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# Coupled coincidence points for monotone operators in partially ordered metric spaces

Fixed Point Theory and Applications20112011:44

https://doi.org/10.1186/1687-1812-2011-44

• Received: 18 March 2011
• Accepted: 30 August 2011
• Published:

## Abstract

Using the notion of compatible mappings in the setting of a partially ordered metric space, we prove the existence and uniqueness of coupled coincidence points involving a (ϕ, ψ)-contractive condition for a mappings having the mixed g-monotone property. We illustrate our results with the help of an example.

## Keywords

• coupled coincidence point
• partially ordered metric space
• mixed g-monotone property

## 1 Introduction

The Banach contraction principle is the most celebrated fixed point theorem. Afterward many authors obtained many important extensions of this principle (cf. ). Recently Bhaskar and Lakshmikantham , Nieto and Lopez [12, 13], Ran and Reurings  and Agarwal et al.  presented some new results for contractions in partially ordered metric spaces. Bhaskar and Lakshmikantham  noted that their theorem can be used to investigate a large class of problems and have discussed the existence and uniqueness of solution for a periodic boundary value problem.

Recently, Luong and Thuan  presented some coupled fixed point theorems for a mixed monotone mapping in a partially ordered metric space which are generalizations of the results of Bhaskar and Lakshmikantham . In this paper, we establish the existence and uniqueness of coupled coincidence point involving a (ϕ,ψ)-contractive condition for mappings having the mixed g-monotone property. We also illustrate our results with the help of an example.

## 2 Preliminaries

A partial order is a binary relation over a set X which is reflexive, antisymmetric, and transitive. Now, let us recall the definition of the monotonic function f : XX in the partially order set (X, ). We say that f is non-decreasing if for x, y X, x y, we have fx fy. Similarly, we say that f is non-increasing if for x, y X, x y, we have fx fy. Any one could read on  for more details on fixed point theory.

Definition 2.1  (Mixed g-Monotone Property)

Let (X, ) be a partially ordered set and F : X × XX. We say that the mapping F has 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 X,
${x}_{1},{x}_{2}\in X,g{x}_{1}\preccurlyeq g{x}_{2}⇒F\left({x}_{1},y\right)\preccurlyeq F\left({x}_{2},y\right)$
(1)
and
${y}_{1},{y}_{2}\in X,g{y}_{1}\preccurlyeq g{y}_{2}⇒F\left(x,{y}_{1}\right)\succcurlyeq F\left(x,{y}_{2}\right).$
(2)

Definition 2.2  (Coupled Coincidence Point)

Let (x, y) X × X, F : X × XX and g : XX. We say that (x, y) is a coupled coincidence point of F and g if F(x, y) = gx and F(y, x) = gy for x, y X.

Definition 2.3  Let X be a non-empty set and let F : X × XX and g : XX. We say F and g are commutative if, for all x, y X,
$g\left(F\left(x,y\right)\right)=F\left(g\left(x\right),g\left(y\right)\right).$
Definition 2.4  The mapping F and g where F : X × XX and g : XX, are said to be compatible if
$\underset{n\to \infty }{lim}d\left(g\left(F\left({x}_{n},{y}_{n}\right)\right),F\left(g{x}_{n},g{y}_{n}\right)\right)=0$
and
$\underset{n\to \infty }{lim}d\left(g\left(F\left({y}_{n},{x}_{n}\right)\right),F\left(g{y}_{n},g{x}_{n}\right)\right)=0,$

whenever {x n } and {y n } are sequences in X, such that lim n →∞ F (x n , y n ) = lim n →∞ gx n = x and lim n →∞ F (y n , x n ) = lim n →∞ gy n = y, for all x, y X are satisfied.

## 3 Existence of coupled coincidence points

As in , let ϕ denote all functions ϕ : [0, ∞) → [0, ∞) which satisfy
1. 1.

ϕ is continuous and non-decreasing,

2. 2.

ϕ (t) = 0 if and only if t = 0,

3. 3.

ϕ (t + s) ≤ ϕ (t) + ϕ (s), t, s [0, ∞)

and let ψ denote all the functions ψ : [0, ∞) → (0, ∞) which satisfy lim t r ψ (t) > 0 for all r > 0 and $\underset{t\to {0}^{+}}{lim}\psi \left(t\right)=0$.

For example , functions ϕ1(t) = kt where k > 0, ${\varphi }_{\mathsf{\text{2}}}\left(t\right)=\frac{t}{t+\mathsf{\text{1}}}$, ϕ3(t) = ln(t + 1), and ϕ4(t) = min{t, 1} are in Φ; ψ1(t) = kt where k > 0, ${\psi }_{\mathsf{\text{2}}}\left(t\right)=\frac{\mathsf{\text{ln}}\left(\mathsf{\text{2}}t+\mathsf{\text{1}}\right)}{2}$, and
${\psi }_{3}\left(t\right)=\left\{\begin{array}{cc}\hfill 1,\hfill & \hfill t=0\hfill \\ \hfill \frac{t}{t+1},\hfill & \hfill 01\hfill \end{array}\right\$

are in Ψ,

Now, let us start proving our main results.

