# Remarks onα,β-Admissible Mappings and Fixed Points under Z-Contraction Mappings

Authors: Bulbul Khomdram; N. Priyobarta; Yumnam Rohen; Thounaojam Indubala
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we discuss about different types ofα,β-admissible mappings and introduce some new α,β-contraction-type mappings under simulation function. Furthermore, we present the definition of S-metric-like space and its topological properties. Some fixed point theorems in this space are established, proved, and verified with examples.


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## Body

## 1. Introduction and Preliminaries

Banach contraction principle is considered as one of the most important tools for examining the existence and nonexistence of fixed point. Due to its simplicity and applicability, it is generalised in different directions.Samet et al. [1] introduced the concept of α-admissible and α−ψ contraction to generalise the Banach contraction principle. These concepts were further generalised by many researchers to α−β admissible [2], β-admissible [3, 4], α-admissible in S-metric space [5], α,β-admissible [6, 7], and γ-admissible [8]. Some results on different types on contractive mappings can be seen in [9–11].Different from theα,β-admissible mapping introduced by Alizadeh et al. [6], Chandok [7] introduced a new type of α,β-admissible mapping to obtain some fixed point results.On the contrary, Khojasteh et al. [12] introduced simulation function and proved several fixed point theorems. Argoubi et al. [13] made some modifications to the definition of Khojasteh et al. [12]. It can be noted that the definition of Khojasteh et al. [12] implies the definition of Argoubi et al. [13], but the converse is not true.Various generalizations of metric space are found in the literature. The concept ofS-metric space [14] is one of them. More results on S-metric space can be found in [15, 16]. Hitzler [17] also introduced the concept of dislocated metric by generalizing metric space. Amini-Harandi [18] rediscovered dislocated metric space under the new name “metric-like.”Mehravaran et al. [19] introduced the concept of dislocated Sb-metric space and dislocated S-metric space as a particular case of dislocated Sb-metric space when the parameter b=1, but the topological properties of neither dislocated Sb-metric space nor dislocated S-metric space were given in [19].In order to fill up the missing gap in [19], in our present study, first of all, we present the definition of dislocated S-metric space and its topological properties. For our convenience, it will be known as S-metric-like space in place of dislocated S-metric space.Definition 1.
(see [14]). In a nonempty set Ω, the function S:Ω3⟶0,+∞ is said to be S-metric if it satisfies the following:(1)
Sθ,μ,ξ≥0,(2)
Sθ,μ,ξ=0 if and only if θ=μ=ξ,(3)
Sθ,μ,ξ≤Sθ,σ,σ+Sμ,σ,σ+Sξ,σ,σ,for all θ,μ,ξ,σ∈Ω. Ω,S is known as S-metric space.Definition 2.
(see [14]). In a S-metric space, we have Sθ,θ,μ=Sμ,μ,θ.Definition 3.
(see [14]). In a nonempty set Ω, the function S:Ω3⟶0,+∞ is said to be S-metric-like if it satisfies the following:(1)
Sθ,μ,ξ=0 implies θ=μ=ξ,(2)
Sθ,μ,ξ≤Sθ,σ,σ+Sμ,σ,σ+Sξ,σ,σ,for all θ,μ,ξ,σ∈Ω. Ω,S is known as S-metric-like space.Remark 1.
It is true that everyS-metric space is a S-metric-like space, but every S-metric-like space may not be a S-metric space.Example 1.
LetΩ=0,2 and(1)Sθ,μ,ξ=3,θ=μ=ξ,1,otherwise.
Here,Ω,S is S-metric-like space. But S0,0,0=3≠0 and hence not S-metric space.
We discuss some topological properties ofS-metric-like space. Let τs be a topology generated on S-metric-like space Ω,S with base as the family of open S-balls:(2)BSθ,ε=μ∈Ω:Sθ,θ,μ−Sθ,θ,θ<ε,for allθ∈Ωandε>0.
MappingR:Ω⟶Ω in a S-metric-like Ω,S is said to be continuous at θ∈Ω if there exists δ>0 for all ε>0 satisfying RBSθ,δ⊆BSRθ,ε.
IfR:Ω⟶Ω is continuous, then limn⟶+∞θn=θ implies limn⟶+∞Rθn=Rθ.
A sequenceθn in Ω is said to be Cauchy if limn,m⟶+∞Sθn,θn,θm exists and is finite.
TheS-metric-like Ω,S is said to be complete if every Cauchy sequence θn in Ω satisfies(3)limn⟶+∞Sθn,θn,θ=Sθ,θ,θ=limn,m⟶+∞Sθn,θn,θm,for some θ∈Ω.
A subsetP⊆Ω is said to be bounded if there is a point ξ∈Ω satisfying(4)Sθ,μ,ξ≤L,for all θ,μ∈P and L is a positive constant.Definition 4.
