The class of $(\alpha,\psi)$-type contractions in b-metric spaces and fixed point theorems

Abstract

We study the existence and uniqueness of fixed points for self-operators defined in a b-metric space and belonging to the class of $(\alpha,\psi)$-type contraction mappings. The obtained results generalize and unify several existing fixed point theorems in the literature.

1 Introduction and preliminaries

Very recently, we studied in [1] the existence and uniqueness of fixed points for self-operators defined in a metric space and belonging to the class of $(\alpha,\psi)$-type contraction mappings (see [2–5] for some works in this direction). We proved that the class of α-ψ-type contractions includes large classes of contraction-type operators, whose fixed points can be obtained by means of the Picard iteration. The aim of this paper is to extend the obtained results in [1] to self-operators defined in a b-metric space.

We start by recalling the following definition.

Definition 1.1

([6])

Let X be a nonempty set. A mapping $d: X\times X \to[0,\infty)$ is called b-metric if there exists a real number $b\geq1$ such that for every $x,y,z\in X$, we have

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

• (ii) $d(x,y)=d(y,x)$;

• (iii) $d(x,z)\leq b[d(x,y)+d(y,z)]$.

In this case, the pair $(X,d)$ is called a b-metric space.

There exist many examples in the literature (see [6–8]) showing that the class of b-metrics is effectively larger than that of metric spaces.

The notions of convergence, compactness, closedness and completeness in b-metric spaces are given in the same way as in metric spaces. For works on fixed point theory in b-metric spaces, we refer to [9–12] and the references therein.

Definition 1.2

([13])

Let $\psi:[0,\infty)\to[0,\infty)$ be a given function. We say that ψ is a comparison function if it is increasing and $\psi^{n}(t)\to 0$, $n\to\infty$, for any $t\geq0$, where $\psi^{n}$ is the nth iterate of ψ.

In [13, 14], several results regarding comparison functions can be found. Among these we recall the following.

Lemma 1.3

If $\psi:[0,\infty)\to[0,\infty)$ is a comparison function, then

• (i) each iterate $\psi^{k}$ of ψ, $k\geq1$, is also a comparison function;

• (ii) ψ is continuous at zero;

• (iii) $\psi(t)< t$ for any $t >0$;

• (iv) $\psi(0)=0$.

The following concept was introduced in [15].

Definition 1.4

Let $b\geq1$ be a real number. A mapping $\psi:[0,\infty)\to[0,\infty )$ is called a b-comparison function if

• (i) ψ is monotone increasing;

• (ii) there exist $k_{0}\in\mathbb{N}$, $a\in(0,1)$ and a convergent series of nonnegative terms $\sum_{k=1}^{\infty}v_{k}$ such that

  $$b^{k+1}\psi^{k+1}(t)\leq a b^{k} \psi^{k}(t)+v_{k} $$

  for $k\geq k_{0}$ and any $t\geq0$.

The following lemma has been proved.

Lemma 1.5

([15, 16])

Let $\psi:[0,\infty)\to[0,\infty)$ be a b-comparison function. Then

• (i) the series $\sum_{k=0}^{\infty}b^{k}\psi ^{k}(t)$ converges for any $t\geq0$;

• (ii) the function $s_{b}:[0,\infty)\to[0,\infty)$ defined by

  $$s_{b}(t)=\sum_{k=0}^{\infty}b^{k}\psi^{k}(t), \quad t\geq0 $$

  is increasing and continuous at 0.

Lemma 1.6

([17])

Any b-comparison function is a comparison function.

Throughout this paper, for $b\geq1$, we denote by $\Psi_{b}$ the set of b-comparison functions.

Definition 1.7

Let $(X,d)$ be a b-metric space with constant $b\geq1$, and let $T: X\to X$ be a given mapping. We say that T is an α-ψ contraction if there exist a b-comparison function $\psi\in\Psi_{b}$ and a function $\alpha:X\times X\to\mathbb{R}$ such that

$$ \alpha(x,y)d(Tx,Ty)\leq\psi\bigl(d(x,y)\bigr) \quad \text{for all } x,y\in X. $$

(1.1)

2 Main results

Let $T: X\to X$ be a given mapping. We denote by $\operatorname{Fix}(T)$ the set of its fixed points; that is,

$$\operatorname{Fix}(T)=\{x\in X: x=Tx\}. $$

For $b\geq1$ and $\psi\in\Psi_{b}$, let $\Sigma_{\psi}^{b}$ be the set defined by

$$\Sigma_{\psi}^{b}=\bigl\{ \sigma\in(0,\infty): \sigma\psi\in \Psi_{b}\bigr\} . $$

We have the following result.

Proposition 2.1

Let $(X,d)$ be a b-metric space with constant $b\geq1$, and let $T: X\to X$ be a given mapping. Suppose that there exist $\alpha: X\times X \to\mathbb{R}$ and $\psi\in\Psi_{b}$ such that T is an α-ψ contraction. Suppose that there exists $\sigma\in\Sigma_{\psi}^{b}$ and for some positive integer p, there exists a finite sequence $\{\xi_{i}\}_{i=0}^{p}\subset X$ such that

$$ \xi_{0}=x_{0}, \qquad \xi_{p}=Tx_{0}, \qquad \alpha\bigl(T^{n}\xi_{i},T^{n} \xi_{i+1}\bigr)\geq\sigma^{-1}, \quad n\in\mathbb {N}, i=0, \ldots,p-1, x_{0}\in X. $$

(2.1)

Then $\{T^{n}x_{0}\}$ is a Cauchy sequence in $(X,d)$.

