ABSTRACT

In this paper, we establish a fixed point theorem for two mappings under a contraction condition in b₂-metric space, and this theorem is related to a Suzuki-type of contraction.

Subject Areas:

Mathematical Analysis

Keywords:

Common Fixed Point, b₂-Metric Space, Generalized Suzuki-Type Contraction

1. Introduction

Banach [1] proved a principle, and this famous Banach contraction principle has many generalizations, see [2] - [7], and in 2008, Suzuki [8] established one of those generalizations, and this generalization is called Suzuki principle.

The aim of this paper is to prove a fixed point result generalized from the above mentioned principle in b₂-metric space [9].

2. Preliminaries

Before giving our results, these definitions and results as follows will be needed to present.

Definition 2.1 [9] Let X be a nonempty set, $s\ge 1$ be a real number and let d: $X\times X\times X\to R$ be a map satisfying the following conditions:

1) For every pair of distinct points $x,y\in X$, there exists a point $z\in X$ such that $d\left(x,y,z\right)\ne 0$.

2) If at least two of three points $x,y,z$ are the same, then $d\left(x,y,z\right)=0$,

3) The symmetry:

$d\left(x,y,z\right)=d\left(x,z,y\right)=d\left(y,x,z\right)=d\left(y,z,x\right)=d\left(z,x,y\right)=d\left(z,x,y\right)$ for all $x,y,z\in X$.

1) The rectangle inequality:

$d\left(x,y,z\right)\le s\left[d\left(x,y,a\right)+d\left(y,z,a\right)+d\left(z,x,a\right)\right]$, for all $x,y,z,a\in X$.

Then d is called a b₂ metric on X and $\left(X,d\right)$ is called a b₂ metric space with parameter s. Obviously, for $s=1$, b₂ metric reduces to 2-metric.

Definition 2.2 [9] Let $\left\{x_n\right\}$ be a sequence in a b₂ metric space $\left(X,d\right)$.

1) A sequence $\left\{x_n\right\}$ is said to be b₂-convergent to $x\in X$, written as ${\lim }_{n\to \infty }{x}_n=x$, if all $a\in X$ ${\lim }_{n\to \infty }d\left(x_n,x,a\right)=0$.

2) $\left\{x_n\right\}$ is Cauchy sequence if and only if $d\left(x_n,x_m,a\right)\to 0$, when $n,m\to \infty$. for all $a\in X$.

3) $\left(X,d\right)$ is said to be complete if every b₂-Cauchy sequence is a b₂-convergent sequence.

Definition 2.3 [9] Let $\left(X,d\right)$ and $\left(X^{\prime },d^{\prime }\right)$ be two b₂-metric spaces and let $f:X\to X^{\prime }$ be a mapping. Then f is said to be b₂-continuous, at a point $z\in X$ if for a given $\epsilon >0$, there exists $\delta >0$ such that $x\in X$ and $d\left(z,x,a\right)<\delta$ for all $a\in X$ imply that $d^{\prime }\left(fz,fx,a\right)<\epsilon$. The mapping f is b₂-continuous on X if it is b₂-continuous at all $z\in X$.

Definition 2.4 [9] Let $\left(X,d\right)$ and $\left(X^{\prime },d^{\prime }\right)$ be two b₂-metric spaces. Then a mapping $f:X\to X^{\prime }$ is b₂-continuous at a point $x\in X^{\prime }$ if and only if it is b₂-sequentially continuous at x; that is, whenever $\left\{x_n\right\}$ is b₂-convergent to x, $\left\{f{x}_n\right\}$ is b₂-convergent to $f\left(x\right)$.

Lemma 2.5 [9] Let $\left(X,d\right)$ be a b₂-metric space and suppose that $\left\{x_n\right\}$ and $\left\{y_n\right\}$ are b₂-convergent to x and y, respectively. Then we have

$\frac{1}{s^2}d\left(x,y,a\right)\le \underset{n\to \infty }{\lim }\inf d\left(x_n,y_n,a\right)\le \underset{n\to \infty }{\lim }\sup d\left(x_n,y_n,a\right)\le s^{2}d\left(x,y,a\right)$, for all a in X. In particular, if $y_n=y$ is a constant, then

$\frac{1}{s}d\left(x,y,a\right)\le \underset{n\to \infty }{\lim }\inf d\left(x_n,y,a\right)\le \underset{n\to \infty }{\lim }\sup d\left(x_n,y,a\right)\le sd\left(x,y,a\right)$, for all a in X.

