# Common Fixed Point Theorems for Weakly Contractions in Rectangularb-Metric Spaces with Supportive Applications

Authors: Hongyan Guan; Jianju Li; Yan Hao
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this manuscript, two new classes of generalized weakly contractions are introduced and common fixed point results concerning the new contractions are proved in the context of rectangularb-metric spaces. Also, some examples are included to present the validity of our theorems. As an application, we provide the existence and uniqueness of solution of an integral equation.


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## Body

## 1. Introduction

In the field of nonlinear analysis, Banach fixed point theorem, which is introduced by Banach [1], is a powerful and classical means to deal with problems on fixed points in metric spaces. It is widely used in many disciplines of mathematics and has been promoted in many aspects. One important extension is to extend the concept of metric spaces. b-metric spaces and rectangular metric spaces are regarded as two well-known generalizations of metric spaces.As a extension of a metric space,b-metric space was firstly introduced by Czerwik [2], by modifying the third condition of metric function. In that paper, the author provided fixed point results for contraction conditions in this type space. Afterwards, some authors have obtained many excellent results concerning fixed point theory of a lot of new types of contractive mappings on b-metric spaces. Generalizing the results of Berinde [3], Zada et al. [4] obtained fixed point results for mappings with rational type and Pacurar [5] got fixed point theorems of φ-contractions. In [6], common fixed point results for weak φ-contraction mappings were proved in this type spaces by Aydi et al. In 2019, problems about periodic common fixed point were studied by Hussain et al. [7]. Recently, in [8], Gopal et al. explored the latest researches and developments on theory of fixed point in the framework of b-metric spaces. Younis et al. [9] introduced new fixed point results for the underlying mappings in the framework of dislocated b-metric spaces. In [10], in b-metric-like spaces, the authors extended the concept of Kannan mappings in view of F-contraction. Lately, Younis et al. [11] presented the notion of graphical extended b-metric spaces and discussed the framework of an open ball in this new type space.In 2000, by changing triangular inequality to quadrilateral inequality, more general inequality, Branciari [12] introduced the concept of rectangular metric spaces. Also, the author extended the Banach contraction mapping principle for this new context. Subsequently, a lot of fixed point theorems of various contractive conditions in rectangular metric spaces were obtained. Lakzian et al. [13] established fixed point theorems dealing with ψ,ϕ-weakly contraction conditions in this type space, which was ulteriorly extended by Erhan et al. in [14]. Bari and Vetro [15] got common fixed point results on given functions with ψ,ϕ-weakly contractive conditions. In [16], George and Rajagopalan studied problems of common fixed points of ψ,ϕ-contractive mappings. Lately, in complete rectangular metric spaces, Wang and Pei-Sheng [17] gave generalised θ-contraction mappings which can be regarded as generalized Suzuki-Berinde type θ-contraction mappings and provided conditions which ensured this type mapping possesses a unique fixed point. By the help of C-functions, in [18], some fixed point results were established by Budhia et al. In graphical rectangular b-metric spaces, some errors from literature [19] were rectified by Younis et al. in [20].Inspired by results of Czerwik [2] and Branciari [12], George et al. [21] extended b-metric space and rectangular metric space by introduced rectangular b-metric space. In that paper, the authors presented an analogue of Banach fixed point theorem and fixed point theorem of Kannan. After that, many researchers had solved problems of fixed point of new type of contractive mappings on this type space. Kadelburg and Radenovic [22] and Mitrovic [23] presented common fixed point theorems in this type space. In the setting of rectangular b-metric spaces, a Boyd-Wong type theorem was studied by Ding et al. in [24]. Sukprasert et al. [25] presented the concept of weak altering distance function and discussed fixed point result of a new generalized contractive mapping. Roshan et al. [26] gave some fixed point theorems concerning almost generalized weakly contractive mappings and rational type contractions. In [27], Mitrovic obtained an analogue of Banach contractive mapping principle and solved an open problem arose in [21]. Recently, Sunarsini et al. [28] introduced a new extension of metric space named as complex valued rectangular b−metric space and gave an example of Banach contractive mapping principle at linear equation system. In [29], common coupled fixed point theorems concerning generalised T−contraction conditions were studied by George and Reshma. Lately, in ordered partial rectangular b−metric spaces, Asim et al. [30] established some ordered-theoretic fixed point results of Geraghty-weak contractive mappings.In 1997, by using the notion of weak contractive mappings, Alber et al. [31] extended Banach contraction mapping principle in Hilbert spaces. In [32], weak contraction principle was generalized to metric spaces by Rhoades. After that, many authors had generalised the weak contraction principle. For example, in [33], the authors obtained the fixed point results involving α-ψ contraction conditions and applied them to solve quadratic integral equations. In [34], Jamal et al. used ψ,ϕ-weak contraction to extend coincidence point theorems obtained in partially ordered b-metric spaces.Set(1)Ψ=ψ:0,+∞⟶0,+∞isacontinuousandincreasingfunction,Φ=ϕ:0,+∞⟶0,+∞isalowersemicontinuousandnondecreasingfunctionandϕt=0ifandonlyift=0.Hao and Guan [35] proved common fixed point result dealing with a new class of generalized weakly contraction conditions in complete b-metric spaces as follows.Theorem 1 (see [35]).
LetE,ρ be a complete b-metric space with coefficient s≥1. Let O,R:E⟶E be self-mappings such that R is injective and OE⊂RE where RE is closed. Assume that p≥2 is a fixed number and φ:E⟶0,+∞ is lower semicontinuous. If there exist ψ∈Ψ and ϕ∈Φ satisfying
(2)ψspρOξ,Oη+φOξ+φOη≤ψIξ,η,ρ,O,R,φ−ϕJξ,η,ρ,O,R,φ,where
(3)Iξ,η,ρ,O,R,φ=maxρRξ,Rη+φRξ+φRη,12ρOξ,Rξ+φOξ+φRξ+ρOη,Rη+φOη+φRη,12sρOξ,Rη+φOξ+φRη+ρOη,Rξ+φOη+φRξ,Jξ,η,ρ,O,R,φ=maxρRξ,Rη+φRξ+φRη,ρOη,Rη+φOη+φRη,then O and R possess a unique coincidence point in E. Further, if O and R are weakly compatible, then O and R have a unique common fixed point.Continuing in the same direction, our aim is to give two new classes of generalized weakly contractions and establish some common fixed point theorems dealing with the new contractions in the setting of rectangularb-metric spaces. Moreover, we present some examples that elaborate the validity of our theorems. Also, as an application, we prove the existence of solution of an integral equation.


## 2. Preliminaries

First, we recall some definitions and lemmas as follows:Definition 2 (see [2]).
LetM be a nonempty set and s≥1 be a constant. A function σ:M×M⟶0,+∞ is said to be a b-metric iff
(i)
σξ,η=0 iff ξ=η for ξ,η∈M(ii)
σξ,η=ση,ξ for ξ,η∈M(iii)
there exists a real numbers≥1 satisfying σξ,η≤sσξ,ϑ+ση,ϑ for ξ,η,ϑ∈MUsually, we callM,σ a b-metric space with coefficient s≥1.Definition 3 (see [12]).
LetM be a nonempty set. A function ρ:M×M⟶0,+∞ is said to be a rectangular metric iff
(i)
ρξ,η=0 iff ξ=η for ξ,η∈M(ii)
ρξ,η=ρη,ξ for ξ,η∈M(iii)
ρξ,η≤ρξ,κ+ρκ,ν+ρν,η for ξ,η∈M and all different points κ,ν∈M−ξ,ηIn general, we callM,ρ a rectangular metric space.Definition 4 (see [21]).