Proof

Let $\varphi=\sigma\psi$. By the definition of $\Sigma_{\psi}^{b}$, we have $\varphi\in\Psi_{b}$. Let $\{\xi_{i}\}_{i=0}^{p}$ be a finite sequence in X satisfying (2.1). Consider the sequence $\{x_{n}\}_{n\in\mathbb{N}}$ in X defined by $x_{n+1}=Tx_{n}$, $n\in\mathbb{N}$. We claim that

$$ d\bigl(T^{r}\xi_{i}, T^{r} \xi_{i+1}\bigr)\leq\varphi^{r}\bigl(d(\xi_{i}, \xi_{i+1})\bigr),\quad r\in\mathbb{N}, i=0,\ldots,p-1. $$

(2.2)

Let $i\in\{0,1,\ldots,p-1\}$. From (2.1), we have

$$\sigma^{-1}d(T\xi_{i}, T\xi_{i+1})\leq\alpha( \xi_{i},\xi_{i+1})d(T\xi_{i}, T\xi_{i+1}) \leq\psi\bigl(d(\xi_{i},\xi_{i+1})\bigr), $$

which implies that

$$ d(T\xi_{i}, T\xi_{i+1})\leq\varphi\bigl(d( \xi_{i},\xi_{i+1})\bigr). $$

(2.3)

Again, we have

$$\sigma^{-1}d\bigl(T^{2}\xi_{i}, T^{2} \xi_{i+1}\bigr) \leq\alpha(T\xi_{i},T\xi_{i+1})d \bigl(T(T\xi_{i}), T(T\xi_{i+1})\bigr) \leq\psi\bigl(d(T \xi_{i},T\xi_{i+1})\bigr), $$

which implies that

$$ d\bigl(T^{2}\xi_{i}, T^{2} \xi_{i+1}\bigr)\leq\varphi\bigl(d(T\xi_{i},T \xi_{i+1})\bigr). $$

(2.4)

Since φ is an increasing function (from Lemma 1.6), from (2.3) and (2.4), we obtain

$$d\bigl(T^{2}\xi_{i}, T^{2}\xi_{i+1} \bigr)\leq\varphi^{2}\bigl(d(\xi_{i},\xi_{i+1}) \bigr). $$

Continuing this process, by induction we obtain (2.2).

Now, using the property (iii) of a b-metric and (2.2), for every $n\in\mathbb{N}$, we have

$$\begin{aligned} d(x_{n},x_{n+1})&= d\bigl(T^{n}x_{0},T^{n+1}x_{0} \bigr) \\ &\leq b d\bigl(T^{n}\xi_{0},T^{n} \xi_{1}\bigr)+b^{2}d\bigl(T^{n} \xi_{1},T^{n}\xi_{2}\bigr) +\cdots+ b^{p}d\bigl(T^{n}\xi_{p-1},T^{n} \xi_{p}\bigr) \\ &= \sum_{i=0}^{p-1}b^{i+1}d \bigl(T^{n}\xi_{i},T^{n}\xi_{i+1}\bigr) \\ &\leq \sum_{i=0}^{p-1} b^{i+1} \varphi^{n}\bigl(d(\xi_{i},\xi_{i+1})\bigr). \end{aligned}$$

Thus we proved that

$$d(x_{n},x_{n+1})\leq\sum_{i=0}^{p-1} b^{i+1}\varphi^{n}\bigl(d(\xi_{i},\xi _{i+1})\bigr),\quad n\in\mathbb{N}, $$

which implies that for $q\geq1$,

$$\begin{aligned} d(x_{n},x_{n+q}) &\leq \sum_{j=n}^{n+q-1} b^{j-n+1}d(x_{j},x_{j+1}) \\ &\leq \sum_{j=n}^{n+q-1} b^{j-n+1}\sum _{i=0}^{p-1} b^{i+1}\varphi ^{j}\bigl(d(\xi_{i},\xi_{i+1})\bigr) \\ &= \frac{1}{b^{n-1}}\sum_{i=0}^{p-1}b^{i+1} \sum_{j=n}^{n+q-1} b^{j} \varphi^{j}\bigl(d(\xi_{i},\xi_{i+1})\bigr) \\ &\leq\frac{1}{b^{n-1}}\sum_{i=0}^{p-1}b^{i+1} \sum_{j=n}^{\infty} b^{j} \varphi^{j}\bigl(d(\xi_{i},\xi_{i+1})\bigr). \end{aligned}$$

Since $b\geq1$, using Lemma 1.5(i), we obtain

$$\frac{1}{b^{n-1}}\sum_{i=0}^{p-1}b^{i+1} \sum_{j=n}^{\infty} b^{j}\varphi ^{j}\bigl(d(\xi_{i},\xi_{i+1})\bigr)\to0\quad \text{as } n \to\infty. $$

This proves that $\{x_{n}\}$ is a Cauchy sequence in the b-metric space $(X,d)$. □

Our first main result is the following fixed point theorem which requires the continuity of the mapping T.

Theorem 2.2

Let $(X,d)$ be a complete b-metric space with constant $b\geq1$, and let $T: X\to X$ be a given mapping. Suppose that there exist $\alpha: X\times X \to\mathbb{R}$ and $\psi \in\Psi_{b}$ such that T is an α-ψ contraction. Suppose also that (2.1) is satisfied. Then $\{T^{n}x_{0}\}$ converges to some $x^{*}\in X$. Moreover, if T is continuous, then $x^{*}\in\operatorname{Fix}(T)$.

Proof

From Proposition 2.1, we know that $\{ T^{n}x_{0}\}$ is a Cauchy sequence. Since $(X,d)$ is a complete b-metric space, there exists $x^{*}\in X$ such that

$$\lim_{n\to\infty} d\bigl(T^{n}x_{0},x^{*}\bigr)=0. $$

The continuity of T yields

$$\lim_{n\to\infty} d\bigl(T^{n+1}x_{0},Tx^{*} \bigr)=0. $$

By the uniqueness of the limit, we obtain $x^{*}=Tx^{*}$, that is, $x^{*}\in \operatorname{Fix}(T)$. □

In the next theorem, we omit the continuity assumption of T.

Theorem 2.3

Let $(X,d)$ be a complete b-metric space with constant $b\geq1$, and let $T: X\to X$ be a given mapping. Suppose that there exist $\alpha: X\times X \to\mathbb{R}$ and $\psi \in\Psi_{b}$ such that T is an α-ψ contraction. Suppose also that (2.1) is satisfied. Then $\{T^{n}x_{0}\}$ converges to some $x^{*}\in X$. Moreover, if there exists a subsequence $\{T^{\gamma(n)}x_{0}\}$ of $\{T^{n}x_{0}\}$ such that

$$\max\bigl\{ \alpha\bigl(T^{\gamma(n)}x_{0},x^{*}\bigr), \alpha \bigl(x^{*},T^{\gamma(n)}x_{0}\bigr)\bigr\} \geq\ell\in(0,\infty),\quad n \textit{ large enough}, $$

then $x^{*}\in\operatorname{Fix}(T)$.

