Retracted Article: Coupled fixed point theorems without continuity and mixed monotone property

Abstract

In this paper, we generalize some coupled fixed point theorems for the mixed monotone operators $F:X\times X\to X$ obtained in (Choudhury and Maity in Math. Comput. Model., 2011, doi:10.1016/j.mcm.2011.01.036) by significantly weakening the contractive condition involved and by replacing the mixed monotone property with another property which is automatically satisfied in the case of a totally ordered space. The proof follows a different and more natural new technique recently introduced by Berinde (Nonlinear Anal. 74:7347-7355, 2011). The example demonstrates that our main result is an actual improvement over the results which are generalized.

MSC:47H10, 54H25.

1 Introduction and preliminaries

Banach’s contraction principle is the most celebrated fixed point theorem. Since this principle, many authors have improved, extended and generalized this principle in many ways. Recently, Mustafa and Sims [1, 2] introduced an improved version of the generalized metric space structure, which they called a G-metric space, and established Banach’s contraction principle in this work. For more details on G-metric spaces, one can refer to the papers [1–15]. Since then, some fixed point theorems in partially ordered G-metric spaces have been considered in [16] and others.

Studies on coupled fixed point problems in partially ordered metric spaces and ordered cone metric spaces have received considerable attention in recent years ([17–24] and others). One of the reasons for this interest is their potential applicability. Specifically, Bhaskar and Lakshmikanthan [25] established coupled fixed point theorems for a mixed monotone operator in partially ordered metric spaces. Afterward, Lakshmikanthan and Ciric [26] extended the results of [25] by furnishing coupled coincidence and a coupled fixed point theorem for two commuting mappings having the mixed g-monotone property. In a subsequent series, Choudhary and Kundu [27] introduced the concept of compatibility and proved the result of [26] under a different set of some conditions. Very recently, Berinde [28] extended the results of [25] by weakening the contractive condition using a different and more natural technique, and Doric et al. [29] and Agarwal et al. [30] established coupled fixed points results without the mixed monotone property.

Recently, Choudhary and Maity [31] published coupled fixed point results in partially ordered G-metric spaces. Following the new technique of Berinde [28], we extend the result of Choudhary and Maity [31] by weakening the contractive condition involving and relaxing the mixed monotone property and continuity requirement. An illustrative example is discussed which shows that the above mentioned improvements are actual.

In what follows, we collect some related definitions and results for our further use. In 2004, Mustafa and Sims [4] introduced the concept of G-metric spaces as follows.

Definition 1.1 (see [1])

Let X be a nonempty set and let $G:X\times X\times X⟶R_{+}$ be a function satisfying the following properties:

(G1) $G(x,y,z)=0$ if $x=y=z$,

(G2) $0<G(x,x,y)$ for all $x,y\in X$ with $x\ne y$,

(G3) $G(x,x,y)\le G(x,y,z)$ for all $x,y,z\in X$ with $y\ne z$,

(G4) $G(x,y,z)=G(x,z,y)=G(y,z,x)=\cdots$ (symmetry in all three variables),

(G5) $G(x,y,z)\le G(x,a,a)+G(a,y,z)$ for all $x,y,z,a\in X$ (rectangle inequality).

Then the function G is called a generalized metric or, more specifically, a G-metric on X and the pair $(X,G)$ is called a G-metric space.

Definition 1.2 (see [1])

Let $(X,G)$ be a G-metric space and let $\{x_n\}$ be a sequence in X. A point $x\in X$ is said to be the limit of the sequence $\{x_n\}$ if

$$\underset{n,m\to +\mathrm{\infty }}{lim}G(x,x_n,x_m)=0.$$

We say that the sequence $\{x_n\}$ is G-convergent to x or $\{x_n\}$ G-converges to x.

Thus $x_n\to x$ in a G-metric space $(X,G)$ if, for any $\epsilon >0$, there exists $k\in \mathbb{N}$ such that $G(x,x_n,x_m)<\epsilon$ for all $m,n\ge k$.

Proposition 1.3 (see [1])

Let $(X,G)$ be a G-metric space. Then the following are equivalent:

• (1) $\{x_n\}$ is G-convergent to x.

