International Journal of Mathematics and Mathematical Sciences

International Journal of Mathematics and Mathematical Sciences / 2014 / Article

Research Article | Open Access

Volume 2014 |Article ID 652925 | 9 pages | https://doi.org/10.1155/2014/652925

Set-Valued Hardy-Rogers Type Contraction in 0-Complete Partial Metric Spaces

Academic Editor: Hernando Quevedo
Received27 Jan 2014
Accepted19 Mar 2014
Published08 Apr 2014

Abstract

In this paper we introduce set-valued Hardy-Rogers type contraction in 0-complete partial metric spaces and prove the corresponding theorem of fixed point. Our results generalize, extend, and unify several known results, in particular the recent Nadler’s fixed point theorem in the context of complete partial metric spaces established by Aydi et al. (2012). As an application of our results, a homotopy theorem for such mappings is derived. Also, some examples are included which show that our generalization is proper.

1. Introduction and Preliminaries

The well-known Banach contraction mapping principle states that, if is a complete metric space and is a self-mapping such that for all , where , then has a fixed point in . Because of simplicity and several applications, Banach principle was generalized by several authors, in different directions.

For instance, Hardy and Rogers [1] used the contractive condition for all , , where are nonnegative constants such that , and proved fixed point result. Note that condition (2) generalizes the contractive conditions of Banach, Kannan, Reich, Chatterjea, and iri (see [2]).

It is well known that the theory of set-valued mappings has application in control theory, convex optimization, differential equations, and economics. Nadler Jr. [3] generalized the Banach contraction mapping principle to set-valued mappings and proved the following fixed point theorem.

Theorem 1. Let be a complete metric space and let be a mapping from into (here denotes the set of all nonempty closed bounded subset of ) such that, for all , where . Then has a fixed point.

In recent years, Matthews [4] introduced the notion of partial metric space as a part of the study of denotational semantics of dataflow networks, with the interesting property of “nonzero self-distance” in space. He showed that the Banach contraction mapping principle can be generalized to the partial metric context for applications in program verification. For a more detailed explanation we refer the reader to Bukatin et al. [5] where the motivation for introducing nonzero distance is explained, which is also leading to interesting research in foundations of topology. Later on, Romaguera [6] introduced the notions of 0-Cauchy sequence and 0-complete partial metric spaces and proved some characterizations of partial metric spaces in terms of completeness and 0-completeness.

Very recently, Aydi et al. [7] introduced the notion of partial Hausdorff metric and extended the Nadler's theorem in partial metric spaces. In this paper, we discuss some properties of partial metric spaces and extend the results of Kadelburg et al. [8], Altun et al. [9], and Aydi et al. [7] for set-valued mappings in 0-complete partial metric spaces. As an application, we prove a homotopy result which extends and generalizes the homotopy result of [7, 9] for set-valued mappings in 0-complete partial metric spaces.

Consistent with [4, 610], the following definitions and results will be needed in the sequel.

Definition 2 (see [4]). A partial metric on a nonempty set is a function ( stands for the set of nonnegative reals) 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 .

It is clear that, if , then from (P1) and (P2) . But, if , may not be . Also every metric space is a partial metric space, with zero self-distance.

Example 3. If is defined by , for all , then is a partial metric space.

Some more examples of partial metric spaces can be seen in [4, 7, 8].

Each partial metric on generates a topology on which has as a base the family of open -balls , where for all and .

Theorem 4 (see [4]). For each partial metric the pair where for all , , is a metric space.

Here is called induced metric space and is called induced metric.

Definition 5 (see [4, 9]). Let be a partial metric space. Then, (1)a sequence in converges to a point if and only if ;(2)a sequence in is called Cauchy sequence if there exists (and is finite) ;(3) is said to be complete if every Cauchy sequence in converges with respect to to a point such that ;(4)a sequence in is called -Cauchy sequence if . The space is said to be -complete if every -Cauchy sequence in converges with respect to to a point such that .

Lemma 6 (see [4, 6, 10]). Let be a partial metric space and be any sequence in . Then, (i) is a Cauchy sequence in if and only if it is a Cauchy sequence in ;(ii) is complete if and only if the metric space is complete. Furthermore, if and only if ;(iii)every -Cauchy sequence in is Cauchy in ;(iv)if is complete, then it is -complete.

