Coupled fixed point theorems for contractions involving altering distances in ordered metric spaces

Abstract

In this paper, we establish a coupled fixed point result of F:X×X→X having the mixed monotone property involving generalized altering distance functions in five variables on ordered metric spaces. An example is given to support the usability of our results. We also give a coupled fixed point result involving a contraction of integral type.

Introduction and preliminaries

The very famous Banach contraction principle [1] can be stated as follows.

Theorem 1

[1]. Let (X,d)be a complete metric space and T be a mapping of X into itself satisfying:

$$d(\mathit{\text{Tx}},\mathit{\text{Ty}})\le \mathit{\text{kd}}(x,y),\forall x,y\in X,$$

(1.1)

where k is a constant in (0,1). Then, T has a unique fixed point x^∗∈X.

In the literature, there is a great number of generalizations of the Banach contraction principle. Khan et al. [2] introduced the notion of an altering distance function, which is a control function that alters distance between two points in a metricspace.

Definition 2

[2]. A function φ:[0,+∞)→[0,+∞) is called an altering distance function if and only if

(i) φ Is continuous,

(ii) φ Is non-decreasing, and

(iii) φ(t)=0⇔t=0.

Afterwards, a number of works have appeared in which altering distances have been used. Altering distances have been generalized to a two-variable function by Choudhury and Dutta [3] and to a three-variable function by Choudhury [4] and was applied for obtaining fixed point results in metric spaces.

Definition 3

[4]. Let Ψ₃ denote the set of all functions φ:[0,+∞)×[0,+∞)×[0,+∞)→[0,+∞). Then φ is said to be a generalized altering distance function if andonly if

(i) φ is continuous,

(ii) φ is nondecreasing in all the three variables, and

(iii) φ(x,y,z)=0⇔x=y=z=0.

In [5], Rao et al. introduced the generalized altering distance function in five variables as a generalization ofthree variables.

Definition 4

[5]. Let Ψ₅ denote the set of all functions φ:[0,+∞)×[0,+∞)×[0,+∞)×[0,+∞)×[0,+∞)→[0,+∞). Then φ is said to be a generalized altering distance function if and only if

(i) φ is continuous,

(ii) φ is non-decreasing in all five variables, and

(iii) φ(t₁,t₂,t₃,t₄,t₅)=0⇔t₁=t₂=t₃=t₄=t₅=0.

On the other hand, the first result on existence of fixed points in partially ordered sets was given by Turinici [6], where he extended the Banach contraction principle in partially ordered sets. Ran and Reurings [7] presented some applications of Turinici’s theorem to matrix equations. The obtained result by Turinici was further extended and refined in [8]-[33]. Subsequently, Harjani and Sadarangani [17] generalized their own results [18] by considering a pair of altering functions (ψ,φ). Nashine and Altun [20] and Nashine and Bessem [22] generalized the results of Harjani and Sadarangani [17],[18]. Nashine, Samet, and Vetro [23] also had fixed point theorems for T-weakly isotone-increasing mappings which satisfy a generalized nonlinear contractive condition in complete ordered metric spaces and gave an application to an existence theorem for a solution of some integral equations. Jachymski[30] established a geometric lemma (Lemma 1 in [30]) giving a list of equivalent conditions for some subsets of the plane. Using this lemma, he proved that some very recent fixed point theorems for generalized contractions on ordered metric spaces obtained by Harjani and Sadarangani [17],[18], and Amini-Harandi and Emami [10] follow from an earlier result of O’Regan and Petruşel (Theorem 3.6 in [28]).

Now, we introduce some known notations and definitions that will be used later.

Definition 5

Let X be a nonempty set. Then (X,d,≼) is called an ordered metric space iff

(i) (X,d) is a metric space, and

(ii) (X,≼) is partially ordered.

Definition 6

Let (X,≼) be a partial ordered set. Then x,y∈X are called comparable if x≼y or y≼x holds.

Let (X,≼) be a partially ordered set. The concept of a mixed monotone property of the mapping F:X×X→X has been introduced by Bhaskar andLakshmikantham [13].

Definition 7

(see Bhaskar and Lakshmikantham [13]). Let (X,≼) be a partially ordered set and F:X×X→X. Then the map F is said to have mixed monotone property if F(x,y) is monotone nondecreasing in x and is monotone non-increasing in y; that is,for any x,y∈X,

$$x_1\preccurlyeq x_2\text{implies}F(x_1,y)\preccurlyeq F(x_2,y)\text{for all}y\in X$$

and

$$y_1\preccurlyeq y_2\text{implies}F(x,y_2)\preccurlyeq F(x,y_1)\text{for all}x\in \mathrm{X.}$$

Definition 8

(see Bhaskar and Lakshmikantham [13]). An element (x,y)∈X×X is called a coupled fixed point of a mapping F:X×X→X if

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

Bhaskar and Lakshmikantham [13] proved the following coupled fixed point theorem. For other coupled fixed point results, see [34]-[46].

