# Fixed Point Results for Generalizedα,ψ-Contraction Mapping in Rectangular b-Metric Spaces

Authors: Mustefa Abduletif Mamud; Kidane Koyas Tola
Journal: Abstract and Applied Analysis
 (2022)

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## Abstract

In this paper, we introduce generalizedα,ψ-contraction mappings in the setting of rectangular b-metric spaces and established existence and uniqueness of fixed points for the mappings introduced. Our results extend and generalize related fixed point results in the existing literature. We derive some consequences and corollaries from our obtained results. Also, we provide examples in support of our main findings. Furthermore, we determined a solution to an integral equation by applying our obtained results.


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## Body

## 1. Introduction

Fixed point theory is one of the most important topics in mathematics, especially in analysis. Due to its application in various disciplines like engineering, computer science, biological sciences, and economics, many researchers took interest in fixed point theory and its application. The Banach contraction principle is a fundamental result in fixed point theory [1]. For one century, due to its importance and simplicity, several authors have obtained many interesting extensions and generalizations of the Banach contraction principle in several direction. One possible direction is the notion of best proximity point results. In this line of research, [2–4] obtained interesting best proximity point results and derived fixed results as consequences of their works in which Banach fixed point is one of them. Also, taking the key role of the notion of the metric in mathematics and hence in quantitative sciences, it has been extended and generalized in several distinct directions by many authors. In particular, b-metric spaces introduced by Bakhtin [5] and Czerwik [6] are one generalization of metric spaces. In 2000, Branciari [7] introduced the concept of rectangular metric space by replacing the sum on the right hand side of the triangular inequality in the definition of a metric space by a three-term expression proved an analogue of the Banach contraction principle in such space. Also, various fixed point results were established on such spaces (see [8–13] and references therein). Recently, George et al. [14] announced the notion of rectangular b-metric space as a generalization of metric, b-metric space, and rectangular metric space; many authors initiated and studied many existing fixed point theorems in such spaces (see [15–20]). Recently, in 2019, Baiya and Kaewcharoen [21] established fixed point theorems for generalized contractions in complete rectangular metric spaces and proved the existence of fixed points, and in 2020, Kari et al. [22] introduced the notion of θ−ϕ-contraction in b-rectangular metric spaces and studied the existence and uniqueness of fixed point for the mappings introduced. Inspired and motivated by the works of Baiya and Kaewcharoen [21] and Kari et al. [22] in this paper, we introduce a new notion of generalized α,ψ-contraction mappings and establish some fixed point results for such mappings in complete rectangular b-metric spaces. Our results extend and generalize related fixed point results in the existing literature.