LetM be a nonempty set and s≥1 be a constant. A function ϱ:M×M⟶0,+∞ is said to be a rectangular b-metric iff
(i)
ϱξ,η=0 iff ξ=η for ξ,η∈M(ii)
ϱξ,η=ϱη,ξ for ξ,η∈M(iii)
there exists a real numbers≥1 satisfying ϱξ,η≤sϱξ,κ+ϱκ,υ+ϱυ,η for ξ,η∈M and all different points κ,υ∈M−ξ,ηAs usual, we callM,ϱ a rectangular b-metric space with coefficient s≥1.Remark 5.
It is obvious that a rectangular metric function becomes a metric function whenκ=ν and a rectangular b-metric function becomes a rectangular metric function when s=1, whereas the converse of this statement may not be true (see [21], Examples 1.4 and 1.5).Example 1.
LetM=A∪B, where A=1/2,1/3,1/4,1/5, B=1,2. Define ϱ:M×M⟶0,+∞ with ϱξ,η=ϱη,ξ for x,y∈X and
(4)ϱ12,13=ϱ13,14=ϱ14,15=0.05,ϱ12,14=ϱ13,15=0.08,ϱ12,15=0.6,ϱξ,η=ξ−η,otherwise.
By calculation, we getM,ϱ is a rectangular b-metric space as s=4, whereas we obtain the following results:
(1)
M,ϱ is not a metric space, as(5)ϱ12,15=0.6>0.13=ϱ12,13+ϱ13,15(2)
M,ϱ is not a rectangular metric space, as(6)ϱ12,15=0.6>0.15=ϱ12,13+ϱ13,14+ϱ14,15(3)
M,ϱ is not a b-metric space with s=4, as(7)ϱ12,15=0.6>0.52=4·ϱ12,13+ϱ13,15Example 2.
AssumeM,ϱ∗ is a metric space. For p≥2, define ϱξ,η=ϱ∗ξ,ηp. Then, M,ϱ is a rectangular b-metric space with parameter s=3p−1.Proof.
One can verify easily the conditions (i) and (ii) hold by definition ofϱξ,η. In order to check (iii), we can infer from the following inequality:
(8)m+n+lp≤3p−1mp+np+lp,foranym,n,l≥0andp≥2.
Then, forξ,η∈M and all different points τ,υ∈M−ξ,η, we have
(9)ϱξ,η=ϱ∗ξ,ηp≤ϱ∗ξ,τ+ϱ∗τ,υ+ϱ∗υ,ηp≤3p−1ϱξ,τ+ϱτ,υ+ϱυ,η.
That is,M,ϱ is a rectangular b-metric space when s=3p−1.☐Definition 6 (see [21]).
LetM,ϱ be a rectangular b-metric space with coefficient s≥1. A sequence ξn in M is called:
(i)
convergent sequence iff there isξ∈M such that ϱξn,ξ⟶0 as n⟶+∞(ii)
a Cauchy sequence iffϱξn,ξm⟶0 when n,m⟶+∞Furthermore, a rectangularb-metric space is called completeness iff every Cauchy sequence is convergent.Remark 9.
In rectangularb-metric spaces, one can show that the limit of a sequence may not unique and every convergent sequence in a rectangular b-metric space may not be a Cauchy sequence(see [21], Example 1.7).Definition 8 (see [36]).
LetO and R be two self-maps defined on a nonempty set M. If ν=Oξ=Rξ, for some ξ∈M, then ν is called the point of coincidence of O and R, where ξis said to be the coincidence point of O and R. Let CO,R represent the collection of all coincidence points of O and R.Definition 9 (see [36]).
LetO and R be two self-maps defined on a nonempty set M. Then, O and R are called weakly compatible mappings when they commute at each coincidence point, i.e., Oξ=Rξ⇒ORξ=ROξ for each ξ∈CO,R.Lemma 10 (see [26]).
LetM,ϱ be a rectangular b-metric space with parameter s≥1. Assume that ξn and ηn are convergent to ξ and η, respectively. Then, one can get
(10)1sϱξ,η≤liminfn⟶+∞ϱξn,ηn≤limsupn⟶+∞ϱξn,ηn≤sϱξ,η.
Moreover, ifξ=η, then we have limn⟶+∞ϱξn,ηn=0. Further, for ζ∈M, we deduce
(11)1sϱξ,ζ≤liminfn⟶+∞ϱξn,ζ≤limsupn⟶+∞ϱξn,ζ≤sϱξ,ζ.


## 3. Main Results

In this section, a few of new common fixed point results on generalized weakly contractive conditions in a complete rectangularb-metric space will be presented. Moreover, two examples will be provided to prove the validity of our theorems.SupposeM,ϱ is a rectangular b-metric space. A mapping O:M⟶0,+∞ is named as a lower semicontinuous mapping if, for ξ∈M and ξn is convergent to ξ, one get
(12)Oξ≤liminfn⟶+∞Oξn.LetΩ represent the set of all functions β:ℝ0+⟶0,1/s. We shall consider the contractive conditions defined by the family Θ:
(13)Θ=θ:0,+∞⟶0,+∞isacontinuousandincreasingfunction,forallκ>0,θκ<κandθκ=0iffκ=0.Lemma 11 (see [37]).
Letθ:0,+∞⟶0,+∞ be a nondecreasing and upper semicontinuous mapping. Then, θx<x for any x>0 iff θnx⟶0 as n⟶∞.Definition 12.
LetM,ρ be a rectangular b-metric space with coefficient s≥1. Let α:M×M⟶0,+∞ and O,R:M⟶M be given functions and p≥2 be a real number. A function O:M⟶M is called R-αsp-admissible function if, for all ξ,η∈M,αRξ,Rη≥sp implies αOξ,Oη≥sp.Definition 13.
LetM,ϱ be a rectangular b-metric space with coefficient s≥1. Let α:M×M⟶0,+∞ and O,R:M⟶M be three given mappings. Suppose that p≥2 is a real number and φ:M⟶0,+∞ is a lower semicontinuous function. A mapping O is called a generalized R−αsp,θ,φ contractive mapping, if there exist θ∈Θ,β∈Ω and L≥0,1/s+L<1 satisfying
(14)θαRξ,RηϱOξ,Oη+φOξ+φOη≤βθhξ,η,ϱ,O,R,φθhξ,η,ϱ,O,R,φ+Lθqξ,η,ϱ,O,R,φ,for all ξ,η∈M with αRξ,Rη≥sp and ϱOξ,Oη+φOξ+φOη≠0, where
(15)hξ,η,ϱ,O,R,φ=maxϱOξ,Oη+φOξ+φOη,ϱOη,Oη+φOη+φRη1+ϱOξ,Rξ+φOξ+φRξ·ϱRξ,Rη+φRξ+φRη,12ϱOξ,Oη+φOξ+φOη+ϱRξ,Rη+φRξ+φRη,qξ,η,ϱ,O,R,φ=12minϱOξ,Oη+φOξ+φOη,ϱRξ,Rη+φRξ+φRη.
Letα:M×M⟶0,+∞ be a mapping. Set
Asp If ξn is a sequence in M satisfying Rξn⟶Rξ as n⟶+∞, then there is a subsequence Rξnk of Rξn with αRξnk,Rξ≥sp for k∈ℕ
Bsp For x,y∈CO,R, one can get the condition of αRx,Ry≥sp and αRx,Ry≥spTheorem 14.
LetM,ϱ be a complete rectangular b-metric space with coefficient s≥1. Let O,R:M⟶M be given self-mappings satisfying OM⊂RM and RM is closed. Assume that φ:M⟶0,+∞ is a lower semicontinuous mapping and α:M×M⟶0,+∞. If
(i)
O is R-αsp-admissible(ii)
O is generalized R−αsp,θ,φ contractive(iii)
there isξ0∈M satisfying αRξ0,Oξ0≥sp(iv)
propertiesAsp and Bsp are fulfilled(v)
α satisfies transitive property, i.e., for ξ,η,ζ∈M(16)αξ,η≥spandαη,ζ≥sp⇒αξ,ζ≥sp,
thenO and R possess a unique point of coincidence. Furthermore, if O and R are weakly compatible, then O and R possess a unique common fixed point in M.Proof.
