Coupled common fixed point theorems for mixed weakly monotone mappings in partially ordered metric spaces

Abstract

In this paper, we introduce the concept of a mixed weakly monotone pair of mappings and prove some coupled common fixed point theorems for a contractive-type mappings with the mixed weakly monotone property in partially ordered metric spaces. Our results are generalizations of the main results of Bhaskar and Lakshmikantham and Kadelburg et al.

Mathematics Subject Classification 2000: 54H25.

1. Introduction

In 1922, Banach gave a theorem, which is well-known as Banach's Fixed Point Theorem (or Banach's Contractive Principle) to establish the existence of solutions for nonlinear operator equations and integral equations. Since then, because of their simplicity and usefulness, it has become a very popular tools in solving the existence problems in many branches of mathematical analysis. Since then, many authors have extended, improved and generalized Banach's theorem in several ways [1–11].

Recently, the existence of coupled fixed points for some kinds of contractive-type mappings in partially ordered metric spaces, (ordered) cone metric spaces, fuzzy metric spaces and other spaces with applications has been investigated by some authors, for example, Bhaskar and Lakshmikantham [5], Cho et al. [12–14], Dhage et al. [15], Gordji et al. [16, 17], Kadelburg et al. [18], Nieto and Lopez [10], Ran and Rarings [11], Sintunavarat et al. [19, 20], Yang et al. [21] and others.

Especially, in [5], Bhaskar and Lakshmikantham introduced the notions of a mixed monotone mapping and a coupled fixed point and proved some coupled fixed point theorems for mixed monotone mappings and discussed the existence and uniqueness of solution for periodic boundary value problems.

Definition 1.1. [5] Let (X, ≤) be a partially ordered set and f: X × X → X be a mapping. We say that f has the mixed monotone property on X if, for any x, y ∈ X,

x_1,x_2\in X,x_1\le x_2\Rightarrow f\left(x_1,y\right)\le f\left(x_2,y\right)

and

y_1,y_2\in X,y_1\le y_2\Rightarrow f\left(x,y_1\right)\ge f\left(x,y_2\right).

Definition 1.2. [5] An element (x, y) ∈ X × X is called a coupled fixed point of a mapping F: X × X → X if x = F (x, y) and y = F (y, x).

Theorem 1.3. [5] Let (X, ≤, d) be a partially ordered complete metric space. Let f: X × X → X be a mapping having the mixed monotone property on X. Assume that there exists k ∈ [0, 1) with

\mathsf{\text{d}}\left(f\left(x,y\right),f\left(u,v\right)\right)\le \frac{k}{2}\left(d\left(x,u\right)+\mathsf{\text{d}}\left(y,v\right)\right)

for all x, y, u, v ∈ X with x ≤ u and y ≥ v. Also, suppose that either

• (1) f is continuous or

• (2) X has the following properties:

• (a) if {x _n } is an increasing sequence with x _n → x, then x _n ≤ x for all n ≥ 1;

• (b) if {y _n } is a decreasing sequence y _n → y, then y _n ≥ y for all n ≥ 1.

If there exist x₀, y₀ ∈ X such that x₀ ≤ f(x₀, y₀) and y₀ ≥ f(y₀, x₀), then f has a coupled fixed point in X.

Very recently, Kadelburg et al. [18] proved the following theorem on cone metric spaces.

Theorem 1.4. [18] Let (X, ≤, d) be an ordered cone metric space. Let (f, g) be a weakly increasing pair of self-mappings on X with respect to ≤. Suppose that the following conditions hold:

• (1) there exist p, q, r, s, t ≥ 0 satisfying p + q + r + s + t < 1 and q = r or s = t such that

  \mathsf{\text{d}}\left(fx,gy\right)\le p\mathsf{\text{d}}\left(x,y\right)+q\mathsf{\text{d}}\left(x,fx\right)+s\mathsf{\text{d}}\left(x,gy\right)+t\mathsf{\text{d}}\left(y,fx\right)

for all comparable x, y ∈ X;

• (2) f or g is continuous or, if a nondecreasing {x _n } converges to a point x ∈ X, then x _n ≤ x for all n ≥ 1.

Then f and g have a common fixed point in X.

Note that a pair (f, g) of self-mappings on partially ordered set (X, ≤) is said to be weakly increasing if fx ≤ gfx and gx ≤ fgx for all x ∈ X.

Now, we introduce the following concept of the mixed weakly increasing property of mappings.

Definition 1.5. Let (X, ≤) be a partially ordered set and f, g: X × X → X be mappings. We say that a pair (f, g) has the mixed weakly monotone property on X if, for any x, y ∈ X,

\begin{array}{cc}\hfill x& \le f\left(x,y\right),y\ge f\left(y,x\right)\hfill \\ \Rightarrow f\left(x,y\right)\le g\left(f\left(x,y\right),f\left(y,x\right)\right),f\left(y,x\right)\ge g\left(f\left(y,x\right),f\left(x,y\right)\right)\hfill \end{array}

and

\begin{array}{cc}\hfill x& \le g\left(x,y\right),y\ge g\left(y,x\right)\hfill \\ \Rightarrow g\left(x,y\right)\le f\left(g\left(x,y\right),g\left(y,x\right)\right),g\left(y,x\right)\ge f\left(g\left(y,x\right),g\left(x,y\right)\right).\hfill \end{array}

Example 1.6. Consider an ordered cone metric space (ℝ, ≤, d), where ≤ represents the usual order relation and d is a usual metric on ℝ and let f, g: ℝ × ℝ → ℝ be two functions defined by

f\left(x,y\right)=x-2y,g\left(x,y\right)=x-y.

Then a pair (f, g) has the mixed weakly monotone property.

Example 1.7. Consider an ordered cone metric space (ℝ, ≤, d), where ≤ represents the usual order relation and d is a usual metric on ℝ and let f, g: ℝ × ℝ → ℝ be two functions defined by

f\left(x,y\right)=x-y+1,g\left(x,y\right)=2x-3y.

