# Almost contractive coupled mapping in ordered complete metric spaces

## Abstract

In this paper, we introduce the notion of almost contractive mapping $F:X×X\to X$ with respect to the mapping $g:X\to X$ and establish some existence and uniqueness theorems of a coupled common coincidence point in ordered complete metric spaces. Also, we introduce an example to support our main results. Our results generalize several well-known comparable results in the literature.

MSC:54H25, 47H10, 34B15.

## 1 Introduction and preliminaries

The existence and uniqueness theorems of a fixed point in complete metric spaces play an important role in constructing methods for solving problems in differential equations, matrix equations, and integral equations. Furthermore, the fixed point theory is a crucial method in numerical analysis to present a way for solving and approximating the roots of many equations in real analysis. One of the main theorems on a fixed point is the Banach contraction theorem . Many authors generalized the Banach contraction theorem in different metric spaces in different ways. For some works on fixed point theory, we refer the readers to . The study of a coupled fixed point was initiated by Bhaskar and Lakshmikantham . Bhaskar and Lakshmikantham  obtained some nice results on a coupled fixed point and applied their results to solve a pair of differential equations. For some results on a coupled fixed point in ordered metric spaces, we refer the reader to .

The following definitions will be needed in the sequel.

Definition 1.1 Let $\left(X,⪯\right)$ be a partially ordered set and $F:X×X\to X$. The mapping F is said to have the mixed monotone property if $F\left(x,y\right)$ is monotone non-decreasing in x and is monotone non-increasing in y, that is, for any

$x,y\in X,{x}_{1},{x}_{2}\in X,\phantom{\rule{1em}{0ex}}{x}_{1}⪯{x}_{2}\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}F\left({x}_{1},y\right)⪯F\left({x}_{2},y\right)$

and

${y}_{1},{y}_{2}\in X,\phantom{\rule{1em}{0ex}}{y}_{1}⪯{y}_{2}\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}F\left(x,{y}_{1}\right)⪰F\left(x,{y}_{2}\right).$

Definition 1.2 We call an element $\left(x,y\right)\in X×X$ a coupled fixed point of the mapping $F:X×X\to X$ if

$F\left(x,y\right)=x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}F\left(y,x\right)=y.$

Definition 1.3 

Let $\left(X,⪯\right)$ be a partially ordered set and $F:X×X\to X$ and $g:X\to X$. The mapping F is said to have the mixed g-monotone property if F is monotone g-non-decreasing in its first argument and is monotone g-non-increasing in its second argument, that is, for any $x,y\in X$,

${x}_{1},{x}_{2}\in X,\phantom{\rule{1em}{0ex}}g\left({x}_{1}\right)⪯g\left({x}_{2}\right)\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}F\left({x}_{1},y\right)⪯F\left({x}_{2},y\right)$
(1)

and

${y}_{1},{y}_{2}\in X,\phantom{\rule{1em}{0ex}}g\left({y}_{1}\right)⪯g\left({y}_{2}\right)\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}F\left(x,{y}_{1}\right)⪰F\left(x,{y}_{2}\right).$
(2)

Definition 1.4 An element $\left(x,y\right)\in X×X$ is called a coupled coincidence point of the mappings $F:X×X\to X$ and $g:X\to X$ if

$F\left(x,y\right)=g\left(x\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}F\left(y,x\right)=g\left(y\right).$

The main results of Bhaskar and Lakshmikantham in  are the following.

Theorem 1.1 

Let $\left(X,⪯\right)$ be a partially ordered set and d be a metric on X such that $\left(X,d\right)$ is a complete metric space. Let $F:X×X\to X$ be a continuous mapping having the mixed monotone property on X. Assume that there exists a $k\in \left[0,1\right)$ with

If there exist two elements ${x}_{0},{y}_{0}\in X$ with

${x}_{0}⪯F\left({x}_{0},{y}_{0}\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{y}_{0}⪰F\left({y}_{0},{x}_{0}\right),$

then there exist $x,y\in X$ such that

$x=F\left(x,y\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}y=F\left(y,x\right).$

Theorem 1.2 

Let $\left(X,⪯\right)$ be a partially ordered set and d be a metric on X such that $\left(X,d\right)$ is a complete metric space. Assume that X has the following property:

1. (i)

if a nondecreasing sequence $\left\{{x}_{n}\right\}$ in X converges to $x\in X$, then ${x}_{n}⪯x$ for all n,

2. (ii)

if a nonincreasing sequence $\left\{{y}_{n}\right\}$ in X converges to $y\in X$, then ${y}_{n}⪰y$ for all n.

Let $F:X×X\to X$ be a mapping having the mixed monotone property on X. Assume that there exists $k\in \left[0,1\right)$ with

If there exist two elements ${x}_{0},{y}_{0}\in X$ with

${x}_{0}⪯F\left({x}_{0},{y}_{0}\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{y}_{0}⪰F\left({y}_{0},{x}_{0}\right),$

then there exist $x,y\in X$ such that

$x=F\left(x,y\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}y=F\left(y,x\right).$

Definition 1.5 Let $\left(X,d\right)$ be a metric space and $F:X×X\to X$ and $g:X\to X$ be mappings. We say that F and g commute if

$F\left(g\left(x\right),g\left(y\right)\right)=g\left(F\left(x,y\right)\right)$

for all $x,y\in X$.

Nashine and Shatanawi  proved the following coupled coincidence point theorems.

Theorem 1.3 

Let $\left(X,d,⪯\right)$ be an ordered metric space. Let $F:X×X\to X$ and $g:X\to X$ be mappings such that F has the mixed g-monotone property on X such that there exist two elements ${x}_{0},{y}_{0}\in X$ with $g\left({x}_{0}\right)⪯F\left({x}_{0},{y}_{0}\right)$ and $g\left({y}_{0}\right)⪰F\left({y}_{0},{x}_{0}\right)$. Suppose that there exist non-negative real numbers α, β, L with $\alpha +\beta <1$ such that

$\begin{array}{rcl}d\left(F\left(x,y\right),F\left(u,v\right)\right)& \le & \alpha min\left\{d\left(F\left(x,y\right),g\left(x\right)\right),d\left(F\left(u,v\right),g\left(x\right)\right)\right\}\\ +\beta min\left\{d\left(F\left(x,y\right),g\left(u\right)\right),d\left(F\left(u,v\right),g\left(u\right)\right)\right\}\\ +Lmin\left\{d\left(F\left(x,y\right),g\left(u\right)\right),d\left(F\left(u,v\right),g\left(x\right)\right)\right\}\end{array}$
(3)

for all $\left(x,y\right),\left(u,v\right)\in X×X$ with $g\left(x\right)⪯g\left(u\right)$ and $g\left(y\right)⪰g\left(v\right)$. Further suppose that $F\left(X×X\right)\subseteq g\left(X\right)$ and $g\left(X\right)$ is a complete subspace of X. Also suppose that X satisfies the following properties:

1. (i)

if a nondecreasing sequence $\left\{{x}_{n}\right\}$ in X converges to $x\in X$, then ${x}_{n}⪯x$ for all n,

2. (ii)

if a nonincreasing sequence $\left\{{y}_{n}\right\}$ in X converges to $y\in X$, then ${y}_{n}⪰y$ for all n.

Then there exist $x,y\in X$ such that

$F\left(x,y\right)=g\left(x\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}F\left(y,x\right)=g\left(y\right),$

that is, F and g have a coupled coincidence point $\left(x,y\right)\in X×X$.