• (2) $G(x_n,x_n,x)\to 0$ as $n\to +\mathrm{\infty }$.

• (3) $G(x_n,x,x)\to 0$ as $n\to +\mathrm{\infty }$.

• (4) $G(x_n,x_m,x)\to 0$ as $n,m\to +\mathrm{\infty }$.

Proposition 1.4 (see [1])

Let $(X,G)$ be a G-metric space. Then $f:X\to X$ is G-continuous at a point $x\in X$ if and only if it is G-sequentially continuous at x, that is, whenever $\{x_n\}$ is G-convergent to x, $\{f(x_n)\}$ is G-convergent to $f(x)$.

Proposition 1.5 (see [1])

Let $(X,G)$ be a G-metric space. Then the function $G(x,y,z)$ is jointly continuous in all three of its variables.

Definition 1.6 (see [31])

Let $(X,G)$ be a G-metric space. A mapping $F:X\times X\to X$ is said to be continuous on $X\times X$ if, for any two G-convergent sequences $\{x_n\}$ and $\{y_n\}$ converging to x and y, respectively, $\{F(x_n,y_n)\}$ is G-convergent to $F(x,y)$.

Definition 1.7 (see [1])

A G-metric space $(X,G)$ is called G-complete if every G-Cauchy sequence is G-convergent in $(X,G)$.

Definition 1.8 A G-metric space $(X,G)$ is called symmetric if $G(x,y,y)=G(y,x,x)$ for all $x,y\in X$.

Proposition 1.9 (see [1])

• (1) Every G-metric space $(X,G)$ defines a metric space $(X,d_G)$ by $d_G(x,y)=G(x,y,y)+G(y,x,x)$ for all $x,y\in X$.

• (2) If a G-metric space $(X,G)$ is symmetric, then $d_G(x,y)=2G(x,y,y)$ for all $x,y\in X$.

• (3) However, if $(X,G)$ is not symmetric, then it follows from G-metric properties that

  $$\frac{3}{2}G(x,y,y)\le d_G(x,y)\le 3G(x,y,y)$$

for all $x,y\in X$.

The concept of a mixed monotone property has been introduced by Bhaskar and Lakshmikantham in [25].

Definition 1.10 (see [25])

Let $(X,⪯)$ be a partially ordered set. A mapping $F:X\times X\to X$ is said to have the mixed monotone property if $F(x,y)$ is monotone nondecreasing in x and is monotone nonincreasing in y, that is, for any $x,y\in X$,

$$\begin{array}{l}{x}_1,x_2\in X,x_1⪯x_2⟹F(x_1,y)⪯F(x_2,y),\\ y_1,y_2\in X,y_1⪯y_2⟹F(x,y_2)⪯F(x,y_1).\end{array}$$

Lakshmikantham and Ćirić in [26] introduced the concept of a g-mixed monotone mapping.

Definition 1.11 (see [26])

Let $(X,⪯)$ be a partially ordered set. Let us consider the mappings $F:X\times X\to X$ and $g:X\to X$. The mapping F is said to have the mixed g-monotone property if $F(x,y)$ is monotone g-nondecreasing in x and is monotone g-nonincreasing in y, that is, for any $x,y\in X$,

$$\begin{array}{l}{x}_1,x_2\in X,g{x}_1⪯g{x}_2⟹F(x_1,y)⪯F(x_2,y),\\ y_1,y_2\in X,g{y}_1⪯g{y}_2⟹F(x,y_2)⪯F(x,y_1).\end{array}$$

Definition 1.12 (see [25])

An element $(x,y)\in X\times X$ is called a coupled fixed point of a mapping $F:X\times X\to X$ if $F(x,y)=x$ and $F(y,x)=y$.

Definition 1.13 (see [26])

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

To relax the mixed monotone property, Doric et al. [29] introduced the following condition.

