Almost contractive coupled mapping in ordered complete metric spaces

Abstract

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

MSC:54H25, 47H10, 34B15.

1 Introduction and preliminaries

The existence and uniqueness theorems of a fixed point in complete metric spaces play an important role in constructing methods for solving problems in differential equations, matrix equations, and integral equations. Furthermore, the fixed point theory is a crucial method in numerical analysis to present a way for solving and approximating the roots of many equations in real analysis. One of the main theorems on a fixed point is the Banach contraction theorem [1]. 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 [2–17]. The study of a coupled fixed point was initiated by Bhaskar and Lakshmikantham [18]. Bhaskar and Lakshmikantham [18] 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 [18–26].

The following definitions will be needed in the sequel.

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

$$x,y\in X,x_1,x_2\in X,x_1⪯x_2\Rightarrow F(x_1,y)⪯F(x_2,y)$$

and

$$y_1,y_2\in X,y_1⪯y_2\Rightarrow F(x,y_1)⪰F(x,y_2).$$

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

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

Definition 1.3 [20]

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

$$x_1,x_2\in X,g(x_1)⪯g(x_2)\Rightarrow F(x_1,y)⪯F(x_2,y)$$

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

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

The main results of Bhaskar and Lakshmikantham in [18] are the following.

Theorem 1.1 [18]

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

$$d(F(x,y),F(u,v))\le \frac{k}{2}[d(x,u)+d(y,v)]\mathrm{\forall }x⪰u\mathit{\text{ and }}y⪯v.$$

If there exist two elements $x_0,y_0\in X$ with

$$x_0⪯F(x_0,y_0)\mathit{\text{and}}{y}_0⪰F(y_0,x_0),$$

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

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

Theorem 1.2 [18]

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

• (i) if a nondecreasing sequence $\{x_n\}$ in X converges to $x\in X$, then $x_n⪯x$ for all n,

• (ii) if a nonincreasing sequence $\{y_n\}$ in X converges to $y\in X$, then $y_n⪰y$ for all n.

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

$$d(F(x,y),F(u,v))\le \frac{k}{2}[d(x,u)+d(y,v)]\mathrm{\forall }x⪰u\mathit{\text{ and }}y⪯v.$$

If there exist two elements $x_0,y_0\in X$ with

$$x_0⪯F(x_0,y_0)\mathit{\text{and}}{y}_0⪰F(y_0,x_0),$$

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

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

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

$$F(g(x),g(y))=g(F(x,y))$$

for all $x,y\in X$.

Nashine and Shatanawi [22] proved the following coupled coincidence point theorems.

Theorem 1.3 [22]

Let $(X,d,⪯)$ be an ordered metric space. Let $F:X\times X\to X$ 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 $g(x_0)⪯F(x_0,y_0)$ and $g(y_0)⪰F(y_0,x_0)$. Suppose that there exist non-negative real numbers α, β, L with $\alpha +\beta <1$ such that

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

(3)

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

• (i) if a nondecreasing sequence $\{x_n\}$ in X converges to $x\in X$, then $x_n⪯x$ for all n,

• (ii) if a nonincreasing sequence $\{y_n\}$ in X converges to $y\in X$, then $y_n⪰y$ for all n.

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

$$F(x,y)=g(x)\mathit{\text{and}}F(y,x)=g(y),$$

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

Theorem 1.4 [22]

Let $(X,⪯)$ be a partially ordered set and suppose that there is a metric d on X such that $(X,d)$ is a complete metric space. Let $F:X\times X\to X$ 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 $g(x_0)⪯F(x_0,y_0)$ and $g(y_0)⪰F(y_0,x_0)$. Suppose that there exist non-negative real numbers α, β, L with $\alpha +\beta <1$ such that

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

(4)

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

• (a) F is continuous, or

• (b) X has the following property:

  • (i) if a nondecreasing sequence $\{x_n\}$ in X converges to $x\in X$, then $x_n⪯x$ for all n,

  • (ii) if a nonincreasing sequence $\{y_n\}$ in X converges to $y\in X$, then $y_n⪰y$ for all n.

