Discrete Dynamics in Nature and Society
Volume 2015 (2015), Article ID 969726, 8 pages
http://dx.doi.org/10.1155/2015/969726

Research Article

Fixed Points of Generalized -Suzuki Type Contraction in Complete -Metric Spaces

Hamed Hamdan Alsulami,¹ Erdal Karapınar,1,2 and Hossein Piri³

¹Nonlinear Analysis and Applied Mathematics Research Group (NAAM), King Abdulaziz University, Jeddah, Saudi Arabia
²Department of Mathematics, Atılım University, Incek, 06836 Ankara, Turkey
³Department of Mathematics, Faculty of Basic Science, University of Bonab, P.O. Box 5551-761167, Bonab, Iran

Received 29 September 2014; Accepted 25 November 2014

Academic Editor: Binggen Zhang

Copyright © 2015 Hamed Hamdan Alsulami 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 introduce the notion of generalized -Suzuki type contraction in -metric spaces and investigate the existence of fixed points of such mappings. The presented results generalize and improve several results of the topics in the literature.

1. Introduction and Preliminaries

The concept of -metric was introduced by Czerwik [1] as a generalization of metric (see also Bakhtin [2, 3]) to extend celebrated Banach contraction mapping principle. Following this initial paper of Czerwik [1], a number of researchers in nonlinear analysis investigate the topology of the paper and proved several fixed point theorems in the context of complete -metric spaces (see, e.g., [4–8] and the related references therein).

Definition 1 (see [1]). Let be a nonempty set and let be a given real number. A mapping is said to be a -metric if for all the following conditions are satisfied:

if and only if ;;.

In this case, the pair is called a -metric space (with constant ).

Remark 2. It is clear that the notion of -metric is a real generalization of usual metric since a -metric space is a metric space when . For more details and examples on -metric, see, for example, [1, 3, 9, 10].

Example 3. Let and let a mapping be defined as follows:

Then, is a -metric space with coefficient . But it is not a metric space since the triangle inequality is not satisfied. Indeed,

Definition 4 (see [11]). 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 .

Remark 5 (see [11]). Notice that in a -metric space the following assertions hold:

(A)a convergent sequence has a unique limit;(B)each convergent sequence is Cauchy;(C)in general, a -metric is not continuous;(D)in general, a -metric does not induce a topology on .

Definition 6 (see [11]). The -metric space is complete if every Cauchy sequence in converges in .

Definition 7. Let and be -metric spaces; a mapping is called

(A)continuous at a point , if for every sequence in such that , then ;(B)continuous on , if it is continuous at each point .

2. Main Result

In this section we state and prove our main results. Throughout the paper, we assume that -metric is continuous.

Inspired by the notion of -contraction, defined by Wardowski [12], we introduce the notion of generalized -Suzuki type contraction as follows. , , denote the set of natural numbers, set of real numbers, and the set of non-negative real numbers.

Definition 8. Let be a -metric space with constant . A mapping is called a generalized -Suzuki type contraction if there exists such that for all with

where and are real numbers with and is a mapping satisfying the following conditions:

is strictly increasing; that is, for all such that ,  ;for every sequence of positive real numbers if and only if .

The following is the main result of this paper.

Theorem 9. Let be a complete -metric space with constant . If is a generalized -Suzuki type contraction, then has a fixed point .

Proof. Fix . We establish an iterative sequence in the following way:

Throughout the proof, we assume that

Indeed, if there exists such that , then the proof is completed trivially.
Due to assumption (5), we have

Thus, by hypotheses of theorem, we have

which is equivalent to

Since , inequality (8) turns into

From , we conclude that

Therefore, is a decreasing sequence of real numbers which is bounded from below. Therefore, converges and

We will show that . Suppose, on the contrary, that . In other words, for every there exists , such that

From , we find that

On the other hand, we have

due to (6). Since is generalized -Suzuki type contraction, we derive

which yields that

Taking the fact that into account, we find

Analogously, again by (6), we have . Owing to the fact that is generalized -Suzuki type contraction, we conclude that

It implies that

since . Furthermore, by combining (17) and (20), we get

Iteratively, we obtain that

By letting , we find that

Consequently, from (), we derive that . Thus, there exists such that

and from (4) we get

This is a contradiction with the definition of . Hence, we have

In what follows, we will prove that

Suppose, on the contrary, that there exist and sequences and of natural numbers such that

From the triangle inequality, we have

Owing to (25), there exists such that

Taking (29) into account, (28) yields that

So from , we obtain

On the other hand, we can easily get that

from (27) and (29). Since is generalized -Suzuki type contraction, for all we find that

Taking (31) into account, (33) turns into

Regarding (25) and , we obtain that

From , we get that

This is a contradiction with relations in (27). Hence, ; that is, is a Cauchy sequence in . On account of the completeness of , there exists such that

