Coupled fixed point results on quasi-Banach spaces with application to a system of integral equations

Abstract

The aim of this paper is to obtain coupled fixed point theorems for self-mappings defined on an ordered closed and convex subset of a quasi-Banach space. Our method of proof is different and constructive in nature. As an application of our coupled fixed point results, we establish corresponding coupled coincidence point results without any type of commutativity of underlying maps. Moreover, an application to integral equations is given to illustrate the usability of the obtained results.

MSC:47H10, 54H25, 55M20.

Dedication

Dedicated to Professor Wataru Takahashi on the occasion of his seventieth birthday

1 Introduction

It is well known that the Banach contraction principle is one of the most important results in classical functional analysis. It is widely considered as the source of metric fixed point theory. Also, its significance lies in its vast applicability in a number of branches of mathematics (see [1–15]). The study of coupled fixed points in partially ordered metric spaces was first investigated in 1987 by Guo and Lakshmikantham [16], and then it attracted many researchers; see, for example, [3, 5] and references therein. Recently, Bhaskar and Lakshmikantham [12] presented coupled fixed point theorems for contractions in partially ordered metric spaces. Luong and Thuan [17] presented nice generalizations of these results. Alsulami et al. [3] further extended the work of Luong and Thuan to coupled coincidences in partial metric spaces. For more related work on coupled fixed points and coincidences, we refer the readers to recent results in [13–30].

In recent years, several authors have obtained coupled fixed point results for various classes of mappings on the setting of many generalized metric spaces. The concept of metric-type space appeared in some works, such as Czerwik [8], Khamsi [9] and Khamsi and Hussain [10]. Metric-type space is a symmetric space with some special properties. A metric-type space can also be regarded as a triplet $(X,d,K)$, where $(X,d)$ is a symmetric space and $K\ge 1$ is a real number such that

$$d(x,z)\le K(d(x,y)+d(y,z))$$

for any $x,y,z\in X$. In this paper, we adopt a different and constructive method to prove some coupled fixed and coincidence point results for contractive mappings defined on an ordered closed convex subset of a quasi-Banach space. Moreover, an application to integral equations is given to illustrate the usability of the obtained results.

2 Preliminaries

The aim of this section is to present some notions and results used in the paper.

Definition 2.1 Let X be a non-empty set and $d:X\times X\to [0,+\mathrm{\infty })$. $(X,d)$ is a symmetric space (also called E-space) if and only if it satisfies the following conditions:

• (W1) $d(x,y)=0$ if and only if $x=y$;

• (W2) $d(x,y)=d(y,x)$ for any $x,y\in X$.

Symmetric spaces differ from more convenient metric spaces in the absence of triangle inequality. Nevertheless, many notions can be defined similar to those in metric spaces. For instance, in a symmetric space $(X,d)$, the limit point of a sequence $\{x_n\}$ is defined by

$$\underset{n\to +\mathrm{\infty }}{lim}d(x_n,x)=0\text{if and only if}\underset{n\to +\mathrm{\infty }}{lim}{x}_n=x.$$

Also, a sequence $\{x_n\}\subset X$ is said to be a Cauchy sequence if, for every given $\epsilon >0$, there exists a positive integer $n(\epsilon )$ such that $d(x_m,x_n)<\epsilon$ for all $m,n\ge n(\epsilon )$. A symmetric space $(X,d)$ is said to be complete if and only if each of its Cauchy sequences converges to some $x\in X$.

Definition 2.2 Let X be a nonempty set and let $K\ge 1$ be a given real number. A function $d:X\times X\to {\mathbb{R}}_{+}$ is said to be of metric type if and only if for all $x,y,z\in X$ the following conditions are satisfied:

• (1) $d(x,y)=0$ if and only if $x=y$;

• (2) $d(x,y)=d(y,x)$;

• (3) $d(x,z)\le K[d(x,y)+d(y,z)]$.

A triplet $(X,d,K)$ is called a metric-type space.

We observe that a metric-type space is included in the class of symmetric spaces. So the notions of convergent sequence, Cauchy sequence and complete space are defined as in symmetric spaces.

Next, we give some examples of metric-type spaces.

Example 2.3 [4]

The space $l_p$ with ($0<p<1$)

$$l_p=\{\{x_n\}\subset \mathbb{R}:\sum _{n=1}^{+\mathrm{\infty }}{|x_n|}^p<+\mathrm{\infty }\},$$

together with the function $d:l_p\times l_p\to \mathbb{R}$, defined by

$$d(x,y)={\left(\sum _{n=1}^{+\mathrm{\infty }}{|x_n-y_n|}^p\right)}^{1/p},$$

where $x=\{x_n\},y=\{y_n\}\in l_p$, is a metric-type space. By an elementary calculation, we obtain $d(x,y)\le 2^{1/p}[d(x,y)+d(y,z)]$.

Example 2.4 [4]

The space $L^p$ ($0<p<1$) of all real functions $x:[0,1]\to \mathbb{R}$ such that

$${\int }_0^1{|x(t)|}^{p}dt<+\mathrm{\infty }$$

is a metric-type space if we take

$$d(x,y)={\left({\int }_0^1{|x(t)-y(t)|}^{p}dt\right)}^{1/p}$$

for each $x,y\in L^p$. The constant K is again equal to $2^{1/p}$.

For more examples of metric-type (or b-metric) spaces, we refer to [8, 10, 11]. We recall that a quasi-norm $\parallel \cdot \parallel$ defined on a real vector space X is a mapping $X\to {\mathbb{R}}_{+}$ such that:

• (1) $\parallel x\parallel >0$ if and only if $x\ne 0$;

• (2) $\parallel \lambda x\parallel =|\lambda |\parallel x\parallel$ for $\lambda \in \mathbb{R}$ and $x\in X$;

• (3) $\parallel x+y\parallel \le K[\parallel x\parallel +\parallel y\parallel ]$ for all $x,y\in X$, where $K\ge 1$ is a constant independent of x, y.

A triplet $(X,\parallel \cdot \parallel ,K)$ is called a quasi-Banach space.

