Extended Partial S_b-Metric Spaces

Abstract

In this paper, we introduce the concept of extended partial $S_b$-metric spaces, which is a generalization of the extended $S_b$-metric spaces. Basically, in the triangle inequality, we add a control function with some very interesting properties. These new metric spaces generalize many results in the literature. Moreover, we prove some fixed point theorems under some different contractions, and some examples are given to illustrate our results.

1. Introduction

Fixed point theory has become the focus of many researchers lately due its applications in many fields see [1,2,3,4,5,6,7,8,9,10,11,12]). The concept of b-metric space was introduced by Bakhtin [13], which is a generalization of metric spaces.

Definition 1

([13]). A b-metric on a non empty set X is a function $d:X^2\to [0,\infty )$ such that, for all $x,y,z\in X\mathrm{and}k\ge 1,$ the following three conditions are satisfied:

(i) $d(x,y)=0$ if and only if $x=y,$

(ii) $d(x,y)=d(y,x),$

(iii) $d(x,y)\le k\left[d\right(x,z)+d(z,y\left)\right].$

As usual, the pair $(X,d)$ is called a b-metric spaces.

A three-dimensional metric space was introduced by Sedghi et al. [14], and it is called S-metric spaces. Later, and as a generalization of the S-metric spaces, $S_b$-metric spaces were introduced. In [15], extended $S_b$-metric spaces were introduced

Definition 2

([15]). Let X be a non empty set and a function $\theta :X^3\to [1,\infty ).$ If the function $S_{\theta }:X^3\to [0,\infty )$ satisfies the following conditions for all $x,y,z,t\in X$:

1.

$S_{\theta }(x,y,z)=0\mathit{implies}x=y=z;$

2.

$S_{\theta }(x,y,z)\le \theta (x,y,z)[S_{\theta }(x,x,t)+S_{\theta }(y,y,t)+S_{\theta }(z,z,t)],$

then the pair $(X,S_{\theta })$ is called an extended $S_b$-metric spaces.

First, note that, if $\theta (x,y,z)=s\ge 1,$ then we have an $S_b$-metric spaces, which leads us to conclude that every $S_b$-metric spaces is an extended $S_b$-metric spaces, but the converse is not always true.

Definition 3

([15]). Let $(X,S_{\theta })$ be an extended $S_b$-metric space. Then,

(i) 

A sequence $\left\{x_n\right\}$ is called convergent if and only if there exists $z\in X$ such that $S_{\theta }(x_n,x_n,z)$ goes to 0 as n goes toward ∞. In this case, we write ${\displaystyle \underset{n⟶\infty }{lim}{x}_n=z}$.

(ii) 

A sequence $\left\{x_n\right\}$ is called a Cauchy sequence if and only if $S_{\theta }(x_n,x_n,x_m)$ goes to 0 as $n,m$ goes toward ∞.

(iii) 

$(X,S_{\theta })$ is said to be a complete extended $S_b$-metric space if every Cauchy sequence $\left\{x_n\right\}$ converges to a point $x\in X$

(iv) 

Define the diameter of a subset Y of X by

$${\displaystyle diam\left(Y\right):=Sup\{S_{\theta }(x,y,z)\mid x,y,z\in Y\}.}$$

In the extended $S_b$-metric spaces, we define a ball as follows:

$$B(x,ϵ)=\{y\in X\mid S_{\theta }(x,x,y)\le ϵ\}.$$

Next, we present some examples of extended $S_b$-metric spaces.

Example 1.

Let $X=C\left(\right[a,b],(-\infty ,\infty \left)\right)$ be the set of all continuous real valued functions on $[a,b].$ Define

$$S_{\theta }:X^3\to [0,\infty );S_{\theta }(x\left(t\right),y\left(t\right),z\left(t\right))=\underset{t\in [a,b]}{sup}{|max\{x\left(t\right),y\left(t\right)\}-z\left(t\right)|}^2,$$

and

$$\theta :X^3\to [1,\infty );\theta (x\left(t\right),y\left(t\right),z\left(t\right))=max\left\{\right|x\left(t\right)|,|y\left(t\right)\left|\right\}+\left|z\left(t\right)\right|+1.$$

It is not difficult to see that $(X,S_{\theta })$ is a complete extended $S_b$-metric spaces.

