# Application of Some Fixed-Point Theorems in Orthogonal Extended S-Metric Spaces

Authors: Mohammad Mahdi Rezaei; Shaban Sedghi; Vahid Parvaneh
Journal: Journal of Mathematics
 (2021)

---

## Abstract

In this study, we obtain some coincidence point theorems for weaklyO-α-admissible contractive mappings in an orthogonal extended S-metric space. An example and an application are provided to illustrate the usability of the obtained results. Our results generalize the results of several studies from metric and S-metric frameworks to the setting of orthogonal extended S-metric spaces.


---

## Body

## 1. Introduction

In 1922, Banach [1] proposed a theorem known as the Banach fixed point theorem (or Banach contraction principle or BCP for short) to prove the existence of solutions for nonlinear operator equations and integral equations. Since then, it has become a very popular tool due to its simplicity and practicality in solving a variety of problems such as control theory, economic theory, nonlinear analysis, and global analysis. Later, there is a large body of literature on the applications, generalizations, and expansions of this theorem. In [2], for example, the authors introduced the definition of a cyclic p-contraction pair for single-valued mappings and presented some best proximity point results for such mappings on a proximally complete pair of subsets in a metric space. In [3], Dehici and Redjel obtained some fixed-point results for nonexpansive mappings defined on an intersection of a finite number of closed bounded and convex nonempty subsets in Banach spaces. These research studies have been done by several authors in different directions, for example, by weakening the hypothesis, using different settings, considering different contractions, and so on.Many mathematical problems require finding distances between two or more objects that are not easy to measure exactly in general. There are different approaches to get the right concept of a metric structure. Due to the need to create an appropriate framework for modeling several salient problems of a practical nature, the study of metric spaces has attracted the attention of many researchers. In the last few decades, a number of generalizations of metric spaces in several articles, such as 2-metric spaces [4], G-metric spaces [5], D∗-metric spaces [4], partial metric spaces [6], and cone metric spaces [7], have been illustrated.Sedghi et al. [8] introduced the concept of an S metric space and proved that this concept is a generalization of G-metric space and D∗-metric space. Also, they have proven some properties on S-metric spaces and some fixed-point theorems for mappings defined on an S-metric space.The concept ofb-metric space, as one of the useful generalizations of standard metric spaces, was first used by Bakhtin in [9] and Czerwik in [10]. This idea has been extended by Parvaneh and Ghoncheh [11] to the p-metric settings.The concept of weak contraction mapping was introduced by Alber and Guerre-Delabrere in [12] in the arrangement of Hilbert spaces. Rhoades [13] proved that every weak contraction mapping on a complete metric space has a unique fixed point. Zhang and Sang [14] have generalized this concept to weakly contractive pairs.In this study, we prove some coincidence point results forα-admissible π,ϖ-contractive mappings in orthogonal extended S-metric spaces where we were motivated by the work in [15]. Using an auxiliary function, commonly referred to as α, in the article [16] by Samet et al., which continued in several articles, we want to define some related concepts such as orthogonal α-compatibility and orthogonal α-regularity.