Theorem 3.1 Let (X, ) be a partially ordered set and suppose there is a metric d on X such that (X, d) is a complete metric space. Let F : X × XX be a mapping having the mixed g-monotone property on X such that there exist two elements x0, y0 X with
$g{x}_{0}\preccurlyeq F\left({x}_{0},{y}_{0}\right)\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}g{y}_{0}\succcurlyeq F\left({y}_{0},{x}_{0}\right).$
Suppose there exist ϕ Φ and ψ Ψ such that
$\varphi \left(d\left(F\left(x,y\right),F\left(u,v\right)\right)\right)\le \frac{1}{2}\varphi \left(d\left(gx,gu\right)+d\left(gy,gv\right)\right)-\psi \left(\frac{d\left(gx,gu\right)+d\left(gy,gv\right)\right)}{2}\right)$
(3)

for all x, y, u, v X with gx gu and gy gv. Suppose F(X × X) g(X), g is continuous and compatible with F and also suppose either

(a) F is continuous or

(b) X has the following property:

(i) if a non-decreasing sequence {x n } → x, then x n x, for all n,

(ii) if a non-increasing sequence {y n } → y, then y y n , for all n.

Then there exists x, y X such that
$gx=F\left(x,y\right)\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}gy=F\left(y,x\right),$

i.e., F and g have a coupled coincidence point in X.

Proof. Let x0, y0 X be such that gx0 F (x0, y0) and gy0 F (y0, x0).

Using F(X × X) g(X), we construct sequences {x n } and {y n } in X as
(4)
We are going to prove that
(5)
and
(6)

To prove these, we are going to use the mathematical induction.

Let n = 0. Since gx0 F(x0, y0) and gy0 F(y0, x0) and as gx1 = F(x0, y0) and gy1 = F (y0, x0), we have gx0 gx1 and gy0 gy1. Thus (5) and (6) hold for n = 0.