(see [12]). A function ζ:0,+∞×0,+∞⟶ℝ is said to be a simulation function if ζ satisfies the following:(1)
ζ0,0=0.(2)
ζυ,μ<μ−υ for all υ,μ>0.(3)
Ifυn and μn are sequences in 0,+∞ such that limn⟶+∞υn=limn⟶+∞μn=l∈0,+∞, then(5)limn⟶+∞supζυn,μn<0.
Simulation function is modified by Argoubi et al. [13] as follows.Definition 5.
(see [13]). A mapping ζ:0,+∞×0,+∞⟶ℝ is said to be a simulation function if satisfying the conditions (2) and (3) of Definition 4.
Now, we present different forms ofα,β-admissible mappings.
Alizadeh et al. [6] defined a type of α,β-admissible mappings by extending definition given by Salimi et al. [20] which they called as cyclic α,β-admissible mapping.Definition 6.
(see [6]). Let R:Ω⟶Ω be a mapping and α,β:Ω⟶ℝ+ be two functions. R is said to be a cyclic α,β-admissible mapping if the following holds:(i)
For someθ∈Ω, αθ≥1 induces βRθ≥1.(ii)
For someθ∈Ω, βθ≥1 induces αRθ≥1.
The following definition was given by Chandok [7].Definition 7.
(see [7]). Let R:Ω⟶Ω be a mapping in a nonempty set Ω and α,β:Ω×Ω⟶ℝ+. R is said to be α,β-admissible if αθ,μ≥1 and βθ,μ≥1 implies αRθ,Rμ≥1 and βRθ,Rμ≥1 for all θ,μ∈Ω.
Shatanawi [2] also introduced a type of α,β-admissible mapping which he named as α,β-admissibility and defined as follows.Definition 8.
(see [2]). Let S,T:Ω⟶Ω be mappings in a nonempty set Ω and α,β:Ω×Ω⟶ℝ+∪0 be functions. Then, R,S is said to be a pair of α,β-admissibility if θ,μ∈Ω and αθ,μ≥βθ,μ implies αRθ,Sμ≥βRθ,Sμ and αSθ,Rμ≥βSθ,Rμ.
For our study, we will consider the definition given by Chandok [7]. This definition will be extended in the framework of S-metric-like space to define some new contractive types under simulation function defined by Khojasteh et al. [12].
Following is the extension of the property of metric-like space toS-metric-like space.


## 2. Main Results

We start our result with the following definitions.Definition 9.
LetR be a self-mapping on a nonempty set Ω and α,β:Ω×Ω×Ω⟶ℝ+. Then, R is said to be α,βs-admissible if αθ,μ,ξ≥1 and βθ,μ,ξ≥1 imply that αRθ,Rμ,Rξ≥1 and βRθ,Rμ,Rξ≥1 for all θ,μ,ξ∈Ω.Definition 10.
LetR be a self-mapping on a nonempty set Ω and α,β:Ω×Ω×Ω⟶ℝ+. Then, R is said to be triangular α,βs-admissible mapping if it is α,βs-admissible and αθ,θ,t≥1, βθ,θ,t≥1, αμ,μ,t≥1, βμ,μ,t≥1, and αξ,ξ,t≥1, βξ,ξ,t≥1 imply αθ,μ,ξ≥1, βθ,μ,ξ≥1.Definition 11.
LetR be a self-mapping on a S-metric-like space Ω,S and α,β:Ω×Ω×Ω⟶ℝ+. R is said to be an α,βs-admissible Z-contraction of type I with respect to ζ if(6)ζαθ,μ,ξβθ,μ,ξSRθ,Rμ,Rξ,Sθ,μ,ξ≥0,for all θ,μ,ξ∈Ω.Definition 12.
LetR be a self-mapping on a S-metric-like space Ω,S and α,β:Ω×Ω×Ω⟶ℝ+. R is said to be an α,βs-admissible Z-contraction of type II with respect to ζ if(7)ζαθ,θ,μβθ,θ,μSRθ,Rθ,Rμ,Sθ,θ,μ≥0,for all θ,μ∈Ω.Lemma 1.
LetΩ,S be a S-metric-like space and θn be a sequence in Ω such that θn⟶θ with Sθ,θ,θ=0. Then, for all μ∈Ω, we have limn⟶+∞Sθn,θn,μ=Sθ,θ,μ.Proof.
From (2) of Definition 3, we have(8)Sθn,θn,μ−Sθ,θ,μ≤2Sθn,θn,θ.
Taking limit asn⟶+∞, we have(9)limn⟶+∞Sθn,θn,μ=Sθ,θ,μ.
Now, we prove the following theorem.Theorem 1.