It follows from condition (iii) that one can choose anξ0∈M with αRξ0,Oξ0≥sp. Define sequences ξn and ηn in M by ηn=Oξn=Rξn+1 for n∈ℕ. If ηn=ηn+1 for some n, then we deduce ηn=ηn+1=Oξn+1=Rξn+1 and O and R possess a point of coincidence. Next, we suppose that ηn≠ηn+1 for n∈ℕ. In light of contraction condition (i), we obtain
(17)αRξ0,Rξ1=αRξ0,Oξ0≥sp,αRξ1,Rξ2=αOξ0,Oξ1≥sp,αRξ2,Rξ3=αOξ1,Oξ2≥sp.
Hence, for alln∈ℕ, we deduce αRξn,Rξn+1=αηn−1,ηn≥sp. Applying (14) with ξ=ξn and η=ξn+1,
(18)θϱηn,ηn+1+φηn+φηn+1≤θspϱηn,ηn+1+φηn+φηn+1≤θαRξn,Rξn+1ϱOξn,Oξn+1+φOξn+φOξn+1≤βθhξn,ξn+1,ϱ,O,R,φθhξn,ξn+1,ϱ,O,R,φ+Lθqξn,ξn+1,ϱ,O,R,φ,where
(19)hξn,ξn+1,ϱ,O,R,φ=maxϱOξn,Oξn+1+φOξn+φOξn+1,ϱOξn+1,Rξn+1+φOξn+1+φRξn+11+ϱOξn,Rξn+φOξn+φRξn·ϱRξn,Rξn+1+φRξn+φRξn+1,12ϱOξn,Oξn+1+φOξn+φOξn+1+ϱRξn,Rξn+1+φRξn+φRξn+1=maxϱηn,ηn+1+φηn+φηn+1,ϱηn+1,ηn+φηn+1+φηn1+ϱηn,ηn−1+φηn+φηn−1·ϱηn−1,ηn+φηn−1+φηn,12ϱηn,ηn+1+φηn+φηn+1+ϱηn−1,ηn+φηn−1+φηn≤maxϱηn−1,ηn+φηn−1+φηn,ϱηn+1,ηn+φηn+1+φηn,(20)qξn,ξn+1,ϱ,O,R,φ=12minϱOξn,Oξn+1+φOξn+φOξn+1,ϱRξn,Rξn+1+φRξn+φRξn+1=12minϱηn,ηn+1+φηn+φηn+1,ϱηn−1,ηn+φηn−1+φηn.
If we assume thatϱηn,ηn+1+φηn+φηn+1>ϱηn,ηn−1+φηn+φηn−1 for some n∈ℕ, according to (18), (19), and (20), we have
(21)θϱηn,ηn+1+φηn+φηn+1≤1sθhξn,ξn+1,ϱ,O,R,φ+Lθqξn,ξn+1,ϱ,O,R,φ≤1sθϱηn+1,ηn+φηn+1+φηn+Lθϱηn+1,ηn+φηn+1+φηn<θϱηn,ηn+1+φηn+φηn+1,which is a contradiction. Thus,
(22)ϱηn,ηn+1+φηn+φηn+1≤ϱηn,ηn−1+φηn+φηn−1,(23)hξn,ξn+1,ϱ,O,R,φ≤ϱηn,ηn−1+φηn+φηn−1,(24)qξn,ξn+1,ϱ,O,R,φ<ϱηn,ηn−1+φηn+φηn−1.
It follows from (22) that ϱηn,ηn+1+φηn+φηn+1 is decreasing. It follows that there exists a real number γ≥0 satisfying
(25)limn⟶+∞ϱηn,ηn+1+φηn+φηn+1=γ.
In view of (18), (23), and (24), one can obtain
(26)θϱηn,ηn+1+φηn+φηn+1≤βθhξn,ξn+1,ϱ,O,R,φθhξn,ξn+1,ϱ,O,R,φ+Lθqξn,ξn+1,ϱ,O,R,φ<θϱηn,ηn−1+φηn+φηn−1.
Ifγ>0, putting n⟶∞ in (26), we obtain
(27)θγ=limn⟶+∞θϱηn,ηn+1+φηn+φηn+1≤limn⟶+∞βθhξn,ξn+1,ϱ,O,R,φθhξn,ξn+1,ϱ,O,R,φ+Llimn⟶+∞θqξn,ξn+1,ϱ,O,R,φ<1slimn⟶+∞θϱηn,ηn−1+φηn+φηn−1+Llimn⟶+∞θϱηn,ηn−1+φηn+φηn−1<θγ,a contradiction. Hence,
(28)limn⟶+∞ϱηn,ηn+1+φηn+φηn+1=γ=0,which implies that limn⟶+∞ϱηn,ηn+1=0 and limn⟶+∞φηn=0. In view of hypothesis (v), we have αηn−2,ηn≥sp. Taking ξ=ξn−1 and η=ξn+1 in (14), we obtain
(29)θϱηn−1,ηn+1+φηn−1+φηn+1≤θαηn−2,ηnϱηn−1,ηn+1+φηn−1+φηn+1≤βθhξn−1,ξn+1,ϱ,O,R,φθhξn−1,ξn+1,ϱ,O,R,φ+Lθqξn−1,ξn+1,ϱ,O,R,φ,where
(30)hξn−1,ξn+1,ϱ,O,R,φ=maxϱOξn−1,Oξn+1+φOξn−1+φOξn+1,ϱOξn+1,Rξn+1+φOξn+1+φRξn+11+ϱOξn−1,Rξn−1+φOξn−1+φRξn−1·ϱRξn−1,Rξn+1+φRξn−1+φRξn+1,12ϱOξn−1,Oξn+1+φOξn−1+φOξn+1+ϱRξn−1,Rξn+1+φRξn−1+φRξn+1≤maxϱηn−1,ηn+1+φηn−1+φηn+1,ϱηn−2,ηn+φηn−2+φηn,(31)qξn−1,ξn+1,ϱ,O,R,φ=12minϱOξn−1,Oξn+1+φOξn−1+φOξn+1,ϱRξn−1,Rξn+1+φRξn−1+φRξn+1=12minϱηn−1,ηn+1+φηn−1+φηn+1,ϱηn−2,ηn+φηn−2+φηn.
If for somen∈ℕ, ϱηn−1,ηn+1+φηn−1+φηn+1>ϱηn−2,ηn+φηn−2+φηn, according to (29), (30), and (31), we get
(32)θϱηn−1,ηn+1+φηn−1+φηn+1<1sθhξn−1,ξn+1,ϱ,O,R,φ+Lθqξn−1,ξn+1,ϱ,O,R,φ<θϱηn−1,ηn+1+φηn−1+φηn+1,which is a contradiction. It follows that
(33)ϱηn−1,ηn+1+φηn−1+φηn+1≤ϱηn−2,ηn+φηn−2+φηn,(34)hξn−1,ξn+1,ϱ,O,R,φ≤ϱηn−2,ηn+φηn−2+φηn,(35)qξn−1,ξn+1,ϱ,O,R,φ<ϱηn−2,ηn+φηn−2+φηn.
Inequality (33) yields that ϱηn−2,ηn+φηn−2+φηn is non-increasing and which yields that there exists ∈≥0 satisfying
(36)limn⟶+∞ϱηn−2,ηn+φηn−2+φηn=∈.
In light of (32), (34), and (35), one can deduce
(37)θϱηn−1,ηn+1+φηn−1+φηn+1≤βθhξn−1,ξn+1,ϱ,O,R,φθhξn−1,ξn+1,ϱ,O,R,φ+Lθqξn−1,ξn+1,ϱ,O,R,φ<θϱηn−2,ηn+φηn−2+φηn.