Suppose now that (5) and (6) hold for some fixed n ≥ 0, Then, since gx n gx n +1 and gy n gy n +1 , and by mixed g-monotone property of F, we have
$g{x}_{n+2}=F\left({x}_{n+1},{y}_{n+1}\right)\succcurlyeq F\left({x}_{n},{y}_{n+1}\right)\succcurlyeq F\left({x}_{n},{y}_{n}\right)=g{x}_{n+1}$
(7)
and
$g{y}_{n+2}=F\left({y}_{n+1},{x}_{n+1}\right)\preccurlyeq F\left({y}_{n},{x}_{n+1}\right)\preccurlyeq F\left({y}_{n},{x}_{n}\right)=g{y}_{n+1}.$
(8)
Using (7) and (8), we get
$g{x}_{n+1}\preccurlyeq g{x}_{n+2}\phantom{\rule{1em}{0ex}}\mathsf{\text{and}}\phantom{\rule{1em}{0ex}}g{y}_{n+1}\succcurlyeq g{y}_{n+2}.$
Hence by the mathematical induction we conclude that (5) and (6) hold for all n ≥ 0. Therefore,
$g{x}_{0}\preccurlyeq g{x}_{1}\preccurlyeq g{x}_{2}\preccurlyeq \cdots \preccurlyeq g{x}_{n}\preccurlyeq g{x}_{n+1}\preccurlyeq \cdots \phantom{\rule{0.3em}{0ex}}$
(9)
and
$g{y}_{0}\succcurlyeq g{y}_{1}\succcurlyeq g{y}_{2}\succcurlyeq \cdots \succcurlyeq g{y}_{n}\succcurlyeq g{y}_{n+1}\succcurlyeq \cdots \phantom{\rule{0.3em}{0ex}}.$
(10)
Since gx n gx n - 1 and gy n gy n - 1 , using (3) and (4), we have
$\begin{array}{lll}\hfill \varphi \left(d\left(g{x}_{n+1},g{x}_{n}\right)\right)& =\varphi \left(d\left(F\left({x}_{n},{y}_{n}\right),F\left({x}_{n-1},{y}_{n-1}\right)\right)\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ \le \frac{1}{2}\varphi \left(d\left(g{x}_{n},g{x}_{n-1}\right)+d\left(g{y}_{n},g{y}_{n-1}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ -\psi \left(\frac{d\left(g{x}_{n},g{x}_{n-1}\right)+d\left(g{y}_{n},g{y}_{n-1}\right)}{2}\right).\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ \hfill \text{(4)}\end{array}$
(11)
Similarly, since gy n - 1 gy n and gx n - 1 gx n , using (3) and (4), we also have
$\begin{array}{ll}\varphi \left(d\left(g{y}_{n},g{y}_{n+1}\right)\right)\hfill & =\varphi \left(d\left(F\left({y}_{n-1},{x}_{n-1}\right),F\left({y}_{n},{x}_{n}\right)\right)\right)\hfill \\ \le \frac{1}{2}\varphi \left(d\left(g{y}_{n-1},g{y}_{n}\right)+d\left(g{x}_{n-1},g{x}_{n}\right)\right)\hfill \\ -\psi \left(\frac{d\left(\left(g{y}_{n-1},g{y}_{n}\right)+d\left(g{x}_{n-1},g{x}_{n}\right)}{2}\right).\hfill \end{array}$
(12)
Using (11) and (12), we have
$\begin{array}{ll}\varphi \left(d\left(g{x}_{n+1},g{x}_{n}\right)\right)+\varphi \left(d\left(g{y}_{n+1},g{y}_{n}\right)\right)\hfill & \le \varphi \left(d\left(g{x}_{n},g{x}_{n-1}\right)+d\left(g{y}_{n},g{y}_{n-1}\right)\right)\hfill \\ -2\psi \left(\frac{d\left(g{x}_{n},g{x}_{n-1}\right)+d\left(g{y}_{n},g{y}_{n-1}\right)}{2}\right).\hfill \end{array}$
(13)
By property (iii) of ϕ, we have
$\varphi \left(d\left(g{x}_{n+1},g{x}_{n}\right)+d\left(g{y}_{n+1},g{y}_{n}\right)\right)\le \varphi \left(d\left(g{x}_{n+1},g{x}_{n}\right)\right)+\varphi \left(d\left(g{y}_{n+1},g{y}_{n}\right)\right).$
(14)
Using (13) and (14), we have
$\begin{array}{lll}\hfill \varphi \left(d\left(g{x}_{n+1},g{x}_{n}\right)+d\left(g{y}_{n+1},g{y}_{n}\right)\right)& \le \varphi \left(d\left(g{x}_{n},g{x}_{n-1}\right)+d\left(g{y}_{n},g{y}_{n-1}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ -2\psi \left(\frac{d\left(g{x}_{n},g{x}_{n-1}\right)+d\left(g{y}_{n},g{y}_{n-1}\right)}{2}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$
(15)
which implies, since ψ is a non-negative function,
$\varphi \left(d\left(g{x}_{n+1},g{x}_{n}\right)+d\left(g{y}_{n+1},g{y}_{n}\right)\right)\le \varphi \left(d\left(g{x}_{n},g{x}_{n-1}\right)+d\left(g{y}_{n},g{y}_{n-1}\right)\right).$
Using the fact that ϕ is non-decreasing, we get
$d\left(g{x}_{n+1},g{x}_{n}\right)+d\left(g{y}_{n+1},g{y}_{n}\right)\le d\left(g{x}_{n},g{x}_{n-1}\right)+d\left(g{y}_{n},g{y}_{n-1}\right).$
Set
${\delta }_{n}=d\left(g{x}_{n+1},g{x}_{n}\right)+d\left(g{y}_{n+1},g{y}_{n}\right).$
Now we would like to show that δ n → 0 as n → ∞. It is clear that the sequence {δ n } is decreasing. Therefore, there is some δ ≥ 0 such that
$\underset{n\to \infty }{lim}{\delta }_{n}=\underset{n\to \infty }{lim}\left[d\left(g{x}_{n+1},g{x}_{n}\right)+d\left(g{y}_{n+1},g{y}_{n}\right)\right]=\delta .$
(16)
We shall show that δ = 0. Suppose, to the contrary, that δ > 0. Then taking the limit as n → ∞ (equivalently, δ n δ) of both sides of (15) and remembering lim t r ψ(t) > 0 for all r > 0 and ϕ is continuous, we have
$\begin{array}{lll}\hfill \varphi \left(\delta \right)& =\underset{n\to \infty }{lim}\varphi \left({\delta }_{n}\right)\le \underset{n\to \infty }{lim}\left[\varphi \left({\delta }_{n-1}\right)-2\psi \left(\frac{{\delta }_{n-1}}{2}\right)\right]\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ =\varphi \left(\delta \right)-2\underset{{\delta }_{n-1}\to \delta }{lim}\psi \left(\frac{{\delta }_{n-1}}{2}\right)<\varphi \left(\delta \right)\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$
a contradiction. Thus δ = 0, that is
$\underset{n\to \infty }{lim}{\delta }_{n}=\underset{n\to \infty }{lim}\left[d\left(g{x}_{n+1},g{x}_{n}\right)+d\left(g{y}_{n+1},g{y}_{n}\right)\right]=0.$
(17)
Now, we will prove that {gx n } and {gy n } are Cauchy sequences. Suppose, to the contrary, that at least one of {gx n } or {gy n } is not Cauchy sequence. Then there exists an ε > 0 for which we can find subsequences {gx n ( k )}, {gx m ( k )} of {gx n } and {gy n ( k )}, {gy m ( k )} of {gy n } with n(k) > m(k) ≥ k such that
$d\left(g{x}_{n\left(k\right)},g{x}_{m\left(k\right)}\right)+d\left(g{y}_{n\left(k\right)},g{y}_{m\left(k\right)}\right)\ge \epsilon .$
(18)
Further, corresponding to m(k), we can choose n(k) in such a way that it is the smallest integer with n(k) > m(k) and satisfying (18). Then
$d\left(g{x}_{n\left(k\right)-1},g{x}_{m\left(k\right)}\right)+d\left(g{y}_{n\left(k\right)-1},g{y}_{m\left(k\right)}\right)<\epsilon .$
(19)
Using (18), (19) and the triangle inequality, we have
Letting k → ∞ and using (17), we get
$\underset{k\to \infty }{\mathrm{lim}}{r}_{k}=\underset{k\to \infty }{\mathrm{lim}}\left[d\left(g{x}_{n\left(k\right)},g{x}_{m\left(k\right)}\right)+d\left(g{y}_{n\left(k\right)},g{y}_{m\left(k\right)}\right]=\epsilon .$
(20)
By the triangle inequality
$\begin{array}{lll}\hfill {r}_{k}& =d\left(g{x}_{n\left(k\right)},g{x}_{m\left(k\right)}\right)+d\left(g{y}_{n\left(k\right)},g{y}_{m\left(k\right)}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ \le d\left(g{x}_{n\left(k\right)},g{x}_{n\left(k\right)+1}\right)+d\left(g{x}_{n\left(k\right)+1},g{x}_{m\left(k\right)+1}\right)+d\left(g{x}_{m\left(k\right)+1},g{x}_{m\left(k\right)}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ +d\left(g{y}_{n\left(k\right)},g{y}_{n\left(k\right)+1}\right)+d\left(g{y}_{n\left(k\right)+1},g{y}_{m\left(k\right)+1}\right)+d\left(g{y}_{m\left(k\right)+1},g{y}_{m\left(k\right)}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ ={\delta }_{n\left(k\right)}+{\delta }_{m\left(k\right)}+d\left(g{x}_{n\left(k\right)+1},g{x}_{m\left(k\right)+1}\right)+d\left(g{y}_{n\left(k\right)+1},g{y}_{m\left(k\right)+1}\right).\phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\\ \hfill \text{(5)}\end{array}$
Using the property of ϕ, we have
$\begin{array}{lll}\hfill \varphi \left({r}_{k}\right)& =\varphi \left({\delta }_{n\left(k\right)}+{\delta }_{m\left(k\right)}+d\left(g{x}_{n\left(k\right)+1},g{x}_{m\left(k\right)+1}\right)+d\left(g{y}_{n\left(k\right)+1},g{y}_{m\left(k\right)+1}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ \le \varphi \left({\delta }_{n\left(k\right)}+{\delta }_{m\left(k\right)}\right)+\varphi \left(d\left(g{x}_{n\left(k\right)+1},g{x}_{m\left(k\right)+1}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ +\varphi \left(d\left(g{y}_{n\left(k\right)+1},g{y}_{m\left(k\right)+1}\right)\right).\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ \hfill \text{(4)}\end{array}$
(21)