2. Extended Partial $S_b$-metric spaces

In this section, as a generalization of the extended $S_b$-metric spaces, we introduce extended partial $S_b$-metric spaces, along with its topology.

Definition 4.

Let X be a non empty set and a function $\theta :X^3\to [1,\infty ).$ If the function $S_{\theta }:X^3\to [0,\infty )$ satisfies the following conditions for all $x,y,z,t\in X$:

1.

$x=y=z\mathit{if}\mathit{and}\mathit{only}\mathit{if}{S}_{\theta }(x,y,z)=S_{\theta }(x,x,x)=S_{\theta }(y,y,y)=S_{\theta }(z,z,z);$

2.

$S_{\theta }(x,x,x)\le S_{\theta }(x,y,z),$

3.

$S_{\theta }(x,y,z)\le \theta (x,y,z)[S_{\theta }(x,x,t)+S_{\theta }(y,y,t)+S_{\theta }(z,z,t)],$

then the pair $(X,S_{\theta })$ is called an extended partial $S_b$-metric spaces.

First, note that, if $\theta (x,y,z)=s\ge 1,$ then we have a partial $S_b$-metric spaces, which leads us to conclude that every $S_b$-metric spaces is an extended $S_b$-metric spaces, but the converse is not always true.

Definition 5.

Let $(X,S_{\theta })$ be a extended partial $S_b$-metric space. Then,

• A sequence $\left\{x_n\right\}$ is called convergent if and only if there exists $z\in X$ such that $S_{\theta }(x_n,x_n,z)$ goes to $S_{\theta }(z,z,z)$ as n goes toward ∞. In this case, we write ${\displaystyle \underset{n⟶\infty }{lim}{x}_n=z}$.
• A sequence ${\left\{x_n\right\}}_{n=0}^{\infty }$ of elements in X is called $S_{\theta }$-Cauchy if ${lim}_{n,m}{S}_{\theta }(x_n,x_n,x_m)$ exists and is finite.
• The extended partial $S_{\theta }$-metric spaces $(X,S_{\theta })$ is called complete if, for each $S_{\theta }$-Cauchy sequence ${\left\{x_n\right\}}_{n=0}^{\infty },$ there exists $z\in X$ such that

  $$S_{\theta }(z,z,z)=\underset{n}{lim}{S}_{\theta }(z,z,x_n)=\underset{n,m}{lim}{S}_{\theta }(x_n,x_n,x_m).$$
• A sequence ${\left\{x_n\right\}}_n$ in an extended partial $S_b$-metric spaces $(X,S_{\theta })$ is called 0-Cauchy if

  $${lim}_{n,m}{S}_{\theta }(x_n,x_n,x_m)=0.$$
• We say that $(X,S_{\theta })$ is 0-complete if every 0-Cauchy in X converges to a point $x\in X$ such that $S_{\theta }(x,x,x)=0.$

Note that every extended $S_b$-metric spaces is an extended partial $S_b$-metric spaces, but the converse is not always true. The following example is an example of an extended partial $S_b$-metric spaces which is not extended $S_b$-metric spaces.

Example 2.

Let $X=C\left(\right[a,b],(-\infty ,\infty \left)\right)$ be the set of all continuous real valued functions on $[a,b].$ Define

$$S_{\theta }:X^3\to [0,\infty );S_{\theta }(x\left(t\right),y\left(t\right),z\left(t\right))=\underset{t\in [a,b]}{sup}{|max\{x\left(t\right),y\left(t\right),z\left(t\right)\}|}^2,$$

and

$$\theta :X^3\to [1,\infty );\theta (x\left(t\right),y\left(t\right),z\left(t\right))=max\left\{\right|x\left(t\right)|,|y\left(t\right)\left|\right\}+\left|z\left(t\right)\right|+1.$$

