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Zabihi, Farzaneh; Razani, Abdolrahman

doi:https://doi.org/10.1155/2014/929821

Journal of Applied Mathematics

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Research Article | Open Access

Volume 2014 | Article ID 929821 | https://doi.org/10.1155/2014/929821

Fixed Point Theorems for Hybrid Rational Geraghty Contractive Mappings in Ordered -Metric Spaces

Farzaneh Zabihi¹and Abdolrahman Razani²

Academic Editor: Vu Ngoc Phat

Received03 Sept 2013

Revised03 Mar 2014

Accepted17 Mar 2014

Published24 Apr 2014

Abstract

We introduce the new notion of a hybrid rational Geraghty contractive mapping and investigate the existence of fixed point and coincidence point for such mappings in ordered -metric spaces. We also provide an example to illustrate the results presented herein. Finally, we establish an existence theorem for a solution of an integral equation.

1. Introduction

Recently, many researchers have focused on different contractive conditions in complete metric spaces endowed with a partial order and obtained many fixed point results in such spaces. For more details on fixed point results, their applications, comparison of different contractive conditions, and related results in ordered metric spaces, we refer the reader to [1–4].

The concept of a -metric space was introduced by Czerwik in [5]. After that, several interesting results about the existence of fixed points for single-valued and multivalued operators in -metric spaces have been obtained (see, e.g., [6–11] and [12–15]).

Definition 1 (see [5]). Let be a (nonempty) set and let be a given real number. A function is a -metric if for all , the following conditions are satisfied: () iff ,(),().
The pair is called a -metric space.

A -metric is a metric if (and only if) . The following example shows that in general a -metric need not to be a metric.

Example 2 (see [16]). Let be a metric space, and , where is a real number. Then, is a -metric with .

However, is not necessarily a metric space. For example, if is the set of real numbers and is the usual metric, then is a -metric on with , but it is not a metric on .

Definition 3 (see [17]). Let be a -metric space. Then, a sequence in is called(a)-convergent if and only if there exists such that , as and in this case, we write .(b)-Cauchy if and only if , as .

Proposition 4 (see Remark 2.1 in [17]). In a -metric space , the following assertions hold. A -convergent sequence has a unique limit. Each -convergent sequence is -Cauchy. In general, a -metric is not continuous.

The -metric space is -complete if every -Cauchy sequence in is -convergent.

Note that a -metric might not be a continuous function. The following example (corrected from [18]) illustrates this fact.

Example 5. Let and let be defined by Then, considering all possible cases, it can be checked that for all , we have Thus, is a -metric space (with ). Let for each . Then that is, , but as .

Lemma 6 (see [16]). Let be a -metric space with , and suppose that and are -convergent to , respectively. Then, we have In particular, if , then we have . Moreover, for each , we have

Let denote the class of all real functions satisfying the condition

In order to generalize the Banach contraction principle, Geraghty in 1973 proved the following.

Theorem 7 (see [19]). Let be a complete metric space, and let be a self-map. Suppose that there exists such that holds for all . Then, f has a unique fixed point and for each Picard's sequence converges to .

In 2010, Amini-Harandi and Emami [20] characterized the result of Geraghty in the framework of a partially ordered complete metric space in the following way.

Theorem 8. Let be a complete partially ordered metric space. Let be an increasing self-map such that there exists with . Suppose that there exists such that holds for all with . Assume that either is continuous or is such that if an increasing sequence in converges to , then for all . Then, has a fixed point in . If, moreover, for each there exists comparable with , then the fixed point of is unique.

Cabellero et al. [21] discussed the existence of a best proximity point of Geraghty contraction. In [22], some fixed point theorems for mappings satisfying Geraghty-type contractive conditions are proved in various generalized metric spaces. As in [22], we will consider the class of functions , where if and has the property

Theorem 9 (see [22]). Let , and let be a complete metric type space. Suppose that a mapping satisfies the condition for all and some . Then f has a unique fixed point , and for each Picard's sequence converges to z in .