The converse assertions of (iii) and (iv) do not hold. Indeed the partial metric space , where denotes the set of rational numbers and the partial metric is given by provides an easy example of a -complete partial metric space which is not complete. It is easy to see that every closed subset of a -complete partial metric space is -complete.

Consistent with [7], we recall the notions of closedness and boundedness in a partial metric space . Let be the family of all nonempty, closed, and bounded subsets of , induced by the partial metric . Note that closedness is taken from ( is the topology induced by ) and boundedness is given as follows: is a bounded subset in if there exist and such that, for all , we have , that is, .

Now, for , and , define

Lemma 7 (see [9]). Let be a partial metric space and be any nonempty set in . Then if and only if , where denotes the closure of with respect to the partial metric . Note that is closed in if and only if .

Proposition 8 (see [7]). Let be a partial metric space. For all , we have the following: (i);(ii);(iii) implies that ;(iv).

Let be a partial metric space. For , , define

Proposition 9 (see [7]). Let be partial metric spaces. For all , , , we have (h1); (h2); (h3). (h4).

In view of Proposition 9, we call the mapping a partial Hausdorff metric induced by .

The following Lemma is crucial for the proof of our main result.

Lemma 10. Let be a partial metric space, be arbitrary, and . For any , there exists such that

Proof. If , then, from (i) of Proposition 8, we have Since , for , we have Therefore the result holds true with . If , suppose there exists such that for all we have . Therefore or . Note that As is arbitrary, we get a contradiction and the result holds true.

Remark 11. Each partial metric on generates a topology on , but it is not necessarily , in the sense that a singleton set in partial metric spaces may not be closed. Consider the partial metric space as in Example 3. Let ; then the singleton set is not a closed set with respect to . Indeed, for any , Therefore, for any , we have and hence is not closed with respect to .

2. Main Results

Firstly, we consider some properties of partial metric spaces which will be useful in further discussion.

The following proposition shows that, for every partial metric space , there exists a partial metric space such that the topological space is a Hausdorff space.

Proposition 12. Let be any nonempty set and and be, respectively, a metric and a partial metric on . Define by for all . Then is a partial metric space such that the topological space is a Hausdorff space.

Proof. Let with . We shall show that and are contained into two disjoint open sets with respect to . Since , we have . Now, let ; then , ,  , and . If , let . Note that that is Using (12) we obtain that is a contradiction (as and ). Therefore, and hence is a Hausdorff space.

Corollary 13. Let be a partial metric space; then there exists a partial metric on such that the topological space is a Hausdorff space.

Proof. Consider the induced metric by and define for all , . By Proposition 12, is the required partial metric.

Corollary 14. Let be a partial metric space; then there exists a partial metric on such that the topological space is a space and therefore every singleton subset of is closed with respect to .

Now we state the fixed point theorem for set-valued Hardy-Rogers type mappings.

Theorem 15. Let be a 0-complete partial metric space and be a mapping such that for all , , where are nonnegative constants with . Then has a fixed point in .

Proof. Let be arbitrary. We construct a sequence in as follows: since , let . Also, since , by Lemma 10, there exists such that where is arbitrary. Similarly there exists such that Continuing this procedure we construct a sequence in such that and Now, if for any , then the proof is finished. Therefore, let for all . Now, as for all , by using (15) and (18), it follows that for all ; that is for all . Again using symmetry of and and similar process as above with interchanging the roles of and , we obtain for all . Thus, from (20) and (21), we deduce that that is Since and is arbitrary, choosing , we obtain From successive application of (24), it follows that Note that ; therefore and .
Consequently, we have Therefore for all , . Thus is a 0-Cauchy sequence in . Since is 0-complete, there exists such that We will show that ; that is is a fixed point of .
Since for all , we obtain that By using (15) we get that is By using (27), from the above inequality, it follows that . Therefore by Lemma 7, we have . Thus is a fixed point of .

Remark 16. In [7], authors used a Banach type contractive condition on . In the above theorem we use a more general Hardy-Rogers type contractive condition on . Also, the completeness of space is replaced by 0-completeness, which is more general than completeness. Our method for the proof of the main result is different from the method used in [7]. Also, if is replaced with a metric on , we obtain Hardy-Rogers theorem for usual metric spaces.

With suitable values of control constants , the following corollaries are obtained.

Corollary 17 (see [7], Banach type). Let be a 0-complete partial metric space and be a mapping such that for all , , where . Then has a fixed point in .

Corollary 18 (Kannan type). Let be a 0-complete partial metric space and be a mapping such that for all , , where . Then has a fixed point in .