## 2. Preliminaries

From now on,(1)Ω=ω:0,∞⟶0,∞:ωis strictly increasing, onto, continuous andω−1t≤t≤ωt.Definition 1 (see [11]).
Letℒ be a nonempty set. A function d¯:ℒ×ℒ⟶0,∞ is an extended b-metric (p-metric, for short) if there exists an ω∈Ω, such that(1)
d¯ℓ,ℓ′=0 iff ℓ=ℓ′(2)
d¯ℓ,ℓ′=d¯ℓ′,ℓ(3)
d¯ℓ,ℓ″≤ωd¯ℓ,ℓ′+d¯ℓ′,ℓ″
for allℓ,ℓ′,ℓ″∈ℒ.
In this case, the pairℒ,d¯ is called an extended b-metric space or, briefly, a p-metric space.
Ab-metric is a p-metric when ωt=st for some fixed s≥1, and a metric is a p-metric when ωt=t. The following example shows that a p-metric need not be a b-metric.Example 1 (See [11]).
Letℒ,d be a b-metric space with coefficient s≥1 and let d¯ℓ,ℓ′=sinhdℓ,ℓ′.
For eachℓ,ℓ′,ℓ″∈ℒ, we have(2)d¯ℓ,ℓ″=sinhdℓ,ℓ″≤sinhsdℓ,ℓ′+sdℓ′,ℓ″≤sinhssinhdℓ,ℓ′+ssinhdℓ′,ℓ″=ωd¯ℓ,ℓ′+d¯ℓ′,ℓ″.
So,d¯ is a p-metric with ωt=sinhs⋅t.
Note thatsinhℓ−ℓ′ is not a metric on ℝ, as, e.g.,(3)sinh5≈74.203≰3.627+10.0179≈sinh2+sinh3.
Similarly, althoughdℓ,ℓ′=ℓ−ℓ′2 is a b-metric on ℝ with s=2, there is no s≥1, such that d¯ℓ,ℓ′=sinhℓ−ℓ′2 is a b-metric with parameter s. Indeed, putting ℓ″=0 and ℓ′=1, we should have sinhℓ2≤ssinhℓ−12+sinh1 which cannot hold for any fixed s and ℓ sufficiently large.Definition 2 (see [8]).
Letℒ be a nonempty set. An S-metric on ℒ is a function S:ℒ3⟶0,∞ so that(1)
Sℓ,ℓ′,ℓ″≥0(2)
Sℓ,ℓ′,ℓ″=0 if and only if ℓ=ℓ′=ℓ″(3)
Sℓ,ℓ′,ℓ″≤Sℓ,ℓ,a+Sℓ′,ℓ′,a+Sℓ″,ℓ″,a
For allℓ,ℓ′,ℓ″,a∈ℒ.
The pairℒ,S is called an S-metric space.Example 2.
(see [8]).(1)
Letℒ=ℝn and . be a norm on ℒ. Then, Sℓ,ℓ′,ℓ″=ℓ′+ℓ″−2ℓ+ℓ′−ℓ″ is an S-metric on ℒ.(2)
Letℒ=ℝn and . be a norm on ℒ. Then, Sℓ,ℓ′,ℓ″=ℓ−ℓ″+ℓ′−ℓ″ is an S-metric on ℒ.(3)
Letℒ be a nonempty set and d be an ordinary metric on ℒ. Then, Sℓ,ℓ′,ℓ″=dℓ,ℓ′+dℓ′,ℓ″ is an S-metric on ℒ.Definition 3 (see [17]).
Letℒ be a nonempty set and s≥1 be a given real number. Suppose that a mapping S:ℒ×ℒ×ℒ⟶0,∞ satisfies(1)
Sℓ,ℓ′,ℓ″≥0(2)
Sℓ,ℓ′,ℓ″=0 if and only if ℓ=ℓ′=ℓ″(3)
Sℓ,ℓ′,ℓ″≤sSℓ,ℓ,a+Sℓ′,ℓ′,a+Sℓ″,ℓ″,a
For allℓ,ℓ′,ℓ″,a∈ℒ.
Then,S is called a Sb-metric, and the pair ℒ,S is called a Sb-metric space.Definition 4 (see [18]).
Letℒ be a nonempty set and ω∈Ω. Suppose that s¯:ℒ×ℒ×ℒ⟶0,∞ satisfies(1)
s¯ℓ,ℓ′,ℓ″≥0(2)
s¯ℓ,ℓ′,ℓ″=0 if and only if ℓ=ℓ′=ℓ″(3)
s¯ℓ,ℓ′,ℓ″≤ωs¯ℓ,ℓ,a+s¯ℓ′,ℓ′,a+s¯ℓ″,ℓ″,a
For allℓ,ℓ′,ℓ″,a∈ℒ.
Then,s¯ is called an extended S-metric (s¯-metric) and the pair ℒ,s¯ is called an extended S-metric space (s¯-metric space).
EachS-metric space is an s¯-metric space (s¯-MS) with ωt=t and every Sb-MS with parameter s≥1 is an s¯-MS with ωt=st.Remark 1.
Ins¯-MS, we have s¯ℓ,ℓ′,ℓ′≤ω2s¯ℓ′,ℓ′,ℓ and s¯ℓ,ℓ,ℓ′≤ωs¯ℓ′,ℓ′,ℓ for all ℓ,ℓ′∈ℒ, i.e, any s¯-metric is not symmetry in general.Proposition 1 (see [18]).
Letℒ,S be Sb-MS with coefficient s≥1 and let s¯ℓ,ℓ′,ℓ″=ζSℓ,ℓ′,ℓ″, where ζ∈Ω. Then, s¯ is an extended S-metric with ωt=ζst.Proof.
For allℓ,ℓ′,ℓ″,a∈ℒ,(4)s¯ℓ,ℓ′,ℓ″=ζSℓ,ℓ′,ℓ″≤ζsSℓ,ℓ,a+sSℓ′,ℓ′,a+sSℓ″,ℓ″,a≤ζsζSℓ,ℓ,a+sζSℓ′,ℓ′,a+sζSℓ″,ℓ″,a=ζss¯ℓ,ℓ,a+ss¯ℓ′,ℓ′,a+ss¯ℓ″,ℓ″,a=ωs¯ℓ,ℓ,a+s¯ℓ′,ℓ′,a+s¯ℓ″,ℓ″,a.
Therefore,s¯ is an extended S-metric with ωt=ζst.Example 3 (see [18]).
Letℒ,S be Sb-MS with coefficient s≥1. Then, s¯ℓ,ℓ′,ℓ″=eSℓ,ℓ′,ℓ″sec−1eSℓ,ℓ′,ℓ″, s¯ℓ,ℓ′,ℓ″=Sℓ,ℓ′,ℓ″+1sec−1Sℓ,ℓ′,ℓ″+1, s¯ℓ,ℓ′,ℓ″=eSℓ,ℓ′,ℓ″tan−1eSℓ,ℓ′,ℓ″−1, s¯ℓ,ℓ′,ℓ″=Sℓ,ℓ′,ℓ″coshSℓ,ℓ′,ℓ″, and s¯ℓ,ℓ′,ℓ″=eSℓ,ℓ′,ℓ″ln1+Sℓ,ℓ′,ℓ″, and s¯ℓ,ℓ′,ℓ″=Sℓ,ℓ′,ℓ″+ln1+Sℓ,ℓ′,ℓ″ are some s¯-metrics.Definition 5 (see [18]).
Letℒ,S be s¯-MS and ω:0,∞⟶0,∞. The open ball Bs¯ℓ,r and the closed ball Bs¯ℓ,r with center ℓ and radius r are as follows:(5)Bs¯ℓ,r=ℓ′∈ℒ:s¯ℓ′,ℓ′,ℓ<r,Bs¯ℓ,r=ℓ′∈ℒ:s¯ℓ′,ℓ′,ℓ≤r.Example 4 (see [18]).
Letℒ=ℝ. Denote by s¯ℓ,ℓ′,ℓ″=ℓ′+ℓ″−2ℓ+ℓ′−ℓ″ for all ℓ,ℓ′,ℓ″∈ℝ and let ωℓ=ℓ. Thus,(6)Bs¯1,2=ℓ′∈ℝ:s¯ℓ′,ℓ′,1<2=ℓ′∈ℝ:ℓ′−1<1=ℓ′∈ℝ:0<ℓ′<2=0,2.Definition 6 (see [18]).
Letℒ be s¯-MS. A sequence ℓn in ℒ is said to be(1)
s¯-Cauchy if, for each 0<ε, there exists a positive integer n0, such that for all n,m≥n0, s¯ℓm,ℓn,ℓn<ε(2)
s¯-convergent to a point ℓ∈ℒ if, for each 0<ε, there exists a positive integer n0, such that for all n≥n0, s¯ℓn,ℓn,ℓ<ε(3)
Ans¯-MS ℒ is called s¯-complete, if every s¯-Cauchy sequence is s¯-convergent in ℒ.
Recently, Eshaghi Gordji et al. [15] introduced the notion of orthogonal sets.Definition 7.
