Research Article | Open Access
A Fixed Point Result for Boyd-Wong Cyclic Contractions in Partial Metric Spaces
A fixed point theorem involving Boyd-Wong-type cyclic contractions in partial metric spaces is proved. We also provide examples to support the concepts and results presented herein.
1. Introduction and Preliminaries
Partial metric spaces were introduced by Matthews  to the study of denotational semantics of data networks. In particular, he proved a partial metric version of the Banach contraction principle . Subsequently, many fixed points results in partial metric spaces appeared (see, e.g., [1, 3–19] for more details).
Throughout this paper, the letters and will denote the sets of all real numbers and positive integers, respectively. We recall some basic definitions and fixed point results of partial metric spaces.
Definition 1.1. A partial metric on a nonempty set is a function such that for all (p1),(p2),(p3), (p4). A partial metric space is a pair such that is a nonempty set and is a partial metric on .
If is a partial metric on , then the function given by is a metric on .
Example 1.3 (see, e.g., ). Let , and define . Then, is a partial metric space.
Each partial metric on generates a topology on , which has as a base the family of open -balls , where for all and .
Definition 1.5. Let be a partial metric space and a sequence in . Then, (i) converges to a point if and only if ,(ii) is called a Cauchy sequence if exists and is finite.
Definition 1.6. A partial metric space is said to be complete if every Cauchy sequence in converges, with respect to , to a point , such that .
Lemma 1.7 (see, e.g., [3, 11, 12]). Let be a partial metric space. Then,(a) is a Cauchy sequence in if and only if it is a Cauchy sequence in the metric space , (b) is complete if and only if the metric space is complete. Furthermore, if and only if
Remark 1.9. If , may not be .
Let be the set of functions such that(i) is upper semicontinuous (i.e., for any sequence in such that as , we have ),(ii) for each .
Theorem 1.11. Let be a complete partial metric space, and let be a map such that for all where and Then, has a unique fixed point.
In 2003, Kirk et al.  introduced the following definition.
Definition 1.12 (see ). Let be a nonempty set, a positive integer, and a mapping. is said to be a cyclic representation of with respect to if (i) are nonempty closed sets, (ii).
Theorem 1.13. Let be a complete partial metric space. Let be a positive integer, nonempty closed subsets of , and . Let be a mapping such that(i) is a cyclic representation of with respect to , (ii)there exists such that is continuous and for each , satisfying for any , , , where and is defined by (1.5). Then, has a unique fixed point .
In the following example, , but it is not continuous.
Example 1.14. Define by for all and for , . Then, is upper semicontinuous on with for all . However, it is not continuous at for all .
Following Example 1.14, the main aim of this paper is to present the analog of Theorem 1.13 for a weaker hypothesis on , that is, with . Our proof is simpler than that in . Also, some examples are given.
2. Main Results
Our main result is the following.
Theorem 2.1. Let be a complete partial metric space. Let be a positive integer, nonempty closed subsets of , and . Let be a mapping such that(1) is a cyclic representation of with respect to ,(2)there exists such that for any , , , where and is defined by (1.5).Then, has a unique fixed point .
Proof. Let . Consider the Picard iteration given by for . If there exists such that , then and the existence of the fixed point is proved.
Assume that , for each . Having in mind that , so for each , there exists such that and . Then, by (2.1) where Therefore, If for some , we have , so by (2.2) which is a contradiction. It follows that Thus, from (2.2), we get that Hence, is a decreasing sequence of positive real numbers. Consequently, there exists such that . Assume that . Letting in the above inequality, we get using the upper semicontinuity of which is a contradiction, so that , that is, By (1.1), we have for all , and then from (2.9) Also, by , In the sequel, we will prove that is a Cauchy sequence in the partial metric space . By Lemma 1.7, it suffices to prove that is Cauchy sequence in the metric space . We argue by contradiction. Assume that is not a Cauchy sequence in . Then, there exists for which we can find subsequences and of with such that Further, corresponding to , we can choose in such a way that it is the smallest integer with and satisfying (2.12). Then, We use (2.13) and the triangular inequality Letting in (2.14) and using (2.10), we find On the other hand Letting in the two above inequalities and using (2.10) and (2.15), Similarly, we have Also, by (1.1), (2.11), and (2.15)–(2.18), we may find On the other hand, for all , there exists , , such that . Then, (for large enough, ) and lie in different adjacently labeled sets and for certain . Using the contractive condition (2.1), we get where As (2.19), using (2.9), we may get By (2.22) and (2.23), we get that Letting in (2.20), we get using (2.22), (2.24), and the upper semicontinuity of which is a contradiction.
This shows that is a Cauchy sequence in the complete subspace equipped with the metric . Thus, there exists . Notice that the sequence has an infinite number of terms in each , , so since is complete, from each , one can extract a subsequence of that converges to . Because , are closed in , it follows that Thus, .
For simplicity, set . Clearly, is also closed in , so it is a complete subspace of and then is a complete partial metric space. Consider the restriction of on , that is, . Then, satisfies the assumptions of Theorem 1.11, and thus has a unique fixed point in .
We give some examples illustrating our results.
Example 3.1. Let and . It is obvious that is a complete partial metric space.
Set , , and . Define by
Notice that and , and hence . Analogously, and , and hence .
Take Clearly, satisfies condition (2.1). Indeed, we have the following cases.
Case 1. and . Inequality (2.1) turns into which is necessarily true.
Case 3. and . Inequality (2.1) turns into which is true again by the fact that for all .
The rest of the assumptions of Theorem 2.1 are also satisfied. The function has as a unique fixed point.
However, since is not a continuous function, we could not apply Theorem 1.13.
Example 3.2. Let and for all . Then, is a complete partial metric space. Take . Define by . Consider given by Example 1.14.
For all , we have Then all the assumptions of Theorem 2.1 are satisfied. The function has as a unique fixed point.
Similarly, Theorem 1.13 is not applicable.
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Copyright © 2012 Hassen Aydi and Erdal Karapinar. 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.