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

$$F(x,y)=g(x)\mathit{\text{and}}F(y,x)=g(y),$$

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

Berinde [27–30] 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 [27]

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

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

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

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

The aim of this paper is to introduce the notion of almost contractive mapping $F:X\times X\to X$ 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

We start with the following definition.

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

$$\begin{array}{r}d(F(x,y),F(u,v))\\ \le \alpha max\{d(g(x),g(u)),d(g(y),g(v)),d(F(x,y),g(x)),d(F(u,v),g(u))\}\\ +Lmin\{d(F(x,y),g(u)),d(F(u,v),g(x))\}\end{array}$$

(5)

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

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

• (1) F is an almost contractive mapping with respect to g.

• (2) F has the mixed g-monotone property on X.

• (3) There exist two elements $x_0,y_0\in X$ with $g(x_0)⪯F(x_0,y_0)$ and $g(y_0)⪰F(y_0,x_0)$.

• (4) $F(X\times X)\subseteq g(X)$ and $g(X)$ is a complete subspace of X.

Also, suppose that X satisfies the following properties:

• (i) if a nondecreasing sequence $\{x_n\}$ in X converges to $x\in X$, then $x_n⪯x$ for all n,

• (ii) if a nonincreasing sequence $\{y_n\}$ in X converges to $y\in X$, then $y_n⪰y$ for all n.

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

$$F(x,y)=g(x)\mathit{\text{and}}F(y,x)=g(y),$$

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

Proof Let $x_0,y_0\in X$ be such that $g(x_0)⪯F(x_0,y_0)$ and $g(y_0)⪰F(y_0,x_0)$. Since $F(X\times X)\subseteq g(X)$, we can choose $x_1,y_1\in X$ such that $g(x_1)=F(x_0,y_0)$ and $g(y_1)=F(y_0,x_0)$.

In the same way, we construct $g(x_2)=F(x_1,y_1)$ and $g(y_2)=F(y_1,x_1)$.

Continuing in this way, we construct two sequences $\{x_n\}$ and $\{y_n\}$ in X such that

$$g(x_{n+1})=F(x_n,y_n)\text{and}g(y_{n+1})=F(y_n,x_n)\mathrm{\forall }n\in \mathbb{N}\cup \{0\}.$$

(6)

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

$$g(x_0)⪯g(x_1)⪯g(x_2)⪯\cdots ⪯g(x_{n+1})⪯\cdots$$

and

$$g(y_0)⪰g(y_1)⪰g(y_2)⪰\cdots ⪰g(y_{n+1})⪰\cdots .$$

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

$$\begin{array}{r}d(g(x_n),g(x_{n+1}))\\ =d(F(x_{n-1},y_{n-1}),F(x_n,y_n))\\ \le \alpha max\{d(g(x_{n-1}),g(x_n)),d(g(y_{n-1}),g(y_n)),d(F(x_n,y_n),g(x_n)),\\ d(F(x_{n-1},y_{n-1}),g(x_{n-1}))\}+Lmin\{d(F(x_n,y_n),g(x_{n-1})),d(F(x_{n-1},y_{n-1}),g(x_n))\}\\ =\alpha max\{d(g(x_{n-1}),g(x_n)),d(g(y_{n-1}),g(y_n)),d(g(x_{n+1}),g(x_n)),d(g(x_n),g(x_{n-1}))\}\\ +Lmin\{d(g(x_{n+1}),g(x_{n-1})),d(g(x_n),g(x_n))\}\\ =\alpha max\{d(g(x_{n-1}),g(x_n)),d(g(y_{n-1}),g(y_n)),d(g(x_{n+1}),g(x_n))\}.\end{array}$$