Lemma 2.6 [10] Let $\left(X,d\right)$ be a b₂ metric space with $s\ge 1$ and let ${\left\{x_n\right\}}_{n=0}^{\infty }$ be a sequence in X such that

$d\left(x_n,x_{n+1},a\right)\le \lambda d\left(x_{n-1},x_n,a\right)$, (2.1)

for all $n\in N$ and all $a\in X$, where $\lambda \in \left[0,\frac{1}{s}\right)$. Then $\left\{x_n\right\}$ is a b₂-Cauchy sequence in $\left(X,d\right)$.

3. Main Results

Theorem 3.1. Let $\left(X,d\right)$ be a complete b₂-metric space. Let $f,g:X\to X$ be two self-maps and $\varphi :\left[0,1\right)\to \left(\frac{1}{2},1\right]$ be defined as follows

$\varphi \left(\rho \right)=\{\begin{array}{l}1,0\le \rho \le \frac{\sqrt{5}-1}{2}\\ \frac{1-\rho }{{\rho }^2},\frac{\sqrt{5}-1}{2}\le \rho \le \frac{1}{\sqrt{2}}\\ \frac{1}{1+\rho },\frac{1}{\sqrt{2}}\le \rho <1\end{array}$ (3.1)

Assume there exists $\rho \in \left[0,1\right)$ such that for every $x,y\in X$, the following condition is satisfied

$\begin{array}{l}\frac{1}{s}\varphi \left(\rho \right)\min \left\{d\left(x,fx,a\right),d\left(fx,fy,a\right)\right\}\le d\left(x,y,a\right)\\ \Rightarrow \max \{d\left(gx,gy,a\right),d\left(gx,fy,a\right),d\left(fx,fy,a\right),d\left(gy,fx,a\right)\}\le \frac{\rho }{s^2}d\left(x,y,a\right).\end{array}$ (3.2)

Then $f,g$ have a unique common fixed point $z\in X$.

Proof in (3.2), we take $y=fx$

$\begin{array}{l}\frac{1}{s}\varphi \left(\rho \right)\min \left\{d\left(x,fx,a\right),d\left(x,gx,a\right)\right\}\le d\left(x,gx,a\right)\\ \Rightarrow \max \{d\left(gx,g^{2}x,a\right),d\left(gx,fgx,a\right),d\left(fx,fgx,a\right),d\left(g^{2}x,fx,a\right)\}\\ \le \frac{\rho }{s^2}d\left(x,gx,a\right),\end{array}$ for $x\in X$.(3.3)

therefore,

$d\left(gx,fgx,a\right)\le \frac{\rho }{s^2}d\left(x,gx,a\right).$ (3.4)

Now we take $y=fx$ in (3.2)

$\begin{array}{l}\frac{1}{s}\varphi \left(\rho \right)\min \left\{d\left(x,fx,a\right),d\left(x,gy,a\right)\right\}\le d\left(x,fy,a\right)\\ \Rightarrow \max \{d\left(gx,gfy,a\right),d\left(gx,f^{2}y,a\right),d\left(fx,f^{2}x,a\right),d\left(gfx,fx,a\right)\}\\ \le \frac{\rho }{s^2}d\left(x,fx,a\right),\end{array}$ for all $x\in X$.(3.5)

therefore,

$d\left(fx,f^{2}x,a\right)\le \frac{\rho }{s^2}d\left(x,fx,a\right),$ (3.6)

and

$d\left(gfx,fx,a\right)\le \frac{\rho }{s^2}d\left(x,fx,a\right).$ (3.7)

Given an arbitrary point $x_0$ in X thenby $x_{2n+1}=g{x}_{2n}$ and $x_{2n+1}=f{x}_{2n+1}$ we construct a sequence $\left\{x_n\right\}$, for $n\in N$.

