New Results in Controlled Metric Type Spaces

Abstract

In this paper, a new class of functions denoted by is introduced which we use to prove new interesting fixed point results in controlled metric type spaces. Also, we present examples to illustrate our work.

1. Introduction

Fixed point theory is one of the most interesting topics that was introduced by Banach [1] in 1922. Since that time and for the last decades, this area of research has become the inspiration for many researchers in the field of nonlinear analysis and its applications. We refer the reader to check these extensions of the Banach theorem in different metric spaces; for example, in [2], the authors proved the existence of a fixed point for a self-mapping on metric spaces that satisfies a more general contraction.

In 1989, Bakhtin [3] introduced an extension of metric spaces, called metric spaces, where many interesting fixed point results for some contractive mappings in metric spaces were studied. Also, in 1993, Czerwik [4] extended the results of b-metric spaces. In 2018, Shatanawi et al. [5] introduced the -contraction on the extended metric spaces. Recently, many research studies were conducted on b-metric space under different contraction conditions. After that, many authors used -contraction mapping on different metric spaces (see [6, 7]). In 2017, Kamran et al. [8] presented a very interesting generalization of the metric spaces, called extended b-metric spaces. An extension of the extended b-metric spaces, called controlled metric type space, was introduced by Mlaiki et al. [9].

In this paper, we generalize the results of Mehmet [10] and Mukheimer [11] by introducing the -contractive mapping on controlled metric type spaces.

2. Preliminaries

The concept of extended -metric spaces was initiated by Kamran et al. [8] in 2017, and their work generalized many results in the literature (see, for example, [12–16]).

Definition 1. (see [8]). Let be an nonempty set and define the mappings and , such that ,(1).(2).(3).Then, we say that the pair is an extended metric space.
A general extension of the metric spaces was considered by Mlaiki et al. in [9] where the concepts of controlled metric type spaces is defined as follows.

Definition 2. (see [9]). On a nonempty set , define the mappings and , such that , the following conditions hold:(d1) if and only if .(d2).(d3).Then, the pair is called a controlled metric type space.
To illustrate the above definition, we present the following examples.

Example 1. (see [9]). Choose . Take such thatConsider asIt is clear that the conditions and are satisfied. Now, we investigate condition .

Case 1. If or , is satisfied.

Case 2. If and , holds when . Now, we may assume that . Then, we have . It is clear that holds in all following possible subcases:(1) are even and .(2) and are odd.(3) are odd and .(4) are even and .(5) are even.(6) are even and .(7) are odd and .(8) are odd.Thus, is a controlled metric type.
Moreover, for , we haveTherefore, is not an extended -metric.

Example 2. (see [9]). Take . Consider the function given asDefine a symmetric function such thatOne can easily verify that is a controlled metric type.
Since is not an extended -metric.
The concepts of Cauchy and convergent sequences in controlled metric type spaces are defined as follows.

Definition 3. Let be a controlled metric type space and be a sequence in .(1)The sequence converges to some , if , such that . We write .(2)We say that is Cauchy, if , such that .(3)If every Cauchy sequence is convergent, then the space is called complete.

Definition 4. (see [9]). Let be a controlled metric type space. Let and .(i)The open ball is defined as(ii)A self-mapping on is said to be continuous at , if , such that .

Remark 1. If for all in , , then is a metric space. Therefore, we conclude that every metric space is a controlled metric type space. However, the converse is not always true.
Clearly, if a mapping is continuous at in the controlled metric type space , then implies that as .
Let denote to the set of all functions such that(1) is nondecreasing.(2) for all , where is the -th iterate of .Now, we recall the following lemma.

Lemma 1. (see [5]). If , then , for all .
Next, we introduce the following class of functions.

Definition 5. (see [5]). Let be a nonempty set and be a mapping. A function is said to be controlled comparison function if satisfies the following conditions:(1) is nondecreasing.(2), and for any sequence in , for all and non-negative integer where is the -th iterate of .The set of all controlled comparison functions is denoted by which is an extension of comparison functions of Berinde.
Note that if , then we have , since , for all . Hence, by Lemma 1, we have .
To show that the family is a nonempty set, we present the following examples.