## 2. Preliminaries

In this section, we recall some basic definitions and results that will be used to prove our main results.Notation: We need the following symbols and class of functions to prove certain results of this section:
(1)
R+=0,∞;(2)
ℕ is the set of all natural numbers;(3)
Ψ1=ψ=0,∞⟶1,∞, such that, ψ is nondecreasing, for each sequence tn⊂0,∞, ψtn⟶1 if and only if tn⟶0 and there exist r∈0,1 and l∈0,∞ such that limt⟶0ψt−1/tr=l;(4)
Ψ2=ψ=0,∞⟶1,∞, such that, ψ is continuous and nondecreasing}.Definition 1 (see [23]).
LetX be a nonempty set and d:X×X⟶R+ be a function satisfying the following conditions:
(i)
dx,y=0 if and only if x=y;(ii)
dx,y=dy,x;(iii)
dx,z≤dx,y+dy,z for all x,y,z∈X.Then,d is called a metric, and the pair X,d is called a metric space.The following is the definition of the notion ofb-metric space.Definition 2 (see [6]).
LetXbe a nonempty set and s≥1 be a given real number. A function d:X×X⟶R+ is said to be a b-metric if and only if for all x,y,z∈X, the following conditions are satisfied:
(i)
dx,y=0 if and only if x=y;(ii)
dx,y=dy,x;(iii)
dx,z≤sdx,y+dy,z.
The pairX,d is called a b-metric space.Definition 3 (see [7]).
LetX be a nonempty set, and let d:X×X⟶R+ be a mapping such that for all x,y∈X and all distinct points u,v∈X, each distinct from x and y:
(i)
dx,y=0 if and only if x=y;(ii)
dx,y=dy,x;(iii)
dx,y≤dx,u+du,v+dv,y (rectangular inequality).
Then,d is called a rectangular metric, and the pair X,d is called a rectangular metric space.Definition 4 (see [14]).
LetX be a nonempty set, s≥1 be a given real number, and d:X×X⟶R+ be a mapping such that for all x,y∈X and all distinct points u,v∈X, each distinct from x and y:
(i)
dx,y=0 if and only if x=y;(ii)
dx,y=dy,x;(iii)
dx,y≤sdx,u+du,v+dv,y (b-rectangular inequality).
Then,d is called a b-rectangular metric, and the pair X,d is called a b-rectangular metric space.Definition 5 (see [14]).
LetX be a b-rectangular metric space and xn be a sequence in X; we say that:
(i)
xnb-rectangular converges to x∈X if dxn,x⟶0 as n⟶∞;(ii)
xn is a b-rectangular Cauchy sequence if dxn,xm⟶0 as n,m⟶∞;(iii)
X,d is a b-rectangular complete if every b-rectangular Cauchy sequence in X is b-rectangular convergent.Lemma 6 (see [22]).
LetX be a rectangular b-metric space, and let xn be a sequence in X. If xn is not a Cauchy sequence, then, there exists ε>0 and two subsequences xnk and xmk of xn, such that nk is the smallest index with nk>mk>k for which dxmk,xnk≥ε and dxmk,xnk−1<ε. Then, the following hold:
(1)
ε≤liminfk⟶∞dxmk,xnk≤limsupk⟶∞dxmk,xnk≤sε.(2)
ε≤liminfk⟶∞dxmk+1,xnk≤limsupk⟶∞dxmk+1,xnk≤sε.(3)
ε≤liminfk⟶∞dxmk,xnk+1≤limsupk⟶∞dxmk,xnk+1≤sε.(4)
ε/s≤liminfk⟶∞dxmk+1,xnk+1≤limsupk⟶∞dxmk+1,xnk+1≤s2ε.Definition 7 (see [24]).
LetX be a nonempty set and α:X×X⟶R+ be a function. A mapping T:X⟶X is said to be α-admissible; if for all x,y∈X, αx,y≥1 implies αTx,Ty≥1.Definition 8 (see [25]).
LetX be a nonempty set and α:X×X⟶R+ be a function. A mapping T:X⟶X is said to be α-orbital admissible; if for all x∈X, αx,Tx≥1 implies αTx,T2x≥1.Definition 9 (see [25]).
LetX be a nonempty set and T:X⟶X and α:X×X⟶R+ be a function. We say that T is triangular α-orbital admissible if
(i)
T is α-orbital admissible;(ii)
For allx,y∈X,αx,y≥1 and αy,Ty≥1implythatαx,Ty≥1.Lemma 10 (see [25]).
LetX be a nonempty set, T:X⟶X, and α:X×X⟶R+. Suppose that T is a triangular α-orbital admissible mapping and assume that there exists x0∈X such that αx0,Tx0≥1. Define a sequence xn in X by xn+1=Txn for all n∈ℕ∪0. Then, αxn,xm≥1 for all m,n∈ℕ with n<m.Theorem 11 (see [26]).
LetX,d be a complete rectangular metric space and T:X⟶X. Suppose that there exist ψ∈Ψ1 and λ∈0,1 such that for all x,y∈X(1)dTx,Ty≠0impliesψdTx,Ty≤ψRx,yλ,where
(2)Rx,y=maxdx,y,dx,Tx,dy,Ty.
Then,T has a fixed point.Theorem 12 (see [21]).
LetX,d be a complete rectangular metric space, T:X⟶X, and α:X×X⟶R+. Suppose that the following conditions hold:
(i)
There existψ∈Ψ2 and λ∈0,1 such that for all x,y∈X(3)dTx,Ty≠0impliesαx,y.ψdTx,Ty≤ψRx,yλ,where
(4)Rx,y=maxdx,y,dx,Tx,dy,Ty;(ii)
There existsx0∈X such that αx0Tx0≥1;(iii)
T is a triangular α-orbital admissible mapping;(iv)
T is continuous.
Then,T has a fixed point.Theorem 13 (see [21]).
LetX,d be a complete rectangular metric space and T:X⟶X and α:X×X⟶R+ be a function. Suppose that the following conditions hold:
(i)
There existψ∈Ψ2 and λ∈0,1 such that for all x,y∈X(5)dTx,Ty≠0impliesαx,y.ψdTx,Ty≤ψRx,yλ,where
(6)Rx,y=maxdx,y,dx,Tx,dy,Ty;(ii)
There existsx0∈X such that αx0Tx0≥1;(iii)
T is a triangular α-orbital admissible mapping;(iv)
Ifxn is a sequence in X such that αxn,xn+1≥1 for all n and xn⟶x∈X as n⟶∞, then there exists a subsequence xnk of xn such that αxnk,x≥1 for all k∈ℕ∪0.
Then,T has a fixed point.Theorem 14 (see [27]).
LetX,d be a complete rectangular metric space and T:X⟶X and α:X×X⟶R+ be a function. Suppose that the following conditions hold:
(i)
There existψ∈Ψ1 and λ∈0,1 such that for all x,y∈X(7)dTx,Ty≠0impliesαx,y.ψdTx,Ty≤ψRx,yλ,where
(8)Rx,y=maxdx,y,dx,Tx,dy,Ty,dx,Txdy,Ty1+dx,y;(ii)
There existsx0∈X such that αx0,Tx0≥1 and αx0,Tx02≥1;(iii)
T is a triangular α-orbital admissible mapping;(iv)
IfTx0n is a sequence in X such that αTx0n,Tx0n+1≥1 for all n and xn⟶x∈X as n⟶∞, then there exists a subsequence Tx0nk of Tx0n such that αTx0nk,x≥1 for all k∈ℕ∪0. Then T has a fixed point z in X and Tx0n converges to z.