LetR:Ω⟶Ω be a mapping on a complete S-metric-like space Ω,S satisfying the following:(a)
R is triangular α,βs-admissible.(b)
There existsθ0∈Ω such that αθ0,θ0,Rθ0≥1 and βθ0,θ0,Rθ0≥1.(c)
R is α,βs-admissible Z-contraction of type I on Ω,S.(d)
R is S-continuous.
Then, there is a unique fixed pointξ∈Ω of R with Sξ,ξ,ξ=0.Proof.
Byb, let θ0∈Ω with αθ0,θ0,Rθ0≥1 and βθ0,θ0,Rθ0≥1. We construct the sequence θn by θn+1=Rθn for all n∈ℕ∪0. For some n, if θn=θn+1, then we have θn=Rθn. This gives that θn is a fixed point of R. In this case, proof is completed. Now, let θn≠θn+1 for all n∈ℕ∪0. By a, we have(10)αθ0,θ0,Rθ0=αθ0,θ0,θ1≥1⇒αRθ0,Rθ0,Rθ1=αθ1,θ1,θ2≥1.
Deductively, we have(11)αθn,θn,θn+1≥1,for alln≥0,(12)βθn,θn,θn+1≥1,for alln≥0.
By (6), (11), and (12), we have(13)0≤ζαθn,θn,θn+1βθn,θn,θn+1SRθn,Rθn,Rθn+1,Sθn,θn,θn+1=ζαθn,θn,θn+1βθn,θn,θn+1Sθn+1,θn+1,θn+2,Sθn,θn,θn+1<Sθn,θn,θn+1−αθn,θn,θn+1βθn,θn,θn+1Sθn+1,θn+1,θn+2.
Hence,(14)Sθn+1,θn+1,θn+2≤αθn,θn,θn+1βθn,θn,θn+1Sθn+1,θn+1,θn+2<Sθn,θn,θn+1,for all n≥0.
This shows thatSθn,θn,θn+1 is a decreasing sequence, then we have(15)limn⟶+∞Sθn,θn,θn+1=r,r≥0.
We prove that(16)limn⟶+∞Sθn,θn,θn+1=0.
Letr>0. From (14), we have(17)limn⟶+∞αθn,θn,θn+1βθn,θn,θn+1Sθn+1,θn+1,θn+2=r.
Takingυn=αθn,θn,θn+1βθn,θn,θn+1Sθn+1,θn+1,θn+2 and μn=Sθn,θn,θn+1 and using (3) of Definition 4, we have(18)0≤limn⟶+∞supζαθn,θn,θn+1βθn,θn,θn+1Sθn+1,θn+1,θn+2,Sθn,θn,θn+1<0,a contradiction. Hence, r=0.
Next, we have to prove thatθn is Cauchy. If possible, let θn is not Cauchy. Then, there exists ε>0 for which θn has subsequences θnk and θmk with nk>mk>k such that for every k,(19)Sθnk,θnk,θmk≥ε,Sθnk−1,θnk−1,θmk<ε.
We have(20)ε≤Sθnk,θnk,θmk≤2Sθnk,θnk,θnk−1+Sθnk−1,θnk−1,θmk<2Sθnk,θnk,θnk−1+ε.
Takingk⟶+∞ and using (16),(21)limn⟶+∞Sθnk,θnk,θmk=ε.
Also,(22)Sθnk+1,θnk+1,θmk−Sθnk,θnk,θmk≤2Sθnk,θnk,θnk+1.
Taking limit ask⟶+∞ and by (16) and (21),(23)limk⟶+∞Sθnk+1,θnk+1,θmk=ε.
Similarly,(24)limk⟶+∞Sθmk+1,θmk+1,θnk=ε,limk⟶+∞Sθnk+1,θnk+1,θmk+1=ε.
By triangularα,βs-admissibility of R,(25)αθnk,θnk,θmk≥1andβθnk,θnk,θmk≥1.
SinceR is α,βs-admissible Z-contraction of type I and using (21), (25), and (3) of Definition 4, we have(26)0≤limk⟶+∞supζαθnk,θnk,θmkβθnk,θnk,θmkSθnk+1,θnk+1,θmk+1,Sθnk,θnk,θmk<0,is a contradiction, and hence, θn is a Cauchy sequence. By completeness of S-metric-like space Ω,S, we know that there exist some ξ∈Ω such that(27)limn⟶+∞Sθn,θn,ξ=Sξ,ξ,ξ=limn,m⟶+∞Sθn,θn,θm=0,and thus, Sξ,ξ,ξ=0. Since R is S-continuous,(28)limn⟶+∞Sθn+1,θn+1,Rξ=limn⟶+∞SRθn,Rθn,Rξ=SRξ,Rξ,Rξ=0.
Using Lemma1 and (27), we have(29)limn⟶+∞Sθn+1,θn+1,Rξ=Sξ,ξ,Rξ.