What makes quasi-Banach spaces different from the more classical Banach spaces is the triangle inequality. In quasi-Banach spaces, the triangle inequality is allowed to hold approximately: $\parallel x+y\parallel \le K(\parallel x\parallel +\parallel y\parallel )$ for some constant $K\ge 1$. This relaxation leads to a broad class of spaces. Lebesgue spaces $L^p$ are Banach spaces for $1\le p\le +\mathrm{\infty }$ and only quasi-Banach spaces for $0<p<1$.

Remark 2.5 Let $(X,\parallel \cdot \parallel ,K)$ be a quasi-Banach space, then the mapping $d:X\times X\to {\mathbb{R}}_{+}$ defined by $d(x,y)=\parallel x-y\parallel$ for all $x,y\in X$ is a metric-type (b-metric). In general, a metric-type (b-metric) function d is not continuous (see [23, 26]).

The following result is useful for some of the proofs in the paper.

Lemma 2.6 Let $(X,d,K)$ be a metric-type space and let ${\{x_k\}}_{k=0}^n\subset X$. Then

$$d(x_n,x_0)\le Kd(x_0,x_1)+\cdots +K^{n-1}d(x_{n-2},x_{n-1})+K^{n-1}d(x_{n-1},x_n).$$

From Lemma 2.6, we deduce the following lemma.

Lemma 2.7 Let $\{y_n\}$ be a sequence in a metric-type space $(X,d,K)$ such that

$$d(x_n,x_{n+1})+d(y_n,y_{n+1})\le \lambda [d(x_{n-1},x_n)+d(y_{n-1},y_n)]$$

for some λ, $0<\lambda <1/K$, and each $n\in \mathbb{N}$. Then $\{x_n\}$ and $\{y_n\}$ are two Cauchy sequences in X.

Definition 2.8 (Mixed monotone property)

Let $(X,⪯)$ be a partially ordered set and $F:X\times X\to X$. We say that the mapping F has the mixed monotone property if F is monotone non-decreasing in its first argument and is monotone non-increasing in its second argument. That is, for any $x,y\in X$,

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

(2.1)

and

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

(2.2)

Definition 2.9 [13]

Let $F:X\times X\to X$. We say that $(x,y)\in X\times X$ is a coupled fixed point of F if $F(x,y)=x$ and $F(y,x)=y$.

Lakshmikantham and Ćirić [13] introduced the following concept of a mixed g-monotone mapping.

Definition 2.10 [13]

Let $(X,⪯)$ be a partially ordered set, $F:X\times X\to X$ and $g:X\to X$. We say that F has the mixed g-monotone property if F is monotone g-non-decreasing in its first argument and is monotone g-non-increasing in its second argument, that is, for any $x,y\in X$,

$$x_1,x_2\in X,g(x_1)\le g(x_2)\text{implies}F(x_1,y)\le F(x_2,y)$$

and

$$y_1,y_2\in X,g(y_1)\le g(y_2)\text{implies}F(x,y_1)\ge F(x,y_2).$$

Note that if g is the identity mapping, then this definition reduces to Definition 2.8.

Definition 2.11 [13]

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

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

3 Main results

Let $(C,⪯)$ be an ordered subset of a quasi-Banach space $(X,\parallel \cdot \parallel ,K)$. Throughout this paper, we assume that the partial order ⪯ have the following properties:

• (A) If $x⪯y$ and $\lambda \in {\mathbb{R}}_{+}$, then $\lambda x⪯\lambda y$;

• (B) If $x⪯y$ and $z\in C$, then $x+z⪯y+z$.

The following theorem is our first main result.

Theorem 3.1 Let $(C,⪯)$ be an ordered closed and convex subset of a quasi-Banach space $(X,\parallel \cdot \parallel ,K)$ where $1\le K<\sqrt{2}$ and $d:X\times X\to {\mathbb{R}}_{+}$ is such that $d(x,y)=\parallel x-y\parallel$. Assume that $F:C\times C\to C$ is a mapping with the mixed monotone property on C and suppose that there exist non-negative real numbers α, β and γ with $0\le \gamma +2\alpha +2\beta +1<2/K^2$ such that

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

(3.1)

for all $x,y,u,v\in C$, for which $u⪯x$ and $y⪯v$. Also suppose that either

• (a) F is continuous, or

• (b) C has the following property:

  $$\begin{array}{c}\mathit{\text{if a non-decreasing sequence }}\{x_n\}\to x,\mathit{\text{ then }}{x}_n⪯x\mathit{\text{ for all }}n\ge 0,\hfill \\ \mathit{\text{if a non-increasing sequence }}\{y_n\}\to y,\mathit{\text{ then }}y⪯y_n\mathit{\text{ for all }}n\ge 0.\hfill \end{array}$$

If there exist $x_0,y_0\in C$ such that $x_0⪯F(x_0,y_0)$ and $F(y_0,x_0)⪯y_0$, then there exist $x,y\in C$ such that $x=F(x,y)$ and $y=F(y,x)$, that is, F has a coupled fixed point in C.

Proof Let $x_0,y_0\in C$ be such that $x_0⪯F(x_0,y_0)$ and $y_0⪰F(y_0,x_0)$. Then

$$x_0=\frac{x_0+x_0}{2}⪯\frac{x_0+F(x_0,y_0)}{2}\text{and}{y}_0=\frac{y_0+y_0}{2}⪰\frac{y_0+F(y_0,x_0)}{2}.$$

Define $x_1,y_1\in X$ such that $x_1=\frac{x_0+F(x_0,y_0)}{2}$ and $y_1=\frac{y_0+F(y_0,x_0)}{2}$. Similarly, $x_2=\frac{x_1+F(x_1,y_1)}{2}$ and $y_2=\frac{y_1+F(y_1,x_1)}{2}$. We construct two sequences $\{x_n\}$ and $\{y_n\}$ such that

$$y_{n+1}=\frac{y_n+F(y_n,x_n)}{2}\text{for all }n\ge 0,$$

(3.2)

and

$$x_{n+1}=\frac{x_n+F(x_n,y_n)}{2}\text{for all }n\ge 0.$$

(3.3)

Let us prove that

$$x_n⪯x_{n+1}\text{and}{y}_n⪰y_{n+1}\text{for all }n\ge 0.$$

(3.4)