Then both mappings f and g have the mixed monotone property, but a pair (f, g) has not the mixed weakly monotone property. To see this, for any \left(\frac{9}{8},\frac{7}{8}\right)\in {ℝ}^2, we have

\frac{9}{8}\le f\left(\frac{9}{8},\frac{7}{8}\right),\frac{7}{8}\ge f\left(\frac{7}{8},\frac{9}{8}\right),

but

f\left(\frac{9}{8},\frac{7}{8}\right)\nleqq g\left(f\left(\frac{9}{8},\frac{7}{8}\right.\right),f\left(\left.\frac{7}{8},\frac{9}{8}\right)\right),f\left(\frac{7}{8},\frac{9}{8}\right)\ge g\left(f\left(\frac{7}{8},\frac{9}{8}\right.\right),f\left(\left.\frac{9}{8},\frac{7}{8}\right)\right).

The purpose of this paper is to present some coupled common fixed point theorems for a pair of mappings with the mixed weakly monotone property in a partially ordered metric space. Our results generalize the main results of Bhaskar and Lakshmikantham [5], Kadelburg et al. [18] and others.

2. Coupled common fixed point theorems

Let (X, ≤, d) be a partially ordered complete metric space. Now, we consider the product space X × X with following partial order: for all (x, y), (u, v) ∈ X × X,

\left(x,y\right)\le \left(u,v\right)\iff x\le u,y\ge v.

Also, let (X × X, D) be a metric space with the following metric:

D\left(\left(x,y\right),\left(u,v\right)\right):=\mathsf{\text{d}}\left(x,u\right)+\mathsf{\text{d}}\left(y,v\right)

for all (x, y), (u, v) ∈ X × X.

Theorem 2.1. Let (X, ≤, d) be a partially ordered complete metric space. Let f, g: X × X → X be the mappings such that a pair (f, g) has the mixed weakly monotone property on X. Suppose that there exist p, q, r, s ≥ 0 with p + q + r + 2s < 1 such that

\begin{array}{cc}\hfill d\left(f\left(x,y\right),g\left(u,v\right)\right)& \le \frac{p}{2}D\left(\left(x,y\right),\left(u,v\right)\right)+\frac{q}{2}D\left(\left(x,y\right),\left(f\left(x,y\right),f\left(y,x\right)\right)\right)\hfill \\ +\frac{r}{2}D\left(\left(u,v\right),\left(g\left(u,v\right),g\left(v,u\right)\right)\right)+\frac{s}{2}D\left(\left(x,y\right),\left(g\left(u,v\right),g\left(v,u\right)\right)\right)\hfill \\ +\frac{s}{2}D\left(\left(u,v\right),\left(f\left(x,y\right),f\left(y,x\right)\right)\right)\hfill \end{array}

(2.1)

for all x, y, u, v ∈ X with x ≤ u and y ≥ v. Let x₀, y₀ ∈ X be such that x₀ ≤ f(x₀, y₀), y₀ ≥ f(y₀, x₀) or x₀ ≤ g(x₀, y₀), y₀ ≥ g(y₀, x₀). If f or g is continuous, then f and g have a coupled common fixed point in X.

Proof. Suppose that x₀ ≤ f(x₀, y₀) and y₀ ≥ f (y₀, x₀) and let

f\left(x_0,y_0\right)=x_1,f\left(y_0,x_0\right)=y_1.

From the mixed weakly monotone property of the pair (f, g), we have

x_1=f\left(x_0,y_0\right)\le g\left(f\left(x_0,y_0\right),f\left(y_0,x_0\right)\right)=g\left(x_1,y_1\right)

and

y_1=f\left(y_0,x_0\right)\ge g\left(f\left(y_0,x_0\right),f\left(x_0,y_0\right)\right)=g\left(y_1,x_1\right).

Let

g\left(x_1,y_1\right)=x_2,g\left(y_1,x_1\right)=y_2.

Then we have

g\left(x_1,y_1\right)\le f\left(g\left(x_1,y_1\right),g\left(y_1,x_1\right)\right)=f\left(x_2,y_2\right)

and

g\left(y_1,x_1\right)\ge f\left(g\left(y_1,x_1\right),g\left(x_1,y_1\right)\right)=f\left(y_2,x_2\right).

Continuously, let

x_{2n+1}=f\left(x_{2n},y_{2n}\right),y_{2n+1}=f\left(y_{2n},x_{2n}\right)

and

x_{2n+2}=g\left(x_{2n+1},y_{2n+1}\right),y_{2n+2}=g\left(y_{2n+1},x_{2n+1}\right)

for all n ≥ 1. Then we can easily verify that

x_0\le x_1\le x_2\le \cdots \le x_n\le x_{n+1}\le \cdots

and

y_0\ge y_1\ge y_2\ge \cdots \ge y_n\ge y_{n+1}\ge \cdots .

Similarly, from the condition x₀ ≤ g(x₀, y₀) and y₀ ≥ g(y₀, x₀), one can show that the sequences {x _n } and {y _n } are increasing and decreasing, respectively. Thus, applying (2.1), we obtain