First, note that, for all $x\left(t\right),y\left(t\right),z\left(t\right)\mathit{and}f\left(t\right)\in X$, we have

$$\begin{array}{ll}\hfill S_{\theta }(x\left(t\right),y\left(t\right),z\left(t\right))& ={\displaystyle \underset{t\in [a,b]}{sup}}{|max\{x\left(t\right),y\left(t\right),z\left(t\right)\}|}^2\hfill \\ & \le {\displaystyle \underset{t\in [a,b]}{sup}}{|max\{x\left(t\right),x\left(t\right),f\left(t\right)\}|}^2+{\displaystyle \underset{t\in [a,b]}{sup}}{|max\{y\left(t\right),y\left(t\right),f\left(t\right)\}|}^2\hfill \\ & +{\displaystyle \underset{t\in [a,b]}{sup}}{|max\{z\left(t\right),z\left(t\right),f\left(t\right)\}|}^2\hfill \\ & \le S_{\theta }(x\left(t\right),x\left(t\right),f\left(t\right))+S_{\theta }(y\left(t\right),y\left(t\right),f\left(t\right))+S_{\theta }(z\left(t\right),z\left(t\right),f\left(t\right))\hfill \\ & \le \theta (x\left(t\right),y\left(t\right),z\left(t\right))[S_{\theta }(x\left(t\right),x\left(t\right),f\left(t\right))+S_{\theta }(y\left(t\right),y\left(t\right),f\left(t\right))+S_{\theta }(z\left(t\right),z\left(t\right),f\left(t\right))].\hfill \end{array}$$

In the last inequality, we used the fact that, for all $x\left(t\right),y\left(t\right),z\left(t\right)\in X$, we have $\theta \left(x\right(t),y(t),z(t\left)\right)\ge 1,$ Hence, $(X,S_{\theta })$ is an extended partial $S_b$-metric spaces, but it is not an extended $S_b$-metric spaces, since the self distance is not zero.

In the extended partial $S_b$-metric spaces, we define a ball as follows:

$$B(x,ϵ)=\{y\in X\mid S_{\theta }(x,x,y)\le ϵ\}.$$

Definition 6.

An extended partial $S_b$-metric spaces $(X,S_{\theta })$ is said to be symmetric if it satisfies the following condition:

$$S_{\theta }(x,x,y)=S_{\theta }(y,y,x)\mathit{for}\mathit{all}x,y\in X.$$

Theorem 1.

Let $(X,S_{\theta })$ be a complete symmetric extended partial $S_b$-metric spaces such that $S_{\theta }$ is continuous, and let T be a continuous self mapping on X satisfying the following condition:

$$S_{\theta }(Tx,Ty,Tz)\le k{S}_{\theta }(x,y,z)\mathit{for}\mathit{all}x,y,z\in X,$$

where $0<k<1$ and for every $x_0\in X$ we have ${lim}_{n,m\to \infty }\theta (T^{n}x,T^{n}x,T^m)<\frac{1}{k}.$ Then, T has a unique fixed point say $u.$ In addition, for every $y\in X$, we have ${lim}_{n\to \infty }{T}^{n}y=u.$

Proof. 

Since X is a nonempty set, pick $x_0\in X$ and define the sequence $\left\{x_n\right\}$ as follows:

$$x_1=T{x}_0,x_2=T{x}_1=T^2{x}_0,\cdots ,x_n=T^n{x}_0,\cdots$$

Note that, by $\left(1\right),$ we have

$$S_{\theta }(x_n,x_n,x_{n+1})\le k{S}_{\theta }(x_{n-1},x_{n-1},x_n)\le \cdots \le k^n{S}_{\theta }(x_0,x_0,x_1).$$

