Suzuki-type fixed point results in b-metric spaces

Abstract

An ingenious approach to generalize Banach contraction principle was adopted by Suzuki in his seminal papers (Proc Am Math Soc 136:1861–1869, 2008, Nonlinear Anal Theory Methods Appl 71:5313–5317, 2009). In this paper we prove certain common fixed point results for generalized Suzuki contractions in the set-up of b-metric spaces, where the b-metric function is not necessarily continuous. Finally, some examples are presented to verify the effectiveness and applicability of our main results.

Introduction

There are a lot of generalizations of Banach fixed point principle in the literature. In 2008 Suzuki introduced an interesting generalization of Banach fixed point principle. This interesting fixed-point result is as follows.

Theorem 1

[26] Let (X, d) be a complete metric space, and let T be a mapping on X. Define a non-increasing function ${\theta }$ from [0, 1) into (1/2, 1] by

$$ \theta (r) = \left\{ {\begin{array}{*{20}l} {1,} \hfill & \,\,\quad{0 \le r \le \frac{{\sqrt 5 - 1}}{2}} \hfill \\ {\frac{{1 - r}}{{r^{2} }},} \hfill & \quad{\frac{{\sqrt 5 - 1}}{2} \le r \le \frac{1}{{\sqrt 2 }}} \hfill \\ {\frac{1}{{1 + r}},} \hfill & \quad{\frac{1}{{\sqrt 2 }} \le r < 1.} \hfill \\ \end{array}} \right. $$

Assume that there exists $r\in [0,1)$, such that

$$\begin{aligned} {\theta (r)}{\mathrm{d}}(x,Tx)\le {\mathrm{d}}(x,y)\Longrightarrow {\mathrm{d}}(Tx,Ty)\le r{\mathrm{d}}(x,y), \end{aligned}$$

for all $\ x,y\in X$, then there exists a unique fixed-point z of T. Moreover, ${\rm lim}_{{\it n}\rightarrow \infty }T^{{\it n}}x=z$ for all $x\in X$.

Suzuki proved also the following version of Nemytckii fixed point theorem.

Theorem 2

Let (X, d) be a compact metric space. Let $T:X\rightarrow X$ be a selfmap, satisfying for all $x,y\in X$ $x\ne y$ the condition

$$\begin{aligned} \frac{1}{2}{\mathrm{d}}(x,Tx)\le {\mathrm{d}}(x,y)\Longrightarrow {\mathrm{d}}(Tx,Ty)<{\mathrm{d}}(x,y) \end{aligned}$$

Then T has a unique fixed point in X.

This theorem was also generalized in [6].

In addition to the above results, Kikkawa and Suzuki [11] provided a Kannan-type version of the theorems mentioned above. In [21], a Chatterjea-type version is provided, whereas Popescu [20] obtained a Cirić-type version. Recently, Kikkawa and Suzuki also provided multivalued versions in [12, 13].

Very recently, Hussain et al. in [8] have extended Suzuki’s Theorems 1 and 2, as well as Popescu’s results from [20] to the case of metric-type spaces and cone metric-type spaces.

Czerwik in [5] introduced the concept of b-metric space. Since then, several papers deal with fixed point theory for single-valued and multivalued operators in b-metric spaces (see also [1–5, 7–10, 14–17, 19, 24, 25]). Pacurar [19] proved results on sequences of almost contractions and fixed points in b-metric spaces. Recently, Hussain and Shah [9] obtained results on KKM mappings in cone b-metric spaces. Khamsi ([14, 15]) also showed that each cone metric space has a b-metric structure.

The aim of this paper is to present some common fixed point results for two mappings under generalized contractive condition in b-metric space, where the b-metric function is not necessarily continuous. Because many of the authors in their works have used the b-metric spaces in which the b-metric function is assumed to be continuous. From this point of view the results obtained in this paper generalize and extend several ones obtained earlier concerning b-metric space.

Consistent with [5] and [25, p. 264], the following definition and results will be needed in the sequel.

Definition 1

[5] Let X be a (nonempty) set and $b\ge 1$ be a given real number. A function $d:X\times X\rightarrow R^{+}$ is a b-metric spaces iff, for all $x,y,z\in X$, the following condition are satisfied:

• (b1) d$(x,y)=0$    iff $x=y,$

• (b2) d$(x,y)={\mathrm{d}}(y,x),$

• (b3) d$(x,z)\le b[{\mathrm{d}}(x,y)+{\mathrm{d}}(y,z)].$

The pair $(X,{\mathrm{d}})$ is called a b-metric space.

It should be noted that, the class of b-metric spaces is effectively larger than that of metric spaces, since a b-metric is a metric only if $b=1.$

We present an example which shows that a b-metric on X need not be a metric on X. (see also [25, p. 264]):

Example 1

[22] Let (X, d) be a metric space, and $\rho (x,y)=(d(x,y))^{p},$ where $p>1$ is a real number. Then $\rho$ is a b-metric with $b=2^{p-1}.$

However, if (X, d) is a metric space, then $(X,\rho )$ is not necessarily a metric space.

For example, if $X= {\mathbb {R}}$ is the set of real numbers and $d(x,y)=\left| x-y\right|$ is the usual Euclidean metric, then $\rho (x,y)=(x-y)^{2}$ is a b-metric on ${\mathbb {R}}$ with $b=2,$ but is not a metric on ${\mathbb {R}}$.

Before stating and proving our results, we present some definition and proposition in b-metric space. We recall first the notions of convergence and completeness in a b-metric space.

