Suppose that, for any , there exists such that implies for all . Then, has a unique fixed point . Moreover, one has .

Theorem 2 (see [8]). Let be a complete metric space, a lower semi-continuous function and a given mapping. Suppose that there exists a constant such that for all . Then, has a unique fixed point . Moreover, one has .

Theorem 3 (see [8]). Let be a complete metric space, a lower semi-continuous function and a given mapping. Suppose that, there exist with such that for all , Then, has a unique fixed point . Moreover, one has .

Establishing the existence of fixed points in partially ordered sets is of recent interest in fixed point theory. In 2004, Ran and Reurings [9] established the existence of fixed points for contraction maps in partially ordered sets.

Definition 4. Let be a partially ordered set. A map is said to be non-decreasing if, for any with , one has .

Theorem 5 (see [9]). Let be a partially ordered set and be a metric on such that is a complete metric space. Suppose that is a continuous map and there exists such that for each with . If there exists such that , then has a fixed point in .

Definition 6. Let be a nonempty set and . Let . The orbit of is defined by .

Definition 7. Let be a metric space and . is said to be T-orbitally complete if every Cauchy sequence in , , converges to a point in .

Definition 8. Let be a metric space and . is said to be orbitally continuous at if as whenever as .

If is continuous at , then is orbitally continuous at . But its converse need not be true.

For more literature on these concepts, we refer to Türkoğlu et al. [10] and for works on ordered orbitally complete metric spaces we refer to Ding et al. [11] and Nashine et al. [12].

Definition 9. Let be a nonempty set. Let be a sequence of self-maps. An element is said to be a common fixed point of if for each .

In this paper, we prove a fixed point theorem in partially ordered metric space , for a self-map , when is -orbitally complete. Our result extends and generalizes the results of Samet et al. [8] to partially ordered sets. Our result also generalizes the results of Ran and Reurings [9].

We use the following lemma in our main result.

Lemma 10 (see [13]). Suppose that is a metric space. Let be a sequence in such that as . If is not a Cauchy sequence, then there exist and sequences of the positive integers and with such that , and and .

2. Main Result

Throughout this paper, we denote that

Theorem 11. Let be a partially ordered set and a metric on . Suppose that is a non-decreasing map and such that . Suppose that there exist a lower semicontinuous function and such that the following condition holds.
“For each , there exists such that implies where for each with .”
Assume that is -orbitally complete. Then, the sequence defined by , is Cauchy in . Let . Suppose that either is orbitally continuous at orif is a nondecreasing sequence converging to , then , for all .Then, is a fixed point of and .

Proof. Let such that . Choose such that . Then, . Since is non-decreasing, we have . Now, we choose such that . Hence, .
Continuing the same procedure, we obtain a sequence such that If there exists an integer such that , then, is a fixed point of and hence we are through. Without loss of generality, we assume that for each .
We denote We show that for each . Suppose, if possible, there exists such that . That is, Then, from (6), there exists such that where Hence, from (11), we get that is, a contradiction since . Therefore, for each . That is, the sequence is a decreasing sequence of real numbers. Hence, there exists such that and for each .
We now show that . Suppose that . Since for all , from (6), there exists such that where Thus, from (15), we get Letting , we get a contradiction. Hence, .
Therefore Thus, we have and .
By the property of , we have
We now show that is a Cauchy sequence in . Suppose if possible that is not a Cauchy sequence in . Then, there exist and sequences of the positive integers and with such that We choose , the least positive integer satisfying (21). Then, we have with Then, by Lemma 10, we have Now, since , there exists a number such that Hence, Since is non-decreasing, we have Now, using (22), we get That is, for each .
Therefore, from (6), there exists such that where Letting , we get Now, letting in (28), we get a contradiction. Hence, is a Cauchy sequence in . Since is -orbitally complete, there exists such that .
Since is lower semi-continuous, we have Hence, .
Now, suppose that holds. That is, is orbitally continuous at . Then, . Hence,.
Now, suppose that holds. Then, we have for all .
If for infinitely many , then . But . Hence, . So, without loss of generality, we assume that for infinitely many . That is, there exists a subsequence of such that for each .
As the sequence is convergent, it is bounded. Hence, there exists such that for each .
Since is non-decreasing, we have Now, for each , we have Hence, from (6), there exists such that where and , since .
Letting in (35), we get Thus, . That is, .

3. Corollaries and Examples

The following are the possible corollaries of Theorem 11.

Corollary 12. Let be a partially ordered set. Suppose that is a non-decreasing selfmap on . Assume that there is a metric on such that is -orbitally complete. Assume that there exists such that . Suppose that there exist a lower semi-continuous function , , and such that where for each with .
Then, the sequence defined by , is Cauchy in . Let . Suppose that either or of Theorem 11 holds, then is a fixed point of and .

Corollary 13. Let be a partially ordered set. Suppose that is a non-decreasing selfmap on . Assume that there is a metric on such that is -orbitally complete metric space. Assume that there exists such that . Suppose that there exist a lower semi-continuous function , such that the following condition holds.
“For each , there exists such that implies for each with .”
Then, the sequence defined by , is Cauchy in . Let . Suppose that either or of Theorem 11 holds, then is a fixed point of and .

