Common coupled fixed point theorems for weakly compatible mappings in fuzzy metric spaces

Abstract

In this paper, we prove a common fixed point theorem for weakly compatible mappings under ϕ-contractive conditions in fuzzy metric spaces. We also give an example to illustrate the theorem. The result is a genuine generalization of the corresponding result of Hu (Fixed Point Theory Appl. 2011:363716, 2011, doi:10.1155/2011/363716). We also indicate a minor mistake in Hu (Fixed Point Theory Appl. 2011:363716, 2011, doi:10.1155/2011/363716).

1 Introduction

In 1965, Zadeh [1] introduced the concept of fuzzy sets. Then many authors gave the important contribution to development of the theory of fuzzy sets and applications. George and Veeramani [2, 3] gave the concept of a fuzzy metric space and defined a Hausdorff topology on this fuzzy metric space, which have very important applications in quantum particle physics, particularly, in connection with both string and E-infinity theory.

Bhaskar and Lakshmikantham [4], Lakshmikantham and Ćirić [5] discussed the mixed monotone mappings and gave some coupled fixed point theorems, which can be used to discuss the existence and uniqueness of solution for a periodic boundary value problem. Sedghi et al. [6] gave a coupled fixed point theorem for contractions in fuzzy metric spaces, and Jin-xuan Fang [7] gave some common fixed point theorems for compatible and weakly compatible ϕ-contractions mappings in Menger probabilistic metric spaces. Xin-Qi Hu [8] proved a common fixed point theorem for mappings under φ-contractive conditions in fuzzy metric spaces. Many authors [9–26] proved fixed point theorems in (intuitionistic) fuzzy metric spaces or probabilistic metric spaces.

In this paper, we give a new coupled fixed point theorem under weaker conditions than in [8] and give an example, which shows that the result is a genuine generalization of the corresponding result in [8].

2 Preliminaries

First, we give some definitions.

Definition 2.1 (see [2])

A binary operation $\ast :[0,1]\times [0,1]\to [0,1]$ is a continuous t-norm if ∗ satisfies the following conditions:

• (1) ∗ is commutative and associative,

• (2) ∗ is continuous,

• (3) $a\ast 1=a$ for all $a\in [0,1]$,

• (4) $a\ast b\le c\ast d$ whenever $a\le c$ and $b\le d$ for all $a,b,c,d\in [0,1]$.

Definition 2.2 (see [27])

Let ${sup}_{0<t<1}\mathrm{\Delta }(t,t)=1$. A t-norm Δ is said to be of H-type if the family of functions ${\{{\mathrm{\Delta }}^m(t)\}}_{m=1}^{\mathrm{\infty }}$ is equicontinuous at $t=1$, where

$${\mathrm{\Delta }}^1(t)=t\mathrm{\Delta }t,{\mathrm{\Delta }}^{m+1}(t)=t\mathrm{\Delta }({\mathrm{\Delta }}^m(t)),m=1,2,\dots ,t\in [0,1].$$

(2.1)

The t-norm ${\mathrm{\Delta }}_M=min$ is an example of t-norm of H-type, but there are some other t-norms Δ of H-type [27].

Obviously, Δ is a t-norm of H-type if and only if for any $\lambda \in (0,1)$, there exists $\delta (\lambda )\in (0,1)$ such that ${\mathrm{\Delta }}^m(t)>1-\lambda$ for all $m\in \mathbb{N}$, when $t>1-\delta$.

Definition 2.3 (see [2])

A 3-tuple $(X,M,\ast )$ is said to be a fuzzy metric space if X is an arbitrary nonempty set, ∗ is a continuous t-norm and M is a fuzzy set on $X^2\times (0,+\mathrm{\infty })$ satisfying the following conditions for each $x,y,z\in X$ and $t,s>0$,

(FM-1) $M(x,y,t)>0$,

(FM-2) $M(x,y,t)=1$ if and only if $x=y$,

(FM-3) $M(x,y,t)=M(y,x,t)$,

(FM-4) $M(x,y,t)\ast M(y,z,s)\le M(x,z,t+s)$,

(FM-5) $M(x,y,\cdot ):(0,\mathrm{\infty })\to [0,1]$ is continuous.

