Some coupled coincidence point theorems for a mixed monotone operator in a complete metric space endowed with a partial order by using altering distance functions

Abstract

In this paper, we present some coupled coincidence point results for mixed g-monotone mappings in partially ordered complete metric spaces involving altering distance functions. Moreover, we present an example to illustrate our main result. Our results extend some results in the field.

MSC:47H09, 47H10, 49M05.

1 Introduction and preliminaries

The existence of a fixed point for contractive mappings in partially ordered metric spaces has attracted the attention of many mathematicians (cf. [1–11] and the references therein). In [3], Bhaskar and Lakshmikantham introduced the notion of a mixed monotone mapping and proved some coupled fixed point theorems for the mixed monotone mapping. Afterwards, Lakshmikantham and Ciric in [11] introduced the concept of a mixed g-monotone mapping and proved coupled coincidence point results for two mappings F and g, where F has the mixed g-monotone property and the functions F and g commute. It is well known that the concept of commuting has been weakened in various directions. One such notion which is weaker than commuting is the concept of compatibility introduced by Jungck [7]. In [5], Choudhury and Kundu defined the concept of compatibility of F and g. The purpose of this paper is to present some coupled coincidence point theorems for a mixed g-monotone mapping in the context of complete metric spaces endowed with a partial order by using altering distance functions which extend some results of [6]. We also present an example which illustrates the results.

Recall that if $(X,⪯)$ is a partially ordered set, then f is said to be non-decreasing if for $x,y\in X$, $x⪯y$, we have $fx⪯fy$. Similarly, f is said to be non-increasing if for $x,y\in X$, $x⪯y$, we have $fx⪰fy$. We also recall the used definitions in the present work.

Definition 1.1 [11] (Mixed g-monotone property)

Let $(X,⪯)$ be a partially ordered set, $g:X\to X$ and $F:X\times X\to X$. 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 $x,y\in X$,

$$x_1,x_2\in X,g{x}_1⪯g{x}_2\Rightarrow F(x_1,y)⪯F(x_2,y)$$

(1)

and

$$y_1,y_2\in X,g{y}_1⪯g{y}_2\Rightarrow F(x,y_1)⪰F(x,y_2).$$

(2)

Definition 1.2 [11] (Coupled coincidence fixed point)

Let $(x,y)\in X\times X$, $F:X\times X\to X$ and $g:X\to X$. 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 $x,y\in X$.

Definition 1.3 [11]

Let X be a non-empty set and let $F:X\times X\to X$ and $g:X\to X$. We say F and g are commutative if, for all $x,y\in X$,

$$g(F(x,y))=F(gx,gy).$$

Definition 1.4 [5]

The mappings F and g, where $F:X\times X\to X$ and $g:X\to X$, are said to be compatible if

$$\underset{n\to \mathrm{\infty }}{lim}d(g(F(x_n,y_n)),F(g{x}_n,g{y}_n))=0$$

and

$$\underset{n\to \mathrm{\infty }}{lim}d(g(F(y_n,x_n)),F(g{y}_n,g{x}_n))=0,$$

whenever $\{x_n\}$ and $\{y_n\}$ are sequences in X such that ${lim}_{n\to \mathrm{\infty }}F(x_n,y_n)={lim}_{n\to \mathrm{\infty }}g{x}_n=x$ and ${lim}_{n\to \mathrm{\infty }}F(y_n,x_n)={lim}_{n\to \mathrm{\infty }}g{y}_n=y$ for all $x,y\in X$.

Definition 1.5 (Altering distance function)

An altering distance function is a function $\psi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ satisfying

• 1. ψ is continuous and non-decreasing.

• 2. $\psi (t)=0$ if and only if $t=0$.

2 Existence of coupled coincidence points

Let $(X,⪯)$ be a partially ordered set and suppose that there exists a metric d in X such that $(X,d)$ is a complete metric space. Also, let φ and ϕ be altering distance functions. Now, we are in a position to state our main theorem.

Theorem 2.1 Let $F:X\times X\to X$ be a mapping having the mixed g-monotone property on X such that

$$\begin{array}{r}\phi \left(d(F(x,y),F(u,v))\right)\\ \le \phi (max(d(gx,gu),d(gy,gv)))-\varphi (max(d(gx,gu),d(gy,gv)))\end{array}$$

(3)

for all $x,y,u,v\in X$ with $gx⪰gu$ and $gy⪯gv$. Suppose that $F(X\times X)\subset g(X)$, g is continuous, monotone increasing and suppose also that F and g are compatible mappings. Moreover, suppose either

• (a) F is continuous, or

• (b) X has the following properties:

• (i) if a non-decreasing sequence $\{x_n\}\to x$, then $x_n⪯x$ for all n,

• (ii) if a non-increasing sequence $\{y_n\}\to y$, then $y⪯y_n$ for all n.

If there exist $x_0,y_0\in X$ with $g{x}_0⪯F(x_0,y_0)$ and $g{y}_0⪰F(y_0,x_0)$, then F and g have a coupled coincidence point.