From (3.4), we get

$d\left(x_{2n+1},x_{2n+2},a\right)=d\left(g{x}_{2n},fg{x}_{2n},a\right)\le \frac{\rho }{s^2}d\left(x_{2n},g{x}_{2n},a\right)=\frac{\rho }{s^2}d\left(x_{2n},x_{2n+1},a\right).$ (3.8)

From (3.7) and (3.8) we get

$d\left(x_{2n+1},x_{2n},a\right)=d\left(gf{x}_{2n-1},f{x}_{2n-1},a\right)\le \frac{\rho }{s^2}d\left(x_{2n},f{x}_{2n-1},a\right)=\frac{\rho }{s^2}d\left(x_{2n-1},x_{2n},a\right)$,

that is,

$d\left(x_{n+1},x_n,a\right)\le \frac{\rho }{s^2}d\left(x_n,x_{n-1},a\right)$, since $\frac{\rho }{s^2}\in \left[0,1\right)$, by Lemma 2.6, we get $\left\{x_n\right\}$ is a Cauchy sequence.

Since X is complete, there exists z in X, such that $\underset{n\to \infty }{\lim }{x}_n=z$, that is $\underset{n\to \infty }{\lim }g{x}_{2n}=\underset{n\to \infty }{\lim }{x}_{2n+1}=z$, and $\underset{n\to \infty }{\lim }f{x}_{2n+1}=\underset{n\to \infty }{\lim }{x}_{2n+2}=z$.

Now let us give that

$d\left(fx,z,a\right)\le \rho d\left(x,z,a\right)$, for every $x\ne z$. For $\left\{d\left(x_{2n},g{x}_{2n},a\right)\right\}$ is convergent to 0, and by Lemma 2.5, we get

$\frac{1}{s}d\left(x,z,a\right)\le \underset{n\to \infty }{\lim }\sup d\left(x_{2n},x,a\right)$, thus we have $\underset{n\to \infty }{\lim }\sup d\left(x_{2n},x,a\right)>0$, thus from the above relation, there exists a point $x_{2n_k}$ in X such that

$\frac{1}{s}\varphi \left(\rho \right)\min \left\{d\left(x_{2n_k},g{x}_{2n_k},a\right),d\left(x_{2n_k},f{x}_{2n_k},a\right)\right\}\le d\left(x_{2n_k},x,a\right).$

For such $x_{2n_k}$, (3.2) implies that

$\begin{array}{l}d\left(g{x}_{2n_k},fx,a\right)\\ \le \max \{d\left(g{x}_{2n_k},gx,a\right),d\left(g{x}_{2n_k},fx,a\right),d\left(f{x}_{2n_k},fx,a\right),d\left(gx,f{x}_{2n_k},a\right)\}\\ \le \frac{\rho }{s_2}d\left(x_{2n_k},x,a\right),\end{array}$

therefore by Lemma 3.5,

$\frac{1}{s}d\left(fx,z,a\right)\le \underset{n\to \infty }{\lim \sup }d\left(g{x}_{2n_k},fx,a\right)\le \frac{\rho }{s^2}\underset{n\to \infty }{\lim \sup }d\left(x_{2n_k},x,a\right)\le \frac{\rho }{s}d\left(x,z,a\right),$

therefore we get

$d\left(fx,z,a\right)\le \rho d\left(x,z,a\right)$, for each $x\ne z$. (3.9)

Now we show that for each $n\in N$,

$d\left(f^{n}z,z,a\right)\le d\left(fz,z,a\right),$ (3.10)

It is obvious that the above inequality is true for $n=1$, assume that the relation holds for some $m\in N$. We get (3.10) is true when we have $f^{m}z=fz$ if $f^{m}z=z$, then if $f^{m}z\ne z$, we get the following relation from (3.9) and induction hypothesis, and that is

$\begin{array}{c}d\left(z,f^{m+1}z,a\right)\le \rho d\left(z,f^{m}z,a\right)\le {\rho }^{2}d\left(z,f^{m-1}z,a\right)\le \cdots \le {\rho }^{m+1}d\left(z,fz,a\right)\\ \le \rho d\left(fz,z,a\right)\le d\left(fz,z,a\right),\end{array}$

then (3.10) is proved.

Now we consider the following two possible cases in order to prove that f has a fixed point z in X, and that is $fz=z$.