Proof

From Proposition 2.1 and the completeness of the b-metric space $(X,d)$, we know that $\{T^{n}x_{0}\}$ converges to some $x^{*}\in X$.

Suppose now that there exists a subsequence $\{T^{\gamma(n)}x_{0}\}$ of $\{T^{n}x_{0}\}$ such that

$$ \max\bigl\{ \alpha\bigl(T^{\gamma(n)}x_{0},x^{*}\bigr), \alpha\bigl(x^{*},T^{\gamma(n)}x_{0}\bigr)\bigr\} \geq\ell\in(0, \infty), \quad n \mbox{ large enough}. $$

(2.5)

Since T is an α-ψ contraction, we have

$$\alpha\bigl(T^{\gamma(n)}x_{0},x^{*}\bigr)d\bigl(T^{\gamma(n)+1}x_{0},Tx^{*} \bigr) \leq\psi\bigl(d\bigl(T^{\gamma(n)}x_{0},x^{*}\bigr)\bigr), \quad n\in\mathbb{N} $$

and

$$\alpha\bigl(x^{*},T^{\gamma(n)}x_{0}\bigr)d\bigl(T^{\gamma(n)+1}x_{0},Tx^{*} \bigr) \leq\psi\bigl(d\bigl(T^{\gamma(n)}x_{0},x^{*}\bigr)\bigr), \quad n\in\mathbb{N}. $$

Thus we have

$$\max\bigl\{ \alpha\bigl(T^{\gamma(n)}x_{0},x^{*}\bigr), \alpha \bigl(x^{*},T^{\gamma(n)}x_{0}\bigr)\bigr\} d\bigl(T^{\gamma(n)+1}x_{0},Tx^{*} \bigr) \leq\psi\bigl(d\bigl(T^{\gamma(n)}x_{0},x^{*}\bigr)\bigr),\quad n\in\mathbb{N}. $$

From (2.5), we get

$$ \ell d\bigl(T^{\gamma(n)+1}x_{0},Tx^{*}\bigr)\leq \psi \bigl(d\bigl(T^{\gamma(n)}x_{0},x^{*}\bigr)\bigr),\quad n \mbox{ large enough}. $$

(2.6)

On the other hand, using the property (iii) of a b-metric, we get

$$ d\bigl(T^{\gamma(n)+1}x_{0},Tx^{*}\bigr)\geq \frac{1}{b} d\bigl(x^{*},Tx^{*}\bigr)-d\bigl(x^{*},T^{\gamma (n)+1}x_{0} \bigr), \quad n\in\mathbb{N}. $$

(2.7)

Now, (2.6) and (2.7) yield

$$\ell \biggl(\frac{1}{b} d\bigl(x^{*},Tx^{*}\bigr)-d\bigl(x^{*},T^{\gamma(n)+1}x_{0} \bigr) \biggr)\leq \psi\bigl(d\bigl(T^{\gamma(n)}x_{0},x^{*}\bigr) \bigr), \quad n \mbox{ large enough}. $$

Letting $n\to\infty$ in the above inequality, using Lemma 1.6 and Lemma 1.3(ii) and (iv), we obtain

$$0\leq\frac{\ell}{b} d\bigl(x^{*},Tx^{*}\bigr)\leq\psi(0)=0, $$

which implies that $d(x^{*},Tx^{*})=0$, that is, $x^{*}\in\operatorname {Fix}(T)$. □

We provide now a sufficient condition for the uniqueness of the fixed point.

Theorem 2.4

Let $(X,d)$ be a b-metric space with constant $b\geq1$, and let $T: X\to X$ be a given mapping. Suppose that there exist $\alpha: X\times X \to\mathbb{R}$ and $\psi \in\Psi_{b}$ such that T is an α-ψ contraction. Suppose also that

• (i) $\operatorname{Fix}(T)\neq\emptyset$;

• (ii) for every pair $(x,y)\in\operatorname{Fix}(T)\times \operatorname{Fix}(T)$ with $x\neq y$, if $\alpha(x,y)<1$, then there exists $\eta\in\Sigma_{\psi}^{b}$ and for some positive integer q, there is a finite sequence $\{\zeta_{i}(x,y)\}_{i=0}^{q}\subset X$ such that

  $$\zeta_{0}(x,y)=x, \qquad \zeta_{q}(x,y)=y,\qquad \alpha \bigl(T^{n}\zeta_{i}(x,y),T^{n} \zeta_{i+1}(x,y)\bigr)\geq\eta^{-1} $$

  for $n\in\mathbb{N}$ and $i=0,\ldots,q-1$.

Then T has a unique fixed point.

Proof

Let $\varphi=\eta\psi\in\Psi_{b}$. Suppose that $u,v\in X$ are two fixed points of T such that $d(u,v)>0$. We consider two cases.

Case 1: $\alpha(u,v)\geq1$. Since T is an α-ψ contraction, we have

$$d(u,v)\leq\alpha(u,v)d(Tu,Tv)\leq\psi\bigl(d(u,v)\bigr). $$

On the other hand, from Lemma 1.6 and Lemma 1.3(iii), we have

$$\psi\bigl(d(u,v)\bigr)< d(u,v). $$

The two above inequalities yield a contradiction.