Definition 2

[3] Let (X, d) be a b-metric space. Then a sequence $\{x_{n}\}$ in X is called:

• (a) convergent if and only if there exists $x\in X$ such that $d(x_{n},x)\rightarrow 0$ as $n\rightarrow \infty$. In this case, we write $\lim _{n\rightarrow \infty }x_{n}=x.$

• (b) Cauchy if and only if $d(x_{n},x_{m})\rightarrow 0$ as $n,m\rightarrow \infty .$

Proposition 1

(see remark 2.1 in [3]) In a b-metric space (X, d) the following assertions hold:

• (i) a convergent sequence has a unique limit,

• (ii) each convergent sequence is Cauchy,

Definition 3

[3] The b-metric space (X, d) is complete if every Cauchy sequence in X converges.

It should be noted that, in general a b-metric function d(x, y) for $b>1$ is not jointly continuous in all two of its variables. Now we present an example of a b-metric which is not continuous.

Example 2

(see Example 3 in [8]) Let $X= {\mathbb {N}} \cup \{\infty \}$ and let $D:X\times X\rightarrow {\mathbb {R}} ^{+}$ be defined by,

$$D(m,n) = \left\{ {\begin{array}{*{20}l} 0 \hfill & {{\text{if}}\,{\it m}= {\it n}}, \\ {\left| {\frac{1}{m} - \frac{1}{n}} \right|,} \hfill & {{\text{if}}\,{\text{one}}\,{\text{of}}\,{\it m,}\,{\it n}\,{\text{is}}\,{\text{even}}\,{\text{and}}\,{\text{the}}\,{\text{other}}\;{\text{is}}\,{\text{even}}\,{\text{or}}\;\infty \,{\text{,}}} \hfill \\ 5 \hfill & {{\text{if}}\,{\text{one}}\,{\text{of}}\,{\it m},\;{\it n}\,{\text{is}}\,{\text{odd}}\,{\text{and}}\,{\text{other}}\;{\text{is}}\,{\text{odd}}\, {\text{(and}}\;{\text{m}}\ne{\text{n) or }}} \;\infty, \hfill \\ 2 \hfill & {{\text{otherwise}}}. \hfill \\ \end{array} } \right.{\text{ }} $$

Then it is easy to see that for all $m,n,p\in X,$ we have

$$ D(m,p)\le \frac{5}{2}(D(m,n)+D(n,p)). $$

Thus, (X, D) is a b-metric space with $b=\frac{5}{2}.$ In [8], it is proved that D(x, y) is not a continuous function.

Since in general a b-metric is not continuous, we need the following simple lemma about the b-convergent sequences.

Lemma 1

[22] Let $(X,{\mathrm d})$ be a b-metric space with $b\ge 1$, and suppose that $\{x_{n}\}$ and $\{y_{n}\}$ b-converge to x, y, respectively. Then, we have

$$ \frac{1}{b^{2}}{\mathrm{d}}(x,y)\le \liminf _{n\rightarrow \infty }{\mathrm{d}}(x_{n},y_{n})\le \limsup _{n\rightarrow \infty }{\mathrm{d}}(x_{n},y_{n})\le b^{2}{\mathrm{d}}(x,y). $$

In particular, if $x=y$, then $\lim _{n\rightarrow \infty }{\mathrm{d}}(x_{n},y_{n})=0$. Moreover, for each $z\in X$ we have

$$ \frac{1}{b}{\mathrm{d}}(x,z)\le \liminf _{n\rightarrow \infty }{\mathrm{d}}(x_{n},z)\le \limsup _{n\rightarrow \infty }{\mathrm{d}}(x_{n},z)\le b{\mathrm{d}}(x,z). $$

Main result

We start our work by proving the following crucial Theorem.

Theorem 3

Let (X, d) be a complete b-metric space. Let $T,S:X \longrightarrow X$ be two self-maps and $\theta :[0,1)\longrightarrow (\frac{ 1}{2},1]$ be defined by

$$ \theta (r) = \left\{ {\begin{array}{*{20}l} {1,} \hfill & {0 \le r \le \frac{{\sqrt 5 - 1}}{2}} \hfill \\ {\frac{{1 - r}}{{r^{2} }},} \hfill & {\frac{{\sqrt 5 - 1}}{2} \le r \le \frac{1}{{\sqrt 2 }}} \hfill \\ {\frac{1}{{1 + r}},} \hfill & {\frac{1}{{\sqrt 2 }} \le r < 1.} \hfill \\ \end{array} } \right. $$

(1)

Suppose there exists $r\in [0,1)$ such that for each $x,y\in X$, the following condition is satisfied

$$\begin{aligned} \frac{1}{b}\theta &(r)\min \{{\mathrm{d}}(x,Tx),d(x,Sx)\}\le {\mathrm{d}}(x,y)\Longrightarrow \nonumber \\ &\max \left\{ \begin{array}{l} {\mathrm{d}}(Sx,Sy),{\mathrm{d}}(Tx,Ty), \\ {\mathrm{d}}(Sx,Ty),{\mathrm{d}}(Sy,Tx) \end{array} \right\}\le \frac{r}{b^{2}}{\rm d}(x,y). \end{aligned}$$

( 2)

Then T, S have a unique common fixed point $z\in X.$

Proof

At first we show that if z is a fixed point of S or T, then z is a common fixed point of T and S. Let z be a fixed point of T that is $Tz=z$ then we show that $Sz=z$. From

$$ 0=\frac{1}{b}\theta (r)\min \{{\mathrm{d}}(z,Tz),d(z,Sz)\}\le {\mathrm{d}}(z,Tz), $$

it follows

$$\begin{aligned} {\mathrm{d}}(Sz,z)\le & {} \max \left\{ \begin{array}{l} {\mathrm{d}}(Sz,STz),\mathrm{d}(Tz,T^{2}z), \\ {\mathrm{d}}(Sz,T^{2}z),{\mathrm{d}}(STz,Tz) \end{array} \right\} \\\le & {} \frac{r}{b^{2}}{\mathrm{d}}(z,Tz)=0, \end{aligned}$$

thus $Sz=z$. Therefore it is enough to show that T have a fixed point. Putting $y=Sx$ in (2)

$$\frac{1}{b}\theta (r)\min \{\mathrm{d}(x,Tx),{\mathrm{d}}(x,Sx)\}\le {\mathrm{d}}(x,Sx), $$

it follows

$$\begin{aligned} \max \left\{ \begin{array}{l} {\mathrm{d}}(Sx,S^{2}x),{\mathrm{d}}(Tx,TSx), \\ {\mathrm{d}}(Sx,TSx),{\mathrm{d}}(S^{2}x,Tx) \end{array} \right\} \le \frac{r}{b^{2}}{\mathrm{d}}(x,Sx), \end{aligned}$$