(see [15]). Let ⊥⊆ℒ×ℒ be a binary relation. If there exists x0∈ℒ, such that y⊥x0 for all y∈ℒ or x0⊥y for all y∈ℒ, then ℒ is called an orthogonal set (briefly, O-set). We denote this O-set by ℒ,⊥.Definition 8 (see [15]).
Letℒ,⊥ be an O-set. A sequence ℓn is called an orthogonal sequence (briefly, O-sequence) if xn⊥xn+1 or xn+1⊥xn for all n∈ℕ.Definition 9 (see [15]).
Letℒ,⊥,d be an orthogonal set with metric d. Then, ℒ is orthogonally complete (briefly, O-complete) if every Cauchy O-sequence is convergent.
Orthogonal continuous mappings and other concepts can be defined as above.
Motivated by [11], we present the following definitions.Definition 10.
Letℒ,s¯,ω,⊥ be an O-s¯-MS, Υ be a self-mapping on ℒ, and let α:ℒ×ℒ×ℒ⟶0,+∞ be a function.(1)
Υ is an O-α-admissible mapping if(7)ℓ⊥ℓ′⊥ℓ″,αℓ,ℓ′,ℓ″≥1⇒ϒℓ⊥ϒℓ′⊥ϒℓ″,αϒℓ,ϒℓ′,ϒℓ″≥1,for allℓ,ℓ′,ℓ″∈ℒ.(2)
The spaceℒ,s¯,ω,⊥ is said to be O-α-complete if every Cauchy O-sequence ℓn in ℒ with αℓn,ℓn+1,ℓn+2≥1 for all n∈ℕ converges in ℒ(3)
ϒ is an O-α-continuous mapping on ℒ,s¯,ω,⊥, if ℓn⟶ℓ as n⟶∞, ℓn⊥ℓn+1⊥ℓn+2, and αℓn,ℓn+1,ℓn+2≥1 for all n∈ℕ⇒ϒℓn⟶ϒℓ, for given ℓ∈ℒ and arbitrary sequence ℓn⊆ℒ(4)
ℒ,s¯,ω,⊥ is O-α-regular if ℓn⟶ℓ, ℓn⊥ℓn+1⊥ℓn+2, and αℓn,ℓn+1,ℓn+2≥1 for all n∈ℕ; then, αℓn,ℓn+1,ℓ≥1 for all n∈ℕ.
Letℒ be a nonempty set and Γ:ℒ⟶ℒ be a given mapping. For every ℓ∈ℒ, let Γ−1ℓ=u∈ℒ:Γu=ℓ.
Motivated by the work in [19], we present the following definition.Definition 11.
Letℒ be a set, Γ1,Γ2,h:ℒ⟶ℒ be mappings such that Γ1ℒ∪Γ2ℒ⊆hℒ, and let α:ℒ×ℒ×ℒ⟶0,∞ be a function. The ordered pair Γ1,Γ2 is said to be(i)
WeaklyO-α-h-admissible if for all ℓ∈ℒ,αΓ1ℓ,Γ1ℓ,Γ2ℓ′≥1 and Γ1ℓ⊥Γ2ℓ′ for all ℓ′∈h−1Γ1ℓ, and αΓ2ℓ,Γ2ℓ,Γ1ℓ′≥1 and Γ2ℓ⊥Γ1ℓ′ for all ℓ′∈h−1Γ2ℓ,(ii)
Partially weaklyO-α-h-admissible if for all ℓ∈ℒ, αΓ1ℓ,Γ1ℓ,Γ2ℓ′≥1 and Γ1ℓ⊥Γ2ℓ′ for all ℓ′∈h−1Γ1ℓ(iii)
The ordered pairΓ1,Γ2 is said to be rectangular weakly O-α-h-admissible (rectangular partially weakly O-α-admissible) if it is weakly O-α-admissible (partially weakly O-α-admissible) with respect to h and αℓ,ℓ,ℓ″≥1, ℓ⊥ℓ′′, αℓ″,ℓ″,ℓ′≥1, and ℓ″⊥ℓ′ imply αℓ,ℓ,ℓ′≥1 and ℓ⊥ℓ′ for all ℓ,ℓ′,ℓ″∈ℒ.
If, in the previous conditions,h=Iℒ (the identity mapping), then we can omit the function h in the respective notions.Definition 12.
Letℒ,s¯,ω,⊥ be an O-s¯-MS and Γ1,Γ2:ℒ⟶ℒ. The pair Γ1,Γ2 is said to be O-α-compatible if limn⟶∞SΓ1Γ2ℓn,Γ1Γ2ℓn,Γ2Γ1ℓn=0, whenever ℓn is an O-sequence in ℒ, such that αℓn,ℓn,ℓn+1≥1 for all n∈ℕ and limn⟶∞Γ1ℓn=limn⟶∞Γ2ℓn=t for some t∈ℒ.
In general, ans¯-metric mapping s¯ℓ,ℓ′,ℓ″ with nontrivial function ω need not be jointly continuous in all its variables [18]. Thus, in some proofs, we will need the following simple lemma about the s¯-convergent sequences.Lemma 1.
Let aℒ,s¯,ω,⊥ be O-s¯-MS.(1)
Suppose thatℓn and ℓn′ are s¯-convergent to ℓ and ℓ′, respectively. Then, we have(8)ω−2s¯ℓ,ℓ,ℓ′≤liminfn⟶∞s¯ℓn,ℓn,ℓn′≤limsupn⟶∞s¯ℓn,ℓn,ℓn′≤ω2s¯ℓ,ℓ,ℓ′.In particular, ifℓ=ℓ′, then we have limn⟶∞s¯ℓn,ℓn′,ℓn′=0.(2)
Suppose thatℓn is s¯-convergent to ℓ and ℓ″∈ℒ is arbitrary. Then, we have(9)ω−2s¯ℓ″,ℓ″,ℓ≤liminfn⟶∞s¯ℓ″,ℓ″,ℓn≤limsupn⟶∞s¯ℓ″,ℓ″,ℓn≤ω2s¯ℓ″,ℓ″,ℓn.(3)
Suppose thatℓn is s¯-convergent to ℓ and ℓ′′∈ℒ is arbitrary. Then, we have(10)ω−2s¯ℓ,ℓ,ℓ″≤lim infn⟶∞s¯ℓn,ℓn,ℓ″≤lim supn⟶∞s¯ℓn,ℓn,ℓ″≤ω2s¯ℓ,ℓ,ℓ″.Proof.
(1)
Using the rectangle inequality in thes¯-metric space, it is easy to see that(11)s¯ℓ,ℓ,ℓ′≤ω2s¯ℓ,ℓ,ℓn+s¯ℓ′,ℓ′,ℓn≤ω2s¯ℓ,ℓ,ℓn+ω2s¯ℓ′,ℓ′,ℓn′+s¯ℓn,ℓn,ℓn′,s¯ℓn,ℓn,ℓn′≤ω2s¯ℓn,ℓn,ℓ+s¯ℓn′,ℓn′,ℓ≤ω2s¯ℓn,ℓn,ℓ+ω2s¯ℓn′,ℓn′,ℓ′+s¯ℓ,ℓ,ℓ′.Makingn⟶∞ in the above inequalities, we acquire the desired result.(2)
According to the rectangle inequality, we see that(12)s¯ℓ″,ℓ″,ℓ≤ωs¯ℓ,ℓ,ℓ″≤ω22s¯ℓ,ℓ,ℓn+s¯ℓ″,ℓ″,ℓn,s¯ℓ″,ℓ″,ℓn≤ω2s¯ℓ″,ℓ″,ℓ+s¯ℓn,ℓn,ℓ.(3)
According to the rectangle inequality, we see that(13)s¯ℓn,ℓn,ℓ″≤ω2s¯ℓn,ℓn,ℓ+s¯ℓ″,ℓ″,ℓ≤ω2s¯ℓn,ℓn,ℓ+ωs¯ℓ,ℓ,ℓ″,s¯ℓ,ℓ,ℓ″≤ω2s¯ℓ,ℓ,ℓn+s¯ℓ″,ℓ″,ℓn≤ω2s¯ℓ,ℓ,ℓn+ωs¯ℓn,ℓn,ℓ″.
Recall that the self-mappingsΓ1,Γ2 of a set ℒ are said to be weakly compatible if Γ1Γ2ℓ=Γ2Γ1ℓ, whenever Γ1ℓ=Γ2ℓ [20].
The functionϖ:0,+∞⟶0,+∞ is called an altering distance function (a.d.f) [21], if(1)
ϖ is continuous and nondecreasing(2)
ϖt=0 iff t=0