The aim of this paper is to present some fixed point and coincidence point theorems for hybrid rational Geraghty contractive mappings in partially ordered -metric spaces. In fact, our results extend Theorems 7, 8, and 9.

2. Main Results

Let denote the class of all functions satisfying the following condition:

Theorem 10. Let be a partially ordered set and suppose that there exists a -metric on such that is a complete -metric space (with parameter ). Let be an increasing mapping with respect to such that there exists an element with . Suppose that for all comparable elements , where , If is continuous, then has a fixed point. Moreover, the set of fixed points of is well ordered if and only if has one and only one fixed point.

Proof. Put . Since and is an increasing function, we obtain by induction that
Step 1. We will show that . Since for each , then by (12), we have because Therefore, is decreasing. Then there exists such that . We will prove that . Suppose on contrary that . Then, letting from (15), we have which implies that , a contradiction. Hence, . That is,
Step 2. Now, we prove that the sequence is a -Cauchy sequence. Suppose the contrary; that is, is not a -Cauchy sequence. Then, there exists for which we can find two subsequences and of such that is the smallest index for which This means that From (19) and using the triangular inequality, we get By taking the upper limit as , we get Using the triangular inequality, we have Taking the upper limit as in the above inequality and using (20), we get From the definition of and and the above limits, we have Now, from (12) and the above inequalities, we have which is a contradiction. So, we conclude that is a -Cauchy sequence. -Completeness of yields that -converges to a point .
Step 3. Now, we show that is a fixed point of . Since is continuous, we have Finally, suppose that the set of fixed points of is well ordered. Assume, on contrary, that , are two fixed points for such that . Then by (12), we have because
So, we get , a contradiction. Hence, , and has a unique fixed point. Conversely, if has a unique fixed point, then the set of fixed points of is a singleton and is well ordered.

Note that the continuity of in Theorem 10 is not necessary and can be dropped.

Theorem 11. Under the same hypotheses of Theorem 10, without the continuity assumption of , assume that whenever is a nondecreasing sequence in such that , one has for all . Then, has a unique fixed point.

Proof. Repeating the proof of Theorem 10, we construct an increasing sequence in such that . Using the assumption on , we have . Now, we show that . By Lemma 6, we have where Therefore, from the above relations, we deduce that , so, .

If in the above theorems we take , where , then we have the following corollary.

Corollary 12. Let be a partially ordered set and suppose that there exists a -metric on such that is a -complete -metric space, and let be an increasing mapping with respect to such that there exists an element with . Suppose that for all comparable elements , where , If is continuous or, for any nondecreasing sequence in such that , one has for all , then, has a fixed point.

Corollary 13. Let be a partially ordered set and suppose that there exists a -metric on such that is a -complete -metric space, and let be an increasing mapping with respect to such that there exists an element with . Suppose that for all comparable elements , where . If is continuous or, for any nondecreasing sequence in such that , one has for all , then, has a fixed point.

Corollary 14. Let be a partially ordered set and suppose that there exists a -metric on such that is a -complete -metric space, and let be an increasing mapping with respect to such that there exists an element with . Suppose that for all comparable elements , where and .
If is continuous or, for any nondecreasing sequence in such that , one has for all , then, has a fixed point.

Proof. Since then from (38) we have where . Hence, all the conditions of Corollary 13 hold and has a fixed point.

Let be the family of all nondecreasing functions such that for all .

Lemma 15 (Berinde [23], Rus [24]). If , then the following are satisfied: (a) for all ;(b).

Example 16. for all , where , and for all are in .

Theorem 17. Let be a partially ordered set and suppose that there exists a -metric on such that is a -complete -metric space, and let be an increasing mapping with respect to such that there exists an element with . Suppose that where for all comparable elements . If is continuous, then has a fixed point.