## 3. Main Results

In this section, we introduce generalizedα,ψ-contraction mapping in the setting of rectangular b-metric spaces and prove fixed point results for the mappings introduced.We start this section by giving the definition of generalizedα,ψ-contraction mapping in the setting of rectangular b-metric space as follows:Definition 15.
LetX,d be a rectangular b-metric space with parameter s≥1. Suppose that ψ∈Ψ2, α:X×X⟶R+ be a function and λ∈0,1/s. A self-mapping T:X⟶X is called generalized α,ψ-contraction mapping if it satisfies for all x,y∈X in the following condition:
(9)dTx,Ty≠0impliesαx,yψs2dTx,Ty≤ψMx,yλ,where
(10)Mx,y=maxdx,y,dx,Tx,dy,Ty,dy,Tx,dx,Tx.dy,Ty1+dx,y,dx,Tx.dy,Ty1+dTx,Ty,dx,Tx1+dy,Tx1+dTx,Ty.
Now, we state and prove the following fixed point theorem.Theorem 16.
LetX,d be a complete rectangular b-metric space, T:X⟶X, and α:X×X⟶R+.
Suppose that the following conditions hold:(i)
T is a triangular α-orbital admissible mapping;(ii)
T is generalized α,ψ-contraction mapping;(iii)
There existsx0∈X such that αx0,Tx0≥1;(iv)
T is continuous.
Then,Thas a fixed point in X.Proof.
By (iii), there existsx0∈X such that αx0,Tx0≥1. We define an iterative sequence xn in X by xn+1=Txn, for n∈ℕ∪0. Suppose that xn0=xn0+1 for some n0∈N. Since Txn0=xn0+1=xn0, the point xn0 forms a fixed point of T that completes the proof.
From now on, we suppose thatxn≠xn+1 for all n∈ℕ∪0. By condition (iii), we have αx0,Tx0≥1. Using Lemma 10, we obtain that
(11)αxn,xn+1≥1foralln∈ℕ∪0.
From (9) to (11), for all n∈ℕ∪0, we have the following:
(12)ψdxn,xn+1=ψdTxn−1,Txn≤αxn−1,xnψs2dTxn−1,Txn≤ψMxn−1,xnλ<ψMxn−1,xn,where
(13)Mxn−1,xn=maxdxn−1,xn,dxn−1,Txn−1,dxn,Txn,dxn,Txn−1,dxn−1,Txn−1.dxn,Txn1+dxn−1,xn,dxn−1,Txn−1.dxn,Txn1+dTxn−1,Txn,dxn−1,Txn−11+dxn,Txn−11+dTxn−1,Txn=maxdxn−1,xn,dxn−1,xn,dxn,xn+1,dxn,xndxn−1,xn.dxn,xn+11+dxn−1,xn,dxn−1,xn.dxn,xn+11+dxn,xn+1,dxn−1,xn1+dxn,xn1+dxn,xn+1=maxdxn−1,xn,dxn,xn+1.
IfMxn−1,xn=dxn,xn+1, then by (12), we get ψdxn,xn+1≤ψdxn,xn+1λ<ψdxn,xn+1, which is a contradiction. Hence, Mxn−1,xn=dxn−1,xn. Using (12), we have ψdxn,xn+1<ψdxn−1,xn. Since ψ is nondecreasing, we have dxn,xn+1<dxn−1,xn for all n≥1.
Hence,dxn,xn+1 is decreasing sequence of nonnegative real numbers. Hence, dxn,xn+1 converges to a nonnegative real number, say r≥0. That is
(14)limn⟶∞dxn,xn+1=randdxn,xn+1≥r.
We will prove thatr=0. Suppose that r>0. Since ψ is nondecreasing, by using (12) and (14), we obtain that
(15)1<ψr≤ψdxn+1,xn+2=ψdTxn,Txn+1≤αxn,xn+1ψs2dTxn,Txn+1≤ψdxn,xn+1λ≤ψdxn−1,xnλ2⋮≤ψdx0,x1λn.
By taking limit asn⟶∞ in the above inequality, we get ψr=1, which contradicts to the assumption that ψt>1 for all t>0. Hence, we have r=0, and therefore
(16)limn⟶∞dxn,xn+1=r=0.
Now, we shall prove thatxn≠xm for all n≠m or xn≠xn+p for all n,p∈ℕ. Assume on the contrary that there exist n,p∈ℕ such that xn=xn+p. Since dxn,xn+1>0, for each n∈ℕ. Without loss of generality, we may assume that p>1. Using (9) and (11), we obtain that
(17)ψdxn,xn+1=ψdxn+p,xn+p+1=ψdTxn+p−1,Txn+p≤αxn+p−1,xn+pψs2dTxn+p−1,Txn+p≤ψMxn+p−1,xn+pλ<ψMxn+p−1,xn+p,where
(18)Mxn+p−1,xn+p=maxdxn+p−1,xn+p,dxn+p−1,Txn+p−1,dxn+p,Txn+p,dxn+p,Txn+p−1,dxn+p−1,Txn+p−1.dxn+p,Txn+p1+dxn+p−1,xn+p,dxn+p−1,Txn+p−1.dxn+p,Txn+p1+dTxn+p−1,Txn+p,dxn+p−1,Txn+p−11+dxn+p,Txn+p−11+dTxn+p−1,Txn+p=maxdxn+p−1,xn+p,dxn+p−1,xn+p,dxn+p,xn+p+1,dxn+p−1,xn+p.dxn+p,xn+p+11+dxn+p−1,xn+p,dxn+p−1,xn+p.dxn+p,xn+p+11+dxn+p,xn+p+1,dxn+p−1,xn+p1+dxn+p,xn+p1+dxn+p,xn+p+1=maxdxn+p−1,xn+p,dxn+p,xn+p+1.
IfMxn+p−1,xn+p=dxn+p,xn+p+1, then from (17), we obtain that