## 3. Main Results

Letℒ,s¯,ω,⊥ be O-s¯-MS, and let Γ1,Γ2,ϒ1,ϒ2:ℒ⟶ℒ be four self-mappings. Throughout this study, unless otherwise stated, for all ℓ,ℓ′∈ℒ, Mℓ,ℓ,ℓ′ will denote an arbitrary element of the set(14)s¯ϒ2ℓ,ϒ2ℓ,ϒ1ℓ′,ω−1s¯ϒ2ℓ,ϒ2ℓ,Γ1ℓ+s¯ϒ1ℓ′,ϒ1ℓ′,Γ2ℓ′2,ω−1s¯ϒ2ℓ,ϒ2ℓ,Γ2ℓ′+s¯ϒ1ℓ′,ϒ1ℓ′,Γ1ℓ2.Theorem 1.
Letℒ,s¯,ω,⊥ be an O-α-complete O-s¯-MS, α:ℒ×ℒ×ℒ⟶0,∞, and let Γ1,Γ2,ϒ1,ϒ2:ℒ⟶ℒ be such that Γ1ℒ⊆ϒ1ℒ and Γ2ℒ⊆ϒ2ℒ.
Suppose that for allℓ,ℓ′∈ℒ with αϒ1ℓ,ϒ1ℓ,ϒ2ℓ′≥1 and ϒ1ℓ⊥ϒ2ℓ′,(15)πωs¯Γ1ℓ,Γ1ℓ,Γ2ℓ′≤πMℓ,ℓ,ℓ′−ϖMℓ,ℓ,ℓ′,where π,ϖ are two a.d.f. Assume that Γ1,Γ2,ϒ1, and ϒ2 are O-α-continuous, the pairs Γ1,ϒ2 and Γ2,ϒ1 are O-α-compatible, and the pairs Γ1,ϒ2 and Γ2,ϒ1 are rectangular partially weakly O-α-ϒ1-admissible and O-α-ϒ2-admissible, respectively. Then, the pairs Γ1,ϒ2 and Γ2,ϒ1 have a coincidence point ℓ″ in ℒ. Moreover, if αϒ2ℓ″,ϒ2ℓ″,ϒ1ℓ″≥1 and ϒ2ℓ″⊥ϒ1ℓ″, then ℓ′′ is a coincidence point of Γ1,Γ2,ϒ1, and Υ2.Proof.
Letℓ0∈ℒ. Choose ℓ1∈ℒ, such that Γ1ℓ0=Υ1ℓ1, and ℓ2∈ℒ, such that Γ2ℓ1=ϒ2ℓ2. Continuing in this way, we construct a sequence ℓn″ defined by(16)ℓ2n+1″=ϒ1ℓ2n+1=Γ1ℓ2n,ℓ2n+2″=ϒ2ℓ2n+2=Γ2ℓ2n+1,for all n≥0. As ℓ1∈Υ1−1Γ1ℓ0 and ℓ2∈Υ2−1Γ2ℓ1 and the pairs Γ1,Γ2 and Γ2,Γ1 are partially weakly O-α-Υ1-admissible and O-α-Υ2-admissible, respectively, we have(17)αϒ1ℓ1,ϒ1ℓ1,ϒ2ℓ2=αΓ1ℓ0,Γ1ℓ0,Γ2ℓ1≥1ϒ1ℓ1⊥ϒ2ℓ2Γ1ℓ0⊥Γ2ℓ1,αϒ2ℓ2,ϒ2ℓ2,ϒ1ℓ3=αΓ2ℓ1,Γ2ℓ1,Γ1ℓ2≥1ϒ2ℓ2⊥ϒ1ℓ3Γ2ℓ1⊥Γ1ℓ2.
Continuing this way, we acquire that(18)αϒ1ℓ2n+1,ϒ1ℓ2n+1,ϒ2ℓ2n+2=αℓ2n+1″,ℓ2n+1″,ℓ2n+2″≥1ϒ1ℓ2n+1⊥ϒ2ℓ2n+2ℓ2n+1″⊥ℓ2n+2″,αϒ2ℓ2n+2,ϒ2ℓ2n+2,ϒ1ℓ2n+3=αℓ2n+2″,ℓ2n+2″,ℓ2n+3″≥1,ϒ2ℓ2n+2⊥ϒ1ℓ2n+3ℓ2n+2″⊥ℓ2n+3″,for all n≥0.
Three steps will be taken to finish the proof.Step 1.
We will prove thatlimk⟶∞s¯ℓk″,ℓk″,ℓk+1″=0.
Lets¯k=s¯ℓk″,ℓk″,ℓk+1″. Suppose that s¯k0=0 for some k0. Then, ℓk0″=ℓk0+1″.
Ifk0=2n, then ℓ2n″=ℓ2n+1″ gives ℓ2n+1″=ℓ2n+2″. Indeed,(19)πωs¯ℓ2n+1′′,ℓ2n+1′′,ℓ2n+2′′=πωs¯Γ1ℓ2n,Γ1ℓ2n,Γ2ℓ2n+1≤πMℓ2n,ℓ2n,ℓ2n+1−ϖMℓ2n,ℓ2n,ℓ2n+1, where(20)Mℓ2n,ℓ2n,ℓ2n+1∈s¯ϒ2ℓ2n,ϒ2ℓ2n,ϒ1ℓ2n+1,ω−1s¯ϒ2ℓ2n,ϒ2ℓ2n,Γ1ℓ2n+s¯ϒ1ℓ2n+1,ϒ1ℓ2n+1,Γ2ℓ2n+12,ω−1s¯ϒ2ℓ2n,ϒ2ℓ2n,Γ2ℓ2n+1+s¯ϒ1ℓ2n+1,ϒ1ℓ2n+1,Γ1ℓ2n2=s¯ℓ2n″,ℓ2n″,ℓ2n+1″,ω−1s¯ℓ2n″,ℓ2n″,ℓ2n+1″+s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″2,ω−1s¯ℓ2n″,ℓ2n″,ℓ2n+2″+s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″2=0,ω−1s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″2,ω−1s¯ℓ2n″,ℓ2n″,ℓ2n+2″2.
IfMℓ2n,ℓ2n,ℓ2n+1=ω−1s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″/2, then (19) becomes(21)πωs¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″≤πω−1s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″2−ϖω−1s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″2≤πωs¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″−ϖω−1s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″2,which implies that ϖω−1s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″/2=0, that is, ℓ2n″=ℓ2n+1″=ℓ2n+2″.
IfMℓ2n,ℓ2n,ℓ2n+1=ω−1s¯ℓ2n″,ℓ2n″,ℓ2n+2″/2, then (19) becomes(22)πωs¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″≤πω−1s¯ℓ2n″,ℓ2n″,ℓ2n+2″2−ϖω−1s¯ℓ2n″,ℓ2n″,ℓ2n+2″2≤πs¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″−ϖω−1s¯ℓ2n″,ℓ2n″,ℓ2n+2″2,which implies that ϖω−1s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″/2=0, that is, ℓ2n″=ℓ2n+1″=ℓ2n+2″.
Similarly, ifk0=2n+1, then ℓ2n+1″=ℓ2n+2″ gives ℓ2n+2″=ℓ2n+3″. Continuing this process, we find that ℓk″ is a constant sequence for k≥k0. Hence, limk⟶∞s¯ℓk″,ℓk″,ℓk+1″=0 is correct.
Now, suppose that(23)s¯k=s¯ℓk″,ℓk″,ℓk+1″>0,for all k. We show that(24)s¯ℓk″,ℓk″,ℓk+1″≤s¯ℓk″,ℓk″,ℓk+1″,for each k=1,2,3,⋯.
Letk=2n and s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″≥s¯ℓ2n″,ℓ2n″,ℓ2n+1″>0, for n≥0. Then, as αΥ2ℓ2n,Υ2ℓ2n,Υ1ℓ2n+1≥1 and Υ2ℓ2n⊥Υ1ℓ2n+1, using (15), we acquire that(25)πωs¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″=πωs¯Γ1ℓ2n,Γ1ℓ2n,Γ2ℓ2n+1≤πMℓ2n,ℓ2n,ℓ2n+1−ϖMℓ2n,ℓ2n,ℓ2n+1,where(26)Mℓ2n,ℓ2n,ℓ2n+1∈s¯ϒ2ℓ2n,ϒ2ℓ2n,ϒ1ℓ2n+1,ω−1s¯ϒ2ℓ2n,ϒ2ℓ2n,Γ1ℓ2n+s¯ϒ1ℓ2n+1,ϒ1ℓ2n+1,Γ2ℓ2n+12,ω−1s¯ϒ2ℓ2n,ϒ2ℓ2n,Γ2ℓ2n+1+s¯ϒ1ℓ2n+1,ϒ1ℓ2n+1,ϒ1ℓ2n2=s¯ℓ2n″,ℓ2n″,ℓ2n+1″,ω−1s¯ℓ2n″,ℓ2n″,ℓ2n+1″+s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″2,ω−1s¯ℓ2n″,ℓ2n″,ℓ2n+2″2.