Proof By using $F(X\times X)\subset g(X)$, we construct sequences $\{x_n\}$ and $\{y_n\}$ as follows:

$$g{x}_{n+1}=F(x_n,y_n)\text{and}g{y}_{n+1}=F(y_n,x_n)\text{for }n\ge 0.$$

(4)

We are going to divide the proof into several steps in order to make it easy to read.

Step 1. We will show that $g{x}_n⪯g{x}_{n+1}$ and $g{y}_n⪰g{y}_{n+1}$ for $n\ge 0$.

We use the mathematical induction to show that. From the assumption of the theorem, it follows that $g{x}_0⪯F(x_0,y_0)=g{x}_1$ and $g{y}_0⪰F(y_0,x_0)=g{y}_1$, so our claim is satisfied for $n=0$. Now, suppose that our claim holds for some fixed $n>0$. Since $g{x}_{n-1}⪯g{x}_n$, $g{y}_n⪯g{y}_{n-1}$ and F has the mixed g-monotone property, then we get

$$g{x}_{n+1}=F(x_n,y_n)⪰F(x_{n-1},y_n)⪰F(x_{n-1},y_{n-1})=g{x}_n$$

and

$$g{y}_{n+1}=F(y_n,x_n)⪯F(y_{n-1},x_n)⪯F(y_{n-1},x_{n-1})=g{y}_n.$$

Thus the claim holds for $n+1$ and by the mathematical induction our claim is proved.

Step 2. We will show that ${lim}_{n\to \mathrm{\infty }}d(g{x}_n,g{x}_{n+1})={lim}_{n\to \mathrm{\infty }}d(g{y}_n,g{y}_{n+1})=0$.

In fact, using (3), $g{x}_n⪰g{x}_{n-1}$ and $g{y}_n⪯g{y}_{n-1}$, we get

$$\begin{array}{rcl}\phi (d(g{x}_{n+1},g{x}_n))& =& \phi \left(d(F(x_n,y_n),F(x_{n-1},y_{n-1}))\right)\\ \le & \phi (max(d(g{x}_n,g{x}_{n-1}),d(g{y}_n,g{y}_{n-1})))\\ -\varphi (max(d(g{x}_n,g{x}_{n-1}),d(g{y}_n,g{y}_{n-1}))).\end{array}$$

(5)

Since ϕ is non-negative, we have

$$\phi (d(g{x}_{n+1},g{x}_n))\le \phi (max(d(g{x}_n,g{x}_{n-1}),d(g{y}_n,g{y}_{n-1}))),$$

and since φ is non-decreasing, we have

$$d(g{x}_{n+1},g{x}_n)\le max(d(g{x}_n,g{x}_{n-1}),d(g{y}_n,g{y}_{n-1})).$$

(6)

In the same way, we get the following:

$$\begin{array}{rcl}\phi (d(g{y}_{n+1},g{y}_n))& =& \phi \left(d(F(y_n,x_n),F(y_{n-1},x_{n-1}))\right)\\ =& \phi \left(d(F(y_{n-1},x_{n-1}),F(y_n,x_n))\right)\\ \le & \phi (max(d(g{y}_{n-1},g{y}_n),d(g{x}_{n-1},g{x}_n)))\\ -\varphi (max(d(g{y}_{n-1},g{y}_n),d(g{x}_{n-1},g{x}_n)))\\ \le & \phi (max(d(g{x}_n,g{x}_{n-1}),d(g{y}_n,g{y}_{n-1}))),\end{array}$$

(7)

and hence

$$d(g{y}_{n+1},g{y}_n)\le max(d(g{x}_n,g{x}_{n-1}),d(g{y}_n,g{y}_{n-1})).$$

(8)

Using (6) and (8), we have

$$max(d(g{x}_{n+1},g{x}_n),d(g{y}_{n+1},g{y}_n))\le max(d(g{x}_n,g{x}_{n-1}),d(g{y}_n,g{y}_{n-1})).$$

From the last inequality, we notice that the sequence $(max(d(g{x}_{n+1},g{x}_n),d(g{y}_{n+1},g{y}_n)))$ is non-negative decreasing. This implies that there exists $r\ge 0$ such that

$$\underset{n\to \mathrm{\infty }}{lim}max(d(g{x}_{n+1},g{x}_n),d(g{y}_{n+1},g{y}_n))=r.$$

(9)

It is easily seen that if $\phi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ is non-decreasing, we have $\phi (max(a,b))=max(\phi (a),\phi (b))$ for $a,b\in [0,\mathrm{\infty })$ for $a,b\in [0,\mathrm{\infty })$. Using this, (5) and (7), we obtain

$$\begin{array}{rcl}max(\phi (d(g{x}_{n+1},g{x}_n)),\phi (d(g{y}_{n+1},g{y}_n))& =& \phi (max(d(g{x}_{n+1},g{x}_n)d(g{y}_{n+1},g{y}_n)))\\ \le & \phi (max(d(g{x}_n,g{x}_{n-1}),d(g{y}_n,g{y}_{n-1})))\\ -\varphi (max(d(g{x}_n,g{x}_{n-1})d(g{y}_n,g{y}_{n-1}))).\end{array}$$

(10)

Letting $n\to \mathrm{\infty }$ in the last inequality and using (6), we have

$$\phi (r)\le \phi (r)-\varphi (r)\le \phi (r),$$

and this implies $\varphi (r)=0$. Thus, using the fact that ϕ is an altering distance function, we have $r=0$. Therefore,

$$\underset{n\to \mathrm{\infty }}{lim}max(d(g{x}_{n+1},g{x}_n),d(g{y}_{n+1},g{y}_n))=0.$$

(11)

Hence, ${lim}_{n\to \mathrm{\infty }}d(g{x}_n,g{x}_{n+1})={lim}_{n\to \mathrm{\infty }}d(g{y}_n,g{y}_{n+1})=0$ and this completes the proof of our claim.