\begin{array}{c}\mathsf{\text{d}}\left(x_{2n+1},x_{2n+2}\right)\hfill \\ =\mathsf{\text{d}}\left(f\left(x_{2n},y_{2n}\right),g\left(x_{2n+1},y_{2n+1}\right)\right)\hfill \\ \le \frac{p}{2}D\left(\left(x_{2n},y_{2n}\right),\left(x_{2n+1},y_{2n+1}\right)\right)+\frac{q}{2}D\left(\left(x_{2n},y_{2n}\right),\left(f\left(x_{2n},y_{2n}\right),f\left(y_{2n},x_{2n}\right)\right)\right)\hfill \\ +\frac{r}{2}D\left(\left(x_{2n+1},y_{2n+1}\right),\left(g\left(x_{2n+1},y_{2n+1}\right),g\left(y_{2n+1},x_{2n+1}\right)\right)\right)\hfill \\ +\frac{s}{2}D\left(\left(x_{2n},y_{2n}\right),\left(g\left(x_{2n+1},y_{2n+1}\right),g\left(y_{2n+1},x_{2n+1}\right)\right)\right)\hfill \\ +\frac{s}{2}D\left(\left(x_{2n+1},y_{2n+1}\right),\left(f\left(x_{2n},y_{2n}\right),f\left(y_{2n},x_{2n}\right)\right)\right)\hfill \\ =\frac{p}{2}D\left(\left(x_{2n},y_{2n}\right),\left(x_{2n+1},y_{2n+1}\right)\right)+\frac{q}{2}D\left(\left(x_{2n},y_{2n}\right),\left(x_{2n+1},y_{2n+1}\right)\right)\hfill \\ +\frac{r}{2}D\left(\left(x_{2n+1},y_{2n+1}\right),\left(x_{2n+2},y_{2n+2}\right)\right)+\frac{s}{2}D\left(\left(x_{2n+1},y_{2n+1}\right),\left(x_{2n+1},y_{2n+1}\right)\right)\hfill \\ +\frac{s}{2}D\left(\left(x_{2n},y_{2n}\right),\left(x_{2n+2},y_{2n+2}\right)\right)\hfill \\ \le \frac{p+q}{2}D\left(\left(x_{2n},y_{2n}\right),\left(x_{2n+1},y_{2n+1}\right)\right)+\frac{r}{2}D\left(\left(x_{2n+1},y_{2n+1}\right),\left(x_{2n+2},y_{2n+2}\right)\right)\hfill \\ +\frac{s}{2}\left[D\left(\left(x_{2n},y_{2n}\right),\left(x_{2n+1},y_{2n+1}\right)\right)+D\left(\left(x_{2n+1},y_{2n+1}\right),\left(x_{2n+2},y_{2n+2}\right)\right)\right]\hfill \\ =\frac{p+q+s}{2}D\left(\left(x_{2n},y_{2n}\right),\left(x_{2n+1},y_{2n+1}\right)\right)+\frac{r+s}{2}D\left(\left(x_{2n+1},y_{2n+1}\right),\left(x_{2n+2},y_{2n+2}\right)\right).\hfill \end{array}

Hence it follows that

\begin{array}{cc}\hfill \mathsf{\text{d}}\left(x_{2n+1},x_{2n+2}\right)& \le \frac{p+q+s}{2}\left(\mathsf{\text{d}}\left(x_{2n},x_{2n+1}\right)+\mathsf{\text{d}}\left(y_{2n},y_{2n+1}\right)\right)\hfill \\ +\frac{r+s}{2}\left(\mathsf{\text{d}}\left(x_{2n+1},x_{2n+2}\right)+\mathsf{\text{d}}\left(y_{2n+1},y_{2n+2}\right)\right)\hfill \end{array}

(2.2)

for all n ≥ 1. Similarly, we have

\begin{array}{cc}\hfill \mathsf{\text{d}}\left(y_{2n+1},y_{2n+2}\right)& \le \frac{p+q+s}{2}\left(\mathsf{\text{d}}\left(y_{2n},y_{2n+1}\right)+\mathsf{\text{d}}\left(x_{2n},x_{2n+1}\right)\right)\hfill \\ +\frac{r+s}{2}\left(\mathsf{\text{d}}\left(y_{2n+1},y_{2n+2}\right)+\mathsf{\text{d}}\left(x_{2n+1},x_{2n+2}\right)\right)\hfill \end{array}

(2.3)

for all n ≥ 1. Thus it follows from (2.2) and (2.3) that

\mathsf{\text{d}}\left(x_{2n+1},x_{2n+2}\right)+\mathsf{\text{d}}\left(y_{2n+1},y_{2n+2}\right)\le \frac{p+q+s}{1-\left(r+s\right)}\left(\left(\mathsf{\text{d}}\left(x_{2n},x_{2n+1}\right)+\mathsf{\text{d}}\left(y_{2n},y_{2n+1}\right)\right)\right.

(2.4)

for all n ≥ 1. Moreover, if we apply (2.1), then we have

\begin{array}{c}\mathsf{\text{d}}\left(x_{2n+2},x_{2n+3}\right)\hfill \\ =\mathsf{\text{d}}\left(g\left(x_{2n+1},y_{2n+1}\right),f\left(x_{2n+2},y_{2n+2}\right)\right)\hfill \\ \le \frac{p}{2}D\left(\left(x_{2n+1},y_{2n+1}\right),\left(x_{2n+2},y_{2n+2}\right)\right)\hfill \\ +\frac{q}{2}D\left(\left(x_{2n+1},y_{2n+1}\right),\left(g\left(x_{2n+1},y_{2n+1}\right),g\left(y_{2n+1},x_{2n+1}\right)\right)\right)\hfill \\ +\frac{r}{2}D\left(\left(x_{2n+2},y_{2n+2}\right),\left(f\left(x_{2n+2},y_{2n+2}\right),f\left(y_{2n+2},x_{2n+2}\right)\right)\right)\hfill \\ +\frac{s}{2}D\left(\left(x_{2n+1},y_{2n+1}\right),\left(f\left(x_{2n+2},y_{2n+2}\right),f\left(y_{2n+2},x_{2n+2}\right)\right)\right)\hfill \\ +\frac{s}{2}D\left(\left(x_{2n+2},y_{2n+2}\right),\left(g\left(x_{2n+1},y_{2n+1}\right),g\left(y_{2n+1},x_{2n+1}\right)\right)\right)\hfill \\ =\frac{p}{2}D\left(\left(x_{2n+1},y_{2n+1}\right),\left(x_{2n+2},y_{2n+2}\right)\right)+\frac{q}{2}D\left(\left(x_{2n+1},y_{2n+1}\right),\left(x_{2n+2},y_{2n+2}\right)\right)\hfill \\ +\frac{r}{2}D\left(\left(x_{2n+2},y_{2n+2}\right),\left(x_{2n+3},y_{2n+3}\right)\right)+\frac{s}{2}D\left(\left(x_{2n+1},y_{2n+1}\right),\left(x_{2n+3},y_{2n+3}\right)\right)\hfill \\ +\frac{s}{2}D\left(\left(x_{2n+2},y_{2n+2}\right),\left(x_{2n+2},y_{2n+2}\right)\right)\hfill \\ \le \frac{p+q}{2}D\left(\left(x_{2n+1},y_{2n+1}\right),\left(x_{2n+2},y_{2n+2}\right)\right)+\frac{r}{2}D\left(\left(x_{2n+2},y_{2n+2}\right),\left(x_{2n+3},y_{2n+3}\right)\right)\hfill \\ +\frac{s}{2}\left[D\left(\left(x_{2n+1},y_{2n+1}\right),\left(x_{2n+2},y_{2n+2}\right)\right)+D\left(\left(x_{2n+2},y_{2n+2}\right),\left(x_{2n+3},y_{2n+3}\right)\right)\right].\hfill \end{array}

