Coupled coincidence points for monotone operators in partially ordered metric spaces

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.

1 Introduction

The Banach contraction principle is the most celebrated fixed point theorem. Afterward many authors obtained many important extensions of this principle (cf. [1–16]). Recently Bhaskar and Lakshmikantham [5], Nieto and Lopez [12, 13], Ran and Reurings [14] and Agarwal et al. [3] presented some new results for contractions in partially ordered metric spaces. Bhaskar and Lakshmikantham [5] 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 [11] 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 [5]. 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 : X → X 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 [9] for more details on fixed point theory.

Definition 2.1 [10] (Mixed g-Monotone Property)

Let (X, ≼) be a partially ordered set and F : X × X → 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 ∈ X,

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

(1)

and

$$y_1,y_2\in X,g{y}_1\preccurlyeq g{y}_2\Rightarrow F\left(x,y_1\right)\succcurlyeq F\left(x,y_2\right).$$

(2)

Definition 2.2 [10] (Coupled Coincidence Point)

Let (x, y) ∈ X × X, F : X × X → X and g : X → 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 ∈ X.

Definition 2.3 [10] Let X be a non-empty set and let F : X × X → X and g : X → X. 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 [6] The mapping F and g where F : X × X → X and g : X → X, 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 [11], let ϕ denote all functions ϕ : [0, ∞) → [0, ∞) which satisfy

• 1. ϕ is continuous and non-decreasing,

• 2. ϕ (t) = 0 if and only if t = 0,

• 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 [11], 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

$${\psi }_3\left(t\right)=\left\{\begin{array}{cc}\hfill 1,\hfill & \hfill t=0\hfill \\ \hfill \frac{t}{t+1},\hfill & \hfill 0<t<1\hfill \\ \hfill 1,\hfill & \hfill t=1\hfill \\ \hfill \frac{1}{2}t,\hfill & \hfill t>1\hfill \end{array}\right.$$

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 × X → X be a mapping having the mixed g-monotone property on X such that there exist two elements x₀, y₀ ∈ X with

$$g{x}_0\preccurlyeq F\left(x_0,y_0\right)andg{y}_0\succcurlyeq F\left(y_0,x_0\right).$$

Suppose there exist ϕ ∈ Φ and ψ ∈ Ψ such that

$$\varphi (d(F(x,y),F(u,v)))\le \frac{1}{2}\varphi (d(gx,gu)+d(gy,gv))-\psi \left(\frac{d(gx,gu)+d(gy,gv))}{2}\right)$$

(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

$$gx=F\left(x,y\right)andgy=F\left(y,x\right),$$

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

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

(4)

We are going to prove that

$$g{x}_n\preccurlyeq g{x}_{n+1}\mathsf{\text{for all }}n\ge 0$$

(5)

and

$$g{y}_n\succcurlyeq g{y}_{n+1}\mathsf{\text{for all }}n\ge 0.$$

(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

$$g{x}_{n+1}\preccurlyeq g{x}_{n+2}\mathsf{\text{and}}g{y}_{n+1}\succcurlyeq g{y}_{n+2}.$$

Hence by the mathematical induction we conclude that (5) and (6) hold for all n ≥ 0. Therefore,

$$g{x}_0\preccurlyeq g{x}_1\preccurlyeq g{x}_2\preccurlyeq \cdots \preccurlyeq g{x}_n\preccurlyeq g{x}_{n+1}\preccurlyeq \cdots$$

(9)

and

$$g{y}_0\succcurlyeq g{y}_1\succcurlyeq g{y}_2\succcurlyeq \cdots \succcurlyeq g{y}_n\succcurlyeq g{y}_{n+1}\succcurlyeq \cdots .$$

(10)

Since gx _n ≽ gx _{n - 1} and gy _n ≼ gy _{n - 1} , using (3) and (4), we have

$$\begin{array}{lll}\hfill \varphi \left(d\left(g{x}_{n+1},g{x}_n\right)\right)& =\varphi \left(d\left(F\left(x_n,y_n\right),F\left(x_{n-1},y_{n-1}\right)\right)\right)& \hfill \text{(1)}\\ \le \frac{1}{2}\varphi \left(d\left(g{x}_n,g{x}_{n-1}\right)+d\left(g{y}_n,g{y}_{n-1}\right)\right)& \hfill \text{(2)}\\ -\psi \left(\frac{d\left(g{x}_n,g{x}_{n-1}\right)+d\left(g{y}_n,g{y}_{n-1}\right)}{2}\right).& \hfill \text{(3)}\\ \hfill \text{(4)}\end{array}$$