(3)

for every $x\in X$. Hence,

$${\mathrm{d}}(Sx,TSx)\le \frac{r}{b^{2}}{\mathrm{d}}(x,Sx). $$

(4)

Now, putting $y=Tx$ in (2)

$$ \frac{1}{b}\theta (r)\min \{{\mathrm{d}}(x,Tx),{\mathrm{d}}(x,Sx)\}\le {\mathrm{d}}(x,Tx), $$

it follows

$$\begin{aligned} \max \left\{ \begin{array}{l} {\mathrm{d}}(Sx,STx),{\mathrm{d}}(Tx,T^{2}x), \\ {\mathrm{d}}(Sx,T^{2}x),{\mathrm{d}}(STx,Tx) \end{array} \right\} \le \frac{r}{b^{2}}{\mathrm{d}}(x,Tx), \end{aligned}$$

(5)

for every $x\in X$. Hence,

$$ {\mathrm{d}}(Tx,T^{2}x)\le \frac{r}{b^{2}}{\mathrm{d}}(x,Tx),$$

(6)

and

$$ {\mathrm{d}}(STx,Tx)\le \frac{r}{b^{2}}{\mathrm{d}}(x,Tx). $$

(7)

Let $x_{0}\in X$ be arbitrary and form the sequence $\{x_{n}\}$ by, $x_{2n+1}=Sx_{2n}$ and $Tx_{2n+1}=x_{2n+2}$ for $n\in {\mathbb {N}}\cup \{0\}$. We show that $\{x_{n}\}$ is a Cauchy sequence.

By (4), we have

$$\begin{aligned} {\mathrm{d}}(x_{2n+1},x_{2n+2})\;=\; & {} {\mathrm{d}}(Sx_{2n},TSx_{2n}) \nonumber \\\le & {} \frac{r}{b^{2}}{\mathrm{d}}(x_{2n},Sx_{2n})=\frac{r}{b^{2}}{\mathrm{d}}(x_{2n},x_{2n+1}). \end{aligned}$$

(8)

By (7), we have

$$\begin{aligned} {\mathrm{d}}(x_{2n+1},x_{2n})\;=\; & {} {\mathrm{d}}(STx_{2n-1},Tx_{2n-1}) \\\le & {} \frac{r}{b^{2}}{\mathrm{d}}(x_{2n-1},Tx_{2n-1})=\frac{r}{b^{2}} {\mathrm{d}}(x_{2n-1},x_{2n}). \end{aligned}$$

Therefore,

$$\begin{aligned} {\mathrm{d}}(x_{n},x_{n+1})\le & {} \frac{r}{b^{2}}{\mathrm{d}}(x_{n-1},x_{n}) \\\le & {} \frac{r^{2}}{b^{4}}{\mathrm{d}}(x_{n-2},x_{n-1}) \\&\vdots \\\le & {} \frac{r^{n}}{b^{2n}}{\mathrm{d}}(x_{0},x_{1}). \end{aligned}$$

Also, by definition of b-metric spaces for all $\ m\ge n,$ we have

$$\begin{aligned} \mathrm{d}(x_{n},x_{m})\le & {} b\mathrm{d}(x_{n},x_{n+1})+b^{2}\;\mathrm{d}(x_{n+1},x_{n+2})+\cdots +b^{m-n-1}\;\mathrm{d}(x_{m-1},x_{m}) \\\le & {} b\frac{r^{n}}{b^{2n}}\;\mathrm{d}(x_{0},x_{1})+b^{2}\frac{r^{n+1}}{b^{2n+2}} \;\mathrm{d}(x_{0},x_{1})+\cdots +b^{m-n-1}\frac{r^{m-1}}{b^{2m-2}}\;\mathrm{d}(x_{0},x_{1}) \\= & {} \frac{r^{n}}{b^{2n-1}}\mathrm{d}(x_{0},x_{1})+\frac{r^{n+1}}{b^{2n}} \mathrm{d}(x_{0},x_{1})+\cdots +\frac{r^{m-1}}{b^{m+n-1}}\;\mathrm{d}(x_{0},x_{1}) \\= & {} \frac{r^{n}}{b^{2n-1}}\left(1+\frac{r}{b}+\cdots +\frac{r^{m-n-1}}{b^{m-n}}\right)\;\mathrm{d}(x_{0},x_{1}) \\\le & {} \frac{r^{n}}{b^{2n-1}}\left(1+\frac{r}{b}+\cdots +(\frac{r}{b} )^{m-n-1}\right)\;\mathrm{d}(x_{0},x_{1}) \\\le & {} \frac{r^{n}}{b^{2n-1}}\mathrm{d}(x_{0},x_{1})\left( \frac{1}{1-\frac{r}{b}} \right) \longrightarrow 0\text { \quad as\quad }{\it n}\rightarrow \infty . \end{aligned}$$

So, we have

$$\begin{aligned} \lim _{n,m\rightarrow \infty }{\mathrm{d}}(x_{{\it n}},x_{m})=0. \end{aligned}$$

