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Journal of Function Spaces
Volume 2016, Article ID 1907676, 6 pages
http://dx.doi.org/10.1155/2016/1907676
Research Article

Some Fixed Point Results for -Type Contractive Mappings

1School of Mathematics, Thapar University, Patiala, Punjab 147004, India
2Department of Mathematics and Computer Science, Cankaya University, Ankara, Turkey
3Department of Mathematics, Payame Noor University, P.O. Box 19395-3697, Tehran, Iran
4Department of Mathematics, Karaj Branch, Islamic Azad University, Karaj, Iran

Received 15 November 2015; Accepted 10 February 2016

Academic Editor: Krzysztof Ciepliński

Copyright © 2016 Sumit Chandok 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 prove some fixed point results for new type of contractive mappings using the notion of cyclic admissible mappings in the framework of metric spaces. Our results extend, generalize, and improve some well-known results from literature. Some examples are given to support our main results.

1. Introduction

In the fixed point theory, a well-known theorem of Banach [1] states that if is a self-mapping on a complete metric space and satisfies , for some and all , then has a unique fixed point. Thereafter, various researchers generalized this result for different type of nonlinear contractive mappings and prove some interesting fixed point results (see [227] and references cited therein).

Recently, Samet et al. [23] introduced the concept of --contractive type mappings and established various fixed point theorems for such mappings in complete metric spaces. Thereafter, a lot of researchers worked on it and generalized the results under certain contractive conditions (see [5, 9, 14, 18, 22] and references cited therein).

Using the concept of Samet et al. [23], we prove some fixed point results for a new type of contractive mappings. Our results extend, generalize, and improve some well-known results from literature. Some examples are given to support our main results.

2. Preliminaries

Let be a nonempty set and let be an arbitrary mapping. We say that is a fixed point for , if . We denote the set of all fixed points of .

Definition 1 (see [18]). Let be a mapping and let be two functions. One can say that is a cyclic ()-admissible mapping if(i) for some implies ,(ii) for some implies .

Example 2 (see [18]). Let be defined by . Suppose that are given by , for all , and , for all . Then, is a cyclic -admissible mapping.

Let denote the set of all monotone increasing continuous functions , with .

Let denote the set of all continuous functions , with .

Lemma 3 (see [19]). Suppose that is a metric space. Let be a sequence in such that as . If is not a Cauchy sequence, then there exist an and sequences of positive integers and with such that , , and(i),(ii),(iii).

Remark 4. In the same way as the proof of Lemma 3, we get .

3. Main Results

In 2014, the concept of -class functions (see Definition 5) was introduced by Ansari in [6] and is important; for example, see numbers (1), (2) from Example 6. Also, see [7, 8, 12, 13].

Definition 5 (see [6]). One can say that is called -class function if it is continuous and satisfies the following axioms:(1).(2) implies that either or .for all .

Note that .

One can denote -class functions as .

Example 6 (see [6]). The following functions are elements of :(1), .(2), , .(3), or .(4), , or .(5), , .(6), , .

Definition 7. Let be a metric space and let be two functions. One can say that is a -contractive mapping iffor , where , , and .

Now, we are ready to prove our first theorem.

Theorem 8. Let be a complete metric space and let be a cyclic -admissible mapping. Assume that is a -contractive mapping. Suppose that there exists such that and and either of the following conditions hold:(a) is continuous.(b)if is a sequence in such that and , for all , then .Then, has a fixed point.
Moreover, if and , for all , then has a unique fixed point.

Proof. Define a sequence by , for all . Since is a cyclic -admissible mapping and , then which implies . By continuing this process, we get and , for all . Again, since is a cyclic ()-admissible mapping and , by the similar method, we have and , for all . That is, and , for all . Equivalently, , for all . From (1), we haveUsing monotonicity of , we get for all . Hence, the sequence is a decreasing sequence. So for the nonnegative decreasing sequence , there exists some , such thatAssume that . On letting in (2), using the continuity of and and (4), we obtainand thus . Now, by using Definition 5, we get that either or ; in both cases, it follows that , which impliesNow, we shall prove that is a Cauchy sequence. If possible, let not be a Cauchy sequence. Then, by Lemma 3 and Remark 4, there exist a and two sequences of positive integers and with such thatNow, by setting and in (1), and using , we obtain On letting , using (7), we obtain, or ; that is, , which is a contradiction. This shows that is a Cauchy sequence. Since is a complete metric space, then there is such that as .
Now, first we suppose that is continuous. Hence, So is a fixed point of .
In the second part, we suppose that condition (b) holds; that is, . So, we haveBy taking the limit and using the properties of , we obtain . Hence, is a fixed point of .
To prove the uniqueness of fixed point, suppose that and are two fixed points of . Since , from (1), we haveHence, by using the properties of , we have .