We shall consider a fuzzy metric space $(X,M,\ast )$, which satisfies the following condition:

$$\underset{t\to +\mathrm{\infty }}{lim}M(x,y,t)=1,\mathrm{\forall }x,y\in X.$$

(2.2)

Let $(X,M,\ast )$ be a fuzzy metric space. For $t>0$, the open ball $B(x,r,t)$ with a center $x\in X$ and a radius $0<r<1$ is defined by

$$B(x,r,t)=\{y\in X:M(x,y,t)>1-r\}.$$

(2.3)

A subset $A\subset X$ is called open if for each $x\in A$, there exist $t>0$ and $0<r<1$ such that $B(x,r,t)\subset A$. Let τ denote the family of all open subsets of X. Then τ is called the topology on X, induced by the fuzzy metric M. This topology is Hausdorff and first countable.

Example 2.4 Let $(X,d)$ be a metric space. Define t-norm $a\ast b=ab$ or $a\ast b=min\{a,b\}$ and for all $x,y\in X$ and $t>0$, $M(x,y,t)=\frac{t}{t+d(x,y)}$. Then $(X,M,\ast )$ is a fuzzy metric space.

Definition 2.5 (see [2])

Let $(X,M,\ast )$ be a fuzzy metric space. Then

• (1) a sequence $\{x_n\}$ in X is said to be convergent to x (denoted by ${lim}_{n\to \mathrm{\infty }}{x}_n=x$) if

  $$\underset{n\to \mathrm{\infty }}{lim}M(x_n,x,t)=1$$

  for all $t>0$.

• (2) A sequence $\{x_n\}$ in X is said to be a Cauchy sequence if for any $\epsilon >0$, there exists $n_0\in \mathbb{N}$, such that

  $$M(x_n,x_m,t)>1-\epsilon$$

  for all $t>0$ and $n,m\ge n_0$.

• (3) A fuzzy metric space $(X,M,\ast )$ is said to be complete if and only if every Cauchy sequence in X is convergent.

Remark 2.6 (see [9])

Let $(X,M,\ast )$ be a fuzzy metric space. Then

• (1) for all $x,y\in X$, $M(x,y,\cdot )$ is non-decreasing;

• (2) if $x_n\to x$, $y_n\to y$, $t_n\to t$, then

  $$\underset{n\to \mathrm{\infty }}{lim}M(x_n,y_n,t_n)=M(x,y,t);$$

• (3) if $M(x,y,t)>1-r$ for x, y in X, $t>0$, $0<r<1$, then we can find a $t_0$, $0<t_0<t$ such that $M(x,y,t_0)>1-r$;

• (4) for any $r_1>r_2$, we can find a $r_3$ such that $r_1\ast r_3\ge r_2$, and for any $r_4$, we can find a $r_5$ such that $r_5\ast r_5\ge r_4$ ($r_1,r_2,r_3,r_4,r_5\in (0,1)$).

Define $\mathrm{\Phi }=\{\varphi :R^{+}\to R^{+}\}$, where $R^{+}=[0,+\mathrm{\infty })$ and each $\varphi \in \mathrm{\Phi }$ satisfies the following conditions:

(ϕ-1) ϕ is non-decreasing,

(ϕ-2) ϕ is upper semi-continuous from the right,

(ϕ-3) ${\sum }_{n=0}^{\mathrm{\infty }}{\varphi }^n(t)<+\mathrm{\infty }$ for all $t>0$, where ${\varphi }^{n+1}(t)=\varphi ({\varphi }^n(t))$, $n\in \mathbb{N}$.

It is easy to prove that if $\varphi \in \mathrm{\Phi }$, then $\varphi (t)<t$ for all $t>0$.