Step 3. We will prove that $\{g{x}_n\}$ and $\{g{y}_n\}$ are Cauchy sequences.

Suppose that one of the sequences $\{g{x}_n\}$ or $\{g{y}_n\}$ is not a Cauchy sequence. This implies that ${lim}_{n,m\to \mathrm{\infty }}d(g{x}_n,g{x}_m)\nrightarrow 0$ or ${lim}_{n,m\to \mathrm{\infty }}d(g{y}_n,g{y}_m)\nrightarrow 0$, and hence

$$\underset{n,m\to \mathrm{\infty }}{lim}max(d(g{x}_n,g{x}_m),d(g{y}_n,g{y}_m))\nrightarrow 0.$$

This means that there exists $ϵ>0$, for which we can find subsequences $\{g{x}_{m(k)}\}$ and $\{g{x}_{n(k)}\}$ with $n(k)>m(k)>k$, such that

$$max(d(g{x}_{m(k)},g{x}_{n(k)}),d(g{y}_{m(k)},g{y}_{n(k)}))\ge ϵ.$$

(12)

Further, we can choose $n(k)$ corresponding to $m(k)$ in such a way that it is the smallest integer with $n(k)>m(k)$ and satisfying (12). Then

$$max(d(g{x}_{m(k)},g{x}_{n(k)-1}),d(g{y}_{m(k)},g{y}_{n(k)-1}))<ϵ.$$

(13)

Using (3), $g{x}_{n(k)-1}⪰g{x}_{m(k)-1}$ and $g{y}_{n(k)-1}⪯g{y}_{m(k)-1}$, we get

$$\begin{array}{rcl}\phi (d(g{x}_{n(k)},g{x}_{m(k)}))& =& \phi \left(d(F(x_{n(k)-1},y_{n(k)-1}),F(x_{m(k)-1},y_{m(k)-1}))\right)\\ \le & \phi (max(d(g{x}_{n(k)-1},g{x}_{m(k)-1}),d(g{y}_{n(k)-1},g{y}_{m(k)-1})))\\ -\varphi (max(d(g{x}_{n(k)-1},g{x}_{m(k)-1}),d(g{y}_{n(k)-1},g{y}_{m(k)-1}))),\end{array}$$

(14)

and also we get

$$\begin{array}{rcl}\phi (d(g{y}_{n(k)},g{y}_{m(k)}))& =& \phi \left(d(F(y_{n(k)-1},x_{n(k)-1}),F(y_{m(k)-1},x_{m(k)-1}))\right)\\ =& \phi \left(d(F(y_{m(k)-1},x_{m(k)-1}),F(y_{n(k)-1},x_{n(k)-1}))\right)\\ \le & \phi (max(d(g{x}_{n(k)-1},g{x}_{m(k)-1}),d(g{y}_{n(k)-1},g{y}_{m(k)-1})))\\ -\varphi (max(d(g{x}_{n(k)-1},g{x}_{m(k)-1}),d(g{y}_{n(k)-1},g{y}_{m(k)-1}))).\end{array}$$

(15)

Combining (14) and (15), we obtain

$$\begin{array}{r}max(\phi (d(g{x}_{n(k)},g{x}_{m(k)})),\phi (d(g{y}_{n(k)},g{y}_{m(k)})))\\ \le \phi (max(d(g{x}_{n(k)-1},g{x}_{m(k)-1}),d(g{y}_{n(k)-1},g{y}_{m(k)-1})))\\ -\varphi (max(d(g{x}_{n(k)-1},g{x}_{m(k)-1}),d(g{y}_{n(k)-1},g{y}_{m(k)-1}))).\end{array}$$

(16)

Using the triangular inequality and (13), we get

$$\begin{array}{rcl}d(g{x}_{n(k)},g{x}_{m(k)})& \le & d(g{x}_{n(k)},g{x}_{n(k)-1})+d(g{x}_{n(k)-1},g{x}_{m(k)})\\ <& d(g{x}_{n(k)},g{x}_{n(k)-1})+ϵ\end{array}$$

(17)

and

$$\begin{array}{rcl}d(g{y}_{n(k)},g{y}_{m(k)})& \le & d(g{y}_{n(k)},g{y}_{n(k)-1})+d(g{y}_{n(k)-1},g{y}_{m(k)})\\ <& d(g{y}_{n(k)},g{y}_{n(k)-1})+ϵ.\end{array}$$