Case 1 $0\le \rho <\frac{1}{\sqrt{2}}$, therefore, $\varphi \left(\rho \right)\le \frac{1-\rho }{{\rho }^2}$. First, we prove the following relation

$d\left(f^{n}z,fz,a\right)\le \frac{\rho }{s}d\left(fz,z,a\right)$, for $n\in N$. (3.11)

When $n=1$ it is obvious, and it follows from (3.6) when $n=2$, from (3.10) and take $a=fz$ we have

$d\left(f^{n}z,z,fz\right)\le d\left(fz,z,fz\right)=0$, then we get $d\left(f^{n}z,fz,z\right)=0$.

Now suppose that (3.11) holds for some $n>2$,

$\begin{array}{c}d\left(fz,z,a\right)\le s\left(d\left(z,f^{n}z,a\right)+d\left(f^{n}z,fz,a\right)+d\left(f^{n}z,fz,z\right)\right)\\ \le sd\left(z,f^{n}z,a\right)+sd\left(z,fz,a\right),\end{array}$

Therefore, we get

$\left(1-\rho \right)d\left(z,fz,a\right)\le sd\left(z,f^{n}z,a\right)$, that is $d\left(z,fz,a\right)\le \frac{s}{1-\rho }sd\left(z,f^{n}z,a\right)$, (3.11.1)

then by taking $x=f^{n-1}z$ in (3.6)

$d\left(f^{n}z,f^{n+1}z,a\right)\le \frac{\rho }{s^2}d\left(f^{n-1}z,f^{n}z,a\right)\le \cdots \le \frac{{\rho }^n}{s^{2n}}d\left(z,fz,a\right)$, (3.11.2)

using the above two relations, (3.11.1) and (3.11.2) we have

$\begin{array}{l}\frac{1}{s}\varphi \left(\rho \right)\min \left\{d\left(g{f}^{n}z,f^{n}z,a\right),d\left(f^{n}z,f^{n+1}z,a\right)\right\}\\ \le \frac{1-\rho }{s{\rho }^2}d\left(f^{n}z,f^{n+1}z,a\right)\le \frac{1-\rho }{s{\rho }^n}d\left(f^{n}z,f^{n+1}z,a\right)\\ \le \frac{1-\rho }{s{\rho }^n}\cdot \frac{{\rho }^n}{s^{2n}}d\left(z,fz,a\right)=\frac{1-\rho }{s^{2n+1}}d\left(z,fz,a\right)\\ \le \frac{1-\rho }{s^{2n+1}}\cdot \frac{s}{1-\rho }d\left(z,f^{n}z,a\right)\le \frac{1}{s^{2n}}d\left(z,f^{n}z,a\right)\le d\left(z,f^{n}z,a\right).\end{array}$

From (3.2) and (3.10) with $x=f^{n}z$ and $y=z$, we have

$\begin{array}{l}\max \left\{d\left(g{f}^{n}z,gz,a\right),d\left(g{f}^{n}z,fz,a\right),d\left(f^{n+1}z,fz,a\right),d\left(gz,f^{n+1}z,a\right)\right\}\\ \le \frac{\rho }{s^2}d\left(z,f^{n}z,a\right)\le \frac{\rho }{s^2}d\left(z,fz,a\right)\le \frac{\rho }{s}d\left(z,fz,a\right).\end{array}$

Therefore,

$d\left(f^{n+1}z,fz,a\right)\le \frac{\rho }{s}d\left(fz,z,a\right).$ (3.12)

So by induction we prove the relation of (3.11).

Now (3.11) and $fz\ne z$ show that for every $n\in N$ $f^{n}z\ne z$, thus, (3.9) shows that

$d\left(z,f^{n+1}z,a\right)\le \rho d\left(z,f^{n}z,a\right)\le {\rho }^{2}d\left(z,f^{n-1}z,a\right)\le \cdots \le {\rho }^{n}d\left(z,fz,a\right).$

Therefore $\underset{n\to \infty }{\lim }d\left(z,f^{n+1}z,a\right)=0$. Furthermore by using Lemma 2.5, we get

$\frac{1}{s}d\left(z,\underset{n\to \infty }{\lim \inf }{f}^{n+1}z,a\right)\le \underset{n\to \infty }{\lim \inf }d\left(z,f^{n+1}z,a\right)=0,$

so

$d\left(z,\underset{n\to \infty }{\lim \inf }{f}^{n+1}z,a\right)=0.$

In the same way,

$d\left(z,\underset{n\to \infty }{\lim \sup }{f}^{n+1}z,a\right)=0$, thus we have $d\left(z,\underset{n\to \infty }{\lim }{f}^{n+1}z,a\right)=0$, that is $f^{n+1}z\to z$, and by using Lemma 2.5 in (3.12), we get

$\frac{1}{s}d\left(z,fz,a\right)\le \underset{n\to \infty }{\lim \sup }d\left(f^{n+1}z,fz,a\right)\le \frac{\rho }{s}d\left(z,fz,a\right)$, which claims that $d\left(z,fz,a\right)=0$, and that is a contraction.