(11)

Similarly, since gy _{n - 1} ≽ gy _n and gx _{n - 1} ≼ gx _n , using (3) and (4), we also have

$$\begin{array}{ll}\varphi (d(g{y}_n,g{y}_{n+1}))\hfill & =\varphi (d(F(y_{n-1},x_{n-1}),F(y_n,x_n)))\hfill \\ \le \frac{1}{2}\varphi (d(g{y}_{n-1},g{y}_n)+d(g{x}_{n-1},g{x}_n))\hfill \\ -\psi \left(\frac{d((g{y}_{n-1},g{y}_n)+d(g{x}_{n-1},g{x}_n)}{2}\right).\hfill \end{array}$$

(12)

Using (11) and (12), we have

$$\begin{array}{ll}\varphi (d(g{x}_{n+1},g{x}_n))+\varphi (d(g{y}_{n+1},g{y}_n))\hfill & \le \varphi (d(g{x}_n,g{x}_{n-1})+d(g{y}_n,g{y}_{n-1}))\hfill \\ -2\psi \left(\frac{d(g{x}_n,g{x}_{n-1})+d(g{y}_n,g{y}_{n-1})}{2}\right).\hfill \end{array}$$

(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

$$\begin{array}{lll}\hfill \varphi \left(d\left(g{x}_{n+1},g{x}_n\right)+d\left(g{y}_{n+1},g{y}_n\right)\right)& \le \varphi \left(d\left(g{x}_n,g{x}_{n-1}\right)+d\left(g{y}_n,g{y}_{n-1}\right)\right)& \hfill \text{(1)}\\ -2\psi \left(\frac{d\left(g{x}_n,g{x}_{n-1}\right)+d\left(g{y}_n,g{y}_{n-1}\right)}{2}\right)& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$$

(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

$$\begin{array}{lll}\hfill \varphi \left(\delta \right)& =\underset{n\to \infty }{lim}\varphi \left({\delta }_n\right)\le \underset{n\to \infty }{lim}\left[\varphi \left({\delta }_{n-1}\right)-2\psi \left(\frac{{\delta }_{n-1}}{2}\right)\right]& \hfill \text{(1)}\\ =\varphi \left(\delta \right)-2\underset{{\delta }_{n-1}\to \delta }{lim}\psi \left(\frac{{\delta }_{n-1}}{2}\right)<\varphi \left(\delta \right)& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$$

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

$$\begin{array}{lll}\hfill \epsilon & \le r_k:=d\left(g{x}_{n\left(k\right)},g{x}_{m\left(k\right)}\right)+d\left(g{y}_{n\left(k\right)},g{y}_{m\left(k\right)}\right)& \hfill \text{(1)}\\ \le d\left(g{x}_{n\left(k\right)},g{x}_{n\left(k\right)-1}\right)+d\left(g{x}_{n\left(k\right)-1},g{x}_{m\left(k\right)}\right)+d\left(g{y}_{n\left(k\right)},g{y}_{n\left(k\right)-1}\right)+d\left(g{y}_{n\left(k\right)-1},g{y}_{m\left(k\right)}\right)& \hfill \text{(2)}\\ \le d\left(g{x}_{n\left(k\right)},g{x}_{n\left(k\right)-1}\right)+d\left(g{y}_{n\left(k\right)},g{y}_{n\left(k\right)-1}\right)+\epsilon .& \hfill \text{(3)}\\ \hfill \text{(4) }\end{array}$$

Letting k → ∞ and using (17), we get

$$\underset{k\to \infty }{\lim }{r}_k=\underset{k\to \infty }{\lim }[d(g{x}_{n(k)},g{x}_{m(k)})+d(g{y}_{n(k)},g{y}_{m(k)}]=\epsilon .$$

(20)

