Journal of Mathematics
Volume 2020 (2020), Article ID 1632526, 8 pages
https://doi.org/10.1155/2020/1632526

Research Article

On -Hybrid Wardowski Contractions

Erdal Karapınar,1,2,3 Hassen Aydi,4,5 and Andreea Fulga⁶

¹ETSI Division of Applied Mathematics, Thu Dau Mot University, Thủ Dầu Một, Binh Duong Province, Vietnam
²Department of Medical Research, China Medical University, Taichung, Taiwan
³Department of Mathematics, Çankaya University, Ankara 06790, Turkey
⁴Nonlinear Analysis Research Group, Ton Duc Thang University, Ho Chi Minh City, Vietnam
⁵Faculty of Mathematics and Statistics, Ton Duc Thang University, Ho Chi Minh City, Vietnam
⁶Department of Mathematics and Computer Sciences, Transilvania University of Brasov, Brasov, Romania

Correspondence should be addressed to Hassen Aydi; hassen.aydi@tdtu.edu.vn

Received 11 May 2020; Accepted 25 July 2020; Published 24 August 2020

Academic Editor: Ljubisa Kocinac

Copyright © 2020 Erdal Karapınar 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.

The goal of this work is to introduce the concept of -hybrid Wardowski contractions. We also prove related fixed-point results. Moreover, some illustrated examples are given.

1. Introduction

Let represent the collection of functions so that

(i) is strictly increasing(ii) for each sequence in , iff (iii) there is so that

Definition 1 (see [1]). A mapping is called a Wardowski contraction if there exist and such that for all ,

Example 1 (see [1]). The functions defined by

(1)(2)(3)(4)

belong to .
Wardowski [1] introduced a new proper generalization of Banach contraction. For other related papers in the literature, see [2–10]. The main result of Wardowski is as follows.

Theorem 1 (see [1]). Let be a complete metric space, and let be an -contraction. Then, has a unique fixed point, say , in and for any point , the sequence converges to .

Theorem 2 (see [11]). Let be a complete metric space and be a given mapping such that

for all , where , , are nonnegative real numbers such that . Then, admits a unique fixed point in .

In the paper [12], the concept of interpolative Hardy–Rogers-type contractions was introduced.

Definition 2. (see [12]). On a metric space , a self-mapping is an interpolative Hardy–Rogers-type contraction if there exist and with , such that

for all , where .

Theorem 3 (see [12]). Let be a complete metric space and be an interpolative Hardy–Rogers-type contraction. Then, has a fixed point in .
The interpolation concept was used in other new papers related to fixed-point theory. For example, see [13–17]. In this paper, we consider new contractive type self-mappings, named as -hybrid Wardowski contractions. Our fixed-point results will be supported by concrete examples.

2. Main Results

Let be a metric space and be a self-mapping on this space. For and , such that , we define the following expression:

On the other hand, let represent the set of functions such that

(i) is strictly increasing(ii) there exists such that , for every

Definition 3. A mapping is called a -hybrid Wardowski contraction, if there is such that

In particular, if inequality (5) holds for , we say the mapping is a 0-hybrid Wardowski contraction.

Theorem 4. A -hybrid Wardowski contraction self-mapping on a complete metric space admits exactly one fixed point in .

Proof. Taking an arbitrary point , we consider the sequence defined by the relation , . According to this construction, it is easy to see that if there is so that , turns into a fixed point of . We shall presume that for all ,

On account of (4), for and , we have that

Denoting by , we have

and from (5), it follows that

which gives us

If , then the above inequality becomes

which is a contradiction. Consequently, and then there exists such that

Supposing that , we have and by , we obtain

which is a contradiction. Therefore,

In order to prove that is a Cauchy sequence in , we suppose that there exist and the sequences of positive integers, with such that

for any .
Thus, we have

When , using (14) and (15), it follows

By using the triangle inequality, we have

So,

Moreover, since

we have

So, the inequality

occurs for all , and using (5), there exists such that

where

Moreover, since the function is increasing, we have

And letting ,

That is a contradiction, so and then, . Consequently, the sequence is Cauchy and by completeness of , it converges to some point .
There exists a subsequence such that for all ; then,

On the contrary, if there is a natural number such that for all , applying (5), for and , we have

where

We suppose that . Inasmuch as

Letting in inequality (29), we find that

which contradicts . Therefore, .
We claim now that admits only one fixed point. If there exists another point , , such that , then and we have

which is a contradiction.

Example 2. Let be endowed with the standard metric . Let the mapping be defined by . Take , , , , , and . Then, we have the following:

 For ,

 For and ,

Thus, all assumptions of Theorem 4 hold, and has a unique fixed point. On the other hand, for and , we have

Thus, it is not a Wardowski contraction, since for every function and

Theorem 5. A 0-hybrid Wardowski contraction self-mapping on a complete metric space admits a fixed point in provided that for each sequence in , iff .

Proof. Following the same reasoning from the proof of the previous theorem, we can assume that for all ,

On account of (4), for and , we have that

Using the same notation, , and taking into account , by (5), we have

We can remark that the case , is not possible since the above inequality becomes

a contradiction. Therefore, for all , and then, there exists such that

We claim that . Indeed, if we suppose that , taking the limit as in (40), we have

which contradicts We conclude that

Let and now; we have

And taking into account (44),

Therefore, and since

we obtain that and so . Thus, is a Cauchy sequence on a complete metric space and there exists such that . Of course, it easy to see that, for and , we have

If we suppose that there is a subsequence such that , then we have

which means that is a fixed point of . Therefore, we can assume that for every , and by (5), we obtain

Letting and taking into account the previous considerations, we have and then . Consequently, is a fixed point of .

Example 3. Let be a set endowed with the metric (Table 1).
And the mapping is defined as .
First, we remark that Theorem 1 is not satisfied, since for and ,

Hence, for any and , we can write

Choosing , , , and , for and , we have

Table 1: Definition of metric d.

0
3
3
2

3
0
3
1

3
3
0
2

2
1
2
0

3. Consequences

  Considering in Theorem 5 and , we obtain Theorem 2.  Considering in Theorem 5 and , we obtain Theorem 3.  Considering in Theorem 4, , and , we obtain Theorem 3.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

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