Case 2. $\frac{1}{\sqrt{2}}\le \rho <1$, and that is when $\varphi \left(\rho \right)=\frac{1}{1+\rho }$. We now prove that we can find a subsequence $\left\{x_{n_k}\right\}$ of $\left\{x_n\right\}$ such that

$\frac{1}{s\left(1+\rho \right)}\min \left\{d\left(x_{n_k},g{x}_{n_k},a\right),d\left(x_{n_k},f{x}_{n_k},a\right)\right\}\le d\left(x_{n_k},z,a\right)$, for $k\in N$. (3.13)

The contraries of the above relation are as follows

$\frac{1}{s\left(1+\rho \right)}d\left(x_n,f{x}_n,a\right)\ge \frac{1}{s\left(1+\rho \right)}\min \left\{d\left(x_n,g{x}_n,a\right),d\left(x_n,f{x}_n,a\right)\right\}>d\left(x_n,z,a\right),$

and

$\frac{1}{s\left(1+\rho \right)}d\left(x_n,f{x}_n,a\right)\ge \frac{1}{s\left(1+\rho \right)}\min \left\{d\left(x_n,g{x}_n,a\right),d\left(x_n,f{x}_n,a\right)\right\}>d\left(x_n,z,a\right),$

for $n\in N$. If n is even we have

$\begin{array}{l}\frac{1}{s\left(1+\rho \right)}d\left(x_{2n},g{x}_{2n},a\right)\\ \ge \frac{1}{s\left(1+\rho \right)}\min \left\{d\left(x_{2n},g{x}_{2n},a\right),d\left(x_{2n},f{x}_{2n},a\right)\right\}\\ >d\left(x_{2n},z,a\right),\end{array}$

if n is odd then we get

$\begin{array}{l}\frac{1}{s\left(1+\rho \right)}d\left(x_{2n+1},f{x}_{2n+1},a\right)\\ \ge \frac{1}{s\left(1+\rho \right)}\min \left\{d\left(x_{2n+1},g{x}_{2n+1},a\right),d\left(x_{2n+1},f{x}_{2n+1},a\right)\right\}\\ >d\left(x_{2n+1},z,a\right),\end{array}$

for $n\in N$. By (3.8) we have

$\begin{array}{l}d\left(x_{2n},x_{2n+1},a\right)\\ \le s\left(d\left(x_{2n},z,a\right)+d\left(x_{2n+1},z,a\right)+d\left(x_{2n},x_{2n+1},z\right)\right)\\ <\frac{s}{s\left(1+\rho \right)}d\left(x_{2n},g{x}_{2n},a\right)+\frac{s}{s\left(1+\rho \right)}d\left(x_{2n+1},f{x}_{2n+1},a\right)\\ +\frac{s}{s\left(1+\rho \right)}d\left(x_{2n},g{x}_{2n},x_{2n+1}\right)\\ =\frac{1}{1+\rho }\left(d\left(x_{2n},x_{2n+1},a\right)+d\left(x_{2n+1},x_{2n+2},a\right)+d\left(x_{2n},x_{2n+1},x_{2n+1}\right)\right)\end{array}$

$\begin{array}{l}\le \frac{1}{1+\rho }d\left(x_{2n},x_{2n+1},a\right)+\frac{\rho }{s^2\left(1+\rho \right)}d\left(x_{2n+1},x_{2n},a\right)\\ \le \frac{1}{1+\rho }d\left(x_{2n},x_{2n+1},a\right)+\frac{\rho }{1+\rho }d\left(x_{2n+1},x_{2n},a\right)\\ =d\left(x_{2n},x_{2n+1},a\right),\end{array}$

this is impossible. Therefore, one of the following relations is true for every $n\in N$,