If $max\{d(g(x_{n-1}),g(x_n)),d(g(y_{n-1}),g(y_n)),d(g(x_{n+1}),g(x_n))\}=d(g(x_{n+1}),g(x_n))$, then $d(g(x_{n+1}),g(x_n))\le \alpha d(g(x_{n+1}),g(x_n))$ and hence $d(g(x_{n+1}),g(x_n))=0$. Thus $d(g(x_{n-1}),g(x_n))=d(g(y_{n-1}),g(y_n))=0$. Therefore $d(g(x_{n-1}),g(y_{n-1}))=d(g(x_n),g(y_n))$, a contradiction. Thus

$$\begin{array}{r}max\{d(g(x_{n-1}),g(x_n)),d(g(y_{n-1}),g(y_n)),d(g(x_{n+1}),g(x_n))\}\\ =max\{d(g(x_{n-1}),g(x_n)),d(g(y_{n-1}),g(y_n))\}.\end{array}$$

Therefore

$$d(g(x_{n+1}),g(x_n))\le \alpha max\{d(g(x_{n-1}),g(x_n)),d(g(y_{n-1}),g(y_n))\}.$$

(7)

Similarly, we may show that

$$d(g(y_n),g(y_{n+1}))\le \alpha max\{d(g(x_{n-1}),g(x_n)),d(g(y_{n-1}),g(y_n))\}.$$

(8)

From (7) and (8), we have

$$\begin{array}{r}max\{d(g(x_{n+1}),g(x_n)),d(g(y_n),g(y_{n+1}))\}\\ \le \alpha max\{d(g(x_{n-1}),g(x_n)),d(g(y_{n-1}),g(y_n))\}.\end{array}$$

(9)

Repeating (9) n-times, we get

$$\begin{array}{r}max\{d(g(x_{n+1}),g(x_n)),d(g(y_n),g(y_{n+1}))\}\\ \le {\alpha }^{n}max\{d(g(x_0),g(x_1)),d(g(y_0),g(y_1))\}.\end{array}$$

(10)

Now, we shall prove that $\{g(x_n)\}$ and $\{g(y_n)\}$ are Cauchy sequences in $g(X)$.

For each $m\ge n$, we have

$$\begin{array}{r}d(g(x_m),g(x_n))\\ \le d(g(x_n),g(x_{n+1}))+d(g(x_{n+1}),g(x_{n+2}))+\cdots \\ +d(g(x_{m-1}),g(x_m))\\ \le {\alpha }^{n}max\{d(g(x_0),g(x_1)),d(g(y_0),g(y_1))\}+\cdots \\ +{\alpha }^{m-1}max\{d(g(x_0),g(x_1)),d(g(y_0),g(y_1))\}\\ \le \frac{{\alpha }^n}{1-\alpha }max\{d(g(x_0),g(x_1)),d(g(y_0),g(y_1))\}.\end{array}$$

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

$$\begin{array}{r}d(g(x_{n+1}),F(x,y))\\ =d(F(x_n,y_n),F(x,y))\\ \le \alpha max\{d(g(x_n),g(x)),d(g(y_n),g(y)),d(g(x_{n+1}),g(x_n)),d(F(x,y),g(x))\}\\ +Lmin\{d(g(x_{n+1}),g(x)),d(F(x,y),g(x_n))\}.\end{array}$$

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

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

• (1) F is an almost contractive mapping with respect to g.

• (2) F has the mixed g-monotone property on X.

• (3) There exist two elements $x_0,y_0\in X$ with $g(x_0)⪯F(x_0,y_0)$ and $g(y_0)⪰F(y_0,x_0)$.

• (4) $F(X\times X)\subseteq g(X)$.

• (5) g is continuous nondecreasing and commutes with F.

Also suppose that either

• (a) F is continuous, or

• (b) X has the following property:

  • (i) if a nondecreasing sequence $\{x_n\}$ in X converges to $x\in X$, then $x_n⪯x$ for all n,

  • (ii) if a nonincreasing sequence $\{y_n\}$ in X converges to $y\in X$, then $y_n⪰y$ for all n.