If(27)Mℓ2n,ℓ2n,ℓ2n+1=ω−1s¯ℓ2n″,ℓ2n″,ℓ2n+1″+s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″2≤ω−1s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″,
Ass¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″≥s¯ℓ2n″,ℓ2n″,ℓ2n+1″, then from (25), we have(28)πωs¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″≤πω−1s¯ℓ2n″,ℓ2n″,ℓ2n+1″+s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″2−ϖω−1s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″+s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″2≤πωs¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″−ϖω−1s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″2,which implies that ϖω−1s¯ℓ2n″,ℓ2n″,ℓ2n+1″+s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″/2≤0. This is possible only if s¯ℓ2n″,ℓ2n″,ℓ2n+1″+s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″/2=0, that is, s¯ℓ2n″,ℓ2n″,ℓ2n+1″=0, a contradiction to (23). Hence, s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″≤s¯ℓ2n″,ℓ2n″,ℓ2n+1″ for all n≥0.
Therefore, (24) is proved for k=2n.
Analogously,(29)s¯ℓ2n+2″,ℓ2n+2″,ℓ2n+3″≤s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″,for all n≥0.
Similarly, for other values ofMℓ2n,ℓ2n,ℓ2n+1, we can see that s¯ℓk″,ℓk″,ℓk+1″ is a nondecreasing sequence of nonnegative real numbers. Therefore,(30)limk⟶∞s¯ℓk″,ℓk″,ℓk+1″=r,for some r≥0.
We know that(31)Mℓ2n,ℓ2n,ℓ2n+1∈s¯ℓ2n″,ℓ2n″,ℓ2n+1″,ω−1s¯ℓ2n″,ℓ2n″,ℓ2n+1″+s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″2,ω−1s¯ℓ2n″,ℓ2n″,ℓ2n+2″+s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+1″2.
Substituting possible values ofMℓ2n,ℓ2n,ℓ2n+1 in (25) and making n⟶∞, we acquire that r=0. For instance, let(32)Mℓ2n,ℓ2n,ℓ2n+1=ω−1s¯ℓ2n″,ℓ2n″,ℓ2n+2″+s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+1″2.
Then, from (25), we have(33)πωs¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″≤πω−1s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″2−ϖω−1s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″2≤πs¯ℓ2n″,ℓ2n″,ℓ2n+2″+s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″2−ϖω−1s¯ℓ2n+1″,ℓ2n+1″,ℓ2n+2″2.
Lettingn⟶∞ in (33), using (30) and the continuity of π and ϖ, we have(34)ϖlimn⟶∞ω−1s¯ℓ2n″,ℓ2n″,ℓ2n+2″2=0.
Hence,limn⟶∞ω−1s¯ℓ2n″,ℓ2n″,ℓ2n+2″/2=0.
Now, taking into account (33) and letting n⟶∞, we find that πωr≤π0−ϖ0. Hence, r=0. Similarly, for the other values of Mℓ2n,ℓ2n,ℓ2n+1, we can show that(35)r=limk⟶∞s¯ℓk″,ℓk″,ℓk+1″=limn⟶∞s¯ℓ2n″,ℓ2n″,ℓ2n+1″=0.Step 2.
We will show thes¯-Cauchyness of ℓn″ in ℒ. Assume that for some ε>0, we have subsequences ℓ2mk″,ℓ2nk″⊆ℓ2n″, such that nk>mk≥k and(36)s¯ℓ2mk″,ℓ2mk″,ℓ2nk″≥ε,in which nk is the least index, i.e.,(37)s¯ℓ2mk″,ℓ2mk″,ℓ2nk−1″<ε.
From the rectangle inequality and (36) and (37), we have(38)ε≤s¯ℓ2mk″,ℓ2mk″,ℓ2nk″≤ωs¯ℓ2mk″,ℓ2mk″,ℓ2nk−1″+s¯ℓ2mk″,ℓ2mk″,ℓ2nk−1″+s¯ℓ2mk″,ℓ2mk″,ℓ2nk−1″.
Taking the limit ask⟶∞ in (38), from (35), we acquire that(39)ε≤limsupk⟶∞s¯ℓ2mk″,ℓ2mk″,ℓ2nk″≤ωε.
Using the rectangle inequality again, we have(40)s¯ℓ2mk″,ℓ2mk″,ℓ2nk″≤ωs¯ℓ2mk″,ℓ2mk″,ℓ2nk−1″+s¯ℓ2mk″,ℓ2mk″,ℓ2nk−1″+s¯ℓ2mk″,ℓ2mk″,ℓ2nk−1″.
Lettingk⟶∞ in the above inequality, we have(41)ω−1ε≤limsupk⟶∞s¯ℓ2mk+1″,ℓ2mk+1″,ℓ2nk″.
Finally,(42)s¯ℓ2mk″,ℓ2mk″,ℓ2nk−1″≤ωs¯ℓ2mk+1″,ℓ2mk+1″,ℓ2mk″+s¯ℓ2mk+1″,ℓ2mk+1″,ℓ2mk″+s¯ℓ2nk−1″,ℓ2nk−1″,ℓ2mk″.
Lettingk⟶∞ in the above and using (39), we have(43)limsupk⟶∞s¯ℓ2mk+1″,ℓ2mk+1″,ℓ2nk−1″≤ωε.
We know that2nk−1≥2mk and(44)αϒ2ℓ2n+2,ϒ2ℓ2n+2,ϒ1ℓ2n+1=αΓ2ℓ2n+1,Γ2ℓ2n+1,Γ1ℓ2n≥1,for all n∈ℕ. On the other hand, the pairs Γ1,Γ2 and Γ2,Γ1 are rectangular partially weakly O-α-admissible with respect to ϒ1 and ϒ2, respectively. So,(45)αϒ1ℓ2nk−1,ϒ1ℓ2nk−1,ϒ2ℓ2nk−2≥1,ϒ1ℓ2nk−1⊥ϒ2ℓ2nk−2,αϒ2ℓ2nk−2,ϒ2ℓ2nk−2,ϒ1ℓ2nk−3≥1,ϒ2ℓ2nk−2⊥ϒ1ℓ2nk−3,
which imply that(46)αϒ1ℓ2nk−1,ϒ1ℓ2nk−1,ϒ1ℓ2nk−3≥1,ϒ1ℓ2nk−1⊥ϒ1ℓ2nk−3.
Also,(47)αϒ1ℓ2nk−1,ϒ1ℓ2nk−1,ϒ1ℓ2nk−3≥1,ϒ1ℓ2nk−1⊥ϒ1ℓ2nk−3,αϒ1ℓ2nk−3,ϒ1ℓ2nk−3,ϒ2ℓ2nk−4≥1,ϒ1ℓ2nk−3⊥ϒ2ℓ2nk−4,
which imply that(48)αϒ1ℓ2nk−1,ϒ1ℓ2nk−1,ϒ2ℓ2nk−4≥1,ϒ1ℓ2nk−1⊥ϒ2ℓ2nk−4.