Since

$$x_0⪯\frac{x_0+F(x_0,y_0)}{2}=x_1\text{and}{y}_0⪰\frac{y_0+F(y_0,x_0)}{2}=y_1,$$

then (3.4) hold for $n=0$. Suppose that (3.4) hold for $n\ge 1$. Since F has the mixed monotone property, so we have

$$\begin{array}{rl}{x}_{n+1}& =\frac{x_n+F(x_n,y_n)}{2}⪯\frac{x_n+F(x_{n+1},y_n)}{2}\\ ⪯\frac{x_{n+1}+F(x_{n+1},y_n)}{2}⪯\frac{x_{n+1}+F(x_{n+1},y_{n+1})}{2}=x_{n+2}\end{array}$$

and

$$\begin{array}{rl}{y}_{n+1}& =\frac{y_n+F(y_n,x_n)}{2}⪰\frac{y_{n+1}+F(y_n,x_n)}{2}\\ ⪰\frac{y_{n+1}+F(y_{n+1},x_n)}{2}⪰\frac{y_{n+1}+F(y_{n+1},x_{n+1})}{2}=y_{n+2}.\end{array}$$

Then, by mathematical induction, it follows that (3.4) hold for all $n\ge 0$.

By (3.2) and (3.3) we have

$$x_{n+1}-x_n=\frac{x_n-x_{n-1}+[F(x_n,y_n)-F(x_{n-1},y_{n-1})]}{2}$$

and

$$y_{n+1}-y_n=\frac{y_n-y_{n-1}+[F(y_n,x_n)-F(y_{n-1},x_{n-1})]}{2}.$$

Thus

$$\parallel x_{n+1}-x_n\parallel \le \frac{K\parallel x_n-x_{n-1}\parallel +K\parallel F(x_n,y_n)-F(x_{n-1},y_{n-1})\parallel }{2}$$

and

$$\parallel y_{n+1}-y_n\parallel \le \frac{K\parallel y_n-y_{n-1}\parallel +K\parallel F(y_n,x_n)-F(y_{n-1},x_{n-1})\parallel }{2}.$$

Therefore

$$\frac{2}{K}d(x_{n+1},x_n)-d(x_n,x_{n-1})\le d(F(x_n,y_n),F(x_{n-1},y_{n-1}))$$

(3.5)

and

$$\frac{2}{K}d(y_{n+1},y_n)-d(y_n,y_{n-1})\le d(F(y_n,x_n),F(y_{n-1},x_{n-1})).$$

(3.6)

Also,

$$x_{n-1}-F(x_{n-1},y_{n-1})=2(x_{n-1}-\frac{x_{n-1}+F(x_{n-1},y_{n-1})}{2})=2(x_{n-1}-x_n).$$

Then

$$d(x_{n-1},F(x_{n-1},y_{n-1}))=2d(x_{n-1},x_n).$$

(3.7)

Similarly,

$$d(y_{n-1},F(y_{n-1},x_{n-1}))=2d(y_{n-1},y_n).$$

(3.8)

On the other hand, by (3.1) and (3.4), we have

$$\begin{array}{rcl}d(F(x_{n-1},y_{n-1}),F(x_n,y_n))& \le & \alpha d(x_{n-1},F(x_{n-1},y_{n-1}))+\beta d(y_{n-1},F(y_{n-1},x_{n-1}))\\ +\frac{\gamma }{2}[d(x_{n-1},x_n)+d(y_{n-1},y_n)]\end{array}$$

and

$$\begin{array}{rcl}d(F(y_{n-1},x_{n-1}),F(y_n,x_n))& \le & \alpha d(y_{n-1},F(y_{n-1},x_{n-1}))+\beta d(x_{n-1},F(x_{n-1},y_{n-1}))\\ +\frac{\gamma }{2}[d(y_{n-1},y_n)+d(x_{n-1},x_n)].\end{array}$$

Hence, by (3.5), (3.6), (3.7) and (3.8), we have

$$\begin{array}{rcl}\frac{2}{K}d(x_{n+1},x_n)-d(x_n,x_{n-1})& \le & 2\alpha d(x_{n-1},x_n)+2\beta d(y_{n-1},y_n)\\ +\frac{\gamma }{2}\left[(d(x_{n-1},x_n)+d(y_{n-1},y_n))\right]\end{array}$$

and

$$\begin{array}{rcl}\frac{2}{K}d(y_{n+1},y_n)-d(y_n,y_{n-1})& \le & 2\alpha d(y_{n-1},y_n)+2\beta d(x_{n-1},x_n)\\ +\frac{\gamma }{2}\left[(d(y_{n-1},y_n)+d(x_{n-1},x_n))\right].\end{array}$$

Thus

$$d(x_{n+1},x_n)+d(y_{n+1},y_n)\le \frac{K}{2}(\gamma +2\alpha +2\beta +1)(d(x_{n-1},x_n)+d(y_{n-1},y_n)).$$

By Lemma 2.7 we conclude that $\{x_n\}$ and $\{y_n\}$ are two Cauchy sequences. Then there exist $x^{\ast },y^{\ast }\in C$ such that $x_n\to x^{\ast }$ and $y_n\to y^{\ast }$.

At first, we assume that F is continuous. Hence

$$x^{\ast }=\underset{n\to \mathrm{\infty }}{lim}{x}_n=\underset{n\to \mathrm{\infty }}{lim}F(x_{n-1},y_{n-1})=F(x^{\ast },y^{\ast }).$$

Similarly,

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

That is, F has a coupled fixed point.