Example 3. Consider the controlled b-metric space which was defined in Example 2. Define the mapping , where . Note that . Then, we have . Therefore, . Similarly, it is not difficult to see that .

3. Main Result

First, we define the --contractive self-mapping in controlled metric type spaces.

Definition 6. Let be a self-mapping on a complete controlled metric type space . We say that is --contractive mapping if there exists a function and such that for all , we have

Definition 7. (see [17]). Let be a controlled metric type space. A mapping is said to be an admissible if the following condition holds: if , with , then .
Now, we prove our first result.

Theorem 1. Let be a complete controlled metric type space and be an contractive mapping for some . Assume that(A)isadmissible.(B)There exists such that .(C) is continuous.Then, has a fixed point. Moreover, if for any two fixed points of in say we have , then has a unique fixed point in .

Proof. Take to be the point in Condition (2) in our theorem. Define the sequence by .
First of all, note if there exists such that , then we are done and is the fixed point of .
So, we may assume that for all . Also, we know from the hypothesis of our theorem that , , and using the fact that is admissible, we can easily deduce that for all , .
Now, using the fact that is an contractive mapping, we deduce thatHence, with , we havewhereand since , we deduce that andThus, the sequence is a Cauchy sequence. The completeness of the controlled metric type space implies that converges to some .
Also, note thatTaking the limit in above inequality and since and using the fact that is continuous, we conclude that ; that is, . Thus, has a fixed point as desired. Now, assume that has two fixed points such that . Hence, using the fact that is an contractive mapping and is -admissible, we obtainSince , taking the limit in the above inequalities, we deduce that which implies that . Thus, has a unique fixed point.
Next, we present the following example as an application of Theorem 1.

Example 4. Let be the controlled metric type space that was defined in Example 2. Define the function such thatDefine the self-mapping on by and the function .
We want to verify that satisfies the conditions of Theorem 1. It is clear that is continuous for . We have . So, is -admissible. Now, we verify that is -contractive mapping. Note that .(i), for all .(ii).(iii).(iv).Therefore, satisfies the conditions in Theorem 1, and hence it has a unique fixed point .
Now, we present the following as an immediate consequence of Theorem 1.

Corollary 1. Let be a complete controlled metric type space and be a mapping satisfying the following conditions:(1) is continuous.(2)There exists such that , for all .Then, has a unique fixed point.

Proof. Define the function via . Note that is admissible. Moreover, satisfies all the conditions of Theorem 1. So, has a unique fixed point.
In our next theorem, we replace the hypothesis of the continuity by a weaker condition.

Theorem 2. Let be a complete, controlled metric type space and be an --contractive mapping for some . Suppose that the following conditions hold:(1) is -admissible.(2)There exists such that .(3)If is a sequence in such that and as , then for all .Then, has a fixed point.

Proof. In proving the result, we follow the same steps as in the proof of Theorem 1 to construct a sequence that converges to a point . The constructed sequence has the property , for all natural numbers . The last assumption of the result implies that . We finally prove that is a fixed point for . The triangle inequality implies thatNote that the first term of the inequality, , converges to 0, since converges to 0. Also, the second term will bewhich converges to 0. Therefore, , and hence is a fixed point for .

4. Conclusion

Note that in Theorem 1, we used the continuity of the self-mapping on the controlled metric type space, which is a strong hypothesis. We want to bring to the reader’s attention that we only used this hypothesis to prove that , and thus is a fixed point. So, the open question is can we omit the hypothesis of the continuity without replacing it with another weaker condition as in Theorem 2. In closing, we present the following open question:

Under what conditions we obtain the same results of Theorems 1 and 2 for self-mappings in double controlled metric type spaces [18].

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

The authors would like to thank Prince Sultan University for their support through NAMAM Research Group.