(18)

Using (12), (17) and (18), we have

$$\begin{array}{rcl}ϵ& \le & max(d(g{x}_{n(k)},g{x}_{m(k)}),d(g{y}_{n(k)},g{y}_{m(k)}))\\ \le & max(d(g{x}_{n(k)},g{x}_{n(k)-1}),d(g{y}_{n(k)},g{y}_{n(k)-1}))+ϵ.\end{array}$$

Letting $k\to \mathrm{\infty }$ in the last inequality and using (11), we have

$$\underset{k\to \mathrm{\infty }}{lim}max(d(g{x}_{n(k)},g{x}_{m(k)}),d(g{y}_{n(k)},g{y}_{m(k)}))=ϵ.$$

(19)

Similarly, using the triangular inequality and (13), we have

$$\begin{array}{rcl}d(g{x}_{n(k)-1},g{x}_{m(k)-1})& \le & d(g{x}_{n(k)-1},g{x}_{m(k)})+d(g{x}_{m(k)},g{x}_{m(k)-1})\\ <& ϵ+d(g{x}_{m(k)},g{x}_{m(k)-1})\end{array}$$

(20)

and

$$\begin{array}{rcl}d(g{y}_{n(k)-1},g{y}_{m(k)-1})& \le & d(g{y}_{n(k)-1},g{y}_{m(k)})+d(g{y}_{m(k)},g{y}_{m(k)-1})\\ <& ϵ+d(g{y}_{m(k)},g{y}_{m(k)-1}).\end{array}$$

(21)

Combining (20) and (21), we obtain

$$\begin{array}{r}max(d(g{x}_{n(k)-1},g{x}_{m(k)-1}),d(g{y}_{n(k)-1},g{y}_{m(k)-1}))\\ <max(d(g{x}_{m(k)},g{x}_{m(k)-1}),d(g{y}_{m(k)},g{y}_{m(k)-1}))+ϵ.\end{array}$$

(22)

Using the triangular inequality, we have

$$\begin{array}{rcl}d(g{x}_{n(k)},g{x}_{m(k)})& \le & d(g{x}_{n(k)},g{x}_{n(k)-1})+d(g{x}_{n(k)-1},g{x}_{m(k)-1})\\ +d(g{x}_{m(k)-1},g{x}_{m(k)})\end{array}$$

and

$$\begin{array}{rcl}d(g{y}_{n(k)},g{y}_{m(k)})& \le & d(g{y}_{n(k)},g{y}_{n(k)-1})+d(g{y}_{n(k)-1},g{y}_{m(k)-1})\\ +d(g{y}_{m(k)-1},g{y}_{m(k)}).\end{array}$$

Using the two last inequalities and (12), we have

$$\begin{array}{rcl}ϵ& \le & max(d(g{x}_{n(k)},g{x}_{m(k)}),d(g{y}_{n(k)},g{y}_{m(k)}))\\ \le & max(d(g{x}_{n(k)},g{x}_{n(k)-1}),d(g{y}_{n(k)},g{y}_{n(k)-1}))\\ +max(d(g{x}_{n(k)-1},g{x}_{m(k)-1}),d(g{y}_{n(k)-1},g{y}_{m(k)-1}))\\ +max(d(g{x}_{m(l)-1},g{x}_{m(k)}),d(g{y}_{m(k)-1},g{y}_{m(k)})).\end{array}$$

(23)

Using (22) and (23), we get

$$\begin{array}{rcl}ϵ& \le & max(d(g{x}_{n(k)},g{x}_{n(k)-1}),d(g{y}_{n(k)},g{y}_{n(k)-1}))\\ \le & max(d(g{x}_{m(k)-1},g{x}_{m(k)}),d(g{y}_{m(k)-1},g{y}_{mk)}))\\ \le & max(d(g{x}_{n(k)-1},g{x}_{m(k)-1}),d(g{y}_{n(k)-1},g{y}_{m(k)-1}))\\ <& max(d(g{x}_{m(k)},g{x}_{m(k)-1}),d(g{y}_{m(k)},g{y}_{m(k)-1}))+ϵ.\end{array}$$

Letting $k\to \mathrm{\infty }$ in the last inequality and using (11), we obtain

$$\underset{k\to \mathrm{\infty }}{lim}max(d(g{x}_{n(k)-1},g{x}_{m(k)-1}),d(g{y}_{n(k)-1},g{y}_{m(k)-1}))=ϵ.$$

(24)

Finally, letting $k\to \mathrm{\infty }$ in (15) and using (18), (23) and the continuity of φ and ϕ, we have

$$\phi (ϵ)\le \phi (ϵ)-\varphi (ϵ)\le \phi (ϵ)$$

and, consequently, $\varphi (ϵ)=0$. Since ϕ is an altering distance function, we get $ϵ=0$, and this is a contradiction. This proves our claim.