Now we assume that (b) holds. Since $x_n\to x^{\ast }$ and $y_n\to y^{\ast }$ as $n\to \mathrm{\infty }$, then (b) implies that $x_n⪯x^{\ast }$ and $y^{\ast }⪯y_n$ for all $n\ge 0$. Now, by (3.1) with $x=x_n$, $y=y_n$, $u=x^{\ast }$, $v=y^{\ast }$, we have

$$d(F(x_n,y_n),F(x^{\ast },y^{\ast }))\le \alpha d(x_n,F(x_n,y_n))+\beta d(y_n,F(y_n,x_n))+\frac{\gamma }{2}[d(x_n,x^{\ast })+d(y_n,y^{\ast })],$$

which implies

$$d(F(x_n,y_n),F(x^{\ast },y^{\ast }))\le 2\alpha d(x_n,x_{n+1})+2\beta d(y_n,y_{n+1})+\frac{\gamma }{2}[d(x_n,x^{\ast })+d(y_n,y^{\ast })].$$

Taking the limit as $n\to \mathrm{\infty }$ in the above inequality, we have

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

Also, by (3.3) we get ${lim}_{n\to \mathrm{\infty }}F(x_n,y_n)=x^{\ast }$. Now we can write

$$d(x^{\ast },F(x^{\ast },y^{\ast }))\le K\underset{n\to \mathrm{\infty }}{lim}d(x^{\ast },F(x_n,y_n))+K\underset{n\to \mathrm{\infty }}{lim}d(F(x_n,y_n),F(x^{\ast },y^{\ast })),$$

and hence $d(x^{\ast },F(x^{\ast },y^{\ast }))=0$. That is, $x^{\ast }=F(x^{\ast },y^{\ast })$. Similarly, $y^{\ast }=F(y^{\ast },x^{\ast })$ as required. □

If in Theorem 3.1 we take $\alpha =\beta =0$, we obtain following result.

Corollary 3.2 Let $(C,⪯)$ be an ordered closed and convex subset of a quasi-Banach space $(X,\parallel \cdot \parallel ,K)$, where $1\le K<\sqrt{2}$, and let $d:X\times X\to {\mathbb{R}}_{+}$ be such that $d(x,y)=\parallel x-y\parallel$. Assume that $F:C\times C\to C$ is a mapping such that F has the mixed monotone property on X and there exists a non-negative real number γ with $0\le \gamma +1<2/K^2$ such that

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

for all $x,y,u,v\in X$, for which $u⪯x$ and $y⪯v$. Also suppose that either

• (a) F is continuous, or

• (b) C has the following property:

  $$\begin{array}{c}\mathit{\text{if a non-decreasing sequence }}\{x_n\}\to x,\mathit{\text{ then }}{x}_n⪯x\mathit{\text{ for all }}n\ge 0,\hfill \\ \mathit{\text{if a non-increasing sequence }}\{y_n\}\to y,\mathit{\text{ then }}y⪯y_n\mathit{\text{ for all }}n\ge 0.\hfill \end{array}$$

If there exist $x_0,y_0\in C$ such that $x_0⪯F(x_0,y_0)$ and $F(y_0,x_0)⪯y_0$, then there exist $x,y\in C$ such that $x=F(x,y)$ and $y=F(y,x)$, that is, F has a coupled fixed point.

The following theorem is our second main result.

Theorem 3.3 Let $(C,⪯)$ be an ordered closed and convex subset of a quasi-Banach space $(X,\parallel \cdot \parallel ,K)$, where $1\le K<\frac{1+\sqrt{17}}{4}$, and let $d:X\times X\to {\mathbb{R}}_{+}$ be such that $d(x,y)=\parallel x-y\parallel$. Assume that $F:C\times C\to C$ is a mapping such that F has the mixed monotone property on X and there exists a non-negative real number α with $0\le \alpha +2K-1<2/K$ such that

$$d(F(x,y),F(u,v))\le \frac{\alpha }{2}\sqrt{d{(x,u)}^2+d{(y,v)}^2+\frac{1}{2K^2}d(x,F(x,y))d(y,F(y,x))}$$

(3.9)

for all $x,y,u,v\in C$, for which $u⪯x$ and $y⪯v$. Also suppose that either

• (a) F is continuous, or

• (b) C has the following property:

  $$\begin{array}{c}\mathit{\text{if a non-decreasing sequence }}\{x_n\}\to x,\mathit{\text{ then }}{x}_n⪯x\mathit{\text{ for all }}n\ge 0,\hfill \\ \mathit{\text{if a non-increasing sequence }}\{y_n\}\to y,\mathit{\text{ then }}y⪯y_n\mathit{\text{ for all }}n\ge 0.\hfill \end{array}$$

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

Proof Let $x_0,y_0\in X$ be such that $x_0⪯F(x_0,y_0)$ and $y_0⪰F(y_0,x_0)$. Then

$$x_0=\frac{(2K-1)x_0+x_0}{2K}⪯\frac{(2K-1)x_0+F(x_0,y_0)}{2K}$$

and

$$y_0=\frac{(2K-1)y_0+y_0}{2K}⪰\frac{(2K-1)y_0+F(y_0,x_0)}{2K}.$$

Define $x_1,y_1\in X$ such that $x_1=\frac{(2K-1)x_0+F(x_0,y_0)}{2}$ and $y_1=\frac{(2K-1)y_0+F(y_0,x_0)}{2K}$. Similarly, $x_2=\frac{x_1+F(x_1,y_1)}{2}$ and $y_2=\frac{(2K-1)y_1+F(y_1,x_1)}{2K}$. We construct two sequences $\{x_n\}$ and $\{y_n\}$ such that

$$y_{n+1}=\frac{(2K-1)y_n+F(y_n,x_n)}{2K}\text{for all }n\ge 0$$

(3.10)

and

$$x_{n+1}=\frac{(2K-1)x_n+F(x_n,y_n)}{2K}\text{for all }n\ge 0.$$

(3.11)

Let us prove that

$$x_n⪯x_{n+1}\text{and}{y}_n⪰y_{n+1}\text{for all }n\ge 0.$$

(3.12)

As

$$x_0⪯\frac{(2K-1)x_0+F(x_0,y_0)}{2K}=x_1\text{and}{y}_0⪰\frac{(2K-1)y_0+F(y_0,x_0)}{2K}=y_1,$$

so (3.12) hold for $n=0$. Suppose that (3.12) hold for $n\ge 1$. As F has the mixed monotone property, so

$$\begin{array}{rcl}{x}_{n+1}=\frac{(2K-1)x_n+F(x_n,y_n)}{2K}& ⪯& \frac{(2K-1)x_n+F(x_{n+1},y_n)}{2K}\\ ⪯& \frac{(2K-1)x_{n+1}+F(x_{n+1},y_n)}{2K}\\ ⪯& \frac{(2K-1)x_{n+1}+F(x_{n+1},y_{n+1})}{2K}\\ =& x_{n+2}\end{array}$$

and

$$\begin{array}{rcl}{y}_{n+2}=\frac{(2K-1)y_{n+1}+F(y_{n+1},x_{n+1})}{2K}& ⪯& \frac{(2K-1)y_{n+1}+F(y_{n+1},x_n)}{2K}\\ ⪯& \frac{(2K-1)y_{n+1}+F(y_n,x_n)}{2K}\\ ⪯& \frac{(2K-1)y_n+F(y_n,x_n)}{2K}\\ =& y_{n+1}.\end{array}$$

Then, by mathematical induction, it follows that (3.12) holds for all $n\ge 0$.

