Fixed point theorems for weakly T-Chatterjea and weakly T-Kannan contractions in b-metric spaces

Abstract

In this work, we obtain some fixed point results for generalized weakly T-Chatterjea-contractive and generalized weakly T-Kannan-contractive mappings in the framework of complete b-metric 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 contractive-type 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 K-contraction and C-contraction 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 K-contraction 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 C-contraction 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 K-contraction on X has a unique fixed point. In 1972, Chatterjea [3] established a fixed point theorem for C-contractions.

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 C-contractive (or a weak C-contraction) 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 K-contractive (or a weak K-contraction) 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 C-contraction on a complete metric space has a unique fixed point.

For more details of weakly C-contractive 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 T-Kannan-contraction 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 T-Chatterjea-contraction if there exists $\alpha \in (0,\frac{1}{2})$ such that for all $x,y\in X$ the following inequality holds:

  $$d(Tfx,Tfy)\le \alpha (d(Tx,Tfy)+d(Ty,Tfx)).$$

T-Kannan-contractions (in short, T-K-contractions) and T-Chatterjea-contractions (in short, T-C-contractions) are special cases of T-Hardy-Rogers 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, one-to-one and subsequentially convergent. If f is a T-K-contraction 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 T-C-contraction 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 T-K-contraction 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 T-C-contraction and weak T-K-contraction.

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 one-to-one and continuous. Suppose that:

• 1. f is a generalized weak T-C-contraction, or

• 2. f is a generalized weak T-K-contraction.

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 b-metric 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 b-metric 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 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 if (and only if) $s=1$. We present an easy example to show that in general a b-metric 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 b-metric with $s=2^{p-1}$.

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

Recently, Hussain et al. [15] have presented an example of a b-metric which is not continuous (see [[15], Example 2]). Thus, while working in b-metric spaces, the following lemma is useful.

Lemma 1 ([14])

Let $(X,d)$ be a b-metric space with $s\ge 1$, and suppose that the sequences $\{x_n\}$ and $\{y_n\}$ are b-convergent to x, y, respectively. Then we have

$$\frac{1}{s^2}d(x,y)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(x_n,y_n)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,y_n)\le s^{2}d(x,y).$$

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,

$$\frac{1}{s}d(x,z)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(x_n,z)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,z)\le sd(x,z).$$

2 Fixed points of weakly T-Chatterjea contractions

From now on, we assume:

$$\mathrm{\Psi }=\{\psi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })\mid \psi \text{ is an altering distance function}\}$$

and

$$\begin{array}{rcl}\mathrm{\Phi }& =& \{\phi :{[0,\mathrm{\infty })}^2\to [0,\mathrm{\infty })|\phi (x,y)=0⟺x=y=0\text{ and}\\ \phi (\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{a}_n,\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{b}_n)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\phi (a_n,b_n)\}.\end{array}$$

Our first result is the following.

Theorem 4 Let $(X,d)$ be a complete b-metric 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 $x,y\in X$,

$$\psi (sd(Tfx,Tfy))\le \psi \left(\frac{d(Tx,Tfy)+d(Ty,Tfx)}{s+1}\right)-\phi (d(Tx,Tfy),d(Ty,Tfx)),$$

(2.1)

and let T be one-to-one 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

$$\begin{array}{rcl}\psi (sd(T{x}_{n+1},T{x}_n))& =& \psi (sd(Tf{x}_n,Tf{x}_{n-1}))\\ \le & \psi \left(\frac{d(T{x}_n,Tf{x}_{n-1})+d(T{x}_{n-1},Tf{x}_n)}{s+1}\right)\\ -\phi (d(T{x}_n,Tf{x}_{n-1}),d(T{x}_{n-1},Tf{x}_n))\\ =& \psi \left(\frac{d(T{x}_n,T{x}_n)+d(T{x}_{n-1},T{x}_{n+1})}{s+1}\right)\\ -\phi (d(T{x}_n,T{x}_n),d(T{x}_{n-1},T{x}_{n+1})).\end{array}$$