Hence it follows that

\begin{array}{cc}\hfill \mathsf{\text{d}}\left(x_{2n+2},x_{2n+3}\right)& \le \frac{p+q+s}{2}\left(\mathsf{\text{d}}\left(x_{2n+1},x_{2n+2}\right)+\mathsf{\text{d}}\left(y_{2n+1},y_{2n+2}\right)\right)\hfill \\ +\frac{r+s}{2}\left(\mathsf{\text{d}}\left(x_{2n+2},x_{2n+2}\right)+\mathsf{\text{d}}\left(y_{2n+3},y_{2n+3}\right)\right)\hfill \end{array}

(2.5)

for all n ≥ 1. Similarly, we have

\begin{array}{cc}\hfill \mathsf{\text{d}}\left(y_{2n+2},y_{2n+3}\right)& \le \frac{p+q+s}{2}\left(\mathsf{\text{d}}\left(y_{2n+1},y_{2n+2}\right)+\mathsf{\text{d}}\left(x_{2n+1},x_{2n+2}\right)\right)\hfill \\ +\frac{r+s}{2}\left(\mathsf{\text{d}}\left(y_{2n+2},y_{2n+2}\right)+\mathsf{\text{d}}\left(x_{2n+3},x_{2n+3}\right)\right).\hfill \end{array}

(2.6)

Thus, using (2.5) and (2.6), we have

\mathsf{\text{d}}\left(x_{2n+2},x_{2n+3}\right)+\mathsf{\text{d}}\left(y_{2n+2},y_{2n+3}\right)\le \frac{p+q+s}{1-\left(r+s\right)}\left(\mathsf{\text{d}}\left(x_{2n+1},x_{2n+2}\right)+\mathsf{\text{d}}\left(y_{2n+1},y_{2n+2}\right)\right)

(2.7)

for all n ≥ 1. Also, it follows from (2.4) and (2.7) that

\mathsf{\text{d}}\left(x_{2n+2},x_{2n+3}\right)+\mathsf{\text{d}}\left(y_{2n+2},y_{2n+3}\right)\le {\left(\frac{p+q+s}{1-\left(r+s\right)}\right)}^2\left(\mathsf{\text{d}}\left(x_{2n},x_{2n+1}\right)+\mathsf{\text{d}}\left(y_{2n},y_{2n+1}\right)\right)

(2.8)

for all n ≥ 1. Let A=\frac{p+q+s}{1-\left(r+s\right)}. Then 0 ≤ A < 1 and

\begin{array}{cc}\hfill \mathsf{\text{d}}\left(x_{2n+1},x_{2n+2}\right)+\mathsf{\text{d}}\left(y_{2n+1},y_{2n+2}\right)& \le A\left(\mathsf{\text{d}}\left(x_{2n},x_{2n+1}\right)+\mathsf{\text{d}}\left(y_{2n},y_{2n+1}\right)\right)\hfill \\ \le A^3\left(\mathsf{\text{d}}\left(x_{2n-2},x_{2n-1}\right)+\mathsf{\text{d}}\left(y_{2n-2},y_{2n-1}\right)\right)\hfill \\ \le A^5\left(\mathsf{\text{d}}\left(x_{2n-4},x_{2n-3}\right)+\mathsf{\text{d}}\left(y_{2n-4},y_{2n-3}\right)\right)\hfill \\ \le \cdots \hfill \\ \le A^{2n+1}\left(\mathsf{\text{d}}\left(x_0,x_1\right)+\mathsf{\text{d}}\left(y_0,y_1\right)\right)\hfill \end{array}

and

\begin{array}{cc}\hfill \mathsf{\text{d}}\left(x_{2n+2},x_{2n+3}\right)+\mathsf{\text{d}}\left(y_{2n+2},y_{2n+3}\right)& \le A^2\left(\mathsf{\text{d}}\left(x_{2n},x_{2n+1}\right)+\mathsf{\text{d}}\left(y_{2n},y_{2n+1}\right)\right)\hfill \\ \le A^4\left(\mathsf{\text{d}}\left(x_{2n-2},x_{2n-1}\right)+\mathsf{\text{d}}\left(y_{2n-2},y_{2n-1}\right)\right)\hfill \\ \le A^6\left(\mathsf{\text{d}}\left(x_{2n-4},x_{2n-3}\right)+\mathsf{\text{d}}\left(y_{2n-4},y_{2n-3}\right)\right)\hfill \\ \le \cdots \hfill \\ \le A^{2n+2}\left(\mathsf{\text{d}}\left(x_0,x_1\right)+\mathsf{\text{d}}\left(y_0,y_1\right)\right)\hfill \end{array}

for all n ≥ 1. Now, for all m, n ≥ 1 with n ≤ m, we have

\begin{array}{c}\mathsf{\text{d}}\left(x_{2n+1},x_{2m+1}\right)+\mathsf{\text{d}}\left(y_{2n+1},y_{2m+1}\right)\hfill \\ \le \left(\mathsf{\text{d}}\left(x_{2n+1},x_{2n+2}\right)+\mathsf{\text{d}}\left(y_{2n+1},y_{2n+2}\right)\right)+\left(\mathsf{\text{d}}\left(x_{2n+2},x_{2n+3}\right)+\mathsf{\text{d}}\left(y_{2n+2},y_{2n+3}\right)\right)\hfill \\ +\cdots \hfill \\ +\left(\mathsf{\text{d}}\left(x_{2m},x_{2m+1}\right)+\mathsf{\text{d}}\left(y_{2m},y_{2m+1}\right)\right)\hfill \\ \le \left(A^{2n+1}+A^{2n+2}+\cdots +A^{2m}\right)\left(\mathsf{\text{d}}\left(x_0,x_1\right)+\mathsf{\text{d}}\left(y_0,y_1\right)\right)\hfill \\ \le \frac{A^{2n+1}}{1-A}\left(\mathsf{\text{d}}\left(x_0,x_1\right)+\mathsf{\text{d}}\left(y_0,y_1\right)\right).\hfill \end{array}