Assume that∈>0. Letting n⟶∞ in (37), we derive
(38)θ∈=limn⟶+∞θϱηn−1,ηn+1+φηn−1+φηn+1≤limn⟶+∞βθhξn−1,ξn+1,ϱ,O,R,φθhξn−1,ξn+1,ϱ,O,R,φ+Llimn⟶+∞θqξn−1,ξn+1,ϱ,O,R,φ<limn⟶+∞θϱηn−2,ηn+φηn−2+φηn=θ∈,which gives a contradiction. This yields that
(39)limn⟶+∞ϱηn−2,ηn+φηn−2+φηn=∈=0.
It follows thatlimn⟶+∞ϱηn−2,ηn=0.
Now, we aim to show thatηn is a Cauchy sequence. Assume on the contrary that, ηn is not Cauchy. So, there exists ε>0 for which we can choose sequences ηmk and ηnk of ηn such that nk is the smallest index for which nk>mk>k,
(40)ε≤ϱηmk,ηnk,(41)ϱηmk,ηnk−1<ε.
In light of the rectangular inequality and (40) and (41), we have
(42)ε≤ϱηmk,ηnk≤sϱηmk,ηnk−1+ϱηnk−1,ηnk+1+ϱηnk+1,ηnk<sε+sϱηnk−1,ηnk+1+sϱηnk+1,ηnk.
Taking the superior limit ask⟶+∞, we have
(43)ε≤limsupk⟶+∞ϱηmk,ηnk≤sε.
Similarly,(44)ϱηmk,ηnk≤sϱηmk,ηmk+1+ϱηmk+1,ηmk−1+ϱηmk−1,ηnk,(45)ϱηmk,ηnk≤sϱηmk,ηmk−1+ϱηmk−1,ηnk−1+ϱηnk−1,ηnk,(46)ϱηmk−1,ηnk≤sϱηmk−1,ηmk+ϱηmk,ηnk−1+ϱηnk−1,ηnk.
It follows from (40), (41), and (42) that
(47)εs≤limsupk⟶+∞ϱηmk,ηnk−1≤ε.
By (40), (41), (44), and (46), we get
(48)εs≤limsupk⟶+∞ϱηmk−1,ηnk≤sε.
By the similar method, we have(49)ϱηmk−1,ηnk−1≤sϱηmk−1,ηmk+ϱηmk,ηnk+ϱηnk,ηnk−1,so is
(50)εs≤limsupk⟶+∞ϱηmk−1,ηnk−1≤s2ε.
According to the definition ofhξ,η,ϱ,O,R,φ, we get
(51)hξmk,ξnk,ϱ,O,R,φ=maxϱOξmk,Oξnk+φOξmk+φOξnk,ϱOξnk,Rξnk+φOξnk+φRξnk1+ρOξmk,Rξmk+φOξmk+φRξmk·ϱRξmk,Rξnk+φRξmk+φRξnk,12ϱOξmk,Oξnk+φOξmk+φOξnk+ϱRξmk,Rξnk+φRξmk+φRξnk=maxϱηmk,ηnk+φηmk+φηnk,ϱηnk,ηnk−1+φηnk+φηnk−11+ϱηmk,ηmk−1+φηmk+φηmk−1·ϱηmk−1,ηnk−1+φηmk−1+φηnk−1,12ϱηmk,ηnk+φηmk+φηnk+ϱηmk−1,ηnk−1+φηmk−1+φηnk−1.
Taking the superior limit ask⟶+∞ in (51), we get
(52)limsupk⟶+∞hξmk,ξnk,ϱ,O,R,φ≤maxsε,0,sε+s2ε2<s2ε.
Also, we have(53)qξmk,ξnk,ϱ,O,R,φ=12minϱOξmk,Oξnk+φOξmk+φOξnk,ϱRξmk,Rξnk+φRξmk+φRξnk=12minϱηmk,ηnk+φηmk+φηnk,ϱηmk−1,ηnk−1+φηmk−1+φηnk−1.
It follows that(54)limsupk⟶+∞qξmk,ξnk,ϱ,O,R,φ<s2ε.
The transitivity property ofα yields that αRξmk,Rξnk≥sp. Taking ξ=ξmk and η=ξnk in (14), one can deduce
(55)θs2ε≤θspε≤θαRξmk,Rξnklimsupn⟶+∞ϱηmk,ηnk+φηmk+φηnk≤limsupn⟶+∞βθhξmk,ξnk,ϱ,O,R,φθhξmk,ξnk,ϱ,O,R,φ+limsupn⟶+∞Lθqξmk,ξnk,ϱ,O,R,φ≤1sθs2ε+Lθs2ε<θs2ε,a contradiction. Hence, ηn is Cauchy. Since M,ρ is complete, there is a ϑ∈M such that
(56)limn⟶+∞ϱηn,ϑ=limn⟶+∞ϱOξn,ϑ=limn⟶+∞ϱRξn+1,ϑ=limn,m⟶+∞ϱηn,ηm=0.
SinceRM is closed, we have ϑ∈RM. Hence,we choose a z∈M satisfying ϑ=Rz. We write (56) as
(57)limn⟶+∞ϱηn,Rz=limn⟶+∞ϱOξn,Rz=limn⟶+∞ϱRξn+1,Rz=0.
It follows from the definition ofφ that
(58)φRz=φϑ≤liminfn⟶+∞φηn=0,which implies that φRz=φϑ=0.
The propertyAsp ensures that there exists a subsequence ηnk of ηn with αηnk−1,Rz≥sp for k∈ℕ. If ϱOz,Rz+φOz≠0, taking ξ=ξnk and η=z in (14), one deduce that
(59)θϱOξnk,Oz+φOξnk+φOz≤θspϱOξnk,Oz+φOξnk+φOz≤θαηnk−1,RzϱOξnk,Oz+φOξnk+φOz≤βθhξnk,z,ϱ,O,R,φθhξnk,z,ϱ,O,R,φ+Lθqξnk,z,ϱ,O,R,φ,where
(60)hξnk,z,ϱ,O,R,φ=maxϱOξnk,Oz+φOξnk+φOz,ϱOz,Rz+φOz+φRz1+ρOξnk,Rξnk+φOξnk+φRξnk·ϱRξnk,Rz+φRξnk+φRz,12ϱOξnk,Oz+φOξnk+φOz+ϱRz,Rξnk+φRz+φRξnk=maxϱηnk,Oz+φηnk+φOz,ϱOz,Rz+φOz+φRz1+ϱηnk,ηnk−1+φηnk+φηnk−1·ϱηnk−1,Rz+φηnk−1+φRz,12ϱηnk,Oz+φηnk+φOz+ϱηnk−1,Rz+φηnk−1+φRz,(61)qξnk,z,ϱ,O,R,φ=12minϱOξnk,Oz+φOξnk+φOz,ϱRξnk,Rz+φRξnk+φRz=12minϱηnk,Oz+φηnk+φOz,ϱηnk−1,Rz+φηnk−1+φRz.
By simple calculation, we get(62)limsupk⟶+∞hξnk,z,ϱ,O,R,φ≤sϱOz,Rz+φOz,(63)limsupk⟶+∞qξnk,z,ϱ,O,R,φ<sρOz,Rz+φOz.
Lettingk⟶+∞ in (59), using (62) and (63), we obtain
(64)θsϱOz,Rz+φOz≤θs2limsupk⟶∞ϱOξnk,Oz+φOξnk+φOz≤θlimsupk⟶∞αηnk−1,RzϱOξnk,Oz+φOξnk+φOz<1sθsϱOz,Rz+φOz+LθsϱOz,Rz+φOz<θsϱOz,Rz+φOz.
It follows thatϱOz,Rz+φOz=0, which implies that Oz=Rz, φOz=0.
Next, we show thatO and R possess the unique point of coincidence ϑ. Assume on the contrary, there exist z,z′∈CO,R and Oz≠Oz′. By the property of Bsp, one can obtain
(65)αRz′,Rz≥sp.