Hence, $\{x_{n}\}$ is a Cauchy sequence. Since X is complete, we conclude $\{x_{n}\}$ converges to z for some $z\in X.$ That is

$$ \lim _{n\rightarrow \infty }Sx_{2n}=\lim _{n\rightarrow \infty }x_{2n+1}=z, $$

and

$$ \lim _{n\rightarrow \infty }Tx_{2n+1}=\lim _{n\rightarrow \infty }x_{2n+2}=z. $$

Let us prove now that

$$ {\mathrm{d}}(z,Tx)\le r{\mathrm{d}}(z,x), $$

holds for each $x\ne z$. Since ${\mathrm{d}}(x_{2n},Sx_{2n})\longrightarrow 0,$ and by Lemma 1

$$\begin{aligned} \frac{1}{b}{\mathrm{d}}(z,x)\le \text { }\underset{ {\it n}\longrightarrow \infty }{\lim \sup \text { }}{\mathrm{d}}(x_{2{\it n}},x), \end{aligned}$$

$\text { thus} \ \underset{{\it n}\longrightarrow \infty }{\lim \sup \text { }} {\mathrm{d}}(x_{2{\it n}},x)>0$, it follows that there exists a $x_{2{\it n}_{k}}\in X$ such that

$$\begin{aligned} \frac{1}{b}\theta (r)\min \{{\mathrm{d}}(x_{2n_{k}},Sx_{2n_{k}}),{\mathrm{d}}(x_{2n_{k}},Tx_{2n_{k}})\}\le {\mathrm{d}}(x_{2n_{k}},x). \end{aligned}$$

Assumption (2) implies that for such $x_{2n_{k}}$

$$\begin{aligned} {\mathrm{d}}(Sx_{2n_{k}},Tx)\le & {} \max \left\{ \begin{array}{l} {\mathrm{d}}(Sx_{2n_{k}},Sx),{\mathrm{d}}(Tx_{2n_{k}},Tx) \\ {\mathrm{d}}(Sx_{2n_{k}},Tx),{\mathrm{d}}(Sx,Tx_{2n_{k}}) \end{array} \right\} \\\le & {} \frac{r}{b^{2}}{\rm d}(x_{2n_{k}},x), \end{aligned}$$

hence by Lemma 1

$$\begin{aligned} \frac{1}{b}{\mathrm{d}}(z,Tx)\le \text { }\lim \sup _{{\it n}\rightarrow \infty }{\mathrm{d}}(Sx_{2n_{k}},Tx)\le \frac{r}{b^{2}}\text { }\lim \sup _{{\it n}\rightarrow \infty }{\mathrm{d}}(x_{2n_{k}},x)\le \frac{r}{b}{\mathrm{d}}(z,x), \end{aligned}$$

thus for each $x\ne z$ we get that

$$ {\mathrm{d}}(z,Tx)\le\; rd({\it z},{\it x}). $$

(9)

We will prove that

$$ {\mathrm{d}}(T^{n}z,z)\le\; {\mathrm{d}}(Tz,z), $$

(10)

for each $n\in {\mathbb {N}}$. For $n=1$ this relation is obvious. Suppose that it holds for some $m\in {\mathbb {N}}$. If $T^{m}z=z$ then $T^{m+1}z=Tz$ it follows that the above inequality is true. If $T^{m}z\ne z$, we can apply (9) and the induction hypothesis, we get that

$$\begin{aligned} {\mathrm{d}}(z,T^{m+1}z)\le & {} r{\mathrm{d}}(z,T^{m}z) \\\le & {} r{\mathrm{d}}(Tz,z)\le {\mathrm{d}}(Tz,z), \end{aligned}$$

and (10) is proved by induction.

In order to prove that $Tz=z$. We consider two possible cases.

Case I. $0\le r<\frac{1}{\sqrt{2}}$ (and hence $\theta (r)\le \frac{1-r}{ r^{2}}$). We will prove first that

$$\begin{aligned} {\mathrm{d}}(T^{n}z,Tz)\le \frac{r}{b}{\mathrm{d}}(Tz,z) \end{aligned}$$

(11)

for each $n\in {\mathbb {N}}$. For $n=1$ it is obvious. For $n=2$ it follows from (6). Suppose that (11) holds for some $n>2$. Since

$$\begin{aligned} {\mathrm{d}}(Tz,z)\le & {} b{\mathrm{d}}(z,T^{n}z)+b{\mathrm{d}}(T^{n}z,Tz) \\\le & {} b{\mathrm{d}}(z,T^{n}z)+r{\mathrm{d}}(z,Tz), \end{aligned}$$

hence $(1-r){\mathrm{d}}(z,Tz)\le b{\mathrm{d}}(z,T^{n}z)$. It follows [using (6) with $x=T^{n-1}z$] that

$$\begin{aligned} \frac{1}{b}\theta (r)\min \{{\mathrm{d}}(ST^{n}z,T^{n}z),{\mathrm{d}}(T^{n}z,T^{n+1}z)\}\le & {} \frac{1-r}{br^{2}}{\mathrm{d}}(T^{n}z,T^{n+1}z) \\\le & {} \frac{1-r}{br^{n}}{\mathrm{d}}(T^{n}z,T^{n+1}z) \\\le & {} \frac{1-r}{br^{n}}.\frac{r^{n}}{b^{2n}}{\mathrm{d}}(z,Tz) \\= & {} \frac{1-r}{b^{2n+1}}{\mathrm{d}}(z,Tz) \\\le & {} \frac{1}{b^{2n}}{\mathrm{d}}(z,T^{n}z) \\\le & {} {\mathrm{d}}(z,T^{n}z). \end{aligned}$$

Assumptions (2) and (10) imply that

$$\begin{aligned}&\max \{{\mathrm{d}}(ST^{n}z,Sz),{\mathrm{d}}(ST^{n}z,Tz), \\ {\mathrm{d}}(T^{n+1}z,Tz),{\mathrm{d}}(Sz,T^{n+1}z)\}\le & {} \frac{r}{b^{2}}\mathrm{d}(z,T^{n}z) \\\le & {} \frac{r}{b^{2}}\mathrm{d}(z,Tz)\le \frac{r}{b}\mathrm{d}(z,Tz). \end{aligned}$$

Thus

$$\begin{aligned} {\mathrm{d}}(T^{n+1}z,Tz)\le \frac{r}{b}{\mathrm{d}}(Tz,z). \end{aligned}$$

(12)

So relation (11) is proved by induction.