Now, pick two natural numbers $n<m.$ Hence, by the triangle inequality of the extended partial $S_b$-metric space, we deduce

$$\begin{array}{ll}\hfill S_{\theta }(x_n,x_n,x_m)& \le \theta (x_n,x_n,x_m)2k^n{S}_{\theta }(x_0,x_0,x_1)+\theta (x_n,x_n,x_m)\theta (x_{n+1},x_{n+1},x_m)2k^{n+1}{S}_{\theta }(x_0,x_0,x_1)\hfill \\ & +\cdots +\theta (x_n,x_n,x_m)\cdots \theta (x_{m-1},x_{m-1},x_m)2k^{m-1}{S}_{\theta }(x_0,x_0,x_1)\hfill \\ & \le S_{\theta }(x_0,x_0,x_1)[\theta (x_1,x_1,x_m)\theta (x_2,x_2,x_m)\cdots \theta (x_{n-1},x_{n-1},x_m)\theta (x_n,x_n,x_m)2k^n\hfill \\ & +\theta (x_1,x_1,x_m)\theta (x_2,x_2,x_m)\cdots \theta (x_n,x_n,x_m)\theta (x_{n+1},x_{n+1},x_m)2k^{n+1}\hfill \\ & +\cdots +\theta (x_1,x_1,x_m)\theta (x_2,x_2,x_m)\cdots \theta (x_{m-2},x_{m-2},x_m)\theta (x_{m-1},x_{m-1},x_m)2k^{m-1}].\hfill \end{array}$$

By the hypothesis of the theorem, we have

$$\underset{n,m\to \infty }{lim}\theta (x_n,x_n,x_m)\left(k\right)<1.$$

Therefore, by the Ratio test, the series ${\sum }_{n=1}^{\infty }2k^n{\prod }_{i=1}^n\theta (x_i,x_i,x_m)$ converges. Now, let

$$A=\sum _{n=1}^{\infty }2k^n\prod _{i=1}^n\theta (x_i,x_i,x_m)\mathrm{and}{A}_n=\sum _{j=1}^{n}2k^j\prod _{i=1}^j\theta (x_i,x_i,x_m).$$

Hence, for $m>n$, we deduce that

$$S_{\theta }(x_n,x_n,x_m)\le S_{\theta }(x_0,x_0,x_1)[A_{m-1}-A_{n-1}].$$

Taking the limit as $n,m\to \infty ,$ we conclude that $\left\{x_n\right\}$ is a Cauchy sequence. Since X is complete, $\left\{x_n\right\}$ converges to some $u\in X,$ such that

$$S_{\theta }(u,u,u)=\underset{n\to \infty }{lim}{S}_{\theta }(u,u,x_n)=\underset{n,m\to \infty }{lim}{S}_{\theta }(x_n,x_n,x_m).$$

Now, using the fact that T and $S_{\theta }$ are continuous, we deduce that

$$S_{\theta }(x_{n+1},x_{n+1},x_{n+1})=S_{\theta }(x_{n+1},x_{n+1},T{x}_n)=S_{\theta }(T{x}_n,T{x}_n,T{x}_n).$$

Taking the limit in the above equalities, we can easily conclude that $Tu=u.$ Hence, u is a fixed point of $T.$ To show uniqueness, assume that there exists $v\ne u\in X$ such that $Tu=u$ and $Tv=v.$ Thus,

$$\begin{array}{ll}\hfill S_{\theta }(u,u,v)& =S_{\theta }(Tu,Tu,Tv)\hfill \\ & \le k{S}_{\theta }(u,u,v)\hfill \\ & <S_{\theta }(u,u,v),\hfill \end{array}$$

which leads us to a contradiction. Therefore, T has a unique fixed point in X as desired. □

Now, we present the following example as an application of Theorem 1.

Example 3.