Theorem 9

(see Bhaskar and Lakshmikantham [13]). 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. Assume that there exists a k∈[0,1)with

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

(1.2)

for all x,y,u,v∈X with x≼u and v≼y. Suppose either F is continuous or X has the following properties:

• 1. If a non-decreasing sequence x _n →x, then x _n ≼x for all n.

• 2. If a non-increasing sequence y _n →y , then y≼y _n for all n.

If there exist x₀,y₀∈X such that x₀≼F(x₀,y₀)and F(y₀,x₀)≼y₀, then there exist x,y∈X such that x=F(x,y)and y=F(y,x), that is, F has a coupledfixed point.

In this paper, we first obtain a coupled fixed point result for F:X×X→X having the mixed monotone property satisfying a contractive condition which involves generalized altering distance functions in five variables in complete ordered metric spaces. An example is also been given to the validity of our results. In particular, in this example, we will show that the result of Bhaskar and Lakshmikantham [13] cannot be applied. Finally, we establish a coupled fixed point result involving a contraction of integral type.

Main results

Before stating our main theorem, the following lemma is needed:

Lemma 10

Let (X,d) be a metric space and let {y _n } be a sequence in X such that

$$\underset{n\to \infty }{lim}d(y_{n+1},y_n)=0.$$

If {y _n } is not a Cauchy sequence, then there exist ε>0 and two sequences {m _k } and {n _k } of positive integers such that the following four sequences tend to ε when k→∞:

$$\begin{array}{ll}d(y_{m_k},y_{n_k}),d(y_{m_k},y_{n_k+1}),& d(y_{m_k-1},y_{n_k}),\\ d(y_{m_k-1},y_{n_k+1}).\end{array}$$

(2.1)

Our main theorem is the following.

Theorem 11

Let (X,≼)be a partially ordered set and suppose that there exists 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

$$\begin{array}{l}{\Phi }_1\left(d\right(F(x,y),F(u,v)\left)\right)\le {\psi }_1\left(d\right(x,u),d(y,v),d(x,F(x,y)),\\ d(u,F(u,v\left)\right),\frac{1}{2}\left[d\right(x,F(u,v))+d(u,F(x,y)\left)\right])\\ -{\psi }_2\left(\right(d(x,u),d(y,v),d(x,F(x,y\left)\right),d(u,F(u,v\left)\right),\\ \frac{1}{2}\left[d\right(x,F(u,v))+d(u,F(x,y)\left)\right]),\end{array}$$

(2.2)

for x,y,u,v∈X with x≽u and y≼v, where ψ₁and ψ₂are generalized altering distance functions and Φ₁(x)=ψ₁(x,x,x,x,x). Assume either

• 1. F is continuous, or

• 2. X has the following properties:

• (a) if a non-decreasing sequence x _n →x, then x _n ≼x for all n,

• (a) if a non-increasing sequence y _n →y, then y≼y _n for all n.

If there exist x₀,y₀∈X such that x₀≼F(x₀,y₀)and y₀≽F(y₀,x₀), then F has a coupled fixed point, that is, there exists (x,y)∈X such that F(x,y)=x and F(y,x)=y.

Proof

Let x₀,y₀∈X be arbitrarily chosen and define x₁:=F(x₀,y₀) and y₁:=F(y₀,x₀). Next, we consider x₂:=F(x₁,y₁) and y₂:=F(y₁,x₁). Continuing in this way, we construct two sequences {x _n } and {y _n } in X such that

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

(2.3)

Using the fact that F has a mixed monotone property, we have as in [13]

$$x_n\preccurlyeq x_{n+1}\text{and}{y}_{n+1}\preccurlyeq y_n\forall n\ge 0.$$

(2.4)

Since x _n ≽x_n−1 and y _n ≼y_n−1, then from (2.2)