(2.2)

Therefore, by the triangular inequality and since φ is nonnegative and ψ is an increasing function,

$$\begin{array}{rl}\psi (sd(T{x}_{n+1},T{x}_n))& \le \psi \left(\frac{d(T{x}_{n-1},T{x}_{n+1})}{s+1}\right)\\ \le \psi \left(\frac{s}{s+1}(d(T{x}_{n-1},T{x}_n)+d(T{x}_n,T{x}_{n+1}))\right).\end{array}$$

Again, since ψ is increasing, we have

$$d(T{x}_{n+1},T{x}_n)\le \frac{1}{s+1}(d(T{x}_{n-1},T{x}_n)+d(T{x}_n,T{x}_{n+1})),$$

wherefrom

$$d(T{x}_{n+1},T{x}_n)\le \frac{1}{s}d(T{x}_n,T{x}_{n-1})\le d(T{x}_n,T{x}_{n-1}).$$

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

$$\begin{array}{rl}sd(T{x}_{n+1},T{x}_n)& \le \frac{1}{s+1}d(T{x}_{n-1},T{x}_{n+1})\\ \le \frac{s}{s+1}(d(T{x}_{n-1},T{x}_n)+d(T{x}_n,T{x}_{n+1}))\\ \le \frac{s}{2}(d(T{x}_{n-1},T{x}_n)+d(T{x}_n,T{x}_{n+1})).\end{array}$$

Passing to the limit when $n\to \mathrm{\infty }$, we obtain

$$\underset{n\to \mathrm{\infty }}{lim}d(T{x}_{n-1},T{x}_{n+1})=s(s+1)r.$$

We have proved in (2.2) that

$$\psi (sd(T{x}_{n+1},T{x}_n))\le \psi \left(\frac{0+d(T{x}_{n-1},T{x}_{n+1})}{s+1}\right)-\phi (0,d(T{x}_{n-1},T{x}_{n+1})).$$

Now, letting $n\to \mathrm{\infty }$ and using the continuity of ψ and the properties of φ we obtain

$$\psi (sr)\le \psi (sr)-\phi (0,s(s+1)r),$$

and consequently, $\phi (0,s(s+1)r)=0$. This yields

$$r=\underset{n\to \mathrm{\infty }}{lim}d(T{x}_n,T{x}_{n+1})=0,$$

(2.3)

by our assumptions about φ.

Step II. $\{T{x}_n\}$ is a b-Cauchy sequence.

Suppose that $\{T{x}_n\}$ is not a b-Cauchy 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

$$d(T{x}_{m(k)},T{x}_{n(k)})\ge \epsilon .$$

(2.4)

This means that

$$d(T{x}_{m(k)},T{x}_{n(k)-1})<\epsilon .$$

(2.5)

From (2.4), (2.5) and the triangular inequality,

$$\begin{array}{rl}\epsilon & \le d(T{x}_{m(k)},T{x}_{n(k)})\le s[d(T{x}_{m(k)},T{x}_{n(k)-1})+d(T{x}_{n(k)-1},T{x}_{n(k)})]\\ <s\epsilon +sd(T{x}_{n(k)-1},T{x}_{n(k)}).\end{array}$$

Letting $k\to \mathrm{\infty }$, and taking into account (2.3), we can conclude that

$$\epsilon \le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(T{x}_{m(k)},T{x}_{n(k)})\le s\epsilon .$$

(2.6)

Further, from

$$d(T{x}_{m(k)},T{x}_{n(k)})\le s[d(T{x}_{m(k)},T{x}_{n(k)-1})+d(T{x}_{n(k)-1},T{x}_{n(k)})]$$

and (2.5), and using (2.3), we get

$$\frac{\epsilon }{s}\le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(T{x}_{n(k)-1},T{x}_{m(k)})\le \epsilon .$$