Continuing in this manner, we obtain thatαϒ1ℓ2nk−1,ϒ1ℓ2nk−1,ϒ2ℓ2mk≥1 and ϒ1ℓ2nk−1⊥ϒ2ℓ2mk. Now, we can apply (15) to acquire that(49)πωs¯ℓ2mk+1″,ℓ2mk+1″,ℓ2nk″=πωs¯Γ1ℓ2mk,Γ1ℓ2mk,Γ2ℓ2nk−1≤πMℓ2mk,ℓ2mk,ℓ2nk−1−ϖMℓ2mk,ℓ2mk,ℓ2nk−1,where(50)Mℓ2mk,ℓ2mk,ℓ2nk−1∈s¯ϒ2ℓ2mk,ϒ2ℓ2mk,ϒ1ℓ2nk−1,ω−1s¯ϒ2ℓ2mk,ϒ2ℓ2mk,Γ1ℓ2mk+s¯ϒ1ℓ2nk−1,ϒ1ℓ2nk−1,Γ2ℓ2nk−12,ω−1s¯ϒ2ℓ2mk,ϒ2ℓ2mk,Γ2ℓ2nk−1+s¯ϒ1ℓ2nk−1,ϒ1ℓ2nk−1,Γ1ℓ2mk2=s¯ℓ2mk″,ℓ2mk″,ℓ2nk−1″,ω−1s¯ℓ2mk″,ℓ2mk″,ℓ2mk+1″+s¯ℓ2nk−1″,ℓ2nk−1″,ℓ2nk″2,ω−1s¯ℓ2mk″,ℓ2mk″,ℓ2nk″+s¯ℓ2nk−1″,ℓ2nk−1″,ℓ2mk−1″2.
If(51)Mℓ2mk,ℓ2mk,ℓ2nk−1=ω−1s¯ℓ2mk″,ℓ2mk″,ℓ2mk+1″+s¯ℓ2nk−1″,ℓ2nk−1″,ℓ2nk″2,then from (35), we get that limk⟶∞Mℓ2mk,ℓ2mk,ℓ2nk−1=0. Hence, according to (49), we have limk⟶∞s¯ℓ2mk+1″,ℓ2mk+1″,ℓ2nk″=0, which contradicts (41).
If(52)Mℓ2mk,ℓ2mk,ℓ2nk−1=ω−1s¯ℓ2mk″,ℓ2mk″,ℓ2nk″+s¯ℓ2nk−1″,ℓ2nk−1″,ℓ2mk+1″2,then from (39) and (43), we get that(53)limsupk⟶∞Mℓ2mk,ℓ2mk,ℓ2nk−1≤ω−1ωε+ωε2=ε.
Taking the limit ask⟶∞ in (49), we have(54)πε=πωω−1ε≤πωlimsupk⟶∞s¯ℓ2mk+1″,ℓ2mk+1″,ℓ2nk″≤πlimsupk⟶∞Mℓ2mk,ℓ2mk,ℓ2nk−1−ϖliminfk⟶∞Mℓ2mk,ℓ2mk,ℓ2nk−1≤πε−ϖliminfk⟶∞Mℓ2mk,ℓ2mk,ℓ2nk−1,which implies that ϖliminfk⟶∞Mℓ2mk,ℓ2mk,ℓ2nk−1≤0. Hence, liminfk⟶∞s¯ℓ2mk,ℓ2mk,ℓ2nk=0, a contradiction to (39).
If(55)Mℓ2mk,ℓ2mk,ℓ2nk−1=s¯ℓ2mk,ℓ2mk,ℓ2nk−1,then from (37), by taking the limit as k⟶∞ in (49), we have(56)πε=πωω−1ε≤πωlimsupk⟶∞s¯ℓmk+1″,ℓmk+1″,ℓnk″≤πlimsupk⟶∞s¯ℓ2mk″,ℓ2mk″,ℓ2nk−1″−ϖliminfk⟶∞s¯ℓ2mk″,ℓ2mk″,ℓ2nk−1″≤πε−ϖliminfk⟶∞s¯ℓ2mk″,ℓ2mk″,ℓ2nk−1″,which implies that ϖliminfk⟶∞s¯ℓ2mk″,ℓ2mk″,ℓ2nk−1″≤0. Hence, liminfk⟶∞s¯ℓ2mk″,ℓ2mk″,ℓ2nk″=0. Therefore, from the rectangular inequality, we can conclude that liminfk⟶∞s¯ℓ2mk″,ℓ2mk″,ℓ2nk″=0 which contradicts (39). Hence, ℓn″ is a s¯-Cauchy O-sequence.Step 3.
We will show thatΓ1,Γ2, Υ1, and Υ2 have a point of coincidence.
Sinceℓn″ is a s¯-Cauchy O-sequence in the O-α-complete O-s¯-MS ℒ and αℓk″,ℓk″,ℓk+1″≥1, then there exists ℓ″∈ℒ, such that(57)limn⟶∞s¯ℓ2n+1″,ℓ2n+1″,ℓ″=limn⟶∞s¯ϒ1ℓ2n+1,ϒ1ℓ2n+1,ℓ″=limn⟶∞s¯Γ1ℓ2n+1,Γ1ℓ2n+1,ℓ″=0,limn⟶∞s¯ℓ2n″,ℓ2n″,ℓ″=limn⟶∞s¯ϒ2ℓ2n,ϒ2ℓ2n,ℓ″=limn⟶∞s¯Γ2ℓ2n−1,Γ2ℓ2n−1,ℓ″=0.
Hence,(58)ϒ2ℓ2n⟶ℓ″andΓ1ℓ2n⟶ℓ″,asn⟶∞.
AsΓ1,Υ2 is O-α-compatible and αℓ2n″,ℓ2n″,ℓ2n+1″≥1, so(59)limn⟶∞s¯ϒ2Γ1ℓ2n,ϒ2Γ1ℓ2n,Γ1ϒ2ℓ2n=0.
Moreover, fromlimn⟶∞s¯Γ1ℓ2n,Γ1ℓ2n,ℓ″=0, limn⟶∞s¯Υ2ℓ2n,Υ2ℓ2n,ℓ″=0, and the O-α-continuity of Υ2 and Γ1, we acquire that(60)limn⟶∞s¯ϒ2Γ1ℓ2n,ϒ2Γ1ℓ2n,ϒ2ℓ″=0=limn⟶∞s¯Γ1ϒ2ℓ2n,Γ1ϒ2ℓ2n,Γ1ℓ″.
By the rectangle inequality, we have(61)s¯ϒ2ℓ″,ϒ2ℓ″,Γ1ℓ″≤ωs¯ϒ2ℓ″,ϒ2ℓ″,ϒ2Γ1ℓ2n+s¯ϒ2ℓ″,ϒ2ℓ″,ϒ2Γ1ℓ2n+s¯Γ1ℓ″,Γ1ℓ″,ϒ2Γ1ℓ2n≤ωs¯ϒ2ℓ″,ϒ2ℓ″,ϒ2Γ1ℓ2n+s¯ϒ2ℓ″,ϒ2ℓ″,ϒ2Γ1ℓ2n+ωs¯Γ1ℓ″,Γ1ℓ″,Γ1ϒ2ℓ2n+s¯Γ1ℓ″,Γ1ℓ″,Γ1ϒ2ℓ2n+s¯ϒ2Γ1ℓ2n,ϒ2Γ1ℓ2n,Γ1ϒ2ℓ2n.
Taking the limit asn⟶∞ in (61), we deduce that(62)s¯ϒ2ℓ″,ϒ2ℓ″,Γ1ℓ″≤0,which yields that Γ1ℓ″=Υ2ℓ″, that is, ℓ″ is a coincidence point of Γ1 and Υ2.
Similarly, it can be proved thatΓ2ℓ″=Υ1ℓ″. Now, let αΥ1ℓ″,Υ1ℓ″,Υ2ℓ″≥1 and Υ1ℓ′′⊥Υ2ℓ′′. From (15), we have(63)πωs¯Γ1ℓ″,Γ1ℓ″,Γ2ℓ″≤πMℓ″,ℓ″,ℓ″−ϖMℓ″,ℓ″,ℓ″,where(64)Mℓ″,ℓ″,ℓ″∈s¯ϒ2ℓ″,ϒ2ℓ″,ϒ1ℓ″,ω−1s¯ϒ2ℓ″,ϒ2ℓ″,ϒ1ℓ″+s¯ϒ1ℓ″,ϒ1ℓ″,Γ2ℓ″2,ω−1s¯ϒ2ℓ″,ϒ2ℓ″,Γ2ℓ″+s¯ϒ1ℓ″,ϒ1ℓ″,Γ1ℓ″2=s¯Γ1ℓ″,Γ1ℓ″,Γ2ℓ″,0,ω−1s¯Γ1ℓ″,Γ1ℓ″,Γ2ℓ″.
In all three cases, the above inequality yields thatΓ1ℓ′′=Γ2ℓ′′=Υ2ℓ′′=Υ1ℓ′′.
In what follows, we replace the obligation ofO-α-continuity of Γ1,Γ2,Υ1, and Υ2 by an hypothesis on the underlying space and replace the O-α-compatibility by weak compatibility.Theorem 2.
Letℒ,s¯,ω,⊥ be an O-α-regular O-α-complete s¯-MS, α:ℒ×ℒ×ℒ⟶0,∞, and let Γ1,Γ2,Υ1,Υ2:ℒ⟶ℒ be such that Γ1ℒ⊆Υ1ℒ, Γ2ℒ⊆Υ2ℒ, and Υ1ℒ and Υ2ℒ are two s¯-closed subsets of ℒ. Suppose that(65)πω2s¯Γ1ℓ,Γ1ℓ,Γ2ℓ′≤πMℓ,ℓ,ℓ′−ϖMℓ,ℓ,ℓ′,for all ℓ and ℓ′ with αΥ2ℓ,Υ2ℓ,Υ1ℓ′≥1 and Υ2ℓ⊥Υ1ℓ′. Then, the pairs Γ1,Γ2 and Γ2,Υ1 admit a coincidence point ℓ′′ in ℒ whenever the pairs Γ1,Υ2 and Γ2,Υ1 are weakly compatible and the pairs Γ1,Γ2 and Γ2,Υ1 are rectangular partially weakly O-α-Υ1-admissible and O-α-Υ2-admissible, respectively. Moreover, if αΥ2ℓ″,Υ2ℓ″,Υ1ℓ″≥1 and Υ2ℓ″⊥Υ1ℓ″, then ℓ″∈ℒ is a coincidence point of Γ1,Γ2,Υ1, and Υ2.Proof.