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

$$F(x,y)=g(x)\mathit{\text{and}}F(y,x)=g(y),$$

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

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

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

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

$$\begin{array}{r}d(g(g{x}_{n+1}),F(x,y))\\ =d(F(g{x}_n,g{y}_n),F(x,y))\\ \le \alpha max\{d(g(g{x}_n),g(x)),d(g(g{y}_n),g(y)),d(g(g{x}_{n+1}),g(g{x}_n)),d(F(x,y),g(x))\}\\ +Lmin\{d(g(g{x}_{n+1}),g(x)),d(F(x,y),g(g{x}_n))\}.\end{array}$$

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

Corollary 2.1 Let $(X,⪯)$ be a partially ordered set and suppose that there is a metric d on X such that $(X,d)$ is a complete metric space. Let $F:X\times X\to X$ 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 $x_0⪯F(x_0,y_0)$ and $y_0⪰F(y_0,x_0)$. Suppose that there exist a real number $\alpha \in [0,1)$ and a nonnegative number L such that

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

(11)

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

• (a) F is continuous, or

• (b) X has the following property:

  • (i) if a nondecreasing sequence $\{x_n\}$ in X converges to $x\in X$, then $x_n⪯x$ for all n,

  • (ii) if a nonincreasing sequence $\{y_n\}$ in X converges to $y\in X$, then $y_n⪰y$ for all n,

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

$$F(x,y)=x\mathit{\text{and}}F(y,x)=y,$$

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

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

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

$$\text{for }(x,y),(u,v)\in X\times X,(u,v)⪯(x,y)\iff x⪰u,y⪯v.$$

Now, we prove some uniqueness theorem of a coupled common fixed point of mappings $F:X\times X\to X$ 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 $(x,y),(y^{\ast },x^{\ast })\in X\times X$, there exists $(u,v)\in X\times X$ such that $(F(u,v),F(v,u))$ is comparable to $(F(x,y),F(y,x))$ and $(F(x^{\ast },y^{\ast }),F(y^{\ast },x^{\ast }))$. Then F and g have a unique coupled common fixed point, that is, there exists a unique $(x,y)\in X\times X$ such that

$$x=g(x)=F(x,y)\mathit{\text{and}}y=g(y)=F(y,x).$$

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

$$g(x)=g\left(x^{\ast }\right)\text{and}g(y)=g\left(y^{\ast }\right).$$

(12)

By the hypotheses, there exists $(u,v)\in X\times X$ such that $(F(u,v),F(v,u))$ is comparable to $(F(x,y),F(y,x))$ and $(F(x^{\ast },y^{\ast }),F(y^{\ast },x^{\ast }))$. Put $u_0=u$, $v_0=v$. Let $u_1,v_1\in X$ be such that $g(u_1)=F(u_0,v_0)$ and $g(v_1)=F(v_0,u_0)$. Then as a similar proof of Theorem 2.1, we construct two sequences $\{g(u_n)\}$, $\{g(v_n)\}$ in $g(X)$, where $g(u_{n+1})=F(u_n,v_n)$ and $g(v_{n+1})=F(v_n,u_n)$ 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 $\{g(x_n)\}$, $\{g(y_n)\}$ in the following way: define $g{x}_1=F(x_0,y_0)=F(x,y)$ and $g{y}_1=F(y_0,x_0)=F(y,x)$. Also, define $g{x}_2=F(x_1,y_1)$ and $g{y}_2=F(y_1,x_1)$. For each $n\in \mathbb{N}$, define $g{x}_{n+1}=F(x_n,y_n)$ and $g{y}_{n+1}=F(y_n,x_n)$. In the same way, we define the sequences $\{g(x_n^{\ast })\}$, $\{g(y_n^{\ast })\}$. Now, we prove that