(2.7)

Moreover, from

$$d(T{x}_{m(k)},T{x}_{n(k)})\le s[d(T{x}_{m(k)},T{x}_{m(k)-1})+d(T{x}_{m(k)-1},T{x}_{n(k)})]$$

and

$$d(T{x}_{m(k)-1},T{x}_{n(k)})\le s[d(T{x}_{m(k)-1},T{x}_{m(k)})+d(T{x}_{m(k)},T{x}_{n(k)})],$$

and using (2.3) and (2.6), we get

$$\frac{\epsilon }{s}\le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(T{x}_{m(k)-1},T{x}_{n(k)})\le s^2\epsilon .$$

(2.8)

Similarly, we can show that

$$\frac{\epsilon }{s}\le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(T{x}_{n(k)-1},T{x}_{m(k)})\le \epsilon$$

(2.9)

and

$$\frac{\epsilon }{s}\le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(T{x}_{m(k)-1},T{x}_{n(k)})\le s^2\epsilon .$$

(2.10)

Using (2.1) and (2.7)-(2.10) we have

$$\begin{array}{rcl}\psi (s\epsilon )& \le & \psi (s\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(T{x}_{m(k)},T{x}_{n(k)}))\\ =& \psi (s\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(Tf{x}_{m(k)-1},Tf{x}_{n(k)-1}))\\ \le & \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi \left(\frac{d(T{x}_{m(k)-1},Tf{x}_{n(k)-1})+d(T{x}_{n(k)-1},Tf{x}_{m(k)-1})}{s+1}\right)\\ -\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\phi (d(T{x}_{m(k)-1},Tf{x}_{n(k)-1}),d(T{x}_{n(k)-1},Tf{x}_{m(k)-1}))\\ \le & \psi \left(\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{d(T{x}_{m(k)-1},T{x}_{n(k)})+d(T{x}_{n(k)-1},T{x}_{m(k)})}{s+1}\right)\\ -\phi (\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(T{x}_{m(k)-1},T{x}_{n(k)}),\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(T{x}_{n(k)-1},T{x}_{m(k)}))\\ \le & \psi \left(\frac{s^2\epsilon +\epsilon }{s+1}\right)-\phi (\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(T{x}_{m(k)-1},T{x}_{n(k)}),\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(T{x}_{n(k)-1},T{x}_{m(k)}))\\ \le & \psi (s\epsilon )-\phi (\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(T{x}_{m(k)-1},T{x}_{n(k)}),\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(T{x}_{n(k)-1},T{x}_{m(k)}))\end{array}$$

since $\frac{s^2+1}{s+1}\le s$. Hence, we have

$$\phi (\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(T{x}_{m(k)-1},T{x}_{n(k)}),\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(T{x}_{n(k)-1},T{x}_{m(k)}))\le 0.$$

By our assumption about φ, we have

$$\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(T{x}_{m(k)-1},T{x}_{n(k)})=\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(T{x}_{n(k)-1},T{x}_{m(k)})=0,$$

which contradicts (2.9) and (2.10).

Since $(X,d)$ is b-complete and $\{T{x}_n\}=\{T{f}^n{x}_0\}$ is a b-Cauchy sequence, there exists $v\in X$ such that

$$\underset{n\to \mathrm{\infty }}{lim}T{f}^n{x}_0=v.$$

(2.11)

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 b-convergent subsequence. Hence, there exist $u\in X$ and a subsequence $\{n_i\}$ such that

$$\underset{i\to \mathrm{\infty }}{lim}{f}^{n_i}{x}_0=u.$$

(2.12)

Since T is continuous, by (2.12) we obtain

$$\underset{i\to \mathrm{\infty }}{lim}T{f}^{n_i}{x}_0=Tu,$$

(2.13)

and by (2.11) and (2.13) we conclude that $Tu=v$.