Takingξ=z′ and η=z in (14), we obtain
(66)θϱOz′,Oz+φOz′+φOz≤θspϱOz′,Oz+φOz′+φOz≤θαRz′,RzϱOz′,Oz+φOz′+φOz≤βθhz′,z,ϱ,O,R,φθhz′,z,ϱ,O,R,φ+Lθqz′,z,ϱ,O,R,φ,where
(67)hz′,z,ϱ,O,R,φ=maxϱOz′,Oz+φOz′+φOz,ϱOz,Rz+φOz+φRz1+ϱOz′,Rz′+φOz′+φRz′·ϱRz′,Rz+φRz′+φRz,12ϱOz′,Oz+φOz′+φOz+ϱRz′,Rz+φRz′+φRz≤ϱRz′,Rz+φRz′,qz′,z,ϱ,O,R,φ=12minϱOz′,Oz+φOz′+φOz,ϱRz′,Rz+φRz′+φRz<ϱRz′,Rz+φRz′.
In view of (66), we have
(68)θϱRz′,Rz+φRz′<1sθϱRz′,Rz+φRz′+LθϱRz′,Rz+φRz′<θϱRz′,Rz+φRz′.
Therefore, one can obtain thatϱRz,Rz′+φRz′=0, that is, Rz=Rz′=ϑ and φRz′=0. Hence, ϑ is a unique point of coincidence for O and R. Furthermore, if O and R are weak compatible mappings, it is easy to prove that O and R have a unique common fixed point z. The proof is complete.☐Example 3.
LetM,ϱ be the same as it is in Example 1. Define mappings O,R:M⟶M by
(69)Oξ=15,ξ∈A,13,ξ∈B,Rξ=15,ξ=15,13,ξ=14,12,ξ=13,1,ξ∈12∪B.
Define mappingsα:M×M⟶0,+∞ by
(70)αξ,η=sp,ξ,η∈15,14,13,12withξ≠η,orξ=η=15,orξ=η=1,orξ=η=13,0,otherwise.
Defineθ:0,+∞⟶0,+∞, φ:M⟶0,+∞ as follows:
(71)θξ=ξ2,φξ=0,ξ∈0,15,0.15ξ−0.03,ξ∈15,13,11.97ξ−3.97,ξ∈13,+∞.Definedβξ=1/5 for all ξ≥0, then β∈Ω. We can show that OM⊂RM, RM is closed. For ξ,η∈M such that αRξ,Rη≥sp, we get that Rξ,Rη∈1/5,1/4,1/3,1/2 with Rξ≠Rη, or Rξ=Rη=1/5, or Rξ=Rη=1. This implies that ξ,η∈1/5,1/4,1/3 with ξ≠η, or ξ=η=1/5, or ξ=η=1/4, or ξ,η∈1/2∪B. So we obtain Oξ,Oη∈1/5,1/3 and αOξ,Oη≥sp. Combining with the condition ϱOξ,Oη+φOξ+φOη≠0, the following cases are considered:Case 1.
ξ=1/2,η∈B (or η=1/2,ξ∈B).
(72)θαRξ,RηϱOξ,Oη+φOξ+φOη=12·16·ϱ15,13+φ15+φ13=0.8,βθhξ,η,ϱ,O,R,φθhξ,η,ϱ,O,R,φ+Lθqξ,η,ϱ,O,R,φ≥βθhξ,η,ϱ,O,R,φθhξ,η,ϱ,O,R,φ≥15·12·12ϱ15,13+φ15+φ13+ϱ1,1+φ1+φ1=120·0.1+8+8>0.8.
In view of above inequalities, one can get that(73)θαRξ,RηϱOξ,Oη+φOξ+φOη≤βθhξ,η,ϱ,O,R,φθhξ,η,ϱ,O,R,φ+Lθqξ,η,ϱ,O,R,φ,with L≥0,1/s+L<1 and s=4,p=2.Case 2.
ξ,η∈B.
(74)θαRξ,RηϱOξ,Oη+φOξ+φOη=12·16·ϱ13,13+φ13+φ13=0.32,βθhξ,η,ϱ,O,R,φθhξ,η,ϱ,O,R,φ+Lθqξ,η,ϱ,O,R,φ≥βθhξ,η,ϱ,O,R,φθhξ,η,ϱ,O,R,φ≥15·12·12ϱ13,13+φ13+φ13+ϱ1,1+φ1+φ1=120·0.04+8+8>0.32.
That is, forξ,η∈B,
(75)θαRξ,RηϱOξ,Oη+φOξ+φOη≤βθhξ,η,ϱ,O,R,φθhξ,η,ϱ,O,R,φ+Lθqξ,η,ϱ,O,R,φ,with L≥0,1/s+L<1 and s=4,p=2.In summary, all requirements of Theorem14 are satisfied. O and R have a unique common fixed point 1/5.In Theorem14, letting φ=0, we can obtain the following result.Corollary 15.
LetM,ρ be a complete rectangular b-metric space with coefficient s≥1. Let α:M×M⟶0,+∞ and O,R:M⟶M be given mappings with OM⊂RM and RM is closed. Assume that p≥2 is a arbitrary constant and φ:M⟶0,+∞ is a lower semi-continuous function. If
(i)
O is R-αsp-admissible(ii)
forξ,η∈M such that αRξ,Rη≥sp and ϱOξ,Oη≠0, there exist θ∈Θ,β∈Ω and L≥0,1/s+L<1 satisfying:(76)θαRξ,RηϱOξ,Oη≤βθmξ,η,ϱ,O,Rθmξ,η,ϱ,O,R+Lθnξ,η,ϱ,O,R,where
(77)mξ,η,ϱ,O,R=maxϱOξ,Oη,ϱOη,Rη1+ϱOξ,Rξ·ϱRξ,Rη,12ϱOξ,Rη+ϱRξ,Rη,and
(78)nξ,η,ϱ,O,R=12minϱOξ,Oη,ϱRξ,Rη(iii)
there isξ0∈M such that αRξ0,Oξ0≥sp(iv)
propertiesAsp and Bsp are fulfilled(v)
α satisfies transitive property, i.e., for ξ,η,z∈M(79)αξ,η≥spandαη,z≥sp⇒αξ,z≥sp,then, O and R possess a unique point of coincidence in M. Further, if O and R are weakly compatible, then O and R possess a unique common fixed pointIfφ=0, R=I, and L=0 in Theorem 14, we have the following.Corollary 16.
LetM,ϱ be a complete rectangular b-metric space with coefficient s≥2 and M be closed. Let α:M×M⟶0,+∞ and O:M⟶M be given mappings. Suppose p≥2 is a arbitrary constant. If
(i)
O is αsp-admissible(ii)
forξ,η∈M such that αξ,η≥sp and ϱOξ,Oη≠0, there exists θ∈Θ satisfying:(80)θαξ,ηϱOξ,Oη≤βθm∗ξ,η,ϱ,O,where β∈0,1/s is a constant and
(81)m∗ξ,η,ϱ,O=maxϱOξ,Oη,ϱOη,η1+ϱOξ,ξ·ϱξ,η,12ϱOξ,η+ϱξ,η(iii)
there isξ0∈M with αξ0,Oξ0≥sp(iv)
propertiesAsp and Bsp are fulfilled when R=I(v)
α satisfies transitive property, i.e., for ξ,η,z∈M(82)αξ,η≥sp,αη,z≥sp⇒αξ,z≥sp,then O possesses a unique fixed pointDefinition 17.
The self-mappingsO,R:M⟶M are called αs orbital admissible mapping, if the following condition holds:
(83)αsξ,Oξ≥sp,αsξ,Rξ≥spimplyαsOξ,ROξ≥sp,αsRξ,ORξ≥sp,for a constant p≥3.Definition 18.
LetO,R:M⟶M be two self-maps and p≥3 be a real number. The pair O,R is said to be triangular αs orbital admissible if
(i)
O,R:M⟶M are αs orbital admissible(ii)
αsξ,η≥sp,αsη,Oη≥spandαsη,Rη≥sp imply αsξ,Oη≥sp,αsξ,Rη≥spLemma 19.
