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Observations on relationtheoretic coincidence theorems under Boyd–Wong type nonlinear contractions
Fixed Point Theory and Applicationsvolume 2019, Article number: 6 (2019)
Abstract
In this article, we carry out some observations on existing metrical coincidence theorems of Karapinar et al. (Fixed Point Theory Appl. 2014:92, 2014) and Erhan et al. (J. Inequal. Appl. 2015:52, 2015) proved for Lakshmikantham–Ćirićtype nonlinear contractions involving $(f,g)$closed transitive sets after proving some coincidence theorems satisfying Boyd–Wongtype nonlinear contractivity conditions employing the idea of $(f,g)$closed locally ftransitive binary relation.
Introduction
In last ten decades, the classical Banach contraction principle [3] has been generalized by numerous authors in the different directions by improving the underlying contraction conditions (e.g., [4,5,6]), enhancing the number of involved mappings [4, 7], weakening the involved metrical notions [7, 8], and enlarging the class of ambient spaces [9,10,11]. In 2004, Ran and Reurings [12] obtained a variant of the classical Banach contraction principle to a complete metric space endowed with partial order relation, which was slightly modified by Nieto and RodríguezLópez [13] in 2005. Later, the trend of utilization of partial order relation in the context of fixed/coincidence point theorems was adopted by various authors [6, 8, 14,15,16,17,18,19]. Such generalizations are carried out by improving either the contraction conditions or the underlying spaces keeping the partial order relation fixed; however, several authors adopted another way of improving the Banach contraction principle by using various binary relations, such as preorder (Turinici [20]), transitive relation (BenElMechaiekh [21]), tolerance (Turinici [22, 23]), strict order (Ghods et al. [24]), and symmetric closure (Samet and Turinici [25]).
In 2015, Alam and Imdad [26] obtained yet another generalization of the classical Banach contraction principle employing an amorphous (arbitrary) binary relation and observed that several wellknown metrical fixed point theorems can further be improved up to arbitrary binary relations (instead of partial order, preorder, transitive relation, tolerance, strict order, and symmetric closure). With this in mind, Alam and Imdad [7] introduced relationtheoretic analogues of certain involved metrical notions such as completeness, continuity, gcontinuity, compatibility, etc. and utilized the same to prove coincidence theorems for relationpreserving contractions. It is worth noticing that under the universal relation, such newly defined notions reduce to their corresponding usual notions, and henceforth relationtheoretic metrical fixed/coincidence point theorems reduce to their corresponding classical fixed/coincidence point theorems. Note that relationpreserving contractions remain relatively weaker than usual contractions as they are required to hold merely for the elements that are related in the underlying relation.
Several metrical fixed point theorems under arbitrary binary relations are proved by various authors such as Khan et al. [10], Ayari et al. [27], RoldánLópezdeHierro [28], RoldánLópezdeHierro and Shahzad [29], and Shahzad et al. [30], which are generalizations of the relationtheoretic contraction principle due to Alam and Imdad [26]. Here we can point out that arbitrary binary relation is general enough and often does not work for certain contractions, so that various fixed/coincidence point theorems are proved in metric spaces equipped with different types of binary relations, for example, preorder (RoldánLópezdeHierro and Shahzad [11]), transitive relations (Shahzad et al. [31]), finitely transitive relations (Berzig and Karapinar [32], Berzig et al. [33]), locally finitely transitive relations (Turinici [34, 35]), locally finitely Ttransitive relations (Alam et al. [36]), and locally Ttransitive relations (see Alam and Imdad [37]).
In 2010, Samet and Vetro [38] proved coupled fixed point theorems without using partial ordering employing the idea of an Finvariant set. In 2013, Kutbi et al. [39] weakened the idea of an Finvariant set by introducing the notion of Fclosed sets. Thereafter Karapinar et al. [1] proved some unidimensional versions of earlier coupled fixed point results involving Fclosed sets and then obtained such coupled fixed point results by using the corresponding unidimensional fixed point results. Furthermore, Karapinar et al. [1] observed that the notion of a transitive Fclosed (or Finvariant) set is equivalent to the concept of a preordered set and thereafter also showed that some recent multidimensional results using Finvariant sets can be reduced to wellknown results on ordered metric spaces.
Here we mention two main results of Karapinar et al. [1]. The reader is required to go through Karapinar et al. [1] for relevant notions, such as an $(f,g)$closed set, an $(f,g)$compatible set, Mcontinuity, $(O,M)$compatibility, regularity, a $(g,\preceq )$increasing mapping, transitive and gtransitive sets. In these results, the authors used the following family of control functions utilized by Lakshmikantham and Ćirić [40]:
Theorem 1
([1])
Let $(X,d)$ be a complete metric space, let $f,g:X\rightarrow X$ be two mappings, and let $M \subseteq X^{2}$ be a subset such that

(i)
$f(X)\subseteq g(X)$,

(ii)
M is $(f,g)$compatible, $(f,g)$closed, and transitive,

(iii)
there exists $x_{0}\in X$ such that $(gx_{0},fx _{0})\in M$,

(iv)
there exists $\varphi \in \varPhi $ such that
$$ d(fx,fy)\leq \varphi \bigl(d(gx,gy) \bigr) \quad \forall x,y \in X\textit{ with }(gx,gy)\in M. $$
Also assume that at least one of the following conditions holds:
 $(a)$ :

f and g are Mcontinuous and $(O,M)$compatible,
 $(b)$ :

f and g are continuous and commuting,
 $(c)$ :

$(X,d,M)$ is regular, and $g(X)$ is closed.
Theorem 2
([1])
Let $(X,d)$ be a complete metric space, let ≼ be a transitive relation on X, and let $f,g:X\rightarrow X$ be two mappings such that

(i)
$f(X)\subseteq g(X)$,

(ii)
f is $(g,\preccurlyeq )$increasing,

(iii)
there exists $x_{0}\in X$ such that $g(x_{0}) \preccurlyeq f(x_{0})$,

(iv)
there exists $\varphi \in \varPhi $ such that
$$ d(fx,fy)\leq \varphi \bigl(d(gx,gy) \bigr)\quad \forall x,y \in X\textit{ with }g(x)\preccurlyeq g(y), $$ 
(v)
$\varphi (0)=0$, or ≼ is antisymmetric.
Assume that either
 $(a)$ :

f and g are continuous and commuting, or
 $(b)$ :

$(X,d,\preccurlyeq )$ is regular, and $g(X)$ is closed.
Erhan et al. [2] slightly modified Theorem 1 by proving the following sharpened version.
Theorem 3
([2])
Let $(X,d)$ be a complete metric space, let $f,g:X\rightarrow X$ be two mappings, and let $M \subseteq X^{2}$ be a subset such that