If the elements x, y of a partially ordered set $(X,⪯)$ are comparable (that is, $x⪯y$ or $y⪯x$), then we write $x\asymp y$. Let $F:X\times X\to X$ be a mapping. Then consider the following condition:

$$\text{if }x,y,v\in X\text{ are such that }x\asymp F(x,y),\text{ then }F(x,y)\asymp F(F(x,y),v).$$

(1.1)

The following example shows that this condition may be satisfied when F does not have the mixed monotone property.

Example 1.14 (see [29])

Let

$$\begin{array}{l}X=\{a,b,c,d\},⪯=\{(a,a),(b,b),(c,c),(d,d),(a,b),(c,d)\},\\ F:\left(\begin{array}{cccc}(a,y)& (b,y)& (c,y)& (d,y)\\ a& b& c& d\end{array}\right)\end{array}$$

for all $y\in X$. Then F does not have the mixed monotone property since $a⪯b$ and $F(a,y)=b⪰a=F(b,y)$, while $c⪯d$ and $F(c,y)=c⪯d=F(d,y)$. But it has the condition (1.1) since $a\asymp F(a,y)=b$ and $F(a,y)=b\asymp a=F(b,v)=F(F(a,y),v)$ and $b\asymp a=F(b,y)$ and $F(b,y)=a\asymp b=F(a,v)=F(F(b,y),v)$ (the other two cases are trivial).

Using the concepts of continuity, mixed monotone property and coupled fixed point, Choudhary and Maity [31] introduced the following theorem.

Theorem 1.15 Let $(X,⪯)$ be a partially ordered set and let G be a G-metric on X such that $(X,G)$ is a complete G-metric space. Let $F:X\times X\to X$ be a continuous mapping having the mixed monotone property on X. Assume that there exists $k\in [0,1)$ such that, for all $x,y,u,v,w,z\in X$,

$$G(F(x,y),F(u,v),F(w,z))\le \frac{k}{2}[G(x,u,w)+G(y,v,z)]$$

(1.2)

for all $x,y,u,v,w,z\in X$ with $x⪰u⪰w$ and $y⪯v⪯z$, where either $u\ne w$ or $v\ne z$. If there exist $x_0$ and $y_0\in X$ such that $x_0⪯F(x_0,y_0)$ and $y_0⪰F(y_0,x_0)$, then F has a coupled fixed point in X, that is, there exist $x,y\in X$ such that $x=F(x,y)$ and $y=F(y,x)$.

In [31], Choudhary and Maity established some coupled fixed point theorems in the setting of G-metric spaces. Starting from the results in [31], our main aim of this paper is to obtain more general coupled fixed point theorems for the mappings having no mixed monotone property and satisfying a contractive condition which is more general than (1.2). Following the same approach as in [28], we weaken the contractive condition satisfied by F. Also, we relax the continuity requirement of F. The techniques of the proofs are simpler and different from those of the results in [29, 31] and others.

2 Main results

Theorem 2.1 Let $(X,⪯)$ be a partially ordered set and let G be a G-metric on X such that $(X,G)$ is a complete G-metric space. Let $F:X\times X\to X$ be a mapping satisfying the property (1.1). Assume that there exists $k\in [0,1)$ such that for $x,y,u,v,w,z\in X$, then the following holds:

$$\begin{array}{l}G(F(x,y),F(u,v),F(w,z))+G(F(y,x),F(v,u),F(z,w))\\ \le k[G(x,u,w)+G(y,v,z)]\end{array}$$

(2.1)

for all $w\asymp u\asymp x$ and $y\asymp v\asymp z$, where either $u\ne w$ or $v\ne z$. If there exist $x_0,y_0\in X$ such that

$$x_0\asymp F(x_0,y_0)\mathit{\text{and}}F(y_0,x_0)\asymp y_0,$$

(2.2)

then there exists $(\overline{x},\overline{y})\in X\times X$ such that $\overline{x}=F(\overline{x},\overline{y})$ and $\overline{y}=F(\overline{y},\overline{x})$.

