# New Results in Controlled Metric Type Spaces

Authors: Muhib Abuloha; Doaa Rizk; Kamaleldin Abodayeh; Nabil Mlaiki; Thabet Abdeljawad
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, a new class of functions denoted byΨν 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.


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## Body

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


## 2. Preliminaries

The concept of extendedb-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 Y be an nonempty set and define the mappings ν:Y×Y⟶1,∞ and ℘¯:Y×Y⟶0,∞, such that ∀g,h,r∈Y,(1)
℘¯g,h=0⇔g=h.(2)
℘¯g,h=℘¯h,g.(3)
℘¯g,h≤νg,h℘¯g,r+℘¯r,h.
Then, we say that the pairY,℘¯ is an extended b−metric space.
A general extension of theb−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 Y, define the mappings ν:Y×Y⟶1,∞ and ℘¯:Y×Y⟶0,∞, such that ∀g,h,r∈Y, the following conditions hold:(d1)
℘¯g,h=0 if and only if g=h.(d2)
℘¯g,h=℘¯h,g.(d3)
℘¯g,h≤νg,r℘¯g,r+νr,h℘¯r,h.
Then, the pairY,℘¯ is called a controlled metric type space.
To illustrate the above definition, we present the following examples.Example 1.
(see [9]). Choose Y=1,2,…,. Take ℘¯:Y×Y⟶0,∞ such that(1)℘¯g,h=0,⇔g=h,1g,ifg=2nandh=2n+1,1h,ifg=2n+1andh=2n,1,otherwise.
Considerν:Y×Y⟶1,∞ as(2)νg,h=g,ifg=2nandh=2n+1,h,ifg=2n+1andh=2n,1,otherwise.
It is clear that the conditionsd1 and d2 are satisfied. Now, we investigate condition d3.Case 1.
Ifr=g or r=h, d3 is satisfied.Case 2.
Ifr≠g and r≠h, d3 holds when g=h. Now, we may assume that g≠h. Then, we have g≠h≠r. It is clear that d3 holds in all following possible subcases:(1)
g,r are even and h=2n+1.(2)
g=2n and h,r are odd.(3)
g,r are odd and h=2n.(4)
g,r are even and h=2n+1.(5)
g,h,r are even.(6)
g,h are even and r=2n+1.(7)
g,h are odd and r=2n.(8)
g,h,r are odd.
Thus,℘¯ is a controlled metric type.
Moreover, forn=2,3,…,, we have(3)℘¯2n+1,4n+1=1>1n=ν2n+1,4n+1℘¯2n+1,2n+℘¯2n,4n+1.
Therefore,℘¯ is not an extended b-metric.Example 2.
(see [9]). Take Y=0,1,2. Consider the function ℘¯ given as(4)℘¯0,0=℘¯1,1=℘¯2,2=0,℘¯0,1=℘¯1,0=1,℘¯0,2=℘¯2,0=12,℘¯1,2=℘¯2,1=25.
Define a symmetric functionν:Y×Y⟶1,∞ such that(5)ν0,0=ν1,1=ν2,2=ν0,2=1,ν1,2=54,ν0,1=1110.
One can easily verify that℘¯ is a controlled metric type.
Since(6)℘¯0,1=1>99100=ν0,1℘¯0,2+℘¯2,1,℘¯ is not an extended b-metric.
The concepts of Cauchy and convergent sequences in controlled metric type spaces are defined as follows.Definition 3.
LetY,℘¯ be a controlled metric type space and gnn≥0 be a sequence in Y.(1)
The sequencegn converges to some g∈Y, if ∀ϵ>0, ∃N=Nɛ∈ℕ such that ℘¯gn,g<ɛ∀n≥N. We write limn⟶∞gn=g.(2)
We say thatgn is Cauchy, if ∀ɛ>0, ∃N=Nɛ∈ℕ such that ℘¯gm,gn<ɛ∀m,n≥N.(3)
If every Cauchy sequence is convergent, then the spaceY,℘¯ is called complete.Definition 4.
(see [9]). Let Y,℘¯ be a controlled metric type space. Let g∈Y and ε>0.(i)
The open ballBg,ε is defined as(7)Bg,ε=h∈Y,℘¯g,h<ε.(ii)
A self-mappingξ on Y is said to be continuous at g∈Y, if ∀ε>0, ∃δ>0 such that ξBg,δ⊆Bξg,ε.Remark 1.
If for allg,y in Y, νg,h=s≥1, then Y,℘¯ is a b−metric space. Therefore, we conclude that every b−metric space is a controlled metric type space. However, the converse is not always true.
Clearly, if a mappingξ is continuous at g in the controlled metric type space Y,℘¯, then gn⟶g implies that ξgn⟶ξg as n⟶∞.
LetΨ denote to the set of all functions ψ:0,∞⟶0,∞ such that(1)
ψ is nondecreasing.(2)
∑n=1∞ψnt<+∞ for all t>0, where ψn is the n-th iterate of ψ.
Now, we recall the following lemma.Lemma 1.
(see [5]). If ψ∈Ψ, then ψt<t, for all t∈0,+∞.
Next, we introduce the following class of functions.Definition 5.
(see [5]). Let Y be a nonempty set and ν:Y×Y⟶1,+∞ be a mapping. A function ψ:0,∞⟶0,∞ is said to be controlled comparison function if ψ satisfies the following conditions:(1)
ψ is nondecreasing.(2)
∑n=1∞ψnt∏i=1nνgi,gmνgn,gn+1<+∞, and limn⟶∞ψntνgn,gn+1<∞ for any sequence gnn=1∞ in Y, for all t>0 and non-negative integer m where ψn is the n-th iterate of ψ.
The set of all controlled comparison functions is denoted byΨν which is an extension of b−comparison functions of Berinde.
Note that ifψ∈Ψν, then we have ∑n=1∞ψnt<∞, since ψnt∏i=1nνgi,gm≥ψnt, for all t>0. Hence, by Lemma 1, we have ψt<t.
To show that the familyΨν is a nonempty set, we present the following examples.Example 3.
Consider the controlled b-metric spaceY,℘¯ which was defined in Example 2. Define the mapping ψt=kt/2, where k<1. Note that νg,h≤2. Then, we have ψnt∏i=1nνgi,gmνgn,gn+1≤knt/2n⋅2n+1=2knt. Therefore, ∑n=1∞ψnt∏i=1nνgi,gmνgn,gn+1≤∑n=1∞2knt<∞. Similarly, it is not difficult to see that limn⟶∞ψntνgn,gn+1<∞.