$$g(x_n)=F(x,y)=g(x)\text{and}g(y_n)=F(y,x)=g(y).$$

Since $(x,y)$ is a coupled coincidence point of F and g, we have $F(x,y)=g(x)$ and $F(y,x)=g(y)$. Thus $g(x_1)=F(x_0,y_0)=F(x,y)=g(x)$ and $g(y_1)=F(y_0,x_0)=F(y,x)=g(y)$. Therefore $g(x_1)⪯g(x)$, $g(x)⪯g(x_1)$, $g(y_1)⪯g(y)$ and $g(y)⪯g(y_1)$. Since F is monotone g-non-decreasing on its first argument, $g(x_1)⪯g(x)$, and $g(x)⪯g(x_1)$, we have $F(x_1,y_1)⪯F(x,y_1)$ and $F(x,y_1)⪯F(x_1,y_1)$. Therefore,

$$F(x_1,y_1)=F(x,y_1).$$

(13)

Also, since F is monotone g-non-increasing on its second argument, $g(y_1)⪯g(y)$ and $g(y)⪯g(y_1)$, we have $F(x,y)⪯F(x,y_1)$ and $F(x,y_1)⪯F(x,y)$. Therefore,

$$F(x,y)=F(x,y_1).$$

(14)

From (13) and (14), we have

$$g(x_2)=F(x_1,y_1)=F(x,y)=g(x).$$

Similarly, we may show that

$$g(y_2)=F(y_1,x_1)=F(y,x)=g(y).$$

Note that $g(x_2)⪯g(x)$, $g(x)⪯g(x_2)$, $g(y_2)⪯g(y)$ and $g(y)⪯g(y_2)$. Since F is monotone g-non-decreasing on its first argument, $g(x_2)⪯g(x)$, and $g(x)⪯g(x_2)$, we have $F(x_2,y_2)⪯F(x,y_2)$ and $F(x,y_2)⪯F(x_2,y_2)$. Therefore,

$$F(x_2,y_2)=F(x,y_2).$$

(15)

Also, since F is monotone g-non-increasing on its second argument, $g(y_2)⪯g(y)$ and $g(y)⪯g(y_2)$, we have $F(x,y)⪯F(x,y_2)$ and $F(x,y_2)⪯F(x,y)$. Therefore,

$$F(x,y)=F(x,y_2).$$

(16)

From (15) and (16), we have

$$g(x_3)=F(x_2,y_2)=F(x,y)=g(x).$$

Similarly, we may show that

$$g(y_3)=F(y_2,x_2)=F(y,x)=g(y).$$

Continuing in the same way, we have that

$$g(x_n)=F(x,y)=g(x)\text{and}g(y_n)=F(y,x)=g(y)$$

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

$$g\left(x_n^{\ast }\right)=F(x^{\ast },y^{\ast })=g\left(x^{\ast }\right)\text{and}g\left(y_n^{\ast }\right)=F(y^{\ast },x^{\ast })=g\left(y^{\ast }\right)\mathrm{\forall }n\in \mathbb{N}$$

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

$$(F(x,y),F(y,x))=(g(x_1),g(y_1))=(g(x),g(y))$$

and

$$(F(u,v),F(v,u))=(g(u_1),g(v_1))$$

are comparable, $g(x)⪯g(u_1)$ and $g(y)⪰g(v_1)$. Since F has the mixed g-monotone property of X, we have $g(x)⪯g(u_n)$ and $g(y)⪰g(v_n)$ for all $n\in \mathbb{N}$. Also, since $(g(x^{\ast }),g(y^{\ast }))$ and $(F(u,v),F(v,u))=(g(u_1),g(v_1))$ are comparable, and F has the g-monotone property, then we can show that for $n\in \mathbb{N}$, we have that $(g(x^{\ast }),g(y^{\ast }))$ and $(g(u_n),g(v_n))$ are comparable. Now, if $(g(x),g(y))=(g(u_k),g(v_k))$ for some $k\in \mathbb{N}$ or $(g(x^{\ast }),g(y^{\ast }))=(g(u_k),g(v_k))$ for some $k\in \mathbb{N}$, then $(g(x),g(y))$ and $(g(x^{\ast }),g(y^{\ast }))$ are comparable, say $g(x)⪯g(x^{\ast })$ and $g(y)⪰g(y^{\ast })$. Thus from (5) we have

$$\begin{array}{r}d(g(x),g\left(x^{\ast }\right))\\ =d(F(x,y),F(x^{\ast },y^{\ast }))\\ \le \alpha max\{d(g(x),g\left(x^{\ast }\right)),d(g(y),g\left(y^{\ast }\right)),d(F(x,y),g(x)),d(F(x^{\ast },y^{\ast }),g\left(x^{\ast }\right))\}\\ =\alpha max\{d(g(x),g\left(x^{\ast }\right)),d(g(y),g\left(y^{\ast }\right))\}\end{array}$$