Proof Consider the functional $G_3:X^2\times X^2\times X^2\to R_{+}$ defined by

$$G_3(Y,U,V)=\frac{1}{2}[G(x,u,w)+G(y,v,z)]$$

(2.3)

for all $Y=(x,y),U=(u,v),V=(w,z)\in X^2$. It is simple to check that $G_3$ is a G-metric on $X^2$ and, moreover, if $(X,G)$ is complete, then $(X^2,G)$ is a complete G-metric space, too. We consider the mapping $T:X^2\to X^2$ defined by

$$T(Y)=(F(x,y),F(y,x))$$

(2.4)

for all $Y=(x,y)\in X^2$. Clearly, for all $Y=(x,y),U=(u,v),V=(w,z)\in X^2$, in view of the definition of $G_3$, we have

$$\begin{array}{l}{G}_3(T(Y),T(U),T(V))\\ =\frac{G(F(x,y),F(u,v),F(w,z))+G(F(y,x),F(v,u),F(z,w))}{2}\end{array}$$

and

$$G_3(Y,U,V)=\frac{G(x,u,w)+G(y,v,z)}{2}.$$

Hence, by the contractive condition (2.1), we obtain the Banach-type contractive condition in a G-metric space as follows:

$$G_3(T(Y),T(U),T(V))\le k{G}_3(Y,U,V)$$

(2.5)

for all $Y,U,V\in X^2$ with $Y\ge U$ and $U\le V$. Assume that (2.2) holds. Then there exist $x_0$ and $y_0$ in X such that

$$x_0\asymp F(x_0,y_0)\text{and}F(y_0,x_0)\asymp y_0.$$

Denote $Z_0=(x_0,y_0)\in X^2$ and consider the Picard iteration associated to T and the initial approximation $Z_0$, that is, the sequence $\{Z_n\}\subset X^2$ is defined by

$$Z_{n+1}=T(Z_n)$$

(2.6)

for all $n\ge 0$, where $Z_n=(x_n,y_n)\in X^2$ for all $n\ge 0$. Since X has the condition (1.1), we have

$$Z_0=(x_0,y_0)\asymp (F(x_0,y_0),F(y_0,x_0))=(x_1,y_1)=Z_1,$$

and so, by induction,

$$Z_n=(x_n,y_n)\asymp (F(x_n,y_n),F(y_n,x_n))=(x_{n+1},y_{n+1})=Z_{n+1},$$

which shows that T is monotone and the sequence $\{Z_n\}$ is nondecreasing. We now follow the steps as in the proof of Banach’s contraction principle in a G-metric space established by Mustafa and Sims [32]. Taking $Y=Z_n\ge U=Z_{n-1}=V$ in (2.6), we have

$$G_3(T(Z_n),T(Z_{n-1}),T(Z_{n-1}))\le k{G}_3(Z_n,Z_{n-1},Z_{n-1})$$

(2.7)

for all $n\ge 1$, which implies that

$$G_3(Z_{n+1},Z_n,Z_n)\le k{G}_3(Z_n,Z_{n-1},Z_{n-1})$$

(2.8)

for all $n\ge 1$. Thus, by induction, we have

$$G_3(Z_{n+1},Z_n,Z_n)\le k^n{G}_3(Z_1,Z_0,Z_0)$$

(2.9)

for all $n\ge 1$.

Now, we claim that $\{Z_n\}$ is a Cauchy sequence in $(X^2,G_3)$. Let $m<n$. Then, by (2.9), we have

$$\begin{array}{rcl}{G}_3(Z_n,Z_m,Z_m)& \le & \sum _{i=m+1}^n{G}_3(Z_i,Z_{i-1},Z_{i-1})\\ \le & (k^m+k^{m+1}+\cdots +k^{n-m-1})G_3(Z_1,Z_0,Z_0)\\ \le & k^n\frac{1-k^{n-m-1}}{1-k}{G}_3(Z_1,Z_0,z_0).\end{array}$$

So, $\{Z_n\}$ is indeed a Cauchy sequence in a complete G-metric space $(X^2,G_3)$ and hence it is convergent. Therefore, there exists $\overline{Z}\in X^2$ such that

$$\underset{n\to \mathrm{\infty }}{lim}{Z}_n=\overline{Z}.$$

Since T is continuous in $(X^2,G_3)$, by virtue of the Lipschitzian type conditions (2.1) and (2.7), it follows that $\overline{Z}$ is a fixed point of T, that is,

$$T(\overline{Z})=\overline{Z}.$$

Let $\overline{Z}=(\overline{x},\overline{y})$. Then, by the definition of T, we obtain