Since n(k) > m(k), hence gx n ( k ) gx m ( k ) and gy n ( k ) gy m ( k ). Using (3) and

(4), we get
$\begin{array}{lll}\hfill \varphi \left(d\left(g{x}_{n\left(k\right)+1},g{x}_{m\left(k\right)+1}\right)\right)& =\varphi \left(d\left(F\left({x}_{n\left(k\right)},{y}_{n\left(k\right)}\right),F\left({x}_{m\left(k\right)},{y}_{m\left(k\right)}\right)\right)\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ \le \frac{1}{2}\varphi \left(d\left(g{x}_{n\left(k\right)},g{x}_{m\left(k\right)}\right)+d\left(g{y}_{n\left(k\right)},g{y}_{m\left(k\right)}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ -\phantom{\rule{2.77695pt}{0ex}}\psi \left(\frac{d\left(g{x}_{n\left(k\right)},g{x}_{m\left(k\right)}\right)+d\left(g{y}_{n\left(k\right)},g{y}_{m\left(k\right)}\right)}{2}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ =\frac{1}{2}\varphi \left({r}_{k}\right)-\psi \left(\frac{{r}_{k}}{2}\right).\phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\\ \hfill \text{(5)}\end{array}$
(22)
By the same way, we also have
$\begin{array}{lll}\hfill \varphi \left(d\left(g{y}_{m\left(k\right)+1},g{y}_{n\left(k\right)+1}\right)\right)& =\varphi \left(d\left(F\left({y}_{m\left(k\right)},{x}_{m\left(k\right)}\right),F\left({y}_{n\left(k\right)},{x}_{n\left(k\right)}\right)\right)\right)\\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ \le \frac{1}{2}\varphi \left(d\left(g{y}_{m\left(k\right)},g{y}_{n\left(k\right)}\right)+d\left(g{x}_{m\left(k\right)},g{x}_{n\left(k\right)}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ -\psi \left(\frac{d\left(g{y}_{m\left(k\right)},g{y}_{n\left(k\right)}\right)+d\left(g{x}_{m\left(k\right)},g{x}_{n\left(k\right)}\right)}{2}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ =\frac{1}{2}\varphi \left({r}_{k}\right)-\psi \left(\frac{{r}_{k}}{2}\right).\phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\\ \hfill \text{(5)}\end{array}$
(23)
Inserting (22) and (23) in (21), we have
$\varphi \left({r}_{k}\right)\le \varphi \left({\delta }_{n\left(k\right)}+{\delta }_{m\left(k\right)}\right)+\varphi \left({r}_{k}\right)-2\psi \left(\frac{{r}_{k}}{2}\right).$
Letting k → ∞ and using (17) and (20), we get
$\varphi \left(\epsilon \right)\le \varphi \left(0\right)+\varphi \left(\epsilon \right)-2\underset{k\to \infty }{lim}\psi \left(\frac{{r}_{k}}{2}\right)=\varphi \left(\epsilon \right)-2\underset{{r}_{k}\to \epsilon }{lim}\psi \left(\frac{{r}_{k}}{2}\right)<\varphi \left(\epsilon \right)$

a contradiction. This shows that {gx n } and {gy n } are Cauchy sequences.

Since X is a complete metric space, there exist x, y X such that
$\underset{n\to \infty }{lim}F\left({x}_{n},{y}_{n}\right)=\underset{n\to \infty }{lim}g{x}_{n}=x\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}\underset{n\to \infty }{lim}F\left({y}_{n},{x}_{n}\right)=\underset{n\to \infty }{lim}g{y}_{n}=y.$
(24)
Since F and g are compatible mappings, we have
$\underset{n\to \infty }{lim}d\left(g\left(F\left({x}_{n},{y}_{n}\right)\right),F\left(g{x}_{n},g{y}_{n}\right)\right)=0$
(25)
and
$\underset{n\to \infty }{lim}d\left(g\left(F\left({y}_{n},{x}_{n}\right)\right),F\left(g{y}_{n},g{x}_{n}\right)\right)=0$
(26)
We now show that gx = F(x, y) and gy = F(y, x). Suppose that the assumption (a) holds. For all n ≥ 0, we have,
$d\left(gx,F\left(g{x}_{n},g{y}_{n}\right)\right)\le d\left(gx,g\left(F\left({x}_{n},{y}_{n}\right)\right)\right)+d\left(g\left(F\left({x}_{n},{y}_{n}\right)\right),F\left(g{x}_{n},g{y}_{n}\right).$

Taking the limit as n → ∞, using (4), (24), (25) and the fact that F and g

are continuous, we have d(gx, F(x, y)) = 0.

Similarly, using (4), (24), (26) and the fact that F and g are continuous, we have d(gy, F(y, x)) = 0.