By (3.10) and (3.11), we have

$$x_{n+1}-x_n=\frac{(2K-1)(x_n-x_{n-1})+[F(x_n,y_n)-F(x_{n-1},y_{n-1})]}{2K}$$

and

$$y_{n+1}-y_n=\frac{(2K-1)(y_n-y_{n-1})+[F(y_n,x_n)-F(y_{n-1},x_{n-1})]}{2K}.$$

Thus

$$\parallel x_{n+1}-x_n\parallel \le \frac{(2K-1)\parallel x_n-x_{n-1}\parallel +\parallel F(x_n,y_n)-F(x_{n-1},y_{n-1})\parallel }{2}$$

and

$$\parallel y_{n+1}-y_n\parallel \le \frac{(2K-1)\parallel y_n-y_{n-1}\parallel +\parallel F(y_n,x_n)-F(y_{n-1},x_{n-1})\parallel }{2}.$$

Therefore

$$2d(x_{n+1},x_n)-(2K-1)d(x_n,x_{n-1})\le d(F(x_n,y_n),F(x_{n-1},y_{n-1}))$$

(3.13)

and

$$2d(y_{n+1},y_n)-(2K-1)d(y_n,y_{n-1})\le d(F(y_n,x_n),F(y_{n-1},x_{n-1})).$$

(3.14)

Also, we have

$$x_{n-1}-F(x_{n-1},y_{n-1})=2K(x_{n-1}-\frac{(2K-1)x_{n-1}+F(x_{n-1},y_{n-1})}{2K})=2K(x_{n-1}-x_n),$$

which implies

$$d(x_{n-1},F(x_{n-1},y_{n-1}))=2Kd(x_{n-1},x_n).$$

(3.15)

Similarly, we have

$$d(y_{n-1},F(y_{n-1},x_{n-1}))=2Kd(y_{n-1},y_n).$$

(3.16)

Now, by (3.9), (3.13) and (3.15), (3.16), we have

$$\begin{array}{r}2d(x_{n+1},x_n)-(2K-1)d(x_n,x_{n-1})\\ \le \frac{\alpha }{2}\sqrt{d{(x_{n-1},x_n)}^2+d{(y_{n-1},y_n)}^2+2d(x_{n-1},x_n)d(y_{n-1},y_n)}\\ =\frac{\alpha }{2}\left[(d(x_{n-1},x_n)+d(y_{n-1},y_n))\right].\end{array}$$

Similarly,

$$2d(y_{n+1},y_n)-(2K-1)d(y_n,y_{n-1})\le \frac{\alpha }{2}[(d(y_{n-1},y_n)+d(x_{n-1},x_n)].$$

Thus

$$d(x_{n+1},x_n)+d(y_{n+1},y_n)\le \frac{1}{2}(\alpha +2K-1)(d(x_{n-1},x_n)+d(y_{n-1},y_n)).$$

By Lemma 2.7, we conclude that $\{x_n\}$ and $\{y_n\}$ are Cauchy sequences. Thus, there exist $x^{\ast },y^{\ast }\in C$ such that $x_n\to x^{\ast }$ and $y_n\to y^{\ast }$.

Now, proceeding as in the proof of Theorem 3.1, we can prove that $(x^{\ast },y^{\ast })$ is a coupled fixed point of F. □

Since

$$\begin{array}{r}\frac{\alpha }{2}\sqrt{\frac{1}{2K^2}d(x,F(x,y))d(y,F(y,x))}\\ \le \frac{\alpha }{2}\sqrt{d{(x,u)}^2+d{(y,v)}^2+\frac{1}{2K^2}d(x,F(x,y))d(y,F(y,x))},\end{array}$$

so by Theorem 3.3 we obtain the following result.

Corollary 3.4 Let $(C,⪯)$ be an ordered closed and convex subset of a quasi-Banach space $(X,\parallel \cdot \parallel ,K)$, where $1\le K<\frac{1+\sqrt{17}}{4}$, and let $d:X\times X\to {\mathbb{R}}_{+}$ be such that $d(x,y)=\parallel x-y\parallel$. Assume that $F:C\times C\to C$, F has the mixed monotone property on C and for a non-negative real number α with $0\le \alpha +2K-1<2/K$, F satisfies following inequality:

$$d(F(x,y),F(u,v))\le \frac{\alpha \sqrt{2}}{4K}\sqrt{d(x,F(x,y))d(y,F(y,x))}$$

for all $x,y,u,v\in X$, for which $u⪯x$ and $y⪯v$. Also suppose that either

• (a) F is continuous, or

• (b) C has the following property:

  $$\begin{array}{c}\mathit{\text{if a non-decreasing sequence }}\{x_n\}\to x,\mathit{\text{ then }}{x}_n⪯x\mathit{\text{ for all }}n\ge 0,\hfill \\ \mathit{\text{if a non-increasing sequence }}\{y_n\}\to y,\mathit{\text{then }}y⪯y_n\mathit{\text{ for all }}n\ge 0.\hfill \end{array}$$

If there exist $x_0,y_0\in C$ such that $x_0⪯F(x_0,y_0)$ and $F(y_0,x_0)⪯y_0$, then F has a coupled fixed point.

By taking $K=1$ in the above proved results, we can obtain the following couple fixed results in Banach spaces.