(19)ψdxn,xn+1=ψdxn+p,xn+p+1<ψdxn+p,xn+p+1,which is a contradiction. Hence, Mxn+p−1,xn+p=dxn+p−1,xn+p.
By (17), we obtain that
(20)ψdxn,xn+1=ψdxn+p,xn+p+1<ψdxn+p−1,xn+p.
Sinceψ is nondecreasing from the above inequality, we get
(21)dxn,xn+1<dxn+p−1,xn+p.
By using (9), we have
(22)ψdxn+p−1,xn+p=ψdTxn+p−2,Txn+p−1≤αxn+p−2,xn+p−1ψs2dTxn+p−2,Txn+p−1≤ψMxn+p−2,xn+p−1λ<ψMxn+p−2,xn+p−1,
where(23)Mxn+p−2,xn+p−1=maxdxn+p−2,xn+p−1,dxn+p−2,Txn+p−2,.dxn+p−1,Txn+p−1,dxn+p−1,Txn+p−2,dxn+p−2,Txn+p−2.dxn+p−1,Txn+p−11+dxn+p−2,xn+p−1,dxn+p−2,Txn+p−2.dxn+p−1,Txn+p−11+dTxn+p−2,Txn+p−1,dxn+p−2,Txn+p−21+dxn+p−1,Txn+p−21+dTxn+p−2,Txn+p−1=maxdxn+p−2,xn+p−1,dxn+p−2,xn+p−1,dxn+p−1,xn+p,dxn+p−2,xn+p−1.dxn+p−1,xn+p1+dxn+p−2,xn+p−1,dxn+p−2,xn+p−1.dxn+p−1,xn+p1+dxn+p−1,xn+p,dxn+p−2,xn+p−11+dxn+p−1,xn+p−11+dxn+p−1,xn+p=maxdxn+p−2,xn+p−1,dxn+p−1,xn+p.
IfMxn+p−2,xn+p−1=dxn+p−1,xn+p, then by (22), we obtain that
(24)ψdxn+p−1,xn+p≤ψdxn+p−1,xn+pλ<ψdxn+p−1,xn+p,which is a contradiction. Hence,
(25)Mxn+p−2,xn+p−1=dxn+p−2,xn+p−1.
By (22), we have
(26)ψdxn+p−1,xn+p≤ψdxn+p−2,xn+p−1λ<ψdxn+p−2,xn+p−1.
Sinceψ is nondecreasing, we get
(27)dxn+p−1,xn+p<dxn+p−2,xn+p−1.
By continuing this process, we obtain the following inequality:(28)dxn,xn+1<dxn+p−1,xn+p<dxn+p−2,xn+p−1<⋯<dxn,xn+1,which is a contradiction. Hence xn≠xm for all n≠m.
We now prove thatdxn,xn+2 is bounded. Since dxn,xn+1 is bounded, there exists B>0 such that
(29)dxn,xn+1≤Bforalln∈ℕ∪0.
Ifdxn,xn+2>B for all n∈ℕ∪0, then from
(30)Mxn−1,xn+1=maxdxn−1,xn+1,dxn−1,Txn−1,dxn+1,Txn+1,dxn+1,Txn−1,dxn−1,Txn−1dxn+1,Txn+11+dxn−1,xn+1,dxn−1,Txn−1dxn+1,Txn+11+dTxn−1,Txn+1,dxn−1,Txn−11+dxn+1,Txn−11+dTxn−1,Txn+1=maxdxn−1,xn+1,dxn−1,xn,dxn+1,xn+2,dxn+1,xn,dxn−1,xndxn+1,xn+21+dxn−1,xn+1,dxn−1,xndxn+1,xn+21+dxn,xn+2,dxn−1,xn1+dxn+1,xn1+dxn,xn+2=dxn−1,xn+1.
and Lemma10, we obtain that
(31)ψdxn,xn+2=ψdTxn−1,Txn+1≤αxn−1,xn+1ψs2dTxn−1,Txn+1≤ψMxn−1,xn+1λ=ψdxn−1,xn+1λ<ψdxn−1,xn+1.
From above the inequality, we get for alln≥1(32)dxn,xn+2<dxn−1,xn+1.
This implies thatdxn,xn+2 is decreasing. Therefore, dxn,xn+2 is bounded.
Ifdxn,xn+2≤B for some n∈ℕ∪0, then from
(33)Mxn,xn+2=maxdxn,xn+2,dxn,Txn,dxn+2,Txn+2,dxn+2,Txn,dxn,Txndxn+2,Txn+21+dxn,xn+2,dxn,Txndxn+2,Txn+21+dTxn,Txn+2,dxn,Txn1+dxn+2,Txn1+dTxn,Txn+2=maxdxn,xn+2,dxn,xn+1,dxn+2,xn+3,dxn+2,xn+1,dxn,xn+1dxn+2,xn+31+dxn,xn+2,dxn,xn+1dxn+2,xn+31+dxn+1,xn+3,dxn,xn+11+dxn+2,xn+11+dxn+1,xn+3≤B
and Lemma10, we obtain that
(34)ψdxn+1,xn+3=ψdTxn,Txn+2≤αxn,xn+2ψs2dTxn,Txn+2≤ψMxn,xn+2λ≤ψBλ<ψB.
Therefore,dxn+1,xn+3<B. This implies that dxn,xn+2 is bounded. We next prove that limn⟶∞dxn,xn+2=0. Suppose that limn⟶∞dxn,xn+2≠0. So there exists a subsequence xnk of xn such that limk⟶∞dxnk,xnk+2=a for some a>0. Using (9) and Lemma 10, we have
(35)ψdxnk,xnk+2=ψdTxnk−1,Txnk+1≤αxnk−1,xnk+1ψs2dTxnk−1,Txnk+1≤ψMxnk−1,xnk+1λ<ψMxnk−1,xnk+1,where
(36)Mxnk−1,xnk+1=maxdxnk−1,xnk+1,dxnk−1,Txnk−1,dxnk+1,Txnk+1,dxnk+1,Txnk−1,dxnk−1,Txnk−1dxnk+1,Txnk+11+dxnk−1,xnk+1,dxnk−1,Txnk−1dxnk+1,Txnk+11+dTxnk−1,Txnk+1,dxnk−1,Txnk−11+dxnk+1,Txnk−11+dTxnk−1,Txnk+1=maxdxnk−1,xnk+1,dxnk−1,xnk,dxnk+1,xnk+2,dxnk+1,xnk,dxnk−1,xnkdxnk+1,xnk+21+dxnk−1,xnk+1,dxnk−1,xnkdxnk+1,xnk+21+dxnk,xnk+2,dxnk−1,xnk1+dxnk+1,xnk1+dxnk,xnk+2.
Lettingk⟶∞ in (35), we obtain that
(37)ψa=limk⟶∞ψdxnk,xnk+2≤limk⟶∞ψMxnk−1,xnk+1λ=ψaλ<ψa,which is a contradiction. Therefore,
(38)limn⟶∞dxn,xn+2=0.
We now prove thatxn is a b-rectangular b-Cauchy sequence in X. Suppose that xn is not a Cauchy sequence. Then, there exists ε>0 and two subsequences xnk and xmk of xn, such that nk is the smallest index with nk>mk>k for which
(39)dxmk,xnk≥ε.
This implies that(40)dxmk,xnk−1<ε.
