# Contraction conditions using comparison functions on *b*-metric spaces

- Wasfi Shatanawi
^{1}, - Ariana Pitea
^{2}Email author and - Rade Lazović
^{3}

**2014**:135

https://doi.org/10.1186/1687-1812-2014-135

© Shatanawi et al.; licensee Springer. 2014

**Received: **7 January 2014

**Accepted: **23 May 2014

**Published: **2 June 2014

## Abstract

In this paper, we consider the setting of *b*-metric spaces to establish results regarding the common fixed points of two mappings, using a contraction condition defined by means of a comparison function. An example is presented to support our results comparing with existing ones.

**MSC:**49H09, 47H10.

## Keywords

*b*-metric spacecommon fixed pointcontraction conditioncomparison function

## 1 Introduction

The contraction principle of Banach [1], proved in 1922, was followed by diverse works about fixed points theory regarding different classes of contractive conditions on some spaces such as: quasi-metric spaces [2, 3], cone metric spaces [4, 5], partially ordered metric spaces [6–8], *G*-metric spaces [9], partial metric spaces [10–13], Menger spaces [14], metric-type spaces [15], and fuzzy metric spaces [16–18]. Also, there have been developed studies on approximate fixed point or on qualitative aspects of numerical procedures for approximating fixed points see, for example [19, 20].

The concept of *b*-metric spaces was introduced by Bakhtin [21] in 1989, who used it to prove a generalization of the Banach principle in spaces endowed with such kind of metrics. Since then, this notion has been used by many authors to obtain various fixed point theorems. Aydi *et al.* in [22] proved common fixed point results for single-valued and multi-valued mappings satisfying a weak *ϕ*-contraction in *b*-metric spaces. Roshan *et al.* in [23] used the notion of almost generalized contractive mappings in ordered complete *b*-metric spaces and established some fixed and common fixed point results. Starting from the results of Berinde [24], Păcurar [25] proved the existence and uniqueness of fixed points of *ϕ*-contractions on *b*-metric spaces. Hussain and Shah in [26] introduced the notion of a cone *b*-metric space, generalizing both notions of *b*-metric spaces and cone metric spaces. In this paper they also considered topological properties of cone *b*-metric spaces and results on KKM mappings in the setting of cone *b*-metric spaces. Fixed point theorems of contractive mappings in cone *b*-metric spaces without the assumption of the normality of a corresponding cone are proved by Huang and Xu in [27]. The setting of partially ordered *b*-metric spaces was used by Hussain *et al.* in [28] to study tripled coincidence points of mappings which satisfy nonlinear contractive conditions, extending those results of Berinde and Borcut [29] for metric spaces to *b*-metric spaces. Using the concept of a *g*-monotone mapping, Shah and Hussain in [30] proved common fixed point theorems involving *g*-non-decreasing mappings in *b*-metric spaces, generalizing several results of Agarwal *et al.* [31] and Ćirić *et al.* [32]. Some results of Suzuki [33] are extended to the case of metric-type spaces and cone metric-type spaces.

The aim of this paper is to consider and establish results on the setting of *b*-metric spaces, regarding common fixed points of two mappings, using a contraction condition defined by means of a comparison function. An example is given to support our results.

## 2 Preliminaries

**Definition 1**Let

*X*be a nonempty set and $d:X\times X\to [0,+\mathrm{\infty})$. A function

*d*is called a

*b*-metric with constant (base) $s\ge 1$ if:

- (1)
$d(x,y)=0$ iff $x=y$.

- (2)
$d(x,y)=d(y,x)$ for all $x,y\in X$.

- (3)
$d(x,y)\le s(d(x,z)+d(z,y))$ for all $x,y,z\in X$.

The pair $(X,d)$ is called a *b*-metric space.

It is obvious that a *b*-metric space with base $s=1$ is a metric space. There are examples of *b*-metric spaces which are not metric spaces (see, *e.g.*, Singh and Prasad [34]).

The notions of a Cauchy sequence and a convergent sequence in *b*-metric spaces are defined by Boriceanu [35].

**Definition 2**Let $\{{x}_{n}\}$ be a sequence in a

*b*-metric space $(X,d)$.

- (1)
A sequence $\{{x}_{n}\}$ is called convergent if and only if there is $x\in X$ such that $d({x}_{n},x)\to 0$ when $n\to +\mathrm{\infty}$.

- (2)
$\{{x}_{n}\}$ is a Cauchy sequence if and only if $d({x}_{n},{x}_{m})\to 0$, when $n,m\to +\mathrm{\infty}$.