LetO,R:M⟶M be two self-mappings satisfying the pair O,R is triangular αs orbital admissible. Suppose that there is ξ0∈M satisfying αsξ0,Oξ0≥sp. Define a sequence ξninMbyξ2i+1=Oξ2i,ξ2i+2=Rξ2i+1 where i=0,1,2,⋯. Then αsξn,ξm≥sp for m,n∈ℕ∪0 with m>n, .Proof.
Sinceαsξ0,Oξ0=αsξ0,ξ1≥sp and O,R is triangular αs orbital admissible, αsξ0,Oξ0≥sp implies αsOξ0,ROξ0=αsξ1,Rξ1=αsξ1,ξ2≥sp, αsξ1,Rξ1≥sp implies αsRξ1,ORξ1=αsξ2,Oξ2=αsξ2,ξ3≥sp, αsξ2,Oξ2≥sp implies αsOξ2,ROξ2=αsξ3,Rξ3=αsξ3,ξ4≥sp. Applying the above argument repeated, we obtain αsξn,ξn+1≥sp for all n∈ℕ∪0. Since O,R is triangular αs orbital admissible, αsξn,ξm≥sp for n,m∈ℕ∪0 with m>n.☐Definition 20.
LetM,ϱ be a rectangular b-metric space with coefficient s≥1. Let O,R:M⟶M be two self-mappings. Suppose that αs:M×M⟶0,+∞ and φ:M⟶0,+∞ is a lower semicontinuous function and p≥3 is an arbitrary constant. The mappings O,R are said to be generalized αs-θ-Geraghty contractions, if there exist θ∈Θ,β,L≥0 and β+L<1,0<λ<1/4 such that
(84)αsξ,ηϱOξ,Rη+φOξ+φRη≤βθsξ,η,ϱ,O,R,φ+Lθtξ,η,ϱ,O,φ,for all ξ,η∈M with αsξ,η≥sp and ϱOξ,Rη+φOξ+φRη≠0, where
(85)sξ,η,ϱ,O,R,φ=λmaxϱξ,η+φξ+φη,1+ϱξ,Oξ+φξ+φOξ1+ϱξ,η+φξ+φη·ϱOξ,Rη+φOξ+φRη,ϱη,Rη+φη+φRη1+ϱξ,η+φξ+φη·ϱξ,Oξ+φξ+φOξ,tξ,η,ϱ,O,φ=λminϱξ,Oξ+φξ+φOξ,ϱη,Oξ+φη+φOξ.
Letαs:M×M⟶0,+∞ be a given mapping. Set.
Csp For all ξ∗∈M, we have αsξ∗,ξ∗≥sp
Dsp For all x,y∈CO,R, one can get the condition of αsx,y≥sp or αsy,x≥spTheorem 21.
LetM,ϱ be a complete rectangular b-metric space with coefficient s≥1. Let O,R:M⟶M be generalized αs-θ-Geraghty contractions and one of O and R is continuous. If
(i)
O,R are triangular αs orbital admissible(ii)
there isξ0∈M with satisfying αsξ0,Oξ0≥sp(iii)
propertiesCsp and Dsp are satisfiedthen O and R possess a unique common fixed pointProof.
Forξ0∈M, define a sequence ξn by ξ2j+1=Oξ2j,ξ2j+2=Rξ2j+1 for j=0,1,2,⋯. We show that O and R have at most one common fixed point. Assume that v≠w are two common fixed points, then Rv=Ov=v≠w=Rw=Ow. Therefore, ϱOv,Rw=ϱv,w>0. It follows from the property of Dsp that αsv,w≥sp or αsw,v≥sp. Without loss of generality, suppose that αsv,w≥sp. Letting ξ=v and η=w in (84), we have
(86)ϱv,w+φv+φw≤spϱOv,Rw+φOv+φRw≤αsv,wϱOv,Rw+φOv+φRw≤βθsv,w,ϱ,O,R,φ+Lθtv,w,ϱ,O,φ,where
(87)sv,w,ϱ,O,R,φ=λmaxϱv,w+φv+φw,1+ϱv,Ov+φv+φOv1+ϱv,w+φv+φw·ϱOv,Rw+φOv+φRw,ϱw,Rw+φw+φRw1+ϱv,w+φv+φw·ϱv,Ov+φv+φOv<14maxϱv,w+φv+φw,2ϱv,w+φv+φw,4ϱv,w+φv+φw=ϱv,w+φv+φw,and
(88)tv,w,ϱ,O,φ=λminϱv,Ov+φv+φOv,ϱw,Ov+φw+φOv<14minϱv,v+φv+φv,ϱw,v+φw+φv<ϱv,w+φv+φw.
In view of (86), we have
(89)ϱv,w+φv+φw<θϱv,w+φv+φw<ϱv,w+φv+φw,which implies that ϱv,w+φv+φw=0. That is, v=w and φv=0. Hence, O,R have at most one common fixed point.
Now, assumeϱξn,ξn+1>0 for n∈ℕ. Otherwise, for some k, ξ2k=ξ2k+1, by assumption (ii) and Lemma 19, we have αsξ2k,ξ2k+1≥sp. According to (84), if ξ2k+1≠ξ2k+2, we obtain
(90)ϱξ2k+1,ξ2k+2+φξ2k+1+φξ2k+2≤spϱOξ2k,Rξ2k+1+φOξ2k+φRξ2k+1≤αsξ2k,ξ2k+1ϱOξ2k,Rξ2k+1+φOξ2k+φRξ2k+1≤βθsξ2k,ξ2k+1,ϱ,O,R,φ+Lθtξ2k,ξ2k+1,ϱ,O,φ,where
(91)sξ2k,ξ2k+1,ϱ,O,R,φ=λmaxϱξ2k,ξ2k+1+φξ2k+φξ2k+1,1+ϱξ2k,Oξ2k+φξ2k+φOξ2k1+ϱξ2k,ξ2k+1+φξ2k+φξ2k+1·ϱOξ2k,Rξ2k+1+φOξ2k+φRξ2k+1,ϱξ2k+1,Rξ2k+1+φξ2k+1+φRξ2k+11+ϱξ2k,ξ2k+1+φξ2k+φξ2k+1·ϱξ2k,Oξ2k+φξ2k+φOξ2k<14maxϱξ2k,ξ2k+1+φξ2k+φξ2k+1,1+ϱξ2k,ξ2k+1+φξ2k+φξ2k+11+ϱξ2k,ξ2k+1+φξ2k+φξ2k+1·ϱξ2k+1,ξ2k+2+φξ2k+1+φξ2k+2,ϱξ2k+1,ξ2k+2+φξ2k+1+φξ2k+21+ϱξ2k,ξ2k+1+φξ2k+φξ2k+1·ϱξ2k,ξ2k+1+φξ2k+φξ2k+1≤ϱξ2k+1,ξ2k+2+φξ2k+1+φξ2k+2,(92)tξ2k,ξ2k+1,ϱ,O,φ=λminϱξ2k,Oξ2k+φξ2k+φOξ2k,ϱξ2k+1,Oξ2k+φξ2k+1+φOξ2k<14minϱξ2k,ξ2k+1+φξ2k+φξ2k+1,ϱξ2k+1,ξ2k+1+φξ2k+1+φξ2k+1≤ϱξ2k+1,ξ2k+2+φξ2k+1+φξ2k+2.
In light of (90) and above inequalities, we have
(93)ϱξ2k+1,ξ2k+2+φξ2k+1+φξ2k+2≤βθsξ2k,ξ2k+1,ϱ,O,R,φ+Lθtξ2k,ξ2k+1,ϱ,O,φ+Lθϱξ2k+1,ξ2k+2+φξ2k+1+φξ2k+2<θϱξ2k+1,ξ2k+2+φξ2k+1+φξ2k+2<ϱξ2k+1,ξ2k+2+φξ2k+1+φξ2k+2,which yields that ϱξ2k+1,ξ2k+2+φξ2k+1+φξ2k+2=0. It follows that ξ2k+1=ξ2k+2.