Combining the above two results we get
$gx=F\left(x,y\right)\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}gy=F\left(y,x\right).$
Finally, suppose that (b) holds. By (5), (6) and (24), we have {gx n } is a non-decreasing sequence, gx n x and {gy n } is a non-increasing sequence, gy n y as n → ∞. Hence, by assumption (b), we have for all n ≥ 0,
$g{x}_{n}\preccurlyeq x\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}g{y}_{n}\preccurlyeq y.$
(27)

Since F and g are compatible mappings and g is continuous, by (25) and (26)

we have
$\underset{n\to \infty }{lim}g\left(g{x}_{n}\right)=gx=\underset{n\to \infty }{lim}g\left(F\left({x}_{n},{y}_{n}\right)\right)=\underset{n\to \infty }{lim}F\left(g{x}_{n},g{y}_{n}\right)$
(28)
and,
$\underset{n\to \infty }{lim}g\left(g{y}_{n}\right)=gy=\underset{n\to \infty }{lim}g\left(F\left({y}_{n},{x}_{n}\right)\right)=\underset{n\to \infty }{lim}F\left(g{y}_{n},g{x}_{n}\right).$
(29)
Now we have
$d\left(gx,F\left(x,y\right)\right)\le d\left(gx,g\left(g{x}_{n+1}\right)\right)+d\left(g\left(g{x}_{n+1}\right),F\left(x,y\right)\right).$
Taking n → ∞ in the above inequality, using (4) and (21) we have,
$\begin{array}{ll}d\left(gx,F\left(x,y\right)\right)\hfill & \le \underset{n\to \infty }{\mathrm{lim}}d\left(gx,g\left(g{x}_{n+1}\right)\right)+\underset{n\to \infty }{\mathrm{lim}}d\left(g\left(F\left({x}_{n},{y}_{n}\right)\right),F\left(x,y\right)\right)\hfill \\ \le \underset{n\to \infty }{\mathrm{lim}}d\left(F\left(g{x}_{n},g{y}_{n}\right)\right),F\left(x,y\right)\right)\hfill \end{array}$
(30)
Using the property of ϕ, we get
$\varphi \left(d\left(gx,F\left(x,y\right)\right)\right)\le \underset{n\to \infty }{\mathrm{lim}}\varphi \left(d\left(F\left(g{x}_{n},g{y}_{n}\right)\right),F\left(x,y\right)\right)\right)$
Since the mapping g is monotone increasing, using (3), (27) and (30), we have for all n ≥ 0,
$\begin{array}{lll}\hfill \varphi \left(d\left(gx,F\left(x,y\right)\right)\right)& \le \underset{n\to \infty }{lim}\frac{1}{2}\varphi \left(d\left(gg{x}_{n},gx\right)+d\left(g{y}_{n},ggy\right)\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ -\underset{n\to \infty }{lim}\psi \left(\frac{d\left(gg{x}_{n},gx\right)+d\left(gg{y}_{n},gy\right)}{2}\right).\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$

Using the above inequality, using (24) and the property of ψ, we get ϕ(d(gx, F(x, y))) = 0, thus d(gx, F(x, y)) = 0. Hence gx = F(x, y).

Similarly, we can show that gy = F(y, x). Thus we proved that F and g have a coupled coincidence point.

Corollary 3.1  Let (X, ) be a partially ordered set and suppose there is a metric d on X such that (X, d) is a complete metric space. Let F : X × XX be a mapping having the mixed monotone property on X such that there exist two elements x0, y0 X with
${x}_{0}\preccurlyeq F\left({x}_{0},{y}_{0}\right)\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}{y}_{0}\succcurlyeq F\left({y}_{0},{x}_{0}\right).$
Suppose there exist ϕ Φ and ψ Ψ such that
$\varphi \left(d\left(F\left(x,y\right),F\left(u,v\right)\right)\right)\le \frac{1}{2}\varphi \left(d\left(x,u\right)+d\left(y,v\right)\right)-\psi \left(\frac{d\left(x,u\right)+d\left(y,v\right)\right)}{2}\right)$

for all x, y, u, v X with xu and yv. Suppose either

(a) F is continuous or

(b) X has the following property.

(i) if a non-decreasing sequence {x n } → x, then x n x, for all n,

(ii) if a non-increasing sequence {y n } → y, then y y n , for all n,

then there exist x, y X such that
$x=F\left(x,y\right)\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}y=F\left(y,x\right)$

that is, F has a coupled fixed point in X.

Corollary 3.2  Let (X, ) be a partially ordered set and suppose there is a metric d on X such that (X, d) is a complete metric space. Let F : X × XX be a mapping having the mixed monotone property on X such that there exist two elements x0, y0 X with
${x}_{0}\preccurlyeq F\left({x}_{0},{y}_{0}\right)\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}{y}_{0}\succcurlyeq F\left({y}_{0},{x}_{0}\right).$
Suppose there exists ψ Ψ such that
$d\left(F\left(x,y\right),F\left(u,v\right)\right)\le \frac{d\left(x,u\right)+d\left(y,v\right)}{2}-\psi \left(\frac{d\left(x,u\right)+d\left(y,v\right)}{2}\right)$

for all x, y, u, v X with xu and yv. Suppose either

(a) F is continuous or

(b) X has the following property:

(i) if a non-decreasing sequence {x n } → x, then x n x, for all n,

(ii) if a non-increasing sequence {y n } → y, then y y n , for all n,

then there exist x, y X such that
$x=F\left(x,y\right)\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}y=F\left(y,x\right)$

that is, F has a coupled fixed point in X.

Proof. Take ϕ(t) = t in Corollary 3.1, we get Corollary 3.2.