Case 2: $\alpha(u,v)<1$. By assumption, there exists a finite sequence $\{\zeta_{i}(u,v)\} _{i=0}^{q}$ in X such that

$$\zeta_{0}(u,v)=u, \qquad \zeta_{q}(u,v)=v,\qquad \alpha \bigl(T^{n}\zeta_{i}(u,v),T^{n} \zeta_{i+1}(u,v)\bigr)\geq\eta^{-1} $$

for $n\in\mathbb{N}$ and $i=0,\ldots,q-1$. As in the proof of Proposition 2.1, we can establish that

$$ d\bigl(T^{r}\zeta_{i}(u,v), T^{r} \zeta_{i+1}(u,v)\bigr) \leq\varphi^{r}\bigl(d\bigl( \zeta_{i}(u,v),\zeta_{i+1}(u,v)\bigr)\bigr),\quad r\in \mathbb{N}, i=0,\ldots,q-1. $$

(2.8)

On the other hand, we have

$$\begin{aligned} d(u,v) &= d\bigl(T^{n}u,T^{n}v\bigr) \\ &\leq \sum_{i=0}^{q-1} b^{i+1}d \bigl(T^{n}\zeta_{i}(u,v),T^{n} \zeta_{i+1}(u,v)\bigr) \\ &\leq \sum_{i=0}^{q-1}b^{i+1} \varphi ^{n}\bigl(d\bigl(\zeta_{i}(u,v),\zeta_{i+1}(u,v) \bigr)\bigr)\to0\quad \text{as } n\to\infty \mbox{ (by Lemma 1.6)}. \end{aligned}$$

Then $u=v$, which is a contradiction. □

3 Particular cases

In this section, we deduce from our main theorems several fixed point theorems in b-metric spaces.

3.1 The class of ψ-type contractions in b-metric spaces

Definition 3.1

Let $(X,d)$ be a b-metric space with constant $b\geq1$. A mapping $T: X\to X$ is said to be a ψ-contraction if there exists $\psi\in\Psi_{b}$ such that

$$ d(Tx,Ty)\leq\psi\bigl(d(x,y)\bigr)\quad \text{for all } x,y\in X. $$

(3.1)

Theorem 3.2

Let $(X,d)$ be a b-metric space with constant $b\geq1$, and let $T: X\to X$ be a given mapping. Suppose that there exists $\psi\in\Psi_{b}$ such that T is a ψ-contraction. Then there exists $\alpha: X\times X \to\mathbb{R}$ such that T is an α-ψ contraction.

Proof

Consider the function $\alpha: X\times X \to\mathbb{R}$ defined by

$$ \alpha(x,y)=1\quad \text{for all } x,y\in X. $$

(3.2)

Clearly, from (3.1), T is an α-ψ contraction. □

Corollary 3.3

([17])

Let $(X,d)$ be a complete b-metric space with constant $b\geq1$, and let $T: X\to X$ be a given mapping. If T is a ψ-contraction for some $\psi\in\Psi_{b}$, then T has a unique fixed point. Moreover, for any $x_{0}\in X$, the Picard sequence $\{T^{n}x_{0}\}$ converges to this fixed point.

Proof

From Lemma 1.6, we have

$$d(Tx,Ty)\leq d(x,y) \quad \text{for all } x,y\in X, $$

which implies that T is a continuous mapping. From Theorem 3.2, T is an α-ψ contraction, where α is defined by (3.2). Clearly, for any $x_{0}\in X$, (2.1) is satisfied with $p=1$ and $\sigma=1$. By Theorem 2.2, $\{T^{n}x_{0}\}$ converges to a fixed point of T. The uniqueness follows immediately from (3.2) and Theorem 2.4. □

Corollary 3.4

Let $(X,d)$ be a complete b-metric space with constant $b\geq1$, and let $T: X\to X$ be a given mapping. Suppose that

$$d(Tx,Ty)\leq k d(x,y)\quad \textit{for all } x,y\in X $$

for some constant $k\in(0,1/b)$. Then T has a unique fixed point. Moreover, for any $x_{0}\in X$, the Picard sequence $\{T^{n}x_{0}\}$ converges to this fixed point.

Proof

It is an immediate consequence of Corollary 3.3 with $\psi(t)=kt$. □

3.2 The class of rational-type contractions in b-metric spaces

3.2.1 Dass-Gupta-type contraction in b-metric spaces

Definition 3.5

Let $(X,d)$ be a b-metric space with constant $b\geq1$. A mapping $T: X\to X$ is said to be a Dass-Gupta contraction if there exist constants $\lambda,\mu\geq0$ with $\lambda b+\mu<1$ such that

$$ d(Tx,Ty)\leq\mu d(y,Ty) \frac{1+d(x,Tx)}{1+d(x,y)}+\lambda d(x,y)\quad \text{for all } x,y\in X. $$

(3.3)

Theorem 3.6

Let $(X,d)$ be a b-metric space with constant $b\geq1$, and let $T: X\to X$ be a given mapping. Suppose that T is a Dass-Gupta contraction. Then there exist $\psi\in \Psi_{b}$ and $\alpha: X\times X \to\mathbb{R}$ such that T is an α-ψ contraction.

Proof

From (3.3), for all $x,y\in X$, we have

$$d(Tx,Ty)-\mu d(y,Ty) \frac{1+d(x,Tx)}{1+d(x,y)}\leq\lambda d(x,y), $$

which yields

$$ \biggl(1-\mu\frac{d(y,Ty)(1+d(x,Tx))}{(1+d(x,y))d(Tx,Ty)} \biggr)d(Tx,Ty) \leq\lambda d(x,y), \quad x,y\in X, Tx\neq Ty. $$

(3.4)

Consider the functions $\psi:[0,\infty)\to[0,\infty)$ and $\alpha:X\times X\to\mathbb{R}$ defined by

$$ \psi(t)=\lambda t,\quad t\geq0 $$

(3.5)

and

$$ \alpha(x,y)= \textstyle\begin{cases} 1- \mu\frac{d(y,Ty)(1+d(x,Tx))}{(1+d(x,y))d(Tx,Ty)}, & \text{if } Tx\neq Ty, \\ 0, & \text{otherwise}. \end{cases} $$

(3.6)

Since $0\leq\lambda b<1$, then $\psi\in\Psi_{b}$. On the other hand, from (3.4) we have

$$\alpha(x,y)d(Tx,Ty)\leq\psi\bigl(d(x,y)\bigr)\quad \text{for all } x,y\in X. $$

Then T is an α-ψ contraction. □

Corollary 3.7

Let $(X,d)$ be a complete b-metric space with constant $b\geq1$, and let $T: X\to X$ be a given mapping. If T is a Dass-Gupta contraction with parameters $\lambda,\mu\geq0$ such that $\lambda b+\mu<1$, then T has a unique fixed point. Moreover, for any $x_{0}\in X$, the Picard sequence $\{ T^{n}x_{0}\}$ converges to this fixed point.