Now $Tz\ne z$ and (11) imply that $T^{n}z\ne z$ for each $n\in {\mathbb {N}}$. Hence, (9) implies that

$$\begin{aligned} {\mathrm{d}}(z,T^{n+1}z)\le & {} r{\mathrm{d}}(z,T^{n}z) \\\le & {} r^{2}{\mathrm{d}}(z,T^{n-1}z) \\&\vdots \\\le & {} r^{n}{\mathrm{d}}(z,Tz). \end{aligned}$$

Hence $\lim _{n\rightarrow \infty }{\mathrm{d}}(z,T^{n+1}z)=0.$ On the other hand using Lemma 1, we have

$$\begin{aligned} \frac{1}{b}{\mathrm{d}}({\it z},\text { }\underset{{\it n}\longrightarrow \infty }{\lim \inf \text { } }{\it T}^{{\it n}+1}z)\le \text { }\underset{{\it n}\longrightarrow \infty }{\lim \inf \text { } }{\mathrm{d}}({\it z},{\it T}^{{\it n}+1}{\it z})=0, \end{aligned}$$

so

$$ {\mathrm{d}}(z,\text { }\underset{{\it n}\longrightarrow \infty }{\lim \inf \text { }} {\it T}^{{\it n}+1}{\it z})=0. $$

Similarly,

$${\mathrm{d}}(z,\text { }\underset{{\it n}\longrightarrow \infty }{\lim \sup \text { }} {\it T}^{{\it n}+1}{\it z})=0, $$

therefore ${\mathrm{d}}({\it z},\text { }\underset{{\it n}\longrightarrow \infty }{\lim \text { }} {\it T}^{{\it n}+1}{\it z})=0.$

Thus $T^{n+1}z\longrightarrow z$ and, using Lemma 1 in (12), we have

$$\begin{aligned} \frac{1}{b}{\mathrm{d}}(z,Tz)\le \text { }\underset{{\it n}\longrightarrow \infty }{\lim \sup \text { }}{\mathrm{d}}({\it T}^{{\it n}+1}{\it z},{\it Tz})\le \frac{r}{b}{\mathrm{d}}({\it z},{\it Tz}), \end{aligned}$$

which implies that ${\mathrm{d}}(z,Tz)=0$, a contradiction.

Case II. $\frac{1}{\sqrt{2}}\le r<1$ (and so $\theta (r)=\frac{1}{1+r})$. We will prove that there exists a subsequence $\{x_{n_{k}}\}$ of $\{x_{n}\}$ such that

$$ \frac{1}{b(1+r)}\min \{{\mathrm{d}}(x_{n_{k}},Sx_{n_{k}}),{\mathrm{d}}(x_{n_{k}},Tx_{n_{k}})\le \;{\mathrm{d}}(x_{n_{k}},z) $$

(13)

holds for each $k\in {\mathbb {N}}$. Suppose the contrary

$$ \frac{1}{b(1+r)}{\mathrm{d}}(x_{n},Tx_{n})\ge \frac{1}{b(1+r)}\min \{{\mathrm{d}}(x_{n},Sx_{n}),{\mathrm{d}}(x_{n},Tx_{n})\}>{\mathrm{d}}(x_{n},z), $$

and

$$ \frac{1}{b(1+r)}{\mathrm{d}}(x_{n},Sx_{n})\ge \frac{1}{b(1+r)}\min \{{\mathrm{d}}(x_{n},Sx_{n}),{\mathrm{d}}(x_{n},Tx_{n})\}>{\mathrm{d}}(x_{n},z),$$

holds for each $n\in {\mathbb {N}}$. Now if n is odd then

$$\begin{aligned} \frac{1}{b(1+r)}\mathrm{d}(x_{2n+1},Tx_{2n+1})&\ge {} \frac{1}{b(1+r)}\min \{\mathrm{d}(x_{2n+1},Sx_{2n+1}),\mathrm{d}(x_{2n+1},Tx_{2n+1}) \\&> {} \mathrm{d}(x_{2n+1},z), \end{aligned}$$

if n is even then

$$\begin{aligned} \frac{1}{b(1+r)}{\mathrm{d}}(x_{2n},Sx_{2n})\ge & {} \frac{1}{b(1+r)}\min \{{\mathrm{d}}(x_{2n},Sx_{2n}),{\mathrm{d}}(x_{2n},Tx_{2n}) \\> & {} {\rm d}(x_{2n},z), \end{aligned}$$

holds for each $n\in {\mathbb {N}}$. Then from (8) we have

$$\begin{aligned} {\mathrm{d}}(x_{2n},x_{2n+1})\le & {} b{\mathrm{d}}(x_{2n},z)+b{\mathrm{d}}(x_{2n+1},z) \\< & {} \frac{b}{b(1+r)}{\mathrm{d}}(x_{2n},Sx_{2n})+\frac{b}{b(1+r)}{\mathrm{d}}(x_{2n+1},Tx_{2n+1}) \\= & {} \frac{1}{1+r}{\mathrm{d}}(x_{2n},x_{2n+1})+\frac{1}{1+r}{\mathrm{d}}(x_{2n+1},x_{2n+2}) \\\le & {} \frac{1}{1+r}{\mathrm{d}}(x_{2n},x_{2n+1})+\frac{r}{b^{2}(1+r)}{\mathrm{d}}(x_{2n},x_{2n+1}) \\\le & {} \frac{1}{1+r}{\mathrm{d}}(x_{2n},x_{2n+1})+\frac{r}{1+r}{\mathrm{d}}(x_{2n},x_{2n+1}) \\\;= \;& {} {\mathrm{d}}(x_{2n},x_{2n+1}), \end{aligned}$$