## 3. Main Result

First, we define theα-ψ-contractive self-mapping in controlled metric type spaces.Definition 6.
Letξ be a self-mapping on a complete controlled metric type space Y,℘¯. We say that ξ is α-ψ-contractive mapping if there exists a function α:Y2⟶0,∞ and ψ∈Ψν such that for all g,y∈Y, we have(8)αg,h℘¯ξg,ξh≤ψ℘¯g,h.Definition 7.
(see [17]). Let Y,℘¯ be a controlled metric type space. A mapping ξ:Y⟶Y is said to be an α−admissible if the following condition holds: if ∀g,y∈Y, with αg,h≥1, then αξg,ξh≥1.
Now, we prove our first result.Theorem 1.
LetY,℘¯ be a complete controlled metric type space and ξ:Y⟶Y be an α−ψ contractive mapping for some ψ∈Ψν. Assume that(A)
ξisα−admissible.(B)
There existsg0∈Y such that αg0,ξg0≥1.(C)
ξ is continuous.
Then,ξ has a fixed point. Moreover, if for any two fixed points of ξ in Y say a,b we have αa,b≥1, then ξ has a unique fixed point in Y.Proof.
Takeg0 to be the point in Condition (2) in our theorem. Define the sequence gnn≥0 by ξgn=ξng0.
First of all, note if there existsn such that gn=gn+1, then we are done and gn is the fixed point of ξ.
So, we may assume thatgn≠gn+1 for all n≥0. Also, we know from the hypothesis of our theorem that αg0,ξg0≥1, αg0,g1≥1, and using the fact that ξ is α− admissible, we can easily deduce that for all n≥0, αgn,gn+1≥1.
Now, using the fact thatξ is an α−ψ contractive mapping, we deduce that(9)℘¯gn,gn+1≤αgn−1,gn℘¯gn,gn+1=αgn−1,gn℘¯ξgn−1,ξgn≤ψ℘¯gn−1,gn...≤ψn℘¯g0,g1.
Hence,∀m,n∈N with m>n, we have(10)℘¯gn,gm≤νgn,gn+1℘¯gn,gn+1+νgn+1,gm℘¯gn+1,gm≤νgn,gn+1℘¯gn,gn+1+νgn+1,gmνgn+1,gn+2℘¯gn+1,gn+2+νgn+1,gmνgn+2,gm℘¯gn+2,gm≤νgn,gn+1℘¯gn,gn+1+νgn+1,gmνgn+1,gn+2℘¯gn+1,gn+2+νgn+1,gmνgn+2,gmνgn+2,gn+3℘¯gn+2,gn+3+νgn+1,gmνgn+2,gmνgn+3,gm℘¯gn+3,gm≤νgn,gn+1ψn℘¯g0,g1+νgn+1,gn+2νgn+1,gmψn+1℘¯g0,g1+νgn+1,gmνgn+2,gmνgn+2,gn+3ψn+2℘¯g0,g1⋮+νgn+1,gmνgn+2,gmνgn+3,gm…νgm−2,gm−1ψm−1℘¯g0,g1=νgn,gn+1ψn℘¯g0,g1+∑j=n+1m−2ψj℘¯g0,g1∏i=n+1jνgi,gmνgj,gj+1=νgn,gn+1ψn℘¯g0,g1+∑j=1m−2ψj℘¯g0,g1∏i=1jνgi,gmνgj,gj+1−∑j=1nψj℘¯g0,g1∏i=1jνgi,gmνgj,gj+1=Sm−2−Sn,where(11)Sm=∑j=1mψj℘¯g0,g1∏i=1jνgi,gmνgj,gj+1,and since ψ∈Ψν, we deduce that limn,m⟶∞Sm−2,Sn=0 and(12)limn⟶∞νgn,gn+1ψn℘¯g0,g1<∞.
Thus, the sequencegnn≥0 is a Cauchy sequence. The completeness of the controlled metric type space g,℘¯ implies that gn converges to some g∈Y.
Also, note that(13)℘¯g,ξg≤νg,gn+1℘¯g,gn+1+νgn+1,ξg℘¯gn+1,ξg=νg,gn+1℘¯g,gn+1+νgn+1,ξg℘¯ξgn,ξg.
Taking the limit in above inequality and sincelimn⟶∞℘¯g,gn+1=0 and using the fact that ξ is continuous, we conclude that ℘¯g,ξg=0; that is, ξg=g. Thus, ξ has a fixed point as desired. Now, assume that ξ has two fixed points a,b such that αa,b≥1. Hence, using the fact that ξ is an α−ψ contractive mapping and ξ is α-admissible, we obtain(14)℘¯a,b=℘¯ξa,ξb≤αa,b℘¯ξa,ξb≤ψ℘¯a,b⋮≤ψn℘¯a,b.