Thus,SRξ,Rξ,Rξ=Sξ,ξ,Rξ, that is, Rξ=ξ. For proving uniqueness of the fixed point, let η∈Ω such that Rη=η and η≠ξ. Then,(30)0≤ζαξ,ξ,ηβξ,ξ,ηSRξ,Rξ,Rη,Sξ,ξ,η<Sξ,ξ,η−αξ,ξ,ηβξ,ξ,ηSRξ,Rξ,η≤0,is a contradiction, so ξ=η.
Next, we remove the continuity condition.Theorem 2.
LetR:Ω⟶Ω be a mapping on a complete S-metric-like space Ω,S satisfying the following:(a)
R is triangular α,βs-admissible.(b)
There existsθ0∈Ω such that αθ0,θ0,Rθ0≥1 and βθ0,θ0,Rθ0≥1.(c)
R is α,βs-admissible Z-contraction of type I on Ω,S.(d)
Ifθn is a sequence in Ω such that αθn,θn,θn+1≥1 and βθn,θn,θn+1≥1 for all n and θn⟶ξ∈Ω as n⟶+∞, then there exists a subsequence θnk of θn such that αθnk,θnk,ξ≥1 and βθnk,θnk,ξ≥1 for all k∈ℕ.
Then, there is a unique fixed pointξ∈Ω of R with Sξ,ξ,ξ=0.Proof.
Proceeding as of Theorem1, let θn be a sequence in Ω given by θn+1=Rθn converges to some ξ∈Ω with αθn,θn,θn+1≥1 and βθn,θn,θn+1≥1 for all n and Sξ,ξ,ξ=0. From d, there exists a subsequence θnk of θn such that αθnk,θnk,ξ≥1 and βθnk,θnk,ξ≥1 for all k∈ℕ. Thus, applying (6) for all k, we have(31)0≤ζαθnk,θnk,ξβθnk,θnk,ξSRθnk,Rθnk,Rξ,Sθnk,θnk,ξ=ζαθnk,θnk,ξβθnk,θnk,ξSθnk+1,θnk+1,Rξ,Sθnk,θnk,ξ<Sθnk,θnk,ξ−αθnk,θnk,ξβθnk,θnk,ξSθnk+1,θnk+1,Rξ,imply(32)Sθnk+1,θnk+1,Rξ≤αθnk,θnk,ξβθnk,θnk,ξSθnk+1,θnk+1,Rξ<Sθnk,θnk,ξ.
Takingk⟶+∞, we have Sξ,ξ,Rξ=0, that is, Rξ=ξ. Proceeding as in Theorem 16, the uniqueness of fixed point of R can be proved. □Definition 13.
LetR:Ω⟶Ω be a mapping on a S-metric-like space Ω,S and α,β:Ω×Ω×Ω⟶ℝ+. R is said to be a generalised α,βs-admissible Z-contraction of type I with respect to ζ if(33)ζαθ,μ,ξβθ,μ,ξMθ,μ,ξ,Sθ,μ,ξ≥0,where(34)Mθ,μ,ξ=maxSθ,μ,ξ,Sθ,θ,Rθ,Sμ,μ,Rμ,Sξ,ξ,Rξ,for all θ,μ,ξ∈Ω.Definition 14.
LetR:Ω⟶Ω be a mapping on a S-metric-like space Ω,S and α,β:Ω×Ω×Ω⟶ℝ+. R is said to be a generalised α,βs-admissible Z-contraction of type II with respect to ζ if(35)ζαθ,θ,μβθ,θ,μMθ,θ,μ,Sθ,θ,μ≥0,where(36)Mθ,θ,μ=Sθ,θ,μ,Sθ,θ,Rθ,Sμ,μ,Rμ,for all θ,μ∈Ω.Theorem 3.
LetR:Ω⟶Ω be a mapping on a complete S-metric-like space Ω,S satisfying the following:(a)
R is triangular α,βs-admissible.(b)
There existsθ0∈Ω such that αθ0,θ0,Rθ0≥1 and βθ0,θ0,Rθ0≥1.(c)
R is a generalised α,βs-admissible Z-contraction of type I on Ω,S.(d)
R is S-continuous.
Then, there is a unique fixed pointξ∈Ω of R with Sξ,ξ,ξ=0.Proof.
Byb, let θ0∈Ω with αθ0,θ0,Rθ0≥1 and βθ0,θ0,Rθ0≥1. We construct the sequence θn by θn+1=Rθn for all n∈ℕ∪0. For some n, if θn=θn+1, then we have θn=Rθn. This gives that θn is a fixed point of R. In this case, proof is completed. Now, let θn≠θn+1 for all n∈ℕ∪0. By a, we have(37)αθ0,θ0,Rθ0=αθ0,θ0,θ1≥1⇒αRθ0,Rθ0,Rθ1=αθ1,θ1,θ2≥1.