Thus,ξ2k is a common fixed point. Similarly, we can prove that ξ2k+1 is a common fixed point of O and R when ξ2k+1=ξ2k+2.
Now, assume thatϱξn,ξn+1>0 for each n∈ℕ. Applying (84) with ξ=ξ2n,η=ξ2n+1, we get
(94)sϱξ2n+1,ξ2n+2+φξ2n+1+φξ2n+2≤βθsξ2n,ξ2n+1,ϱ,O,R,φ+Lθtξ2n,ξ2n+1,ϱ,O,φ,where
(95)sξ2n,ξ2n+1,ϱ,O,R,φ=λmaxϱξ2n,ξ2n+1+φξ2n+φξ2n+1,1+ϱξ2n,ξ2n+1+φξ2n+φξ2n+11+ϱξ2n,ξ2n+1+φξ2n+φξ2n+1·ϱξ2n+1,ξ2n+2+φξ2n+1+φξ2n+2,ϱξ2n+1,ξ2n+2+φξ2n+1+φξ2n+21+ϱξ2n,ξ2n+1+φξ2n+φξ2n+1·ϱξ2n,ξ2n+1+φξ2n+φξ2n+1<14maxϱξ2n,ξ2n+1+φξ2n+φξ2n+1,ϱξ2n+1,ξ2n+2+φξ2n+1+φξ2n+2,(96)tξ2n,ξ2n+1,ϱ,O,φ=λminϱξ2n,ξ2n+1+φξ2n+φξ2n+1,ϱξ2n+1,ξ2n+1+φξ2n+1+φξ2n+1<ϱξ2n,ξ2n+1+φξ2n+φξ2n+1.
Ifϱξ2n+1,ξ2n+2+φξ2n+1+φξ2n+2>ρξ2n,ξ2n+1+φξ2n+φξ2n+1 for some n, then by (94), (95), and (96), we have
(97)sϱξ2n+1,ξ2n+2+φξ2n+1+φξ2n+2≤βθϱξ2n+1,ξ2n+2+φξ2n+1+φξ2n+2+Lθϱξ2n+1,ξ2n+2+φξ2n+1+φξ2n+2<θϱξ2n+1,ξ2n+2+φξ2n+1+φξ2n+2<ϱξ2n+1,ξ2n+2+φξ2n+1+φξ2n+2,which yields that
(98)ϱξ2n+1,ξ2n+2+φξ2n+1+φξ2n+2=0.
That is,ϱξ2n+1,ξ2n+2=0, a contradiction. Therefore,
(99)ϱξ2n+1,ξ2n+2+φξ2n+1+φξ2n+2≤ϱξ2n,ξ2n+1+φξ2n+φξ2n+1,for n∈ℕ. It follows from (94),(95), and (96) that
(100)sϱξ2n+1,ξ2n+2+φξ2n+1+φξ2n+2<ϱξ2n,ξ2n+1+φξ2n+φξ2n+1.
Using the same technique, we have(101)sϱξ2n+2,ξ2n+3+φξ2n+2+φξ2n+3<ϱξ2n+1,ξ2n+2+φξ2n+1+φξ2n+2,which implies that ϱξn,ξn+1+φξn+φξn+1 is a nonincreasing sequence satisfying
(102)sϱξn+1,ξn+2+φξn+1+φξn+2<ϱξn,ξn+1+φξn+φξn+1.
So there exists aλ≥0 satisfying
(103)limn⟶+∞ϱξn,ξn+1+φξn+φξn+1=λ.
Now, we supposeλ>0. By virtue of (94), (95), (96), and (99), one can get that
(104)ϱξ2n+1,ξ2n+2+φξ2n+1+φξ2n+2≤βθϱξ2n,ξ2n+1+φξ2n+φξ2n+1+Lθϱξ2n,ξ2n+1+φξ2n+φξ2n+1.
Lettingn⟶+∞ in (104), we have
(105)λ=limn⟶+∞ϱξ2n+1,ξ2n+2+φξ2n+1+φξ2n+2≤βlimn⟶+∞θϱξ2n,ξ2n+1+φξ2n+φξ2n+1+Llimn⟶+∞θϱξ2n,ξ2n+1+φξ2n+φξ2n+1<limn⟶+∞θϱξ2n+1,ξ2n+2+φξ2n+1+φξ2n+2=θλ<λ,a contradiction. It follows that
(106)limn⟶+∞ϱξn,ξn+1+φξn+φξn+1=0.
Hence,(107)limn⟶+∞ϱξn,ξn+1=0,limn⟶+∞φξn=0.Now, we will show thatξn is Cauchy. It is sufficient to show that ξ3n,ξ3n+1, and ξ3n+2 are Cauchy. First of all, we prove ξ3n is Cauchy. We take into consideration the following.:Case 1.
k=2m+1, where m≥1 and 3n is an odd number. By means of rectangular inequality and (102), we deduce that
(108)ϱξ3n,ξ3n+3k≤sϱξ3n,ξ3n+1+ϱξ3n+1,ξ3n+2+ϱξ3n+2,ξ3n+3k≤sϱξ3n,ξ3n+1+ϱξ3n+1,ξ3n+2+s2ϱξ3n+2,ξ3n+3+ϱξ3n+3,ξ3n+4+ϱξ3n+4,ξ3n+3k≤sϱξ3n,ξ3n+1+ϱξ3n+1,ξ3n+2+s2ϱξ3n+2,ξ3n+3+ϱξ3n+3,ξ3n+4+s3ϱξ3n+4,ξ3n+5+ϱξ3n+5,ξ3n+6+⋯⋯+s3m+1ϱξ3n+32m+1−3,ξ3n+32m+1−2+ϱξ3n+32m+1−2,ξ3n+32m+1−1+ϱξ3n+32m+1−1,ξ3n+32m+1≤s1s3n+1s3n+1+s21s3n+2+1s3n+3+⋯+s·s3m+1·1s3n+6m+2ϱξ0,ξ1+φξ0+φξ1≤s1s3n1+1s+1s2+⋯ϱξ0,ξ1+φξ0+φξ1+s1s3n+11+1s+1s2+⋯ϱξ0,ξ1+φξ0+φξ1=1s3n·1+s1−1/sϱξ0,ξ1+φξ0+φξ1⟶0asn⟶∞.The case that3n is an even number is similar to the case that 3n is an odd number.Case 2.
k=2m, where m≥1 and 3n is an odd number. In view of rectangular inequality again, we get that
(109)ϱξ3n,ξ3n+3k≤sϱξ3n,ξ3n+1+ϱξ3n+1,ξ3n+2+ϱξ3n+2,ξ3n+3k≤sϱξ3n,ξ3n+1+ϱξ3n+1,ξ3n+2+s2ϱξ3n+2,ξ3n+3+ϱξ3n+3,ξ3n+4+ϱξ3n+4,ξ3n+3k≤sϱξ3n,ξ3n+1+ϱξ3n+1,ξ3n+2+s2ϱξ3n+2,ξ3n+3+ϱξ3n+3,ξ3n+4+s3ϱξ3n+4,ξ3n+5+ϱξ3n+5,ξ3n+6+⋯⋯+s3mϱξ3n+6m−3,ξ3n+6m−2+ϱξ3n+6m−2,ξ3n+6m−1+ϱξ3n+6m−1,ξ3n+6m≤s1s3n+1s3n+1+s21s3n+2+1s3n+3+⋯+s·s3m+1·1s3n+6m+2ϱξ0,ξ1+φξ0+φξ1≤s1s3n1+1s+1s2+⋯ϱξ0,ξ1+φξ0+φξ1+s1s3n+11+1s+1s2+⋯ϱξ0,ξ1+φξ0+φξ1=1s3n·1+s1−1/sϱξ0,ξ1+φξ0+φξ1⟶0asn⟶∞.
The case that3n is an even number is similar to the case that 3n is an odd number.