Lemma 2.7 (see [7])

Let $(X,M,\ast )$ be a fuzzy metric space, where ∗ is a continuous t-norm of H-type. If there exists $\varphi \in \mathrm{\Phi }$ such that

$$M(x,y,\varphi (t))\ge M(x,y,t)$$

(2.4)

for all $t>0$, then $x=y$.

Definition 2.8 (see [4])

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

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

(2.5)

Definition 2.9 (see [5])

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

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

(2.6)

Definition 2.10 (see [5])

An element $(x,y)\in X\times X$ is called a common coupled fixed point of the mappings $F:X\times X\to X$ and $g:X\to X$ if

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

(2.7)

Definition 2.11 (see [5])

An element $x\in X$ is called a common fixed point of the mappings $F:X\times X\to X$ and $g:X\to X$ if

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

(2.8)

Definition 2.12 (see [8])

The mappings $F:X\times X\to X$ and $g:X\to X$ are said to be compatible if

$$\underset{n\to \mathrm{\infty }}{lim}M(gF(x_n,y_n),F(g(x_n),g(y_n)),t)=1$$

(2.9)

and

$$\underset{n\to \mathrm{\infty }}{lim}M(gF(y_n,x_n),F(g(y_n),g(x_n)),t)=1$$

(2.10)

for all $t>0$ whenever $\{x_n\}$ and $\{y_n\}$ are sequences in X, such that

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

(2.11)

for all $x,y\in X$ are satisfied.

Definition 2.13 (see [20])

The mappings $F:X\times X\to X$ and $g:X\to X$ are called weakly compatible mappings if $F(x,y)=g(x)$, $F(y,x)=g(y)$ implies that $gF(x,y)=F(gx,gy)$, $gF(y,x)=F(gy,gx)$ for all $x,y\in X$.

Remark 2.14 It is easy to prove that if F and g are compatible, then they are weakly compatible, but the converse need not be true. See the example in the next section.

3 Main results

For simplicity, denote

for all $n\in \mathbb{N}$.

Xin-Qi Hu [8] proved the following result.

Theorem 3.1 (see [8])

Let $(X,M,\ast )$ be a complete FM-space, where ∗ is a continuous t-norm of H-type satisfying (2.2). Let $F:X\times X\to X$ and $g:X\to X$ be two mappings, and there exists $\varphi \in \mathrm{\Phi }$ such that

$$M(F(x,y),F(u,v),\varphi (t))\ge M(g(x),g(u),t)\ast M(g(y),g(v),t)$$

(3.1)

for all $x,y,u,v\in X$, $t>0$.

Suppose that $F(X\times X)\subseteq g(X)$, g is continuous, F and g are compatible. Then there exist $x,y\in X$ such that $x=g(x)=F(x,x)$; that is, F and g have a unique common fixed point in X.

Now we give our main result.

Theorem 3.2 Let $(X,M,\ast )$ be a FM-space, where ∗ is a continuous t-norm of H-type satisfying (2.2). Let $F:X\times X\to X$ and $g:X\to X$ be two weakly compatible mappings, and there exists $\varphi \in \mathrm{\Phi }$ satisfying (3.1).

Suppose that $F(X\times X)\subseteq g(X)$ and $F(X\times X)$ or $g(X)$ is complete. Then F and g have a unique common fixed point in X.

Proof Let $x_0,y_0\in X$ be two arbitrary points in X. Since $F(X\times X)\subseteq g(X)$, we can choose $x_1,y_1\in X$ such that $g(x_1)=F(x_0,y_0)$ and $g(y_1)=F(y_0,x_0)$. Continuing this process, we can construct two sequences $\{x_n\}$ and $\{y_n\}$ in X such that

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

(3.2)

The proof is divided into 4 steps.

Step 1: We shall prove that $\{g{x}_n\}$ and $\{g{y}_n\}$ are Cauchy sequences.

Since ∗ is a t-norm of H-type, for any $\lambda >0$, there exists an $\mu >0$ such that

for all $k\in \mathbb{N}$.