By applyingb-rectangular inequality, using (39) and (40), we get that
(41)ε≤dxmk,xnk≤sdxmk,xnk−1+sdxnk−1,xnk−2+sdxk−2,xnk<sε+sdxnk−1,xnk−2+sdxk−2,xnk.
Lettingk⟶∞ in above inequality, using (16) and (38), we get that
(42)ε≤liminfk⟶∞dxmk,xnk≤limsupk⟶∞dxmk,xnk≤sε.
Also,(43)ε≤dxmk,xnk≤sdxmk,xmk+1+sdxmk+1,xnk+1+sdxnk+1,xnk.
Lettingk⟶∞ in above inequality, using (16) and Lemma 6, we obtain
(44)εs≤liminfk⟶∞dxmk+1,xnk+1≤limsupk⟶∞dxmk+1,xnk+1≤s2ε.
For eachk∈ℕ, we have
(45)Mxmk,xnk=maxdxmk,xnk,dxmk,Txmk,dxnk,Txnk,dxnk,Txmk,dxmk,Txmkdxnk,Txnk1+dxmk,xnk,dxmk,Txmkdxnk,Txnk1+dTxmk,Txnk,dxmk,Txmk1+dxnk,Txmk1+dTxmk,Txnk=maxdxmk,xnk,dxmk,xmk+1,dxnk,xnk+1,dxnk,xmk+1,dxmk,xnkdxnk,xnk+11+dxmk,xnk,dxmk,xmk+1dxnk,xnk+11+dxmk+1,xnk+1,dxmk,xmk+11+dxnk,xmk+11+dxmk+1,xnk+1.
By using (16) and (42), we obtain that
(46)limsupk⟶∞Mxmk,xnk≤sε.
By (42) and (46), there exists a positive integer k0 such that dxnk+1,xmk+1>0 and Mxnk,xmk>0, for all k≥k0.
By Lemma10 and using (9), for all nk>mk>k>k0, we get that
(47)ψs2dxnk+1,xmk+1=ψs2dTxnk,Txmk≤αxnk,xmkψs2dTxnk,Txmk≤ψMxnk,xmkλ<ψMxnk,xmk.
Lettingk⟶∞ in the above inequality, by (42), (44), and (46) and the continuity of ψ, we obtain that
(48)ψsε≤limsupk⟶∞ψs2dxmk+1,xnk+1≤limsupk⟶∞ψdxnk,xmkλ≤ψsελ<ψsε,which is a contradiction. Therefore, xn is a rectangular b-Cauchy sequence in X. Since X is complete rectangular b-metric space, it follows that xn converges to some x∈X. Since T is continuous, we have
(49)x=limn⟶∞xn+1=limn⟶∞Txn=Tx.
Therefore,x is a fixed point of T.By removing the continuity assumption in Theorem16, we get the following fixed point result.Theorem 17.
LetX,d be a complete rectangular b-metric spaces, T:X⟶X and α:X×X⟶R+.
Suppose that the following conditions hold:(i)
T is generalized α,ψ-contraction mapping;(ii)
T is a triangular α-orbital admissible mapping;(iii)
There existsx0∈X such that αx0,Tx0≥1;(iv)
Ifxn is a sequence in X such that αxn,xn+1≥1 for all n∈ℕ∪0 and xn⟶x∈X as n⟶∞, then there exists a subsequence xnk of xn such that αxnk,x≥1 for all k∈ℕ.
Then,T has a fixed point.Proof.
As in the proof of Theorem16, we can construct a sequence xn in X such that xn+1=Txn, for all n∈ℕ∪0 and αxn,xn+1≥1 for all n∈ℕ∪0. The sequence xn is rectangular b-Cauchy, and since X is complete, there exists x∈X such that xn⟶x as n⟶∞. By (iv), there exists a subsequence xnk of xn such that αxnk,x≥1 for all k∈ℕ∪0. We can suppose that xnk≠Tx for all k∈ℕ∪0. By (9), we obtain that
(50)ψsdxnk+1,Tx=ψsdTxnk,Tx≤αxnk,xψs2dTxnk,Tx≤ψMxnk,xλ<ψMxnk,x,where
(51)Mxnk,x=maxdxnk,x,dxnk,Txnk,dx,Tx,dx,Txnk,dxnk,Txnkdx,Tx1+dxnk,x,dxnk,Txnkdx,Tx1+dTxnk,Tx,dxnk,Txnk1+dx,Txnk1+dTxnk,Tx=maxdxnk,x,dxnk,xnk+1,dx,Tx,dx,xnk+1,dxnk,xnk+1dx,Tx1+dxnk,x,dxnk,xnk+1dx,Tx1+dxnk+1,Tx,dxnk,xnk+11+dx,xnk+11+dxnk+1,Tx.
Taking limit ask⟶∞(52)limk⟶∞Mxnk,x=dx,Tx.
Now, we prove thatx=Tx. Suppose that x≠Tx. Therefore,
(53)dx,Tx≤sdx,xnk+sdxnk,xnk+1+sdxnk+1,Tx.
It follows that(54)dx,Tx≤limk⟶∞sdxnk+1,Tx.
By using (50), (52), and (54) and ψ, we obtain that
(55)ψdx,Tx≤limk⟶∞ψsdxnk+1,Tx=limk⟶∞ψsdTxnk,Tx≤limk⟶∞αxnk,xψs2dTxnk,Tx≤limk⟶∞ψMxnk,xλ=ψlimk⟶∞Mxnk,xλ=ψdx,Txλ<ψdx,Tx,which is a contradiction. Thus, x=Tx, and hence, x is a fixed point of T.Now, we prove the uniqueness of fixed point. For this, we need the following additional condition.(U): For allx,y∈FixT, we have αx,y≥1, where FixT denotes the set of all fixed points of T.Theorem 18.
Adding condition (U) to the hypothesis of Theorem 16 (or Theorem 17), we obtain the uniqueness of fixed point of T.Proof.