(i)
$f(X)\subseteq g(X)$,

(ii)
M is $(f,g)$closed and gtransitive,

(iii)
there exists $x_{0}\in X$ such that $(gx_{0},fx _{0})\in M$,

(iv)
there exists $\varphi \in \varPhi $ such that
$$ d(fx,fy)\leq \varphi \bigl(d(gx,gy) \bigr)\quad \forall x,y \in X\textit{ with }(gx,gy)\in M. $$
Also assume that at least one of the following conditions holds:
 $(a)$ :

f and g are Mcontinuous and $(O,M)$compatible,
 $(b)$ :

f and g are continuous and $(O,M)$compatible,
 $(c)$ :

f and g are continuous and commuting,
 $(d)$ :

$(X,d,M)$ is regular, $g(X)$ is closed, and M is $(f,g)$compatible.
Alam and Imdad [7] observed that relationtheoretic metrical fixed/coincidence point results combine the idea contained in Karapinar et al. [1] as the set M (utilized by Karapinar et al. [1]) being subset of $X^{2}$ is in fact a binary relation on X. The aim of this paper is to prove relatively more sharpened and improved versions of foregoing results using a relationtheoretic approach.
Relationtheoretic notions and auxiliary results
In this paper, $\mathbb{N}$ and $\mathbb{N}_{0}$ denote the sets of natural numbers and whole numbers, respectively (i.e., $\mathbb{N} _{0}=\mathbb{N} \cup \{0\}$). In this section, to make our exposition selfcontained, we give some definitions and basic results related to our main results.
Definition 1
([41])
Let X be a nonempty set. A subset $\mathcal{R}$ of $X^{2}$ is called a binary relation on X.
Trivially, $X^{2}$ and ∅ are binary relations on X, which are respectively called the universal relation (or full relation) and empty relation. Another important relation of this kind is the relation $\triangle _{X}=\{(x,x):x\in X\}$, called the identity relation (or the diagonal relation) on X.
In this paper, $\mathcal{R}$ stands for a nonempty binary relation, but for simplicity, we write only “binary relation” instead of “nonempty binary relation.”
Definition 2
([26])
Let $\mathcal{R}$ be a binary relation defined on a nonempty set X, and let $x,y\in X$. We say that x and y are $\mathcal{R}$comparative and write $[x,y]\in \mathcal{R}$ if either $(x,y)\in \mathcal{R}$ or ${(y,x)\in \mathcal{R}}$.
Definition 3
A binary relation $\mathcal{R}$ defined on a nonempty set X is called

reflexive if $(x,x)\in \mathcal{R}$ $\forall x\in X$,

irreflexive if $(x,x)\notin \mathcal{R}$ $\forall x\in X$,

symmetric if $(x,y)\in \mathcal{R}$ implies $(y,x)\in \mathcal{R}$,

antisymmetric if $(x,y)\in \mathcal{R}$ and $(y,x)\in \mathcal{R}$ imply $x=y$,

transitive if $(x,y)\in \mathcal{R}$ and $(y,z)\in \mathcal{R}$ imply $(x,z)\in \mathcal{R}$,

complete or connected or dichotomous if $[x,y]\in \mathcal{R}$ $\forall x,y\in X$,

weakly complete or weakly connected or trichotomous if $[x,y]\in \mathcal{R}$ or $x=y$ $\forall x,y\in X$.
Definition 4
A binary relation $\mathcal{R}$ defined on a nonempty set X is called

amorphous if it has no specific properties at all,

a strict order or sharp order if $\mathcal{R}$ is irreflexive and transitive,

a nearorder if $\mathcal{R}$ is antisymmetric and transitive,

a pseudoorder if $\mathcal{R}$ is reflexive and antisymmetric,

a quasiorder or preorder if $\mathcal{R}$ is reflexive and transitive,

a partial order if $\mathcal{R}$ is reflexive, antisymmetric, and transitive,

a simple order if $\mathcal{R}$ is a weakly complete strict order,

a weak order if $\mathcal{R}$ is a complete preorder,

a total order or linear order or chain if $\mathcal{R}$ is a complete partial order,

a tolerance if $\mathcal{R}$ is reflexive and symmetric,

an equivalence if $\mathcal{R}$ is reflexive, symmetric, and transitive.
Remark 1
Clearly, the universal relation $X^{2}$ on a nonempty set X remains a complete equivalence relation.
Definition 5
([41])
Let X be a nonempty set, and let $\mathcal{R}$ be a binary relation on X.

(1)
The inverse or transpose or dual relation of $\mathcal{R}$, denoted by $\mathcal{R}^{1}$, is defined by $\mathcal{R}^{1}=\{(x,y)\in X^{2}:(y,x)\in \mathcal{R}\}$.

(2)
The reflexive closure of $\mathcal{R}$, denoted by $\mathcal{R}^{\#}$, is defined as the set $\mathcal{R}\cup \triangle _{X}$ (i.e., $\mathcal{R}^{\#}:=\mathcal{R}\cup \triangle _{X}$). In fact, $\mathcal{R}^{\#}$ is the smallest reflexive relation on X containing $\mathcal{R}$.