Let $X=C\left(\right[a,b],(-\infty ,\infty \left)\right)$ be the set of all continuous real valued functions on $[a,b].$ Define

$$S_{\theta }:X^3\to [0,\infty );S_{\theta }(x\left(t\right),y\left(t\right),z\left(t\right))=\underset{t\in [a,b]}{sup}{|max\{x\left(t\right),y\left(t\right),z\left(t\right)\}|}^2,$$

and

$$\theta :X^3\to [1,\infty );\theta (x\left(t\right),y\left(t\right),z\left(t\right))=max\left\{\right|x\left(t\right)|,|y\left(t\right)\left|\right\}+\left|z\left(t\right)\right|+1.$$

Now, let T be a self mapping on X defined by

$${\displaystyle Tx=\frac{x}{2}.}$$

Note that

$$S_{\theta }(Tx,Ty,Tz)\le \frac{1}{3}{S}_{\theta }(x,y,z).$$

In this case, we have $k=\frac{1}{3}.$ On the other hand, it is not difficult to see that, for every $x\in X,$ we have

$${\displaystyle T^{n}x=\frac{x}{2^n}.}$$

Thus, it is not difficult to see that

$$\underset{n,m\to \infty }{lim}\theta (T^{n}x,T^{n}x,T^{m}x)<\frac{3}{2}.$$

Therefore, all the conditions of Theorem 1 are satisfied and hence T has a unique fixed point which is in this case 0.

Theorem 2.

Let $(X,S_{\theta })$ be a symmetric complete extended partial $S_b$-metric spaces such that $S_{\theta }$ is continuous and T be a continuous self mapping on X satisfying

$$S_{\theta }(Tx,Ty,Tz)\le \psi \left[S_{\theta }(x,y,z)\right]forallx,y,z\in X,$$

where ψ is a comparison function (i.e., $\psi :[0,+\infty )⟶[0,+\infty )$ is an increasing function such that ${\displaystyle \underset{n\to \infty }{lim}{\psi }^n\left(t\right)=0}$ for each fixed $t>0$.) In addition, assume that there exists $s>1$ such that, for every $x_0\in X$ and $x\in X$, we have

$$\underset{n\to \infty }{lim}\theta (x_n,x_n,x)<\frac{s}{2}.$$

Then, T has a unique fixed point in X.

Proof. 

Let $x\in X$ and $ϵ>0$. Let n be a natural number such that ${\displaystyle {\psi }^n\left(ϵ\right)<\frac{ϵ}{2s}}$.

Let $F=T^n$ and $x_k=F^k\left(x\right)$ for $k\in \mathbb{N}$. Then, for $x,y\in X$ and $\alpha ={\psi }^n,$ we have

$$\begin{array}{ll}\hfill S_{\theta }(Fx,Fx,Fy)& \le {\psi }^n\left(S_{\theta }(x,x,y)\right)\\ & =\alpha \left(S_{\theta }(x,x,y)\right).\end{array}$$

Hence, for $k\in \mathbb{N}$ $S_{\theta }(x_{k+1},x_{k+1},x_k)$ goes to 0 as k goes toward ∞. Therefore, let k be such that

$${\displaystyle S_{\theta }(x_{k+1},x_{k+1},x_k)<\frac{ϵ}{2s}.}$$

Note that $x_k\in B(x_k,ϵ)$. Therefore, $B(x_k,ϵ)\ne \varnothing$. Hence, for all $z\in B(x_k,ϵ),$ we have

$$\begin{array}{lll}\hfill S_{\theta }(Fz,Fz,F{x}_k)& \le & \alpha \left(S_{\theta }(x_k,x_k,z)\right)\hfill \\ & \le & {\displaystyle \alpha \left(ϵ\right)={\psi }^n\left(ϵ\right)<\frac{ϵ}{2s}<\frac{ϵ}{s}.}\hfill \end{array}$$

Since ${\displaystyle S_{\theta }(F{x}_k,F{x}_k,F{x}_k)=S_{\theta }(x_{k+1},x_{k+1},x_k)<\frac{ϵ}{2s}}$, thus

$$\begin{array}{lll}\hfill S_{\theta }(x_k,x_k,Fz)& \le & \theta (x_k,x_k,Fz)[S_{\theta }(x_k,x_k,x_{k+1})+S_{\theta }(x_k,x_k,x_{k+1})+S_{\theta }(Fz,Fz,x_{k+1})]\hfill \\ & =& \theta (x_k,x_k,Fz)[2S_{\theta }(x_k,x_k,x_{k+1})+S_{\theta }(Fz,Fz,x_{k+1})]\hfill \\ & \le & {\displaystyle \theta (x_k,x_k,Fz)[2\frac{ϵ}{2s}+\frac{ϵ}{s}].}\hfill \end{array}$$