(17)

and

$$\begin{array}{r}d(g\left(y^{\ast }\right),g(y))\\ =d(F(y^{\ast },x^{\ast }),F(y,x))\\ \le \alpha max\{d(g(y),g\left(y^{\ast }\right)),d(g(x),g\left(x^{\ast }\right)),d(F(y^{\ast },x^{\ast }),g\left(y^{\ast }\right)),d(F(y,x),g(y))\}\\ =\alpha max\{d(g(y),g\left(y^{\ast }\right)),d(g(x),g(y))\}.\end{array}$$

(18)

From (17) and (18), we have

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

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

$$\begin{array}{r}d(g(x),g(u_{n+1}))\\ =d(F(x,y),F(u_n,v_n))\\ \le \alpha max\{d(g(x),g(u_n)),d(g(y),g(v_n)),d(F(x,y),g(x)),d(F(u_n,v_n),g(u_n))\}\\ =\alpha max\{d(g(x),g(u_n)),d(g(y),g(v_n)),d(g(u_{n+1}),g(u_n))\}\\ \le \alpha max\{d(g(x),g(u_n)),d(g(y),g(v_n)),d(g(u_{n+1}),g(x))+d(g(x),g(u_n))\}\\ \le \alpha max\{d(g(x),g(u_n)),d(g(y),g(v_n)),2d(g(u_{n+1}),g(x)),2d(g(x),g(u_n))\}\\ =\alpha max\{2d(g(x),g(u_n)),d(g(y),g(v_n)),2d(g(u_{n+1}),g(x))\}.\end{array}$$

If

$$max\{2d(g(x),g(u_n)),d(g(y),g(v_n)),2d(g(u_{n+1}),g(x))\}=2d(g(u_{n+1}),g(x))$$

then $d(g(u_{n+1}),g(x))\le 2\alpha d(g(u_{n+1}),g(x))$. Since $2\alpha <1$, we have $d(g(u_{n+1}),g(x))=0$. Therefore $d(g(x),g(u_n))=0$ and $d(g(y),g(v_n))=0$ and hence $(g(x),g(y))=(g(u_n),g(v_n))$, a contradiction. Thus

$$\begin{array}{rcl}d(g(x),g(u_{n+1}))& \le & \alpha max\{2d(g(x),g(u_n)),d(g(y),g(v_n))\}\\ \le & 2\alpha max\{d(g(x),g(u_n)),d(g(y),g(v_n))\}.\end{array}$$

(19)

Similarly, we may show that

$$d(g(v_{n+1}),g(y))\le 2\alpha max\{d(g(x),g(u_n)),d(g(y),g(v_n))\}.$$

(20)

From (19) and (20), we have

$$\begin{array}{r}max\{d(g(x),g(u_{n+1})),d(g(v_{n+1}),g(y))\}\\ \le 2\alpha max\{d(g(v_n),g(y)),d(g(u_n),g(x)),d(g(u_{n+1})\}.\end{array}$$

(21)

By repeating (21) n-times, we have

$$\begin{array}{r}max\{d(g(x),g(u_{n+1})),d(g(v_{n+1}),g(y))\}\\ \le 2\alpha max\{d(g(v_n),g(y)),d(g(u_n),g(x))\}\\ ⋮\\ \le {(2\alpha )}^{n+1}max\{d(g(x),g(u_0)),d(g(v_0),g(y))\}.\end{array}$$