Since X is a complete metric space, there exist $x,y\in X$ such that

$$\underset{n\to \mathrm{\infty }}{lim}F(x_n,y_n)=\underset{n\to \mathrm{\infty }}{lim}g{x}_n=x\text{and}\underset{n\to \mathrm{\infty }}{lim}F(y_n,x_n)=\underset{n\to \mathrm{\infty }}{lim}g{y}_n=y.$$

(25)

Since F and g are compatible mappings, we have

$$\underset{n\to \mathrm{\infty }}{lim}d(g(F(x_n,y_n)),F(g{x}_n,g{y}_n))=0$$

(26)

and

$$\underset{n\to \mathrm{\infty }}{lim}d(g(F(y_n,x_n)),F(g{y}_n,g{x}_n))=0.$$

(27)

We now show that $gx=F(x,y)$ and $gy=F(y,x)$. Suppose that assumption (a) holds. For all $n\ge 0$, we have

$$d(gx,F(g{x}_n,g{y}_n))\le d(gx,g(F(x_n,y_n)))+d(g(F(x_n,y_n)),F(g{x}_n,g{y}_n)).$$

Taking the limit as $n\to \mathrm{\infty }$, using (3), (25), (26) and the fact that F and g are continuous, we have $d(gx,F(x,y))=0$. Similarly, using (3), (25), (27) and the fact that F and g are continuous, we have $d(gy,F(y,x))=0$. Hence, we get

$$gx=F(x,y)\text{and}gy=F(y,x).$$

Finally, suppose that (b) holds. In fact, since $\{g{x}_n\}$ is non-decreasing and $g{x}_n\to x$ and $\{g{y}_n\}$ is non-increasing and $g{y}_n\to y$, by our assumption, $g{x}_n⪯x$ and $g{y}_n⪰y$ for every $n\in N$.

Applying (3), we have

$$\begin{array}{rcl}\phi \left(d(F(x,y),F(x_n,y_n))\right)& \le & \phi (max(d(gx,g{x}_n),d(gy,g{y}_n)))-\varphi (max(d(gx,g{x}_n),d(gy,g{y}_n))\\ \le & \phi (max(d(gx,g{x}_n),d(gy,g{y}_n))),\end{array}$$

and as φ is non-decreasing, we obtain

$$d(F(x,y),F(x_n,y_n))\le max(d(gx,g{x}_n),d(gy,g{y}_n)).$$

(28)

Using the triangular inequality and (28), we get

$$\begin{array}{rcl}d(gx,F(x,y))& \le & \underset{n\to \mathrm{\infty }}{lim}d(gx,gg{x}_{n+1})+d(gg{x}_{n+1},F(x,y))\\ =& \underset{n\to \mathrm{\infty }}{lim}d(gx,gg{x}_{n+1})+d(F(x,y),gF(x_n,y_n))\\ =& \underset{n\to \mathrm{\infty }}{lim}d(gx,gg{x}_{n+1})+d(F(x,y),F(g{x}_n,g{y}_n))\\ \le & d(gx,gg{x}_{n+1})+max(d(gx,gg{x}_n),d(gy,gg{y}_n)).\end{array}$$

As $x_n\to x$ and $y_n\to y$, taking $n\to \mathrm{\infty }$ in the last inequality, we have

$$d(gx,F(x,y))=0,$$

and, consequently, $F(x,y)=gx$.

Using a similar argument, it can be proved that $gy=F(y,x)$ and this completes the proof. □

Corollary 2.1 [6]

Let $(X,⪯)$ be a partially ordered set and suppose that there exists a metric d in X such that $(X,d)$ is a complete metric space. Let $F:X\times X\to X$ be a mapping having the mixed monotone property on X such that

$$\phi \left(d(F(x,y),F(u,v))\right)\le \phi (max(d(x,u),d(y,v)))-\varphi (max(d(x,u),d(y,v)))$$

for all $x,y,u,v\in X$ with $x⪰u$ and $y⪯v$, where φ and ϕ are altering distance functions. Moreover, suppose either

• (a) F is continuous, or

• (b) X has the following properties:

• (i) if a non-decreasing sequence $\{x_n\}\to x$, then $x_n⪯x$ for all n,

• (ii) if a non-increasing sequence $\{y_n\}\to y$, then $y⪯y_n$ for all n.

If there exist $x_0,y_0\in X$ with $x_0⪯F(x_0,y_0)$ and $y_0⪰F(y_0,x_0)$, then F has a coupled fixed point.

Corollary 2.2 [3]

Let $(X,⪯)$ be a partially ordered set and suppose that there exists a metric d in X such that $(X,d)$ is a complete metric space. Let $F:X\times X\to X$ be a mapping having the mixed monotone property on X such that

$$d(F(x,y),F(u,v))\le \frac{k}{2}[d(x,u)+d(y,v)]$$

for all $x,y,u,v\in X$ with $x⪰u$ and $y⪯v$. Moreover, suppose either

• (a) F is continuous, or

• (b) X has the following properties:

• (i) if a non-decreasing sequence $\{x_n\}\to x$, then $x_n⪯x$ for all n,

• (ii) if a non-increasing sequence $\{y_n\}\to y$, then $y⪯y_n$ for all n.

If there exist $x_0,y_0\in X$ with $x_0⪯F(x_0,y_0)$ and $y_0⪰F(y_0,x_0)$, then F has a coupled fixed point.