Now, taking the limit of the above inequality as $k\to \infty$, we get

$$S_{\theta }(x_k,x_k,F_z)\le ϵ.$$

Hence, F maps $B(x_k,ϵ)$ to itself. Since $x_k\in B(x_k,ϵ)$, we have $F{x}_k\in B(x_k,ϵ)$. By repeating this process, we get

$$F_{x_k}^m\in B(x_k,ϵ)\mathrm{for} \mathrm{all}m\in \mathbb{N}.$$

That is, $x_l\in B(x_k,ϵ)$ for all $l\ge k$. Hence,

$$S_{\theta }(x_m,x_m,x_l)<ϵ\mathrm{for} \mathrm{all}m,l>k.$$

Therefore, $\left\{x_k\right\}$ is a Cauchy sequence and, by the completeness of X, there exists $u\in X$ such that $x_k$ converges to u as k goes toward ∞. Moreover, ${\displaystyle u=\underset{k\to \infty }{lim}{x}_{k+1}=\underset{k\to \infty }{lim}{x}_k=F\left(u\right)}$.

Thus, F has u as a fixed point. We prove now the uniqueness of the fixed point for F. Since $\alpha \left(t\right)={\psi }^n\left(t\right)<t$ for any $t>0$, let u and $u_1$ be two fixed points of F:

$$\begin{array}{lll}\hfill S_{\theta }(u,u,u_1)& =& S_{\theta }(F_u,F_u,F_{u_1})\hfill \\ & \le & {\psi }^n\left(S_{\theta }(u,u,u_1)\right)\hfill \\ & =& \alpha \left(S_{\theta }(u,u,u_1)\right)\hfill \\ & <& S_{\theta }(u,u,u_1).\hfill \end{array}$$

Thus, $S_{\theta }(u,u,u_1)=0,$ that is $u=u_1$ and hence, F has a unique fixed point in X. On the other hand, $T^{nk+r}\left(x\right)=F^k\left(T^r\left(x\right)\right)$ goes to u as k goes toward ∞. Hence, $T^{m}x$ goes to u as m goes toward ∞ for every x. That is ${\displaystyle u=\underset{m\to \infty }{lim}T{x}_m=T\left(u\right)}$. Therefore, T has a fixed point. □

Theorem 3.

Let $(X,S_{\theta })$ be a complete symmetric extended partial $S_b$-metric spaces such that $S_{\theta }$ is continuous, and let T be a continuous self mapping on X satisfying the following condition:

$$S_{\theta }(Tx,Ty,Tz)\le \lambda [S_{\theta }(x,x,Tx)+S_{\theta }(y,y,Ty)+S_{\theta }(z,z,Tz)]\mathit{for}\mathit{all}x,y,z\in X,$$

(1)

where $\lambda \in [0,{\displaystyle \frac{1}{3}})$, $\lambda \ne {\displaystyle \frac{1}{3\theta (T^{n}x,T^{n}x,T^m)}}$ for every $x_0\in X$. Then, T has a unique fixed point $u\in X$ and $S_{\theta }(u,u,u)=0$.

Proof. 

We first prove that if T has a fixed point, then it is unique. We must show that, if $u\in X$ is a fixed point of T, that is, $Tu=u$ then $S_{\theta }(u,u,u)=0$.