which is impossible. Hence one of the following inequalities is satisfied for each $n\in {\mathbb {N}}$:

$$\begin{aligned} \frac{1}{b}\theta (r)\min \{{\mathrm{d}}(x_{2n},Sx_{2n}),{\mathrm{d}}(x_{2n},Tx_{2n})\le {\mathrm{d}}(x_{2n},z). \end{aligned}$$

or

$$\begin{aligned} \frac{1}{b}\theta (r)\min \{{\mathrm{d}}(x_{2n+1},Sx_{2n+1}),{\mathrm{d}}(x_{2n+1},Tx_{2n+1})\le {\mathrm{d}}(x_{2n+1},z). \end{aligned}$$

In other words, there is a subsequence $\{x_{n_{k}}\}$ of $\{x_{n}\}$ such that (13) holds for each $k\in {\mathbb {N}}$. Hence assumption (2) implies that

$$\begin{aligned} {\mathrm{d}}(Sx_{2n},Tz)\le & {} \max \left\{ \begin{array}{l} {\mathrm{d}}(Sx_{2n},Sz),{\mathrm{d}}(Tx_{2n},Tz), \\ {\mathrm{d}}(Tz,Sx_{2n}),{\mathrm{d}}(Sz,Tx_{2n}) \end{array} \right\} \\\le & {} \frac{r}{b^{2}}{\mathrm{d}}(x_{2n},z). \end{aligned}$$

or

$$\begin{aligned} {\mathrm{d}}(Tx_{2n+1},Tz)\le & {} \max \left\{ \begin{array}{l} {\mathrm{d}}(Sx_{2n+1},Sz),{\mathrm{d}}(Tx_{2n+1},Tz), \\ {\mathrm{d}}(Tz,Sx_{2n+1}),{\mathrm{d}}(Sz,Tx_{2n+1}) \end{array} \right\} \\\le & {} \frac{r}{b^{2}}{\mathrm{d}}(x_{2n+1},z). \end{aligned}$$

By Lemma 1, we get

$$\frac{1}{b}{\mathrm{d}}(z,Tz)\le \text { }\underset{n\longrightarrow \infty }{\lim \sup \text { }}\text { }{\mathrm{d}}(Sx_{2{\it n}},Tz)\le \frac{r}{b^{2}}\text { }\underset{{\it n}\longrightarrow \infty }{\lim \sup \text { }}{\mathrm{d}}(x_{2{\it n}},z)\le \frac{r}{b} {\mathrm{d}}({\it z},{\it z})=0, $$

or

$$\frac{1}{b}{\mathrm{d}}(z,Tz)\le \text { }\underset{{\it n}\longrightarrow \infty }{\lim \sup \text { }}\text { }{\mathrm{d}}({\it T}x_{2{\it n}+1},Tz)\le \frac{r}{b^{2}}\text { }\underset{{\it n}\longrightarrow \infty }{\lim \sup \text { }}{\mathrm{d}}(x_{2{\it n}+1},z)\le \frac{r}{b} {\mathrm{d}}(z,z)=0,$$

hence ${\mathrm{d}}(z,Tz)\le 0$, which is possible only if $Tz=z$.

Thus, we have proved that z is a fixed point of T. The uniqueness of the common fixed point follows easily from (2). Indeed, if $z,z^{\prime }$ are two common fixed points of T,

$$\begin{aligned} \frac{1}{b}\theta (r)\min \{{\mathrm{d}}(z,Tz),{\mathrm{d}}(z,Sz)\}\le {\mathrm{d}}(z,z^{\prime }), \end{aligned}$$

then (2) implies that

$$ {\mathrm{d}}(z,z^{\prime })=\max \{{\mathrm{d}}(Sz,Sz^{\prime }),{\mathrm{d}}(Tz,Tz^{\prime }),{\mathrm{d}}(Sz,Tz^{\prime }),{\mathrm{d}}(Sz^{\prime },Tz)\}\le \frac{r}{b^{2}}{\mathrm{d}}(z,z^{\prime }),$$

which is possible only if $z=z^{\prime }$. This proves that z is a unique common fixed point of T and S. $\square$

According to Theorem 3 we get the following result.

Corollary 1

Let $(X,\mathrm{d})$ be a complete b-metric space, and let T be a mapping on X. Define a non-increasing function $\theta$ from [0, 1) into (1/2, 1] by (1).

Suppose there exists $r\in [0,1)$ such that for each $x,y\in X$, the following condition is satisfied

$$\begin{aligned} \frac{1}{b}\theta (r)\mathrm{d}(x,Tx)\le \mathrm{d}(x,y)\Longrightarrow \mathrm{d}(Tx,Ty)\le \frac{r}{ b^{2}}\mathrm{d}(x,y), \end{aligned}$$

then there exists a unique fixed-point z of T. Moreover, $lim_{n\rightarrow \infty }T^{n}x=z$ for all $x\in X$.

Proof

It is enough set $S=T$ in the Theorem 3 then the desired result is obtained. $\square$

Remark 1

Note that for $b=1,$ Corollary 1 reduces to Theorem.

Corollary 2

Let (X, d) be a complete b-metric space, and $f,S,T:X\longrightarrow X$ be three self-maps and $\theta :[0,1)\longrightarrow (\frac{1}{2},1]$ be defined by (1).