Proof Let $\phi =\text{identity}$ and $\varphi =(1-\frac{1}{k})\phi$ and g is the identity function. Then applying Theorem 2.1, we get Corollary 2.2. □

3 Uniqueness of the coupled coincidence point

In this section, we prove the uniqueness of the coupled coincidence point. Note that if $(X,⪯)$ is a partially ordered set, then we endow the product $X\times X$ with the following partial order relation, for all $(x,y),(u,v)\in X\times X$,

$$(x,y)⪯(u,v)\iff x⪯u,y⪰v.$$

Theorem 3.1 In addition to the hypotheses of Theorem 2.1, suppose that for every $(x,y)$, $(z,t)$ in $X\times X$, there exists a $(u,v)$ in $X\times X$ that is comparable to $(x,y)$ and $(z,t)$, then F and g have a unique coupled coincidence point.

Proof Suppose that $(x,y)$ and $(z,t)$ are coupled coincidence points of F, that is, $gx=F(x,y)$, $gy=F(x,y)$, $gz=f(z,t)$ and $gt=F(t,z)$.

Let $(u,v)$ be an element of $X\times X$ comparable to $(x,y)$ and $(z,t)$. Suppose that $(x,y)⪰(u,v)$ (the proof is similar in the other case).

We construct the sequences $\{g{u}_n\}$ and $\{g{v}_n\}$ as follows:

$$u_0=u,v_0=v,g{u}_{n+1}=F(u_n,v_n),g{v}_{n+1}=F(v_n,u_n).$$

We claim that $(x,y)⪰(u_n,v_n)$ for each $n\in N$. In fact, we will use mathematical induction.

For $n=0$, as $(x,y)⪰(u,v)$, this means $u_0=u⪯x$ and $y⪰v=v_0$ and, consequently, $(u_0,v_0)⪯(x,y)$. Suppose that $(x,y)⪰(u_n,v_n)$, then since F has the mixed g-monotone property and since g is monotone increasing, we get

$$\begin{array}{c}g{u}_{n+1}=F(u_n,v_n)⪯F(x,v_n)⪯F(x,y)=gx,\hfill \\ g{v}_{n+1}=F(v_n,u_n)⪯F(y,u_n)⪰F(y,x)=gy,\hfill \end{array}$$

and this proves our claim.

Now, since $u_n⪯x$ and $u_n⪰y$, using (3), we get

$$\begin{array}{rcl}\phi (d(gx,g{u}_n))& =& \phi \left(d(F(x,y),F(u_{n-1},v_{n-1}))\right)\\ \le & \phi (max(d(gx,g{u}_{n-1}),d(gy,g{v}_{n-1})))\\ -\varphi (max(d(gx,g{u}_{n-1}),d(gy,g{v}_{n-1})))\\ \le & \phi (max(d(gx,g{u}_{n-1}),d(gy,g{v}_{n-1}))).\end{array}$$

(29)

In the same way, we have

$$\begin{array}{rcl}\phi (d(gy,g{v}_n))& =& \phi \left(d(F(y,x),F(v_{n-1},u_{n-1}))\right)\\ =& \phi \left(d(F(v_{n-1},u_{n-1}),F(y,x))\right)\\ \le & \phi (max(d(gx,g{u}_{n-1}),d(gy,g{v}_{n-1})))\\ -\varphi (max(d(gx,g{u}_{n-1}),d(gy,g{v}_{n-1})))\\ \le & \phi (max(d(gx,g{u}_{n-1}),d(gy,g{v}_{n-1}))).\end{array}$$

(30)

Using (29) and (30) and the fact that ϕ is non-decreasing, we get

$$\begin{array}{rcl}\phi (max(d(gx,g{u}_n),d(gy,g{v}_n)))& =& max\left(\phi (d(gx,g{u}_n),d(gy,g{v}_n))\right)\\ \le & \phi (max\left(d(gx,g{u}_{n-1},d(gy,g{v}_{n-1}))\right)\\ -\varphi (max(d(gx,g{u}_{n-1}),d(gy,g{v}_{n-1})))\\ \le & \phi (max(d(gx,g{u}_{n-1}),d(gy,g{v}_{n-1}))).\end{array}$$

(31)

Using the last inequality and the fact that φ is non-decreasing, we have

$$max(d(gx,g{u}_n),d(gy,g{v}_n))\le max(d(gx,g{u}_{n-1}),d(gy,g{v}_{n-1})).$$

Thus the sequence $(max(d(gx,g{u}_n),d(gy,g{v}_n)))$ is decreasing and non-negative, and hence, for certain $r\ge 0$,

$$\underset{n\to \mathrm{\infty }}{lim}(max(d(gx,g{u}_n),d(gy,g{v}_n)))=r.$$

(32)

Using (32) and letting $n\to \mathrm{\infty }$ in (31), we have

$$\phi (r)\le \phi (r)-\varphi (r)<\phi (r).$$

This gives $\varphi (r)=0$ and hence $r=0$.