As usual, a *b*-metric space is said to be complete if and only if each Cauchy sequence in this space is convergent.

Regarding the properties of a *b*-metric space, we recall that if the limit of a convergent sequence exists, then it is unique. Also, each convergent sequence is a Cauchy sequence. But note that a *b*-metric, in the general case, is not continuous (see Roshan *et al.* [23]).

The continuity of a mapping with respect to a *b*-metric is defined as follows.

**Definition 3** Let $(X,d)$ and $({X}^{\prime},{d}^{\prime})$ be two *b*-metric spaces with constant *s* and ${s}^{\prime}$, respectively. A mapping $T:X\to {X}^{\prime}$ is called continuous if for each sequence $\{{x}_{n}\}$ in *X*, which converges to $x\in X$ with respect to *d*, then $T{x}_{n}$ converges to *Tx* with respect to ${d}^{\prime}$.

**Definition 4**Let $s\ge 1$ be a constant. A mapping $\phi :[0,+\mathrm{\infty})\to [0,+\mathrm{\infty})$ is called comparison function with base $s\ge 1$, if the following two axioms are fulfilled:

- (a)
*φ*is non-decreasing, - (b)
${lim}_{n\to +\mathrm{\infty}}{\phi}^{n}(t)=0$ for all $t>0$.

Clearly, if *φ* is a comparison function, then $\phi (t)<t$ for each $t>0$.

For different properties and applications of comparison functions on partial metric spaces, we refer the reader to [36].

## 3 Main results

Now we are ready to prove our main results.

**Theorem 1**

*Let*$(X,d)$

*be a complete*

*b*-

*metric space with a constant*

*s*

*and*$T,S:X\to X$

*two mappings on*

*X*.

*Suppose that there is a constant*$L<\frac{1}{1+s}$

*and a comparison function*

*φ*

*such that the inequality*

*holds for each* $x,y\in X$. *Suppose that one of the mappings* *T* *or* *S* *is continuous*. *Then* *T* *and* *S* *have a unique common fixed point*.

*Proof*Let ${x}_{0}\in X$ be arbitrary. We define a sequence $\{{x}_{n}\}$ as follows:

*φ*is non-decreasing,

a contradiction. Therefore, $d({x}_{2k+1},{x}_{2k+2})=0$. Hence ${x}_{2k+1}={x}_{2k+2}$. Thus we have ${x}_{2k}={x}_{2k+1}={x}_{2k+2}$. By (3.2), it means ${x}_{2k}=T{x}_{2k}=S{x}_{2k}$, that is, ${x}_{2k}$ is a common fixed point of *T* and *S*.

If $n=2k+1$, then using the same arguments as in the case ${x}_{2k}={x}_{2k+1}$, it can be shown that ${x}_{2k+1}$ is a common fixed point of *T* and *S*.

From now on, we suppose that ${x}_{n}\ne {x}_{n+1}$ for all $n\in \mathbb{N}$.

There are two cases which we have to consider.

Case I. $n=2k$, $k\in \mathbb{N}$.

*φ*in Definition 4 we get

Thus we proved that (3.3) holds for $n=2k$.

Case II. $n=2k+1$, $k\in \mathbb{N}$.

From (3.4) and (3.5) we conclude that the inequality (3.3) holds for all $n\in \mathbb{N}$.

for all $n\ge {n}_{0}$.

*m*, we shall prove that

Thus (3.9) holds for $m=n+1$.

Assume now that (3.9) holds for some $m\ge n+1$. We have to prove that (3.9) holds for $m+1$.

We have to consider four cases.

Case I. *n* is odd, $m+1$ is even.

Thus we proved that in this case (3.9) holds for $m+1$. Therefore, by induction, we conclude that in Case I the inequality (3.9) holds for all $m>n$.

Case II. *n* is even, $m+1$ is odd. The proof of (3.9) in this case is similar to one given in Case I.

Case III. *n* is even, $m+1$ is even.

Thus we proved that (3.9) holds for $m+1$. Therefore, by induction, we conclude that in Case III the inequality (3.9) holds for all $m>n$.

Case IV. *n* is odd, $m+1$ is odd. The proof of (3.9) in this case is similar to one given in Case III.

Therefore, we proved that in all of four cases the inequality (3.9) holds.

From (3.9) it follows that $\{{x}_{n}\}$ is a Cauchy sequence. Since $(X,d)$ is a complete *b*-metric space, then $\{{x}_{n}\}$ converges to some $u\in X$ as $n\to +\mathrm{\infty}$.