Since $M(x,y,\cdot )$ is continuous and ${lim}_{t\to +\mathrm{\infty }}M(x,y,t)=1$ for all $x,y\in X$, there exists $t_0>0$ such that

$$M(g{x}_0,g{x}_1,t_0)\ge 1-\mu ,M(g{y}_0,g{y}_1,t_0)\ge 1-\mu .$$

(3.3)

On the other hand, since $\varphi \in \mathrm{\Phi }$, by condition (ϕ-3), we have ${\sum }_{n=1}^{\mathrm{\infty }}{\varphi }^n(t_0)<\mathrm{\infty }$. Then for any $t>0$, there exists $n_0\in \mathbb{N}$ such that

$$t>\sum _{k=n_0}^{\mathrm{\infty }}{\varphi }^k(t_0).$$

(3.4)

From condition (3.1), we have

$$\begin{array}{c}\begin{array}{rl}M(g{x}_1,g{x}_2,\varphi (t_0))& =M(F(x_0,y_0),F(x_1,y_1),\varphi (t_0))\\ \ge M(g{x}_0,g{x}_1,t_0)\ast M(g{y}_0,g{y}_1,t_0),\end{array}\hfill \\ \begin{array}{rl}M(g{y}_1,g{y}_2,\varphi (t_0))& =M(F(y_0,x_0),F(y_1,x_1),\varphi (t_0))\\ \ge M(g{y}_0,g{y}_1,t_0)\ast M(g{x}_0,g{x}_1,t_0).\end{array}\hfill \end{array}$$

Similarly, we have

$$\begin{array}{c}\begin{array}{rl}M(g{x}_2,g{x}_3,{\varphi }^2(t_0))& =M(F(x_1,y_1),F(x_2,y_2),{\varphi }^2(t_0))\\ \ge M(g{x}_1,g{x}_2,\varphi (t_0))\ast M(g{y}_1,g{y}_2,\varphi (t_0))\\ \ge {[M(g{x}_0,g{x}_1,t_0)]}^2\ast {[M(g{y}_0,g{y}_1,t_0)]}^2,\end{array}\hfill \\ \begin{array}{rl}M(g{y}_2,g{y}_3,{\varphi }^2(t_0))& =M(F(y_1,x_1),F(y_2,x_2),{\varphi }^2(t_0))\\ \ge {[M(g{y}_0,g{y}_1,t_0)]}^2\ast {[M(g{x}_0,g{x}_1,t_0)]}^2.\end{array}\hfill \end{array}$$

From the inequalities above and by induction, it is easy to prove that

$$\begin{array}{c}M(g{x}_n,g{x}_{n+1},{\varphi }^n(t_0))\ge {[M(g{x}_0,g{x}_1,t_0)]}^{2^{n-1}}\ast {[M(g{y}_0,g{y}_1,t_0)]}^{2^{n-1}},\hfill \\ M(g{y}_n,g{y}_{n+1},{\varphi }^n(t_0))\ge {[M(g{y}_0,g{y}_1,t_0)]}^{2^{n-1}}\ast {[M(g{x}_0,g{x}_1,t_0)]}^{2^{n-1}}.\hfill \end{array}$$

So, from (3.3) and (3.4), for $m>n\ge n_0$, we have

which implies that

$$M(g{x}_n,g{x}_m,t)>1-\lambda$$

(3.5)

for all $m,n\in \mathbb{N}$ with $m>n\ge n_0$ and $t>0$. So $\{g(x_n)\}$ is a Cauchy sequence.

Similarly, we can prove that $\{g(y_n)\}$ is also a Cauchy sequence.

Step 2: Now, we prove that g and F have a coupled coincidence point.