By the triangle inequality

$$\begin{array}{lll}\hfill r_k& =d\left(g{x}_{n\left(k\right)},g{x}_{m\left(k\right)}\right)+d\left(g{y}_{n\left(k\right)},g{y}_{m\left(k\right)}\right)& \hfill \text{(1)}\\ \le d\left(g{x}_{n\left(k\right)},g{x}_{n\left(k\right)+1}\right)+d\left(g{x}_{n\left(k\right)+1},g{x}_{m\left(k\right)+1}\right)+d\left(g{x}_{m\left(k\right)+1},g{x}_{m\left(k\right)}\right)& \hfill \text{(2)}\\ +d\left(g{y}_{n\left(k\right)},g{y}_{n\left(k\right)+1}\right)+d\left(g{y}_{n\left(k\right)+1},g{y}_{m\left(k\right)+1}\right)+d\left(g{y}_{m\left(k\right)+1},g{y}_{m\left(k\right)}\right)& \hfill \text{(3)}\\ ={\delta }_{n\left(k\right)}+{\delta }_{m\left(k\right)}+d\left(g{x}_{n\left(k\right)+1},g{x}_{m\left(k\right)+1}\right)+d\left(g{y}_{n\left(k\right)+1},g{y}_{m\left(k\right)+1}\right).& \hfill \text{(4)}\\ \hfill \text{(5)}\end{array}$$

Using the property of ϕ, we have

$$\begin{array}{lll}\hfill \varphi \left(r_k\right)& =\varphi \left({\delta }_{n\left(k\right)}+{\delta }_{m\left(k\right)}+d\left(g{x}_{n\left(k\right)+1},g{x}_{m\left(k\right)+1}\right)+d\left(g{y}_{n\left(k\right)+1},g{y}_{m\left(k\right)+1}\right)\right)& \hfill \text{(1)}\\ \le \varphi \left({\delta }_{n\left(k\right)}+{\delta }_{m\left(k\right)}\right)+\varphi \left(d\left(g{x}_{n\left(k\right)+1},g{x}_{m\left(k\right)+1}\right)\right)& \hfill \text{(2)}\\ +\varphi \left(d\left(g{y}_{n\left(k\right)+1},g{y}_{m\left(k\right)+1}\right)\right).& \hfill \text{(3)}\\ \hfill \text{(4)}\end{array}$$

(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

$$\begin{array}{lll}\hfill \varphi \left(d\left(g{x}_{n\left(k\right)+1},g{x}_{m\left(k\right)+1}\right)\right)& =\varphi \left(d\left(F\left(x_{n\left(k\right)},y_{n\left(k\right)}\right),F\left(x_{m\left(k\right)},y_{m\left(k\right)}\right)\right)\right)& \hfill \text{(1)}\\ \le \frac{1}{2}\varphi \left(d\left(g{x}_{n\left(k\right)},g{x}_{m\left(k\right)}\right)+d\left(g{y}_{n\left(k\right)},g{y}_{m\left(k\right)}\right)\right)& \hfill \text{(2)}\\ -\psi \left(\frac{d\left(g{x}_{n\left(k\right)},g{x}_{m\left(k\right)}\right)+d\left(g{y}_{n\left(k\right)},g{y}_{m\left(k\right)}\right)}{2}\right)& \hfill \text{(3)}\\ =\frac{1}{2}\varphi \left(r_k\right)-\psi \left(\frac{r_k}{2}\right).& \hfill \text{(4)}\\ \hfill \text{(5)}\end{array}$$

(22)

By the same way, we also have

$$\begin{array}{lll}\hfill \varphi \left(d\left(g{y}_{m\left(k\right)+1},g{y}_{n\left(k\right)+1}\right)\right)& =\varphi \left(d\left(F\left(y_{m\left(k\right)},x_{m\left(k\right)}\right),F\left(y_{n\left(k\right)},x_{n\left(k\right)}\right)\right)\right)\backslash & \hfill \text{(1)}\\ \le \frac{1}{2}\varphi \left(d\left(g{y}_{m\left(k\right)},g{y}_{n\left(k\right)}\right)+d\left(g{x}_{m\left(k\right)},g{x}_{n\left(k\right)}\right)\right)& \hfill \text{(2)}\\ -\psi \left(\frac{d\left(g{y}_{m\left(k\right)},g{y}_{n\left(k\right)}\right)+d\left(g{x}_{m\left(k\right)},g{x}_{n\left(k\right)}\right)}{2}\right)& \hfill \text{(3)}\\ =\frac{1}{2}\varphi \left(r_k\right)-\psi \left(\frac{r_k}{2}\right).& \hfill \text{(4)}\\ \hfill \text{(5)}\end{array}$$