Deductively, we have(38)αθn,θn,θn+1≥1,for alln≥0,(39)βθn,θn,θn+1≥1,for alln≥0.
By (33), (38), and (39),(40)0≤ζαθn,θn,θn+1βθn,θn,θn+1Mθn,θn,θn+1,Sθn,θn,θn+1,where(41)Mθn,θn,θn+1=maxSθn,θn,θn+1,Sθn,θn,Rθn,Sθn+1,θn+1,Rθn+1=maxSθn,θn,θn+1,Sθn+1,θn+1,θn+2.
Thus,(42)0≤ζαθn,θn,θn+1βθn,θn,θn+1maxSθn,θn,θn+1,Sθn+1,θn+1,θn+2,Sθn,θn,θn−1<Sθn,θn,θn+1−αθn,θn,θn+1βθn,θn,θn+1maxSθn,θn,θn+1,Sθn+1,θn+1,θn+2.
Consequently, we have(43)Sθn,θn,θn+1>αθn,θn,θn+1βθn,θn,θn+1maxSθn,θn,θn+1,Sθn+1,θn+1,θn+2.
IfmaxSθn,θn,θn+1,Sθn+1,θn+1,θn+2=Sθn,θn,θn+1 for all n≥0, then(44)Sθn,θn,θn+1>αθn,θn,θn+1βθn,θn,θn+1Sθn,θn,θn+1≥Sθn,θn,θn+1,is a contradiction. Thus, maxSθn,θn,θn+1,Sθn+1,θn+1,θn+2=Sθn+1,θn+1,θn+2 for all n≥0. Hence,(45)Sθn+1,θn+1,θn+2≤αθn,θn,θn+1βθn,θn,θn+1Sθn+1,θn+1,θn+2<Sθn,θn,θn+1,for all n≥0.
This shows thatSθn,θn,θn+1 is a decreasing sequence, and then, we have(46)limn⟶+∞Sθn,θn,θn+1=r,r≥0.
We prove that(47)limn⟶+∞Sθn,θn,θn+1=0.
Letr>0. From (45), we have(48)limn⟶+∞αθn,θn,θn+1βθn,θn,θn+1Sθn+1,θn+1,θn+2=r.
Takingυn=αθn,θn,θn+1βθn,θn,θn+1Sθn+1,θn+1,θn and μn=Sθn,θn,θn+1 and using (3) of Definition 4, we have(49)0≤limn⟶+∞supζαθn,θn,θn+1βθn,θn,θn+1Sθn+1,θn+1,θn+2,Sθn,θn,θn+1<0,is a contradiction. Hence, r=0.
Next, we have to prove thatθn is Cauchy. If possible, let θn is not Cauchy. Then, there exists ε>0 for which subsequences θnk and θmk of θn can be obtained with nk>mk>k such that, for every k,(50)Sθnk,θnk,θmk≥ε,Sθnk−1,θnk−1,θmk<ε.
We have(51)ε≤Sθnk,θnk,θmk≤2Sθnk,θnk,θnk−1+Sθnk−1,θnk−1,θmk<2Sθnk,θnk,θnk−1+ε.
Takingk⟶+∞ and by (47),(52)limk⟶+∞Sθnk,θnk,θmk=ε.
Also,(53)Sθnk+1,θnk+1,θmk−Sθnk,θnk,θmk≤2Sθnk,θnk,θnk+1.
Taking limit ask⟶+∞ and by (47) and (52),(54)limk⟶+∞Sθnk+1,θnk+1,θmk=ε.
Similarly,(55)limk⟶+∞Sθmk+1,θmk+1,θnk=ε,limk⟶+∞Sθnk+1,θnk+1,θmk+1=ε.
By triangularα,βs-admissibility of R,(56)αθnk,θnk,θmk≥1andβθnk,θnk,θmk≥1.
SinceR is generalised α,βs-admissible Z-contraction of type I and using (52), (56), and (3) of Definition 4, we have(57)0≤limk⟶+∞supζαθnk,θnk,θmkβθnk,θnk,θmkMθnk,θnk,θmk,Sθnk,θnk,θmk<0,is a contradiction, and hence, θn is Cauchy. By completeness of S-metric-like space Ω,S, we know that there exist some ξ∈Ω such that(58)limn⟶+∞Sθn,θn,ξ=Sξ,ξ,ξ=limn,m⟶+∞Sθn,θn,θm=0,which implies that Sξ,ξ,ξ=0. The continuity of R implies that(59)limn⟶+∞Sθn+1,θn+1,Rξ=limn⟶+∞SRθn,Rθn,Rξ=SRξ,Rξ,Rξ=0.
Using Lemma1 and (58), we have(60)limn⟶+∞Sθn+1,θn+1,Rξ=Sξ,ξ,Rξ.