From the proof of Theorem (15), for some ℓ″∈ℒ, we have(66)limk⟶∞s¯ℓk″,ℓk″,ℓ″=0.
SinceΥ1ℒ is s¯-closed and ℓ2n+1″⊆Υ1ℒ, therefore, ℓ″∈Υ1ℒ. Hence, for some u∈ℒ, one has ℓ″=Υ1u, and(67)limn⟶∞s¯ℓ2n+1″,ℓ2n+1″,Υ1u=limn⟶∞s¯Υ1ℓ2n+1,Υ1ℓ2n+1,Υ1u=0.
Similarly, forv∈ℒ, we have ℓ″=Υ1u=Υ2v and(68)limn⟶∞s¯ℓ2n″,ℓ2n″,Υ2v=limn⟶∞s¯Υ2ℓ2n,Υ2ℓ2n,Υ2v=0.
We shall show thatv is a coincidence point of Γ1 and Υ2.
SinceΥ1ℓ2n+1⟶ℓ″=Υ2v, as n⟶∞, from O-α-regularity of ℒ, αΥ1ℓ2n+1,Υ1ℓ2n+1,Υ2v≥1. Therefore, from (65), we have(69)πω2s¯Γ1v,Γ1v,Γ2ℓ2n+1≤πMv,v,ℓ2n+1−ϖMv,v,ℓ2n+1,where(70)Mv,v,ℓ2n+1∈s¯ϒ2v,ϒ2v,ϒ1ℓ2n+1,ω−1s¯ϒ2v,ϒ2v,Γ1v+s¯ϒ1ℓ2n+1,ϒ1ℓ2n+1,Γ2ℓ2n+12,ω−1s¯ϒ2v,ϒ2v,Γ2ℓ2n+1+s¯ϒ1ℓ2n+1,ϒ1ℓ2n+1,Γ1v2=s¯ℓ″,ℓ″,ℓ2n+1″,ω−1s¯ℓ″,ℓ″,Γ1v+s¯ℓ2n+1″,ℓ2n+1″,ℓ2n″2,ω−1s¯ℓ″,ℓ″,ℓ2n″+s¯ℓ2n+1″,ℓ2n+1″,Γ1v2.
From Lemma1,(71)ω−1ω−2s¯ℓ″,ℓ″,Γ1v2≤liminfn⟶∞Mv,v,ℓ2n+1≤limsupn⟶∞Mv,v,ℓ2n+1≤ωs¯ℓ″,ℓ″,Γ1v.
Taking the upper limit asn⟶∞ in (69), using Lemma 1 and the continuity of π and ϖ, we acquire that(72)πω2s¯Γ1v,Γ1v,ℓ″≤πs¯Γ1v,Γ1v,Γ2ℓ2n+1≤πω−1ω2s¯ℓ″,ℓ″,Γ1v2−ϖliminfn⟶∞Mv,v,ℓ2n+1≤πωs¯ℓ″,ℓ″,Γ1v−ϖliminfn⟶∞Mv,v,ℓ2n+1≤πω2s¯Γ1v,Γ1v,ℓ″−ϖliminfn⟶∞Mv,v,ℓ2n+1,which implies that ω−1ω−2s¯ℓ″,ℓ″,Γ1v/2=0. Hence, Γ1v=ℓ″=Υ2v.
AsΓ1 and Υ2 are weakly compatible, we have Γ1ℓ″=Γ1Υ2v=Υ2Γ1v=Υ2ℓ″. Thus, ℓ″ is a coincidence point of Γ1 and Υ2.
Analogously, it can be shown thatℓ′′ is a coincidence point of the pair Γ2,Υ1.Example 5.
Letℒ=0,84, and s¯ on ℒ be given by s¯ℓ,ℓ′,ℓ″=eℓ−ℓ″+ℓ′−ℓ″−1, for all ℓ,ℓ′,ℓ″∈ℒ, with ωt=et−1. Let α:ℒ×ℒ×ℒ⟶0,∞ be given by αℓ,ℓ′,ℓ″=eℓ+ℓ′−ℓ″ and ℓ⊥ℓ′, iff ℓ≤ℓ′. Define self-maps Γ1,Γ2,Υ1, and Υ2 on ℒ by(73)Γ1ℓ=ln1+ℓ1002,Υ1ℓ=eℓ−1,Γ2ℓ=ln1+ℓ42,Υ2ℓ=e25ℓ−1.
Letℓ,ℓ′∈ℒ be such that ℓ′∈Υ1−1Γ1ℓ, that is, Υ1ℓ′=Γ1ℓ. By the definition of Γ1 and Υ1, we have eℓ′−1=ln1+ℓ/1002, and so, ℓ′=ln1+ln1+ℓ/1002. Therefore,(74)Γ1ℓ=ln1+ℓ1002≥ln1+ln1+ln1+ℓ/100242=ln1+ℓ′42=Γ2ℓ′.
Thus,αΓ1ℓ,Γ1ℓ,Γ2ℓ′≥1 and Γ2ℓ′⊥Γ1ℓ. Hence, Γ1,Γ2 is partially weakly O-α-Υ1-admissible.
To prove thatΓ2,Γ1 is partially weakly O-α-Υ2-admissible, let ℓ,ℓ′∈ℒ be such that ℓ′∈Υ2−1Γ2ℓ, that is, Υ2ℓ′=Γ2ℓ. Hence, we have e25ℓ′−1=ln1+ℓ/42, and so, ℓ′=ln1+ln1+ℓ/42/25. Therefore,(75)Γ2ℓ=ln1+ℓ42≥ln1+ln1+ln1+ℓ/42/251002=ln1+ℓ′1002=Γ1ℓ′.
Thus,αΓ2ℓ,Γ2ℓ,Γ1ℓ′≥1 and Γ1ℓ′⊥Γ2ℓ.
Furthermore,(76)Γ1ℒ=Υ1ℒ=Γ2ℒ=Υ2ℒ.
Defineπ,ϖ:0,∞⟶0,∞ as πt=ln1+ln1+t and ϖt=1/100t for all t∈0,∞.
Using the mean value theorem, for allℓ and ℓ′ with αΥ2ℓ,Υ2ℓ,Υ1ℓ′≥1, we have(77)πωs¯Γ1ℓ,Γ1ℓ,Γ2ℓ′=ln1+ln1+ee2Γ1ℓ−Γ2ℓ′−1−1=2ln1+ℓ1002−ln1+ℓ′42≤2ℓ100−ℓ′4≤2ℓ−25ℓ′100≤21100eℓ−1−e25ℓ′−1≤21100Υ2ℓ−Υ1ℓ′≤1100e2Υ2ℓ−Υ1ℓ′−1≤ln1+ln1+e2Υ2ℓ−Υ1ℓ′−1−1100e2Υ2ℓ−Υ1ℓ′−1=πs¯Υ2ℓ,Υ2ℓ,Υ1ℓ′−ϖs¯Υ2ℓ,Υ2ℓ,Υ1ℓ′.
Thus, (15) is true for all ℓ,ℓ′∈ℒ and Mℓ,ℓ,ℓ′=s¯Υ2ℓ,Υ2ℓ,Υ1ℓ′. Therefore, all the conditions of Theorem 1 are satisfied. Moreover, 0 is a coincidence point of Γ1,Γ2,Υ1, and Υ2.
We arrange the following example, which supports Theorem2, based on Example 2.8 of [22].Example 6.
Consider the setX=1,2,3,4 and choose(78)Γ1=12342133,Γ2=12342313,Υ1=12342314,Υ2=12342134,dx,y=0,x=yx+y,otherwise,Sx,y,z=dx,y+dy,z.
Define the mappingα:X×X×X⟶ℝ and the relation ⊥ on X×X by(79)αx,x,y=1,x,y∈1,1,2,2,3,3,4,4,1,4,2,4,3,40,otherwise,,x⊥y⇔x≥y.
Letπ,ϖ:0,∞⟶0,∞ be defined by πt=tet and ϖt=tanht.
Obviously,(1)
ℒ,s¯,ωt=t,⊥ is an O-α-regular O-α-complete s¯-MS. Indeed, let ℓn⟶ℓ, ℓn⊥ℓn+1 and αℓn,ℓn,ℓn+1≥1 for all n∈ℕ. Then, ℓn=ℓn+1 for all n∈ℕ. So, ℒ,s¯,ωt=t,⊥ is O-α-regular.(2)
Γ1ℒ⊆Υ1ℒ, Γ2ℒ⊆Υ2ℒ, and Υ1ℒ and Υ2ℒ are two s¯-closed subsets of ℒ(3)
(80)πω2s¯Γ1ℓ,Γ1ℓ,Γ2ℓ′≤πMℓ,ℓ,ℓ′−ϖMℓ,ℓ,ℓ′,for allℓ and ℓ′ with αΥ2ℓ,Υ2ℓ,Υ1ℓ′≥1 and Υ2ℓ⊥Υ1ℓ′, i.e., for all ℓ and ℓ′ with Υ2ℓ=Υ1ℓ′.(4)
The pairsΓ2,Γ1 and Γ2,Γ1 are weakly compatible(5)
The pairΓ2,Γ1 is rectangular partially weakly O-α-Υ1-admissible(6)
The pairΓ2,Γ1 is rectangular partially weakly O-α-Υ2-admissible
Note that in parts 5 and 6, it is sufficient to takeℓ=1.
Therefore,Γ1,Γ2,Υ1, and Υ2 satisfy all the conditions given in Theorem 2. Thus, 1, 2, and 3 are coincidence points of the pairs Γ1,Υ2 and Γ2,Υ1. Moreover, αΥ21,Υ21,Υ11≥1 and Υ21⊥Υ11, so 1∈ℒ is a coincidence point of Γ1,Γ2,Υ1, and Υ2.