Theorem 1.4 

Let $\left(X,⪯\right)$ be a partially ordered set and suppose that there is a metric d on X such that $\left(X,d\right)$ is a complete metric space. Let $F:X×X\to X$ and $g:X\to X$ be mappings such that F has the mixed g-monotone property on X such that there exist two elements ${x}_{0},{y}_{0}\in X$ with $g\left({x}_{0}\right)⪯F\left({x}_{0},{y}_{0}\right)$ and $g\left({y}_{0}\right)⪰F\left({y}_{0},{x}_{0}\right)$. Suppose that there exist non-negative real numbers α, β, L with $\alpha +\beta <1$ such that

$\begin{array}{rl}d\left(F\left(x,y\right),F\left(u,v\right)\right)\le & \alpha min\left\{d\left(F\left(x,y\right),g\left(x\right)\right),d\left(F\left(u,v\right),g\left(x\right)\right)\right\}\\ +\beta min\left\{d\left(F\left(x,y\right),g\left(u\right)\right),d\left(F\left(u,v\right),g\left(u\right)\right)\right\}\\ +Lmin\left\{d\left(F\left(x,y\right),g\left(u\right)\right),d\left(F\left(u,v\right),g\left(x\right)\right)\right\}\end{array}$
(4)

for all $\left(x,y\right),\left(u,v\right)\in X×X$ with $g\left(x\right)⪯g\left(u\right)$ and $g\left(y\right)⪰g\left(v\right)$. Further suppose that $F\left(X×X\right)\subseteq g\left(X\right)$, g is continuous nondecreasing and commutes with F, and also suppose that either

1. (a)

F is continuous, or

2. (b)

X has the following property:

1. (i)

if a nondecreasing sequence $\left\{{x}_{n}\right\}$ in X converges to $x\in X$, then ${x}_{n}⪯x$ for all n,

2. (ii)

if a nonincreasing sequence $\left\{{y}_{n}\right\}$ in X converges to $y\in X$, then ${y}_{n}⪰y$ for all n.

Then there exist $x,y\in X$ such that

$F\left(x,y\right)=g\left(x\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}F\left(y,x\right)=g\left(y\right),$

that is, F and g have a coupled coincidence point $\left(x,y\right)\in X×X$.

Berinde  initiated the concept of almost contractions and studied many interesting fixed point theorems for a Ćirić strong almost contraction. So, it is fundamental to recall the following definition.

Definition 1.6 

A single-valued mapping $f:X×X$ is called a Ćirić strong almost contraction if there exist a constant $\alpha \in \left[0,1\right)$ and some $L\ge 0$ such that

$d\left(fx,fy\right)\le \alpha M\left(x,y\right)+Ld\left(y,fx\right)$

for all $x,y\in X$, where

$M\left(x,y\right)=max\left\{d\left(x,y\right),d\left(x,fx\right),d\left(y,fy\right),\frac{d\left(x,fy\right)+d\left(y,fx\right)}{2}\right\}.$

The aim of this paper is to introduce the notion of almost contractive mapping $F:X×X\to X$ with respect to the mapping $g:X\to X$ and present some uniqueness and existence theorems of coupled fixed and coincidence point. Our results generalize Theorems 1.1-1.4.

## 2 Main theorems

Definition 2.1 Let $\left(X,d,⪯\right)$ be an ordered metric space. We say that the mapping $F:X×X\to X$ is an almost contractive mapping with respect to the mapping $g:X\to X$ if there exist a real number $\alpha \in \left[0,1\right)$ and a nonnegative number L such that

$\begin{array}{r}d\left(F\left(x,y\right),F\left(u,v\right)\right)\\ \phantom{\rule{1em}{0ex}}\le \alpha max\left\{d\left(g\left(x\right),g\left(u\right)\right),d\left(g\left(y\right),g\left(v\right)\right),d\left(F\left(x,y\right),g\left(x\right)\right),d\left(F\left(u,v\right),g\left(u\right)\right)\right\}\\ \phantom{\rule{2em}{0ex}}+Lmin\left\{d\left(F\left(x,y\right),g\left(u\right)\right),d\left(F\left(u,v\right),g\left(x\right)\right)\right\}\end{array}$
(5)

for all $\left(x,y\right),\left(u,v\right)\in X×X$ with $g\left(x\right)⪯g\left(u\right)$ and $g\left(y\right)⪰g\left(v\right)$.

Theorem 2.1 Let $\left(X,d,⪯\right)$ be an ordered metric space. Let $F:X×X\to X$ and $g:X\to X$ be mappings such that

1. (1)

F is an almost contractive mapping with respect to g.

2. (2)

F has the mixed g-monotone property on X.

3. (3)

There exist two elements ${x}_{0},{y}_{0}\in X$ with $g\left({x}_{0}\right)⪯F\left({x}_{0},{y}_{0}\right)$ and $g\left({y}_{0}\right)⪰F\left({y}_{0},{x}_{0}\right)$.

4. (4)

$F\left(X×X\right)\subseteq g\left(X\right)$ and $g\left(X\right)$ is a complete subspace of X.

Also, suppose that X satisfies the following properties:

1. (i)

if a nondecreasing sequence $\left\{{x}_{n}\right\}$ in X converges to $x\in X$, then ${x}_{n}⪯x$ for all n,

2. (ii)

if a nonincreasing sequence $\left\{{y}_{n}\right\}$ in X converges to $y\in X$, then ${y}_{n}⪰y$ for all n.

Then there exist $x,y\in X$ such that

$F\left(x,y\right)=g\left(x\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}F\left(y,x\right)=g\left(y\right),$

that is, F and g have a coupled coincidence point $\left(x,y\right)\in X×X$.

Proof Let ${x}_{0},{y}_{0}\in X$ be such that $g\left({x}_{0}\right)⪯F\left({x}_{0},{y}_{0}\right)$ and $g\left({y}_{0}\right)⪰F\left({y}_{0},{x}_{0}\right)$. Since $F\left(X×X\right)\subseteq g\left(X\right)$, we can choose ${x}_{1},{y}_{1}\in X$ such that $g\left({x}_{1}\right)=F\left({x}_{0},{y}_{0}\right)$ and $g\left({y}_{1}\right)=F\left({y}_{0},{x}_{0}\right)$.

In the same way, we construct $g\left({x}_{2}\right)=F\left({x}_{1},{y}_{1}\right)$ and $g\left({y}_{2}\right)=F\left({y}_{1},{x}_{1}\right)$.

Continuing in this way, we construct two sequences $\left\{{x}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ in X such that

$g\left({x}_{n+1}\right)=F\left({x}_{n},{y}_{n}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g\left({y}_{n+1}\right)=F\left({y}_{n},{x}_{n}\right)\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N}\cup \left\{0\right\}.$
(6)

Since F has the mixed g-monotone property, by induction we may show that

$g\left({x}_{0}\right)⪯g\left({x}_{1}\right)⪯g\left({x}_{2}\right)⪯\cdots ⪯g\left({x}_{n+1}\right)⪯\cdots$

and

$g\left({y}_{0}\right)⪰g\left({y}_{1}\right)⪰g\left({y}_{2}\right)⪰\cdots ⪰g\left({y}_{n+1}\right)⪰\cdots .$

If $\left(g\left({x}_{n+1}\right),g\left({y}_{n+1}\right)\right)=\left(g\left({x}_{n}\right),g\left({y}_{n}\right)\right)$ for some $n\in \mathbb{N}$, then $F\left({x}_{n},{y}_{n}\right)=g\left({x}_{n}\right)$ and $F\left({y}_{n},{x}_{n}\right)=g\left({y}_{n}\right)$, that is, $\left({x}_{n},{y}_{n}\right)$ is a coincidence point of F and g. So we may assume that $\left(g\left({x}_{n+1}\right),g\left({y}_{n+1}\right)\right)\ne \left(g\left({x}_{n}\right),g\left({y}_{n}\right)\right)$ for all $n\in \mathbb{N}$. Let $n\in \mathbb{N}$. Since $g\left({x}_{n}\right)⪰g\left({x}_{n-1}\right)$ and $g\left({y}_{n}\right)⪯g\left({y}_{n-1}\right)$, from (5) and (6), we have