Suppose there exists $r\in [0,1)$ such that for each $x,y\in X$, the following condition is satisfied

$$\begin{aligned}&\frac{1}{b}\theta (r)\min \{\mathrm{d}(x,fTx),d(x,fSx)\}\le \mathrm{d}(x,y)\Longrightarrow \\&\quad \max \left\{ \begin{array}{l} \mathrm{d}(fSx,fSy),\mathrm{d}(fTx,fTy), \\ \mathrm{d}(fSx,fTy),\mathrm{d}(fSy,fTx) \end{array} \right\} \le \frac{r}{b^{2}}\mathrm{d}(x,y). \end{aligned}$$

Also, if f is one to one, $fS=Sf$ and $fT=Tf$, then we have f, T and S have a unique common fixed point $z\in X.$

Proof

By Theorem 3, fT, fS have a unique common fixed point $z\in X$. That is $fSz=fTz=z$, since f is one to one it follows that $Sz=Tz$. From

$$\begin{aligned} 0=\frac{1}{b}\theta (r)\min \{\mathrm{d}(z,fTz),\mathrm{d}(z,fSz)\}\le \mathrm{d}(z,Tz), \end{aligned}$$

it follows that

$$\begin{aligned} \mathrm{d}(z,Tz)\le & {} \max \left\{ \begin{array}{l} \mathrm{d}(fSz,fSTz),\mathrm{d}(fTz,fT^{2}z), \\ \mathrm{d}(fSz,fT^{2}z),\mathrm{d}(fSTz,fTz) \end{array} \right\} \\= & {} \max \left\{ \begin{array}{l} \mathrm{d}(fSz,SfTz),\mathrm{d}(fTz,TfTz), \\ \mathrm{d}(fSz,TfTz),\mathrm{d}(SfTz,fTz) \end{array} \right\} \\= & {} \max \left\{ \begin{array}{l} \mathrm{d}(z,Sz),\mathrm{d}(z,Tz), \\ \mathrm{d}(z,Tz),\mathrm{d}(Sz,z) \end{array} \right\} \\\le & {} \frac{r}{b^{2}}\mathrm{d}(z,Tz), \end{aligned}$$

it follows that $Tz=Sz=z$, hence $fz=fTz=z.$ $\square$

Corollary 3

Let $(X,\mathrm{d})$ be a complete metric space, and $f,S,T:X\longrightarrow X$ be three self-maps and $\theta :[0,1)\longrightarrow (\frac{1}{2},1]$ be defined by (1).

Suppose there exists $r\in [0,1)$ such that for each $x,y\in X$, the following condition is satisfied

$$\begin{aligned}&\theta (r)\min \{\mathrm{d}(x,fTx),\mathrm{d}(x,fSx)\}\le \mathrm{d}(x,y)\Longrightarrow \\&\quad \max \left\{ \begin{array}{l} \mathrm{d}(fSx,fSy),\mathrm{d}(fTx,fTy), \\ \mathrm{d}(fSx,fTy),\mathrm{d}(fSy,fTx) \end{array} \right\} \le r\mathrm{d}(x,y). \end{aligned}$$

Also, if f is one to one, $fS=Sf$ and $fT=Tf$, then we have f, T and S have a unique common fixed point $z\in X$.

Proof

It is enough to set $b=1$ in the Corollary 2 then the desired result is obtained. $\square$

Now, in order to support the useability of our results, let us introduce the following examples.

Let $X=[0,\infty )$. Define $d:X\times X\rightarrow \mathbb {R^{+}}$ by

$$ {\text{d}}(x,y) = \left\{ {\begin{array}{*{20}l} {0,} \hfill & \quad {x = y} \hfill \\ {(x + y)^{2} ,} \hfill & \quad {x \ne y.} \hfill \\ \end{array} } \right. $$

for all $x,y\in X$. Then $(X,\mathrm{d})$ is a complete b-metric space for $b=2$. Define two maps $T,S:X\rightarrow X$ by

$$\begin{aligned} T(x)\;=\; & {} \ln (1+\frac{1}{4\sqrt{2}}x), \\ S(x)\;=\; & {} \ln (1+\frac{1}{8\sqrt{2}}x) \end{aligned}$$

for $x\in X$. Then for each $x,y\in X$ we have

$$\begin{aligned} \dfrac{1}{2}\theta (r)\min \left\{ \mathrm{d}(x,Tx),\mathrm{d}(x,Sx)\right\}= & {} \frac{1}{4} \min \left\{ \begin{array}{l} (x+\ln (1+\frac{1}{4\sqrt{2}}x))^{2}, \\ (x+\ln (1+\frac{1}{8\sqrt{2}}x))^{2} \end{array} \right\} \\= & {} \frac{1}{4}(x+\ln (1+\frac{1}{8\sqrt{2}}x))^{2} \\\le & {} \frac{1}{4}(x+\frac{1}{8\sqrt{2}}x)^{2}=\frac{1}{4}(1+\frac{1}{8\sqrt{ 2}})^{2}x^{2} \\\le & {} x^{2}\le (x+y)^{2}=\mathrm{d}(x,y). \end{aligned}$$