Proof

Let $x_{0}$ be an arbitrary point in X. If for some $r\in\mathbb{N}$, $T^{r}x_{0}=T^{r+1}x_{0}$, then $T^{r}x_{0}$ will be a fixed point of T. So we can suppose that $T^{r}x_{0}\neq T^{r+1}x_{0}$ for all $r\in\mathbb{N}$. From (3.6), for all $n\in\mathbb{N}$, we have

$$\begin{aligned} \alpha\bigl(T^{n}x_{0},T^{n+1}x_{0} \bigr) &= 1- \mu\frac{d(T^{n+1}x_{0},T^{n+2}x_{0}) (1+d(T^{n}x_{0},T^{n+1}x_{0}))}{(1+d(T^{n}x_{0},T^{n+1}x_{0}))d(T^{n+1}x_{0},T^{n+2}x_{0})} \\ &= 1-\mu>0. \end{aligned}$$

On the other hand, from (3.5) we have

$$(1-\mu)^{-1}\psi(t)=\frac{\lambda}{1-\mu}t, \quad t\geq0. $$

From the condition $\lambda b+\mu<1$, clearly we have $(1-\mu)^{-1}\psi \in\Psi_{b}$, which is equivalent to $(1-\mu)^{-1}\in\Sigma_{\psi}^{b}$. Then (2.1) is satisfied with $p=1$ and $\sigma=(1-\mu)^{-1}$. From the first part of Theorem 2.3, the sequence $\{T^{n}x_{0}\}$ converges to some $x^{*}\in X$.

Suppose that $x^{*}$ is not a fixed point of T, that is, $d(x^{*},Tx^{*})>0$. Then

$$T^{n+1}x_{0}\neq Tx^{*},\quad n \mbox{ large enough}. $$

From (3.6), we have

$$\alpha\bigl(x^{*},T^{n}x_{0}\bigr) = 1-\mu\frac{d(T^{n}x_{0},T^{n+1}x_{0})(1+d(x^{*},Tx^{*}))}{(1+d(T^{n}x_{0},x^{*})) d(T^{n+1}x_{0},Tx^{*})}, \quad n \mbox{ large enough}. $$

On the other hand, using the property (iii) of a b-metric, we have

$$d\bigl(T^{n+1}x_{0},Tx^{*}\bigr)\geq\frac{1}{b}d \bigl(x^{*},Tx^{*}\bigr)-d\bigl(x^{*},T^{n+1}x_{0}\bigr)>0,\quad n \mbox{ large enough}. $$

Thus we have

$$\alpha\bigl(x^{*},T^{n}x_{0}\bigr)\geq1-\mu \frac{d(T^{n}x_{0},T^{n+1}x_{0})(1+d(x^{*},Tx^{*}))}{(1+d(T^{n}x_{0},x^{*})) (\frac{1}{b}d(x^{*},Tx^{*})-d(x^{*},T^{n+1}x_{0}) )}, \quad n \mbox{ large enough}. $$

Since

$$\lim_{n\to\infty} 1-\mu \frac{d(T^{n}x_{0},T^{n+1}x_{0})(1+d(x^{*},Tx^{*}))}{(1+d(T^{n}x_{0},x^{*})) (\frac{1}{b}d(x^{*},Tx^{*})-d(x^{*},T^{n+1}x_{0}) )}=1, $$

we have

$$\alpha\bigl(x^{*},T^{n}x_{0}\bigr)>\frac{1}{2}, \quad n \mbox{ large enough}. $$

By Theorem 2.3, we deduce that $x^{*}\in \operatorname{Fix}(T)$, which is a contradiction. Thus $\operatorname {Fix}(T)\neq\emptyset$.

For the uniqueness, observe that for every pair $(x,y)\in\operatorname {Fix}(T)\times\operatorname{Fix}(T)$ with $x\neq y$, we have $\alpha(x,y)=1$. By Theorem 2.4, $x^{*}$ is the unique fixed point of T. □

If $b=1$, Corollary 3.7 recovers the Dass-Gupta fixed point theorem [18].

3.2.2 Jaggi-type contraction in b-metric spaces

Definition 3.8

Let $(X,d)$ be a b-metric space with constant $b\geq1$. A mapping $T: X\to X$ is said to be a Jaggi contraction if there exist constants $\lambda,\mu\geq0$ with $\lambda b+\mu<1$ such that

$$ d(Tx,Ty)\leq\mu\frac{d(x,Tx)d(y,Ty)}{d(x,y)}+\lambda d(x,y)\quad \text{for all } x,y\in X, x\neq y. $$

(3.7)

Theorem 3.9

Let $(X,d)$ be a b-metric space with constant $b\geq1$, and let $T: X\to X$ be a given mapping. Suppose that T is a Jaggi contraction. Then there exist $\psi\in\Psi_{b}$ and $\alpha: X\times X \to\mathbb{R}$ such that T is an α-ψ contraction.

Proof

From (3.7), for all $x,y\in X$ with $x\neq y$, we have

$$d(Tx,Ty)-\mu\frac{d(x,Tx)d(y,Ty)}{d(x,y)}\leq\lambda d(x,y), $$

which yields

$$ \biggl(1-\mu\frac{d(x,Tx)d(y,Ty)}{d(x,y)d(Tx,Ty)} \biggr)d(Tx,Ty)\leq \lambda d(x,y), \quad x,y\in X, Tx\neq Ty. $$

(3.8)

Consider the functions $\psi:[0,\infty)\to[0,\infty)$ and $\alpha:X\times X\to\mathbb{R}$ defined by

$$ \psi(t)=\lambda t,\quad t\geq0 $$

(3.9)

and

$$ \alpha(x,y)= \textstyle\begin{cases} 1- \mu\frac{d(x,Tx)d(y,Ty)}{d(x,y)d(Tx,Ty)}, & \text{if } Tx\neq Ty, \\ 0, & \text{otherwise}. \end{cases} $$

(3.10)

Since $\lambda b<1$, we have $\psi\in\Psi_{b}$. From (3.8), we have

$$\alpha(x,y)d(Tx,Ty)\leq\psi\bigl(d(x,y)\bigr) \quad \text{for all } x,y\in X. $$

Then T is an α-ψ contraction. □

Corollary 3.10

Let $(X,d)$ be a complete b-metric space with constant $b\geq1$, and let $T: X\to X$ be a continuous mapping. If T is a Jaggi contraction with parameters $\lambda,\mu\geq0$ such that $\lambda b+\mu<1$, then T has a unique fixed point. Moreover, for any $x_{0}\in X$, the Picard sequence $\{T^{n}x_{0}\}$ converges to this fixed point.