Corollary 3.5 Let $(C,⪯)$ be an ordered closed and convex subset of a Banach space $(X,\parallel \cdot \parallel )$, and let $F:C\times C\to C$ be a mapping such that F has the mixed monotone property on C. Suppose that there exist non-negative real numbers α, β and γ with $0\le \gamma +2\alpha +2\beta <1$ such that

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

for all $x,y,u,v\in C$ with $u⪯x$ and $y⪯v$. Also suppose that either

• (a) F is continuous, or

• (b) C has the following property:

  $$\begin{array}{c}\mathit{\text{if a non-decreasing sequence }}\{x_n\}\to x,\mathit{\text{ then }}{x}_n⪯x\mathit{\text{ for all }}n\ge 0,\hfill \\ \mathit{\text{if a non-increasing sequence }}\{y_n\}\to y,\mathit{\text{ then }}y⪯y_n\mathit{\text{ for all }}n\ge 0.\hfill \end{array}$$

If there exist $x_0,y_0\in C$ such that $x_0⪯F(x_0,y_0)$ and $F(y_0,x_0)⪯y_0$, then F has a coupled fixed point.

Corollary 3.6 Let $(C,⪯)$ be an ordered closed and convex subset of a Banach space $(X,\parallel \cdot \parallel )$, and let $F:C\times C\to C$ be a mapping such that F has the mixed monotone property on C. Suppose that there exists a non-negative real number γ with $0\le \gamma <1$ such that

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

for all $x,y,u,v\in C$ with $u⪯x$ and $y⪯v$. Also suppose that either

• (a) F is continuous, or

• (b) C has the following property:

  $$\begin{array}{c}\mathit{\text{if a non-decreasing sequence }}\{x_n\}\to x,\mathit{\text{ then }}{x}_n⪯x\mathit{\text{ for all }}n\ge 0,\hfill \\ \mathit{\text{if a non-increasing sequence }}\{y_n\}\to y,\mathit{\text{ then }}y⪯y_n\mathit{\text{ for all }}n\ge 0.\hfill \end{array}$$

If there exist $x_0,y_0\in C$ such that $x_0⪯F(x_0,y_0)$ and $F(y_0,x_0)⪯y_0$, then F has a coupled fixed point.

Corollary 3.7 Let $(C,⪯)$ be an ordered closed and convex subset of a Banach space $(X,\parallel \cdot \parallel )$, and let $F:C\times C\to C$ be a mapping such that F has the mixed monotone property on C. Suppose that there exists a non-negative real number α with $0\le \alpha <1$ such that

$$d(F(x,y),F(u,v))\le \frac{\alpha }{2}\sqrt{d{(x,u)}^2+d{(y,v)}^2+\frac{1}{2}d(x,F(x,y))d(y,F(y,x))}$$

for all $x,y,u,v\in C$ with $u⪯x$ and $y⪯v$. Also suppose that either

• (a) F is continuous, or

• (b) C has the following property:

  $$\mathit{\text{if a non-decreasing sequence }}\{x_n\}\to x,\mathit{\text{ then }}{x}_n⪯x\mathit{\text{ for all }}n\ge 0,$$

  (3.17)
  $$\mathit{\text{if a non-increasing sequence }}\{y_n\}\to y,\mathit{\text{ then }}y⪯y_n\mathit{\text{ for all }}n\ge 0.$$

  (3.18)

If there exist $x_0,y_0\in C$ such that $x_0⪯F(x_0,y_0)$ and $F(y_0,x_0)⪯y_0$, then F has a coupled fixed point.

Corollary 3.8 Let $(C,⪯)$ be an ordered closed and convex subset of a Banach space $(X,\parallel \cdot \parallel )$, and let $F:C\times C\to C$ be a mapping such that F has the mixed monotone property on C. Suppose that there exists a non-negative real number α with $0\le \alpha <1$ such that

$$d(F(x,y),F(u,v))\le \frac{\alpha \sqrt{2}}{4}\sqrt{d(x,F(x,y))d(y,F(y,x))}$$

for all $x,y,u,v\in C$ with $u⪯x$ and $y⪯v$. Also suppose that either

• (a) F is continuous, or

• (b) C has the following property:

  $$\begin{array}{c}\mathit{\text{if a non-decreasing sequence }}\{x_n\}\to x,\mathit{\text{ then }}{x}_n⪯x\mathit{\text{ for all }}n\ge 0,\hfill \\ \mathit{\text{if a non-increasing sequence }}\{y_n\}\to y,\mathit{\text{ then }}y⪯y_n\mathit{\text{ for all }}n\ge 0.\hfill \end{array}$$

If there exist $x_0,y_0\in C$ such that $x_0⪯F(x_0,y_0)$ and $F(y_0,x_0)⪯y_0$, then F has a coupled fixed point.

The following lemma is an easy consequence of the axiom of choice (see p.5 [25], AC5: For every function $f:X\to X$, there is a function g such that $D(g)=R(f)$ and for every $x\in D(g)$ , $f(gx)=x$).

Lemma 3.9 Let X be a nonempty set and $g:X\to X$ be a mapping. Then there exists a subset $E\subseteq X$ such that $g(E)=g(X)$ and $g:E\to X$ is one-to-one.

As an application of Theorem 3.1, we now establish a coupled coincidence point result.

Theorem 3.10 Let $(C,⪯)$ be a nonempty ordered subset of a quasi-Banach space $(X,\parallel \cdot \parallel ,K)$, where $1\le K<\sqrt{2}$, and let $d:X\times X\to {\mathbb{R}}_{+}$ be such that $d(x,y)=\parallel x-y\parallel$. Assume that $g:C\to C$ and $F:C\times C\to C$ are mappings where F has the mixed g-monotone property on C, $g(C)$ is closed and convex and $F(C\times C)\subseteq g(C)$. Suppose that there exist non-negative real numbers α, β and a real number γ with $0\le \gamma +2\alpha +2\beta +1<2/K^2$ such that

$$d(F(x,y),F(u,v))\le \alpha d(gx,F(x,y))+\beta d(gy,F(y,x))+\frac{\gamma }{2}[d(gx,gu)+d(gy,gv)]$$

(3.19)

for all $x,y,u,v\in C$, for which $gu⪯gx$ and $gy⪯gv$. Also suppose that either

• (a) F is continuous, or

• (b) C has the following property:

  $$\begin{array}{c}\mathit{\text{if a non-decreasing sequence }}\{g{x}_n\}\to gx,\mathit{\text{ then }}g{x}_n⪯gx\mathit{\text{ for all }}n\ge 0,\hfill \\ \mathit{\text{if a non-increasing sequence }}\{g{y}_n\}\to gy,\mathit{\text{ then }}gy⪯g{y}_n\mathit{\text{ for all }}n\ge 0.\hfill \end{array}$$

If there exist $x_0,y_0\in C$ such that $g(x_0)⪯F(x_0,y_0)$ and $F(y_0,x_0)⪯g(y_0)$, then there exist $x,y\in C$ such that $gx=F(x,y)$ and $gy=F(y,x)$, that is, F and g have a coupled coincidence point in C.