Similarly, we have

\begin{array}{c}\mathsf{\text{d}}\left(x_{2n},x_{2m+1}\right)+\mathsf{\text{d}}\left(y_{2n},y_{2m+1}\right)\hfill \\ \le \left(A^{2n}+A^{2n+1}+A^{2n+2}+\cdots +A^{2m}\right)\left(\mathsf{\text{d}}\left(x_0,x_1\right)+\mathsf{\text{d}}\left(y_0,y_1\right)\right)\hfill \\ \le \frac{A^{2n}}{1-A}\left(\mathsf{\text{d}}\left(x_0,x_1\right)+\mathsf{\text{d}}\left(y_0,y_1\right)\right),\hfill \\ \mathsf{\text{d}}\left(x_{2n},x_{2m}\right)+\mathsf{\text{d}}\left(y_{2n},y_{2m}\right)\hfill \\ \le \left(A^{2n}+A^{2n+1}+A^{2n+2}+\cdots +A^{2m-1}\right)\left(\mathsf{\text{d}}\left(x_0,x_1\right)+\mathsf{\text{d}}\left(y_0,y_1\right)\right)\hfill \\ \le \frac{A^{2n}}{1-A}\left(\mathsf{\text{d}}\left(x_0,x_1\right)+\mathsf{\text{d}}\left(y_0,y_1\right)\right)\hfill \end{array}

and

\begin{array}{c}\mathsf{\text{d}}\left(x_{2n+1},x_{2m}\right)+\mathsf{\text{d}}\left(y_{2n+1},y_{2m}\right)\hfill \\ \le \left(A^{2n+1}+A^{2n+1}+A^{2n+2}+\cdots +A^{2m-1}\right)\left(\mathsf{\text{d}}\left(x_0,x_1\right)+\mathsf{\text{d}}\left(y_0,y_1\right)\right)\hfill \\ \le \frac{A^{2n+1}}{1-A}\left(\mathsf{\text{d}}\left(x_0,x_1\right)+\mathsf{\text{d}}\left(y_0,y_1\right)\right).\hfill \end{array}

Hence, for all m, n ≥ 1 with n ≤ m, it follows that

\mathsf{\text{d}}\left(x_n,x_m\right)+\mathsf{\text{d}}\left(y_n,y_m\right)\le \frac{A^{2n}}{1-A}\left(\mathsf{\text{d}}\left(x_0,x_1\right)+\mathsf{\text{d}}\left(y_0,y_1\right)\right)

and so, since 0 ≤ A < 1, we can conclude that

\mathsf{\text{d}}\left(x_n,x_m\right)+\mathsf{\text{d}}\left(y_n,y_m\right)\to 0

as n → ∞, which implies that d(x _n , x _m ) → 0 and d(y _n , y _m ) → 0 as m, n → ∞. Therefore, the sequences {x _n } and {y _n } are Cauchy sequences in X. Since (X, d) is a complete metric space, then there exist x, y ∈ X such that x _n → x and y _n → y as n → ∞.

Suppose that f is a continuous. Then we have

x=\underset{k\to \infty }{\mathsf{\text{lim}}}{x}_{2k+1}=\underset{k\to \infty }{\mathsf{\text{lim}}}f\left(x_{2k},y_{2k}\right)=f\left(\underset{k\to \infty }{\mathsf{\text{lim}}}{x}_{2k},\underset{k\to \infty }{\mathsf{\text{lim}}}{y}_{2k}\right)=f\left(x,y\right)

and

y=\underset{k\to \infty }{\mathsf{\text{lim}}}{y}_{2k+1}=\underset{k\to \infty }{\mathsf{\text{lim}}}f\left(y_{2k},x_{2k}\right)=f\left(\underset{k\to \infty }{\mathsf{\text{lim}}}{y}_{2k},\underset{k\to \infty }{\mathsf{\text{lim}}}x2k\right)=f\left(y,x\right).

Taking x = u and y = v in (2.1), we have

\begin{array}{c}\mathsf{\text{d}}\left(f\left(x,y\right),g\left(x,y\right)\right)+\mathsf{\text{d}}\left(f\left(y,x\right),g\left(y,x\right)\right)\hfill \\ \le \frac{p}{2}D\left(\left(x,y\right),\left(x,y\right)\right)+\frac{q}{2}D\left(\left(x,y\right),f\left(x,y\right),f\left(y,x\right)\right)\hfill \\ +\frac{r}{2}D\left(\left(x,y\right),g\left(x,y\right),g\left(y,x\right)\right)+\frac{s}{2}D\left(\left(x,y\right),g\left(x,y\right),g\left(y,x\right)\right)\hfill \\ +\frac{s}{2}D\left(\left(x,y\right),f\left(x,y\right),f\left(y,x\right)\right)\frac{p}{2}D\left(\left(y,x\right),\left(y,x\right)\right)\hfill \\ +\frac{q}{2}D\left(\left(y,x\right),f\left(y,x\right),f\left(x,y\right)\right)+\frac{r}{2}D\left(\left(y,x\right),g\left(y,x\right),g\left(x,y\right)\right)\hfill \\ +\frac{s}{2}D\left(\left(y,x\right),g\left(y,x\right),g\left(x,y\right)\right)+\frac{s}{2}D\left(\left(y,x\right),f\left(y,x\right),f\left(x,y\right)\right).\hfill \end{array}

Hence we have

\mathsf{\text{d}}\left(x,g\left(x,y\right)\right)+\mathsf{\text{d}}\left(y,g\left(y,x\right)\right)\le \left(r+s\right)\left(\mathsf{\text{d}}\left(x,g\left(x,y\right)\right)+\mathsf{\text{d}}\left(y,g\left(y,x\right)\right)\right)

and so, since r + s < 1, we can get that

\mathsf{\text{d}}\left(x,g\left(x,y\right)\right)=0,\mathsf{\text{d}}\left(y,g\left(y,x\right)\right)=0.