Corollary 3.3  eses of Corollary 3.1, suppose that for Let (X, ) be a partially ordered set and suppose there is a metric d on X such that (X, d) is a complete metric space. Let F : X × XX be a mapping having the mixed monotone property on X such that there exist two elements x0, y0 X with
${x}_{0}\preccurlyeq F\left({x}_{0},{y}_{0}\right)\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}{y}_{0}\succcurlyeq F\left({y}_{0},{x}_{0}\right).$
Suppose there exists a real number k [0, 1) such that
$d\left(F\left(x,y\right),F\left(u,v\right)\right)\le \frac{k}{2}\left[d\left(x,u\right)+d\left(y,v\right)\right]$

for all x, y, u, v X with xu and yv. Suppose either

(a) F is continuous or

(b) X has the following property.

(i) if a non-decreasing sequence {x n } → x, then x n x, for all n,

(ii) if a non-increasing sequence {y n } → y, then y y n , for all n,

then there exist x, y X such that
$x=F\left(x,y\right)\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}y=F\left(y,x\right)$

that is, F has a coupled fixed point in X.

Proof. Taking $\psi \left(t\right)\phantom{\rule{0.5em}{0ex}}=\frac{1-k}{2}t$ in Corollary 3.2.

## 4 Uniqueness of coupled coincidence point

In this section, we will prove the uniqueness of the coupled coincidence point. Note that if (X, ) is a partially ordered set, then we endow the product X × X with the following partial order relation, for all (x, y), (u, v) X × X,
$\left(x,y\right)\preccurlyeq \left(u,v\right)\phantom{\rule{1em}{0ex}}⇔\phantom{\rule{1em}{0ex}}x\preccurlyeq u,y\succcurlyeq v.$

Theorem 4.1 In addition to hypotheses of Theorem 3.1, suppose that for every (x, y), (z, t) in X × X, if there exists a (u, v) in X ×X that is comparable to (x, y) and (z, t), then F has a unique coupled coincidence point.

Proof. From Theorem 3.1, the set of coupled coincidence points of F and g is non-empty. Suppose (x, y) and (z, t) are coupled coincidence points of F and g, that is gx = F(x, y), gy = F(y, x), gz = F(z, t) and gt = F(t, z). We are going to show that gx = gz and gy = gt. By assumption, there exists (u, v) X × X that is comparable to (x, y) and (z, t). We define sequences {gu n }, {gv n } as follows
Since (u, v) is comparable with (x, y), we may assume that (x, y) (u, v) = (u0, v0). Using the mathematical induction, it is easy to prove that
(31)
Using (3) and (31), we have
$\begin{array}{lll}\hfill \phi \left(d\left(gx,g{u}_{n+1}\right)\right)& =\phi \left(d\left(F\left(x,y\right),F\left({u}_{n},{v}_{n}\right)\right)\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ <\frac{1}{2}\phi \left(d\left(x,{u}_{n}\right)+d\left(y,{v}_{n}\right)\right)-\psi \left(\frac{d\left(x,{u}_{n}\right)+d\left(y,{v}_{n}\right)}{2}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$
(32)
Similarly
$\begin{array}{lll}\hfill \phi \left(d\left(g{v}_{n+1},gy\right)\right)& =\phi \left(d\left(F\left({v}_{n},{u}_{n}\right),F\left(y,x\right)\right)\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ <\frac{1}{2}\phi \left(d\left({v}_{n},y\right)+d\left({u}_{n},x\right)\right)-\psi \left(\frac{d\left({v}_{n},y\right)+d\left({u}_{n},x\right)}{2}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$
(33)
Using (32), (33) and the property of φ, we have
$\begin{array}{lll}\hfill \phi \left(d\left(gx,g{u}_{n+1}\right)+d\left(gy,g{v}_{n+1}\right)\right)& \le \phi \left(d\left(gx,g{u}_{n+1}\right)\right)+\phi \left(d\left(gy,g{v}_{n+1}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ \le \phi \left(d\left(gx,g{u}_{n}\right)+d\left(gy,g{v}_{n}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ -2\psi \left(\frac{d\left(gx,g{u}_{n}\right)+d\left(gy,g{v}_{n}\right)}{2}\right).\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ \hfill \text{(4)}\end{array}$
(34)
which implies, using the property of ψ,
$\phi \left(d\left(gx,g{u}_{n+1}\right)+d\left(gy,g{v}_{n+1}\right)\right)\le \phi \left(d\left(gx,g{u}_{n}\right)+d\left(gy,g{v}_{n}\right)\right).$
Thus, using the property of ϕ,
$d\left(gx,g{u}_{n+1}\right)+d\left(gy,g{v}_{n+1}\right)\le d\left(gx,g{u}_{n}\right)+d\left(gy,g{v}_{n}\right).$
That is the sequence {d(gx, gu n )+ d(gy, gv n )} is decreasing. Therefore, there exists α ≥ 0 such that
$\underset{n\to \infty }{lim}\left[d\left(gx,g{u}_{n}\right)+d\left(gy,g{v}_{n}\right)\right]=\alpha .$
(35)
We will show that α = 0. Suppose, to the contrary, that α > 0. Taking the limit as n → ∞ in (34), we have, using the property of ψ,
$\phi \left(\alpha \right)\le \phi \left(\alpha \right)-2\underset{n\to \infty }{lim}\psi \left(\frac{d\left(gx,g{u}_{n}\right)+d\left(gy,g{v}_{n}\right)}{2}\right)<\phi \left(\alpha \right)$
a contradiction. Thus. α = 0, that is,
$\underset{n\to \infty }{lim}\left[d\left(gx,g{u}_{n}\right)+d\left(gy,g{v}_{n}\right)\right]=0.$
It implies
$\underset{n\to \infty }{lim}d\left(gx,g{u}_{n}\right)=\underset{n\to \infty }{lim}d\left(gy,g{v}_{n}\right)=0.$
(36)
Similarly, we show that
$\underset{n\to \infty }{lim}d\left(gz,g{u}_{n}\right)=\underset{n\to \infty }{lim}d\left(gt,g{v}_{n}\right)=0.$
(37)

Using (36) and (37) we have gx = gz and gy = gt.