$\frac{1}{s}\varphi \left(\rho \right)\min \left\{d\left(x_{2n},g{x}_{2n},a\right),d\left(x_{2n},f{x}_{2n},a\right)\right\}\le d\left(x_{2n},z,a\right),$

or

$\frac{1}{s}\varphi \left(\rho \right)\min \left\{d\left(x_{2n+1},g{x}_{2n+1},a\right),d\left(x_{2n+1},f{x}_{2n+1},a\right)\right\}\le d\left(x_{2n+1},z,a\right).$

That means there exists a subsequence $\left\{x_{n_k}\right\}$ of $\left\{x_n\right\}$ such that (3.13) is true for every $k\in N$. Thus (3.2) shows that

$\begin{array}{l}d\left(g{x}_{2n},fz,a\right)\\ \le \max \{d\left(f{x}_{2n},gz,a\right),d\left(fz,g{x}_{2n},a\right),d\left(f{x}_{2n},fz,a\right),d\left(gz,f{x}_{2n},a\right)\}\\ \le \frac{\rho }{s^2}d\left(x_{2n},z,a\right).\end{array}$

or

$\begin{array}{l}d\left(f{x}_{2n+1},fz,a\right)\\ \le \max \{d\left(g{x}_{2n+1},gz,a\right),d\left(fz,g{x}_{2n+1},a\right),d\left(f{x}_{2n+1},fz,a\right),d\left(gz,f{x}_{2n+1},a\right)\}\\ \le \frac{\rho }{s^2}d\left(x_{2n+!},z,a\right).\end{array}$

From Lemma 2.5, we have

$\frac{1}{s}d\left(z,fz,a\right)\le \underset{n\to \infty }{\lim \sup }d\left(g{x}_{2n},fz,a\right)\le \frac{\rho }{s^2}\underset{n\to \infty }{\lim \sup }d\left(x_{2n},z,a\right)\le \frac{\rho }{s}d\left(z,z,a\right)=0,$

or

$\frac{1}{s}d\left(z,fz,a\right)\le \underset{n\to \infty }{\lim \sup }d\left(f{x}_{2n+1},fz,a\right)\le \frac{\rho }{s^2}\underset{n\to \infty }{\lim \sup }d\left(x_{2n+1},z,a\right)\le \frac{\rho }{s}d\left(z,z,a\right)=0,$

Therefore $d\left(z,fz,a\right)\le 0$, which is impossible unless $fz=z$. hence z in X is a fixed point of f. From the process of the above proof, we know $fz=z$, then by

$0=\frac{1}{s}\varphi \left(\rho \right)\min \left\{d\left(z,fz,a\right),d\left(z,gz,a\right)\right\}\le d\left(z,fz,a\right)$,

it implies

$\begin{array}{c}d\left(gz,z,a\right)\le \max \left\{d\left(gz,gfz,a\right),d\left(gz,f^{2}z,a\right),d\left(fz,f^{2}z,a\right),d\left(gfz,fz,a\right)\right\}\\ \le \frac{\rho }{s^2}d\left(fz,z,a\right)=0,\end{array}$

this proves that $gz=z$. By (3.2) we can prove the uniqueness of the common fixed point z,

$\frac{1}{s}\varphi \left(\rho \right)\min \left\{d\left(z,fz,a\right),d\left(z,gz,a\right)\right\}\le d\left(z,z^{\prime },a\right)$, so (3.2) shows that

$\begin{array}{c}d\left(z,z^{\prime },a\right)=\max \left\{d\left(gz,g{z}^{\prime },a\right),d\left(fz,f{z}^{\prime },a\right),d\left(gz,f{z}^{\prime },a\right),d\left(g{z}^{\prime },fz,a\right)\right\}\\ \le \frac{\rho }{s^2}d\left(z,z^{\prime },a\right),\end{array}$

which is impossible unless $z=z^{\prime }$. □

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

Cite this paper

Wu, C., Cui, J.X. and Zhong, L.N. (2019) Suzuki-Type Fixed Point Theorem in b₂-Metric Spaces. Open Access Library Journal, 6: e5974. https://doi.org/10.4236/oalib.1105974

References

NOTES

*Corresponding author.