$$\begin{array}{l}{\Phi }_1\left(d\right(x_{n+1},x_n\left)\right)={\Phi }_1\left(d\right(F(x_n,y_n),F(x_{n-1},y_{n-1})\left)\right)\\ \le {\psi }_1\left(d\right(x_n,x_{n-1}),d(y_n,y_{n-1}),d(x_n,F(x_n,y_n)),d(x_{n-1},\\ F(x_{n-1},y_{n-1})),\\ \frac{1}{2}\left[d\right(x_n,F(x_{n-1},y_{n-1}))+d(x_{n-1},F(x_n,y_n)\left)\right])\\ -{\psi }_2\left(d\right(x_n,x_{n-1}),d(y_n,y_{n-1}),d(x_n,F(x_n,y_n)),\\ d(x_{n-1},F(x_{n-1},y_{n-1}\left)\right),\\ \frac{1}{2}\left[d\right(x_n,F(x_{n-1},y_{n-1}))+d(x_{n-1},F(x_n,y_n)\left)\right])\\ ={\psi }_1\left(d\right(x_n,x_{n-1}),d(y_n,y_{n-1}),d(x_n,x_{n+1}),d(x_{n-1},x_n),\\ \frac{1}{2}d(x_{n-1},x_{n+1}))\\ -{\psi }_2\left(d\right(x_n,x_{n-1}),d(y_n,y_{n-1}),d(x_n,x_{n+1}),d(x_{n-1},x_n),\\ \frac{1}{2}d(x_{n-1},x_{n+1}))\\ \le {\psi }_1\left(d\right(x_n,x_{n-1}),d(y_n,y_{n-1}),d(x_n,x_{n+1}),d(x_{n-1},x_n),\\ \frac{1}{2}d(x_{n-1},x_{n+1})).\end{array}$$

(2.5)

Since ψ₁ is monotone increasing with respect to the first variable, we have for all n≥1

$$d(x_{n+1},x_n)\le d(x_n,x_{n-1}).$$

(2.6)

Since x _n ≽x_n−1 and y _n ≼y_n−1, then from (2.2)

$$\begin{array}{l}{\Phi }_1\left(d\right(y_n,y_{n+1}\left)\right)={\Phi }_1\left(d\right(F(y_{n-1},x_{n-1}),F(y_n,x_n)\left)\right)\\ ={\Phi }_1\left(d\right(F(y_n,x_n),F(y_{n-1},x_{n-1})\left)\right)\\ \le {\psi }_1\left(d\right(y_n,y_{n-1}),d(x_n,x_{n-1}),d(y_n,F(y_n,x_n)),d(y_{n-1},\\ F(y_{n-1},x_{n-1})),\\ \frac{1}{2}\left[d\right(y_n,F(y_{n-1},x_{n-1}))+d(y_{n-1},F(y_n,x_n)\left)\right])\\ -{\psi }_2\left(d\right(y_n,y_{n-1}),d(x_n,x_{n-1}),d(y_n,F(y_n,x_n)),d(y_{n-1},\\ F(y_{n-1},x_{n-1})),\\ \frac{1}{2}\left[d\right(y_n,F(y_{n-1},x_{n-1}))+d(y_{n-1},F(y_n,x_n)\left)\right])\\ ={\psi }_1\left(d\right(y_n,y_{n-1}),d(x_n,x_{n-1}),d(y_n,y_{n+1}),d(y_{n-1},y_n),\\ \frac{1}{2}d(y_{n-1},y_{n+1}))\\ -{\psi }_2\left(d\right(y_n,y_{n-1}),d(x_n,x_{n-1}),d(y_n,y_{n+1}),d(y_{n-1},y_n),\\ \frac{1}{2}d(y_{n-1},y_{n+1}))\\ \le {\psi }_1\left(d\right(y_n,y_{n-1}),d(x_n,x_{n-1}),d(y_n,y_{n+1}),d(y_{n-1},y_n),\\ \frac{1}{2}d(y_{n-1},y_{n+1})).\end{array}$$

(2.7)

Since ψ₁ is monotone increasing with respect to the first variable, we have for all n≥1

$$d(y_{n+1},y_n)\le d(y_n,y_{n-1}).$$

(2.8)

In view of (2.6) and (2.8), the sequences {d(x_n+1,x _n )} and {d(y_n+1,y _n )} are nonincreasing, so there exist r≥0 and γ≥0 such that

$$\underset{n\to \infty }{lim}d(x_{n+1},x_n)=r\text{and}\underset{n\to \infty }{lim}d(y_{n+1},y_n)=\mathrm{\gamma .}$$

Again, since ψ₁ is monotone increasing with respect to the fifth variable, from (2.5), we have by triangularinequality

$$\begin{array}{r}d(x_{n+1},x_n)\le \frac{1}{2}d(x_{n-1},x_{n+1})\le \frac{1}{2}d(x_{n-1},x_n)\\ +\frac{1}{2}d(x_n,x_{n+1});\end{array}$$

so in the limit, we have

$$\underset{n\to \infty }{lim}d(x_{n-1},x_{n+1})=2\mathrm{r.}$$

Similarly, from (2.7)

$$\underset{n\to \infty }{lim}d(y_{n-1},y_{n+1})=2\mathrm{\gamma .}$$

Passing on the limit n→∞ in (2.5) and (2.7) respectively and using the continuities of ψ₁,ψ₂, we get

$${\Phi }_1\left(r\right)\le {\psi }_1(r,\gamma ,r,r,r)-{\psi }_2(r,\gamma ,r,r,r),$$

and

$${\Phi }_1\left(\gamma \right)\le {\psi }_1(\gamma ,r,\gamma ,\gamma ,\gamma )-{\psi }_2(\gamma ,r,\gamma ,\gamma ,\gamma ).$$