Hence (x, y) is a coupled common fixed point of f and g.

Similarly, we can prove that (x, y) is a coupled common fixed point of f and g when g is a continuous mapping. This completes the proof.   □

Theorem 2.2. Let (X, ≤, d) be a partially ordered complete metric space. Assume that X has the following property:

• (1) if {x _n } is an increasing sequence with x _n → x, then x _n ≤ x for all n ≥ 1;

• (2) if {y _n } is a decreasing sequence with y _n → y, then y _n ≥ y for all n ≥ 1.

Let f, g: X × X → X be the mappings such that a pair (f, g) has the mixed weakly monotone property on X. Also, suppose that there exist p, q, r, s ≥ 0 with p + q + r + 2s < 1 such that

\begin{array}{cc}\hfill \mathsf{\text{d}}\left(f\left(x,y\right),g\left(u,v\right)\right)& \le \frac{p}{2}D\left(\left(x,y\right),\left(u,v\right)\right)+\frac{q}{2}D\left(\left(x,y\right),\left(f\left(x,y\right),f\left(y,x\right)\right)\right)\hfill \\ +\frac{r}{2}D\left(\left(u,v\right),\left(g\left(u,v\right),g\left(v,u\right)\right)\right)+\frac{s}{2}D\left(\left(x,y\right),\left(g\left(u,v\right),g\left(v,u\right)\right)\right)\hfill \\ +\frac{s}{2}D\left(\left(u,v\right),\left(f\left(x,y\right),f\left(y,x\right)\right)\right)\hfill \end{array}

for all x, y, u, v ∈ X with x ≤ u and y ≥ v. If there exist x₀, y₀ ∈ X such that x₀ ≤ f(x₀, y₀), y₀ ≥ f(y₀, x₀) or x₀ ≤ g(x₀, y₀), y₀ ≥ g(y₀, x₀), then f and g have a coupled common fixed point in X.

Proof. Following the proof of Theorem 2.1, we only have to show that

f\left(x,y\right)=g\left(x,y\right)=x,f\left(y,x\right)=g\left(y,x\right)=y.

It is clear that

\begin{array}{c}D\left(\left(x,y\right),\left(f\left(x,y\right),f\left(y,x\right)\right)\right)\hfill \\ \le D\left(\left(x,y\right),\left(x_{2k+2},y_{2k+2}\right)\right)+D\left(\left(x_{2k+2},y_{2k+2}\right),\left(f\left(x,y\right),f\left(y,x\right)\right)\right)\hfill \\ =D\left(\left(x,y\right),\left(x_{2k+2},y_{2k+2}\right)\right)+D\left(\left(g\left(x_{2k+1},y_{2k+1}\right),g\left(y_{2k+1},x_{2k+1}\right)\right),\left(f\left(x,y\right),f\left(y,x\right)\right)\right)\hfill \\ =D\left(\left(x,y\right),\left(x_{2k+2},y_{2k+2}\right)\right)+\mathsf{\text{d}}\left(g\left(x_{2k+1},y_{2k+1}\right),f\left(x,y\right)\right)+\mathsf{\text{d}}\left(f\left(y,x\right),g\left(y_{2k+1},x_{2k+1}\right)\right)\hfill \\ \le D\left(\left(x,y\right),\left(x_{2k+2},y_{2k+2}\right)\right)+\frac{p}{2}D\left(\left(x_{2k+1},y_{2k+1}\right),\left(x,y\right)\right)\hfill \\ +\frac{q}{2}D\left(\left(x_{2k+1},y_{2k+1}\right),\left(g\left(x_{2k+1},y_{2k+1}\right),g\left(y_{2k+1},x_{2k+1}\right)\right)\right)+\frac{r}{2}D\left(\left(x,y\right),\left(f\left(x,y\right),f\left(y,x\right)\right)\right)\hfill \\ +\frac{s}{2}D\left(\left(x_{2k+1},y_{2k+1}\right),\left(f\left(x,y\right),f\left(y,x\right)\right)\right)+\frac{s}{2}D\left(\left(x,y\right),\left(g\left(x_{2k+1},y_{2k+1}\right),g\left(y_{2k+1},x_{2k+1}\right)\right)\right)\hfill \\ +\frac{p}{2}D\left(\left(y,x\right),\left(y_{2k+1},x_{2k+1}\right)\right)+\frac{q}{2}D\left(\left(y,x\right),\left(f\left(y,x\right),f\left(x,y\right)\right)\right)\hfill \\ +\frac{r}{2}D\left(\left(y_{2k+1},x_{2k+1}\right),\left(g\left(y_{2k+1},x_{2k+1}\right),g\left(x_{2k+1},y_{2k+1}\right)\right)\right)\hfill \\ +\frac{s}{2}D\left(\left(y,x\right),\left(g\left(y_{2k+1},x_{2k+1}\right),g\left(x_{2k+1},y_{2k+1}\right)\right)\right)+\frac{s}{2}D\left(\left(y_{2k+1},x_{2k+1}\right),\left(f\left(y,x\right),f\left(x,y\right)\right)\right)\hfill \end{array}