Finally, since ${lim}_{n\to \mathrm{\infty }}(max(d(gx,g{u}_n),d(gy,g{v}_n)))=0$, we have $g{u}_n\to gx$ and $g{v}_n\to gy$. Using a similar argument for $(z,t)$, we can get $g{u}_n\to gz$ and $g{v}_n\to gt$, and the uniqueness of the limit gives $gx=gz$ and $gy=gt$. This completes the proof. □

Theorem 3.2 Under the assumptions of Theorem 2.1, suppose that $x_0$ and $y_0$ are comparable, then the coupled coincidence point $(x,y)\in X\times X$ satisfies $x=y$.

Proof Assume $x_0⪯y_0$ (a similar argument applies to $y_0⪯x_0$).

We claim that $x_n⪯y_n$ for all n, where $g{x}_{n_1}=F(x_n,y_n)$ and $g{y}_{n+1}=F(y_n,x_n)$.

Obviously, the inequality is satisfied for $n=0$. Suppose $x_n⪯y_n$. Using the mixed g-monotone property of F, we have

$$g{x}_{n+1}=F(x_n,y_n)⪯F(y_n,y_n)⪯F(y_n,x_n)=g{y}_{n+1},$$

and since g is non-decreasing, this proves our claim.

Now, using (3) and $x_n⪯y_n$, we get

$$\begin{array}{rcl}\phi (d(g{x}_{n+1},g{y}_{n+1}))& =& \phi (d(g{y}_{n+1},g{x}_{n+1}))=\phi \left(d(F(y_n,x_n),F(x_n,y_n))\right)\\ \le & \phi (d(g{x}_n,g{y}_n))-\varphi (d(g{x}_n,g{y}_n))\le \phi (d(g{x}_n,g{y}_n)),\end{array}$$

(33)

and since φ is non-decreasing, we get

$$d(g{x}_{n+1},g{y}_{n+1})\le d(g{x}_n,g{y}_n).$$

We notice that the sequence $d(g{x}_n,g{y}_n)$ is decreasing. Thus, ${lim}_{n\to \mathrm{\infty }}d(g{x}_n,g{y}_n)=r$ for certain $r>0$. Hence,

$$\phi (r)\le \phi (r)-\varphi (r)\le \phi (r),$$

and this gives us $r=0$.

Since $g{x}_n\to x$, $g{y}_n\to y$ and ${lim}_{n\to \mathrm{\infty }}d(g{x}_n,g{y}_n)=0$, we have

$$0=\underset{n\to \mathrm{\infty }}{lim}d(g{x}_n,g{y}_n)=d(g{x}_n,g{y}_n)=d(\underset{n\to \mathrm{\infty }}{lim}g{x}_n,\underset{n\to \mathrm{\infty }}{lim}g{y}_n)=d(x,y)$$

and thus $x=y$. This completes the proof. □

4 Example

The following example illustrates our main result.

Example 4.1 Let $X=[0,1]$. Then $(X,\le )$ is a partially ordered set with the natural ordering of real numbers. Let

$$d(x,y)=|x-y|\text{for }x,y\in [0,1].$$

Then $(X,d)$ is a complete metric space. Let $g:X\to X$ be defined as

$$g(x)=x^2\text{for all }x\in X,$$

and let $F:X\times X\to X$ be defined as

$$F(x,y)=\{\begin{array}{cc}\frac{x^2-y^2}{5}\hfill & \text{if }x,y\in [0,1],x\ge y,\hfill \\ 0\hfill & \text{if }x<y.\hfill \end{array}$$

Then, F satisfies the mixed g-monotone property.

Let $\phi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ be defined as

$$\phi (t)=\frac{1}{3}t\text{for }t\in [0,\mathrm{\infty }),$$

and let $\varphi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ be defined as

$$\varphi (t)=\frac{1}{5}t\text{for }t\in [0,\mathrm{\infty }).$$

Let $\{x_n\}$ and $\{y_n\}$ be two sequences in X such that ${lim}_{n\to \mathrm{\infty }}F(x_n,y_n)=a$, ${lim}_{n\to \mathrm{\infty }}g{x}_n=a$, ${lim}_{n\to \mathrm{\infty }}F(y_n,x_n)=b$ and ${lim}_{n\to \mathrm{\infty }}g{y}_n=b$. Then, obviously, $a=0$ and $b=0$. Now, for all $n\ge 0$,

$$\begin{array}{c}g(x_n)=x_n^2,g(y_n)=y_n^2,\hfill \\ F(x_n,y_n)=\{\begin{array}{cc}\frac{x_n^2-y_n^2}{5}\hfill & \text{if }{x}_n\ge y_n,\hfill \\ 0\hfill & \text{if }{x}_n<y_n.\hfill \end{array}\hfill \end{array}$$

and

$$F(y_n,x_n)=\{\begin{array}{cc}\frac{y_n^2-x_n^2}{5}\hfill & \text{if }{y}_n\ge x_n,\hfill \\ 0\hfill & \text{if }{y}_n<x_n.\hfill \end{array}$$