(3)
The symmetric closure of $\mathcal{R}$, denoted by $\mathcal{R}^{s}$, is defined as the set $\mathcal{R}\cup \mathcal{R} ^{1}$ (i.e., $\mathcal{R}^{s}:=\mathcal{R}\cup \mathcal{R}^{1}$). In fact, $\mathcal{R}^{s}$ is the smallest symmetric relation on X containing $\mathcal{R}$.
Proposition 1
([26])
For a binary relation $\mathcal{R}$ defined on a nonempty set X,
Definition 6
([46])
Let X be a nonempty set, let $E\subseteq X$, and let $\mathcal{R}$ be a binary relation on X. Then the restriction of $\mathcal{R}$ to E, denoted by $\mathcal{R}\vert _{E}$, is defined as the set $\mathcal{R}\cap E^{2}$ (i.e., $\mathcal{R}\vert _{E}:= \mathcal{R}\cap E^{2}$). In fact, $\mathcal{R}\vert _{E}$ is a relation on E induced by $\mathcal{R}$.
Definition 7
([26])
Let X be a nonempty set, and let $\mathcal{R}$ be a binary relation on X. A sequence $\{x_{n}\} \subset X$ is called $\mathcal{R}$preserving if
Definition 8
([7])
Let X be a nonempty set, and let f and g be selfmappings on X. A binary relation $\mathcal{R}$ defined on X is called $(f,g)$closed if for all $x,y\in X$,
Note that under the restriction $g=I$, the identity mapping on X, Definition 8 reduces to the notion of fclosedness of $\mathcal{R}$ defined in [26].
Proposition 2
([7])
Let X be a nonempty set, let $\mathcal{R}$ be a binary relation on X, and let f and g be selfmappings on X. If $\mathcal{R}$ is $(f,g)$closed, then so is $\mathcal{R}^{s}$.
Definition 9
([7])
Let $(X,d)$ be a metric space, and let $\mathcal{R}$ be a binary relation on X. We say that $(X,d)$ is $\mathcal{R}$complete if every $\mathcal{R}$preserving Cauchy sequence in X converges.
Remark 2
Every complete metric space is $\mathcal{R}$complete for any binary relation $\mathcal{R}$. Particularly, under the universal relation, the notion of $\mathcal{R}$completeness coincides with usual completeness.
Definition 10
([7])
Let $(X,d)$ be a metric space, and let $\mathcal{R}$ be binary relation on X. A subset E of X is called $\mathcal{R}$closed if every $\mathcal{R}$preserving convergent sequence in E converges to a point of E.
Remark 3
Every closed subset of a metric space is $\mathcal{R}$closed for any binary relation $\mathcal{R}$. Particularly, under the universal relation, the notion of $\mathcal{R}$closedness coincides with usual closedness.
Proposition 3
([7])
An $\mathcal{R}$complete subspace of a metric space is $\mathcal{R}$closed.
Proposition 4
([7])
An $\mathcal{R}$closed subspace of an $\mathcal{R}$complete metric space is $\mathcal{R}$complete.
Definition 11
([7])
Let $(X,d)$ be a metric space, let $\mathcal{R}$ be a binary relation on X, let g be a selfmapping on X, and let $x\in X$. A mapping $f:X\rightarrow X$ is called $(g,\mathcal{R})$continuous at x if for any sequence $\{x_{n}\}$ such that $\{gx_{n}\}$ is $\mathcal{R}$preserving and $g(x_{n})\stackrel{d}{ \longrightarrow } g(x)$, we have $f(x_{n})\stackrel{d}{\longrightarrow } f(x)$. Moreover, f is called $(g,\mathcal{R})$continuous if it is $(g,\mathcal{R})$continuous at each point of X.
Note that under the restriction $g=I$, the identity mapping on X, Definition 11 reduces to the notion of $\mathcal{R}$continuity of f defined in [7].
Remark 4
Every continuous (respectively, gcontinuous) mapping is $\mathcal{R}$continuous (respectively, $(g,\mathcal{R})$continuous) for any binary relation $\mathcal{R}$. Particularly, under the universal relation, the notion of $\mathcal{R}$continuity (respectively, $(g,\mathcal{R})$continuity) coincides with usual continuity (respectively, gcontinuity).
Definition 12
([7])
Let $(X,d)$ be a metric space, let $\mathcal{R}$ be a binary relation on X, and let f and g be selfmappings on X. We say that the mappings f and g are $\mathcal{R}$compatible if for any sequence $\{x_{n}\}\subset X$ such that $\{fx_{n}\}$ and $\{gx_{n}\}$ are $\mathcal{R}$preserving and $\lim_{n\to \infty }g(x_{n})=\lim_{n\to \infty }f(x_{n})$, we have
Remark 5
In a metric space $(X,d)$ endowed with a binary relation $\mathcal{R}$,
Particularly, under the universal relation, the notion of $\mathcal{R}$compatibility coincides with usual compatibility.
Definition 13
([7])
Let $(X,d)$ be a metric space, and let g be a selfmapping on X. A binary relation $\mathcal{R}$ defined on X is called $(g,d)$selfclosed if for any $\mathcal{R}$preserving sequence $\{x_{n}\}$ such that $x_{n}\stackrel{d}{ \longrightarrow } x$, there exists a subsequence $\{x_{n_{k}}\}$ of $\{x_{n}\}$ with $[gx_{n_{k}},gx]\in \mathcal{R} $ for all $k \in \mathbb{N}_{0}$.
Note that under the restriction $g=I$, the identity mapping on X, Definition 13 reduces to the notion of dselfclosedness of $\mathcal{R}$ defined in [26].
Definition 14
([47])
Given a mapping $f:X\rightarrow X$, a binary relation $\mathcal{R}$ defined on X is called ftransitive if for any $x,y,z\in X$,
Inspired by Turinici [34, 35], Alam and Imdad [37] introduced the following notion by localizing the notion of ftransitivity.
Definition 15
([37])
Let X be a nonempty set, and let f be a selfmapping on X. A binary relation $\mathcal{R}$ on X is called locally ftransitive if for each (effectively) $\mathcal{R}$preserving sequence $\{x_{n}\}\subset f(X)$ (with range $E:=\{x_{n}:n \in \mathbb{N}_{0}\}$), the binary relation $\mathcal{R}\vert _{E}$ is transitive.
Clearly, for a given a selfmapping f and a binary relation $\mathcal{R}$ on a nonempty set X,
Definition 16
([25])
Let X be a nonempty set, and let $\mathcal{R}$ be a binary relation on X. A subset E of X is called $\mathcal{R}$directed if for each pair $x,y\in E$, there exists $z\in X$ such that $(x,z)\in \mathcal{R}$ and $(y,z)\in \mathcal{R}$.
Definition 17
([46])
Let X be a nonempty set, and let $\mathcal{R}$ be a binary relation on X. For $x,y\in X$, a path of length k (where k is a natural number) in $\mathcal{R}$ from x to y is a finite sequence $\{z_{0},z_{1},z_{2},\ldots,z_{k}\}\subset X$ satisfying the following conditions:

(i)
$z_{0}=x$ and $z_{k}=y$,

(ii)
$(z_{i},z_{i+1})\in \mathcal{R}$ for each i ($0\leq i\leq k1$).
Note that a path of length k involves $k+1$ elements of X although they are not necessarily distinct.
Definition 18
([7])
Let X be a nonempty set, and let $\mathcal{R}$ be a binary relation on X. A subset E of X is called $\mathcal{R}$connected if for each pair $x,y\in E$, there exists a path in $\mathcal{R}$ from x to y.
Inspired by the notion of an $(f,g)$compatible subset of $X^{2}$ utilized by Karapinar et al. [1] and RoldánLópezdeHierro et al. [47], we introduce the following notion.
Definition 19
Let X be a nonempty set, and let f and g be selfmappings on X. A binary relation $\mathcal{R}$ defined on X is called $(f,g)$compatible if for all $x,y\in X$,
Next, we propose the following fact.
Proposition 5
Let X be a nonempty set, let $\mathcal{R}$ be a binary relation on X, and let f and g be selfmappings on X. If $\mathcal{R}$ is antisymmetric and $(f,g)$closed, then $\mathcal{R}$ is $(f,g)$compatible.
Proof
Take $x,y\in X$ such that $(gx,gy)\in \mathcal{R}$ and $g(x)=g(y)$, which yields that
As $\mathcal{R}$ is $(f,g)$closed, we have
which by the antisymmetry of $\mathcal{R}$ implies that
For a given binary relation $\mathcal{R}$ and two selfmappings f and g defined on a nonempty set X, $X(f,g,\mathcal{R})$ denotes the subset $\{x\in X:(gx,fx)\in \mathcal{R}\}$ of X. □
Recently, Alam and Imdad [7] proved the following relationtheoretic coincidence theorem under linear contraction.
Theorem 4
([7])
Let $(X,d)$ be a metric space, let $\mathcal{R}$ be a binary relation on X, and let f and g be selfmappings on X. Let Y be an $\mathcal{R}$complete subspace of X. Suppose that the following conditions hold:
 $(a)$ :