and so

\begin{array}{c}\mathsf{\text{d}}\left(x,f\left(x,y\right)\right)+\mathsf{\text{d}}\left(y,f\left(y,x\right)\right)\hfill \\ \le \mathsf{\text{d}}\left(x,x_{2k+2}\right)+\mathsf{\text{d}}\left(y,y_{2k+2}\right)+p\left(\mathsf{\text{d}}\left(x_{2k+1},x\right)+\mathsf{\text{d}}\left(y_{2k+1},y\right)\right)\hfill \\ +\frac{q}{2}\left(\mathsf{\text{d}}\left(x_{2k+1},x_{2k+2}\right)+\mathsf{\text{d}}\left(y_{2k+1},y_{2k+2}\right)+\mathsf{\text{d}}\left(x,f\left(x,y\right)\right)+\mathsf{\text{d}}\left(y,f\left(y,x\right)\right)\right)\hfill \\ +\frac{r}{2}\left(\mathsf{\text{d}}\left(x,f\left(x,y\right)\right)+\mathsf{\text{d}}\left(y,f\left(y,x\right)\right)+\mathsf{\text{d}}\left(y_{2k+1},y_{2k+2}\right)+\mathsf{\text{d}}\left(x_{2k+1},x_{2k+2}\right)\right)\hfill \\ +s\left(\mathsf{\text{d}}\left(x_{2k+2},x\right)+\mathsf{\text{d}}\left(y_{2k+2},y\right)+\mathsf{\text{d}}\left(x_{2k+1},f\left(x,y\right)\right)+\mathsf{\text{d}}\left(y_{2k+1},f\left(y,x\right)\right)\right).\hfill \end{array}

(2.9)

Letting k → ∞ in (2.9), we obtain

\mathsf{\text{d}}\left(x,f\left(x,y\right)\right)+\mathsf{\text{d}}\left(y,f\left(y,x\right)\right)\le \frac{q+r+2s}{2}\left[\mathsf{\text{d}}\left(x,f\left(x,y\right)\right)+\mathsf{\text{d}}\left(y,f\left(y,x\right)\right)\right].

Since , we have

\mathsf{\text{d}}\left(x,f\left(x,y\right)\right)+\mathsf{\text{d}}\left(y,f\left(y,x\right)\right)=0

and so f(x, y) = x and f(y, x) = y. Similarly, we can show that g(x, y) = x and g(y, x) = y. Therefore, (x, y) is a coupled common fixed point of f and g. This completes the proof.   □

Now, we give an example to illustrate Theorem 2.1 as follows:

Example 2.3. Consider (ℝ, ≤, d), where ≤ represents the usual order relation and d is a usual metric on ℝ and let f, g: ℝ × ℝ → ℝ be two functions defined by

f\left(x,y\right)=\frac{6x-3y+33}{36},g\left(x,y\right)=\frac{8x-4y+44}{48}.

Then a pair (f, g) has the mixed weakly monotone property and

\begin{array}{cc}\hfill \mathsf{\text{d}}\left(f\left(x,y\right),g\left(u,v\right)\right)=|f\left(x,y\right)-g\left(u,v\right)|& =\left|\frac{6x-3y+33}{36}-\frac{8x-4y+44}{48}\right|\hfill \\ \le \frac{1}{6}|x-u|+\frac{1}{12}|y-v|\hfill \\ \le \frac{1}{6}\left(|x-u|+|y-v|\right).\hfill \end{array}

By putting p=\frac{1}{3} and q = r = s = 0 in (2.1), we see that (1, 1) is a unique coupled common fixed point of f and g.

Corollary 2.4. In Theorems 2.1 and 2.2, if X is a total ordered set, then a coupled common fixed point of f and g is unique and x = y.

Proof. If (x*, y*) ∈ X × X is another coupled common fixed point of f and g, then, by the use of (2.1), we have

\begin{array}{c}\mathsf{\text{d}}\left(x,x^{*}\right)+\mathsf{\text{d}}\left(y,y^{*}\right)\hfill \\ =\mathsf{\text{d}}\left(f\left(x,y\right),g\left(x^{*},y^{*}\right)\right)+\mathsf{\text{d}}\left(f\left(y,x\right),g\left(y^{*},x^{*}\right)\right)\hfill \\ \le \frac{p}{2}D\left(\left(x,y\right),\left(x^{*},y^{*}\right)\right)+\frac{q}{2}D\left(\left(x,y\right),\left(f\left(x,y\right),f\left(y,x\right)\right)\right)\hfill \\ +\frac{r}{2}D\left(\left(x^{*},y^{*}\right),\left(g\left(x^{*},y^{*}\right),g\left(y^{*},x^{*}\right)\right)\right)+\frac{s}{2}D\left(\left(x,y\right),\left(g\left(x^{*},y^{*}\right),g\left(y^{*},x^{*}\right)\right)\right)\hfill \\ +\frac{s}{2}D\left(\left(x^{*},y^{*}\right),\left(f\left(x,y\right),f\left(y,x\right)\right)\right)+\frac{p}{2}D\left(\left(y,x\right),\left(y^{*},x^{*}\right)\right)\hfill \\ +\frac{q}{2}D\left(\left(y,x\right),\left(f\left(y,x\right),f\left(x,y\right)\right)\right)+\frac{r}{2}D\left(\left(y^{*},x^{*}\right),\left(g\left(y^{*},x^{*}\right),g\left(x^{*},y^{*}\right)\right)\right)\hfill \\ +\frac{s}{2}D\left(\left(y,x\right),\left(g\left(y^{*},x^{*}\right),g\left(x^{*},y^{*}\right)\right)\right)+\frac{s}{2}D\left(\left(\left(y^{*},x^{*}\right),\left(f\left(y,x\right),f\left(x,y\right)\right)\right)\right.\hfill \\ =p\left(\mathsf{\text{d}}\left(x,x^{*}\right)\right)+\left.\mathsf{\text{d}}\left(y,y^{*}\right)\right)+2s\left(\mathsf{\text{d}}\left(x,x^{*}\right)\right)+\left.\mathsf{\text{d}}\left(y,y^{*}\right)\right)\hfill \end{array}

and hence

\mathsf{\text{d}}\left(x,x^{*}\right)+\mathsf{\text{d}}\left(y,y^{*}\right)=\left(p+2s\right)\left(\mathsf{\text{d}}\left(x,x^{*}\right)+\mathsf{\text{d}}\left(y,y^{*}\right)\right).