Without loss of generality, we can assume that $g(X)$ is complete, then there exist $x,y\in g(X)$, and exist $a,b\in X$ such that

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}g(x_n)=\underset{n\to \mathrm{\infty }}{lim}F(x_n,y_n)=g(a)=x,\\ \underset{n\to \mathrm{\infty }}{lim}g(y_n)=\underset{n\to \mathrm{\infty }}{lim}F(y_n,x_n)=g(b)=y.\end{array}$$

(3.6)

From (3.1), we get

$$M(F(x_n,y_n),F(a,b),\varphi (t))\ge M(g{x}_n,g(a),t)\ast M(g{y}_n,g(b),t).$$

Since M is continuous, taking limit as $n\to \mathrm{\infty }$, we have

$$M(g(a),F(a,b),\varphi (t))=1,$$

which implies that $F(a,b)=g(a)=x$.

Similarly, we can show that $F(b,a)=g(b)=y$.

Since F and g are weakly compatible, we get that $gF(a,b)=F(g(a),g(b))$ and $gF(b,a)=F(g(b),g(a))$, which implies that $g(x)=F(x,y)$ and $g(y)=F(y,x)$.

Step 3: We prove that $g(x)=y$ and $g(y)=x$.

Since ∗ is a t-norm of H-type, for any $\lambda >0$, there exists an $\mu >0$ such that

for all $k\in \mathbb{N}$.

Since $M(x,y,\cdot )$ is continuous and ${lim}_{t\to +\mathrm{\infty }}M(x,y,t)=1$ for all $x,y\in X$, there exists $t_0>0$ such that $M(gx,y,t_0)\ge 1-\mu$ and $M(gy,x,t_0)\ge 1-\mu$.

On the other hand, since $\varphi \in \mathrm{\Phi }$, by condition (ϕ-3), we have ${\sum }_{n=1}^{\mathrm{\infty }}{\varphi }^n(t_0)<\mathrm{\infty }$. Thus, for any $t>0$, there exists $n_0\in \mathbb{N}$ such that $t>{\sum }_{k=n_0}^{\mathrm{\infty }}{\varphi }^k(t_0)$. Since

$$\begin{array}{rl}M(gx,g{y}_{n+1},\varphi (t_0))& =M(F(x,y),F(y_n,x_n),\varphi (t_0))\\ \ge M(gx,g{y}_n,t_0)\ast M(gy,g{x}_n,t_0),\end{array}$$

letting $n\to \mathrm{\infty }$, we get

$$M(gx,y,\varphi (t_0))\ge M(gx,y,t_0)\ast M(gy,x,t_0).$$

(3.7)

Similarly, we can get

$$M(gy,x,\varphi (t_0))\ge M(gx,y,t_0)\ast M(gy,x,t_0).$$

(3.8)

From (3.7) and (3.8), we have

$$M(gx,y,\varphi (t_0))\ast M(gy,x,\varphi (t_0))\ge {[M(gx,y,t_0)]}^2\ast {[M(gy,x,t_0)]}^2.$$

From this inequality, we can get

$$\begin{array}{rl}M(gx,y,{\varphi }^n(t_0))\ast M(gy,x,{\varphi }^n(t_0))& \ge {\left[M(gx,y,{\varphi }^{n-1}(t_0))\right]}^2\ast {\left[M(gy,x,{\varphi }^{n-1}(t_0))\right]}^2\\ \ge {[M(gx,y,t_0)]}^{2^n}\ast {[M(gy,x,t_0)]}^{2^n}\end{array}$$

for all $n\in \mathbb{N}$. Since $t>{\sum }_{k=n_0}^{\mathrm{\infty }}{\varphi }^k(t_0)$, then, we have

Therefore, for any $\lambda >0$, we have

$$M(gx,y,t)\ast M(gy,x,t)\ge 1-\lambda$$

(3.9)

for all $t>0$. Hence conclude that $gx=y$ and $gy=x$.

Step 4: Now, we prove that $x=y$.

Since ∗ is a t-norm of H-type, for any $\lambda >0$, there exists an $\mu >0$ such that

for all $k\in \mathbb{N}$.

Since $M(x,y,\cdot )$ is continuous and ${lim}_{t\to +\mathrm{\infty }}M(x,y,t)=1$, there exists $t_0>0$ such that $M(x,y,t_0)\ge 1-\mu$.