Assume that γ≠r. Without loss of generality, suppose that γ<r, so

$$\begin{array}{ll}{\Phi }_1\left(r\right)\le {\psi }_1(r,\gamma ,r,r,r)& -{\psi }_2(r,\gamma ,r,r,r)\le {\Phi }_1\left(r\right)\\ -{\psi }_2(r,\gamma ,r,r,r),\end{array}$$

which holds unless ψ₂(r,γ,r,r,r)=0, that is, r=γ (which is equal to 0 using the same idea), a contradiction. We deduce that

$$\underset{n\to \infty }{lim}d(x_{n+1},x_n)=\underset{n\to \infty }{lim}d(y_{n+1},y_n)=0.$$

(2.9)

Now we shall show that {x _n } and {y _n } are Cauchy sequences in (X,d).

Suppose that {x _n } and {y _n } are not Cauchy sequences. Then, Lemma 10 implies that there exist ε>0 and two sequences {m _k } and {n _k } of positive integers (with for all positive integer k, m(k)>n(k)) such that the sequences

$$\begin{array}{ll}d(x_{m_k},x_{n_k}),d(x_{m_k},x_{n_k+1}),& d(x_{m_k-1},x_{n_k}),\\ d(x_{m_k-1},x_{n_k+1}),\end{array}$$

(2.10)

and

$$\begin{array}{ll}d(y_{m_k},y_{n_k}),d(y_{m_k},y_{n_k+1}),& d(y_{m_k-1},y_{n_k}),\\ d(y_{m_k-1},y_{n_k+1})\end{array}$$

(2.11)

tend to ε (from above) when k→∞. It follows that

$$\begin{array}{c}{limsup}_{k\to \infty }d(x_{n_k-1},x_{m_k+1})\le \epsilon \\ \text{and}\\ {limsup}_{k\to \infty }d(y_{n_k-1},y_{m_k+1})\le \mathrm{\epsilon .}\end{array}$$

(2.12)

Since m(k)≥n(k)−1, so from (2.4), x_m(k)≽x_n(k)−1 and y_m(k)≼y_n(k)−1, we have by (2.2)

$$\begin{array}{l}{\Phi }_1\left(d\right(x_{m\left(k\right)+1},x_{n\left(k\right)}\left)\right)\\ ={\Phi }_1\left(d\right(F(x_{m\left(k\right)},y_{m\left(k\right)}),F(x_{n\left(k\right)-1},y_{n\left(k\right)-1})\left)\right)\\ \le {\psi }_1\left(d\right(x_{m\left(k\right)},x_{n\left(k\right)-1}),d(y_{m\left(k\right)},y_{n\left(k\right)-1}),d(x_{m\left(k\right)},\\ F(x_{m\left(k\right)},y_{m\left(k\right)})),d(x_{n\left(k\right)-1},F(x_{n\left(k\right)-1},y_{n\left(k\right)-1})),\\ \frac{1}{2}\left[d\right(x_{m\left(k\right)},F(x_{n\left(k\right)-1},y_{n\left(k\right)-1}))+d(x_{n\left(k\right)-1},\\ F(x_{m\left(k\right)},y_{m\left(k\right)})\left)\right])\\ -{\psi }_2\left(d\right(x_{m\left(k\right)},x_{n\left(k\right)-1}),d(y_{m\left(k\right)},y_{n\left(k\right)-1}),d(x_{m\left(k\right)},\\ F(x_{m\left(k\right)},y_{m\left(k\right)})),d(x_{n\left(k\right)-1},F(x_{n\left(k\right)-1},y_{n\left(k\right)-1})),\\ \frac{1}{2}\left[d\right(x_{m\left(k\right)},F(x_{n\left(k\right)-1},y_{n\left(k\right)-1}))+d(x_{n\left(k\right)-1},\\ F(x_{m\left(k\right)},y_{m\left(k\right)})\left)\right]).\end{array}$$

(2.13)