(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

$$\underset{n\to \infty }{lim}F\left(x_n,y_n\right)=\underset{n\to \infty }{lim}g{x}_n=xand\underset{n\to \infty }{lim}F\left(y_n,x_n\right)=\underset{n\to \infty }{lim}g{y}_n=y.$$

(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,

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

Taking the limit as n → ∞, 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

$$gx=F\left(x,y\right)andgy=F\left(y,x\right).$$

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,

$$g{x}_n\preccurlyeq xandg{y}_n\preccurlyeq y.$$

(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,

$$\begin{array}{ll}d(gx,F(x,y))\hfill & \le \underset{n\to \infty }{\lim }d(gx,g(g{x}_{n+1}))+\underset{n\to \infty }{\lim }d(g(F(x_n,y_n)),F(x,y))\hfill \\ \le \underset{n\to \infty }{\lim }d(F(g{x}_n,g{y}_n)),F(x,y))\hfill \end{array}$$

(30)

Using the property of ϕ, we get

$$\varphi (d(gx,F(x,y)))\le \underset{n\to \infty }{\lim }\varphi (d(F(g{x}_n,g{y}_n)),F(x,y)))$$

Since the mapping g is monotone increasing, using (3), (27) and (30), we have for all n ≥ 0,

$$\begin{array}{lll}\hfill \varphi \left(d\left(gx,F\left(x,y\right)\right)\right)& \le \underset{n\to \infty }{lim}\frac{1}{2}\varphi \left(d\left(gg{x}_n,gx\right)+d\left(g{y}_n,ggy\right)\right)& \hfill \text{(1)}\\ -\underset{n\to \infty }{lim}\psi \left(\frac{d\left(gg{x}_n,gx\right)+d\left(gg{y}_n,gy\right)}{2}\right).& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$$

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 [11] 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 × X → X be a mapping having the mixed monotone property on X such that there exist two elements x₀, y₀ ∈ X with

$$x_0\preccurlyeq F\left(x_0,y_0\right)and{y}_0\succcurlyeq F\left(y_0,x_0\right).$$

Suppose there exist ϕ ∈ Φ and ψ ∈ Ψ such that

$$\varphi (d(F(x,y),F(u,v)))\le \frac{1}{2}\varphi (d(x,u)+d(y,v))-\psi \left(\frac{d(x,u)+d(y,v))}{2}\right)$$

for all x, y, u, v ∈ X with x ≥ u and y ≤ v. 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

$$x=F\left(x,y\right)andy=F\left(y,x\right)$$

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

Corollary 3.2 [11] 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 × X → X be a mapping having the mixed monotone property on X such that there exist two elements x₀, y₀ ∈ X with

$$x_0\preccurlyeq F\left(x_0,y_0\right)and{y}_0\succcurlyeq F\left(y_0,x_0\right).$$

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 x ≥ u and y ≤ v. 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

$$x=F\left(x,y\right)andy=F\left(y,x\right)$$

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 [5] 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 × X → X be a mapping having the mixed monotone property on X such that there exist two elements x₀, y₀ ∈ X with

$$x_0\preccurlyeq F\left(x_0,y_0\right)and{y}_0\succcurlyeq F\left(y_0,x_0\right).$$

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 x ≥ u and y ≥ v. 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

$$x=F\left(x,y\right)andy=F\left(y,x\right)$$

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

Proof. Taking $\psi \left(t\right)=\frac{1-k}{2}t$ 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,

$$\left(x,y\right)\preccurlyeq \left(u,v\right)\iff x\preccurlyeq u,y\succcurlyeq v.$$

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

$$u_0=u{v}_0=v.g{u}_{u+1}=F\left(u_n,v_n\right)\mathsf{\text{and}}g{v}_{n+1}=F\left(v_n,u_n\right)\mathsf{\text{for all }}n.$$

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

$$\left(x,y\right)\succcurlyeq \left(u_n,v_n\right)\mathsf{\text{for all }}n.$$

(31)

Using (3) and (31), we have

$$\begin{array}{lll}\hfill \phi \left(d\left(gx,g{u}_{n+1}\right)\right)& =\phi \left(d\left(F\left(x,y\right),F\left(u_n,v_n\right)\right)\right)& \hfill \text{(1)}\\ <\frac{1}{2}\phi \left(d\left(x,u_n\right)+d\left(y,v_n\right)\right)-\psi \left(\frac{d\left(x,u_n\right)+d\left(y,v_n\right)}{2}\right)& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$$