Thus,SRξ,Rξ,Rξ=Sξ,ξ,Rξ, that is, Rξ=ξ. To prove uniqueness of fixed point, let η∈Ω such that Rη=η and η≠ξ. Then,(61)0≤ζαξ,ξ,ηβξ,ξ,ηSRξ,Rξ,Rη,Sξ,ξ,η<Sξ,ξ,η−αξ,ξ,ηβξ,ξ,ηSRξ,Rξ,η≤0,is a contradiction, so ξ=η.
By removing the continuity condition, we have the following result.Theorem 4.
LetR:Ω⟶Ω be a mapping on a complete S-metric-like space Ω,S satisfying the following:(a)
R is triangular α,βs-admissible.(b)
There existsθ0∈Ω such that αθ0,θ0,Rθ0≥1 and βθ0,θ0,Rθ0≥1.(c)
R is generalised α,βs-admissible Z-contraction of type I on Ω,S.(d)
Ifθn is a sequence in Ω such that αθn,θn,θn+1≥1 and βθn,θn,θn+1≥1 for all n and θn⟶ξ∈Ω as n⟶+∞, then there exists a subsequence θnk of θn such that αθnk,θnk,ξ≥1 and βθnk,θnk,ξ≥1 for all k∈ℕ.
Then, there is a unique fixed pointξ∈Ω of R with Sξ,ξ,ξ=0.Proof.
Proceeding as of Theorem3, let θn be a sequence in Ω given by θn+1=Rθn converges to some ξ∈Ω with αθn,θn,θn+1≥1 and βθn,θn,θn+1≥1 for all n and Sξ,ξ,ξ=0. From d, there exists a subsequence θnk of θn such that αθnk,θnk,ξ≥1 and βθnk,θnk,ξ≥1 for all k∈ℕ. Thus, applying (33) for all k, we have(62)0≤ζαθnk,θnk,ξβθnk,θnk,ξMθnk,θnk,ξ,Sθnk,θnk,ξ<Sθnk,θnk,ξ−αθnk,θnk,ξβθnk,θnk,ξMθnk,θnk,ξ,where(63)Mθnk,θnk,ξ=maxSθnk,θnk,ξ,Sθnk,θnk,Rθnk,Sξ,ξ,Rξ=maxSθnk,θnk,ξ,Sθnk,θnk,θnk+1,Sξ,ξ,Rξ.
From (62), we get(64)Mθnk,θnk,ξ≤αθnk,θnk,ξβθnk,θnk,ξMθnk,θnk,ξ<Sθnk,θnk,ξ.
Takingk⟶+∞, we have Sξ,ξ,Rξ=0, that is, Rξ=ξ. Similar to Theorem 3 uniqueness of fixed point of R can be proved. □
Next, we give the following definition.Definition 15.
LetR:Ω⟶Ω be a mapping on a S-metric-like space Ω,S and α,β:Ω×Ω×Ω⟶ℝ+. R is said to be a generalised rational α,βs-admissible Z-contraction of type I with respect to ζ if(65)ζαθ,μ,ξβθ,μ,ξMrθ,μ,ξ,Sθ,μ,ξ≥0,where(66)Mrθ,μ,ξ=maxSθ,μ,ξ,SRθ,Rμ,Rξ,Sθ,θ,RθSμ,μ,Rμ1+Sθ,μ,ξ+SRθ,Rμ,Rξ,Sμ,μ,RμSξ,ξ,Rξ1+Sθ,μ,ξ+SRθ,Rμ,Rξ,Sξ,ξ,RξSθ,θ,Rθ1+Sθ,μ,ξ+SRθ,Rμ,Rξ,for all θ,μ,ξ∈Ω.Definition 16.
LetR:Ω⟶Ω be a mapping on a S-metric-like space Ω,S and α,β:Ω×Ω×Ω⟶ℝ+. R is said to be a generalised rational α,βs-admissible Z-contraction of type II with respect to ζ if(67)ζαθ,θ,μβθ,θ,μMrθ,θ,μ,Sθ,θ,μ≥0,where(68)Mrθ,θ,μ=maxSθ,θ,μ,SRθ,Rθ,Rμ,Sθ,θ,RθSθ,θ,Rθ1+Sθ,θ,μ+SRθ,Rθ,Rμ,Sθ,θ,RθSμ,μ,Rμ1+Sθ,θ,μ+SRθ,Rθ,Rμ,for all θ,μ,ξ∈Ω.Theorem 5.
LetR:Ω⟶Ω be a mapping on a complete S-metric-like space Ω,S satisfying the following:(a)
R is triangular α,βs-admissible.(b)
There existsθ0∈Ω such that αθ0,θ0,Rθ0≥1 and βθ0,θ0,Rθ0≥1.(c)
R is a generalised rational α,βs-admissible Z-contraction of type I on Ω,S.(d)
R is S-continuous.
Then, there is a unique fixed pointξ∈Ω of R with Sξ,ξ,ξ=0.Proof.