From Lemma 1 and (2.1) we have

$$\begin{array}{rcl}\psi (s\cdot \frac{1}{s}d(Tfu,Tu))& \le & \psi \left(\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}sd(Tfu,T{f}^{n+1}{x}_0)\right)\\ =& \psi (\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}sd(Tfu,Tf{x}_n))\\ \le & \psi \left(\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{d(Tu,Tf{x}_n)+d(T{x}_n,Tfu)}{s+1}\right)\\ -\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\phi (d(Tu,Tf{x}_n),d(T{x}_n,Tfu))\\ \le & \psi \left(\frac{sd(Tu,Tu)+sd(Tu,Tfu)}{s+1}\right)\\ -\phi (\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(Tu,Tf{x}_n),\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(T{x}_n,Tfu))\\ \le & \psi (d(Tu,Tfu))-\phi (0,\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(T{x}_n,Tfu)),\end{array}$$

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

$$d(Tfu,Tu)\le s[d(Tfu,T{x}_n)+d(T{x}_n,Tu)].$$

Letting $n\to \mathrm{\infty }$ we can conclude that $d(Tfu,Tu)=0$. Hence, $Tfu=Tu$. As T is one-to-one, $fu=u$. Consequently, f has a fixed point.

If we assume that w is another fixed point of f, because of (2.1), we have

$$\begin{array}{rl}\psi (sd(Tu,Tw))& =\psi (sd(Tfu,Tfw))\\ \le \psi \left(\frac{d(Tu,Tfw)+d(Tw,Tfu)}{s+1}\right)-\phi (d(Tu,Tfw),d(Tw,Tfu))\\ =\psi \left(\frac{d(Tu,Tw)+d(Tw,Tu)}{s+1}\right)-\phi (d(Tu,Tw),d(Tw,Tu))\\ \le \psi (sd(Tu,Tw))-\phi (d(Tu,Tw),d(Tw,Tu)),\end{array}$$

since $\frac{2}{s+1}\le s$ and ψ is increasing. Hence, $d(Tu,Tw)=0$. Since T is one-to-one, 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 b-metric spaces is obtained.

Corollary 1 Let $(X,d)$ be a complete b-metric space and $T,f:X\to X$ be mappings such that T is continuous, one-to-one and subsequentially convergent. If $\alpha \in [0,\frac{1}{s+1})$ and

$$d(Tfx,Tfy)\le \frac{\alpha }{s}(d(Tx,Tfy)+d(Ty,Tfx)),$$

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 b-metric 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 b-metric 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 $d(x,y)={(x-y)}^2$ for $x,y\in X$. Then d is a b-metric with the parameter $s=2$ and $(X,d)$ is a complete b-metric space. Consider the mappings $f,T:X\to X$ given by

$$f(0)=0,f\left(\frac{1}{n}\right)=\frac{1}{n+1},T(0)=0,T\left(\frac{1}{n}\right)=\frac{1}{n^n},n\in \mathbb{N}.$$

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

$$d(Tf\frac{1}{n},Tf\frac{1}{m})={[\frac{1}{{(n+1)}^{n+1}}-\frac{1}{{(m+1)}^{m+1}}]}^2<{\left[\frac{1}{{(n+1)}^{n+1}}\right]}^2.$$

It is easy to prove that, for $n\in \mathbb{N}$,

$$\frac{1}{{(n+1)}^{n+1}}<\frac{1}{3}[\frac{1}{n^n}-\frac{1}{{(n+2)}^{n+2}}].$$

It follows that

$$d(Tf\frac{1}{n},Tf\frac{1}{m})<\frac{1}{9}{[\frac{1}{n^n}-\frac{1}{{(n+2)}^{n+2}}]}^2.$$

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

$$\begin{array}{rl}d(Tf\frac{1}{n},Tf\frac{1}{m})& <\frac{1}{9}{[\frac{1}{n^n}-\frac{1}{{(m+1)}^{m+1}}]}^2\\ \le \frac{\alpha }{s}[d(T\frac{1}{n},Tf\frac{1}{m})+d(T\frac{1}{m},TF\frac{1}{n})].\end{array}$$

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 T-Kannan contractions

Our second main result is the following.

