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Fixed point theorems for weakly TChatterjea and weakly TKannan contractions in bmetric spaces
Journal of Inequalities and Applications volume 2014, Article number: 46 (2014)
Abstract
In this work, we obtain some fixed point results for generalized weakly TChatterjeacontractive and generalized weakly TKannancontractive mappings in the framework of complete bmetric spaces. Examples are provided in order to distinguish these results from the known ones.
MSC:47H10, 54H25.
1 Introduction and preliminaries
The theoretical framework of metric fixed point theory has been an active research field over the last nine decades. Of course, the Banach contraction principle [1] is the first important result on fixed points for contractivetype mappings. So far, there have been a lot of fixed point results dealing with mappings satisfying various types of contractive inequalities. In particular, the concepts of Kcontraction and Ccontraction were introduced by Kannan [2], respectively, Chatterjea [3] as follows.
Definition 1 Let $(X,d)$ be a metric space and $f:X\to X$.

1.
([2]) The mapping f is said to be a Kcontraction if there exists $\alpha \in (0,\frac{1}{2})$ such that for all $x,y\in X$ the following inequality holds:
$$d(fx,fy)\le \alpha (d(x,fx)+d(y,fy)).$$ 
2.
([3]) The mapping f is said to be a Ccontraction if there exists $\alpha \in (0,\frac{1}{2})$ such that for all $x,y\in X$ the following inequality holds:
$$d(fx,fy)\le \alpha (d(x,fy)+d(y,fx)).$$
In 1968, Kannan [2] proved that if $(X,d)$ is a complete metric space, then every Kcontraction on X has a unique fixed point. In 1972, Chatterjea [3] established a fixed point theorem for Ccontractions.
Definition 2 Let $(X,d)$ be a metric space, $f:X\to X$ and $\phi :{[0,\mathrm{\infty})}^{2}\to [0,\mathrm{\infty})$ be a continuous function such that $\phi (x,y)=0$ if and only if $x=y=0$.

1.
([4]) f is said to be weakly Ccontractive (or a weak Ccontraction) if for all $x,y\in X$,
$$d(fx,fy)\le \frac{1}{2}(d(x,fy)+d(y,fx))\phi (d(x,fy),d(y,fx)).$$ 
2.
([5]) f is said to be weakly Kcontractive (or a weak Kcontraction) if for all $x,y\in X$,
$$d(fx,fy)\le \frac{1}{2}(d(x,fx)+d(y,fy))\phi (d(x,fx),d(y,fy)).$$
In 2009, Choudhury [4] proved the following theorem.
Theorem 1 ([[4], Theorem 2.1])
Every weak Ccontraction on a complete metric space has a unique fixed point.
For more details of weakly Ccontractive mappings we refer to [6] and [7].
Definition 3 Let $(X,d)$ be a metric space and $T,f:X\to X$ be two mappings.

1.
([8]) $f:X\to X$ is said to be a TKannancontraction if there exists $\alpha \in (0,\frac{1}{2})$ such that for all $x,y\in X$ the following inequality holds:
$$d(Tfx,Tfy)\le \alpha (d(Tx,Tfx)+d(Ty,Tfy)).$$ 
2.
([5]) $f:X\to X$ is said to be a TChatterjeacontraction if there exists $\alpha \in (0,\frac{1}{2})$ such that for all $x,y\in X$ the following inequality holds:
$$d(Tfx,Tfy)\le \alpha (d(Tx,Tfy)+d(Ty,Tfx)).$$
TKannancontractions (in short, TKcontractions) and TChatterjeacontractions (in short, TCcontractions) are special cases of THardyRogers contractions [9]. Recently, existence and uniqueness of fixed points for these types of contractions in cone metric spaces have been investigated in [9] and [10].
Definition 4 ([11])
Let $(X,d)$ be a metric space. A mapping $T:X\to X$ is said to be sequentially convergent (respectively, subsequentially convergent) if, for a sequence $\{{x}_{n}\}$ in X for which $\{T{x}_{n}\}$ is convergent, $\{{x}_{n}\}$ is also convergent (respectively, $\{{x}_{n}\}$ has a convergent subsequence).
In [8], Moradi has extended Kannan’s theorem [2] as follows.
Theorem 2 (Extended Kannan’s theorem [8])
Let $(X,d)$ be a complete metric space and $T,f:X\to X$ be mappings such that T is continuous, onetoone and subsequentially convergent. If f is a TKcontraction then f has a unique fixed point. Moreover, if T is sequentially convergent then, for every ${x}_{0}\in X$, the sequence of iterates $\{{f}^{n}{x}_{0}\}$ converges to this fixed point.
The notion of an altering distance function was introduced by Khan et al. as follows.
Definition 5 ([12])
The function $\psi :[0,\mathrm{\infty})\to [0,\mathrm{\infty})$ is called an altering distance function, if the following properties are satisfied:

1.
ψ is continuous and strictly increasing.