On the other hand, since $\varphi \in \mathrm{\Phi }$, by condition (ϕ-3), we have ${\sum }_{n=1}^{\mathrm{\infty }}{\varphi }^n(t_0)<\mathrm{\infty }$. Then, for any $t>0$, there exists $n_0\in \mathbb{N}$ such that $t>{\sum }_{k=n_0}^{\mathrm{\infty }}{\varphi }^k(t_0)$.

From (3.1), we have

$$\begin{array}{rl}M(g{x}_{n+1},g{y}_{n+1},\varphi (t_0))& =M(F(x_n,y_n),F(y_n,x_n),\varphi (t_0))\\ \ge M(g{x}_n,g{y}_n,t_0)\ast M(g{y}_n,g{x}_n,t_0).\end{array}$$

Letting $n\to \mathrm{\infty }$ yields

$$M(x,y,\varphi (t_0))\ge M(x,y,t_0)\ast M(y,x,t_0).$$

Thus, we have

which implies that $x=y$.

Thus, we proved that F and g have a common fixed point in X.

The uniqueness of the fixed point can be easily proved in the same way as above. This completes the proof of Theorem 3.2. □

Taking $g=I$ (the identity mapping) in Theorem 3.2, we get the following consequence.

Corollary 3.3 Let $(X,M,\ast )$ be a FM-space, where ∗ is a continuous t-norm of H-type satisfying (2.2). Let $F:X\times X\to X$, and there exists $\varphi \in \mathrm{\Phi }$ such that

$$M(F(x,y),F(u,v),\varphi (t))\ge M(x,u,t)\ast M(y,v,t)$$

(3.10)

for all $x,y,u,v\in X$, $t>0$. $F(X)$ is complete.

Then there exist $x\in X$ such that $x=F(x,x)$; that is, F admits a unique fixed point in X.

Remark 3.4 Comparing Theorem 3.2 with Theorem 3.1 in [8], we can see that Theorem 3.2 is a genuine generalization of Theorem 3.2.

• (1) We only need the completeness of $g(X)$ or $F(X\times X)$.

• (2) The continuity of g is relaxed.

• (3) The concept of compatible has been replaced by weakly compatible.

Remark 3.5 The Example 3 in [8] is wrong, since the t-norm $a\ast b=ab$ is not the t-norm of H-type.

Next, we give an example to support Theorem 3.2.

Example 3.6 Let $X=\{0,1,\frac{1}{2},\frac{1}{3},\dots ,\frac{1}{n},\dots \}$, $\ast =min$, $M(x,y,t)=\frac{t}{|x-y|+t}$, for all $x,y\in X$, $t>0$. Then $(X,M,\ast )$ is a fuzzy metric space.

Let $\varphi (t)=\frac{t}{2}$. Let $g:X\to X$ and $F:X\times X\to X$ be defined as

$$g(x)=\{\begin{array}{cc}0,\hfill & x=0,\hfill \\ 1,\hfill & x=\frac{1}{2n+1},\hfill \\ \frac{1}{2n+1},\hfill & x=\frac{1}{2n},\hfill \end{array}F(x,y)=\{\begin{array}{cc}\frac{1}{{(2n+1)}^4},\hfill & (x,y)=(\frac{1}{2n},\frac{1}{2n}),\hfill \\ 0,\hfill & \text{others}.\hfill \end{array}$$

Let $x_n=y_n=\frac{1}{2n}$. We have $g{x}_n=\frac{1}{2n+1}\to 0$, $F(x_n,y_n)=\frac{1}{{(2n+1)}^4}\to 0$, but

$$M(F(g{x}_n,g{y}_n),gF(x_n,y_n),t)=M(0,1,t)\nrightarrow 0,$$

so g and F are not compatible. From $F(x,y)=g(x)$, $F(y,x)=g(y)$, we can get $(x,y)=(0,0)$, and we have $gF(0,0)=F(g0,g0)$, which implies that F and g are weakly compatible.