$\begin{array}{r}d\left(g\left({x}_{n}\right),g\left({x}_{n+1}\right)\right)\\ \phantom{\rule{1em}{0ex}}=d\left(F\left({x}_{n-1},{y}_{n-1}\right),F\left({x}_{n},{y}_{n}\right)\right)\\ \phantom{\rule{1em}{0ex}}\le \alpha max\left\{d\left(g\left({x}_{n-1}\right),g\left({x}_{n}\right)\right),d\left(g\left({y}_{n-1}\right),g\left({y}_{n}\right)\right),d\left(F\left({x}_{n},{y}_{n}\right),g\left({x}_{n}\right)\right),\\ \phantom{\rule{2em}{0ex}}d\left(F\left({x}_{n-1},{y}_{n-1}\right),g\left({x}_{n-1}\right)\right)\right\}+Lmin\left\{d\left(F\left({x}_{n},{y}_{n}\right),g\left({x}_{n-1}\right)\right),d\left(F\left({x}_{n-1},{y}_{n-1}\right),g\left({x}_{n}\right)\right)\right\}\\ \phantom{\rule{1em}{0ex}}=\alpha max\left\{d\left(g\left({x}_{n-1}\right),g\left({x}_{n}\right)\right),d\left(g\left({y}_{n-1}\right),g\left({y}_{n}\right)\right),d\left(g\left({x}_{n+1}\right),g\left({x}_{n}\right)\right),d\left(g\left({x}_{n}\right),g\left({x}_{n-1}\right)\right)\right\}\\ \phantom{\rule{2em}{0ex}}+Lmin\left\{d\left(g\left({x}_{n+1}\right),g\left({x}_{n-1}\right)\right),d\left(g\left({x}_{n}\right),g\left({x}_{n}\right)\right)\right\}\\ \phantom{\rule{1em}{0ex}}=\alpha max\left\{d\left(g\left({x}_{n-1}\right),g\left({x}_{n}\right)\right),d\left(g\left({y}_{n-1}\right),g\left({y}_{n}\right)\right),d\left(g\left({x}_{n+1}\right),g\left({x}_{n}\right)\right)\right\}.\end{array}$

If $max\left\{d\left(g\left({x}_{n-1}\right),g\left({x}_{n}\right)\right),d\left(g\left({y}_{n-1}\right),g\left({y}_{n}\right)\right),d\left(g\left({x}_{n+1}\right),g\left({x}_{n}\right)\right)\right\}=d\left(g\left({x}_{n+1}\right),g\left({x}_{n}\right)\right)$, then $d\left(g\left({x}_{n+1}\right),g\left({x}_{n}\right)\right)\le \alpha d\left(g\left({x}_{n+1}\right),g\left({x}_{n}\right)\right)$ and hence $d\left(g\left({x}_{n+1}\right),g\left({x}_{n}\right)\right)=0$. Thus $d\left(g\left({x}_{n-1}\right),g\left({x}_{n}\right)\right)=d\left(g\left({y}_{n-1}\right),g\left({y}_{n}\right)\right)=0$. Therefore $d\left(g\left({x}_{n-1}\right),g\left({y}_{n-1}\right)\right)=d\left(g\left({x}_{n}\right),g\left({y}_{n}\right)\right)$, a contradiction. Thus

$\begin{array}{r}max\left\{d\left(g\left({x}_{n-1}\right),g\left({x}_{n}\right)\right),d\left(g\left({y}_{n-1}\right),g\left({y}_{n}\right)\right),d\left(g\left({x}_{n+1}\right),g\left({x}_{n}\right)\right)\right\}\\ \phantom{\rule{1em}{0ex}}=max\left\{d\left(g\left({x}_{n-1}\right),g\left({x}_{n}\right)\right),d\left(g\left({y}_{n-1}\right),g\left({y}_{n}\right)\right)\right\}.\end{array}$

Therefore

$d\left(g\left({x}_{n+1}\right),g\left({x}_{n}\right)\right)\le \alpha max\left\{d\left(g\left({x}_{n-1}\right),g\left({x}_{n}\right)\right),d\left(g\left({y}_{n-1}\right),g\left({y}_{n}\right)\right)\right\}.$
(7)

Similarly, we may show that

$d\left(g\left({y}_{n}\right),g\left({y}_{n+1}\right)\right)\le \alpha max\left\{d\left(g\left({x}_{n-1}\right),g\left({x}_{n}\right)\right),d\left(g\left({y}_{n-1}\right),g\left({y}_{n}\right)\right)\right\}.$
(8)

From (7) and (8), we have

$\begin{array}{r}max\left\{d\left(g\left({x}_{n+1}\right),g\left({x}_{n}\right)\right),d\left(g\left({y}_{n}\right),g\left({y}_{n+1}\right)\right)\right\}\\ \phantom{\rule{1em}{0ex}}\le \alpha max\left\{d\left(g\left({x}_{n-1}\right),g\left({x}_{n}\right)\right),d\left(g\left({y}_{n-1}\right),g\left({y}_{n}\right)\right)\right\}.\end{array}$
(9)

Repeating (9) n-times, we get

$\begin{array}{r}max\left\{d\left(g\left({x}_{n+1}\right),g\left({x}_{n}\right)\right),d\left(g\left({y}_{n}\right),g\left({y}_{n+1}\right)\right)\right\}\\ \phantom{\rule{1em}{0ex}}\le {\alpha }^{n}max\left\{d\left(g\left({x}_{0}\right),g\left({x}_{1}\right)\right),d\left(g\left({y}_{0}\right),g\left({y}_{1}\right)\right)\right\}.\end{array}$
(10)

Now, we shall prove that $\left\{g\left({x}_{n}\right)\right\}$ and $\left\{g\left({y}_{n}\right)\right\}$ are Cauchy sequences in $g\left(X\right)$.

For each $m\ge n$, we have

$\begin{array}{r}d\left(g\left({x}_{m}\right),g\left({x}_{n}\right)\right)\\ \phantom{\rule{1em}{0ex}}\le d\left(g\left({x}_{n}\right),g\left({x}_{n+1}\right)\right)+d\left(g\left({x}_{n+1}\right),g\left({x}_{n+2}\right)\right)+\cdots \\ \phantom{\rule{2em}{0ex}}+d\left(g\left({x}_{m-1}\right),g\left({x}_{m}\right)\right)\\ \phantom{\rule{1em}{0ex}}\le {\alpha }^{n}max\left\{d\left(g\left({x}_{0}\right),g\left({x}_{1}\right)\right),d\left(g\left({y}_{0}\right),g\left({y}_{1}\right)\right)\right\}+\cdots \\ \phantom{\rule{2em}{0ex}}+{\alpha }^{m-1}max\left\{d\left(g\left({x}_{0}\right),g\left({x}_{1}\right)\right),d\left(g\left({y}_{0}\right),g\left({y}_{1}\right)\right)\right\}\\ \phantom{\rule{1em}{0ex}}\le \frac{{\alpha }^{n}}{1-\alpha }max\left\{d\left(g\left({x}_{0}\right),g\left({x}_{1}\right)\right),d\left(g\left({y}_{0}\right),g\left({y}_{1}\right)\right)\right\}.\end{array}$

Letting $n,m\to +\mathrm{\infty }$ in the above inequalities, we get that $\left\{g\left({x}_{n}\right)\right\}$ is a Cauchy sequence in $g\left(X\right)$. Similarly, we may show that $\left\{g\left({y}_{n}\right)\right\}$ is a Cauchy sequence in $g\left(X\right)$. Since $g\left(X\right)$ is a complete subspace of X, there exists $\left(x,y\right)\in X×X$ such that $g\left({x}_{n}\right)\to g\left(x\right)$ and $g\left({y}_{n}\right)\to g\left(y\right)$. Since $\left\{g\left({x}_{n}\right)\right\}$ is a non-decreasing sequence and $g\left({x}_{n}\right)\to g\left(x\right)$ and as $\left\{g\left({y}_{n}\right)\right\}$ is a non-increasing sequence and $g\left({y}_{n}\right)\to g\left(y\right)$, by the assumption we have $g\left({x}_{n}\right)⪯g\left(x\right)$ and $g\left({y}_{n}\right)⪰g\left(y\right)$ for all n. Since