$f(X)\subseteq g(X)\cap Y$,
 $(b)$ :

$\mathcal{R}$ is $(f,g)$closed,
 $(c)$ :

$X(f,g,\mathcal{R})$ is nonempty,
 $(d)$ :

there exists $\alpha \in [0,1)$ such that
$$ d(fx,fy)\leq \alpha d(gx,gy)\quad \forall x,y\in X \textit{ with }(gx,gy)\in \mathcal{R}, $$  $(e)$ :

 $(e1)$ :

f and g are $\mathcal{R}$compatible,
 $(e2)$ :

g is $\mathcal{R}$continuous,
 $(e3)$ :

either f is $\mathcal{R}$continuous, or $\mathcal{R}$ is $(g,d)$selfclosed,
 $(e^{\prime })$ :

 $(e^{\prime }1)$ :

$Y \subseteq g(X)$,
 $(e^{\prime }2)$ :

either f is $(g,\mathcal{R})$continuous, or f and g are continuous, or $\mathcal{R}\vert _{Y}$ is dselfclosed.
The following family of control functions is indicated by Boyd and Wong [48] but was later used by Jotic [49]:
Clearly, the family Ω enlarges the family Φ, i.e., $\varPhi \subset \varOmega $. By the symmetry of d we have the following:
Proposition 6
If $(X,d)$ is a metric space, $\mathcal{R}$ is a binary relation on X, f and g are selfmappings on X, and $\varphi \in \varOmega $, then the following contractivity conditions are equivalent:
Finally, we state the following known results, which are needed in the proofs of our main results.
Lemma 1
([18])
Let $\varphi \in \varOmega $. If $\{a_{n}\}\subset (0,\infty )$ is a sequence such that $a_{n+1}\leq \varphi (a_{n})$ for all $n\in \mathbb{N}_{0}$, then $\lim_{n\to \infty }a_{n}=0$.
Lemma 2
Let $(X,d)$ be a metric space, and let $\{x_{n}\}$ be a sequence in X. If $\{x_{n}\}$ is not a Cauchy sequence, then there exist $\epsilon >0$ and two subsequences $\{x_{n_{k}}\}$ and $\{x_{m_{k}}\}$ of $\{x_{n}\}$ such that

(i)
$k\leq m_{k}< n_{k} $ $\forall k\in \mathbb{N}$,

(ii)
$d(x_{m_{k}},x_{n_{k}})> \epsilon $ $\forall k \in \mathbb{N}$,

(iii)
$d(x_{m_{k}},x_{n_{k1}})\leq \epsilon $ $\forall k\in \mathbb{N}$.
Moreover, suppose that $\lim_{n\to \infty }d(x_{n},x_{n+1})=0$. Then

(iv)
$\lim_{k\to \infty }d(x_{m_{k}},x_{n_{k}})= \epsilon $,

(v)
$\lim_{k\to \infty } d(x_{m_{k}+1},x_{n_{k}+1})= \epsilon $.
Lemma 3
([50])
Let X be a nonempty set, and let g be a selfmapping on X. Then there exists a subset $E\subseteq X$ such that $g(E)=g(X)$ and $g: E \rightarrow X$ is onetoone.
Main results
Now, we are equipped to prove the following result on the existence of a coincidence point under the φcontractivity condition.
Theorem 5
Let $(X,d)$ be a metric space, let $\mathcal{R}$ be a binary relation on X, and let Y be an $\mathcal{R}$complete subspace of X. Let f and g be selfmappings on X. Suppose that the following conditions hold:
 $(a)$ :

$f(X)\subseteq g(X)\cap Y$,
 $(b)$ :

$\mathcal{R}$ is $(f,g)$closed and locally ftransitive,
 $(c)$ :

$X(f,g,\mathcal{R})$ is nonempty,
 $(d)$ :

there exists $\varphi \in \varOmega $ such that
$$ d(fx,fy)\leq \varphi \bigl(d(gx,gy) \bigr)\quad \forall x,y \in X\textit{ with }(gx,gy)\in \mathcal{R}, $$  $(e)$ :

 $(e1)$ :

f and g are $\mathcal{R}$compatible,
 $(e2)$ :

g is $\mathcal{R}$continuous,
 $(e3)$ :

either f is $\mathcal{R}$continuous, or $\mathcal{R}$ is $(f,g)$compatible and $(g,d)$selfclosed,
 $(e^{\prime })$ :

 $(e^{\prime }1)$ :

$Y \subseteq g(X)$,
 $(e^{\prime }2)$ :