Proof

Let $x_{0}$ be an arbitrary point in X. Without loss of generality, we can suppose that $T^{r}x_{0}\neq T^{r+1}x_{0}$ for all $r\in\mathbb{N}$. From (3.10), for all $n\in\mathbb{N}$, we have

$$\alpha\bigl(T^{n}x_{0},T^{n+1}x_{0} \bigr) = 1- \mu\frac {d(T^{n}x_{0},T^{n+1}x_{0})d(T^{n+1}x_{0},T^{n+2}x_{0})}{d(T^{n}x_{0},T^{n+1}x_{0}) d(T^{n+1}x_{0},T^{n+2}x_{0})} = 1-\mu>0. $$

On the other hand, from (3.9), for all $t\geq0$, we have

$$(1-\mu)^{-1}\psi(t)=\frac{\lambda}{1-\mu} t. $$

Since $\lambda b+\mu<1$, we have $(1-\mu)^{-1}\psi\in\Psi_{b}$, that is, $(1-\mu)^{-1}\in\Sigma_{\psi}^{b}$. Then (2.1) is satisfied with $p=1$ and $\sigma=(1-\mu)^{-1}$. By the first part of Theorem 2.2, $\{T^{n}x_{0}\}$ converges to some $x^{*}\in X$. Since T is continuous, by the second part of Theorem 2.2, $x^{*}$ is a fixed point of T. Moreover, for every pair $(x,y)\in\operatorname{Fix}(T)\times\operatorname{Fix}(T)$ with $x\neq y$, we have $\alpha(x,y)=1$. Then, by Theorem 2.4, $x^{*}$ is the unique fixed point of T. □

If $b=1$, Corollary 3.10 recovers the Jaggi fixed point theorem [19].

3.3 The class of Berinde-type mappings in b-metric spaces

Definition 3.11

Let $(X,d)$ be a b-metric space with constant $b\geq1$. A mapping $T: X\to X$ is said to be a Berinde-type contraction if there exist $\lambda\in(0,1/b)$ and $L\geq0$ such that

$$ d(Tx,Ty)\leq\lambda d(x,y)+L d(y,Tx) \quad \text{for all } x,y\in X. $$

(3.11)

Theorem 3.12

Let $(X,d)$ be a b-metric space with constant $b\geq1$, and let $T: X\to X$ be a given mapping. If T is a Berinde-type contraction, then there exist $\alpha: X\times X \to\mathbb{R}$ and $\psi\in\Psi_{b}$ such that T is an α-ψ contraction.

Proof

From (3.11), we have

$$d(Tx,Ty)-L d(y,Tx) \leq\lambda d(x,y) \quad \text{for all } x,y\in X, $$

which yields

$$ \biggl(1-L \frac{d(y,Tx)}{d(Tx,Ty)} \biggr)d(Tx,Ty)\leq\lambda d(x,y), \quad x,y\in X, Tx\neq Ty. $$

(3.12)

Consider the functions $\psi:[0,\infty)\to[0,\infty)$ and $\alpha:X\times X\to\mathbb{R}$ defined by

$$\psi(t)=\lambda t, \quad t\geq0 $$

and

$$ \alpha(x,y)= \textstyle\begin{cases} 1- L \frac{d(y,Tx)}{d(Tx,Ty)}, &\text{if } Tx\neq Ty, \\ 0, &\text{otherwise}. \end{cases} $$

(3.13)

Since $\lambda b<1$, then $\psi\in\Psi_{b}$. From (3.12), we have

$$\alpha(x,y)d(Tx,Ty)\leq\psi\bigl(d(x,y)\bigr)\quad \text{for all } x,y\in X. $$

Then T is an α-ψ contraction. □

Corollary 3.13

Let $(X,d)$ be a complete b-metric space with constant $b\geq1$, and let $T: X\to X$ be a given mapping. If T is a Berinde-type contraction with parameters $\lambda,L \geq0$ such that $0<\lambda b<1$, then for any $x_{0}\in X$, the Picard sequence $\{T^{n}x_{0}\}$ converges to a fixed point of T.

Proof

Let $x_{0}$ be an arbitrary point in X. Without loss of generality, we can suppose that $T^{r}x_{0}\neq T^{r+1}x_{0}$ for all $r\in\mathbb{N}$. From (3.13), for all $n\in\mathbb{N}$, we have

$$\alpha\bigl(T^{n}x_{0},T^{n+1}x_{0} \bigr) = 1- L \frac{d(T^{n+1}x_{0},T^{n+1}x_{0})}{d(T^{n+1}x_{0},T^{n+2}x_{0})} = 1. $$

Then (2.1) holds with $\sigma=1$ and $p=1$. From the first part of Theorem 2.3, the sequence $\{T^{n}x_{0}\}$ converges to some $x^{*}\in X$.