2.
$\psi (0)=0$.
In the following definitions and theorems, ψ is an altering distance function and $\phi :{[0,\mathrm{\infty})}^{2}\to [0,\mathrm{\infty})$ is a continuous function such that $\phi (x,y)=0$ if and only if $x=y=0$.
Definition 6 ([5])
Let $(X,d)$ be a metric space and let $T,f:X\to X$ be two mappings.

1.
f is said to be a generalized weak TCcontraction if, for all $x,y\in X$,
$$\psi (d(Tfx,Tfy))\le \psi \left(\frac{d(Tx,Tfy)+d(Ty,Tfx)}{2}\right)\phi (d(Tx,Tfy),d(Ty,Tfx)).$$ 
2.
f is said to be a generalized weak TKcontraction if, for all $x,y\in X$,
$$\psi (d(Tfx,Tfy))\le \psi \left(\frac{d(Tx,Tfx)+d(Ty,Tfy)}{2}\right)\phi (d(Tx,Tfx),d(Ty,Tfy)).$$
Putting $\psi (t)=t$ in the above definition, we obtain the concepts of weak TCcontraction and weak TKcontraction.
The following are the main results of [5].
Theorem 3 [5]
Let $(X,d)$ be a complete metric space and let $T,f:X\to X$ be two mappings such that T is onetoone and continuous. Suppose that:

1.
f is a generalized weak TCcontraction, or

2.
f is a generalized weak TKcontraction.
Then we have the following.

(i)
For every ${x}_{0}\in X$ the sequence $\{T{f}^{n}{x}_{0}\}$ is convergent.

(ii)
If T is subsequentially convergent then f has a unique fixed point.

(iii)
If T is sequentially convergent then for each ${x}_{0}\in X$ the sequence $\{{f}^{n}{x}_{0}\}$ converges to the fixed point of f.
The aim of this article is to extend the stated results to the framework of bmetric spaces, introduced in 1993 by Czerwik [13]. These form a nontrivial generalization of metric spaces and several fixed point results for single and multivalued mappings in such spaces have been obtained since then (see, e.g., [14–17] and the references cited therein). We recall the following.
Definition 7 ([13])
Let X be a (nonempty) set and $s\ge 1$ be a given real number. A function $d:X\times X\to [0,\mathrm{\infty})$ is a bmetric if, for all $x,y,z\in X$, the following conditions are satisfied:
(${b}_{1}$) $d(x,y)=0$ iff $x=y$,
(${b}_{2}$) $d(x,y)=d(y,x)$,
(${b}_{3}$) $d(x,z)\le s[d(x,y)+d(y,z)]$.
The pair $(X,d)$ is called a bmetric space.
It should be noted that the class of bmetric spaces is effectively larger than that of metric spaces, since a bmetric is a metric if (and only if) $s=1$. We present an easy example to show that in general a bmetric need not be a metric.
Example 1 Let $(X,\rho )$ be a metric space, and $d(x,y)={(\rho (x,y))}^{p}$, where $p\ge 1$ is a real number. Then d is a bmetric with $s={2}^{p1}$.
However, $(X,d)$ is not necessarily a metric space. For example, if $X=\mathbb{R}$ is the set of real numbers and $\rho (x,y)=xy$ is the usual Euclidean metric, then $d(x,y)={(xy)}^{2}$ is a bmetric on ℝ with $s=2$, but it is not a metric on ℝ.
Recently, Hussain et al. [15] have presented an example of a bmetric which is not continuous (see [[15], Example 2]). Thus, while working in bmetric spaces, the following lemma is useful.
Lemma 1 ([14])
Let $(X,d)$ be a bmetric space with $s\ge 1$, and suppose that the sequences $\{{x}_{n}\}$ and $\{{y}_{n}\}$ are bconvergent to x, y, respectively. Then we have
In particular, if $x=y$, then we have ${lim}_{n\to \mathrm{\infty}}d({x}_{n},{y}_{n})=0$. Moreover, for each $z\in X$, we have,
2 Fixed points of weakly TChatterjea contractions
From now on, we assume:
and
Our first result is the following.
Theorem 4 Let $(X,d)$ be a complete bmetric space with parameter $s\ge 1$, $T,f:X\to X$ be such that, for some $\psi \in \mathrm{\Psi}$, $\phi \in \mathrm{\Phi}$ and all $x,y\in X$,
and let T be onetoone and continuous. Then we have the following.