Similar to Theorem3.Theorem 6.
LetR:Ω⟶Ω be a mapping on a complete S-metric-like space Ω,S satisfying the following:(a)
R is triangular α,βs-admissible.(b)
There existsθ0∈Ω such that αθ0,θ0,Rθ0≥1 and βθ0,θ0,Rθ0≥1.(c)
R is generalised rational α,βs-admissible Z-contraction of type I on Ω,S.(d)
Ifθn is a sequence in Ω such that αθn,θn,θn+1≥1 and βθn,θn,θn+1≥1 for all n and θn⟶ξ∈Ω as n⟶+∞, then there exists a subsequence θnk of θn such that αθnk,θnk,ξ≥1 and βθnk,θnk,ξ≥1 for all k∈ℕ.
Then, there is a unique fixed pointξ∈Ω of R with Sξ,ξ,ξ=0.Proof.
Similar to Theorem4.


## 3. Consequences

In this section, we give various results as consequences of the above results. First, we give a Banach type.Corollary 1.
LetR:Ω⟶Ω be a mapping on a complete S-metric-like space Ω,S satisfying the following:(a)
R is triangular α,βs-admissible.(b)
There existsθ0∈Ω such that αθ0,θ0,Rθ0≥1 and βθ0,θ0,Rθ0≥1.(c)
αθ,μ,ξβθ,μ,ξSRθ,Rμ,Rξ≤kSθ,μ,ξ for all θ,μ,ξ∈Ω and k∈0,1.(d)
R is S-continuous.
Then, there is a unique fixed pointξ∈Ω of R with Sξ,ξ,ξ=0.Proof.
Similar to Theorem1, considering(69)ζt,s=ks−t.Corollary 2.
LetR:Ω⟶Ω be a mapping on a complete S-metric-like space Ω,S satisfying the following:(a)
R is triangular α,βs-admissible.(b)
There existsθ0∈Ω such that αθ0,θ0,Rθ0≥1 and βθ0,θ0,Rθ0≥1.(c)
There exists a lower semicontinuous functionϕ:ℝ+⟶ℝ+ with ϕ−10=0 such that(70)αθ,μ,ξβθ,μ,ξSRθ,Rμ,Rξ≤Sθ,μ,ξ−ϕSθ,μ,ξ,for allθ,μ,ξ∈Ω.(d)
R is S-continuous.
Then, there is a unique fixed pointξ∈Ω of R with Sξ,ξ,ξ=0.Proof.
Consider(71)ζt,s=s−ϕs−t.
Takingαθ,μ,ξ=βθ,μ,ξ=1 for all θ,μ,ξ∈Ω in Theorem 1, we have the following.Corollary 3.
LetR:Ω⟶Ω be a mapping on a complete S-metric-like space Ω,S. Assume that there exists a simulation function ζ such that(72)ζSRθ,Rμ,Rξ,Sθ,μ,ξ≥0,for all θ,μ,ξ∈Ω. Then, ξ∈Ω is a unique fixed point of R with Sξ,ξ,ξ=0.Corollary 4.
LetR:Ω⟶Ω be a mapping on a complete S-metric-like space Ω,S satisfying the following:(a)
R is triangular α,βs-admissible.(b)
There existsθ0∈Ω such that αθ0,θ0,Rθ0≥1 and βθ0,θ0,Rθ0≥1.(c)
αθ,μ,ξβθ,μ,ξMRθ,Rμ,Rξ≤kSθ,μ,ξ for all θ,μ,ξ∈Ω and k∈0,1.(d)
R is S-continuous.
Then, there is a unique fixed pointξ∈Ω of R with Sξ,ξ,ξ=0.Proof.
Similar to Theorem3, considering, (73)ζt,s=ks−t.Corollary 5.
LetR:Ω⟶Ω be a mapping on a complete S-metric-like space Ω,S satisfying the following:(a)
R is triangular α,βs-admissible.(b)
There existsθ0∈Ω such that αθ0,θ0,Rθ0≥1 and βθ0,θ0,Rθ0≥1.(c)
There exists a lower semicontinuous functionϕ:ℝ+⟶ℝ+ with ϕ−10=0 such that(74)αθ,μ,ξβθ,μ,ξMRθ,Rμ,Rξ≤Sθ,μ,ξ−ϕSθ,μ,ξ,(i)
for allθ,μ,ξ∈Ω.(d)
R is S-continuous.
Then, there is a unique fixed pointξ∈Ω of R with Sξ,ξ,ξ=0.Proof.
Consider(75)ζt,s=s−ϕs−t.
If we consider in Theorem 20,αθ,μ,ξ=βθ,μ,ξ=1 for all θ,μ,ξ∈Ω, we haveCorollary 6.