Corollary 4.1  In addition to hypotheses of Corollary 3.1, suppose that for every (x, y), (z, t) in X × X, if there exists a (u, v) in X × X that is comparable to (x, y) and (z, t), then F has a unique coupled fixed point.

## 5 Example

Example 5.1 Let X = [0, 1]. Then (X, ≤) is a partially ordered set with the natural ordering of real numbers. Let
$d\left(x,y\right)=\left|x-y\right|\phantom{\rule{1em}{0ex}}for\phantom{\rule{1em}{0ex}}x,y\in \left[0,1\right].$

Then (X, d) is a complete metric space.

Let g : XX be defined as
$gx={x}^{2},\phantom{\rule{1em}{0ex}}for\phantom{\rule{2.77695pt}{0ex}}all\phantom{\rule{1em}{0ex}}x\in X,$
and let F : X × XX be defined as
$F\left(x,y\right)=\left\{\begin{array}{cc}\hfill \frac{{x}^{2}-{y}^{2}}{3},\hfill & \hfill if\phantom{\rule{2.77695pt}{0ex}}x\ge y,\hfill \\ \hfill 0,\hfill & \hfill if\phantom{\rule{2.77695pt}{0ex}}x

F obeys the mixed g-monotone property.

Let ϕ : [0, ∞) → [0, ∞) be defined as
$\varphi \left(t\right)=\frac{3}{4}t,\phantom{\rule{1em}{0ex}}for\phantom{\rule{1em}{0ex}}t\in \left[0,\infty \right).$
and let ψ : [0, ∞) → [0, ∞) be defined as
$\psi \left(t\right)=\frac{1}{4}t,\phantom{\rule{1em}{0ex}}for\phantom{\rule{1em}{0ex}}t\in \left[0,\infty \right).$
Let {x n } and {y n } be two sequences in X such that lim n →∞ F (x n , y n ) = a, lim n →∞ gx n = a, lim n →∞ F (y n , x n ) = b and lim n →∞ gy n = b Then obviously, a = 0 and b = 0. Now, for all n ≥ 0,
$\begin{array}{c}g{x}_{n}={x}_{n}^{2},g{y}_{n}={y}_{n}^{2},\\ F\left({x}_{n},{y}_{n}\right)=\left\{\begin{array}{cc}\hfill \frac{{x}_{n}^{2}-{y}_{n}^{2}}{3},\hfill & \hfill if\phantom{\rule{2.77695pt}{0ex}}{x}_{n}\ge {y}_{n},\hfill \\ \hfill 0,\hfill & \hfill if\phantom{\rule{2.77695pt}{0ex}}{x}_{n}<{y}_{n}.\hfill \end{array}\right\\end{array}$
and
$F\left({y}_{n},{x}_{n}\right)=\left\{\begin{array}{cc}\hfill \frac{{y}_{n}^{2}-{x}_{n}^{2}}{3},\hfill & \hfill if\phantom{\rule{2.77695pt}{0ex}}{y}_{n}\ge {x}_{n},\hfill \\ \hfill 0,\hfill & \hfill if\phantom{\rule{2.77695pt}{0ex}}{y}_{n}<{x}_{n}.\hfill \end{array}\right\$
Then it follows that,
$\underset{n\to \infty }{lim}d\left(g\left(F\left({x}_{n},{y}_{n}\right)\right),F\left(g{x}_{n},g{y}_{n}\right)\right)=0$
and
$\underset{n\to \infty }{lim}d\left(g\left(F\left({y}_{n},{x}_{n}\right)\right),F\left(g{y}_{n},g{x}_{n}\right)\right)=0,$
Hence, the mappings F and g are compatible in X. Also, x0 = 0 and y0 = c(> 0) are two points in X such that
$g{x}_{0}=g\left(0\right)=0=F\left(0,c\right)=F\left({x}_{0},{y}_{0}\right)$
and
$g{y}_{0}=g\left(c\right)={c}^{2}\ge \frac{{c}^{2}}{3}=F\left(c,0\right)=F\left({y}_{0},{x}_{0}\right).$

We next verify the contraction (3). We take x, y, u, v, X, such that gxgu and gygv, that is, x2u2 and y2v2.

We consider the following cases:

Case 1. xy, uv. Then,
$\begin{array}{ll}\varphi \left(d\left(F\left(x,y\right),F\left(u,v\right)\right)\right)\hfill & =\frac{3}{4}\left[d\left(F\left(x,y\right),F\left(u,v\right)\right]\hfill \\ =\frac{3}{4}\left[d\left(\frac{{x}^{2}-{y}^{2}}{3},\frac{{u}^{2}-{v}^{2}}{3}\right)\right]\hfill \\ =\frac{3}{4}|\frac{\left({x}^{2}-{y}^{2}\right)-\left({u}^{2}-{v}^{2}\right)}{3}|\hfill \\ =\frac{3}{4}\frac{|{x}^{2}-{u}^{2}|+|{y}^{2}-{v}^{2}|}{3}\hfill \\ =\frac{1}{2}\left(\frac{d\left(gx,gu\right)+d\left(gy,gv\right)}{2}\right)\hfill \\ =\frac{3}{4}\left(\frac{d\left(gx,gu\right)+d\left(gy,gv\right)}{2}\right)\hfill \\ -\frac{1}{4}\left(\frac{d\left(gx,gu\right)+d\left(gy,gv\right)}{2}\right)\hfill \\ =\frac{3}{8}\left(d\left(gx,gu\right)+d\left(gy,gv\right)\right)\hfill \\ -\frac{1}{4}\left(\frac{d\left(gx,gu\right)+d\left(gy,gv\right)}{2}\right)\hfill \\ =\frac{1}{2}\varphi \left(d\left(gx,gu\right)+d\left(gy,gv\right)\right)\hfill \\ -\psi \left(\frac{d\left(gx,gu\right)+d\left(gy,gv\right)}{2}\right)\hfill \end{array}$
Case 2. xy, u < v.Then
$\begin{array}{ll}\varphi \left(d\left(F\left(x,y\right),F\left(u,v\right)\right)\hfill & =\frac{3}{4}\left[d\left(F\left(x,y\right),F\left(u,v\right)\right]\hfill \\ =\frac{3}{4}\left[d\left(\frac{{x}^{2}-{y}^{2}}{3},0\right)\right]\hfill \\ =\frac{3}{4}\frac{|{x}^{2}-{y}^{2}|}{3}\hfill \\ =\frac{3}{4}\frac{|{v}^{2}+{x}^{2}-{y}^{2}-{u}^{2}|}{3}\hfill \\ =\frac{3}{4}\frac{|\left({v}^{2}-{y}^{2}\right)-\left({u}^{2}-{x}^{2}\right)|}{3}\hfill \\ \le \frac{3}{4}\frac{|{v}^{2}-{y}^{2}|+|{u}^{2}-{x}^{2}|}{3}\hfill \\ =\frac{3}{4}\left(\frac{|{u}^{2}-{x}^{2}|+|{y}^{2}-{v}^{2}|}{3}\right)\hfill \\ =\frac{1}{2}\left(\frac{|{u}^{2}-{x}^{2}|+|{y}^{2}-{v}^{2}|}{2}\right)\hfill \\ =\frac{1}{2}\left(\frac{d\left(gx,gu\right)+d\left(gy,gv\right)}{2}\right)\hfill \\ =\frac{3}{4}\left(\frac{d\left(gx,gu\right)+d\left(gy,gv\right)}{2}\right)\hfill \\ -\frac{1}{4}\left(\frac{d\left(gx,gu\right)+d\left(gy,gv\right)}{2}\right)\hfill \\ =\frac{3}{8}\left(d\left(gx,gu\right)+d\left(gy,gv\right)\right)\hfill \\ -\frac{1}{4}\left(\frac{d\left(gx,gu\right)+d\left(gy,gv\right)}{2}\hfill \\ =\frac{1}{2}\varphi \left(d\left(gx,gu\right)+d\left(gy,gv\right)\right)\hfill \\ -\psi \left(\frac{d\left(gx,gu\right)+d\left(gy,gv\right)}{2}\right)\hfill \end{array}$
Case 3. x < y and uv. Then
$\begin{array}{lll}\hfill \varphi \left(d\left(F\left(x,y\right),F\left(u,v\right)\right)\right)& =\frac{3}{4}\left[d\left(0,\frac{{u}^{2}-{v}^{2}}{3}\right)\right]\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ =\frac{3}{4}\frac{|{u}^{2}-{v}^{2}|}{3}\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ =\frac{3}{4}\frac{|{u}^{2}+{x}^{2}-{v}^{2}-{x}^{2}|}{3}\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ =\frac{3}{4}\frac{|\left({x}^{2}-{v}^{2}\right)+\left({u}^{2}-{x}^{2}\right)|}{3}\left(since\phantom{\rule{2.77695pt}{0ex}}y>x\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\\ \le \frac{3}{4}\frac{|{y}^{2}-{v}^{2}|+|{u}^{2}-{x}^{2}|}{3}\phantom{\rule{2em}{0ex}}& \hfill \text{(5)}\\ =\frac{1}{2}\left(\frac{|{u}^{2}-{x}^{2}|+|{y}^{2}-{v}^{2}|}{2}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(6)}\\ =\frac{1}{2}\left(\frac{d\left(gx,gu\right)+d\left(gy,gv\right)}{2}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(7)}\\ =\frac{3}{4}\left(\frac{d\left(gx,gu\right)+d\left(gy,gv\right)}{2}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(8)}\\ -\frac{1}{4}\left(\frac{d\left(gx,gu\right)+d\left(gy,gv\right)}{2}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(9)}\\ =\frac{3}{8}\left(d\left(gx,gu\right)+d\left(gy,gv\right)\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(10)}\\ -\frac{1}{4}\left(\frac{d\left(gx,gu\right)+d\left(gy,gv\right)}{2}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(11)}\\ =\frac{1}{2}\varphi \left(d\left(gx,gu\right)+d\left(gy,gv\right)\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(12)}\\ -\psi \left(\frac{d\left(gx,gu\right)+d\left(gy,gv\right)}{2}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(13)}\\ \hfill \text{(14)}\end{array}$
Case 4. x < y and u < v with x2u2 and y2v2. Then, F(x, y) = 0 and F(u, v) = 0, that is,
$\varphi \left(d\left(F\left(x,y\right),F\left(u,v\right)\right)\right)=\varphi \left(d\left(0,0\right)\right)=\varphi \left(0\right)=0.$

Therefore all conditions of Theorem 3.1 are satisfied. Thus the conclusion follows.

## Declarations

### Acknowledgements

The authors would like to thank the referees for the invaluable comments that improved this paper.

## Authors’ Affiliations

(1)
Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia

## References 