From the proof of Theorem16 (or Theorem 17), FixT≠∅. We argue by contradiction that there exist z,w be in FixT such that z=Tz and w=Tw with z≠w. By condition (U), we have αz,w≥1. So, by (9), we get
(56)ψdz,w=ψdTz,Tw≤αz,wψs2dTz,Tw≤ψMz,wλ<ψMz,w,where
(57)Mz,w=maxdz,w,dz,Tz,dw,Tw,dw,Tz,dz,Tzdw,Tw1+dz,w,dz,Tzdw,Tw1+dTz,Tw,dz,Tz1+dw,Tz1+dTz,Tw=maxdz,w,dz,z,dw,w,dw,z,dz,zdw,w1+dz,w,dz,zdw,w1+dz,w,dz,z1+dw,z1+dz,w=maxdz,w,0=dz,w.
From (56), we have
(58)ψdz,w≤ψdz,wλ<ψdz,w,which is a contradiction. Therefore, z=w.The following are corollaries to our main results.Corollary 19.
LetX,d be a complete rectangular b-metric space, T:X⟶X, ψ∈Ψ2, α:X×X⟶R+, and λ∈0,1. Suppose that the following conditions hold:
(i)
dTx,Ty≠0impliesαx,yψdTx,Ty≤ψMx,yλ,where
(59)Mx,y=maxdx,y,dx,Tx,dy,Ty,dy,Tx,dx,Tx.dy,Ty1+dx,y,dx,Tx.dy,Ty1+dTx,Ty,dx,Tx1+dy,Tx1+dTx,Ty;(ii)
T is a triangular α-orbital admissible mapping;(iii)
There existsx0∈X such that αx0,Tx0≥1;(iv)
For every pairx and y of fixed points of T, αx,y≥1;(v)
EitherT is continuous or xn is a sequence in X such that αxn,xn+1≥1 for all n∈ℕ∪0 and xn⟶x∈X as n⟶∞, then there exists a subsequence xnk of xn such that αxnk,x≥1 for all k∈ℕ
Then,T has a unique fixed point.proof The result follows by takings=1 in Theorem 16 (or Theorem 17).Remark 20.
By takingMx,y=maxdx,y,dx,Tx,dy,Ty in Corollary 19, we get Thoerem 2.1 (or Theorem 2.2) of Baiya and Kaewcharoen [21]. Thus, this work generalizes the work of Baiya and Kaewcharoen [21].Corollary 21.
LetX,dbe a complete rectangularbmetric spaceT:X⟶X, ψ∈Ψ2, andλ∈0,1. Suppose that the following condition hold:(60)dTx,Ty≠0impliesψdTx,Ty≤ψMx,yλ,where
(61)Mx,y=maxdx,y,dx,Tx,dy,Ty,dy,Tx.
Then,T has a unique fixed point.proof. The result follows by takingMx,y=maxdx,y,dx,Tx,dy,Ty,dy,Tx, αx,y=1 for all x,y∈X and s=1 in Theorem 16 (or Theorem 17).Corollary 22.
LetX,d be a complete rectangular b-metric space, T:X⟶X, ψ∈Ψ2 and λ∈0,1/s. Suppose that the following condition hold:
(62)dTx,Ty≠0impliesψs2dTx,Ty≤ψMx,yλ,where
(63)Mx,y=maxdx,y,dx,Tx,dy,Ty,dy,Tx,dx,Tx.dy,Ty1+dx,y,dx,Tx.dy,Ty1+dTx,Ty,dx,Tx1+dy,Tx1+dTx,Ty.
Then,T has a unique fixed point.
proof The result follows by takingαx,y=1 for all x,y∈X in Theorem 16 (or Theorem 17).Remark 23.
By takingMx,y=maxdx,y,dx,Tx,dy,Ty,dy,Tx and λ∈0,1 in Corollary 22, we get Theorem 11 of Kari et al. [22].Corollary 24.
LetX,d be a complete rectangularb-metric space, T:X⟶X, ψ∈Ψ2, α:X×X⟶R+ and λ∈0,1. Suppose that the following conditions hold:
(i)
dTx,Ty≠0impliesαx,yψdTx,Ty≤ψdx,yλ;(ii)
T is a triangular α-orbital admissible mapping;(iii)
There existsx0∈X such that αx0,Tx0≥1;(iv)
For every pairx and y of fixed points of T, αx,y≥1;(v)
EitherT is continuous or xn is a sequence in X such that αxn,xn+1≥1 for all n∈ℕ∪0 and xn⟶x∈X as n⟶∞, then there exists a subsequence xnk of xn such that αxnk,x≥1 for all k∈ℕ.
Then,T has a unique fixed point.proof. The result follows by takings=1 and Mx,y=dx,y in Theorem 16 (or Theorem 17).We now present examples for supporting our main results.Example 1.
LetX be a finite set defined as X=1,2,3,4.
Definedd:X×X⟶R+ as
dx,x=0forallx∈X
d1,2=d2,1=20,
d2,3=d3,2=d1,3=d3,1=2,
d1,4=d4,1=d2,4=d4,2=d3,4=d4,3=4.
Clearly,X,d is complete rectangular b-metric spaces with s=2. But X,d is neither a metric nor a rectangular metric space because
(64)20=d1,2≥d1,3+d3,2=2+2=4,20=d1,2≥d1,3+d3,4+d4,2=2+4+4=10.
LetT:X⟶X be the mapping defined by
(65)Tx=2ifx≠43ifx=4.
Defineα:X×X⟶0,∞, ψ:0,∞⟶1,∞ by
(66)αx,y=1ifx,y∈X\411200otherwise.;and ψt=et.We next illustrate that all conditions in Theorem16 and Theorem 17 hold. Taking x0=1, we have αx0,Tx0=α1,T1=α1,2=1≥1. We next prove that T is an α-orbital admissible.Letx∈X such that αx,Tx≥1. Therefore, x,Tx∈X\4 and then x∈1,2,3. By the definition of α, we obtain that
