Single and Multivalued Maps on Parametric Metric Spaces Endowed with an Equivalence Relation

Abstract

This article presents the -parametric metric space, which is a generalized concept of parametric metric space. After that, the discussion is concerned with the existence of fixed points of single and multivalued maps on -parametric metric spaces satisfying some contractive inequalities defined by an auxiliary function.

1. Introduction and Preliminaries

Metric fixed point theory is a very fruitful area of research belonging to nonlinear analysis and operator theory. It has much usage in other fields of mathematics. Numerous generalizations of Banach contraction principle have flourished this field in several aspects through using generalized forms of metric spaces or contraction conditions. It is not wrong to say that the contraction condition given by Salimi et al. in [1] is the most renowned condition of this decade, whereas on the account of metric spaces, we have many abstract forms like -metric spaces [2–5], partial metric spaces [6, 7], and partial -metric spaces [8, 9]. Recently, the notions of -metric [10] and parametric metric spaces [11] have been initiated.

In the following, we recall the concept of parametric metric spaces [11] and the convergence of sequences.

A map (where ) is said to be a parametric metric on if it satisfies the following axioms: (i) for all (ii) for all (iii) for all and for all

Then, is a parametric metric space. If be a parametric metric space and , then (1) for all (2) for all

Note that is complete if each Cauchy sequence in converges in .

One of the ways to generalize the contraction condition is to define it via some auxiliary functions. For example, an auxiliary type function was given in [12], where represents the collection of functions satisfying the below axioms: (i) is nondecreasing and continuous with respect to its each coordinate(ii)If with and , then ;(iii)If with and , then (iv)If with or or , then and is a nondecreasing map with for all and for all .

As examples of , we cite the following: (i)Let with , where .(ii)Let with , where .(iii)Let with , where and .

Khalehoghli et al. [10] took a binary relation on and a simple metric on to define an -metric space, denoted as . They also said that a self map on is a -contraction, if for all with , where . Along with this definition, the authors also defined the concepts of -continuity and -preserving property in order to extend the result of Banach on -metric spaces.

2. Main Results

Let and be an equivalence relation on . We introduce the following: a function is an -parametric-metric on , provided that the following axioms hold: (i)(ii)(iii)

Then, is an -parametric metric space.

Example 1. Take as the collection of all functions . Consider an equivalence relation on defined by if for all (note that means that divides ). Define by

Then, is an -parametric metric space.

Below, stands for an -parametric metric space.

Definition 1. Let be an -sequence in , that is, for each . Then, we say that
(i) is a convergent sequence in and converges to if for all and for each (for some value of )
(ii) is Cauchy if for all .

Definition 2. is -complete if each Cauchy -sequence in is convergent in .

Definition 3. is called regular whenever if for some , then we have for all .

We now present the first main theorem of the section.

Theorem 4. Let be an -complete regular -parametric metric space. Let and be two maps such that for all and for all with , where . Also, consider the following assertions: (a) is -preserving, that is, implies (b)There is such that and (c)For every such that and , we have (d)For each -sequence in with for all and , we have for all Then, has a fixed point.

Proof. By (b), there must be of the form and . Since is -preserving, we have , and by (c), . The repetition of these steps yields that and for all . Take a sequence with the terms as for all . Thus, and for all . Before doing the next part of the proof, we take for all . From (3), for each , we get for all .☐

Here, our claim is, for each , is a strictly decreasing sequence. Suppose this is not true in general, then we have at least one such that for some . Then, from (4) and property (i) of , we get

Property (iii) of together with the above inequality implies that

This fact and regularity condition give us which is not possible under the assumption. Thus, for all and . Again, we use (4) and property (i) of to obtain for all and .

Property (ii) of together with (8) implies

After a few simplification, we get

Hence, is a sequence in with and for all and Now, we will show that this -sequence is also Cauchy. For each , we can obtain

The convergence of for each implies that is a Cauchy -sequence in . The -completeness of leads to the existence of a point with for all and for each , for some . By (d), we also conclude that for each . Now, from (3), we get for all

By triangle property and (12), we get for all

Applying the limit when in (13), we get for all

Thus, by property (iv) of , we get for all . Therefore, .

In the following result, (15) is a special case of (3), but with (15) we do not need the condition of the regularity of the -parametric metric space.

Theorem 5. Let be an -complete -parametric metric space. Let and be two maps such that for all and for all with , where is a continuous and nondecreasing map with for all and for all . Also, consider the following assertions: (a) is -preserving, that is, implies (b)There is such that and (c)For every such that and , we have (d)For each -sequence in with for all and , we have for all Then, has a fixed point.

Proof. By (b), we know that there is of the form and . In view of the fact that has -preserving nature, thus and by (c), . Continuing this process, we get and for all . Take a sequence with the terms as for all . Thus, and for all . Now, we take for all . From (15), one writes for each , for all .

Since for all , (16) implies that

Therefore,

Now, by following the proof of the above result, we may ensure that is a Cauchy -sequence in , and there must exists with for all and for each , for some . Also, using (d), we conclude that for each . Now, from (15), we get for all and for all

By triangle property and (19), we get for all and for all

Applying the limit when in (20) and using the continuity of , we get for all

The above inequality is only possible when for all . Therefore, .

Example 6. Take with equivalence relation on defined as if . Define by

Define the maps and by

Here, one can verify that all the axioms of Theorem 5 are valid, and there is a fixed point of in .

Let be an -parametric metric space. Let denotes the collection of all nonempty subsets of having the following properties: (i)For any , exists for each (ii)If and , then

The Hausdroff -parametric distance is a mapping from into defined by for all where

Here, everyone should know that for each , and for each value of there must exists with , where . Subsequently, we take .

In the following theorem, we will discuss the case of the existence of fixed points for multivalued maps.

Theorem 7. Let be an -complete regular -parametric metric space. Let and be two maps such that for all and for all with , where and . Also, consider the following assertions: (a) is -preserving, that is, implies for each and (b)There is and such that and (c)For every such that and , we have (d)For each -sequence in with for all and , we have for all (e)For each , and there is with for all , where .Then, has a fixed point.

Proof. By (b) and (c), we observe that , since and are of the form and . Here, we take . Otherwise, . By (26), we get for all .

Since , by (e), there is such that for all Here, again, we take . Otherwise, . Thus, by using it in (27), we get for all .

Here, our claim is . Suppose this is not true. Then, for some . Then, from (29) and by property (i) of , for , we get

From property (iii) of and the above inequality, we get

This fact together with regularity condition implies and it is not possible under the assumption. Thus, for all . Again, we use (29) and property (i) of to obtain

The property (ii) of and (33) imply

Clearly, due to the fact that , and . Also, the -preserving characteristic of yields . Thus, by (c), we get . The repeated application of the above steps yields the sequence with , , and for all and for all and for all

The steps given below will show that is a Cauchy -sequence. For each , the triangle inequality and (35) imply that for all

The convergence of together with the above inequality confirm that is a Cauchy -sequence in . The -completeness of ensures the existence of a point with for all and for each , for some . We now able to conclude that, by (d), for each . Since and for each , using (c), we have . Thus, by (26), we get for all

By triangle property and (37), we get