either f is $(g,\mathcal{R})$continuous, or f and g are continuous, or $\mathcal{R}$ and $\mathcal{R}\vert _{Y}$ are $(f,g)$compatible and dselfclosed, respectively.
Proof
Firstly, we notice that assumption $(a)$ is equivalent to $f(X)\subseteq g(X)$ and $f(X)\subseteq Y$. Now, in view of assumption $(c)$, let $x_{0}$ be an arbitrary element of $X(f,g,\mathcal{R})$ such that $(gx_{0},fx_{0})\in \mathcal{R}$. If $g(x_{0})=f(x_{0})$, then $x_{0}$ is a coincidence point of f and g, and hence the proof is finished. Otherwise, using the assumption $f(X)\subseteq g(X)$, we construct a sequence $\{x_{n}\}\subset X$ of joint iteration of f and g based at point $x_{0}$, that is,
Now we assert that $\{gx_{n}\}$ is an $\mathcal{R}$preserving sequence, that is,
We prove this fact by mathematical induction. Using Eq. (1) (with $n=0$) and the fact that $x_{0}\in X(f,g,\mathcal{R})$, we have
Thus (2) holds for $n=0$. Assume that (2) holds for $n=r>0$, that is,
Since $\mathcal{R}$ is $(f,g)$closed, we have
which by (1) implies
that is, (2) holds for $n=r+1$. Hence, by induction, (2) holds for all $n \in \mathbb{N}_{0}$.
By (1) and (2) the sequence $\{fx_{n}\}$ is also $\mathcal{R}$preserving, that is,
If $g(x_{n_{0}})=g(x_{n_{0}+1})$ for some $n_{0}\in \mathbb{N}$, then using (1), we have $g(x_{n_{0}})=f(x_{n_{0}})$, that is, $x_{n_{0}}$ is a coincidence point of f and g, so that we are done. On the other hand, if $g(x_{n})\neq g(x_{n+1})$ for each $n\in \mathbb{N}_{0}$, then we can define a sequence $\{d_{n}\}_{n=0}^{\infty }\subset (0,\infty )$ by
Applying (1), (2), (4), and assumption $(d)$, we deduce that
so that
which by Lemma 1 implies
Now we show that $\{gx_{n}\}$ is a Cauchy sequence. On the contrary, assume that $\{gx_{n}\}$ is not Cauchy. Therefore by Lemma 2 there exist $\epsilon >0$ and two subsequences $\{gx_{n_{k}}\}$ and $\{gx_{m_{k}} \}$ of $\{gx_{n}\}$ such that
Further, in view of (5) and Lemma 2, we have
Denote $\eta _{k}:=d(gx_{m_{k}}, gx_{n_{k}})$. As $\{gx_{n}\}$ is $\mathcal{R}$preserving (owing to (2)) and $\{gx_{n}\}\subseteq f(X)$ (owing to (1)), by the local ftransitivity of $\mathcal{R}$ we have $(gx_{m_{k}}, gx_{n_{k}})\in \mathcal{R}$. Hence, applying contractivity condition $(d)$, we obtain
so that
Since $\eta _{k}\rightarrow \epsilon $ in the real line as $k\rightarrow \infty $ (owing to (7)) and $\eta _{k} > \epsilon $ for all $k \in \mathbb{N}$ (owing to (6)), by the definition of Ω, we have
Taking the limit as $k\rightarrow \infty $ in (8) and using (7) and (9), we obtain
which is a contradiction, so that $\{gx_{n}\}$ is a Cauchy sequence. By (1), $\{gx_{n}\}\subset f(X)\subseteq Y$, so that $\{gx_{n}\}$ is an $\mathcal{R}$preserving Cauchy sequence in Y. As Y is $\mathcal{R}$complete, there exists $z\in Y$ such that $\lim_{n\to \infty } g(x_{n})=z$, which, together with (2), implies
Applying (1) and (10), we obtain
Now we complete the proof by using $(e)$ and $(e^{\prime })$. Assume that $(e)$ holds. Using (2), (10), and assumption $(e2)$ (i.e., the $\mathcal{R}$continuity of g), we have
Now, utilizing (3), (11), and assumption $(e2)$ (i.e., the $\mathcal{R}$continuity of g), we have
Since $\{fx_{n}\}$ and $\{gx_{n}\}$ are $\mathcal{R}$preserving (owing to (2) and (3)) and $\lim_{n\to \infty } f(x_{n})=\lim_{n\to \infty } g(x_{n})=z$ (owing to (10) and (11)), assumption $(e1)$ (i.e., the $\mathcal{R}$compatibility of f and g), implies
Now we prove that z is a coincidence point of f and g. To substantiate this, we use assumption $(e3)$. Assume that f is $\mathcal{R}$continuous. Using (2), (10), and the $\mathcal{R}$continuity of f, we obtain
Applying (13), (14), (15), and the continuity of d, we obtain
so that
Thus z is a coincidence point of f and g, and hence we are through. Alternately, assume that $\mathcal{R}$ is $(g,d)$selfclosed and $(f,g)$compatible. As $\{gx_{n}\}$ is $\mathcal{R}$preserving (due to (2)) and $g(x_{n})\stackrel{d}{\longrightarrow } z$ (in view of (10)), due to the $(g,d)$selfclosedness of $\mathcal{R}$, there exists a subsequence $\{gx_{n_{k}}\}$ of $\{gx_{n}\}$ such that
Since $g(x_{n_{k}})\stackrel{d}{\longrightarrow } z$, Eqs. (10)–(14) also hold for $\{x_{n_{k}}\}$ (instead of $\{x_{n}\}$). Using (16), assumption $(d)$, and Proposition 6, we obtain
Now we claim that
Since there arise two different possibilities, we consider a partition $\{\mathbb{N}^{0},\mathbb{N}^{+}\}$ of $\mathbb{N}$ (i.e., $\mathbb{N}^{0}\cup \mathbb{N}^{+}=\mathbb{N}$ and $\mathbb{N}^{0}\cap \mathbb{N}^{+}=\emptyset$) such that

(i)
$d(ggx_{n_{k}},gz)=0$ $\forall k\in \mathbb{N} ^{0}$, and

(ii)
$d(ggx_{n_{k}},gz)>0$ $\forall k\in \mathbb{N}^{+}$.
In case (i), using (16) and the $(f,g)$compatibility of $\mathcal{R}$, we get $d(fgx_{n_{k}},fz)=0$ for all $k\in \mathbb{N}^{0}$, so that (17) holds for all $k\in \mathbb{N}^{0}$. In case (ii), by the definition of Ω we have $d(fgx_{n_{k}},fz)\leq \varphi (d(ggx _{n_{k}},gz))< d(ggx_{n_{k}},gz)$ for all $k\in \mathbb{N} ^{+}$, and hence (17) holds for all $k\in \mathbb{N}^{+}$, so that (17) holds for all $k\in \mathbb{N}$. Using the triangle inequality, (12), (13), (14), and (17), we get
so that
Thus z is a coincidence point of f and g, and hence we are done.
Now, assume that $(e^{\prime })$ holds. By assumption $(e^{\prime }1)$ (i.e., $Y\subseteq g(X)$) we can find $u\in X$ such that $z=g(u)$. Hence (10) and (11), respectively, reduce to
Now we prove that u is a coincidence point of f and g. To do this, we use assumption $(e^{\prime }2)$. Firstly, assume that f is $(g,\mathcal{R})$continuous. Then using (18), we get
and hence we are done. Secondly, assume that f and g are continuous. By Lemma 3 there exists a subset $E\subseteq X$ such that $g(E)=g(X)$ and $g: E \rightarrow X$ is onetoone. Now define $T: g(E) \rightarrow g(X)$ by
Since $g: E \rightarrow X$ is onetoone and $f(X)\subseteq g(X)$, T is well defined. Again, since f and g are continuous, it follows that T is continuous. Using $g(X)=g(E)$, assumptions $(a)$ and $(e^{\prime }1)$, respectively, reduce to $f(X)\subseteq g(E)\cap Y$ and $Y\subseteq g(E)$, which implies that, without loss of generality, we are able to construct $\{x_{n}\}_{n=1}^{\infty }\subset E$ satisfying (1) and enabling us to choose $u\in E$. Using (18), (19), (21), and the continuity of T, we get
Thus u is a coincidence point of f and g, and hence we are through. Finally, assume that $\mathcal{R}$ and $\mathcal{R}\vert _{Y}$ are $(f,g)$compatible and dselfclosed, respectively. Since $\{gx_{n}\}$ is $\mathcal{R}\vert _{Y}$preserving (due to (2)) and $g(x_{n})\stackrel{d}{\longrightarrow } g(u)\in Y$ (due to (18)), by the dselfcloseness of $\mathcal{R}\vert _{Y}$ there exists a subsequence $\{gx_{n_{k}}\}$ of $\{gx_{n}\}$ such that
Applying (1), (22), assumption $(d)$, and Proposition 6, we obtain
We claim that
Since there arise two different possibilities, we consider a partition $\{\mathbb{N}^{0},\mathbb{N}^{+}\}$ of $\mathbb{N}$ (i.e., $\mathbb{N}^{0}\cup \mathbb{N}^{+}=\mathbb{N}$ and $\mathbb{N} ^{0}\cap \mathbb{N}^{+}=\emptyset $) such that