(67)αT1,T21=α2,2=1≥1,αT2,T22=α2,2=1≥1,αT3,T23=α2,2=1≥1.It follows thatT is an α-orbital admissible. Let x,y∈X such that αx,y≥1 and αy,Ty≥1. By the definition of α, we have x,y,Ty∈X\4. This yieldsα1,2≥1 and α2,T2≥1 implies α1,T2≥1,α1,3≥1 and α3,T3≥1 implies α1,T3≥1,α2,3≥1 and α3,T3≥1 implies α2,T3≥1,α2,1≥1 and α1,T1≥1 implies α2,T1≥1,α3,1≥1 and α1,T1≥1 implies α3,T1≥1,α3,2≥1 and α2,T2≥1 implies α3,T2≥1.This implies thatT is triangular α-orbital admissible. Let xn be a sequence such that αxn,xn+1≥1 for all n∈N and xn⟶x as n⟶∞. By the definition of α, for each n∈N, we get that xn∈X\4=1,2,3. We obtain that x∈1,2,3. Thus, we have αxn,x≥1 for each n∈N. We next prove that T is generalized α,ψ-contraction mapping. Let x,y∈X be such that dTx,Ty≠0. So we consider the following cases: x∈1,2,3 and y=4 or y∈1,2,3 and x=4. Since dx,y = dy,x, we divide the proof into three cases as follows:Case 1 Forx=1 and y=4, we have
(68)M1,4=maxd1,4,d1,T1,d4,T4,d4,T1,d1,T1d4,T41+d1,4,d1,T1d4,T41+dT1,T4,d1,T11+d4,T11+dT1,T4=max4,20,4,16,803,1003=1003.This implies that(69)α1,4ψs2dT1,T4=11200ψ4d2,3=11200e8≤e100/9=ψ10031/3=ψM1,41/3.Case 2 Forx=2 and y=4, we have
(70)M2,4=maxd2,4,d2,T2,d4,T4,d4,T2,d2,T2d4,T41+d2,4,d2,T2d4,T41+dT2,T4,d2,T21+d4,T21+dT2,T4=max4,0=4.This implies that(71)α2,4ψs2dT2,T4=11200ψ4d2,3=11200e8≤ψ41/3=ψM2,41/3.Case 3: Forx=3 and y=4, we have
(72)M3,4=maxd3,4,d3,T3,d4,T4,d4,T3,d3,T3d4,T41+d3,4,d3,T3d4,T41+dT3,T4,d3,T31+d4,T31+dT3,T4=max4,2,85,83,103=4.This implies that(73)α3,4ψs2dT3,T4=11200ψ4d2,3=11200e8≤e4/3=ψ41/3=ψM3,41/3.Hence, all assumptions in Theorem16 and Theorem 17 are satisfied, and thus, T has a fixed point which is x=2. Finally let x,y∈FixT. Clearly x=y=2, therefore, by the definition of α, we have αx,y=α2,2=1≥1. So, all assumptions in Theorem 18 are satisfied, and thus, T has a unique fixed point which is x=2Example 2.
LetX=A∪1,2, where A=0,1/2,1/3,1/4,1/5,1/6.
Defined:X×X⟶R+ by.
dx,y=0ifandonlyifx=y
dx,y=dy,xforallx,y∈Xwith
d0,1/2=d1/3,1/4=d1/5,1/6=0.09,
d0,1/3=d1/2,1/5=d1/4,1/5=0.04,
d0,1/4=d1/2,1/3=d1/4,1/6=0.16,
d0,1/5=d1/2,1/6=d1/3,1/6=0.25,
d0,1/6=d1/2,1/4=d1/3,1/5=0.36,
dx,y=x−y2.Then,X,d is a complete rectangular b-metric space with parameter s=3. But, d is neither metric nor rectangular metric because
(74)d0,16=0.36>0.29=d0,13+d13,16,d13,15=0.36>0.34=d13,14+d14,16+d16,15.Define mappingT:X⟶X by
(75)Tx=16ifx∈1,213ifx∈A.Defineα:X×X⟶0,∞, ψ:0,∞⟶1,∞ by
(76)αx,y=1ifx,y∈A12otherwise.;and ψt=et with λ=1/5∈0,1/3.We next illustrate that all conditions in Theorem17 hold.Clearly,T is a triangular α-orbital admissible mapping. Taking x0=1/2, we have αx0,Tx0=α1/2,T1/2=α1/2,1/3=1≥1. Let xn be a sequence such that αxn,xn+1≥1 for all n∈N and xn⟶x as n⟶∞. By the definition of α, for each n∈N, we get that xn∈A=0,1/2,1/3,1/4,1/5,1/6. We obtain that x∈0,1/2,1/3,1/4,1/5,1/6. Thus, we have αxn,x≥1 for each n∈N. We next prove T is generalized α,ψ-contraction mapping. Let x,y∈X be such that dTx,Ty≠0. So, we consider only the following case: x∈A and y∈1,2 or y∈A and x∈1,2. Since dx,y = dy,x, we have
(77)αx,yψs2dTx,Ty=12ψ9dTx,Ty=12e1/2≤e1/6=ψdx,Tx1/5=ψMx,y1/5.Hence, all assumptions in Theorem17 are satisfied, and thus, T has a fixed point which is x=1/3. Finally let x,y∈FixT. Clearly x=y=1/3; therefore, by the definition of α, we have αx,y=α1/3,1/3=1≥1. So, all assumptions in Theorem 18 are satisfied, and thus, T has a unique fixed point which is x=1/3.