(1)
For every ${x}_{0}\in X$ the sequence $\{T{f}^{n}{x}_{0}\}$ is convergent.

(2)
If T is subsequentially convergent, then f has a unique fixed point.

(3)
If T is sequentially convergent, then for each ${x}_{0}\in X$ the sequence $\{{f}^{n}{x}_{0}\}$ converges to the fixed point of f.
Proof Let ${x}_{0}\in X$ be arbitrary. Consider the sequence ${\{{x}_{n}\}}_{n=0}^{\mathrm{\infty}}$ given by ${x}_{n+1}=f{x}_{n}={f}^{n+1}{x}_{0}$, $n=0,1,2,\dots $ . We will complete the proof in three steps.
Step I. We will prove that ${lim}_{n\to \mathrm{\infty}}d(T{x}_{n},T{x}_{n+1})=0$.
Using condition (2.1), we obtain
Therefore, by the triangular inequality and since φ is nonnegative and ψ is an increasing function,
Again, since ψ is increasing, we have
wherefrom
Thus, $\{d(T{x}_{n+1},T{x}_{n})\}$ is a decreasing sequence of nonnegative real numbers and hence it is convergent.
Assume that ${lim}_{n\to \mathrm{\infty}}d(T{x}_{n+1},T{x}_{n})=r\ge 0$. From the above argument we have
Passing to the limit when $n\to \mathrm{\infty}$, we obtain
We have proved in (2.2) that
Now, letting $n\to \mathrm{\infty}$ and using the continuity of ψ and the properties of φ we obtain
and consequently, $\phi (0,s(s+1)r)=0$. This yields
by our assumptions about φ.
Step II. $\{T{x}_{n}\}$ is a bCauchy sequence.
Suppose that $\{T{x}_{n}\}$ is not a bCauchy sequence. Then there exists $\epsilon >0$ for which we can find subsequences $\{T{x}_{m(k)}\}$ and $\{T{x}_{n(k)}\}$ of $\{T{x}_{n}\}$ such that $n(k)$ is the smallest index for which $n(k)>m(k)>k$ and
This means that
From (2.4), (2.5) and the triangular inequality,
Letting $k\to \mathrm{\infty}$, and taking into account (2.3), we can conclude that
Further, from
and (2.5), and using (2.3), we get
Moreover, from
and
and using (2.3) and (2.6), we get
Similarly, we can show that
and
Using (2.1) and (2.7)(2.10) we have
since $\frac{{s}^{2}+1}{s+1}\le s$. Hence, we have
By our assumption about φ, we have
which contradicts (2.9) and (2.10).
Since $(X,d)$ is bcomplete and $\{T{x}_{n}\}=\{T{f}^{n}{x}_{0}\}$ is a bCauchy sequence, there exists $v\in X$ such that
Step III. f has a unique fixed point, assuming that T is subsequentially convergent.
As T is subsequentially convergent, $\{{f}^{n}{x}_{0}\}$ has a bconvergent subsequence. Hence, there exist $u\in X$ and a subsequence $\{{n}_{i}\}$ such that
Since T is continuous, by (2.12) we obtain
and by (2.11) and (2.13) we conclude that $Tu=v$.
From Lemma 1 and (2.1) we have
since ψ is increasing. By the properties of $\phi \in \mathrm{\Phi}$, it follows that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty}}d(T{x}_{n},Tfu)=0$. By the triangular inequality we have