## 4. Application

In this part of the research, we study the following boundary value problem:(81)ν″μ=G1μ,νμ,μ∈0,1,ν″μ=G2μ,νμ,μ∈0,1,ν0=ν1=0,where G1,G2:0,1×ℝ⟶ℝ are the continuous functions.The above system is metamorphosed to the following system of Fredholm integral equations:(82)νμ=−∫01Kμ,tG1t,νtdt,νμ=−∫01Kμ,tG2t,νtdt,where the kernel is given by(83)Kμ,t=t1−μ,ift∈0,μ,μ1−t,ift∈μ,1.See [23] for details.Now, we present an existence result for a common solution of (82) in Q=CI,ℝ (the set of continuous real functions defined on I=0,1). Note that we consider the space Q, with the s¯-metric given by(84)s¯ζ,η,θ=maxt∈I|ζt−θt|+ηt−θtp,for all ζ,η,θ∈Q which is O-α-complete O-s¯-MS. Note that ωt=2p−1t, αℓ,ℓ,ℓ′=1, and ℓ⊥ℓ′ iff ℓ≤ℓ′.DefineΓ1,Γ2:Q⟶Q by(85)Γ1νμ=−∫01Kμ,tG1t,νtdtΓ2νμ=−∫01Kμ,tG2t,νtdt,for all ν∈Q and μ∈I.Clearly, a functionu∈Q is a common solution of (82) if and only if it is a common fixed point of Γ1,Γ2.Consider the following obligations:C1 For all u,v∈Q and for all t∈I,(86)G1t,u−G2t,v≤β22p−2ut−vtp.C2G1t,.,G2t,.:ℝ⟶ℝ are continuous for all t∈0,1Theorem 3.
Assume that the above presumptionsC1 and C2 hold. Then, the system (82) possesses a common solution in Q.Proof.
To show that all assumptions of Theorem2 are satisfied, it remains to prove that Γ1,Γ2 satisfy the contractive condition in Theorem 2. Let ν,ϱ∈Q. For each μ∈I, we have(87)Γ1νμ−Γ2ϱμp=∫01Kμ,tG1t,νtdt−∫01Kμ,tG2t,ϱtdtp≤∫01Kμ,tqdt1/q∫01G1t,νt−G2t,ϱtpdt1/pp≤∫01Kμ,tqdtp/q∫01G1t,νt−G2t,ϱtpdt≤∫01Kμ,tqdtp/q∫01β22p−2νt−ϱtpdt≤∫01Kμ,tqdtp/q∫01β22p−2s¯ν,ν,ϱdt.
Via a careful calculation, we get that(88)∫01Kμ,tqdt=1−μqμq+1+μq1−μq+1q+1,μ∈0,1.
So, we acquire that(89)|Γ1νμ−Γ2ϱμ|p≤β22p−21−μqμq+1+μq1−μq+1q+1s¯ν,ν,ϱ.
Taking the supremum onμ∈0,1, we deduce that(90)s¯Γ1ν,Γ1ν,Γ2ϱ≤β22p−2s¯ν,ν,ρq+1.
Now, by takingπt=t, ωt=2p−1t, and ϖt=1−β/22p−2t, we infer that(91)πω2s¯Γ1ν,Γ1ν,Γ2ϱ≤πs¯ν,ν,ϱ−ϖs¯ν,ν,ϱ.
Thus, all the hypotheses of Theorem2 are fulfilled, and we deduce the existence of an element u∈Q, such that u=Γ1u=Γ2u. Note that regularity of the space Q can be done as in [24].


## 5. Conclusions

In this study, we prove certain coincidence point theorems in an orthogonal extendedS-metric space for four weakly O-α-admissible contractive mappings. Many authors consider spaces with an extrastructure, such as partially ordered spaces, graphical spaces, and orthogonal spaces, in addition to working on different spaces and different contractive conditions to avoid exploring the contractive conditions with all elements in the space. In reality, they significantly reduce the calculations in contractive conditions. Only elements that are orthogonal to each other are included in this study. Also, α-admissibility is a generalization of monotonicity in the regular case.



---

*Source: 3040469-2021-10-29.xml*