Proof. Since and is an increasing function, we obtain by induction that Putting , we have If there exists such that , then and we have nothing to prove. Hence, for all , we assume .
Step 1. We will prove that Using condition (41), we obtain Here, Hence, By induction, As , we conclude
Step 2. We prove is a -Cauchy sequence. Suppose the contrary; that is, is not a -Cauchy sequence. Then, there exists for which we can find two subsequences and of such that is the smallest index for which This means that From (51) and using the triangular inequality, we have By taking the upper limit as , we have Using the triangular inequality, we have Taking the upper limit as in the above inequality and using (52), we get From the definition of , If , by (50) and (52), then we have Now, from (41), we have Again, if , by (52) and (54), we obtain which is a contradiction. Thus, is a -Cauchy sequence. Completeness of yields that converges to a point ; that is, as .
Step . Now, we show that is a fixed point of . Since, is continuous, we have So, is a fixed point of .

Theorem 18. Under the same hypotheses of Theorem 17, without the continuity assumption of , assume that whenever is a nondecreasing sequence in such that , for all . Then, has a fixed point.

Proof. Repeating the proof of Theorem 17, we construct an increasing sequence in such that . Using the assumption on , we have . Now, we show that . By (41), we have where Letting in (63), we get Again, taking the upper limit as in (62) and using Lemma 6 and (64), we get So, we get . That is, .

3. Coincidence Point Results

Lemma 19 (see [25]). Let be a nonempty set and let be a function. Then, there exists a subset such that and is one-to-one.

Theorem 20. Let be a partially ordered set and suppose that there exists a -metric on such that is a -complete -metric space, and let be such that is an increasing mapping with respect to , and there exists an element with . Suppose that satisfies the following condition for all comparable elements , where , If is continuous, then has a coincidence point.

Proof. Let . From Lemma 19 there exists such that and is one-to-one. Since , we can define the mapping by , for all . Since is one-to-one, then is well defined. Now, since satisfies (66), we have for all , where This shows that satisfies the condition of Theorem 10. So, there exists fixed point of ; that is, . Since , then there exists such that That is, is a coincidence point of and .

Theorem 21. Under the same hypotheses of Theorem 20, without the continuity assumption of , assume that whenever is a nondecreasing sequence in such that , for all . Then, has a coincidence point.

Theorem 22. Let be a partially ordered set and suppose that there exists a -metric on such that is a -complete -metric space, and let be such that is an increasing mapping with respect to , and there exists an element with . Suppose that where for all comparable elements . If is continuous, then has a coincidence point.

Theorem 23. Under the same hypotheses of Theorem 22, without the continuity assumption of , assume that whenever is a nondecreasing sequence in such that , for all . Then has a fixed point.

Remark 24. Taking , our obtained results coincide with the results in usual metric case. Also, Theorems 10 and 11 are extensions of Theorem 3.8 of [22] and Theorem 8.

Example 25. Let and define the partial order on by Consider the function given as which is increasing with respect to . Let . Hence, , so . Define first the -metric on by , , , and . Then, is a -complete -metric space with and the function given by and and is arbitrary. Then, because Also, because Also, Hence, satisfies all the assumptions of Theorem 10 and thus it has a fixed point (which is ).

4. Existence Theorem for a Solution of an Integral Equation

Consider the integral equation where . The purpose of this section is to give an existence theorem for a solution of (81) that belongs to (the set of continuous real functions defined on , by using the obtained result in Theorem 17). Obviously, this space with the -metric given by for all , is a -complete -metric space with and .

We endow with the partial order given by for all . It was proved in [3] that is regular (the proof is valid also in the -metric case). We will consider (81) under the following assumptions: (i) are continuous,(ii) is continuous,(iii)there exist such that for all with (iv),(v)there exists continuous function such that

Theorem 26. Under assumptions (i)–(v), (81) has a solution in , where .

Proof. We define by The mapping is increasing since, for , and from , we have Now, we have where Let be the function appearing in assumption (v). Then, we get Thus, from Theorem 17, by , we deduce the existence of such that .

Conflict of Interests

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

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Copyright

Copyright © 2014 Farzaneh Zabihi and Abdolrahman Razani. 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.

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