Proof. Since inequality (40) implies inequality (6); the conclusion of the corollary follows from Theorem 11.

Corollary 14. Let be a partially ordered set. Suppose that is a non-decreasing selfmap on . Assume that there is a metric on such that is -orbitally complete. Assume that there exists such that . Suppose that there exist a lower semi-continuous function , such that the following condition holds.
“For each , there exist with such that implies for each with .”
Then, the sequence defined by , is Cauchy in . Let . Suppose that either or of Theorem 11 holds, then is a fixed point of and .

Proof. Inequality (41) implies inequality (6). Hence, the conclusion follows from Theorem 11.

As an immediate consequence, we obtain the following corollary.

Corollary 15. Let be a partially ordered set. Suppose that is a non-decreasing selfmap on . Assume that there is a metric on such that is -orbitally complete. Assume that there exists such that . Suppose that there exist a lower semi-continuous function , , and with such that for each with .
Then, the sequence defined by , is Cauchy in . Let . Suppose that either or of Theorem 11 holds, then is a fixed point of and .

Corollary 16. Let be a partially ordered set. Suppose that is a non-decreasing selfmap on . Assume that there is a metric on such that is -orbitally complete. Assume that there exists such that . Suppose that there exist a lower semi-continuous function , such that the following condition holds.
“For each there exist , and with such that implies for each with .”
Then, the sequence defined by , is Cauchy in . Let . Suppose that either or of Theorem 11 holds, then is a fixed point of and .

Proof. The conclusion follows from Theorem 11, since inequality (43) implies inequality (6).

We now obtain the following corollary from Corollary 16.

Corollary 17. Let be a partially ordered set and a metric on . Suppose that is a non-decreasing selfmap on . Assume that there is a metric on such that is -orbitally complete. Assume that there exists such that . Suppose that there exist a lower semi-continuous function , and there exist , and with such that for each with .
Then, the sequence defined by , is Cauchy in . Let . Suppose that either or of Theorem 11 holds, then is a fixed point of and .

Remark 18. By choosing in Corollaries 13, 15, and 17, we obtain partially ordered versions of Theorems 1, 2, and 3, respectively.

Remark 19. By choosing and , , in Corollary 12, we obtain Theorem 5 as a corollary to Corollary 12, which in turn Theorem 5 follows as a corollary to Theorem 11.

Theorem 20. Let be a partially ordered set. Suppose that is a non-decreasing selfmap on . Assume that there is a metric on such that is -orbitally complete. Assume that there exists such that . Let be a sequence of maps and suppose that there exist a lower semi-continuous function , , and such that where for each with and .
If either or of Theorem 11 holds, then the family of maps has a common fixed point.

Proof. If , then has a fixed point by Corollary 12. Let that fixed point be (say). Then, . We show that for each . Let be fixed such that .
Now, using (45), we get where Then, from (47), we obtain a contradiction. Hence, for each .
Thus, the conclusion of the theorem follows.

The following is an example in support of Theorem 11.

Example 21. Let with the usual metric. We define a partial order on by We define , , and by Let ; then, and .
And so that .
We now verify inequality (6).
In this example, we choose for each , satisfying for all .
Let .
Case (i) (, , and ). Suppose that . That is, . That is, . Now,
Case (ii) (, , and ). Suppose that ; that is, . Now,
Case (iii) (). Suppose that ; that is, . Now, Hence the verification of inequality (6) is the same as in Case (ii).
Case (iv) (). In this case, so that inequality (6) holds trivially.

From all the above cases, it is clear that inequality (6) holds for all . Hence, all the hypotheses of Theorem 11 hold and 0 is the fixed point of in . Also, .

Now, for and , suppose that there exists such that . Then, we have so that inequality (1) fails to hold for any and . Hence Theorem 1 is not applicable.

Example 22. Let with the usual metric. We define a partial order on by
if and only if for each .”
We define , , and by Let . Then, so that . Consider We now verify inequality (6).
We choose for each satisfying for all .
Let .
Case (i) (). Suppose that ; that is, . Now,
Case (ii) (). Suppose that ; that is, . Now, so that
Case (iii) (, ). Suppose that ; that is, . Now,
Hence, (6) holds trivially.
Case (iv) (). Suppose that ; that is, . Now, Then,
Case (v). (let ). Suppose that ; that is, . Now, Hence, (6) holds trivially.

Hence, all the hypotheses of Theorem 11 hold and is the fixed point of in . Also, .

We observe that at and , Hence, inequality (2) does not hold for any , and . Therefore, Theorem 2 is not applicable.

Also, and . Hence, does not hold for any and . Thus, inequality (4) fails to hold, and hence Theorem 5 is not applicable.

Thus, by Remark 19, we conclude that Theorem 11 is a generalization of Theorem 5.

Acknowledgment

The authors thank the referee for his/her valuable suggestions, which improved the presentation of the paper.

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