Example 9. Let be endowed with the usual metric , for all , and let be defined by and let be given byAlso, define as , as , and as .
Now, first we prove that is a cyclic -admissible mapping.
If , then and . Therefore, . Similarly, if , then . Then, is a cyclic -admissible mapping.
Now, we check the hypotheses (b) of Theorem 8.
Let such that and . Therefore, . Hence, ,
Let . Then, and and so we have = . Hence, inequality (1) is satisfied. Therefore, by Theorem 8, has a fixed point.

Corollary 10. Let be a complete metric space and let be a cyclic -admissible mapping. Assume that is an -contractive mapping; that is, for all ,Suppose that there exists such that and and either of the following conditions hold:(a) is continuous.(b)if is a sequence in such that and , for all , then .Then, has a fixed point.
Moreover, if and , for all , then has a unique fixed point.

Proof. Let , for . Hence, by using (15), we have the fact that is a -contractive mapping. Therefore, by applying Theorm 8, we have the result.

Definition 11. Let be a metric space and let be two functions. A mapping is called a weak -rational contraction if , for some , implieswhere , , and

Theorem 12. Let be a complete metric space and let be a cyclic -admissible mapping. Suppose that is a weak -rational contraction. Assume that there exists such that and and one of the following assertions holds:(a) is continuous.(b)if is a sequence in such that and , for all , then .Then, has a fixed point.
Moreover, if and , for all , then has a unique fixed point.

Proof. Define a sequence by , for all . Since is a cyclic -admissible mapping and , then which implies . By continuing this process, we get and , for all . Again, since is a cyclic ()-admissible mapping and , by the similar method, we have and , for all . That is, and , for all . Equivalently, , for all . Therefore, by (16), we havewhere
Now, suppose that there exists such that . Therefore, and so, from (18), we getThis implies that , or ; that is, , which is a contradiction. Hence, , for all . Hence, the sequence is a decreasing sequence. So for the nonnegative decreasing sequence , there exists some , such thatAssume that . On letting in (19), using the continuity of and (21), we obtainwhich implies that either or ; that is, in both cases, it follows that , which impliesNow, we shall prove that is a Cauchy sequence. If possible, let not be a Cauchy sequence. Then, by Lemma 3 and Remark 4, there exist and two sequences of positive integers and with such thatNow, by setting and in (16), and using , we obtainwhereOn letting , using (24) and (25), we obtainSo, ; that is, , which is a contradiction. This shows that is a Cauchy sequence. Since is a complete metric space, then there exists such that as .
First, we consider that is continuous. Hence, Therefore, is a fixed point of .
In the second part, we suppose that condition (b) holds; that is, . So, we havewhereBy taking the limit and using the properties of , we obtain . Hence, is a fixed point of .
To prove the uniqueness of fixed point, suppose that and are two fixed points of . Since , from (16), we have where This implies that or and hence

Example 13. Let be endowed with the usual metric , for all and let be defined by and let be given byAlso, define as and as .
It is easy to verify that is a cyclic -admissible mapping.
Now, we check the hypotheses (b) of Theorem 12.
Let such that and . Therefore, . Hence, ,
Let . Then, and and so we have Hence, inequality (16) is satisfied. Therefore, by Theorem 12, has a fixed point; that is, is a fixed point of .

4. Some Cyclic Contraction via Cyclic -Admissible Mapping

In this section, in a natural way, we apply Theorem 8 for proving a fixed point theorem involving a cyclic mapping.

Theorem 14. Let and be two closed subsets of complete metric space such that and let be a mapping such that and . Assume thatfor all and , where , , and . Then, has a unique fixed point in .

Proof. Define byLet Then, and . Hence, by (36), we have for all .
Let for some ; then, . Hence, and so . Now, let for some , so . Hence, and then . Therefore, is a cyclic -admissible mapping. Since is nonempty, then there exists such that and
Now, let be a sequence in such that , for all and ; then, , for all . Therefore, . This implies that . So the condition (b) of Theorem 8 holds. Therefore, has a fixed point in , for example, . Since , then , and since , then . Therefore, . The uniqueness of the fixed point follows easily from (36).

Example 15. Let be endowed with the usual metric , for all , and let be defined by , where and . Also, define , by and Indeed, for all and all , we have Therefore, the conditions of Theorem 14 hold and has a unique fixed point; that is, is a fixed point of .

Corollary 16. Let and be two closed subsets of complete metric space such that , and let be a mapping such that and . Assume thatfor all and , where , and . Then, has a unique fixed point in .

Conflict of Interests

The authors declare that they have no competing interests.

Authors’ Contribution

All authors contributed equally and significantly to writing of this paper. All authors read and approved the final paper.

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