(32)

Similarly

$$\begin{array}{lll}\hfill \phi \left(d\left(g{v}_{n+1},gy\right)\right)& =\phi \left(d\left(F\left(v_n,u_n\right),F\left(y,x\right)\right)\right)& \hfill \text{(1)}\\ <\frac{1}{2}\phi \left(d\left(v_n,y\right)+d\left(u_n,x\right)\right)-\psi \left(\frac{d\left(v_n,y\right)+d\left(u_n,x\right)}{2}\right)& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$$

(33)

Using (32), (33) and the property of φ, we have

$$\begin{array}{lll}\hfill \phi \left(d\left(gx,g{u}_{n+1}\right)+d\left(gy,g{v}_{n+1}\right)\right)& \le \phi \left(d\left(gx,g{u}_{n+1}\right)\right)+\phi \left(d\left(gy,g{v}_{n+1}\right)\right)& \hfill \text{(1)}\\ \le \phi \left(d\left(gx,g{u}_n\right)+d\left(gy,g{v}_n\right)\right)& \hfill \text{(2)}\\ -2\psi \left(\frac{d\left(gx,g{u}_n\right)+d\left(gy,g{v}_n\right)}{2}\right).& \hfill \text{(3)}\\ \hfill \text{(4)}\end{array}$$

(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 [11] 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

$$d\left(x,y\right)=\left|x-y\right|forx,y\in \left[0,1\right].$$

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

Let g : X → X be defined as

$$gx=x^2,forallx\in X,$$

and let F : X × X → X be defined as

$$F\left(x,y\right)=\left\{\begin{array}{cc}\hfill \frac{x^2-y^2}{3},\hfill & \hfill ifx\ge y,\hfill \\ \hfill 0,\hfill & \hfill ifx<y.\hfill \end{array}\right.$$

F obeys the mixed g-monotone property.

Let ϕ : [0, ∞) → [0, ∞) be defined as

$$\varphi \left(t\right)=\frac{3}{4}t,fort\in \left[0,\infty \right).$$

and let ψ : [0, ∞) → [0, ∞) be defined as

$$\psi \left(t\right)=\frac{1}{4}t,fort\in \left[0,\infty \right).$$

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,

$$\begin{array}{c}g{x}_n=x_n^2,g{y}_n=y_n^2,\\ F\left(x_n,y_n\right)=\left\{\begin{array}{cc}\hfill \frac{x_n^2-y_n^2}{3},\hfill & \hfill if{x}_n\ge y_n,\hfill \\ \hfill 0,\hfill & \hfill if{x}_n<y_n.\hfill \end{array}\right.\end{array}$$

and

$$F\left(y_n,x_n\right)=\left\{\begin{array}{cc}\hfill \frac{y_n^2-x_n^2}{3},\hfill & \hfill if{y}_n\ge x_n,\hfill \\ \hfill 0,\hfill & \hfill if{y}_n<x_n.\hfill \end{array}\right.$$

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 gx ≥ gu and gy ≤ gv, that is, x² ≥ u² and y² ≤ v².

We consider the following cases:

Case 1. x ≥ y, u ≥ v. Then,

$$\begin{array}{ll}\varphi (d(F(x,y),F(u,v)))\hfill & =\frac{3}{4}[d(F(x,y),F(u,v)]\hfill \\ =\frac{3}{4}\left[d\left(\frac{x^2-y^2}{3},\frac{u^2-v^2}{3}\right)\right]\hfill \\ =\frac{3}{4}\left|\frac{(x^2-y^2)-(u^2-v^2)}{3}\right|\hfill \\ =\frac{3}{4}\frac{|x^2-u^2|+|y^2-v^2|}{3}\hfill \\ =\frac{1}{2}\left(\frac{d(gx,gu)+d(gy,gv)}{2}\right)\hfill \\ =\frac{3}{4}\left(\frac{d(gx,gu)+d(gy,gv)}{2}\right)\hfill \\ -\frac{1}{4}\left(\frac{d(gx,gu)+d(gy,gv)}{2}\right)\hfill \\ =\frac{3}{8}(d(gx,gu)+d(gy,gv))\hfill \\ -\frac{1}{4}\left(\frac{d(gx,gu)+d(gy,gv)}{2}\right)\hfill \\ =\frac{1}{2}\varphi (d(gx,gu)+d(gy,gv))\hfill \\ -\psi \left(\frac{d(gx,gu)+d(gy,gv)}{2}\right)\hfill \end{array}$$