Sinceψ∈Ψν, taking the limit in the above inequalities, we deduce that ℘¯a,b=0 which implies that a=b. Thus, ξ has a unique fixed point.
Next, we present the following example as an application of Theorem1.Example 4.
LetY,℘¯ be the controlled metric type space that was defined in Example 2. Define the function α:Y×Y⟶−∞,∞ such that(15)αg,h=1,ifg,h=1,1,120,ifg,h≠1,1.
Define the self-mappingξ on Y=0,1,2 by ξ1=1,ξ0=2,ξ2=1 and the function ψt=1/8t.
We want to verify thatξ satisfies the conditions of Theorem 1. It is clear that ξ is continuous for g0=1. We have α1,ξ1=α1,1=1≥1. So, ξ is α-admissible. Now, we verify that ξ is α−ψ-contractive mapping. Note that αg,h=αh,g.(i)
αg,g℘¯ξg,ξg=0≤ψ℘¯g,g=ψ0=0, for all g∈Y.(ii)
α1,2℘¯ξ1,ξ2=1/20℘¯1,1=0≤ψ℘¯1,2=ψ2/5=1/20.(iii)
α1,0℘¯ξ1,ξ0=1/20℘¯1,2=1/50≤ψ℘¯1,0=ψ1=1/8.(iv)
α2,0℘¯ξ2,ξ0=1/20℘¯1,2=1/100≤ψ℘¯2,0=ψ1/2=1/16.
Therefore,ξ satisfies the conditions in Theorem 1, and hence it has a unique fixed point g=1.
Now, we present the following as an immediate consequence of Theorem1.Corollary 1.
LetY,℘¯ be a complete controlled metric type space and ξ:Y⟶Y be a mapping satisfying the following conditions:(1)
ξ is continuous.(2)
There existsψ∈Ψν such that ℘¯ξg,ξh≤ψ℘¯g,h, for all g,h∈Y.
Then,ξ has a unique fixed point.Proof.
Define the functionα:Y×Y⟶0,+∞ via αg,h=1. Note that ξ is α−admissible. Moreover, ξ satisfies all the conditions of Theorem 1. So, ξ has a unique fixed point.
In our next theorem, we replace the hypothesis of the continuity by a weaker condition.Theorem 2.
LetY,℘¯ be a complete, controlled metric type space and ξ:Y⟶Y be an α-ψ-contractive mapping for some ψ∈Ψs. Suppose that the following conditions hold:(1)
ξ is α-admissible.(2)
There existsg0∈Y such that αg0,ξg0≥1.(3)
Ifgnn=1∞ is a sequence in Y such that αgn,gn+1≥1 and gn⟶x as n⟶∞, then αgn,g≥1 for all n.
Then,ξ has a fixed point.Proof.
In proving the result, we follow the same steps as in the proof of Theorem1 to construct a sequence gnn=1∞ that converges to a point g∈Y. The constructed sequence has the property αgn,gn+1≥1, for all natural numbers n. The last assumption of the result implies that αgn,g≥1. We finally prove that g is a fixed point for ξ. The triangle inequality implies that(16)℘¯g,ξg≤νg,ξg℘¯g,gn+1+℘¯gn+1,ξg=νg,ξg℘¯g,gn+1+νg,ξg℘¯gn+1,ξg.
Note that the first term of the inequality,νg,ξg℘¯g,gn+1, converges to 0, since gn converges to 0. Also, the second term will be(17)νg,ξg℘¯gn+1,ξg≤νg,ξg℘¯ξgn,ξgαgn,g≤νg,ξgψ℘¯gn,g≤νg,ξg℘¯gn,g,which converges to 0. Therefore, ℘¯g,ξg≤0, and hence g is a fixed point for ξ.


## 4. Conclusion

Note that in Theorem1, 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 ℘¯a,Ta=0, and thus a 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 Theorems1 and 2 for self-mappings in double controlled metric type spaces [18].



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