Proof Using Lemma 3.9, there exists $E\subseteq C$ such that $g(E)=g(C)$ and $g:E\to C$ is one-to-one. We define a mapping $G:g(E)\times g(E)\to g(E)$ by

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

(3.20)

for all $gx,gy\in g(E)$. As g is one-to-one on $g(E)$ and $F(C\times C)\subseteq g(C)$, so G is well defined. Thus, it follows from (3.19) and (3.20) that

$$\begin{array}{r}d(G(gx,gy),F(gu,gv))+\alpha d(gx,G(gx,gy))+\beta d(gy,G(gy,gx))\\ \le \frac{\gamma }{2}[d(gx,gu)+d(gy,gv)]\end{array}$$

(3.21)

for all $gx,gy,gu,gv\in g(C)$, for which $g(x)⪯g(u)$ and $g(y)⪰g(v)$. Since F has the mixed g-monotone property, for all $gx,gy\in g(C)$,

$$g{x}_1,g{x}_2\in g(C),g(x_1)⪯g(x_2)\text{implies}G(g{x}_1,gy)⪯G(g{x}_2,gy)$$

(3.22)

and

$$g{y}_1,g{y}_2\in g(C),g(y_1)⪯g(y_2)\text{implies}G(gx,g{y}_1)⪰G(gx,g{y}_2),$$

(3.23)

which imply that G has the mixed monotone property. Also, there exist $x_0,y_0\in C$ such that

$$g(x_0)⪯F(x_0,y_0)\text{and}g(y_0)⪰F(y_0,x_0).$$

This implies that there exist $g{x}_0,g{y}_0\in g(C)$ such that

$$g(x_0)⪯G(g{x}_0,g{y}_0)\text{and}g(y_0)⪰G(g{y}_0,g{x}_0).$$

Suppose that assumption (a) holds. Since F is continuous, G is also continuous. Using Theorem 3.1 to the mapping G, it follows that G has a coupled fixed point $(u,v)\in g(C)\times g(C)$.

Suppose that assumption (b) holds. We conclude similarly that the mapping G has a coupled fixed point $(u,v)\in g(C)\times g(C)$. Finally, we prove that F and g have a coupled coincidence point. Since $(u,v)$ is a coupled fixed point of G, we get

$$u=G(u,v)\text{and}v=G(v,u).$$

(3.24)

Since $(u,v)\in g(C)\times g(C)$, there exists a point $(u_0,v_0)\in C\times C$ such that

$$u=g{u}_0\text{and}v=g{v}_0.$$

(3.25)

It follows from (3.24) and (3.25) that

$$g{u}_0=G(g{u}_0,g{v}_0)\text{and}g{v}_0=G(g{v}_0,g{u}_0).$$

(3.26)

Combining (3.20) and (3.26), we get

$$g{u}_0=F(u_0,v_0)\text{and}g{v}_0=F(v_0,u_0).$$

Thus, $(u_0,v_0)$ is a required coupled coincidence point of F and g. This completes the proof. □

Similarly, as an application of Theorem 3.3, we can prove the following coupled coincidence point result.

Theorem 3.11 Let $(C,⪯)$ be a nonempty ordered subset of a quasi-Banach space $(X,\parallel \cdot \parallel ,K)$, where $1\le K<\frac{1+\sqrt{17}}{4}$, and let $d:X\times X\to {\mathbb{R}}_{+}$ be such that $d(x,y)=\parallel x-y\parallel$. Assume that $g:C\to C$ and $F:C\times C\to C$ are mappings where F has the mixed g-monotone property on C, $g(C)$ is closed and convex and $F(C\times C)\subseteq g(C)$. Suppose that there exists a real number α with $0\le \alpha +2K-1<2/K$ such that

$$\begin{array}{r}d(F(x,y),F(u,v))\\ \le \frac{\alpha }{2}\sqrt{d{(gx,gu)}^2+d{(gy,gv)}^2+\frac{1}{2K^2}d(gx,F(x,y))d(gy,F(y,x))}\end{array}$$

(3.27)

for all $x,y,u,v\in C$, for which $gu⪯gx$ and $gy⪯gv$. Also suppose that either

• (a) F is continuous, or

• (b) C has the following property:

  $$\begin{array}{c}\mathit{\text{if a non-decreasing sequence }}\{g{x}_n\}\to gx,\mathit{\text{ then }}g{x}_n⪯gx\mathit{\text{ for all }}n\ge 0,\hfill \\ \mathit{\text{if a non-increasing sequence }}\{g{y}_n\}\to gy,\mathit{\text{ then }}gy⪯g{y}_n\mathit{\text{ for all }}n\ge 0.\hfill \end{array}$$

If there exist $x_0,y_0\in C$ such that $g(x_0)⪯F(x_0,y_0)$ and $F(y_0,x_0)⪯g(y_0)$, then F and g have a coupled coincidence point in C.