From $\left(1\right)$, we obtain

$$\begin{array}{lll}\hfill S_{\theta }(u,u,u)=S_{\theta }(Tu,Tu,Tu)& \le & \lambda [S_{\theta }(u,u,Tu)+S_{\theta }(u,u,Tu)+S_{\theta }(u,u,Tu)]\hfill \\ & =& 3\lambda S_{\theta }(u,u,Tu)\mathrm{since}\lambda \in [0,{\displaystyle \frac{1}{3}}),\mathrm{we} \mathrm{have}\hfill \\ & <& S_{\theta }(u,u,u),\hfill \end{array}$$

which implies that we must have $S_{\theta }(u,u,u)=0$ Suppose $u,v\in X$ be two fixed points, that is, $Tu=u$ and $Tv=v$. Then, we have $S_{\theta }(u,u,u)=S_{\theta }(v,v,v)=0$. Relation $\left(1\right)$ gives

$$\begin{array}{lll}\hfill S_{\theta }(u,u,v)& =& S_{\theta }(Tu,Tu,Tv)\hfill \\ & \le & \lambda [S_{\theta }(u,u,Tu)+S_{\theta }(u,u,Tu)+S_{\theta }(v,v,Tv)]\hfill \\ & =& 2\lambda S_{\theta }(u,u,u)+\lambda S_{\theta }(v,v,v)\hfill \\ & =& 0.\hfill \end{array}$$

Therefore, $u=v$. Thereby, we get the uniqueness of the fixed point if it exists. For the existence of the fixed point, let $x_0\in X$ arbitrary, set $x_n=T^n{x}_0$ and $S_{b_n}=S_{\theta }(x_n,x_n,x_{n+1})$. We can assume $S_{b_n}>0$ for all $n\in \mathbb{N}$; otherwise, $x_n$ is a fixed point of T for at least one $n\ge 0$. For all n, we obtain from $\left(1\right)$

$$\begin{array}{lll}\hfill S_{b_n}& =& S_{\theta }(x_n,x_n,x_{n+1})=S_{\theta }(T{x}_{n-1},T{x}_{n-1},T{x}_n)\hfill \\ & \le & \lambda [2S_{\theta }(x_{n-1},x_{n-1},T{x}_{n-1})+S_{\theta }(x_n,x_n,T{x}_n)]\hfill \\ & =& \lambda [2S_{\theta }(x_{n-1},x_{n-1},x_n)+S_{\theta }(x_n,x_n,x_{n+1})]\hfill \\ & =& \lambda [2S_{b_{n-1}}+S_{b_n}].\hfill \end{array}$$

Therefore, $(1-\lambda )S_{b_n}\le 2\lambda S_{b_{n-1}}$. Thus,

$$S_{b_n}\le {\displaystyle \frac{2\lambda }{1-\lambda }}{S}_{b_{n-1}},\lambda \in [0,\frac{1}{3}).$$

(2)

Let $\beta ={\displaystyle \frac{2\lambda }{1-\lambda }}<1$. By repeating this process, we obtain

$$S_{b_n}\le {\beta }^n{b}_0.$$

Therefore, ${\displaystyle \underset{n\to \infty }{lim}{S}_{b_n}=0}$. Let us prove that $\left\{x_n\right\}$ is a Cauchy sequence. It follows from $\left(1\right)$ that, for $n,m\in \mathbb{N}$:

$$\begin{array}{lll}\hfill S_{\theta }(x_n,x_n,x_m)& =& S_{\theta }(T{x}_{n-1},T{x}_{n-1},T{x}_{m-1})\hfill \\ & \le & \lambda [2S_{\theta }(x_{n-1},x_{n-1},T{x}_{n-1})+S_{\theta }(x_{m-1},x_{m-1},T{x}_{m-1})]\hfill \\ & =& \lambda [2S_{\theta }(x_{n-1},x_{n-1},x_n)+S_{\theta }(x_{m-1},x_{m-1},x_m)]\hfill \\ & =& \lambda [2S_{b_{n-1}}+S_{b_{m-1}}].\hfill \end{array}$$