Case 2. x ≥ y, u < v.Then

$$\begin{array}{ll}\varphi (d(F(x,y),F(u,v))\hfill & =\frac{3}{4}[d(F(x,y),F(u,v)]\hfill \\ =\frac{3}{4}\left[d(\frac{x^2-y^2}{3},0)\right]\hfill \\ =\frac{3}{4}\frac{|x^2-y^2|}{3}\hfill \\ =\frac{3}{4}\frac{|v^2+x^2-y^2-u^2|}{3}\hfill \\ =\frac{3}{4}\frac{|(v^2-y^2)-(u^2-x^2)|}{3}\hfill \\ \le \frac{3}{4}\frac{|v^2-y^2|+|u^2-x^2|}{3}\hfill \\ =\frac{3}{4}\left(\frac{|u^2-x^2|+|y^2-v^2|}{3}\right)\hfill \\ =\frac{1}{2}\left(\frac{|u^2-x^2|+|y^2-v^2|}{2}\right)\hfill \\ =\frac{1}{2}\left(\frac{d(gx,gu)+d(gy,gv)}{2}\right)\hfill \\ =\frac{3}{4}\left(\frac{d(gx,gu)+d(gy,gv)}{2}\right)\hfill \\ -\frac{1}{4}\left(\frac{d(gx,gu)+d(gy,gv)}{2}\right)\hfill \\ =\frac{3}{8}(d(gx,gu)+d(gy,gv))\hfill \\ -\frac{1}{4}(\frac{d(gx,gu)+d(gy,gv)}{2}\hfill \\ =\frac{1}{2}\varphi (d(gx,gu)+d(gy,gv))\hfill \\ -\psi \left(\frac{d(gx,gu)+d(gy,gv)}{2}\right)\hfill \end{array}$$

Case 3. x < y and u ≥ v. Then

$$\begin{array}{lll}\hfill \varphi \left(d\left(F\left(x,y\right),F\left(u,v\right)\right)\right)& =\frac{3}{4}\left[d\left(0,\frac{u^2-v^2}{3}\right)\right]& \hfill \text{(1)}\\ =\frac{3}{4}\frac{|u^2-v^2|}{3}& \hfill \text{(2)}\\ =\frac{3}{4}\frac{|u^2+x^2-v^2-x^2|}{3}& \hfill \text{(3)}\\ =\frac{3}{4}\frac{|\left(x^2-v^2\right)+\left(u^2-x^2\right)|}{3}\left(sincey>x\right)& \hfill \text{(4)}\\ \le \frac{3}{4}\frac{|y^2-v^2|+|u^2-x^2|}{3}& \hfill \text{(5)}\\ =\frac{1}{2}\left(\frac{|u^2-x^2|+|y^2-v^2|}{2}\right)& \hfill \text{(6)}\\ =\frac{1}{2}\left(\frac{d\left(gx,gu\right)+d\left(gy,gv\right)}{2}\right)& \hfill \text{(7)}\\ =\frac{3}{4}\left(\frac{d\left(gx,gu\right)+d\left(gy,gv\right)}{2}\right)& \hfill \text{(8)}\\ -\frac{1}{4}\left(\frac{d\left(gx,gu\right)+d\left(gy,gv\right)}{2}\right)& \hfill \text{(9)}\\ =\frac{3}{8}\left(d\left(gx,gu\right)+d\left(gy,gv\right)\right)& \hfill \text{(10)}\\ -\frac{1}{4}\left(\frac{d\left(gx,gu\right)+d\left(gy,gv\right)}{2}\right)& \hfill \text{(11)}\\ =\frac{1}{2}\varphi \left(d\left(gx,gu\right)+d\left(gy,gv\right)\right)& \hfill \text{(12)}\\ -\psi \left(\frac{d\left(gx,gu\right)+d\left(gy,gv\right)}{2}\right)& \hfill \text{(13)}\\ \hfill \text{(14)}\end{array}$$

Case 4. x < y and u < v with x² ≤ u² and y² ≥ v². 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.