In addition, we have

$$\begin{array}{l}{\Phi }_1\left(d\right(y_{n\left(k\right)},y_{m\left(k\right)+1}\left)\right)\\ ={\Phi }_1\left(d\right(F(y_{n\left(k\right)-1},x_{n\left(k\right)-1}),F(y_{m\left(k\right)},x_{m\left(k\right)})\left)\right)\\ \le {\psi }_1\left(d\right(y_{n\left(k\right)-1},y_{m\left(k\right)}),d(x_{n\left(k\right)-1},x_{m\left(k\right)}),d(y_{n\left(k\right)-1},\\ F(y_{n\left(k\right)-1},x_{n\left(k\right)-1})),d(y_{m\left(k\right)},F(y_{m\left(k\right)},x_{m\left(k\right)})),\\ \frac{1}{2}\left[d\right(y_{n\left(k\right)-1},F(y_{m\left(k\right)},x_{m\left(k\right)}))+d(y_{m\left(k\right)},\\ F(y_{n\left(k\right)-1},x_{n\left(k\right)-1})\left)\right])\\ -{\psi }_2\left(d\right(y_{n\left(k\right)-1},y_{m\left(k\right)}),d(x_{n\left(k\right)-1},x_{m\left(k\right)}),d(y_{n\left(k\right)-1},\\ F(y_{n\left(k\right)-1},x_{n\left(k\right)-1})),d(y_{m\left(k\right)},F(y_{m\left(k\right)},x_{m\left(k\right)})),\\ \frac{1}{2}\left[d\right(y_{n\left(k\right)-1},F(y_{m\left(k\right)},x_{m\left(k\right)}))+d(y_{m\left(k\right)},\\ F(y_{n\left(k\right)-1},x_{n\left(k\right)-1})\left)\right]).\end{array}$$

(2.14)

Taking the limsup as k→∞, using (2.10 to 2.12) and the continuity of ψ₁ and ψ₂ in (2.13), we get

$$\begin{array}{l}{\Phi }_1\left(\epsilon \right)\le {\psi }_1(\epsilon ,\epsilon ,0,0,\epsilon )-{\psi }_2(\epsilon ,\epsilon ,0,0,\frac{1}{2}\underset{k\to \infty }{liminf}\\ \left[d\right(x_{n\left(k\right)-1},F(x_{m\left(k\right)},x_{m\left(k\right)}))+d(x_{m\left(k\right)},\\ F(x_{n\left(k\right)-1},y_{n\left(k\right)-1})\left)\right])\\ \le \Phi \left(\epsilon \right)-{\psi }_2(\epsilon ,\epsilon ,0,0,\frac{1}{2}\underset{k\to \infty }{liminf}[d(x_{n\left(k\right)-1},\\ F(x_{m\left(k\right)},y_{m\left(k\right)}))+d(x_{m\left(k\right)},\\ F(x_{n\left(k\right)-1},y_{n\left(k\right)-1})\left)\right]),\end{array}$$

which implies that ${\psi }_2(\epsilon ,\epsilon ,0,0,\frac{1}{2}{liminf}_{k\to \infty }[d(y_{n\left(k\right)-1},F(y_{m\left(k\right)},x_{m\left(k\right)}\left)\right)+d(y_{m\left(k\right)},F(y_{n\left(k\right)-1},x_{n\left(k\right)-1}\left)\right)\left]\right)=0$, is a contradiction since ε>0. We deduce that {x _n } is a Cauchy sequence.

Similarly, taking the limsup as k→+∞ and using the continuity of φ and ψ in (2.14), we get

$$\begin{array}{r}{\psi }_2(\epsilon ,\epsilon ,0,0,\frac{1}{2}\underset{k\to \infty }{liminf}[d(y_{n\left(k\right)-1},F(y_{m\left(k\right)},x_{m\left(k\right)}\left)\right)\\ +d(y_{m\left(k\right)},F(y_{n\left(k\right)-1},x_{n\left(k\right)-1}\left)\right)\left]\right)=0,\end{array}$$

which implies that ε=0, is a contradiction since ε>0. We deduce that {y _n } is a Cauchy sequence. Since (X,d) is a complete metric space, so there exist points x and y in X such that

$$\underset{n\to \infty }{lim}d(x_n,x)=0\mathit{\text{and}}\underset{n\to \infty }{lim}d(y_n,y)=0.$$

(2.15)

Now, we prove that x=F(x,y), which becomes

$$\begin{array}{ll}d\left(F\right(x,y),x)& \le d\left(F\right(x,y),x_{n+1})+d(x_{n+1},x)\\ \le d\left(F\right(x,y),x_{n+1})+d(x_{n+1},x).\end{array}$$

It follows by (2.15) that

$$d\left(F\right(x,y),x)\le \underset{n\to \infty }{limsup}d\left(F\right(x,y),x_{n+1}).$$

Then, since Φ₁ is nondecreasing and continuous, we get that

$$\begin{array}{r}{\Phi }_1\left(d\right(F(x,y),x\left)\right)\le {limsup}_{n\to \infty }{\Phi }_1\left(d\right(F(x,y),x_{n+1}\left)\right)\\ ={\Phi }_1\left({limsup}_{n\to \infty }d\right(F(x,y),x_{n+1}\left)\right).\end{array}$$

(2.16)

