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Journal of Function Spaces and Applications
Volume 2013 (2013), Article ID 143686, 9 pages
http://dx.doi.org/10.1155/2013/143686
Research Article

Suzuki-Type Fixed Point Results in Metric-Like Spaces

1Department of Mathematics, Babol Branch, Islamic Azad University, Babol, Iran
2Department of Mathematics, Qaemshahr Branch, Islamic Azad University, Qaemshahr, Iran
3Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia

Received 16 May 2013; Accepted 17 July 2013

Academic Editor: Pei De Liu

Copyright © 2013 Nabiollah Shobkolaei 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

We demonstrate a fundamental lemma for the convergence of sequences in metric-like spaces, and by using it we prove some Suzuki-type …fixed point results in the setup of metric-like spaces. As an immediate consequence of our results we obtain certain recent results in partial metric spaces as corollaries. Finally, three examples are presented to verify the effectiveness and applicability of our main results.

1. Introduction

There are a lot of generalizations of Banach fixed-point principle in the literature. So far several authors have studied the problem of existence and uniqueness of a fixed point for mappings satisfying different contractive conditions (e.g., [120]). In 2008, Suzuki introduced an interesting generalization of Banach fixed-point principle. This interesting fixed-point result is as follows.

Theorem 1 (see [19]). Let be a complete metric space, and let be a mapping on . Define a nonincreasing function from into by

Assume that there exists , such that for all , then there exists a unique fixed-point of . Moreover, for all .

Suzuki proved also the following version of Edelstein's fixed point theorem.

Theorem 2. Let be a compact metric space. Let be a self-map, satisfying for all , the condition
Then has a unique fixed point in .

This theorem was generalized in [3].

In addition to the above results, Kikkawa and Suzuki [8] provided a Kannan type version of the theorems mentioned before. In [14], Chatterjea type version is provided. Popescu [15] presented a Cirić type version. Recently, Kikkawa and Suzuki also provided multivalued versions which can be found in [9, 10].

Very recently Hussain et al. [4] have extended Suzuki's Theorems 1 and 2, as well as Popescu's results from [15] to the case of metric type spaces and cone metric type spaces (see also [57, 11]).

The aim of this paper is to generalize the above-mentioned results. Indeed we prove a fixed point theorem in the set up of metric-like spaces and derive certain new results as corollaries. Finally, three examples are presented to verify the effectiveness and applicability of our main results.

In the rest of this section, we recall some definitions and facts which will be used throughout the paper. First, we present some known definitions and propositions in partial metric and metric-like spaces.

A partial metric on a nonempty set is a mapping such that for all , if and only if ,, , .

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 () and () . But if , may not be . A basic example of a partial metric space is the pair , where for all .

Lemma 3 (see [17]). Let and be a metric space and partial metric space, respectively. Then(i)the function defined by is a partial metric;(ii)let be defined by ; then is a partial metric on , where is an arbitrary function;(iii)Let be defined by ; then is a partial metric on ;(iv)Let be defined by ; then is a partial metric on , where .

Other examples of the partial metric spaces which are interesting from a computational point of view may be found in [7, 11, 12, 18].

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

Let be a partial metric.

A sequence in a partial metric space converges to a point if and only if .

A sequence in a partial metric space is called a Cauchy sequence if there exists (and is finite) .

A partial metric space is said to be complete if every Cauchy sequence in converges, with respect to , to a point such that .

Suppose that is a sequence in partial metric space ; then we define .

The following example shows that every convergent sequence in a partial metric space may not be a Cauchy sequence. In particular, it shows that the limit is not unique.

Example 4 (see [17]). Let and . Let Then clearly it is a convergent sequence, and for every we have , hence . But does not exist; that is, it is not a Cauchy sequence.

Definition 5 (see [2]). A metric-like on a nonempty set is a mapping such that for all ,, , .

The pair is called a metric-like space. Then a metric-like on satisfies all of the conditions of a metric except that may be positive for . Each metric-like on generates a topology on whose base is the family of open -ball, , where for all and .

A sequence in a metric-like space converges to a point if and only if .

A sequence in a metric-like space is called a -Cauchy sequence if there exists (and is finite) .

A metric-like space is said to be complete if every -Cauchy sequence in converges, with respect to , to a point such that

Every partial metric space is a metric-like space. Below we give some examples of a metric-like space.

Example 6. Let ; then mapping defined by is a metric-like on .

Example 7. Let ; then mappings defined by are metric-like space on , where and .

2. Main Results

We start our work by proving the following crucial lemma.

Lemma 8. Let be a metric-like space, and suppose that is -convergent to . Then for every , one has In particular, if , then one has .

Proof. Using the triangle inequality in a metric-like space, it is easy to see that
Taking the upper limit as in the first inequality and the lower limit as in the second inequality, we obtain the desired result.

Theorem 9. Let be a complete metric-like space. Let be a self-map, and let be defined by
If there exists such that for each Then has a unique fixed point , and for each , the sequence converges to .