The following result is easy to verify

$$\frac{t}{X+t}\ge min\{\frac{t}{Y+t},\frac{t}{Z+t}\}\iff X\le max\{Y,Z\},\mathrm{\forall }X,Y,Z\ge 0,t>0.$$

By the definition of M and ϕ and the result above, we can get that inequality (3.1)

$$M(F(x,y),F(u,v),\varphi (t))\ge M(g(x),g(u),t)\ast M(g(y),g(v),t)$$

is equivalent to the following

$$2|F(x,y)-F(u,v)|\le max\{|g(x)-g(u)|,|g(y)-g(v)|\}.$$

(3.11)

Now, we verify inequality (3.11). Let $A=\{\frac{1}{2n},n\in \mathbb{N}\}$, $B=X-A$. By the symmetry and without loss of generality, $(x,y)$, $(u,v)$ have 6 possibilities.

Case 1: $(x,y)\in B\times B$, $(u,v)\in B\times B$. It is obvious that (3.11) holds.

Case 2: $(x,y)\in B\times B$, $(u,v)\in B\times A$. It is obvious that (3.11) holds.

Case 3: $(x,y)\in B\times B$, $(u,v)\in A\times A$. If $u\ne v$, (3.11) holds. If $u=v$, let $u=v=\frac{1}{2n}$, then

$$2|F(x,y)-F(u,v)|=\frac{2}{{(2n+1)}^4},max\{|g(x)-g(u)|,|g(y)-g(v)|\}=\frac{2n}{2n+1},$$

which implies that (3.11) holds.

Case 4: $(x,y)\in B\times A$, $(u,v)\in B\times A$. It is obvious that (3.11) holds.

Case 5: $(x,y)\in B\times A$, $(u,v)\in A\times A$. If $u\ne v$, (3.11) holds. If $u=v$, let $x\in B$, $y=\frac{1}{2j}$, $u=v=\frac{1}{2n}$, then

$$\begin{array}{c}2|F(x,y)-F(u,v)|=\frac{2}{{(2n+1)}^4},\hfill \\ max\{|g(x)-g(u)|,|g(y)-g(v)|\}=max\{\frac{1}{2n+1},|\frac{1}{2j+1}-\frac{1}{2n+1}\left|\right\},\hfill \end{array}$$

or

$$max\{|g(x)-g(u)|,|g(y)-g(v)|\}=max\{\frac{2n}{2n+1},|\frac{1}{2j+1}-\frac{1}{2n+1}\left|\right\},$$

(3.11) holds.

Case 6: $(x,y)\in A\times A$, $(u,v)\in A\times A$.

If $x\ne y$, $u\ne v$, (3.11) holds.

If $x\ne y$, $u=v$, let $x=\frac{1}{2i}$, $y=\frac{1}{2j}$, $i\ne j$, $u=v=\frac{1}{2n}$. Then

$$\begin{array}{c}2|F(x,y)-F(u,v)|=\frac{2}{{(2n+1)}^4},\hfill \\ max\{|g(x)-g(u)|,|g(y)-g(v)|\}=max\left\{\right|\frac{1}{2i+1}-\frac{1}{2n+1}|,|\frac{1}{2j+1}-\frac{1}{2n+1}\left|\right\},\hfill \end{array}$$

(3.11) holds.

If $x=y$, $u=v$, let $x=y=\frac{1}{2i}$, $u=v=\frac{1}{2n}$. Then

$$\begin{array}{c}2|F(x,y)-F(u,v)|=2|\frac{1}{{(2i+1)}^4}-\frac{1}{{(2n+1)}^4}|,\hfill \\ max\{|g(x)-g(u)|,|g(y)-g(v)|\}=|\frac{1}{2i+1}-\frac{1}{2n+1}|,\hfill \end{array}$$

(3.11) holds.

Then all the conditions in Theorem 3.2 are satisfied, and 0 is the unique common fixed point of g and F.