$\begin{array}{r}d\left(g\left({x}_{n+1}\right),F\left(x,y\right)\right)\\ \phantom{\rule{1em}{0ex}}=d\left(F\left({x}_{n},{y}_{n}\right),F\left(x,y\right)\right)\\ \phantom{\rule{1em}{0ex}}\le \alpha max\left\{d\left(g\left({x}_{n}\right),g\left(x\right)\right),d\left(g\left({y}_{n}\right),g\left(y\right)\right),d\left(g\left({x}_{n+1}\right),g\left({x}_{n}\right)\right),d\left(F\left(x,y\right),g\left(x\right)\right)\right\}\\ \phantom{\rule{2em}{0ex}}+Lmin\left\{d\left(g\left({x}_{n+1}\right),g\left(x\right)\right),d\left(F\left(x,y\right),g\left({x}_{n}\right)\right)\right\}.\end{array}$

Letting $n\to \mathrm{\infty }$ in the above inequality, we get $d\left(g\left(x\right),F\left(x,y\right)\right)=0$. Hence $g\left(x\right)=F\left(x,y\right)$. Similarly, one can show that $g\left(y\right)=F\left(y,x\right)$. Thus we proved that F and g have a coupled coincidence point. □

Theorem 2.2 Let $\left(X,⪯\right)$ be a partially ordered set and suppose that there is a metric d on X such that $\left(X,d\right)$ is a complete metric space. Let $F:X×X\to X$ and $g:X\to X$ be mappings such that

1. (1)

F is an almost contractive mapping with respect to g.

2. (2)

F has the mixed g-monotone property on X.

3. (3)

There exist two elements ${x}_{0},{y}_{0}\in X$ with $g\left({x}_{0}\right)⪯F\left({x}_{0},{y}_{0}\right)$ and $g\left({y}_{0}\right)⪰F\left({y}_{0},{x}_{0}\right)$.

4. (4)

$F\left(X×X\right)\subseteq g\left(X\right)$.

5. (5)

g is continuous nondecreasing and commutes with F.

Also suppose that either

1. (a)

F is continuous, or

2. (b)

X has the following property:

1. (i)

if a nondecreasing sequence $\left\{{x}_{n}\right\}$ in X converges to $x\in X$, then ${x}_{n}⪯x$ for all n,

2. (ii)

if a nonincreasing sequence $\left\{{y}_{n}\right\}$ in X converges to $y\in X$, then ${y}_{n}⪰y$ for all n.

Then there exist $x,y\in X$ such that

$F\left(x,y\right)=g\left(x\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}F\left(y,x\right)=g\left(y\right),$

that is, F and g have a coupled coincidence point $\left(x,y\right)\in X×X$.

Proof As in the proof of Theorem 2.1, we construct two Cauchy sequences $\left(g{x}_{n}\right)$ and $\left(g{y}_{n}\right)$ in X such that $\left(g{x}_{n}\right)$ is a nondecreasing sequence in X and $\left(g{y}_{n}\right)$ is a nonincreasing sequence in X. Since X is a complete metric space, there is $\left(x,y\right)\in X×X$ such that $g{x}_{n}\to x$ and $g{y}_{n}\to y$. Since g is continuous, we have $g\left(g{x}_{n}\right)\to gx$ and $g\left(g{y}_{n}\right)\to gy$.

Suppose that (a) holds. Since F is continuous, we have $F\left(g{x}_{n},g{y}_{n}\right)\to F\left(x,y\right)$ and $F\left(g{y}_{n},g{x}_{n}\right)\to F\left(y,x\right)$. Also, since g commutes with F and g is continuous, we have $F\left(g{x}_{n},g{y}_{n}\right)=gF\left({x}_{n},{y}_{n}\right)=g\left(g{x}_{n+1}\right)\to gx$ and $F\left(g{y}_{n},g{x}_{n}\right)=gF\left({y}_{n},{x}_{n}\right)=g\left(g{y}_{n+1}\right)\to gy$. By uniqueness of limit, we get $gx=F\left(x,y\right)$ and $gy=F\left(y,x\right)$.

Second, suppose that (b) holds. Since $g\left({x}_{n}\right)$ is a nondecreasing sequence such that $g\left({x}_{n}\right)\to x$, $g\left({y}_{n}\right)$ is a nonincreasing sequence such that $g\left({y}_{n}\right)\to y$, and g is a nondecreasing function, we get that $g\left(g{x}_{n}\right)⪯gx$ and $g\left(g{y}_{n}\right)⪰g\left(y\right)$ hold for all $n\in \mathbb{N}$. By (5), we have

$\begin{array}{r}d\left(g\left(g{x}_{n+1}\right),F\left(x,y\right)\right)\\ \phantom{\rule{1em}{0ex}}=d\left(F\left(g{x}_{n},g{y}_{n}\right),F\left(x,y\right)\right)\\ \phantom{\rule{1em}{0ex}}\le \alpha max\left\{d\left(g\left(g{x}_{n}\right),g\left(x\right)\right),d\left(g\left(g{y}_{n}\right),g\left(y\right)\right),d\left(g\left(g{x}_{n+1}\right),g\left(g{x}_{n}\right)\right),d\left(F\left(x,y\right),g\left(x\right)\right)\right\}\\ \phantom{\rule{2em}{0ex}}+Lmin\left\{d\left(g\left(g{x}_{n+1}\right),g\left(x\right)\right),d\left(F\left(x,y\right),g\left(g{x}_{n}\right)\right)\right\}.\end{array}$

Letting $n\to +\mathrm{\infty }$, we get $d\left(g\left(x\right),F\left(x,y\right)\right)=0$ and hence $g\left(x\right)=F\left(x,y\right)$. Similarly, one can show that $g\left(y\right)=F\left(y,x\right)$. Thus $\left(x,y\right)$ is a coupled coincidence point of F and g. □

Corollary 2.1 Let $\left(X,⪯\right)$ be a partially ordered set and suppose that there is a metric d on X such that $\left(X,d\right)$ is a complete metric space. Let $F:X×X\to X$ be a mapping such that F has the mixed monotone property on X such that there exist two elements ${x}_{0},{y}_{0}\in X$ with ${x}_{0}⪯F\left({x}_{0},{y}_{0}\right)$ and ${y}_{0}⪰F\left({y}_{0},{x}_{0}\right)$. Suppose that there exist a real number $\alpha \in \left[0,1\right)$ and a nonnegative number L such that

$\begin{array}{rcl}d\left(F\left(x,y\right),F\left(u,v\right)\right)& \le & \alpha max\left\{d\left(x,u\right),d\left(y,v\right),d\left(F\left(x,y\right),x\right),d\left(F\left(u,v\right),u\right)\right\}\\ +Lmin\left\{d\left(F\left(x,y\right),u\right),d\left(F\left(u,v\right),x\right)\right\}\end{array}$
(11)

for all $\left(x,y\right),\left(u,v\right)\in X×X$ with $x⪯u$ and $y⪰v$ and also suppose that either

1. (a)

F is continuous, or

2. (b)

X has the following property:

1. (i)

if a nondecreasing sequence $\left\{{x}_{n}\right\}$ in X converges to $x\in X$, then ${x}_{n}⪯x$ for all n,

2. (ii)

if a nonincreasing sequence $\left\{{y}_{n}\right\}$ in X converges to $y\in X$, then ${y}_{n}⪰y$ for all n,

then there exist $x,y\in X$ such that

$F\left(x,y\right)=x\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}F\left(y,x\right)=y,$

that is, F has a coupled fixed point $\left(x,y\right)\in X×X$.

Proof Follows from Theorem 2.2 by taking $g=I$, the identity mapping. □

Let $\left(X,⪯\right)$ be a partially ordered set. Then we define a partial order on the product space $X×X$ as follows:

Now, we prove some uniqueness theorem of a coupled common fixed point of mappings $F:X×X\to X$ and $g:X\to X$.

