Suzuki-Type Fixed Point Theorem in b2-Metric Spaces

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.

Share and Cite:

Wu, C. , Cui, J. and Zhong, L. (2019) Suzuki-Type Fixed Point Theorem in b2-Metric Spaces. Open Access Library Journal, 6, 1-9. doi: 10.4236/oalib.1105974.

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: $X×X×X\to R$ 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 ${\mathrm{lim}}_{n\to \infty }{x}_{n}=x$, if all $a\in X$ ${\mathrm{lim}}_{n\to \infty }d\left({x}_{n},x,a\right)=0$.

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

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

$\frac{1}{s}d\left(x,y,a\right)\le \underset{n\to \infty }{\mathrm{lim}}\mathrm{inf}d\left({x}_{n},y,a\right)\le \underset{n\to \infty }{\mathrm{lim}}\mathrm{sup}d\left({x}_{n},y,a\right)\le sd\left(x,y,a\right)$, 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

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

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

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

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

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

$\begin{array}{l}\frac{1}{s}\varphi \left(\rho \right)\mathrm{min}\left\{d\left(x,fx,a\right),d\left(x,gx,a\right)\right\}\le d\left(x,gx,a\right)\\ ⇒\mathrm{max}\left\{d\left(gx,{g}^{2}x,a\right),d\left(gx,fgx,a\right),d\left(fx,fgx,a\right),d\left({g}^{2}x,fx,a\right)\right\}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\le \frac{\rho }{{s}^{2}}d\left(x,gx,a\right),\end{array}$ 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)

$\begin{array}{l}\frac{1}{s}\varphi \left(\rho \right)\mathrm{min}\left\{d\left(x,fx,a\right),d\left(x,gy,a\right)\right\}\le d\left(x,fy,a\right)\\ ⇒\mathrm{max}\left\{d\left(gx,gfy,a\right),d\left(gx,{f}^{2}y,a\right),d\left(fx,{f}^{2}x,a\right),d\left(gfx,fx,a\right)\right\}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\le \frac{\rho }{{s}^{2}}d\left(x,fx,a\right),\end{array}$ 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 $\underset{n\to \infty }{\mathrm{lim}}{x}_{n}=z$, that is $\underset{n\to \infty }{\mathrm{lim}}g{x}_{2n}=\underset{n\to \infty }{\mathrm{lim}}{x}_{2n+1}=z$, and $\underset{n\to \infty }{\mathrm{lim}}f{x}_{2n+1}=\underset{n\to \infty }{\mathrm{lim}}{x}_{2n+2}=z$.

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

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

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

For such ${x}_{2{n}_{k}}$, (3.2) implies that

$\begin{array}{l}d\left(g{x}_{2{n}_{k}},fx,a\right)\\ \le \mathrm{max}\left\{d\left(g{x}_{2{n}_{k}},gx,a\right),d\left(g{x}_{2{n}_{k}},fx,a\right),d\left(f{x}_{2{n}_{k}},fx,a\right),d\left(gx,f{x}_{2{n}_{k}},a\right)\right\}\\ \le \frac{\rho }{{s}_{2}}d\left({x}_{2{n}_{k}},x,a\right),\end{array}$

therefore by Lemma 3.5,

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

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

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

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

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

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

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

so

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

In the same way,

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

$\frac{1}{s}d\left(z,fz,a\right)\le \underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}d\left({f}^{n+1}z,fz,a\right)\le \frac{\rho }{s}d\left(z,fz,a\right)$, 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

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

The contraries of the above relation are as follows

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

and

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

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

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

if n is odd then we get

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

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

$\begin{array}{l}d\left({x}_{2n},{x}_{2n+1},a\right)\\ \le s\left(d\left({x}_{2n},z,a\right)+d\left({x}_{2n+1},z,a\right)+d\left({x}_{2n},{x}_{2n+1},z\right)\right)\\ <\frac{s}{s\left(1+\rho \right)}d\left({x}_{2n},g{x}_{2n},a\right)+\frac{s}{s\left(1+\rho \right)}d\left({x}_{2n+1},f{x}_{2n+1},a\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }+\frac{s}{s\left(1+\rho \right)}d\left({x}_{2n},g{x}_{2n},{x}_{2n+1}\right)\\ =\frac{1}{1+\rho }\left(d\left({x}_{2n},{x}_{2n+1},a\right)+d\left({x}_{2n+1},{x}_{2n+2},a\right)+d\left({x}_{2n},{x}_{2n+1},{x}_{2n+1}\right)\right)\end{array}$

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

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

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

or

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

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

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

or

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

From Lemma 2.5, we have

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

or

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

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

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

it implies

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

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

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

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

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

NOTES

*Corresponding author.

Conflicts of Interest

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

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