### 3.1. Application to Integral Equation

In this section, we use Corollary22 to show that there is a solution to the following integral equation:
(78)xt=gt+∫01Lt,r,xrdr,where L:0,1×0,1×R⟶R+ and g:0,1⟶R are continuous function. Let X=C0,1,R be the set of all continuous functions defined on 0,1. Define d:X×X⟶R+ by
(79)dx,y=maxt∈0,1∣xt−yt∣2,which for all x,y∈X, then X,d is a complete rectangular b-metric space with parameter s=2. Now, we prove the following result.Theorem 25.
Suppose the following hypothesis hold:(i)
there is continuous functionβ:0,1×0,1⟶R+ such that
(80)Lt,r,xr−Lt,r,yr∣≤14βt,rxr−yr;(ii)
There isλ=1/3∈0,1/2 such that
(81)ψ∫01βt,rdr2dxt,yt≤ψdxt,yt1/3.Then, the integral Equation (78) has a unique solution in X.Proof.
Forx∈X and t∈0,1, define the mapping T:X⟶X by Txt=gt+∫01Lt,r,xrdr..
Forx,y∈X, from condition (i) and (ii), for all t∈0,1, we have
(82)ψs2dTxt,Tyt=ψ4∣Txt−Tyt∣2=ψ4∣gt+∫01Lt,r,xrdr−gt+∫01Lt,r,yrdr∣2=ψ4∣∫01Lt,r,xr−Lt,r,yrdr∣2≤ψ4∫01∣Lt,r,xr−Lt,r,yr∣dr2≤ψ∫01βt,r∣xr−yr∣dr2=ψ∫01βt,rxr−yr21/2dr2≤ψ∫01βt,rdxt,yt1/2dr2=ψdxt,yt∫01βt,rdr2≤ψdxt,yt1/3≤ψMxt,yt1/3.
Therefore, all conditions of Corollary22 are satisfied and as a result the mapping T has a unique fixed point in X, which is a solution of the integral equation in (78).


## 3.1. Application to Integral Equation

In this section, we use Corollary22 to show that there is a solution to the following integral equation:
(78)xt=gt+∫01Lt,r,xrdr,where L:0,1×0,1×R⟶R+ and g:0,1⟶R are continuous function. Let X=C0,1,R be the set of all continuous functions defined on 0,1. Define d:X×X⟶R+ by
(79)dx,y=maxt∈0,1∣xt−yt∣2,which for all x,y∈X, then X,d is a complete rectangular b-metric space with parameter s=2. Now, we prove the following result.Theorem 25.
Suppose the following hypothesis hold:(i)
there is continuous functionβ:0,1×0,1⟶R+ such that
(80)Lt,r,xr−Lt,r,yr∣≤14βt,rxr−yr;(ii)
There isλ=1/3∈0,1/2 such that
(81)ψ∫01βt,rdr2dxt,yt≤ψdxt,yt1/3.Then, the integral Equation (78) has a unique solution in X.Proof.
Forx∈X and t∈0,1, define the mapping T:X⟶X by Txt=gt+∫01Lt,r,xrdr..
Forx,y∈X, from condition (i) and (ii), for all t∈0,1, we have
(82)ψs2dTxt,Tyt=ψ4∣Txt−Tyt∣2=ψ4∣gt+∫01Lt,r,xrdr−gt+∫01Lt,r,yrdr∣2=ψ4∣∫01Lt,r,xr−Lt,r,yrdr∣2≤ψ4∫01∣Lt,r,xr−Lt,r,yr∣dr2≤ψ∫01βt,r∣xr−yr∣dr2=ψ∫01βt,rxr−yr21/2dr2≤ψ∫01βt,rdxt,yt1/2dr2=ψdxt,yt∫01βt,rdr2≤ψdxt,yt1/3≤ψMxt,yt1/3.
Therefore, all conditions of Corollary22 are satisfied and as a result the mapping T has a unique fixed point in X, which is a solution of the integral equation in (78).


## 4. Conclusion

In 2020, Kari et al. [22] introduced the notion of θ−ϕ-contraction in b-rectangular metric spaces and study the existence and uniqueness of fixed point for the mappings introduced. In 2019, Baiya and Kaewcharoen [21] established fixed point theorems for generalized contractions in complete rectangular metric spaces and proved the existence of fixed points. Inspired and motivated by the works of Kari et al. [22] and Baiya and Kaewcharoen [21] in this paper, we introduce generalized α,ψ-contraction mappings in rectangular b-metric spaces and prove the existence and uniqueness of fixed point for the mappings introduced. Our results extend and generalize related fixed point results in the existing literature. We have also supported the main result of this paper by applicable examples. Furthermore, we determined a solution to an integral equation by applying our obtained results.



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*Source: 9370083-2022-04-05.xml*