Proof. Putting in (10), hence from it follows for every . Let be arbitrary and form the sequence by and for . By (12), we have Also, by the condition of the definition of metric-like space, for all , we have
Hence, is a -Cauchy sequence.
Since is -complete, there exists such that That is, . We prove that . Putting in (12), we get that holds for each (where ). It follows by induction that
Let us prove now that holds for each . Since and by Lemma 8, it follows that there exists such that holds for every . Assumption (10) implies that for such , thus as , we get that We will prove that for each . For , this relation is obvious. Suppose that it holds for some . If , then and . If , then applying (18) and the induction hypothesis; we get that and (21) is proved by induction.
In order to prove that , we consider two possible cases.
Case I. (and hence ). We will prove first that for . For , it follows from (16). Suppose that (23) holds for some . Then which implies . Using (17) we obtain Assumption (10) and relation (21) imply that So relation (23) is proved by induction.
Now and (23) implies that for each . Hence, (18) imply that Hence , thus and; using Lemma 8 in (23), we have as which implies that , a contradiction.
Case II. (and so ). We will prove that there exists a subsequence of such that holds for each . From (12) we know that holds for each . Suppose that hold for some . Then which is impossible. Hence one of the following holds for each : or In particular, or In other words, there is a subsequence of such that (28) holds for each . But then assumption (10) implies that or Passing to the limit when , we get that , which is possible only if .
Thus, we have proved that is a fixed point of . The uniqueness of the fixed point follows easily from (10). Indeed, if and are two fixed points of such that , then from (18) we have which is a contradiction.

According to Theorem 9, we get the following result.

Corollary 10 (see [19]). Let be a complete metric space, and let be a mapping on . Define a nonincreasing function from into by
Assume that there exists , such that for all ; then there exists a unique fixed-point of . Moreover, for all .

Proof. Using a similar argument given in Theorem 9 for , the desired result is obtained.

Now, in order to support the useability of our results, let us introduce the following example.

Example 11. Let . Define by for all . Then is a complete metric-like space. Define a map by

for . Then for each , we have

On the other hand, we have Thus satisfies all the hypotheses of Theorem 9, and hence has a unique fixed point. Indeed, , , and is the unique fixed point of .

Theorem 12. Let be a complete metric-like space. Let , be two self-mappings. Suppose that there exists such that for every and that for every with that is not a common fixed point of and . Then there exists such that . Moreover, if , then .

Proof. Let be arbitrary, and define a sequence by Then if is odd, we have
If is even, then by (44), we have Thus for any positive integer , it must be the case that
Repeating (49), we obtain So, if , then
Thus .
That is, is a -Cauchy sequence in the metric-like space . Since is -complete, there exist such that
Assume that is not a common fixed point of and . Then by hypothesis which is a contradiction. Therefore, .
If for some , then which gives that .

Example 13. Let be a metric-like space where and . Define by , , and and , , and . Then for , we have It is easy to see that the above inequality is true for and for . Also, for every with y is not a common fixed point of and . This shows that all conditions of Theorem 12 are satisfied and 0 is a common fixed point for and .

Corollary 14. Let be a complete metric-like space, and let be a mapping. Suppose that there exists such that for every and that for every with . Then there exists such that . Moreover, if , then .

Proof. Taking in Theorem 12, the conclusion of the corollary follows.

Theorem 15. Let be a complete metric-like space. Let , be mappings from onto itself. Suppose that there exists such that for every and that for every with that is not a common fixed point of and . Then there exists such that . Moreover, if , then .

Proof. Let be arbitrary. Since is onto, there is an element satisfying . Since is also onto, there is an element satisfying . Proceeding in the same way, we can find that and for . Therefore, and for . If , then using (59) If , then using (59) Thus for any positive integer , it must be the case that which implies that Let ; then since .
Now, (64) becomes So, if , then
Thus . That is, is a -Cauchy sequence in the metric-like space . Since is -complete, there exists such that Assume that is not a common fixed point of and . Then by hypothesis which is a contradiction. Therefore, .
If for some , then which gives that .

Corollary 16. Let be a complete metric-like space, and let be an onto mapping. Suppose that there exists such that for every and that for every with . Then there exists such that . Moreover, if , then .

Proof. Taking in Theorem 15, we have the desired result.

Definition 17. Let and be metric-like spaces. Then is said to be a continuous mapping, if implies that .

Corollary 18. Let be a complete metric-like space, and let be a mapping of into itself. If there is a real number with satisfying for every and is onto and continuous, then has a fixed point.

Proof. Assume that there exists with and Then there exists a sequence such that So, we have and as . Since, , hence as . Now, Since is continuous, we have This is a contradiction. Hence if , then which is condition (71) of Corollary 16. By Corollary 16, there exists such that .

Now we give an example to support our result.

Example 19. Let and . Define by .
Obviously is onto and continuous. Also for each , we have where . Thus satisfies the conditions given in Corollary 18, and is the fixed point of .

Corollary 20. Let be a complete metric-like space, and be a mapping of into itself. If there is a real number with satisfying for every and is onto and continuous, then has a fixed point.

Proof. Replacing by in (79), we obtain for all .
Without loss of generality, we may assume that . Otherwise has a fixed point. Since , it follows from (80) that for every . By the argument similar to that used in Corollary 18, we can prove that if , then which is condition (71) of Corollary 16. So, Corollary 16 applies to obtain a fixed point of .

According to Theorem 12, we get the following result.

Corollary 21 (see [17, Theorem 1]). Let be a complete partial metric space. Let , be two self-mappings. Suppose that there exists such that for every and that for every with that is not a common fixed point of and . Then there exists such that . Moreover, if , then .

Proof. Using a similar argument given in the Theorem 12 for , the desired result is obtained, where is a partial metric on .

Also, according to Theorem 15, we get Theorem 2 from [17].

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