We claim that, for every ,

We will prove the claim above by the method of reductio ad absurdum. Suppose, on the contrary, that there exists such that

From (10) and , we have

It follows from (39) and (40) that

This is a contradiction. Hence, (38) holds. Since is generalized -Suzuki type contraction, (38) yields that, for every , either

or

holds. On account of , the limits in (25) and (37) imply that

Thus, letting in (42), we conclude that

Again by using , we observe that

By regarding the triangle inequality with (4), we derive that

By letting in the inequality above together with the limits in (37) and (46), we conclude that . Thus, is a fixed point of ; that is, .
Let us analyze the second case (42). Regarding (4), we have

As it was discussed above, from (25), (37), and , we conclude that

From equivalently, we get

Again by the triangle inequality together with (4), we find that

By letting in the inequality above together with the limits in (37) and (50), we obtain . Thus, is a fixed point of and that completes the proof.

Definition 10. Let be a -metric space with constant . A mapping is said to be a -Suzuki type contraction if there exists such that for all with

where is a mapping satisfying the following conditions.

() is strictly increasing; that is, for all such that ,  .()For every sequence of positive real numbers if and only if .

Theorem 11. Let be complete -metric space with constant and let be a -Suzuki type contraction mapping. Then, has a fixed point.

Proof. By taking and in Theorem 9, the proof is complete.

Corollary 12. Let be a self-mapping on a complete metric space . Let and be real numbers such that . Assume that there exists such that, for all with ,

where is satisfied in conditions and . Then, has a fixed point ; that is, .

Proof. Since any metric space is a -metric space with constant , so from Theorem 9 the proof is complete.

Corollary 13. Let be a self-mapping on a complete metric space . Assume that there exists such that, for all with ,

where is satisfied in conditions and . Then, has a fixed point ; that is, .

Proof. Since any metric space is a -metric space with constant , so by taking and in Theorem 9 the proof is complete.

Theorem 14. Let be complete -metric space with constant and let be a continuous self-mapping on . Let and be real numbers such that . Assume that there exists such that for all

where is satisfied in conditions and . Then, has a fixed point ; that is, .

Proof. Choose . Set

If there exists such that , the proof is complete. So we assume that

So from the assumption of theorem, we have

and hence

Since , we get

So from , we conclude that

Therefore, is a decreasing sequence of real numbers which is bounded from below. Therefore, converges and

We will show that . Arguing by contradiction, we assume . For every , there exists , such that

Hence, from , we get

On the other hand (57), we have

So from assumption of theorem, we obtain

and hence

Since , we get

Also from (57), we have , and thus, by assumption of theorem, we have

and therefore,

Since , we get

Now by using (64) and continuing similar method as used in (68) and (71), we obtain

This implies that

So from (), we have , so that there exists such that

and so from (56) we get

This is a contradiction with definition of . So, and from (62) we have

Now, we claim that

Arguing by contradiction, we assume that there exist and sequences and of natural numbers such that

So

From (76), there exists such that

It follows from (79) and (80) that

So, from , we obtain

On the other hand from (78), we have

Hence, from (82) and assumption of theorem, we have

Using (79) and , we obtain

Thus, from this and , we get

This is contradiction with relation (78). Hence, . By completeness of , there exists such that

Since is continuous, we get

Since , therefore and hence .

Theorem 15. Let be complete -metric space with constant and let be a continuous self-mapping on . Assume that there exists such that for all

where is satisfied in conditions and . Then, has a fixed point ; that is, .

Proof. By taking and in Theorem 14, the proof is complete.

Theorem 16. Let be complete metric space and let be a continuous self-mapping on . Let and be real numbers such that . Assume that there exists such that for all

where is satisfied in conditions and . Then, has a fixed point ; that is, .

Proof. It is sufficient to take in Theorem 14.

Theorem 17. Let be complete metric space and let be a continuous self-mapping on . Assume that there exists such that for all

where is satisfied in conditions and . Then, has a unique fixed point ; that is, .

Proof. Since every metric space is a -metric space with constant , from Theorem 15   has a fixed point . Indeed, if there is another fixed point of , such that , therefore . Since , from assumption of theorem we obtain

This is contradiction. So has a unique fixed point.

Remark 18. Theorem 17 gives all consequence of Theorem 2.1 of [12] without assumption used by [12]. Notice also that the results in [12] can be also concluded from the main theorem in [13] that is also a proper extension of the results in [12] in different aspect, more precisely, in complete metric-like spaces.

Conflict of Interests

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

Acknowledgment

This project was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, under Grant no. 5513035HiCi. The authors, therefore, acknowledge the technical and financial support of KAU.

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