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

$$\underset{n\to }{lim}max\{d(g(x),g(u_{n+1})),d(g(v_{n+1}),g(y))\}=0.$$

Hence

$$\underset{n\to \mathrm{\infty }}{lim}d(g(x),g(u_{n+1}))=0$$

(22)

and

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

(23)

Similarly, we may show that

$$\underset{n\to \mathrm{\infty }}{lim}d(g(x),g(u_{n+1}))=0$$

(24)

and

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

(25)

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

$$\begin{array}{c}d(g(x),g\left(x^{\ast }\right))\le d(g(x),g(u_{n+1}))+d(g\left(x^{\ast }\right),g(u_{n+1}))\to 0\text{as }n\to \mathrm{\infty },\hfill \\ d(g(y),g\left(y^{\ast }\right))\le d(g(y),g(v_{n+1}))+d(g\left(y^{\ast }\right),g(v_{n+1}))\to 0\text{as }n\to \mathrm{\infty },\hfill \end{array}$$

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

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

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

(26)

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

$$g(z)=F(z,w)\text{and}g(w)=F(w,z).$$

(27)

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

$$g(z)=z\text{and}g(w)=w.$$

(28)

From (27) and (28),

$$z=g(z)=F(z,w)\text{and}w=g(w)=F(w,z).$$

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

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

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

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 $(x,y)$ such that $gx=F(x,y)=F(y,x)=gy$.

Proof Follow the proof of Theorem 2.1 step by step until constructing two sequences $\{x_n\}$ and $\{y_n\}$ in X such that $g{x}_n\to gx$ and $g{y}_n\to gy$, where $(x,y)$ is a coincidence point of F and g. Suppose $g{x}_0⪯g{y}_0$, then it is an easy matter to show that

$$g{x}_n⪯g{y}_n\text{and }\mathrm{\forall }n\in \mathbb{N}\cup \{0\}.$$

Thus, by (5) we have

$$\begin{array}{r}d(g{x}_n,g{y}_n)\\ =d(F(x_{n-1},y_{n-1}),F(y_{n-1},x_{n-1}))\\ \le \alpha max\{d(g(x_{n-1}),g(y_{n-1}),d(F(x_{n-1},y_{n-1}),g(x_{n-1})),d(F(y_{n-1},x_{n-1}),g(y_{n-1}))\}\\ =\alpha max\{d(g(x_{n-1}),g(y_{n-1})),d(g(x_n),g(x_{n-1})),d(g(y_n),g(y_{n-1}))\}.\end{array}$$

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

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

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 $(x,x)$.

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

$$d(x,y)=\{\begin{array}{ll}max\{x,y\}& \text{if }x\ne y;\\ 0& \text{if }x=y.\end{array}$$

Define $g:X\to X$ by $g(x)=x^2$ and $F:X\times X\to X$ by

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

Then

• (1) $g(X)$ is a complete subset of X.

• (2) $F(X\times X)\subseteq g(X)$.

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

• (4) F has the mixed g-monotone property.

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

  $$\begin{array}{r}d(F(x,y),F(u,v))\\ \le \frac{3}{4}max\{d(g(x),g(u)),d(g(y),g(v)),d(F(x,y),g(x)),d(F(u,v),g(u))\}\\ +Lmin\{d(F(x,y),g(u)),d(F(u,v),g(x))\}\end{array}$$

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

Thus, by Theorem 2.1, F has a coupled fixed point. Moreover, $(0,0)$ 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 $g(x)\le g(y)$ and $g(u)\le g(v)$, then $x\le y$ and $u\le v$. Hence

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

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

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

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

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

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

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

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

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

 □

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 $(0,0)$ is a coupled coincidence point of F and g.

Remarks

• (1) Theorem 1.1 is a special case of Corollary 2.1.

• (2) Theorem 1.2 is a special case of Corollary 2.1.

• (3) Theorem 1.3 is a special case of Theorem 2.1.

• (4) Theorem 1.4 is a special case of Theorem 2.2.