#### 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 ). A mapping is called a Wardowski contraction if there exist and such that for all ,

Example 1 (see ). The functions defined by(1)(2)(3)(4)belong to .
Wardowski  introduced a new proper generalization of Banach contraction. For other related papers in the literature, see . The main result of Wardowski is as follows.

Theorem 1 (see ). 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 ). Let be a complete metric space and be a given mapping such thatfor all , where , , are nonnegative real numbers such that . Then, admits a unique fixed point in .

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

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

Theorem 3 (see ). 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 . 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 thatIn 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 thatDenoting by , we haveand from (5), it follows thatwhich gives usIf , then the above inequality becomeswhich is a contradiction. Consequently, and then there exists such thatSupposing that , we have and by , we obtainwhich 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 thatfor any .
Thus, we haveWhen , using (14) and (15), it followsBy using the triangle inequality, we haveSo,Moreover, sincewe haveSo, the inequalityoccurs for all , and using (5), there exists such thatwhereMoreover, since the function is increasing, we haveAnd 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 havewhereWe suppose that . Inasmuch asLetting in inequality (29), we find thatwhich contradicts . Therefore, .
We claim now that admits only one fixed point. If there exists another point , , such that , then and we havewhich 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 haveThus, 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 thatUsing the same notation, , and taking into account , by (5), we haveWe can remark that the case , is not possible since the above inequality becomesa contradiction. Therefore, for all , and then, there exists such thatWe claim that . Indeed, if we suppose that , taking the limit as in (40), we havewhich contradicts We conclude thatLet and now; we haveAnd taking into account (44),Therefore, and sincewe 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 haveIf we suppose that there is a subsequence such that , then we havewhich means that is a fixed point of . Therefore, we can assume that for every , and by (5), we obtainLetting 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 writeChoosing , , , and , for and , we have

#### 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.