Suppose now that $x^{*}$ is not a fixed point of T, that is, $d(x^{*},Tx^{*})>0$. Then

$$T^{n+1}x_{0}\neq Tx^{*}, \quad n \mbox{ large enough}. $$

From (3.13), we have

$$\alpha\bigl(T^{n}x_{0},x^{*}\bigr)= 1-L \frac{d(x^{*},T^{n+1}x_{0})}{d(T^{n+1}x_{0},Tx^{*})}, \quad n \mbox{ large enough}. $$

Using the property (iii) of a b-metric, we have

$$d\bigl(T^{n+1}x_{0},Tx^{*}\bigr)\geq\frac{1}{b}d \bigl(x^{*},Tx^{*}\bigr)-d\bigl(x^{*},T^{n+1}x_{0}\bigr)>0,\quad n \mbox{ large enough}. $$

Thus we have

$$\alpha\bigl(T^{n}x_{0},x^{*}\bigr)\geq1 -L\frac{d(x^{*},T^{n+1}x_{0})}{\frac {1}{b}d(x^{*},Tx^{*})-d(x^{*},T^{n+1}x_{0})}, \quad n \mbox{ large enough}. $$

Since

$$\lim_{n\to\infty}1 -L\frac{d(x^{*},T^{n+1}x_{0})}{\frac {1}{b}d(x^{*},Tx^{*})-d(x^{*},T^{n+1}x_{0})}=1, $$

then

$$\alpha\bigl(T^{n}x_{0},x^{*}\bigr)>\frac{1}{2}, \quad n \mbox{ large enough}. $$

By Theorem 2.3, we deduce that $x^{*}\in \operatorname{Fix}(T)$, which is a contradiction.

Thus $x^{*}$ is a fixed point of T. □

If $b=1$, Corollary 3.13 recovers the Berinde fixed point theorem [20].

Note that a Berinde mapping need not have a unique fixed point (see [21], Example 2.11).

Corollary 3.14

Let $(X,d)$ be a complete b-metric space with constant $b\geq1$, and let $T: X\to X$ be a given mapping. Suppose that there exists a constant $k\in(0,1/b(b+1))$ such that

$$ d(Tx,Ty)\leq k \bigl(d(x,Tx)+d(y,Ty)\bigr) \quad \textit{for all } x,y\in X. $$

(3.14)

Then, for any $x_{0}\in X$, the Picard sequence $\{T^{n}x_{0}\}$ converges to a fixed point of T.

Proof

At first, observe that from (3.14), for all $x,y\in X$, we have

$$d(Tx,Ty)\leq\lambda d(x,y)+L d(y,Tx), $$

where

$$\lambda=\frac{kb}{1-kb} \quad \mbox{and} \quad L=\frac{2kb}{1-kb}\cdot $$

With the condition $k\in(0,1/b(b+1))$, we have $0<\lambda<1/b$ and $L\geq0$. Then T is a Berinde-type contraction. From Corollary 3.13, if $x_{0}\in X$, then $\{T^{n}x_{0}\}$ converges to a fixed point of T. □

If $b=1$, Corollary 3.14 recovers the Kannan fixed point theorem [22].

Corollary 3.15

Let $(X,d)$ be a complete b-metric space with constant $b\geq1$, and let $T: X\to X$ be a given mapping. Suppose that there exists a constant $k\in(0,1/2b^{2})$ such that

$$ d(Tx,Ty)\leq k \bigl(d(x,Ty)+d(y,Tx)\bigr) \quad \textit{for all } x,y\in X. $$

(3.15)

Then, for any $x_{0}\in X$, the Picard sequence $\{T^{n}x_{0}\}$ converges to a fixed point of T.

Proof

From (3.15), we have

$$d(Tx,Ty)\leq\lambda d(x,y)+L d(y,Tx), $$

where

$$\lambda=\frac{kb}{1-kb^{2}}\quad \mbox{and}\quad L=\frac {k(b^{2}+1)}{1-kb^{2}}\cdot $$

With the condition $k\in(0,1/2b^{2})$, we have $0<\lambda<1/b$ and $L\geq0$. Then T is a Berinde-type contraction. From Corollary 3.13, if $x_{0}\in X$, then $\{T^{n}x_{0}\}$ converges to a fixed point of T. □

If $b=1$, Corollary 3.15 recovers the Chatterjee fixed point theorem [23].

3.4 Ćirić-type mappings in b-metric spaces

Definition 3.16

Let $(X,d)$ be a b-metric space with constant $b\geq1$. A mapping $T: X\to X$ is said to be a Ćirić-type mapping if there exists $\lambda\in(0,1/b)$ such that for all $x,y\in X$, we have

$$ \min\bigl\{ d(Tx,Ty),d(x,Tx),d(y,Ty)\bigr\} -\min\bigl\{ d(x,Ty),d(y,Tx)\bigr\} \leq\lambda d(x,y). $$

(3.16)

Theorem 3.17

Let $(X,d)$ be a b-metric space with constant $b\geq1$, and let $T: X\to X$ be a given mapping. If T is a Ćirić-type mapping with parameter $\lambda\in(0,1/b)$, then there exist $\alpha: X\times X \to\mathbb{R}$ and $\psi\in\Psi _{b}$ such that T is an α-ψ contraction.

Proof

Consider the functions $\psi:[0,\infty)\to[0,\infty)$ and $\alpha:X\times X\to\mathbb{R}$ defined by

$$ \psi(t)=\lambda t,\quad t\geq0 $$

(3.17)

and

$$ \alpha(x,y)= \textstyle\begin{cases} \min \{1,\frac{d(x,Tx)}{d(Tx,Ty)},\frac{d(y,Ty)}{d(Tx,Ty)} \} -\min \{\frac{d(x,Ty)}{d(Tx,Ty)},\frac{d(y,Tx)}{d(Tx,Ty)} \} , &\text{if } Tx\neq Ty, \\ 0, &\text{otherwise}. \end{cases} $$

(3.18)

From (3.16), we have

$$ \alpha(x,y)d(Tx,Ty)\leq\psi\bigl(d(x,y)\bigr)\quad \text{for all } x,y\in X, $$

(3.19)

which implies that T is an α-ψ contraction. □

Corollary 3.18

Let $(X,d)$ be a complete b-metric space with constant $b\geq1$, and let $T: X\to X$ be a continuous mapping. If T is a Ćirić-type mapping with parameter $\lambda\in(0,1/b)$, then for any $x_{0}\in X$, the Picard sequence $\{T^{n}x_{0}\}$ converges to a fixed point of T.