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

that is, $(\overline{x},\overline{y})$ is a coupled fixed point of F. This completes the proof. □

Remark 2.2 Theorem 2.1 is more general than Theorem 1.15 which was established by Choudhary and Maity [31] since the contractive condition (2.1) is more general than the contractive condition (1.2) of Theorem 1.15. This fact is clearly illustrated by the following example.

Example 2.3 Let $X=R$ and let $G(x,y,z)=(|x-y|+|y-z|+|z-x|)$ for all $x,y\in X$ be a G-metric defined on X. Also, let $F:X\times X\to X$ be a mapping defined by

$$F(x,y)=\frac{x+3y}{5}$$

for all $(x,y)\in X^2$. Then F satisfies the conditions (2.1) and (1.1), but not (1.2) of Theorem 1.15 of [31]. Indeed, assume that there exists k, $0\le k<1$, such that (1.2) holds. This means

$$\begin{array}{l}G(F(x,y),F(u,v),F(w,z))\\ =G(\frac{x+3y}{5},\frac{u+3v}{5},\frac{w+3z}{5})\\ =|\frac{x+3y}{5}-\frac{u+3v}{5}|+|\frac{u+3v}{5}-\frac{w+3z}{5}|+|\frac{w+3z}{5}-\frac{x+3y}{5}|\\ \le \frac{k}{2}[|x-u|+|y-v|+|u-w|+|v-z|+|w-x|+|z-y|]\end{array}$$

for all $x,y,u,v,w,z\in X$ with $x\ge u\ge w$ and $y\le v\le z$. From this, in particular, for $x=u=w$ and $y=v\ne z$, we get

$$\frac{6}{5}|v-z|\le k|v-z|.$$

Thus we have $\frac{6}{5}\le k<1$, which is a contradiction.

Now, we show that (2.1) holds. Indeed, since we have, for $x=u$ and $y=v$,

$$\frac{2}{5}|u+3v-w-3z|\le \frac{2}{5}|u-w|+\frac{6}{5}|v-z|$$

and

$$\frac{2}{5}|v+3u-z-3w|\le \frac{2}{5}|v-z|+\frac{6}{5}|u-w|,$$

by adding up the above two inequalities, we get exactly (2.1) with $k=\frac{8}{11}<1$. Also, by Theorem 2.1, we obtain that F has a unique coupled fixed point, that is, $(0.0)$, but Theorem 1.16 cannot be applied to this example.

Now, to ensure the uniqueness of a coupled fixed point, we impose an additional condition used by Bhaskar and Lakshmikantham [25] and Ran and Reurings [33]:

Every pair of elements in $X^2$ has either a lower bound or an upper bound, i.e., for all $Y=(x,y)$, $\overline{Y}=(\overline{x},\overline{y})\in X^2$,

$$\text{there exists }Z=(z_1,z_2)\in X^2\text{ that is comparable to }Y\text{ and }\overline{Y}.$$

(2.10)

Theorem 2.4 Adding the condition (2.9) to the hypothesis of Theorem (2.1), we obtain the uniqueness of a coupled fixed point of F.

Proof Assume that $Z^{\ast }=(x^{\ast },y^{\ast })\in X^2$ is a coupled fixed point of F different from $\overline{Z}=(\overline{x},\overline{y})$. This means, by (G2), that $G_3(Z^{\ast },\overline{Z},\overline{Z})>0$.

Now, we discuss two cases.

Case 1. $Z^{\ast }$ is comparable to $\overline{Z}$. Since $Z^{\ast }$ is comparable to $\overline{Z}$ with respect to the ordering in $X^2$, by taking $Y=Z^{\ast }$ and $V=U=\overline{Z}$ (or $U=V=Z^{\ast }$ and $Y=\overline{Z}$) in (2.6), we obtain

$$G_3(T\left(Z^{\ast }\right),T(\overline{Z}),T(\overline{Z}))=G_3(Z^{\ast },\overline{Z},\overline{Z})\le k\cdot G_3(Z^{\ast },\overline{Z},\overline{Z}),$$

which is a contradiction since $0\le k<1$.