Theorem 2.3 In addition to the hypotheses of Theorem  2.1, suppose that $L=0$, $\alpha <\frac{1}{2}$, F and g commute and for every $\left(x,y\right),\left({y}^{\ast },{x}^{\ast }\right)\in X×X$, there exists $\left(u,v\right)\in X×X$ such that $\left(F\left(u,v\right),F\left(v,u\right)\right)$ is comparable to $\left(F\left(x,y\right),F\left(y,x\right)\right)$ and $\left(F\left({x}^{\ast },{y}^{\ast }\right),F\left({y}^{\ast },{x}^{\ast }\right)\right)$. Then F and g have a unique coupled common fixed point, that is, there exists a unique $\left(x,y\right)\in X×X$ such that

$x=g\left(x\right)=F\left(x,y\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}y=g\left(y\right)=F\left(y,x\right).$

Proof The existence of coupled coincidence points of F and g follows from Theorem 2.1. To prove the uniqueness, let $\left(x,y\right)$ and $\left({x}^{\ast },{y}^{\ast }\right)$ be coupled coincidence points of F and g; that is, $g\left(x\right)=F\left(x,y\right)$, $g\left(y\right)=F\left(y,x\right)$, $g\left({x}^{\ast }\right)=F\left({x}^{\ast },{y}^{\ast }\right)$ and $g\left({y}^{\ast }\right)=F\left({y}^{\ast },{x}^{\ast }\right)$. Now, we prove that

$g\left(x\right)=g\left({x}^{\ast }\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g\left(y\right)=g\left({y}^{\ast }\right).$
(12)

By the hypotheses, there exists $\left(u,v\right)\in X×X$ such that $\left(F\left(u,v\right),F\left(v,u\right)\right)$ is comparable to $\left(F\left(x,y\right),F\left(y,x\right)\right)$ and $\left(F\left({x}^{\ast },{y}^{\ast }\right),F\left({y}^{\ast },{x}^{\ast }\right)\right)$. Put ${u}_{0}=u$, ${v}_{0}=v$. Let ${u}_{1},{v}_{1}\in X$ be such that $g\left({u}_{1}\right)=F\left({u}_{0},{v}_{0}\right)$ and $g\left({v}_{1}\right)=F\left({v}_{0},{u}_{0}\right)$. Then as a similar proof of Theorem 2.1, we construct two sequences $\left\{g\left({u}_{n}\right)\right\}$, $\left\{g\left({v}_{n}\right)\right\}$ in $g\left(X\right)$, where $g\left({u}_{n+1}\right)=F\left({u}_{n},{v}_{n}\right)$ and $g\left({v}_{n+1}\right)=F\left({v}_{n},{u}_{n}\right)$ for all $n\in \mathbb{N}$. Further, set ${x}_{0}=x$, ${y}_{0}=y$, ${x}_{0}^{\ast }={x}^{\ast }$, ${y}_{0}^{\ast }={y}^{\ast }$. Define the sequences $\left\{g\left({x}_{n}\right)\right\}$, $\left\{g\left({y}_{n}\right)\right\}$ in the following way: define $g{x}_{1}=F\left({x}_{0},{y}_{0}\right)=F\left(x,y\right)$ and $g{y}_{1}=F\left({y}_{0},{x}_{0}\right)=F\left(y,x\right)$. Also, define $g{x}_{2}=F\left({x}_{1},{y}_{1}\right)$ and $g{y}_{2}=F\left({y}_{1},{x}_{1}\right)$. For each $n\in \mathbb{N}$, define $g{x}_{n+1}=F\left({x}_{n},{y}_{n}\right)$ and $g{y}_{n+1}=F\left({y}_{n},{x}_{n}\right)$. In the same way, we define the sequences $\left\{g\left({x}_{n}^{\ast }\right)\right\}$, $\left\{g\left({y}_{n}^{\ast }\right)\right\}$. Now, we prove that

$g\left({x}_{n}\right)=F\left(x,y\right)=g\left(x\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g\left({y}_{n}\right)=F\left(y,x\right)=g\left(y\right).$

Since $\left(x,y\right)$ is a coupled coincidence point of F and g, we have $F\left(x,y\right)=g\left(x\right)$ and $F\left(y,x\right)=g\left(y\right)$. Thus $g\left({x}_{1}\right)=F\left({x}_{0},{y}_{0}\right)=F\left(x,y\right)=g\left(x\right)$ and $g\left({y}_{1}\right)=F\left({y}_{0},{x}_{0}\right)=F\left(y,x\right)=g\left(y\right)$. Therefore $g\left({x}_{1}\right)⪯g\left(x\right)$, $g\left(x\right)⪯g\left({x}_{1}\right)$, $g\left({y}_{1}\right)⪯g\left(y\right)$ and $g\left(y\right)⪯g\left({y}_{1}\right)$. Since F is monotone g-non-decreasing on its first argument, $g\left({x}_{1}\right)⪯g\left(x\right)$, and $g\left(x\right)⪯g\left({x}_{1}\right)$, we have $F\left({x}_{1},{y}_{1}\right)⪯F\left(x,{y}_{1}\right)$ and $F\left(x,{y}_{1}\right)⪯F\left({x}_{1},{y}_{1}\right)$. Therefore,

$F\left({x}_{1},{y}_{1}\right)=F\left(x,{y}_{1}\right).$
(13)

Also, since F is monotone g-non-increasing on its second argument, $g\left({y}_{1}\right)⪯g\left(y\right)$ and $g\left(y\right)⪯g\left({y}_{1}\right)$, we have $F\left(x,y\right)⪯F\left(x,{y}_{1}\right)$ and $F\left(x,{y}_{1}\right)⪯F\left(x,y\right)$. Therefore,

$F\left(x,y\right)=F\left(x,{y}_{1}\right).$
(14)

From (13) and (14), we have

$g\left({x}_{2}\right)=F\left({x}_{1},{y}_{1}\right)=F\left(x,y\right)=g\left(x\right).$

Similarly, we may show that

$g\left({y}_{2}\right)=F\left({y}_{1},{x}_{1}\right)=F\left(y,x\right)=g\left(y\right).$

Note that $g\left({x}_{2}\right)⪯g\left(x\right)$, $g\left(x\right)⪯g\left({x}_{2}\right)$, $g\left({y}_{2}\right)⪯g\left(y\right)$ and $g\left(y\right)⪯g\left({y}_{2}\right)$. Since F is monotone g-non-decreasing on its first argument, $g\left({x}_{2}\right)⪯g\left(x\right)$, and $g\left(x\right)⪯g\left({x}_{2}\right)$, we have $F\left({x}_{2},{y}_{2}\right)⪯F\left(x,{y}_{2}\right)$ and $F\left(x,{y}_{2}\right)⪯F\left({x}_{2},{y}_{2}\right)$. Therefore,

$F\left({x}_{2},{y}_{2}\right)=F\left(x,{y}_{2}\right).$
(15)

Also, since F is monotone g-non-increasing on its second argument, $g\left({y}_{2}\right)⪯g\left(y\right)$ and $g\left(y\right)⪯g\left({y}_{2}\right)$, we have $F\left(x,y\right)⪯F\left(x,{y}_{2}\right)$ and $F\left(x,{y}_{2}\right)⪯F\left(x,y\right)$. Therefore,

$F\left(x,y\right)=F\left(x,{y}_{2}\right).$
(16)

From (15) and (16), we have

$g\left({x}_{3}\right)=F\left({x}_{2},{y}_{2}\right)=F\left(x,y\right)=g\left(x\right).$

Similarly, we may show that

$g\left({y}_{3}\right)=F\left({y}_{2},{x}_{2}\right)=F\left(y,x\right)=g\left(y\right).$