*T*or

*S*is continuous, then $Tu=Su=u$. Without loss of generality, we can suppose that

*S*is continuous. Clearly, as ${x}_{n}\to u$, then by (3.2) we have $S{x}_{2n+1}={x}_{2n+2}\to u$ as $n\to +\mathrm{\infty}$. Since ${x}_{2n+1}\to u$ and

*S*is continuous, then $S{x}_{2n+1}\to Su$. Thus, by the uniqueness of the limit in a

*b*-metric space, we have $Su=u$. Now, from the contraction condition (3.1),

a contradiction. Therefore, $d(u,Tu)=0$. Hence $Tu=u$. Thus we proved that *u* is a common fixed point of *T* and *S*.

*u*and

*v*are different common fixed points of

*T*and

*S*, that is, $d(u,v)>0$. Then

Since $2L<1\le s$, then we get $sd(u,v)\le \phi (sd(u,v))<sd(u,v)$, a contradiction. Thus we proved that *S* and *T* have a unique common fixed point in *X*. □

If $S=T$ in Theorem 1, then we have the following result.

**Corollary 1**

*Let*$(X,d)$

*be a complete*

*b*-

*metric space with a constant*

*s*

*and*$T:X\to X$

*two mappings on*

*X*.

*Suppose that there is a constant*$L<\frac{1}{2}$

*and a comparison function*

*φ*

*such that the inequality*

*holds for each* $x,y\in X$. *Suppose that a mapping* *T* *is continuous*. *Then* *T* *has a unique fixed point*.

Omitting the continuity assumption of mapping *T* or *S* in Theorem 1, modifying the contraction condition (3.1) and imposing on a comparison function *φ* a corresponding condition, then we can prove the following theorem.

**Theorem 2**

*Let*$(X,d)$

*be a complete*

*b*-

*metric space with a constant*

*s*

*and*$T,S:X\to X$

*two mappings on*

*X*.

*Suppose that there is a constant*$L<\frac{1}{1+s}$

*and a comparison function*

*φ*

*such that the inequality*

*holds for all*$x,y\in X$.

*If in addition a comparison function*

*φ*

*satisfies the following condition*:

*then* *T* *and* *S* *have a unique common fixed point*.

*Proof*Since the contraction condition (3.13) implies the contraction condition (3.1) in Theorem 1, then from the proof of Theorem 1 it follows that a sequence $\{{x}_{n}\}$, defined as in (3.3), converges to some $u\in X$, that is,

*φ*we obtain

*φ*is non-decreasing and $L<1$, from (3.16) we get

*φ*, we have

*b*-metric space is unique, it follows that $Su=u$. Now, by (3.13),

If we suppose that $d(u,Tu)>0$, then we have $sd(Tu,u)\le $ $\phi (sd(u,Tu))<sd(u,Tu)$, a contradiction. Therefore, $d(Tu,u)=0$, that is, $Tu=u$. Thus we proved that $Tu=Su=u$. □

If $S=T$ in Theorem 2, then we get the following result.

**Corollary 2**

*Let*$(X,d)$

*be a complete*

*b*-

*metric space with a constant*

*s*

*and*$T:X\to X$

*a mapping on*

*X*.

*Suppose that there is a constant*$L<\frac{1}{1+s}$

*and a comparison function*

*φ*

*such that the inequality*

*holds for all* $x,y\in X$. *If in addition a comparison function* *φ* *satisfies the inequality* (3.14), *then* *T* *has a unique fixed point*.

Now we give an example to support our results.

**Example 1**Let $X=[0,1]$ endowed with the

*b*-metric

with constant $s=2$. Consider mappings $T,S:X\to X$, $Tx=\frac{1}{4}x$, $Sx=\frac{1}{8}x$, and the comparison function $\phi :[0,+\mathrm{\infty})\to [0,+\mathrm{\infty})$, $\phi (t)=\frac{t}{t+1}$. Clearly, $(X,d)$ is a complete metric space, and *S* is continuous with respect to *d*, so we have to verify the contraction condition (3.1). There are three cases to be considered.

Case I. $y=2x$. Hence $Tx=Sy$, $d(Tx,Sy)=0$, and, therefore, the inequality (3.1) holds.

Thus in this case the contraction condition (3.1) holds.

Therefore, we showed that the contraction condition (3.1) is satisfied in all cases. Thus we can apply our Theorem 1, and *T* and *S* have a unique common fixed point $u=0$.

## Declarations

### Acknowledgements

Rade Lazović was supported by Grant No. 174025 of the Ministry of Science, Technology and Development, Republic of Serbia.

## Authors’ Affiliations

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