On the other hand, we have

$$\begin{aligned} \max \left\{ \begin{array}{l} \mathrm{d}(Sx,Sy),\mathrm{d}(Tx,Ty), \\ \mathrm{d}(Sx,Ty),\mathrm{d}(Sy,Tx) \end{array} \right\}= & {} \max \left\{ \begin{array}{l} (\ln (1+\frac{1}{8\sqrt{2}}x)+\ln (1+\frac{1}{8\sqrt{2}}y))^{2}, \\ (\ln (1+\frac{1}{4\sqrt{2}}x)+\ln (1+\frac{1}{4\sqrt{2}}y))^{2}, \\ (\ln (1+\frac{1}{8\sqrt{2}}x)+\ln (1+\frac{1}{4\sqrt{2}}y))^{2}, \\ (\ln (1+\frac{1}{4\sqrt{2}}x)+\ln (1+\frac{1}{8\sqrt{2}}y))^{2} \end{array} \right\} \\\le & {} \max \left\{ \begin{array}{l} (\frac{1}{8\sqrt{2}}x+\frac{1}{8\sqrt{2}}y)^{2}, \\ (\frac{1}{4\sqrt{2}}x+\frac{1}{4\sqrt{2}}y)^{2}, \\ (\frac{1}{8\sqrt{2}}x+\frac{1}{4\sqrt{2}}y)^{2}, \\ (\frac{1}{4\sqrt{2}}x+\frac{1}{8\sqrt{2}}y)^{2} \end{array} \right\} \\\le & {} (\frac{1}{4\sqrt{2}}x+\frac{1}{4\sqrt{2}}y)^{2}=[\frac{1}{4\sqrt{2}} (x+y)]^{2} \\\le & {} \frac{1}{4}.\frac{1}{2\sqrt{2}}(x+y)^{2}=\frac{r}{b^{2}}\mathrm{d}(x,y). \end{aligned}$$

Thus T and S satisfy all the hypotheses of Theorem 3 and hence T and S have a unique common fixed point. Indeed, $r=\dfrac{1}{2\sqrt{2 }}<\frac{\sqrt{5}-1}{2},$ $\theta (r)=\frac{1}{2}$ and 0 is the unique common fixed point of T and S.

Inspired by [8, Example 4] and [26, Example 1], we present the following example:

Example 3

Let $X=\{(0,0),(10,12),(12,10),(40,42),(42,40)\}\subset \mathbb {R}^{2}$. Define $d:X\times X\rightarrow \mathbb {R}^{+}$ by

$$\begin{aligned} d((x_{1},y_{1}),(x_{2},y_{2}))=(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}, \end{aligned}$$

for all $x=(x_{1},y_{1}),y=(x_{2},y_{2})\in X$. Then (X, d) is a complete b-metric space for $b=2$. Define two maps $T,S:X\rightarrow X$ by

$$\begin{aligned} \left\{ \begin{array}{l} T(0,0)=T(10,12)=T(12,10)=(0,0), \\ T(40,42)=(10,12), \\ T(42,40)=(12,10). \end{array} \right. \end{aligned}$$

$$\begin{aligned} \left\{ \begin{array}{l} S(0,0)=S(10,12)=S(12,10)=(0,0), \\ S(40,42)=(12,10), \\ S(42,40)=(10,12). \end{array} \right. \end{aligned}$$

Then for each $x,y\in X$, if

$$\begin{aligned} \dfrac{1}{2}\frac{\sqrt{2}}{\sqrt{2}+1}\min \{d(x,Tx),d(x,Sx)\}\le d(x,y), \end{aligned}$$

 this implies that

$$\begin{aligned} \max \left\{ \begin{array}{l} \mathrm{d}(Sx,Sy),\mathrm{d}(Tx,Ty), \\ \mathrm{d}(Sx,Ty),\mathrm{d}(Sy,Tx) \end{array} \right\} \le \dfrac{r}{4}\mathrm{d}(x,y). \end{aligned}$$

Because,

• i) $\frac{1}{2}\frac{\sqrt{2}}{\sqrt{2}+1}\min \left\{ \mathrm{d}((0,0),T(0,0)),\mathrm{d}((0,0),S(0,0))\right\} \le \mathrm{d}(0,0),y),\quad \forall y\in X$.

• ii) $\frac{1}{2}\frac{\sqrt{2}}{\sqrt{2}+1}\min \left\{ \mathrm{d}((10,12),T(10,12)),\mathrm{d}((10,12),S(10,12))\right\} \le \mathrm{d}(10,12),y),\quad \forall y=(0,0),(40,42),(42,40)$.

• iii) $\frac{1}{2}\frac{\sqrt{2}}{\sqrt{2}+1}\min \left\{ \mathrm{d}((12,10),T(12,10)),\mathrm{d}((12,10),S(12,10))\right\} \le \mathrm{d}(12,10),y),\quad \forall y=(0,0),(40,42),(42,40)$.

• iv) $\frac{1}{2}\frac{\sqrt{2}}{\sqrt{2}+1}\min \left\{ \mathrm{d}((40,42),T(40,42)),\mathrm{d}((40,42),S(40,42))\right\} \le \mathrm{d}(40,42),y),\quad \forall y=(0,0),(10,12),(12,10)$.

• v) $\frac{1}{2}\frac{\sqrt{2}}{\sqrt{2}+1}\min \left\{ \mathrm{d}((42,40),T(42,40)),\mathrm{d}((42,40),S(42,40))\right\} \le \mathrm{d}(42,40),y),\quad \forall y=(0,0),(10,12),(12,10)$.

On the other hand, in all of the cases we have

$$ \max \left\{ \begin{array}{l} \mathrm{d}(Sx,Sy),\mathrm{d}(Tx,Ty),\\ \mathrm{d}(Sx,Ty), \mathrm{d}(Sy,Tx) \end{array} \right\} \le \frac{1}{4\sqrt{2}}\mathrm{d}(x,y). $$

Thus T satisfy all the hypotheses of Theorem 3 and hence T has a unique fixed point. Indeed, $r=\dfrac{1}{\sqrt{2}},$ $\theta (r)=\frac{ \sqrt{2}}{\sqrt{2}+1}$ and (0, 0) is the unique common fixed point of T and S.

But,

$$\begin{aligned} \mathrm{d}(T(40,42),T(42,40))\le \frac{r}{4}\mathrm{d}((40,42),(42,40)), \end{aligned}$$

that is $2^{2}+2^{2}\le \frac{r}{4}(2^{2}+2^{2})$ this implies that $r\ge 4$. It is contradiction. This proves that Theorem 1 is not applicable to T.