Continuing in the same way, we have that

$g\left({x}_{n}\right)=F\left(x,y\right)=g\left(x\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g\left({y}_{n}\right)=F\left(y,x\right)=g\left(y\right)$

hold for all $n\in \mathbb{N}$. Similarly, we can show that

$g\left({x}_{n}^{\ast }\right)=F\left({x}^{\ast },{y}^{\ast }\right)=g\left({x}^{\ast }\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g\left({y}_{n}^{\ast }\right)=F\left({y}^{\ast },{x}^{\ast }\right)=g\left({y}^{\ast }\right)\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N}$

hold for all $n\in \mathbb{N}$. Since

$\left(F\left(x,y\right),F\left(y,x\right)\right)=\left(g\left({x}_{1}\right),g\left({y}_{1}\right)\right)=\left(g\left(x\right),g\left(y\right)\right)$

and

$\left(F\left(u,v\right),F\left(v,u\right)\right)=\left(g\left({u}_{1}\right),g\left({v}_{1}\right)\right)$

are comparable, $g\left(x\right)⪯g\left({u}_{1}\right)$ and $g\left(y\right)⪰g\left({v}_{1}\right)$. Since F has the mixed g-monotone property of X, we have $g\left(x\right)⪯g\left({u}_{n}\right)$ and $g\left(y\right)⪰g\left({v}_{n}\right)$ for all $n\in \mathbb{N}$. Also, since $\left(g\left({x}^{\ast }\right),g\left({y}^{\ast }\right)\right)$ and $\left(F\left(u,v\right),F\left(v,u\right)\right)=\left(g\left({u}_{1}\right),g\left({v}_{1}\right)\right)$ are comparable, and F has the g-monotone property, then we can show that for $n\in \mathbb{N}$, we have that $\left(g\left({x}^{\ast }\right),g\left({y}^{\ast }\right)\right)$ and $\left(g\left({u}_{n}\right),g\left({v}_{n}\right)\right)$ are comparable. Now, if $\left(g\left(x\right),g\left(y\right)\right)=\left(g\left({u}_{k}\right),g\left({v}_{k}\right)\right)$ for some $k\in \mathbb{N}$ or $\left(g\left({x}^{\ast }\right),g\left({y}^{\ast }\right)\right)=\left(g\left({u}_{k}\right),g\left({v}_{k}\right)\right)$ for some $k\in \mathbb{N}$, then $\left(g\left(x\right),g\left(y\right)\right)$ and $\left(g\left({x}^{\ast }\right),g\left({y}^{\ast }\right)\right)$ are comparable, say $g\left(x\right)⪯g\left({x}^{\ast }\right)$ and $g\left(y\right)⪰g\left({y}^{\ast }\right)$. Thus from (5) we have

$\begin{array}{r}d\left(g\left(x\right),g\left({x}^{\ast }\right)\right)\\ \phantom{\rule{1em}{0ex}}=d\left(F\left(x,y\right),F\left({x}^{\ast },{y}^{\ast }\right)\right)\\ \phantom{\rule{1em}{0ex}}\le \alpha max\left\{d\left(g\left(x\right),g\left({x}^{\ast }\right)\right),d\left(g\left(y\right),g\left({y}^{\ast }\right)\right),d\left(F\left(x,y\right),g\left(x\right)\right),d\left(F\left({x}^{\ast },{y}^{\ast }\right),g\left({x}^{\ast }\right)\right)\right\}\\ \phantom{\rule{1em}{0ex}}=\alpha max\left\{d\left(g\left(x\right),g\left({x}^{\ast }\right)\right),d\left(g\left(y\right),g\left({y}^{\ast }\right)\right)\right\}\end{array}$
(17)

and

$\begin{array}{r}d\left(g\left({y}^{\ast }\right),g\left(y\right)\right)\\ \phantom{\rule{1em}{0ex}}=d\left(F\left({y}^{\ast },{x}^{\ast }\right),F\left(y,x\right)\right)\\ \phantom{\rule{1em}{0ex}}\le \alpha max\left\{d\left(g\left(y\right),g\left({y}^{\ast }\right)\right),d\left(g\left(x\right),g\left({x}^{\ast }\right)\right),d\left(F\left({y}^{\ast },{x}^{\ast }\right),g\left({y}^{\ast }\right)\right),d\left(F\left(y,x\right),g\left(y\right)\right)\right\}\\ \phantom{\rule{1em}{0ex}}=\alpha max\left\{d\left(g\left(y\right),g\left({y}^{\ast }\right)\right),d\left(g\left(x\right),g\left(y\right)\right)\right\}.\end{array}$
(18)

From (17) and (18), we have

$max\left\{d\left(g\left(x\right),g\left({x}^{\ast }\right)\right),d\left(g\left(y\right),g\left({y}^{\ast }\right)\right)\right\}\le \alpha max\left\{d\left(g\left(y\right),g\left({y}^{\ast }\right)\right),d\left(g\left(x\right),g\left(y\right)\right)\right\}.$

Since $\alpha <1$, we have $d\left(g\left(x\right),g\left({x}^{\ast }\right)\right)=0$ and $d\left(g\left(y\right),g\left({y}^{\ast }\right)\right)=0$. Therefore (12) is satisfied. Now, suppose that $\left(g\left(x\right),g\left(y\right)\right)\ne \left(g\left({u}_{n}\right),g\left({v}_{n}\right)\right)$ for all $n\in \mathbb{N}$ and $\left(g\left({x}^{\ast }\right),g\left({y}^{\ast }\right)\right)\ne \left(g\left({u}_{n}\right),g\left({v}_{n}\right)\right)$ for all $n\in \mathbb{N}$. Let $n\in \mathbb{N}$. Since $g\left(x\right)⪯g\left({u}_{n}\right)$ and $g\left(y\right)⪰g\left({v}_{n}\right)$, then from (5) we have

$\begin{array}{r}d\left(g\left(x\right),g\left({u}_{n+1}\right)\right)\\ \phantom{\rule{1em}{0ex}}=d\left(F\left(x,y\right),F\left({u}_{n},{v}_{n}\right)\right)\\ \phantom{\rule{1em}{0ex}}\le \alpha max\left\{d\left(g\left(x\right),g\left({u}_{n}\right)\right),d\left(g\left(y\right),g\left({v}_{n}\right)\right),d\left(F\left(x,y\right),g\left(x\right)\right),d\left(F\left({u}_{n},{v}_{n}\right),g\left({u}_{n}\right)\right)\right\}\\ \phantom{\rule{1em}{0ex}}=\alpha max\left\{d\left(g\left(x\right),g\left({u}_{n}\right)\right),d\left(g\left(y\right),g\left({v}_{n}\right)\right),d\left(g\left({u}_{n+1}\right),g\left({u}_{n}\right)\right)\right\}\\ \phantom{\rule{1em}{0ex}}\le \alpha max\left\{d\left(g\left(x\right),g\left({u}_{n}\right)\right),d\left(g\left(y\right),g\left({v}_{n}\right)\right),d\left(g\left({u}_{n+1}\right),g\left(x\right)\right)+d\left(g\left(x\right),g\left({u}_{n}\right)\right)\right\}\\ \phantom{\rule{1em}{0ex}}\le \alpha max\left\{d\left(g\left(x\right),g\left({u}_{n}\right)\right),d\left(g\left(y\right),g\left({v}_{n}\right)\right),2d\left(g\left({u}_{n+1}\right),g\left(x\right)\right),2d\left(g\left(x\right),g\left({u}_{n}\right)\right)\right\}\\ \phantom{\rule{1em}{0ex}}=\alpha max\left\{2d\left(g\left(x\right),g\left({u}_{n}\right)\right),d\left(g\left(y\right),g\left({v}_{n}\right)\right),2d\left(g\left({u}_{n+1}\right),g\left(x\right)\right)\right\}.\end{array}$

If

$max\left\{2d\left(g\left(x\right),g\left({u}_{n}\right)\right),d\left(g\left(y\right),g\left({v}_{n}\right)\right),2d\left(g\left({u}_{n+1}\right),g\left(x\right)\right)\right\}=2d\left(g\left({u}_{n+1}\right),g\left(x\right)\right)$

then $d\left(g\left({u}_{n+1}\right),g\left(x\right)\right)\le 2\alpha d\left(g\left({u}_{n+1}\right),g\left(x\right)\right)$. Since $2\alpha <1$, we have $d\left(g\left({u}_{n+1}\right),g\left(x\right)\right)=0$. Therefore $d\left(g\left(x\right),g\left({u}_{n}\right)\right)=0$ and $d\left(g\left(y\right),g\left({v}_{n}\right)\right)=0$ and hence $\left(g\left(x\right),g\left(y\right)\right)=\left(g\left({u}_{n}\right),g\left({v}_{n}\right)\right)$, a contradiction. Thus