Now, from (2.2)

$$\begin{array}{l}{\Phi }_1\left(d\right(F(x,y),x_{n+1}\left)\right)\\ ={\Phi }_1\left(d\right(F(x,y),F(x_n,y_n)\left)\right)\\ \le {\psi }_1\left(d\right(x,x_n),d(y,y_n),d(x,F(x,y)),d(x_n,F(x_n,y_n)),\\ \frac{1}{2}\left[d\right(x,F(x_n,y_n))+d(x_n,F(x,y)\left)\right])\\ -{\psi }_2\left(d\right(x,x_n),d(y,y_n),d(x,F(x,y)),d(x_n,F(x_n,y_n)),\\ \frac{1}{2}\left[d\right(x,F(x_n,y_n))+d(x_n,F(x,y)\left)\right]).\end{array}$$

(2.17)

Passing to the upper limit as n→∞ in (2.17), we obtain using the continuity of ψ₁,ψ₂ that is

$$\begin{array}{l}\underset{n\to \infty }{limsup}{\Phi }_1\left(d\right(F(x,y),x_{n+1}\left)\right)\\ \le {\psi }_1(0,0,d(x,F(x,y)),0,\frac{1}{2}d(x,F(x,y)\left)\right)\\ -{\psi }_2(0,0,d(x,F(x,y)),0,\frac{1}{2}d(x,F(x,y)\left)\right).\end{array}$$

Therefore, from (2.16) we have

$$\begin{array}{ll}\hfill {\Phi }_1\left(d\right(F(x,y),x\left)\right)& \le {\Phi }_1\left(d\right(F(x,y),x\left)\right)\\ -{\psi }_2(0,0,d(x,F(x,y)),0,\frac{1}{2}d(x,F(x,y)\left)\right),\end{array}$$

which implies that d(x,F(x,y))=0. Thus, we deduce that

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

(2.18)

Similarly, we may show that F(y,x)=y. Thus (x,y) is a coupled fixed point of F. Suppose that assumption 2 holds. Since {x _n } is a nondecreasing sequence that converges to x in (X,d); by the assumption on X, we get that x _n ≼x for all $n\in \mathbb{N}$. Similarly, {y _n } is a nonincreasing sequence convergent to y in (X,d); by the assumption on X, we get that y _n ≽y for all $n\in \mathbb{N}$. By (2.2), we obtain

$$\begin{array}{l}{\Phi }_1\left(d\right(F(x,y),x_{n+1}\left)\right)\\ ={\Phi }_1\left(d\right(F(x,y),F(x_n,y_n)\left)\right)\\ \le {\psi }_1\left(d\right(x,x_n),d(y,y_n),d(x,F(x,y)),d(x_n,F(x_n,y_n)),\\ \frac{1}{2}\left[d\right(x,F(x_n,y_n))+d(x_n,F(x,y)\left)\right])\\ -{\psi }_2\left(d\right(x,x_n),d(y,y_n),d(x,F(x,y)),d(x_n,F(x_n,y_n)),\\ \frac{1}{2}\left[d\right(x,F(x_n,y_n))+d(x_n,F(x,y)\left)\right])\\ ={\psi }_1\left(d\right(x,x_n),d(y,y_n),d(x,F(x,y)),d(x_n,x_{n+1}),\\ \frac{1}{2}\left[d\right(x,x_{n+1})+d(x_n,F(x,y)\left)\right])\\ -{\psi }_2\left(d\right(x,x_n),d(y,y_n),d(x,F(x,y)),d(x_n,x_{n+1}),\\ \frac{1}{2}\left[d\right(x,x_{n+1})+d(x_n,F(x,y)\left)\right]).\end{array}$$

(2.19)

Passing to the limit as n→∞ in (2.19) and using the continuity of ψ₁,ψ₂, we obtain

$$\begin{array}{rcl}{\Phi }_1\left(d\right(F(x,y),x\left)\right)& \le & {\psi }_1(0,0,d(x,F(x,y)),0,\frac{1}{2}d(x,F(x,y)\left)\right)\\ -& {\psi }_2(0,0,d(x,F(x,y)),0,\frac{1}{2}d(x,F(x,y)\left)\right)\\ \le & {\Phi }_1\left(d\right(F(x,y),x\left)\right)-{\psi }_2(0,0,d(x,F(x,y)),\\ 0,\frac{1}{2}d(x,F(x,y\left)\right)),\end{array}$$

which implies that

$$d(x,F(x,y)=0\text{and so}F(x,y)=\mathrm{x.}$$

(2.20)

Similarly, we may show that F(y,x)=y. Thus, (x,y) is a coupled fixed point of F. □

A number of coupled fixed point results may be obtained by assuming different forms for the functions ψ₁ and ψ₂. In particular, fixed point results under various contractive conditions may be derived from the above theorems.

Here, for example, we derive the following corollaries from our Theorem 11.