(i)
$d(gx_{n_{k}},gu)=0$ $\forall k\in \mathbb{N} ^{0}$, and

(ii)
$d(gx_{n_{k}},gu)>0$ $\forall k\in \mathbb{N} ^{+}$.
In case (i), using (22) and the $(f,g)$compatibility of $\mathcal{R}$, we get $d(fx_{n_{k}},fu)=0$ for all $k\in \mathbb{N}^{0}$, which by (1) implies $d(gx_{n_{k}+1},fu)=0$ for all $k\in \mathbb{N}^{0}$, and hence (23) holds for all $k\in \mathbb{N}^{0}$. In case (ii), by the definition of Ω we have $d(gx_{n_{k}+1},fu) \leq \varphi (d(gx_{n_{k}},gu))< d(gx_{n_{k}},gu)$ for all $k \in \mathbb{N}^{+}$, and hence (23) holds for all $k\in \mathbb{N} ^{+}$. Thus (23) holds for all $k\in \mathbb{N}$.
Applying (18), (23), and the continuity of d, we get
so that
Hence u is a coincidence point of f and g, which completes the proof. □
Corollary 1
Let X be a nonempty set endowed with a binary relation $\mathcal{R}$ and a metric d such that the metric space $(X,d)$ is $\mathcal{R}$complete. Let and f and g be selfmappings on X. Assume that the following conditions hold:
 $(a)$ :

$f(X)\subseteq g(X)$,
 $(b)$ :

$\mathcal{R}$ is $(f,g)$closed and locally ftransitive,
 $(c)$ :

$X(f,g,\mathcal{R})$ is nonempty,
 $(d)$ :

there exists $\varphi \in \varOmega $ such that
$$ d(fx,fy)\leq \varphi \bigl(d(gx,gy) \bigr)\quad \forall x,y \in X\textit{ with }(gx,gy)\in \mathcal{R}, $$  $(e)$ :

 $(e1)$ :

f and g are $\mathcal{R}$compatible,
 $(e2)$ :

g is $\mathcal{R}$continuous,
 $(e3)$ :

either f is $\mathcal{R}$continuous, or $\mathcal{R}$ is $(g,d)$selfclosed and $(f,g)$compatible,
 $(e^{\prime })$ :

 $(e^{\prime }1)$ :

there exists an $\mathcal{R}$closed subspace Y of X such that $f(X)\subseteq Y \subseteq g(X)$,
 $(e^{\prime }2)$ :

either f is $(g,\mathcal{R})$continuous, or f and g are continuous, or $\mathcal{R}$ and $\mathcal{R}\vert _{Y}$ are $(f,g)$compatible and dselfclosed, respectively.
Proof
The result corresponding to part $(e)$ and alternating part $(e^{\prime })$ follows by taking $Y=X$ in Theorem 5 and using Proposition 4, respectively. □
Remark 6
If g is onto in Corollary 1, then we can remove assumption $(a)$ as in this case it trivially holds. Also, we can omit assumption $(e^{\prime }1)$ as it trivially hods for $Y=g(X)=X$ using Proposition 3. Whenever f is onto, in view of assumption $(a)$, g must be onto, and hence again the same conclusion is immediate.
Remark 7
Firstly, we notice that Theorem 3 is an improved version of Theorem 1, Also, in Theorem 3, assumptions $(b)$ and $(c)$ directly follow from $(a)$. With $\mathcal{R}=M$, Corollary 1 remains a sharpened version of Theorem 3, which is worth noting with the following respects:

The notion of “locally ftransitive binary relation” is weaker than that of “transitive/gtransitive binary relation.”

The notion of “regularity of $(X,d, M)$” (in the context of hypothesis $(d)$ of Theorem 3) can be replaced by a relatively weaker notion, namely “$(g,d)$selfclosedness of M.” Further, the notion of “$(g,d)$selfclosedness of M” is not necessary as it can also alternatively be replaced by either “$(g,M)$continuity of f” or “continuity of f and g.”

The notion of “$({O}, M)$compatibility of f and g” is replaced by a relatively weaker notion, namely “Mcompatibility of f and g.”

There is no need to impose the closedness requirement on $g(X)$ as it suffices to take an Mclosed subspace Y of X such that $f(X)\subseteq Y\subseteq g(X)$.
Corollary 2
Theorem 5 (also Corollary 1) remains true if we replace the assumption “$(f,g)$compatibility of $\mathcal{R}$” (utilized in assumptions $(e3)$ and $(e^{\prime }2)$) by one of the following conditions while retaining the rest of the hypotheses:

(i)
$\varphi (0)=0$,

(ii)
g is onetoone,

(iii)
$\mathcal{R}$ is antisymmetric.
Proof
Suppose that (i) holds. Take $x,y\in X$ such that $(gx,gy) \in \mathcal{R}$ and $g(x)=g(y)$. Utilizing the contractivity condition $(d)$, we get $d(fx,fy)\leq \varphi (0)=0$, which implies that $f(x)=f(y)$. It follows that $\mathcal{R}$ is $(f,g)$compatible.
Suppose that (ii) holds. Take $x,y\in X$ such that $(gx,gy)\in \mathcal{R}$ and $g(x)=g(y)$. As g is onetoone, we get $x=y$, which implies that $f(x)=f(y)$. It follows that $\mathcal{R}$ is $(f,g)$compatible.
Finally, in case (iii), our result follows from Proposition 5. □
Remark 8
Corollary 2 with $\mathcal{R}:=\preccurlyeq $ remains an improved version of Theorem 2 in the following respects:

“$(f,g)$closedness of ≼” is equivalent to “$(g, \preccurlyeq )$increasingness of f”;

“locally ftransitivity of ≼” is weaker than “transitivity of ≼”;

hypothesis (v) (of Theorem 2) can also be replaced by the injectivity of g;

“commutativity of f and g”(in the context of hypothesis $(a)$ of Theorem 2) is replaced by “≼compatibility of f and g”;

“continuity of f and g” is replaced by the relatively weaker notion “≼continuity of f and g”;

“continuity of f” (in the context of hypothesis $(a)$ of Theorem 2) is replaced by the relatively weaker notion“≼continuity of f”. Further, this notion is also not necessary as it can alternatively be replaced by “$(g,d)$selfclosedness of ≼”;

“regularity of $(X,d,M)$” (in the context of hypothesis $(b)$ of Theorem 2) can be replaced by relatively weaker notion namely: “$(g,d)$selfclosedness of ≼”. Further, this notion is also not necessary as it can alternatively be replaced by “$(g, \preccurlyeq )$continuity of f” or “continuity of f and $g''$;

“closedness of whole subspace $g(X)$” (in the context of hypothesis $(b)$ of Theorem 2) is also replaced by “≼closedness of any subset Y with $f(X)\subseteq Y\subseteq g(X)$”.
The following consequence of Theorem 5 and Corollary 1 is immediate.
Corollary 3
Theorem 5 (also Corollary 1) remains true if the local ftransitivity of $\mathcal{R}$ is replaced by one of the following conditions while retaining the rest of the hypotheses:

(i)
$\mathcal{R}$ is transitive,

(ii)
$\mathcal{R}$ is ftransitive,

(iii)
$\mathcal{R}$ is gtransitive,

(iv)
$\mathcal{R}$ is locally transitive.
In view of Remarks 2–5, we conclude that Theorem 5 (also Corollaries 1, 2, and 3) remains true if the usual metrical terms of completeness, closedness, compatibility, continuity, and gcontinuity are used instead of their respective $\mathcal{R}$analogues.
Setting $g = I$, the identity mapping on X, in Theorem 5, we obtain the corresponding fixed point result contained in [7].
Here we can point out that the proof of Theorem 4 can be carried out on the lines of Theorem 5 with $\varphi (t)=\alpha t$ ($\alpha \in [0,1)$) even without using the following hypotheses (utilized in Theorem 5):

(i)
$\mathcal{R}$ is locally ftransitive (in assumption $(b)$),

(ii)
$\mathcal{R}$ is $(f,g)$compatible (in assumptions $(e3)$ and $(e^{\prime }2)$).
Recall that condition (i) is utilized to prove the Cauchy property of $\{gx_{n}\}$ (in Theorem 5). In this case, proceeding on the lines of the proof of Theorem 5, we have
so that
Now, using the standard techniques, we can show that $\{gx_{n}\}$ is a Cauchy sequence. Here there is no need to use condition (i), as contractivity condition is not necessary for $d(gx_{m_{k}}, gx_{n_{k}})$.
In this case, $\varphi (0)=0$, and therefore by Corollary 2, $\mathcal{R}$ is $(f,g)$compatible. Hence condition (ii) is vacuously met out. Finally, we can accomplish the proof of Theorem 4 by proceeding on the lines of the proof of Theorem 5 (see the proof of Theorem 4.1 in [7]).
Uniqueness results
In this section, we present the results regarding the uniqueness of a point of coincidence, coincidence point, and a common fixed point corresponding to some earlier results. Recall that two selfmappings f and g defined on a nonempty set X are called weakly compatible if $f(x)=g(x)$ implies $f(gx)=g(fx)$ for all $x\in X$.
Theorem 6
In addition to the hypotheses of Theorem 5, assume that the following conditions hold:
 $(u_{1})$::

$f(X)$ is $\mathcal{R}\vert ^{s}_{g(X)}$ connected, and
 $(u_{2})$::

$\mathcal{R}$ is $(f,g)$compatible.
Proof
In view of Theorem 5, let x̅ and y̅ be two points of coincidence of f and g. Then there exist $x,y\in X$ such that
We need to prove that $\overline{x}=\overline{y}$. Since $f(x),f(y) \in f(X)\subseteq g(X)$, by assumption $(u_{1})$ there exists a path (say $\{gz_{0},gz_{1},gz_{2},\ldots,gz_{k}\}$) of some finite length k in $\mathcal{R}\vert _{g(X)}^{s}$ from $f(x)$ to $f(y)$ (where $z_{0},z_{1},z_{2},\ldots,z_{k}\in X$). By (24), without loss of generality, we may set $z_{0}=x$ and $z_{k}=y$. Thus we obtain
Define the constant sequences $z_{n}^{0}=x$ and $z_{n}^{k}=y$. Using (24), we have $g(z^{0}_{n+1})=f(z^{0}_{n})=\overline{x}$ and $g(z^{k}_{n+1})=f(z^{k}_{n})=\overline{y}$ for $n\in \mathbb{N}_{0}$. Put $z_{0}^{1}=z_{1},z_{0}^{2}=z_{2},\ldots, z_{0}^{k1}=z_{k1}$. As $f(X)\subseteq g(X)$, on the lines similar to proof of Theorem 5, we can define sequences $\{z_{n}^{1}\},\{z_{n}^{2}\},\ldots,\{z_{n}^{k1}\}$ in X such that $g(z^{1}_{n+1})=f(z^{1}_{n})$ and $g(z^{2}_{n+1})=f(z ^{2}_{n}),\ldots, g(z^{k1}_{n+1})=f(z^{k1}_{n})$ for $n\in \mathbb{N} _{0}$. Hence we obtain
Now we assert that
We prove this assertion by induction. It follows from (25) that (27) holds for $n=0$. Assume that (27) holds for $n=r>0$, that is,
Since $\mathcal{R}$ is $(f,g)$closed, by Proposition 2, we obtain
which by (26) implies
It follows that (27) holds for $n=r+1$. Therefore, by induction, (27) holds for all $n \in \mathbb{N}_{0}$. Now, for all $n \in \mathbb{N} _{0}$ and i ($0\leq i\leq k1$), define $t_{n}^{i}:=d(gz_{n}^{i},gz _{n}^{i+1})$. We claim that
Fix i. Then two cases arise. Firstly, assume that $t_{n_{0}}^{i}:=d(gz _{n_{0}}^{i},gz_{n_{0}}^{i+1})=0$ for some $n_{0}\in \mathbb{N}_{0}$. Then by assumption $(u_{2})$ we obtain $d(fz_{n_{0}}^{i},fz_{n_{0}} ^{i+1})=0$. Consequently, using (26), we get $t_{n_{0}+1}^{i}=d(gz _{n_{0}+1}^{i},gz_{n_{0}+1}^{i+1})=d(fz_{n_{0}}^{i},fz_{n_{0}}^{i+1})=0$. Thus by induction we get $t_{n}^{i}=0$ for all $n\geq n_{0}$, yielding thereby $\lim_{n\to \infty }t_{n}^{i}=0$. On the other hand, assume that $t_{n}>0$ for all $n\in \mathbb{N}_{0}$. Then, using (26), (27), assumption $(d)$, and Proposition 6, we obtain
so that
Hence by Lemma 1 we obtain $\lim_{n\to \infty }t_{n}^{i}=0$. Therefore, in both the cases, (28) is proved for each i ($0\leq i \leq k1$). Using the triangle inequality and (28), we obtain
□
Corollary 4
Theorem 6 remains true if we replace condition $(u_{1})$ by one of the following conditions while retaining the rest of the hypotheses:
 $(u_{1}^{\prime })$ :