$\begin{array}{rcl}d\left(g\left(x\right),g\left({u}_{n+1}\right)\right)& \le & \alpha max\left\{2d\left(g\left(x\right),g\left({u}_{n}\right)\right),d\left(g\left(y\right),g\left({v}_{n}\right)\right)\right\}\\ \le & 2\alpha max\left\{d\left(g\left(x\right),g\left({u}_{n}\right)\right),d\left(g\left(y\right),g\left({v}_{n}\right)\right)\right\}.\end{array}$
(19)

Similarly, we may show that

$d\left(g\left({v}_{n+1}\right),g\left(y\right)\right)\le 2\alpha max\left\{d\left(g\left(x\right),g\left({u}_{n}\right)\right),d\left(g\left(y\right),g\left({v}_{n}\right)\right)\right\}.$
(20)

From (19) and (20), we have

$\begin{array}{r}max\left\{d\left(g\left(x\right),g\left({u}_{n+1}\right)\right),d\left(g\left({v}_{n+1}\right),g\left(y\right)\right)\right\}\\ \phantom{\rule{1em}{0ex}}\le 2\alpha max\left\{d\left(g\left({v}_{n}\right),g\left(y\right)\right),d\left(g\left({u}_{n}\right),g\left(x\right)\right),d\left(g\left({u}_{n+1}\right)\right\}.\end{array}$
(21)

By repeating (21) n-times, we have

$\begin{array}{r}max\left\{d\left(g\left(x\right),g\left({u}_{n+1}\right)\right),d\left(g\left({v}_{n+1}\right),g\left(y\right)\right)\right\}\\ \phantom{\rule{1em}{0ex}}\le 2\alpha max\left\{d\left(g\left({v}_{n}\right),g\left(y\right)\right),d\left(g\left({u}_{n}\right),g\left(x\right)\right)\right\}\\ \phantom{\rule{1em}{0ex}}⋮\\ \phantom{\rule{1em}{0ex}}\le {\left(2\alpha \right)}^{n+1}max\left\{d\left(g\left(x\right),g\left({u}_{0}\right)\right),d\left(g\left({v}_{0}\right),g\left(y\right)\right)\right\}.\end{array}$

Letting $n\to +\mathrm{\infty }$ in the above inequalities, we get that

$\underset{n\to }{lim}max\left\{d\left(g\left(x\right),g\left({u}_{n+1}\right)\right),d\left(g\left({v}_{n+1}\right),g\left(y\right)\right)\right\}=0.$

Hence

$\underset{n\to \mathrm{\infty }}{lim}d\left(g\left(x\right),g\left({u}_{n+1}\right)\right)=0$
(22)

and

$\underset{n\to \mathrm{\infty }}{lim}d\left(g\left(y\right),g\left({v}_{n+1}\right)\right)=0.$
(23)

Similarly, we may show that

$\underset{n\to \mathrm{\infty }}{lim}d\left(g\left(x\right),g\left({u}_{n+1}\right)\right)=0$
(24)

and

$\underset{n\to \mathrm{\infty }}{lim}d\left(g\left(y\right),g\left({v}_{n+1}\right)\right)=0.$
(25)

By the triangle inequality, (22), (23), (24) and (25),

we have $g\left(x\right)=g\left({x}^{\ast }\right)$ and $g\left(y\right)=g\left({y}^{\ast }\right)$. Thus we have (12). This implies that $\left(g\left(x\right),g\left(y\right)\right)=\left(g\left({x}^{\ast }\right),g\left({y}^{\ast }\right)\right)$.

Since $g\left(x\right)=F\left(x,y\right)$ and $g\left(y\right)=F\left(y,x\right)$, by commutativity of F and g, we have

$g\left(g\left(x\right)\right)=g\left(F\left(x,y\right)\right)=F\left(g\left(x\right),g\left(y\right)\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g\left(g\left(y\right)\right)=g\left(F\left(y,x\right)\right)=F\left(g\left(y\right),g\left(x\right)\right).$
(26)

Denote $g\left(x\right)=z$, $g\left(y\right)=w$. Then from (26)

$g\left(z\right)=F\left(z,w\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g\left(w\right)=F\left(w,z\right).$
(27)

Thus $\left(z,w\right)$ is a coupled coincidence point. Then from (26) with ${x}^{\ast }=z$ and ${y}^{\ast }=w$ it follows $g\left(z\right)=g\left(x\right)$ and $g\left(w\right)=g\left(y\right)$, that is,

$g\left(z\right)=z\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g\left(w\right)=w.$
(28)

From (27) and (28),

$z=g\left(z\right)=F\left(z,w\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}w=g\left(w\right)=F\left(w,z\right).$

Therefore, $\left(z,w\right)$ is a coupled common fixed point of F and g. To prove the uniqueness, assume that $\left(p,q\right)$ is another coupled common fixed point. Then by (26) we have $p=g\left(p\right)=g\left(z\right)=z$ and $q=g\left(q\right)=g\left(w\right)=w$. □

Corollary 2.2 In addition to the hypotheses of Corollary  2.1, suppose that $L=0$, $\alpha <\frac{1}{2}$, and for every $\left(x,y\right),\left({y}^{\ast },{x}^{\ast }\right)\in X×X$, there exists $\left(u,v\right)\in X×X$ such that $u⪯F\left(u,v\right)$, $v⪰F\left(v,u\right)$, and $\left(F\left(u,v\right),F\left(v,u\right)\right)$ is comparable to $\left(F\left(x,y\right),F\left(y,x\right)\right)$ and $\left(F\left({x}^{\ast },{y}^{\ast }\right),F\left({y}^{\ast },{x}^{\ast }\right)\right)$. Then F has a unique coupled fixed point, that is, there exist a unique $\left(x,y\right)\in X×X$ such that

$x=F\left(x,y\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}y=F\left(y,x\right).$

Proof Follows from Theorem 2.3 by taking $g=I$, the identity mapping. □

Theorem 2.4 In addition to the hypotheses of Theorem  2.1, if $g{x}_{0}$ and $g{y}_{0}$ are comparable and $L=0$, then F and g have a coupled coincidence point $\left(x,y\right)$ such that $gx=F\left(x,y\right)=F\left(y,x\right)=gy$.

Proof Follow the proof of Theorem 2.1 step by step until constructing two sequences $\left\{{x}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ in X such that $g{x}_{n}\to gx$ and $g{y}_{n}\to gy$, where $\left(x,y\right)$ is a coincidence point of F and g. Suppose $g{x}_{0}⪯g{y}_{0}$, then it is an easy matter to show that

Thus, by (5) we have

$\begin{array}{r}d\left(g{x}_{n},g{y}_{n}\right)\\ \phantom{\rule{1em}{0ex}}=d\left(F\left({x}_{n-1},{y}_{n-1}\right),F\left({y}_{n-1},{x}_{n-1}\right)\right)\\ \phantom{\rule{1em}{0ex}}\le \alpha max\left\{d\left(g\left({x}_{n-1}\right),g\left({y}_{n-1}\right),d\left(F\left({x}_{n-1},{y}_{n-1}\right),g\left({x}_{n-1}\right)\right),d\left(F\left({y}_{n-1},{x}_{n-1}\right),g\left({y}_{n-1}\right)\right)\right\}\\ \phantom{\rule{1em}{0ex}}=\alpha max\left\{d\left(g\left({x}_{n-1}\right),g\left({y}_{n-1}\right)\right),d\left(g\left({x}_{n}\right),g\left({x}_{n-1}\right)\right),d\left(g\left({y}_{n}\right),g\left({y}_{n-1}\right)\right)\right\}.\end{array}$

On taking the limit as $n\to +\mathrm{\infty }$, we get $d\left(gx,gy\right)=0$. Hence

$F\left(x,y\right)=gx=gy=F\left(y,x\right).$

A similar argument can be used if $g{y}_{0}⪯g{x}_{0}$. □

Corollary 2.3 In addition to the hypotheses of Corollary  2.1, if ${x}_{0}$ and ${y}_{0}$ are comparable and $L=0$, then F has a coupled fixed point of the form $\left(x,x\right)$.