Since q + 2s < 1, we have d(x, x*) + d(y, y*) = 0, which implies that x = x* and y = y*.

On the other hand, we have

\begin{array}{cc}\hfill \mathsf{\text{d}}\left(x,y\right)& =\mathsf{\text{d}}\left(f\left(x,y\right),g\left(y,x\right)\right)\hfill \\ \le \frac{p}{2}D\left(\left(x,y\right),\left(y,x\right)\right)+sD\left(\left(x,y\right),\left(y,x\right)\right)\hfill \\ =\left(p+2s\right)\mathsf{\text{d}}\left(x,y\right).\hfill \end{array}

Since p + 2s < 1, we have d(x, y) = 0 and x = y. This completes the proof.   □

Let f: X × X → X be a mapping. Now, we denote

f^{n+1}\left(x,y\right)=f\left(f^n\left(x,y\right),f^n\left(y,x\right)\right)

for all x, y ∈ X and n ≥ 1.

Remark 2.5. Let (X, ≤, d) be a partially ordered complete metric space. Let f: X × X → X be a mapping with the mixed monotone property on X. Then, for each n ≥ 1, a pair (fⁿ, fⁿ) has the mixed weakly monotone property on X. In fact, let x ≤ fⁿ(x, y) and y ≤ fⁿ(y, x). Then it follows from the mixed monotone property of f that

\begin{array}{c}f\left(x,y\right)\le f\left(f^n\left(x,y\right),y\right)\le f\left(f^n\left(x,y\right),f^n\left(y,x\right)\right)=f^{n+1}\left(x,y\right),\\ f\left(y,x\right)\ge f\left(\left(f^n\left(y,x\right),x\right)\right.\ge f\left(f^n\left(y,x\right),f^n\left(x,y\right)\right)=f^{n+1}\left(y,x\right)\end{array}

and

\begin{array}{c}{f}^2\left(x,y\right)=f\left(f\left(x,y\right),f\left(y,x\right)\right)\le f\left(f^{n+1}\left(x,y\right),f^{n+1}\left(y,x\right)\right)=f^{n+2}\left(x,y\right),\\ f^2\left(y,x\right)=f\left(f\left(y,x\right),f\left(x,y\right)\right)\ge f\left(f^{n+1}\left(y,x\right),f^{n+1}\left(x,y\right)\right)=f^{n+2}\left(y,x\right).\end{array}

Continuously, we have

f^n\left(x,y\right)\le f^{n+n}\left(x,y\right),f^n\left(y,x\right)\ge f^{n+n}\left(y,x\right).

Hence we have

f^n\left(x,y\right)\le f^n\left(f^n\left(x,y\right),f^n\left(y,x\right)\right),f^n\left(y,x\right)\ge f^n\left(f^n\left(y,x\right),f^n\left(x,y\right)\right),

which implies that the pair (fⁿ, fⁿ) has the mixed weakly monotone property on X.

Corollary 2.6. Let (X, ≤, d) be a partially ordered complete metric space. Let f: X × X → X be a mapping with the mixed monotone property on X. Assume that there exist p, q, r, s ≥ 0 with p + q + r + 2s < 1 such that

\begin{array}{cc}\hfill \mathsf{\text{d}}\left(f\left(x,y\right),f\left(u,v\right)\right)& \le \frac{p}{2}D\left(\left(x,y\right),\left(u,v\right)\right)+\frac{q}{2}D\left(\left(x,y\right),\left(f\left(x,y\right),f\left(y,x\right)\right)\right)\hfill \\ +\frac{r}{2}D\left(\left(u,v\right),\left(f\left(u,v\right),f\left(v,u\right)\right)\right)+\frac{s}{2}D\left(\left(x,y\right),\left(f\left(u,v\right),f\left(v,u\right)\right)\right)\hfill \\ +\frac{s}{2}D\left(\left(u,v\right),\left(f\left(x,y\right),f\left(y,x\right)\right)\right)\hfill \end{array}

for all x, y, u, v ∈ X with x ≤ u and y ≥ v. Moreover, suppose that either

• (1) f is continuous or

• (2) X has the following properties:

• (a) if {x _n } is an increasing sequence with x _n → x, then x _n ≤ x for all n ≥ 1;

• (b) if {y _n } is a decreasing sequence with y _n → y, then y _n ≥ y for all n ≥ 1.

If there exist x₀, y₀ ∈ X such that x₀ ≤ f(x₀, y₀) and y₀ ≥ f(y₀, x₀), then f has a coupled fixed point in X.

Proof. Taking f = g in Theorems 2.1, 2.2 and using Remark 2.5, we can get the conclusion.   □

Corollary 2.7. Let (X, ≤, d) be a partially ordered complete metric space. Let f: X × X → X be a mapping with the mixed monotone property on X. Assume that there exists k ∈ [0, 1) with

\mathsf{\text{d}}\left(f\left(x,y\right),f\left(u,v\right)\right)\le \frac{k}{2}\left(\mathsf{\text{d}}\left(x,u\right)+\mathsf{\text{d}}\left(y,v\right)\right)

for all x, y, u, v ∈ X with x ≤ u and y ≥ v. Also, suppose that either

• (1) f is continuous or

• (2) X has the following properties:

• (a) if {x _n } is an increasing sequence with x _n → x, then x _n ≤ x for all n ≥ 1;

• (b) if {y _n } is a decreasing sequence with y _n → y, then y _n ≥ y for all n ≥ 1.

If there exist x₀, y₀ ∈ X such that x₀ ≤ f(x₀, y₀) and y₀ ≥ f(y₀, x₀), then f has a coupled fixed point in X.

Proof. Taking f = g, p = k and q = r = s = 0 in Theorems 2.1, 2.2 and using Remark 2.5, we can get the conclusion.   □