Case 2. $Z^{\ast }$ and $\overline{Z}$ are not comparable. In this case, there exists an upper bound or a lower bound $Z=(z_1,z_2)\in X^2$ of $Z^{\ast }$ and $\overline{Z}$. Then, in view of the monotonicity of T, $T^n(Z)$ is comparable to $T^n(Z^{\ast })=Z^{\ast }$ and $T^n(\overline{Z})=\overline{Z}$. Now, again, by the contractive condition (2.6), we have

$$\begin{array}{rcl}{G}_3(Z^{\ast },\overline{Z},\overline{Z})& =& G_3(T^n\left(Z^{\ast }\right),T^n(\overline{Z}),T^n(\overline{Z}))\\ \le & G_3(T^n\left(Z^{\ast }\right),T^n(Z),T^n(Z))+G_3(T^n(Z),T^n(\overline{Z}),T^n(\overline{Z}))\\ \le & k^n\left[G_3(Z^{\ast },Z,Z)\right)+G_3(Z,\overline{Z},\overline{Z})\to 0\end{array}$$

as $n\to \mathrm{\infty }$, which leads to a contradiction. This completes the proof. □

Next, as in [28], we show that even the components of coupled fixed points are equal.

Theorem 2.5 In addition to the hypothesis of Theorem  2.1, suppose that $x_0,y_0\in X$ are comparable. Then, for a coupled fixed point $(\overline{x},\overline{y})$, we have $\overline{x}=\overline{y}$, that is, F has a fixed point such that $F(\overline{x},\overline{x})=\overline{x}$.

Proof Consider the condition (2.2), that is,

$$x_0\asymp F(x_0,y_0)\text{and}{y}_0\asymp F(y_0,x_0).$$

Since $x_0$ and $y_0$ are comparable, we have $x_0\asymp y_0$. Then, by the condition (1.1) of F, we have

$$x_1=F(x_0,y_0)\asymp F(y_0,x_0)=y_1$$

and hence, by induction,

$$x_n\asymp y_n$$

(2.11)

for all $n\ge 0$. Now, since

$$\overline{x}=\underset{n\to \mathrm{\infty }}{lim}F(x_n,y_n);\overline{y}=\underset{n\to \mathrm{\infty }}{lim}F(y_n,x_n),$$

by the continuity of the G-metric G, we have

$$\begin{array}{rcl}G(\overline{x},\overline{y},\overline{y})& =& G(\underset{n\to \mathrm{\infty }}{lim}F(x_n,y_n),\underset{n\to \mathrm{\infty }}{lim}F(y_n,x_n),\underset{n\to \mathrm{\infty }}{lim}F(y_n,x_n))\\ =& \underset{n\to \mathrm{\infty }}{lim}G(F(x_n,y_n),F(y_n,x_n),F(y_n,x_n))\\ =& \underset{n\to \mathrm{\infty }}{lim}G(F(x_{n+1},y_{n+1}),y_{n+1}).\end{array}$$

On the other hand, by taking $Y=(x_n,y_n)$ and $U=V=(y_n,x_n)$ in (2.4), we have

$$G(F(x_n,y_n),F(y_n,x_n),F(y_n,x_n))\le kG(x_n,y_n,y_n)$$

for all $n\ge 0$, which actually means that

$$G(x_{n+1},y_{n+1},y_{n+1})\le kG(x_n,y_n,y_n)$$

for all $n\ge 0$. Therefore, we have

$$G(\overline{x},\overline{y},\overline{y})=\underset{n\to \mathrm{\infty }}{lim}G(x_{n+1},y_{n+1},y_{n+1})\le \underset{n\to \mathrm{\infty }}{lim}{k}^{n}G(x_1,y_1,y_1)=0.$$

This completes the proof. □