Corollary 2.1

Let (X,≼) be a partially ordered set and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Let F:X×X→X be a given mapping having the mixed monotone property. Assume there exists k∈ [0,1) such that

$$\begin{array}{ll}\hfill d\left(F\right(x,y),F(u,v\left)\right)& \le \frac{k}{5}\left[d\right(x,u)+d(y,v)+d(x,F(x,y))\\ +d(u,F(u,v\left)\right)+\frac{1}{2}\left[d\right(x,F(u,v))\\ +d(u,F(x,y\left)\right)\left]\right]\end{array}$$

(2.21)

for x,y,u,v∈X with x≽u and y≼v. Assume either F is continuous, or X has the following properties:

• (i) If a nondecreasing sequence x _n →x, then x _n ≼x for all n.

• (ii) If a nonincreasing sequence y _n →y, then y≼y _n for all n.

If there exist x₀,y₀∈X such that x₀≼F(x₀,y₀) and y₀≽F(y₀,x₀). Then, F has a coupled fixed point, that is, there exists (x,y)∈X such that F(x,y)=x and F(y,x)=y.

Proof

Let ${\psi }_1(t_1,t_2,t_3,t_4,t_5)=\frac{1}{5}[t_1+t_2+t_3+t_4+t_5]$ and ${\psi }_2(t_1,t_2,t_3,t_4,t_5)=\frac{1-k}{5}[t_1+t_2+t_3+t_4+t_5],$ where k∈[0,1). Then Φ₁(t)=t. Now, the corollary follows from Theorem 11. □

Corollary 2.2

Let (X,≼) be a partially ordered set and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Let F:X×X→X be a given mapping having the mixed monotone property such that there exists k∈[0,1),

$$\begin{array}{ll}d\left(F\right(x,y),& F(u,v))\le kmax\{d(x,u),d(y,v),d(x,F(x,y\left)\right),\\ d(u,F(u,v\left)\right),\frac{1}{2}\left[d\right(x,F(u,v))+d(u,F(x,y)\left)\right]\left)\right\}\end{array}$$

for x,y,u,v∈X with x≽u and y≼v. Assume either F is continuous, or X has the following properties:

• (i) If a nondecreasing sequence x _n →x, then x _n ≼x for all n.

• (ii) If a nonincreasing sequence y _n →y, then y≼y _n for all n.

If there exist x₀,y₀∈X such that x₀≼F(x₀,y₀) and y₀≽F(y₀,x₀). Then, F has a coupled fixed point, that is, there exists (x,y)∈X such that F(x,y)=x and F(y,x)=y.

Proof

It suffices to take

$$\begin{array}{ll}{\psi }_1& (t_1,t_2,t_3,t_4,t_5)=max\{t_1,t_2,t_3,t_4,t_5\}\mathit{\text{and}}{\psi }_2\\ (t_1,t_2,t_3,t_4,t_4)=(1-k){\psi }_1(t_1,t_2,t_3,t_4,t_5),\end{array}$$

where k∈(0,1). Then Φ₁(t)=t. □

Now, a consequence of Corollary 2.2 by taking F(x,y)=f x where f:X→X, is the following:

Corollary 2.3

Let (X,≼) be a partially ordered set and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Let f:X→X be a nondecreasing given mapping such that

$$\begin{array}{ll}d(\mathit{\text{fx}},\mathit{\text{fu}})& \le kmax\left\{d\right(x,y),d(x,\mathit{\text{fx}}),d(y,\mathit{\text{fy}}),\\ \frac{1}{2}\left[d\right(x,\mathit{\text{fy}})+d(y,\mathit{\text{fx}}\left)\right]\left)\right\}\end{array}$$

for x,y∈X with x≽y and k∈[0,1). Assume either f is continuous, or X has the following property if a nondecreasing sequence x _n →x, then x _n ≼x for all n.If there exists x₀∈X such that x₀≼f x₀, then f has a fixed point.

Remark 12

Corollary 2.3 is the ordered version of $\acute{\mathrm{C}}\mathit{\text{iri}}\acute{\mathrm{c}}$’s Theorem [14].

Now, we illustrate our results by an example.

Example 2.4

Let X=[0,+∞) be endowed with its Euclidian metric d(x,y)=|x−y| and its usual ordering order ≤. Take F:X×X→X defined by

$$F(x,y)=\left\{\begin{array}{l}\frac{x-3y}{5}\text{if}x\ge 3y\\ 0\text{if not}.\end{array}\right.$$

X satisfies the properties (i) and (ii) in Corollary 2.1. Take $k=\frac{5}{6}$. We claim that (2.21) holds for each x≥u and y≤v. We divide the proof into the followingfour cases:

• If x≥3y and u≥3v, here we have $F(x,y)=\frac{x-3y}{5}$ and $F(u,v)=\frac{u-3v}{5}$,

  $$\begin{array}{ll}\hfill d\left(F\right(x,y),F(u,v\left)\right)& =|\frac{x-3y}{5}-\frac{u-3v}{5}|\\ =\frac{x-u}{5}+\frac{3(v-y)}{5}\\ =\frac{x-u}{6}+\frac{v-y}{6}+\frac{x-u}{30}+\frac{12(v-y)}{30}+\frac{v-y}{30}\\ \le \frac{x-u}{6}+\frac{v-y}{6}+\frac{x}{30}+\frac{4u}{30}+\frac{v}{30}\text{since}v\le \frac{u}{3}\\ \le \frac{1}{6}\left[\right(x-u)+(v-y)+(\frac{4x+3y}{5})+\frac{4u}{5}+\frac{3v}{5}]\\ =\frac{k}{5}\left[d(x,u)+d(y,v)+d(x,F(x,y\left)\right)+d(u,F(u,v\left)\right)\right]\\ \le \frac{k}{5}\left[d(x,u)+d(y,v)+d(x,F(x,y\left)\right)\right.\\ \left.+d(u,F(u,v\left)\right)+\frac{1}{2}\left[d\right(x,F(u,v))+d(u,F(x,y)\left)\right]\right].\end{array}$$

• If x≥3y and u<3v, here we have $F(x,y)=\frac{x-3y}{5}$ and F(u,v)=0,

  $$\begin{array}{ll}\hfill d\left(F\right(x,y),F(u,v\left)\right)& =\frac{x-3y}{5}\le \frac{x}{5}\\ =\frac{x-u}{6}+\frac{u}{6}+\frac{x}{30}\\ \le \frac{x-u}{6}+\frac{u}{6}+\frac{4x+3y}{30}\\ =\frac{k}{5}\left[d(x,u)+d(u,F(u,v\left)\right)+d(x,F(x,y\left)\right)\right]\\ \le \frac{k}{5}\left[d(x,u)+d(y,v)+d(x,F(x,y\left)\right)+d(u,F(u,v\left)\right)\right.\\ \left.+\frac{1}{2}\left[d\right(x,F(u,v))+d(u,F(x,y)\left)\right]\right].\end{array}$$

• If x<3y and u≥3v, here we have $F(u,v)=\frac{u-3v}{5}$ and F(x,y)=0,

  $$\begin{array}{ll}\hfill d\left(F\right(x,y),F(u,v\left)\right)& =\frac{u-3v}{5}\\ \le \frac{u}{5}=\frac{u}{6}+\frac{u}{30}\\ \le \frac{x}{6}+\frac{4u+3v}{30}\\ =\frac{k}{5}\left[d(x,F(x,y\left)\right)+d(u,F(u,v\left)\right)]\right]\\ \le \frac{k}{5}\left[d(x,u)+d(y,v)+d(x,F(x,y\left)\right)+d(u,F(u,v\left)\right)\right.\\ \left.+\frac{1}{2}\left[d\right(x,F(u,v))+d(u,F(x,y)\left)\right]\right].\end{array}$$

• If x<3y and u<3v, here we have F(u,v)=F(x,y)=0, so (2.21) holds.

Moreover, it is easy that the other hypotheses of Corollary 2.1 are verified, so F has a coupled fixed point which is (0,0).

On the other hand, Theorem 9 of Bhaskar and Lakshmikantham could not be applied in this case. Indeed, assume there exists k∈[0,1) such that (1.2) holds for x≥u and y≤v, that is,

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

If we take x=u=7, y=1 and v=2, that is, 3v≤u=x, x≥3y and y≤v, so we get that

$$\frac{3}{5}\le \frac{k}{2};$$

hence, $k\ge \frac{6}{5}>1$, it is a contradiction.

Now, as an application, it is easy to state a corollary of Theorem 11 involving a contraction of integraltype.

Corollary 2.5

Let F satisfy the conditions of Theorem 11, except that condition (2.2) is replaced by the following: there exists a positive Lebesgue integrable function ϕ on ${ℝ}_{+}$ such that $\underset{0}{\overset{\epsilon }{\int }}\varphi \left(t\right)\mathit{\text{dt}}>0$ for each ε>0 and that Then, F has a coupled fixed point.

$$\begin{array}{l}{\int }_0^{{\Phi }_1\left(d\right(F(x,y),F(u,v)\left)\right)}\varphi \left(t\right)\mathit{\text{dt}}\\ \le {\int }_0^{{\psi }_1\left(d\right(x,u),d(y,v),d(x,F(x,y)),d(u,F(u,v)),\frac{1}{2}[d(x,F(u,v\left)\right)+d(u,F(x,y\left)\right)\left]\right)}\varphi \left(t\right)\mathit{\text{dt}}\\ -{\int }_0^{{\psi }_2\left(d\right(x,u),d(y,v),d(x,F(x,y)),d(u,F(u,v)),\frac{1}{2}[d(x,F(u,v\left)\right)+d(u,F(x,y\left)\right)\left]\right)}\varphi \left(t\right)\mathrm{dt.}\end{array}$$

(2.22)