Letting $n\to \mathrm{\infty}$ we can conclude that $d(Tfu,Tu)=0$. Hence, $Tfu=Tu$. As T is onetoone, $fu=u$. Consequently, f has a fixed point.
If we assume that w is another fixed point of f, because of (2.1), we have
since $\frac{2}{s+1}\le s$ and ψ is increasing. Hence, $d(Tu,Tw)=0$. Since T is onetoone, it follows that $u=w$. Consequently, f has a unique fixed point.
Finally, if T is sequentially convergent, replacing $\{n\}$ with $\{{n}_{i}\}$ we conclude that ${lim}_{n\to \mathrm{\infty}}{f}^{n}{x}_{0}=u$. □
Taking $\psi (t)=t$ and $\phi (t,u)=(\frac{1}{s+1}\alpha )(t+u)$, where $\alpha \in [0,\frac{1}{s+1})$ in Theorem 4, the extended Chatterjea’s theorem in the setting of bmetric spaces is obtained.
Corollary 1 Let $(X,d)$ be a complete bmetric space and $T,f:X\to X$ be mappings such that T is continuous, onetoone and subsequentially convergent. If $\alpha \in [0,\frac{1}{s+1})$ and
for all $x,y\in X$, then f has a unique fixed point. Moreover, if T is sequentially convergent, then for every ${x}_{0}\in X$ the sequence of iterates ${f}^{n}{x}_{0}$ converges to this fixed point.
Remark 1 In the case when $Tx=x$, this corollary reduces to [[18], Corollary 3.8.3^{∘}] (the case $g=f$), which is Chatterjea’s theorem [3] in the framework of bmetric spaces.
By taking $Tx=x$ and $\psi (t)=t$ in Theorem 4, we derive an extension of Choudhury’s theorem (Theorem 1) to the setup of bmetric spaces.
If $s=1$, Theorem 4 reduces to Theorem 3 (case (1)).
We demonstrate the use of the obtained results by the following.
Example 2 (Inspired by [8])
Let $X=\{0\}\cup \{1/n\mid n\in \mathbb{N}\}$, and let $d(x,y)={(xy)}^{2}$ for $x,y\in X$. Then d is a bmetric with the parameter $s=2$ and $(X,d)$ is a complete bmetric space. Consider the mappings $f,T:X\to X$ given by
We will show that the mappings f, T satisfy the conditions of Corollary 1 with $\alpha =\frac{2}{9}<\frac{1}{3}=\frac{1}{s+1}$. Indeed, for $m,n\in \mathbb{N}$, $m>n$, we have
It is easy to prove that, for $n\in \mathbb{N}$,
It follows that
Now, $m>n$ implies that $m\ge n+1$ and $n+2\le m+1$. It follows that $1/{(n+2)}^{n+2}\ge 1/{(m+1)}^{m+1}$, and hence
If one of the points is equal to 0, the proof is even simpler.
By Corollary 1, it follows that f has a unique fixed point (which is $u=0$).
3 Fixed points of weakly TKannan contractions
Our second main result is the following.
Theorem 5 Let $(X,d)$ be a complete bmetric space with the parameter $s\ge 1$, $T,f:X\to X$ be such that for some $\psi \in \mathrm{\Psi}$, $\phi \in \mathrm{\Phi}$ and all $x,y\in X$,
and let T be onetoone and continuous. Then:

(1)
For every ${x}_{0}\in X$ the sequence $\{T{f}^{n}{x}_{0}\}$ is convergent.

(2)
If T is subsequentially convergent, then f has a unique fixed point.

(3)
If T is sequentially convergent then, for each ${x}_{0}\in X$, the sequence $\{{f}^{n}{x}_{0}\}$ converges to the fixed point of f.
Proof Let ${x}_{0}\in X$ be arbitrary. Consider the sequence ${\{{x}_{n}\}}_{n=0}^{\mathrm{\infty}}$ given by ${x}_{n+1}=f{x}_{n}={f}^{n+1}{x}_{0}$, $n=0,1,2,\dots $ . At first, we will prove that
Using condition (3.1), we obtain
Since φ is nonnegative and ψ is increasing, it follows that
that is,
Thus, $\{d(T{x}_{n+1},T{x}_{n})\}$ is a decreasing sequence of nonnegative real numbers and hence it is convergent.
Assume that ${lim}_{n\to \mathrm{\infty}}d(T{x}_{n+1},T{x}_{n})=r$. If in (3.2) $n\to \mathrm{\infty}$, using the properties of ψ and φ we obtain
which is possible only if
Now, we will show that $\{T{x}_{n}\}$ is a bCauchy sequence.
Suppose that this is not true. Then there exists $\epsilon >0$ for which we can find subsequences $\{T{x}_{m(k)}\}$ and $\{T{x}_{n(k)}\}$ of $\{T{x}_{n}\}$ such that $n(k)$ is the smallest index for which $n(k)>m(k)>k$ and $d(T{x}_{m(k)},T{x}_{n(k)})\ge \epsilon $. This means that
Again, as in Step II of Theorem 4 one can prove that
Using (3.1) we have
Passing to the upper limit as $k\to \mathrm{\infty}$ in the above inequality and taking into account (3.3), we have
and so $\psi (\epsilon )=0$. By our assumptions about ψ, we have $\epsilon =0$, which is a contradiction.
Since $(X,d)$ is bcomplete and $\{T{x}_{n}\}=\{T{f}^{n}{x}_{0}\}$ is a bCauchy sequence, there exists $v\in X$ such that
Now, if T is subsequentially convergent, then $\{{f}^{n}{x}_{0}\}$ has a convergent subsequence. Hence, there exist a point $u\in X$ and a sequence $\{{n}_{i}\}$ such that
Since T is continuous, by (3.5) we obtain
and by (3.4) and (3.6) we conclude that $Tu=v$.
From Lemma 1 and (3.1) we have
By the properties of $\phi \in \mathrm{\Phi}$, it follows that
Since T is onetoone, we obtain $fu=u$. Consequently, f has a fixed point.
Uniqueness of the fixed point can be proved in the same manner as in Theorem 4.
Finally, if T is sequentially convergent, replacing $\{n\}$ with $\{{n}_{i}\}$ we conclude that ${lim}_{n\to \mathrm{\infty}}{f}^{n}{x}_{0}=u$. □
Taking $\psi (t)=t$ and $\phi (t,u)=(\frac{1}{s+1}\alpha )(t+u)$, where $\alpha \in [0,\frac{1}{s+1})$ in Theorem 5, the extended Kannan’s theorem in the setting of bmetric spaces is obtained.
Corollary 2 Let $(X,d)$ be a complete bmetric space with the parameter $s\ge 1$, $T,f:X\to X$ be such that for some $\alpha <\frac{1}{s+1}$ and all $x,y\in X$,
and let T be onetoone and continuous. Then we have the following.

(1)
For every ${x}_{0}\in X$ the sequence $\{T{f}^{n}{x}_{0}\}$ is convergent.

(2)
If T is subsequentially convergent then f has a unique fixed point.

(3)
If T is sequentially convergent then, for each ${x}_{0}\in X$, the sequence $\{{f}^{n}{x}_{0}\}$ converges to the fixed point of f.
Remark 2 In the case when $Tx=x$, this corollary reduces to [[18], Corollary 3.8.2^{∘}] (the case $g=f$). If $s=1$, Corollary 2 reduces to Theorem 2 (i.e., [[8], Theorem 2.1]). Of course, if both of these conditions are fulfilled, we get just the classical Kannan’s theorem [2].
The following example distinguishes our results from the previously known ones.
Example 3 Let $X=\{a,b,c\}$ and $d:X\times X\to \mathbb{R}$ be defined by $d(x,x)=0$ for $x\in X$, $d(a,b)=d(b,c)=1$, $d(a,c)=\frac{9}{4}$, $d(x,y)=d(y,x)$ for $x,y\in X$. It is easy to check that $(X,d)$ is a bmetric space (with $s=\frac{9}{8}>1$) which is not a metric space. Consider the mapping $f:X\to X$ given by
We first note that the bmetric version of classical weak Kannan’s theorem is not satisfied in this example. Indeed, for $x=b$, $y=c$, we have $d(fx,fy)=d(a,b)=1$ and $d(x,fx)+d(y,fy)=d(b,a)+d(c,b)=2$, hence the inequality
cannot hold, whatever $\psi \in \mathrm{\Psi}$ and $\phi \in \mathrm{\Phi}$ are chosen.
Take now $T:X\to X$ defined by
Obviously, all the properties of T given in Corollary 2 are fulfilled. We will check that the contractive condition (3.7) holds true if α is chosen such that
Only the following cases are nontrivial:
1^{∘} $x=a$, $y=c$. Then (3.7) reduces to
2^{∘} $x=b$, $y=c$. Then (3.7) reduces to
All the conditions of Corollary 2 are satisfied and f has a unique fixed point ($u=a$).
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Acknowledgements
The authors are grateful to the referees for valuable remarks that helped them to improve the exposition of the paper. The fourth author is thankful to the Ministry of Education, Science and Technological Development of Serbia.
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Mustafa, Z., Roshan, J.R., Parvaneh, V. et al. Fixed point theorems for weakly TChatterjea and weakly TKannan contractions in bmetric spaces. J Inequal Appl 2014, 46 (2014). https://doi.org/10.1186/1029242X201446
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Keywords
 fixed point
 complete metric space
 bmetric space
 weak Ccontraction
 altering distance function