Thus, for every $ϵ>0$, as ${\displaystyle \underset{n\to \infty }{lim}{S}_{b_n}=0}$, we can find $n_0\in \mathbb{N}$ such that $S_{b_{n-1}}<{\displaystyle \frac{ϵ}{4}}$ and $S_{b_{m-1}}<{\displaystyle \frac{ϵ}{2}}$ for all $n,m>n_0$. Then, we obtain $2S_{b_{n-1}}+S_{b_{m-1}}\le 2{\displaystyle \frac{ϵ}{4}}+{\displaystyle \frac{ϵ}{2}}=ϵ$. As $\lambda <1$, it follows that $S_{\theta }(x_n,x_n,x_m)<ϵ$ $\mathrm{for}\mathrm{all}n,m>n_0$. Thus, $\left\{x_n\right\}$ is a Cauchy sequence in X and ${\displaystyle \underset{n\to \infty }{lim}{S}_{\theta }(x_n,x_n,x_m)=0}$. By completeness of X, there exists $u\in X$ such that

$${\displaystyle \underset{n,m\to \infty }{lim}{S}_{\theta }(x_n,x_n,u)=S_{\theta }(u,u,u)=0.}$$

Now, we shall prove that $Tu=u$. For any $n\in \mathbb{N}$,

$$\begin{array}{lll}\hfill S_{\theta }(u,u,Tu)& \le & \theta (u,u,Tu)[2S_{\theta }(u,u,x_{n+1})+S_{\theta }(Tu,Tu,T{x}_n)]\hfill \\ & \le & \theta (u,u,Tu)[2S_{\theta }(u,u,x_{n+1})+\lambda (2S_{\theta }(u,u,Tu)+S_{\theta }(x_n,x_n,T{x}_n))].\hfill \end{array}$$

Therefore, $(1-2\theta (u,u,Tu)\lambda )S_{\theta }(u,u,Tu)\le 2\theta (u,u,Tu)S_{\theta }(u,u,x_{n+1})+\theta (u,u,Tu)\lambda S_{\theta }(x_n,x_n,T{x}_n),$ giving

$$S_{\theta }(u,u,Tu)\le {\displaystyle \frac{2\theta (u,u,Tu)}{1-2\theta (u,u,Tu)\lambda }}{S}_{\theta }(u,u,x_{n+1})+{\displaystyle \frac{\theta (u,u,Tu)\lambda }{1-2\theta (u,u,Tu)\lambda }}{S}_{\theta }(x_n,x_n,T{x}_n).$$

Since $S_{\theta }$ and T are continuous, we have $S_{\theta }(x_n,x_n,T{x}_n)$ goes to $S_{\theta }(u,u,Tu)$, and n goes toward $\infty .$ Therefore, we obtain

$$\begin{array}{lll}\hfill S_{\theta }(u,u,Tu)& \le & {\displaystyle \frac{2\theta (u,u,Tu)}{1-2\theta (u,u,Tu)\lambda }}{S}_{\theta }(u,u,u)+{\displaystyle \frac{\theta (u,u,Tu)\lambda }{1-2\theta (u,u,Tu)\lambda }}{S}_{\theta }(u,u,Tu)\hfill \\ \hfill (1-{\displaystyle \frac{\theta (u,u,Tu)\lambda }{1-2\theta (u,u,Tu)\lambda }})S_{\theta }(u,u,Tu)& \le & {\displaystyle \frac{2\theta (u,u,Tu)}{1-2\theta (u,u,Tu)\lambda }}{S}_{\theta }(u,u,u)\hfill \\ \hfill S_{\theta }(u,u,Tu)& \le & {\displaystyle \frac{2\theta (u,u,Tu)}{1-3\theta (u,u,Tu)\lambda }}{S}_{\theta }(u,u,u).\hfill \end{array}$$

As $\lambda \ne {\displaystyle \frac{1}{3\theta (u,u,Tu)}}$ and from $\left(2\right)$, we obtain $S_{\theta }(u,u,Tu)=0$ and then $Tu=u$. □

In closing, we would like to present to the reader the following open questions:

Question 1.

Is it possible to omit the continuity of T in Theorem 1, and obtain a unique fixed point?

Question 2.

If we omit the symmetry condition of the extended partial $S_b$-metric spaces in Theorem 2, is it possible to prove the existence of a fixed point?