$\mathcal{R}\vert _{f(X)}$ is complete, and
 $(u_{1}^{\prime \prime })$ :

$f(X)$ is $\mathcal{R}\vert ^{s}_{g(X)}$directed.
Theorem 7
In addition to the hypotheses of Theorem 6, suppose that the following condition holds:
 $(u_{3})$::

one of f and g is onetoone.
Theorem 8
In addition to hypotheses $(e^{\prime })$ of Theorem 6, suppose that the following condition holds:
 $(u_{4})$::

f and g are weakly compatible.
The proofs of Corollary 4 and of Theorems 7 and 8 are similar to those of Corollary 4.6 and of Theorems 4.7 and 4.8, respectively, which are contained in [7].
An illustrative example
In this section, we construct an example to highlight the worth and realized improvements in our newly proved results.
Example 1
Let $X=[0,\infty )$ with usual metric d and binary relation $\mathcal{R}= \{(0,0),(0,1),(1,0), (1,1),(3,0)\}$. Notice that, $\mathcal{R}$ is neither transitive nor gtransitive, but it is locally ftransitive. Define the pair of selfmappings f and g on X by
Clearly, $\mathcal{R}$ is $(f,g)$closed. Let $Y=\{0,1\}$. Then Y is $\mathcal{R}$complete, and $f(X)=\{0,1\}\subseteq Y\subseteq g(X)=\{0,1,3 \}$. Define the function $\varphi : [0,\infty )\to [0,\infty )$ by $\varphi (t)= \frac{1}{2}t$. Then $\varphi \in \varOmega $.
Take any $\mathcal{R}\vert _{Y}$preserving sequence $\{x_{n}\}$ such that $x_{n}\stackrel{d}{\longrightarrow } x$. As $(x_{n},x_{n+1})\in \mathcal{R}\vert _{Y}$ for all $n\in \mathbb{N}$, there exists $N \in \mathbb{N}$ such that $x_{n}=x \in \{0,1\} $ for all $n \geq N$. Therefore we can choose a subsequence $\{x_{n_{k}}\}$ of the sequence $\{x_{n}\}$ such that $x_{n_{k}}=x$ for all $k\in \mathbb{N}$, which amounts to saying that $[x_{n_{k}},x]\in \mathcal{R}\vert _{Y}$ for all $k\in \mathbb{N}$. Hence $\mathcal{R}\vert _{Y}$ is dselfclosed. We can easily see that contraction condition $(d)$ and the remaining hypotheses of Theorem 5 are also satisfied. Consequently, in view of Theorem 5, f and g have a coincidence point (namely, $x=0$).
Furthermore, hypotheses $(u_{1})$, $(u_{2})$, and $(u_{4})$ of Theorem 8 also hold. Thus all the hypotheses of Theorem 8 are satisfied, and hence f and g have a unique common fixed point (namely, $x=0$).
Note that the present example cannot be covered by Karapinar et al. [1] and Erhan et al. [2] (i.e., Theorems 1 and 3, respectively), which substantiate the utility of Theorem 5 over Theorems 1 and 3.
Conclusions
In view of our newly proved results, we conclude that under relationtheoretic linear contraction, merely an arbitrary binary relation is required. If we extend such results to φcontractions (under the family Ω), then a weaker version of nearorder is required, namely, a “locally ftransitive antisymmetric” (or, more appropriately, “locally ftransitive $(f.g)$compatible)” binary relation. Particularly, in case $\varphi (0)=0$ or g is onetoone, a “locally ftransitive binary relation” is sufficient.
Karapinar et al. [1] observed that the notion of a transitive Fclosed (or Finvariant) set is equivalent to the concept of a preordered set. To do this, given a nonempty subset $M^{n}$ of $X^{2n}$, they defined the following relations:

$n =2$: $(x, y)\sqsubseteq _{M^{2}} (u, v) \Longleftrightarrow [(x,y) =(u,v)\text{ or }(u,v,x,y)\in {M^{2}}]$;

$n=3$: $(x, y, z)\sqsubseteq _{M^{3}} (u,v,w) \Longleftrightarrow [(x,y,z)=(u,v,w)\text{ or }(u,v,w,x,y,z) \in {M^{3}}]$.
For brevity, we consider $n=1$ and denote $M^{1}=M$. Then we analogously have
Using this relation, they proved that (see Lemma 39 in [1])

(1)
$\sqsubseteq _{M}$ is reflexive for whatever M,

(2)
M satisfies the transitive property if and only if $\sqsubseteq _{M}$ is a preorder on X.
Actually, Karapinar et al. [1] defined $\sqsupseteq _{M}$ (the dual relation of $\sqsubseteq _{ M}$) as the reflexive closure of M, which enlarges M, not equivalent, that is,
Using this relation, they succeeded to prove such results in Lemma 39 of [1].
Therefore, if we redefine $x\sqsubseteq u\Longleftrightarrow (x,u) \in M $, then it is equivalent to say that $\sqsubseteq :=M$.
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Acknowledgements
The last author is thankful to University Grant Commission, New Delhi, Government of India, for the financial support in the form of MANF (Moulana Azad National Fellowship). All the authors are thankful to three anonymous learned referees for their encouraging comments on the earlier version of the manuscript.
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Correspondence to Aftab Alam.
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MSC
 $\mathcal{R}$complete metric spaces
 Locally ftransitive binary relations
 $(g,d)$selfclosedness
 $\mathcal{R}$connectedness
Keywords
 47H10
 54H25