Hence,ξ3n is a Cauchy sequence. Similarly, ξ3n+1,ξ3n+2 are also Cauchy sequences. That is, ξn is Cauchy. According to the completeness of M,ρ, we obtain that there is a ξ∗ in M satisfying
(110)limn⟶+∞Oξ2n=limn⟶+∞Rξ2n+1=ξ∗.
In view of the definition ofφ, we have
(111)φξ∗≤liminfn⟶+∞φξn=0.
Next, we show thatOξ∗=Rξ∗=ξ∗ provided O or R is continuous. Without loss of generality, assume that O is continuous. By (110), we get
(112)ξ∗=limn⟶+∞Oξ2n=Olimn⟶+∞ξ2n=Oξ∗.
This implies thatξ∗ is a fixed point of O.
Using propertyCsp, we have αsξ∗,ξ∗≥sp. If ξ∗≠Rξ∗, from condition (84), one can deduce
(113)ϱξ∗,Rξ∗+φξ∗+φRξ∗≤sϱOξ∗,Rξ∗+φOξ∗+φRξ∗≤αsξ∗,ξ∗ϱOξ∗,Rξ∗+φOξ∗+φRξ∗≤βθsξ∗,ξ∗,ϱ,O,R,φ+Lθtξ∗,ξ∗,ϱ,O,φ,where
(114)sξ∗,ξ∗,ϱ,O,R,φ=λmaxϱξ∗,ξ∗+φξ∗+φξ∗,1+ϱξ∗,Oξ∗+φξ∗+φOξ∗1+ϱξ∗,ξ∗+φξ∗+φξ∗·ϱOξ∗,Rξ∗+φOξ∗+φRξ∗,ϱξ∗,Rξ∗+φξ∗+φRξ∗1+ϱξ∗,ξ∗+φξ∗+φξ∗·ϱξ∗,Oξ∗+φξ∗+φOξ∗<ϱξ∗,Rξ∗+φRξ∗,(115)tξ∗,ξ∗,ϱ,O,φ=0≤ϱξ∗,Rξ∗+φRξ∗.
It follows from (113) that
(116)ϱξ∗,Rξ∗+φRξ∗≤βθρξ∗,Rξ∗+φRξ∗+Lθϱξ∗,Rξ∗+φRξ∗<θϱξ∗,Rξ∗+φRξ∗<ϱξ∗,Rξ∗+φRξ∗,which implies that ϱξ∗,Rξ∗+φRξ∗=0, that is, ξ∗=Rξ∗ and φRξ∗=0. It follows that O and R possess the unique common fixed point ξ∗. This completes the proof.Example 4.
LetM=0,+∞ and ϱξ,η=ξ−η2. Define mappings O,R:M⟶M by
(117)Oξ=ξ72,Rξ=ξ63,ξ∈0,+∞.
Define mappingsαs:M×M⟶0,+∞ by
(118)αsξ,η=s3,ξ,η∈0,+∞.
Defineθ:0,+∞⟶0,+∞ and φ:M⟶M with θξ=ξ/2,φξ=ξ2. Let β=1/3 and λ=1/6.
Forξ,η∈M with αsξ,η≥s3, we can know that for ξ,η∈0,+∞ such that ξ≠0 or η≠0,
(119)αsξ,ηϱOξ,Rη+φOξ+φRη=33·ξ72−η632+ξ722+η632=27·181·34949ξ264+η2≤149ξ2+η2,βθsξ,η,ϱ,O,R,φ+Lθtξ,η,ϱ,O,φ≥βθsξ,η,ϱ,O,R,φ≥13·12·16ϱξ,η+φξ+φη=136ξ−η2+ξ2+η2≥136ξ2+η2.
In view of above inequalities, one can obtain that(120)αsξ,ηϱOξ,Rη+φOξ+φRη≤βθsξ,η,ϱ,O,R,φ+Lθtξ,η,ϱ,O,φ,for ξ,η∈M with αsξ,η≥sp and ϱOξ,Rη+φOξ+φRη≠0. Hence, all requirements of Theorem 21 are fulfilled with p=3, s=3, and L<2/3. It is obvious that O and R possess the unique common fixed point 0.Takingφ=0 in Theorem 21, one can obtain the following result.Corollary 22.
LetM,ϱ be a complete rectangular b-metric space with coefficient s≥1. Suppose O,R:M⟶M are two given mappings and one of O and R is a continuous mapping. Assume that αs:M×M⟶0,+∞ and p≥1 is a constant and there exist θ∈Θ,β,L≥0 and β+L<1,0<λ<1/4 such that
(121)αsξ,ηϱOξ,Rη≤βθs∗ξ,η,ϱ,O,R+Lθt∗ξ,η,ϱ,O,for ξ,η∈M with αsξ,η≥sp and ϱOξ,Rη≠0, where
(122)s∗ξ,η,ϱ,O,R=λmaxϱξ,η,1+ϱξ,Oξ1+ϱξ,η·ϱOξ,Rη,ϱη,Rη1+ϱξ,η·ϱξ,Oξ,t∗ξ,η,ϱ,O=λminϱξ,Oξ,ϱη,Oξ.
If(i)
O,R are triangular αs orbital admissible(ii)
there isξ0∈M with satisfying αsξ0,Oξ0≥sp(iii)
propertiesCsp and Dsp are satisfiedthen O and R possess a unique common fixed point


## 4. Application

In this part, we shall prove the existence and uniqueness of solution to the integral equation:(123)ξϑ=∫0KFϑ,λ,ξλdλ.LetM=C0,K denote the collection of all continuous mappings on 0,K. For p≥2, define
(124)ϱξ,η=supϑ∈0,Kξϑ−ηϑpforallξ,η∈M.Hence,M,ρ is a complete rectangular b-metric space with s=3p−1.Theorem 23.
LetO:M⟶M by Oξϑ=∫0KFϑ,λ,ξλdλ and δ:−∞,+∞×−∞,+∞⟶−∞,+∞ be a given mapping. Suppose that
(i)
F:0,K×0,K×−∞,+∞⟶0,+∞ is continuous(ii)
for allϑ∈0,K and ξ,η∈M, δξϑ,ηϑ≥0 implies δOξϑ,Oηϑ≥0(iii)
there existsξ0∈M satisfying δξ0ϑ,Oξ0ϑ≥0 for ϑ∈0,K,(iv)
propertiesAsp and Bsp are fulfilled when R=I(v)
there is a continuous mappingγ:0,K×0,K⟶0,+∞ such that(125)supϑ∈0,K∫0Kγϑ,λdλ≤1(vi)
there exists a real numberβ∈0,1/s satisfying for ϑ,λ∈0,K×0,K,(126)∣Fϑ,λ,ξλ−Fϑ,λ,ηλ∣≤β2sppγϑ,λ∣ξλ−ηλ∣Then, the integral equation (123) possesses a unique solution ξϑ∈M.Proof.
Defineα:M×M⟶0,+∞ by
(127)αξ,η=sp,ifδξϑ,ηϑ≥0,0,otherwise.
Forξ,η∈M, according to assumptions (i)-(vi), we obtain
(128)spϱOξϑ,Oηϑ=spsupϑ∈0,KOξϑ−Oηϑp=spsupϑ∈0,K∫0KFϑ,λ,ξλdλ−∫0KFϑ,λ,ηλdλp≤spsupϑ∈0,K∫0KFϑ,λ,ξλ−Fϑ,λ,ηλdλp≤spsupϑ∈0,K∫0Kβ2sppγϑ,λξλ−ηλdλp≤spsupϑ∈0,K∫0Kβ2sppγϑ,λdλp·supϑ∈0,Kξϑ−ηϑp=β·12ϱξ,η≤βθm∗ξ,η,ϱ,O.
Lettingθϑ=ϑ/2, one can verify that all the conditions of Corollary 16 hold. As a result, O possesses a unique fixed point ξ∈M, that is, ξϑ is the unique solution of integral equation (123). This completes the proof.☐



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*Source: 8476040-2022-01-13.xml*