Then it follows that

$$\underset{n\to \mathrm{\infty }}{lim}d(g(F(x_n,y_n)),F(g{x}_n,g{y}_n))=0$$

and

$$\underset{n\to \mathrm{\infty }}{lim}d(g(F(y_n,x_n)),F(g{y}_n,g{x}_n))=0.$$

Hence, the mappings F and g are compatible in X. Also, $x_0=0$ and $y_0=c$ ($c>0$) are two points in X such that

$$g(x_0)=g(0)=0=F(0,c)=F(x_0,y_0)$$

and

$$g(y_0)=g(c)=c^2\ge \frac{c^2}{5}=F(c,0)=F(y_0,x_0).$$

We next verify the contraction of Theorem 2.1. We take $x,y,u,v,\in X$ such that $gx\ge gu$ and $gy\le gv$, that is, $x^2\ge u^2$ and $y^2\le v^2$.

We consider the following cases.

Case 1. $x\ge y$, $u\ge v$. Then

$$\begin{array}{rcl}\phi \left(d(F(x,y),F(u,v))\right)& =& \frac{1}{3}[d(F(x,y),F(u,v)]\\ =& \frac{1}{3}\left[d(\frac{x^2-y^2}{5},\frac{u^2-v^2}{5})\right]\\ =& \frac{1}{3}\left|\frac{(x^2-y^2)-(u^2-v^2)}{5}\right|\\ \le & \frac{1}{3}\frac{|x^2-u^2|+|y^2-v^2|}{5}\\ =& \frac{2}{15}\left(\frac{d(gx,gu)+d(gy,gv)}{2}\right)\\ \le & \frac{2}{15}[max(d(gx,gu),d(gy,gv)]\\ \le & \frac{1}{3}[max(d(gx,gu),d(gy,gv)]\\ -\frac{1}{5}[max(d(gx,gu),d(gy,gv)]\\ =& \phi (max\{d(gx,gu),d(gy,gv)\})\\ -\varphi (max\{d(gx,gu),d(gy,gv)\}).\end{array}$$

Case 2. $x\ge y$, $u<v$ Then

$$\begin{array}{rcl}\phi \left(d(F(x,y),F(u,v))\right)& =& \frac{1}{3}[d(F(x,y),F(u,v)]\\ =& \frac{1}{3}\left[d(\frac{x^2-y^2}{5},0)\right]\\ =& \frac{1}{3}\frac{|x^2-y^2|}{5}\\ \le & \frac{1}{3}\frac{|v^2+x^2-y^2-u^2|}{5}\\ =& \frac{1}{3}\frac{|(v^2-y^2)-(u^2-x^2)|}{5}\\ \le & \frac{1}{3}\frac{|v^2-y^2|+|u^2-x^2|}{5}\\ =& \frac{1}{3}\frac{|u^2-x^2|+|y^2-v^2|}{5}\\ =& \frac{2}{15}\left(\frac{|u^2-x^2|+|y^2-v^2|}{2}\right)\\ =& \frac{2}{15}\left(\frac{d(gx,gu)+d(gy,gv)}{2}\right)\\ \le & \frac{2}{15}(max\{d(gx,gu),d(gy,gv)\})\\ =& \frac{1}{3}(max\{d(gx,gu),d(gy,gv)\})\\ -\frac{1}{5}(max\{d(gx,gu),d(gy,gv)\})\\ =& \phi (max\{d(gx,gu),d(gy,gv)\})\\ -\varphi (max\{d(gx,gu),d(gy,gv)\}).\end{array}$$

Case 3. $x<y$ and $u\ge v$. Then

$$\begin{array}{rcl}\phi \left(d(F(x,y),F(u,v))\right)& =& \frac{1}{3}\left[d(0,\frac{u^2-v^2}{5})\right]\\ =& \frac{1}{3}\frac{|u^2-v^2|}{5}\\ =& \frac{1}{3}\frac{|u^2+x^2-v^2-x^2|}{5}\\ =& \frac{1}{3}\frac{|(x^2-v^2)+(u^2-x^2)|}{5}(\text{since }y>x)\\ \le & \frac{1}{3}\frac{|y^2-v^2|+|u^2-x^2|}{5}\\ =& \frac{2}{15}\left(\frac{|u^2-x^2|+|y^2-v^2|}{2}\right)\\ =& \frac{2}{15}\left(\frac{d(gx,gu)+d(gy,gv)}{2}\right)\\ \le & \frac{2}{15}(max\{d(gx,gu),d(gy,gv)\})\\ =& \frac{1}{3}(max\{d(gx,gu),d(gy,gv)\})\\ -\frac{1}{5}(max\{d(gx,gu),d(gy,gv)\})\\ =& \phi (max\{d(gx,gu),d(gy,gv)\})\\ -\varphi (max\{d(gx,gu),d(gy,gv)\}).\end{array}$$

Case 4. $x<y$ and $u<v$ with $x^2\le u^2$ and $y^2\ge v^2$. Then $F(x,y)=0$ and $F(u,v)=0$, that is,

$$\phi \left(d(F(x,y),F(u,v))\right)=\phi (d(0,0))=\phi (0)=0.$$

Obviously, the contraction of Theorem 2.1 is satisfied.