Corollary 19 (Reich type). Let be a 0-complete partial metric space and be a mapping such that for all , , where , , are nonnegative constants such that . Then has a fixed point in .

Corollary 20 (Chatterjea type). Let be a 0-complete partial metric space and be a mapping such that for all , , where . Then has a fixed point in .

Corollary 21 (iri type). Let be a 0-complete partial metric space and be a mapping such that for all , , where , , , are nonnegative constants such that . Then has a fixed point in .

The following examples illustrate the cases when the results of Aydi et al. [7] and Nadler Jr. [3] are not applicable, while the new results are applicable.

Example 22. Let be endowed with the partial metric defined by Obviously, is not a metric on and is a 0-complete partial metric space. Note that where is a partial metric space and where is the usual metric on is a metric space; also for all . Therefore, in view of Corollary 14, is a space and hence all subsets of are closed (also bounded).
Define by We will show that is not a contraction in the sense of Aydi et al. [7] and Nadler Jr. [3]. Note that, for , we have and . Therefore, there is no real such that with . Thus, is not a contraction in the sense of Aydi et al. [7] and we cannot conclude the existence of fixed point of with the assumptions of [7].
Note that is not a set-valued contraction in usual metric space . Indeed, if represents the Hausdorff metric induced by , then for and we have and ; therefore there is no real such that . Therefore, the result of Nadler Jr. [3] is not applicable on .
Again, if represents the Hausdorff metric induced by , where (induced by ) is given by then for and we have and . Also in this case, there is no real such that . Therefore, again the result of Nadler Jr. is not applicable on .
Now by careful calculations, one can see that satisfies all the conditions of Corollary 21, with , . Thus, there must be at least one fixed point of ; here is a fixed point of .

Example 23. Let be endowed with the partial metric defined by Then is a partial metric space. Note that for all , , where is the usual metric on and , defined by , for all , , is a partial metric on . In view of Corollary 14, is a space. Also, is 0-complete but not complete, because the induced metric space , where for all , , is not complete. Define by Then, satisfies all the conditions of Corollary 17 with and has a fixed point, . Note that the result of Aydi et al. [7] is not applicable (since the partial metric space is not complete).

In Theorem 15, the fixed point of is a limit of sequence , and, due to convergence in 0-complete space, this fixed point has zero self-distance. The following remark is helpful in proving the next homotopy result; it shows that, if has a fixed point in , then its self-distance must be zero.

Remark 24. Let be a partial metric space and be a mapping satisfying (15). If for some , then for all , and hence .

Proof. Let , then, by Lemma 7, and = . Suppose ; then from (15) we have Since , the above inequality yields a contradiction and so the result holds true.

3. Application to Homotopy Results

In this section, as an application of our main result, we derive a homotopy result.

Let be a function such that one of the following conditions holds:for all , , , , , we have and if ;for all , and some , we have .

Theorem 25. Let be a 0-complete partial metric space, be a closed subset of , and be an open subset of with . Let , and be an operator such that the following conditions hold: (a) for each and each ;(b)there exist nonnegative constants such that and, for all , and each , we have (c)there exists such that (d)if , then .
If has a fixed point in for at least one , then has a fixed point in for all . Furthermore, for any fixed , the fixed point of is unique.

Proof. Define As has a fixed point in for at least one ; that is there exists such that for at least one , and (a) holds and therefore . We will show that is both open and closed in and so, by connectedness of ,.
(I) is closed. Let be a sequence in and as . We must show that . Since for all , there exists with for all . For , with , using (b), (c), and Remark 24, we obtain that is Therefore, using (48), (d), and (h3) of Proposition 9, we obtain that is Using the properties of and the fact that , we get easily that as . Since , from (50), we deduce that Thus is a 0-Cauchy sequence in ,   is 0-complete, and is closed; therefore there exists such that For any , we have and, hence, using the properties of , . Also Using the properties of , the above inequality implies . Now or equivalently Using the above inequality we obtain that is Since ,   as , and, using the properties of , it follows from the above inequality that . Therefore , and from (a) we deduce that . Thus and hence is closed in .
(II) is open. Let and with . Note that, for this , Remark 24 is applicable and so Since is open, there exists such that . Suppose , where . By the definition of we can choose such that for all .
Let ; then, for all = = (as ), we shall show that , and so . Let ; then we have Since , we obtain Again, since and are symmetric, applying a similar process as above, we get