LetR:Ω⟶Ω be a S-continuous mapping on a complete S-metric-like space Ω,S. Suppose that there exists a simulation function ζ such that(76)ζMRθ,Rμ,Rξ,Sθ,μ,ξ≥0,for all θ,μ,ξ∈Ω. Then, there is a unique fixed point ξ∈Ω of R with Sξ,ξ,ξ=0.Corollary 7.
LetR:Ω⟶Ω be a mapping on a complete S-metric-like space Ω,S satisfying the following:(a)
R is triangular α,βs-admissible.(b)
There existsθ0∈Ω such that αθ0,θ0,Rθ0≥1 and βθ0,θ0,Rθ0≥1.(c)
αθ,μ,ξβθ,μ,ξMrRθ,Rμ,Rξ≤kSθ,μ,ξ for all θ,μ,ξ∈Ω and θ∈0,1.(d)
R is S-continuous.
Then, there is a unique fixed pointξ∈Ω of R with Sξ,ξ,ξ=0.Proof.
Similar to Theorem5, considering,(77)ζt,s=ks−t.Corollary 8.
LetR:Ω⟶Ω be a mapping on a complete S-metric-like space Ω,S satisfying the following:(a)
R is triangular α,βs-admissible.(b)
There existsθ0∈Ω such that αθ0,θ0,Rθ0≥1 and βθ0,θ0,Rθ0≥1.(c)
There exists a lower semicontinuous functionϕ:ℝ+⟶ℝ+ with ϕ−10=0 such that(78)αθ,μ,ξβθ,μ,ξMrRθ,Rμ,Rξ≤Sθ,μ,ξ−ϕSθ,μ,ξ,for allθ,μ,ξ∈Ω.(d)
R is S-continuous.
Then, there is a unique fixed pointξ∈Ω of R with Sξ,ξ,ξ=0.Proof.
It suffices to take(79)ζt,s=s−ϕs−t.
If we consider in Theorem5, αθ,μ,ξ=βθ,μ,ξ=1 for all θ,μ,ξ∈Ω, we have the following.Corollary 9.
LetR:Ω⟶Ω be a S-continuous mapping on a complete S-metric-like space Ω,S. Suppose that there exists a simulation function ζ such that(80)ζMrRθ,Rμ,Rξ,Sθ,μ,ξ≥0,for all θ,μ,ξ∈Ω. Then, there is a unique fixed point ξ∈Ω of R with Sξ,ξ,ξ=0.
We take the following example.Example 2.
SupposeΩ=0,+∞, Sθ,μ,ξ=θ+μ+μ+ξ for all θ,μ,ξ∈Ω and R:Ω⟶Ω as(81)Rθ=θ4,if0≤θ≤1,4θ,otherwise.
Consider(82)ζt,s=cs−t,where0≤14<c<1.
Letα,β:Ω×Ω×Ω⟶ℝ+ be defined as(83)αθ,μ,ξ=43,if0≤θ,μ,ξ≤1,0,otherwise,βθ,μ,ξ=32,if0≤θ,μ,ξ≤1,0,otherwise.
We will verify Corollary1. Here, Ω,S is a complete S-metric-like space. Let θ,μ,ξ∈Ω such that αθ,μ,ξ≥1 and βθ,μ,ξ≥1. Then, θ,μ,ξ∈0,1 and so Rθ∈0,1, Rμ∈0,1, Rξ∈0,1, and αRθ,Rμ,Rξ≥1 and βRθ,Rμ,Rξ≥1. Hence, R is triangular α,βs-admissible. When θ0=1, condition b is true, θ0=Rnθ1=1/n satisfies condition d.
If0≤θ≤1, then αθ,μ,ξ=4/3 and βθ,μ,ξ=3/2. We have(84)ζαθ,μ,ξβθ,μ,ξSRθ,fμ,fξ,Sθ,μ,ξ=cSθ,μ,ξ−αθ,μ,ξβθ,μ,ξSfθ,fμ,fξ=34θ+μ+μ+ξ−2.14θ+μ+μ+ξ=14θ+μ+μ+ξ.
If0≤μ≤1 and ξ>1, then ζαθ,μ,ξβθ,μ,ξSfθ,fμ,fξ,Sθ,μ,ξ≥0 since αθ,μ,ξ=βθ,μ,ξ=0. Thus, ξ=0 is the unique fixed point of f.
We also notice that (72) is not satisfied. In fact, for θ=1, μ=2, ξ=3, we get(85)SR1,R2,R3=S14,8,12=14+8+8+12=334+20=1134>8=S1,2,3.


## 4. Conclusion

In this paper, we presentS-metric-like space and some of its topological properties. Also, we present some new α,β-contraction-type mappings under simulation function by extending the definition of α,β-admissible mappings and prove some fixed point results in the setting of S-metric-like space. We also open the scope for extending various contractions in S-metric-like space.



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