Proof Follows from Theorem 2.4 by taking $g=I$, the identity mapping. □

Now, we introduce the following example to support our results.

Example 2.1 Let $X=\left[0,1\right]$. Then $\left(X,\le \right)$ is a partially ordered set with the natural ordering of real numbers. Define the metric d on X by

Define $g:X\to X$ by $g\left(x\right)={x}^{2}$ and $F:X×X\to X$ by

$F\left(x,y\right)=\left\{\begin{array}{ll}\frac{3\left({x}^{2}-{y}^{2}\right)}{4},& x>y;\\ 0,& x\le y.\end{array}$

Then

1. (1)

$g\left(X\right)$ is a complete subset of X.

2. (2)

$F\left(X×X\right)\subseteq g\left(X\right)$.

3. (3)

X satisfies (i) and (ii) of Theorem 2.1.

4. (4)

F has the mixed g-monotone property.

5. (5)

For any $L\in \left[0,+\mathrm{\infty }\right)$, F and g satisfy that

$\begin{array}{r}d\left(F\left(x,y\right),F\left(u,v\right)\right)\\ \phantom{\rule{1em}{0ex}}\le \frac{3}{4}max\left\{d\left(g\left(x\right),g\left(u\right)\right),d\left(g\left(y\right),g\left(v\right)\right),d\left(F\left(x,y\right),g\left(x\right)\right),d\left(F\left(u,v\right),g\left(u\right)\right)\right\}\\ \phantom{\rule{2em}{0ex}}+Lmin\left\{d\left(F\left(x,y\right),g\left(u\right)\right),d\left(F\left(u,v\right),g\left(x\right)\right)\right\}\end{array}$

for all $g\left(x\right)\le g\left(u\right)$ and $g\left(y\right)\ge g\left(v\right)$ holds for all $x,y,u,v\in X$ with $g\left(x\right)\le g\left(u\right)$ and $g\left(y\right)\ge g\left(v\right)$.

Thus, by Theorem 2.1, F has a coupled fixed point. Moreover, $\left(0,0\right)$ is a coupled coincidence point of F.

Proof The proof of (1)-(4) is clear. We divide the proof of (5) into the following cases.

Case 1: If $g\left(x\right)\le g\left(y\right)$ and $g\left(u\right)\le g\left(v\right)$, then $x\le y$ and $u\le v$. Hence

$\begin{array}{rl}d\left(F\left(x,y\right),F\left(u,v\right)\right)=& d\left(0,0\right)=0\\ \le & \frac{3}{4}max\left\{d\left(g\left(x\right),g\left(u\right)\right),d\left(g\left(y\right),g\left(v\right)\right),d\left(F\left(x,y\right),g\left(x\right)\right),d\left(F\left(u,v\right),g\left(u\right)\right)\right\}\\ +Lmin\left\{d\left(F\left(x,y\right),g\left(u\right)\right),d\left(F\left(u,v\right),g\left(x\right)\right)\right\}.\end{array}$

Case 2: If $g\left(x\right)\le g\left(y\right)$ and $g\left(u\right)>g\left(v\right)$, then $x\le y$ and $u>v$. Hence

$\begin{array}{rcl}d\left(F\left(x,y\right),F\left(u,v\right)\right)& =& d\left(0,\frac{3\left({u}^{2}-{v}^{2}\right)}{4}\right)\\ =& \frac{3}{4}\left({u}^{2}-{v}^{2}\right)\\ \le & \frac{3}{4}{u}^{2}\\ =& \frac{3}{4}max\left\{\frac{3}{4}\left({u}^{2}-{v}^{2}\right),{u}^{2}\right\}\\ =& \frac{3}{4}max\left\{F\left(u,v\right),g\left(u\right)\right\}\\ =& \frac{3}{4}d\left(F\left(u,v\right),g\left(u\right)\right)\\ \le & \frac{3}{4}max\left\{d\left(g\left(x\right),g\left(u\right)\right),d\left(g\left(y\right),g\left(v\right)\right),d\left(F\left(x,y\right),g\left(x\right)\right),d\left(F\left(u,v\right),g\left(u\right)\right)\right\}\\ +Lmin\left\{d\left(F\left(x,y\right),g\left(u\right)\right),d\left(F\left(u,v\right),g\left(x\right)\right)\right\}.\end{array}$

Case 3: If $g\left(x\right)>g\left(y\right)$ and $g\left(u\right)\le g\left(v\right)$, then $x>y$ and $u\le v$. Hence $v\le y. Therefore $v, which is impossible.

Case 4: If $g\left(x\right)>g\left(y\right)$ and $g\left(u\right)>g\left(v\right)$, then $x>y$ and $u>v$. Thus $v\le y.

Subcase I: $x=u$ and $y=v$. Here, we have

$\begin{array}{rl}d\left(F\left(x,y\right),F\left(u,v\right)\right)=& d\left(0,0\right)=0\\ \le & \frac{3}{4}max\left\{d\left(g\left(x\right),g\left(u\right)\right),d\left(g\left(y\right),g\left(v\right)\right),d\left(F\left(x,y\right),g\left(x\right)\right),d\left(F\left(u,v\right),g\left(u\right)\right)\right\}\\ +Lmin\left\{d\left(F\left(x,y\right),g\left(u\right)\right),d\left(F\left(u,v\right),g\left(x\right)\right)\right\}.\end{array}$

Subcase II: $x\ne u$ or $y\ne v$. Here, we have ${u}^{2}-{v}^{2}>{x}^{2}-{y}^{2}$. Therefore

$\begin{array}{rcl}d\left(F\left(x,y\right),F\left(u,v\right)\right)& =& d\left(\frac{3\left({x}^{2}-{y}^{2}\right)}{4},\frac{3\left({u}^{2}-{v}^{2}\right)}{4}\right)\\ =& \frac{3}{4}\left({u}^{2}-{v}^{2}\right)\\ \le & \frac{3}{4}{u}^{2}\\ =& \frac{3}{4}max\left\{\frac{3}{4}\left({u}^{2}-{v}^{2}\right),{u}^{2}\right\}\\ =& \frac{3}{4}max\left\{F\left(u,v\right),g\left(u\right)\right\}\\ =& \frac{3}{4}d\left(F\left(u,v\right),g\left(u\right)\right)\\ \le & \frac{3}{4}max\left\{d\left(g\left(x\right),g\left(u\right)\right),d\left(g\left(y\right),g\left(v\right)\right),d\left(F\left(x,y\right),g\left(x\right)\right),d\left(F\left(u,v\right),g\left(u\right)\right)\right\}\\ +Lmin\left\{d\left(F\left(x,y\right),g\left(u\right)\right),d\left(F\left(u,v\right),g\left(x\right)\right)\right\}.\end{array}$

□

Note that the mappings F and g satisfy all the hypotheses of Theorem 2.1 for $\alpha =\frac{3}{4}$ and any $L\ge 0$. Thus F and g have a coupled coincidence point. Here $\left(0,0\right)$ is a coupled coincidence point of F and g.

Remarks

1. (1)

Theorem 1.1 is a special case of Corollary 2.1.

2. (2)

Theorem 1.2 is a special case of Corollary 2.1.

3. (3)

Theorem 1.3 is a special case of Theorem 2.1.

4. (4)

Theorem 1.4 is a special case of Theorem 2.2.

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## Acknowledgements

The authors would like to thank the editor and the referees for helpful comments.

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Correspondence to Choonkil Park.

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Shatanawi, W., Saadati, R. & Park, C. Almost contractive coupled mapping in ordered complete metric spaces. J Inequal Appl 2013, 565 (2013). https://doi.org/10.1186/1029-242X-2013-565 