Theorem 5 Let $(X,d)$ be a complete b-metric 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 $x,y\in X$,

$$\psi (d(Tfx,Tfy))\le \psi \left(\frac{d(Tx,Tfx)+d(Ty,Tfy)}{s+1}\right)-\phi (d(Tx,Tfx),d(Ty,Tfy)).$$

(3.1)

and let T be one-to-one 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

$$\underset{n\to \mathrm{\infty }}{lim}d(T{x}_n,T{x}_{n+1})=0.$$

Using condition (3.1), we obtain

$$\begin{array}{rcl}\psi (d(T{x}_{n+1},T{x}_n))& =& \psi (d(Tf{x}_n,Tf{x}_{n-1}))\\ \le & \psi \left(\frac{d(T{x}_n,Tf{x}_n)+d(T{x}_{n-1},Tf{x}_{n-1})}{s+1}\right)\\ -\phi (d(T{x}_n,Tf{x}_n),d(T{x}_{n-1},Tf{x}_{n-1}))\\ =& \psi \left(\frac{d(T{x}_n,T{x}_{n+1})+d(T{x}_{n-1},T{x}_n)}{s+1}\right)\\ -\phi (d(T{x}_n,Tf{x}_n),d(T{x}_{n-1},Tf{x}_{n-1})).\end{array}$$

(3.2)

Since φ is nonnegative and ψ is increasing, it follows that

$$d(T{x}_{n+1},T{x}_n)\le \frac{d(T{x}_n,T{x}_{n+1})+d(T{x}_{n-1},T{x}_n)}{s+1},$$

that is,

$$d(T{x}_{n+1},T{x}_n)\le \frac{1}{s}d(T{x}_n,T{x}_{n-1})\le d(T{x}_n,T{x}_{n-1}).$$

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

$$\psi (r)\le \psi \left(\frac{2r}{s+1}\right)-\phi (r,r)\le \psi (r)-\phi (r,r),$$

which is possible only if

$$r=\underset{n\to \mathrm{\infty }}{lim}d(T{x}_n,T{x}_{n+1})=0.$$

Now, we will show that $\{T{x}_n\}$ is a b-Cauchy 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

$$d(T{x}_{m(k)},T{x}_{n(k)-1})<\epsilon .$$

Again, as in Step II of Theorem 4 one can prove that

$$\epsilon \le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(T{x}_{m(k)},T{x}_{n(k)})\le s\epsilon .$$

(3.3)

Using (3.1) we have

$$\begin{array}{rcl}\psi (d(T{x}_{m(k)},T{x}_{n(k)}))& =& \psi (d(Tf{x}_{m(k)-1},Tf{x}_{n(k)-1}))\\ \le & \psi \left(\frac{d(T{x}_{m(k)-1},Tf{x}_{m(k)-1})+d(T{x}_{n(k)-1},Tf{x}_{n(k)-1})}{s+1}\right)\\ -\phi (d(T{x}_{m(k)-1},Tf{x}_{m(k)-1}),d(T{x}_{n(k)-1},Tf{x}_{n(k)-1}))\\ =& \psi \left(\frac{d(T{x}_{m(k)-1},T{x}_{m(k)})+d(T{x}_{n(k)-1},T{x}_{n(k)})}{s+1}\right)\\ -\phi (d(T{x}_{m(k)-1},T{x}_{m(k)}),d(T{x}_{n(k)-1},T{x}_{n(k)})).\end{array}$$

Passing to the upper limit as $k\to \mathrm{\infty }$ in the above inequality and taking into account (3.3), we have

$$\psi (\epsilon )\le \psi (0)-\phi (0,0)=0,$$

and so $\psi (\epsilon )=0$. By our assumptions about ψ, we have $\epsilon =0$, which is a contradiction.