4 Existence of a solution for a system of integral equations

We consider the space $\mathcal{X}=C([0,T],\mathbb{R})$ of continuous functions defined on $I=[0,T]$ endowed with the structure $(\mathcal{X},\parallel \cdot \parallel )$ given by

$$\parallel u\parallel =\underset{t\in [0,T]}{sup}|u(t)|$$

for all $u\in \mathcal{X}$. We endow $\mathbb{X}$ with the partial order ⪯ given by

$$x⪯y⟺x(t)\le y(t)\text{for all }t\in [0,T].$$

Clearly, the partial order ⪯ satisfies conditions A and B. Further, it is known that $(\mathcal{X},d,⪯)$ is regular [24], that is,

$$\begin{array}{c}\text{if a non-decreasing sequence }\{x_n\}\to x,\text{ then }{x}_n⪯x\text{ for all }n\ge 0,\hfill \\ \text{if a non-increasing sequence }\{y_n\}\to y,\text{then }y⪯y_n\text{ for all }n\ge 0.\hfill \end{array}$$

Motivated by the work in [1, 17, 18, 26], we study the existence of solutions for a system of nonlinear integral equations using the results proved in the previous section.

Consider the integral equations in the following system.

$$\begin{array}{r}x(t)=P(t)+{\int }_0^{T}S(t,r)[f(r,x(r))+k(r,y(r))]dr,\\ y(t)=P(t)+{\int }_0^{T}S(t,r)[f(r,y(r))+k(r,x(r))]dr.\end{array}$$

(4.1)

We will consider system (4.1) under the following assumptions:

• (i) $f,k:[0,T]\times \mathbb{R}\to \mathbb{R}$ are continuous;

• (ii) $P:[0,T]\to \mathbb{R}$ is continuous;

• (iii) $S:[0,T]\times \mathbb{R}\to [0,\mathrm{\infty })$ is continuous;

• (iv) there exist $a,b,c>0$ with $0\le 2a+2b+c<1$ such that for all $r\in [0,T]$ and all $x(r),y(r),u(r),v(r)\in \mathcal{X}$ with $u(r)⪯x(r)⪯y(r)⪯v(r)$, we have

  $$\begin{array}{c}\begin{array}{rl}0\le & f(r,y(r))-f(r,x(r))\le a|x(r)-F(x(r),y(r))|+b|y(r)-F(y(r),x(r))|\\ +\frac{c}{2}[(x(r)-u(r))+(y(r)-v(r))],\end{array}\hfill \\ \begin{array}{rl}0\le & k(r,x(r))-k(r,y(r))\le a|x(r)-F(x(r),y(r))|+b|y(r)-F(y(r),x(r))|\\ +\frac{c}{2}[(x(r)-u(r))+(y(r)-v(r))],\end{array}\hfill \end{array}$$

where

$$F(x,y)(t)=P(t)+{\int }_0^{T}S(t,r)[f(r,x(r))+k(r,y(r))]dr;$$

• (v) there exist continuous functions $\alpha ,\gamma :[0,T]\to \mathbb{R}$ such that

  $$\begin{array}{c}\alpha (t)\le P(t)+{\int }_0^{T}S(t,r)[f(r,\alpha (r))+k(r,\gamma (r))]dr,\hfill \\ \gamma (t)\ge P(t)+{\int }_0^{T}S(t,r)[f(r,\gamma (r))+k(r,\alpha (r))]dr;\hfill \end{array}$$

• (vi) assume that

  $$\underset{t\in [0,T]}{sup}{\int }_0^{T}S(t,r)dr\le 1/2.$$

Theorem 4.1 Under assumptions (i)-(vi), system (4.1) has a solution in ${\mathcal{X}}^2$, where $\mathcal{X}=(C([0,T],\mathbb{R}))$ is defined above.

Proof We consider the operator $F:{\mathcal{X}}^2\to \mathcal{X}$ defined by

$$F(x_1,x_2)(t)=P(t)+{\int }_0^{T}S(t,r)[f(r,x_1(r))+k(r,x_2(r))]dr$$

for all $t,r\in [0,T]$, $x_1,x_2\in \mathcal{X}$.

Clearly, F has the mixed monotone property [26].

Let $x,y,u,v\in \mathcal{X}$ with $u⪯x⪯y⪯v$. Since F has the mixed monotone property, we have

$$F(u,v)⪯F(x,y).$$

Notice that

$$\begin{array}{r}|F(x,y)(t)-F(u,v)(t)|\\ =|{\int }_0^{T}S(t,r)[f(r,x(r))-f(r,u(r))]dr+{\int }_0^{T}S(t,r)[k(r,y(r))-k(r,v(r))]dr|\\ \le {\int }_0^{T}S(t,r)\left[|f(r,x(r))-f(r,u(r))|\right]dr+{\int }_0^{T}S(t,r)\left[|k(r,y(r))-k(r,v(r))|\right]dr\\ \le {\int }_0^{T}S(t,r)[a|x(r)-F(x(r),y(r))|+b|y(r)-F(y(r),x(r))|\\ +\frac{c}{2}[|x(r)-u(r)|+|y(r)-v(r)|]]dr\\ +{\int }_0^{T}S(t,r)[a|x(r)-F(x(r),y(r))|+b|y(r)-F(y(r),x(r))|\\ +\frac{c}{2}[|x(r)-u(r)|+|y(r)-v(r)|]]dr\\ =2{\int }_0^{T}S(t,r)[a|x(r)-F(x(r),y(r))|+b|y(r)-F(y(r),x(r))|\\ +\frac{c}{2}[|x(r)-u(r)|+|y(r)-v(r)|]]dr\\ \le 2\underset{t\in [0,T]}{sup}{\int }_0^{T}S(t,r)dr[a\parallel x-F(x,y)\parallel +b\parallel y-F(y,x)\parallel +\frac{c}{2}(\parallel x-u\parallel +\parallel y-v\parallel )]\\ \le a\parallel x-F(x,y)\parallel +b\parallel y-F(y,x)\parallel +\frac{c}{2}(\parallel x-u\parallel +\parallel y-v\parallel ).\end{array}$$

Thus,

$$\parallel F(x,y)-F(u,v)\parallel \le a\parallel x-F(x,y)\parallel +b\parallel y-F(y,x)\parallel +\frac{c}{2}(\parallel x-u\parallel +\parallel y-v\parallel ).$$

Further, by (v), we get

$$\alpha \le F(\alpha ,\gamma ),\gamma \ge F(\gamma ,\alpha ).$$

All of the conditions of Corollary 3.5 are satisfied, so we deduce the existence of $x_1,x_2\in \mathcal{X}$ such that $x_1=F(x_1,x_2)$ and $x_2=F(x_2,x_1)$. □