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# Almost contractive coupled mapping in ordered complete metric spaces

Journal of Inequalities and Applications20132013:565

https://doi.org/10.1186/1029-242X-2013-565

• Received: 11 July 2013
• Accepted: 29 October 2013
• Published:

## 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.

## Keywords

• coupled fixed point
• partially ordered set
• mixed monotone property

## 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.

## Declarations

### Acknowledgements

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

## Authors’ Affiliations

(1)
Department of Mathematics, Hashemite University, P.O. Box 150459, Zarqa, 13115, Jordan
(2)
Department of Mathematics, Iran University of Science and Technology, Tehran, Iran
(3)
Department of Mathematics, Research Institute for Natural Sciences, Hanyang University, Seoul, 133-791, Korea

## References 