Since $(X,d)$ is b-complete and $\{T{x}_n\}=\{T{f}^n{x}_0\}$ is a b-Cauchy sequence, there exists $v\in X$ such that

$$\underset{n\to \mathrm{\infty }}{lim}T{f}^n{x}_0=v.$$

(3.4)

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

$$\underset{i\to \mathrm{\infty }}{lim}{f}^{n_i}{x}_0=u.$$

(3.5)

Since T is continuous, by (3.5) we obtain

$$\underset{i\to \mathrm{\infty }}{lim}T{f}^{n_i}{x}_0=Tu,$$

(3.6)

and by (3.4) and (3.6) we conclude that $Tu=v$.

From Lemma 1 and (3.1) we have

$$\begin{array}{rcl}\psi (\frac{1}{s}d(Tfu,Tu))& \le & \psi \left(\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(Tfu,T{f}^{n+1}{x}_0)\right)\\ =& \psi (\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(Tfu,Tf{x}_n))\\ \le & \psi \left(\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{d(Tu,Tfu)+d(T{x}_n,Tf{x}_n)}{s+1}\right)\\ -\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\phi (d(Tu,Tfu),d(T{x}_n,Tf{x}_n))\\ =& \psi \left(\frac{d(Tu,Tfu)+0}{s+1}\right)-\phi (d(Tu,Tfu),0)\\ \le & \psi \left(\frac{d(Tu,Tfu)}{s}\right)-\phi (d(Tu,Tfu),0).\end{array}$$

By the properties of $\phi \in \mathrm{\Phi }$, it follows that

$$d(Tu,Tfu)=0.$$

Since T is one-to-one, 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 b-metric spaces is obtained.

Corollary 2 Let $(X,d)$ be a complete b-metric space with the parameter $s\ge 1$, $T,f:X\to X$ be such that for some $\alpha <\frac{1}{s+1}$ and all $x,y\in X$,

$$d(Tfx,Tfy)\le \alpha (d(Tx,Tfx)+d(Ty,Tfy))$$

(3.7)

and let T be one-to-one 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 b-metric space (with $s=\frac{9}{8}>1$) which is not a metric space. Consider the mapping $f:X\to X$ given by

$$f=\left(\begin{array}{ccc}a& b& c\\ a& a& b\end{array}\right).$$

We first note that the b-metric 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

$$\psi (d(fx,fy))\le \psi \left(\frac{d(x,fx)+d(y,fy)}{s+1}\right)-\phi (d(x,fx),d(y,fy))$$

cannot hold, whatever $\psi \in \mathrm{\Psi }$ and $\phi \in \mathrm{\Phi }$ are chosen.

Take now $T:X\to X$ defined by

$$T=\left(\begin{array}{ccc}a& b& c\\ b& c& a\end{array}\right).$$

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

$$\frac{4}{9}<\alpha <\frac{8}{17}=\frac{1}{s+1}.$$

Only the following cases are nontrivial:

1^∘ $x=a$, $y=c$. Then (3.7) reduces to

$$d(Tfa,Tfc)=d(b,c)=1=\frac{4}{9}\cdot \frac{9}{4}<\alpha (d(b,b)+d(a,c))=\alpha (d(Ta,Tfa)+d(Tc,Tfc)).$$

2^∘ $x=b$, $y=c$. Then (3.7) reduces to

$$d(Tfb,Tfc)=d(b,c)=1<\frac{4}{9}\cdot \frac{13}{4}<\alpha (d(c,b)+d(a,c))=\alpha (d(Tb,Tfb)+d(Tc,Tfc)).$$

All the conditions of Corollary 2 are satisfied and f has a unique fixed point ($u=a$).