Proof

Let $x_{0}\in X$ be an arbitrary point. Without loss of generality, we can suppose that $T^{r}x_{0}\neq T^{r+1}x_{0}$ for all $r\in\mathbb{N}$. From (3.18), for all $n\in\mathbb{N}$, we have

$$\begin{aligned} \alpha\bigl(T^{n}x_{0},T^{n+1}x_{0} \bigr) =& \min \biggl\{ 1,\frac{d(T^{n}x_{0},T^{n+1}x_{0})}{d(T^{n+1}x_{0},T^{n+2}x_{0})}, \frac{d(T^{n+1}x_{0},T^{n+2}x_{0})}{d(T^{n+1}x_{0},T^{n+2}x_{0})} \biggr\} \\ &{} -\min \biggl\{ \frac{d(T^{n}x_{0},T^{n+2}x_{0})}{d(T^{n+1}x_{0},T^{n+2}x_{0})}, \frac{d(T^{n+1}x_{0},T^{n+1}x_{0})}{d(T^{n+1}x_{0},T^{n+2}x_{0})} \biggr\} \\ =& \min \biggl\{ 1,\frac{d(T^{n}x_{0},T^{n+1}x_{0})}{d(T^{n+1}x_{0},T^{n+2}x_{0})} \biggr\} . \end{aligned}$$

Suppose that for some $n\in\mathbb{N}$, we have

$$\alpha\bigl(T^{n}x_{0},T^{n+1}x_{0} \bigr)=\frac{d(T^{n}x_{0},T^{n+1}x_{0})}{d(T^{n+1}x_{0},T^{n+2}x_{0})}. $$

In this case, from (3.17) and (3.19), we have

$$d\bigl(T^{n}x_{0},T^{n+1}x_{0}\bigr) \leq\lambda d\bigl(T^{n}x_{0},T^{n+1}x_{0} \bigr). $$

This implies (from the assumption $T^{r}x_{0}\neq T^{r+1}x_{0}$ for all $r\in\mathbb{N}$) that $\lambda\geq1$, which is a contradiction. Then

$$\alpha\bigl(T^{n}x_{0},T^{n+1}x_{0} \bigr)=1 \quad \text{for all } n\in\mathbb{N}. $$

Then (2.1) is satisfied with $p=1$ and $\sigma=1$. By Theorem 2.3, we deduce that the sequence $\{T^{n}x_{0}\}$ converges to a fixed point of T. □

If $b=1$, Corollary 3.18 recovers Ćirić’s fixed point theorem [24].

3.5 Edelstein fixed point theorem in b-metric spaces

Another consequence of our main results is the following generalized version of Edelstein fixed point theorem [25] in b-metric spaces.

Corollary 3.19

Let $(X,d)$ be a complete b-metric space with constant $b\geq1$, and ε-chainable for some $\varepsilon>0$; i.e., given $x,y\in X$, there exist a positive integer N and a sequence $\{x_{i}\}_{i=0}^{N}\subset X$ such that

$$ x_{0}=x,\qquad x_{N}=y,\qquad d(x_{i},x_{i+1})< \varepsilon \quad \textit{for } i=0, \ldots,N-1. $$

(3.20)

Let $T: X\to X$ be a given mapping such that

$$ x,y\in X, \quad d(x,y)< \varepsilon\quad \Longrightarrow \quad d(Tx,Ty)\leq\psi\bigl(d(x,y)\bigr) $$

(3.21)

for some $\psi\in\Psi_{b}$. Then T has a unique fixed point.

Proof

It is clear from (3.21) that the mapping T is continuous. Now, consider the function $\alpha:X\times X\to\mathbb{R}$ defined by

$$ \alpha(x,y)= \textstyle\begin{cases} 1, &\text{if } d(x,y)< \varepsilon, \\ 0, &\text{otherwise}. \end{cases} $$

(3.22)

From (3.21), we have

$$\alpha(x,y)d(Tx,Ty)\leq\psi\bigl(d(x,y)\bigr) \quad \text{for all } x,y\in X. $$

Let $x_{0}\in X$. For $x=x_{0}$ and $y=Tx_{0}$, from (3.20) and (3.22), for some positive integer p, there exists a finite sequence $\{\xi_{i}\}_{i=0}^{p}\subset X$ such that

$$x_{0}=\xi_{0}, \qquad \xi_{p}=Tx_{0}, \qquad \alpha(\xi_{i},\xi_{i+1})\geq1\quad \text{for } i=0, \ldots,p-1. $$

Now, let $i\in\{0,\ldots,p-1\}$ be fixed. From (3.22) and (3.21), we have

$$\begin{aligned} \alpha(\xi_{i},\xi_{i+1})\geq1 &\quad \Longrightarrow\quad d(\xi_{i},\xi_{i+1})< \varepsilon \\ &\quad \Longrightarrow\quad d(T\xi_{i},T\xi_{i+1})\leq\psi \bigl(d(\xi_{i},\xi_{i+1})\bigr) \leq d(\xi_{i}, \xi_{i+1})< \varepsilon \\ &\quad \Longrightarrow\quad \alpha(T\xi_{i},T\xi_{i+1}) \geq1. \end{aligned}$$

Again,

$$\begin{aligned} \alpha(T\xi_{i},T\xi_{i+1})\geq1 &\quad \Longrightarrow\quad d(T\xi_{i},T\xi_{i+1})< \varepsilon \\ &\quad \Longrightarrow\quad d\bigl(T^{2}\xi_{i},T^{2} \xi_{i+1}\bigr)\leq\psi\bigl(d(T\xi_{i},T\xi _{i+1}) \bigr)\leq d(T\xi_{i},T\xi_{i+1})< \varepsilon \\ &\quad \Longrightarrow \quad \alpha\bigl(T^{2}\xi_{i},T^{2} \xi_{i+1}\bigr)\geq1. \end{aligned}$$

By induction, we obtain

$$\alpha\bigl(T^{n}\xi_{i},T^{n+1}\xi_{i+1} \bigr)\geq1 \quad \text{for all } n\in\mathbb{N}. $$

Then (2.1) is satisfied with $\sigma=1$. From Theorem 2.2, the sequence $\{T^{n}x_{0}\}$ converges to a fixed point of T. Using a similar argument, we can see that condition (ii) of Theorem 2.4 is satisfied, which implies that T has a unique fixed point. □

3.6 Contractive mapping theorems in b-metric spaces with a partial order

Let $(X,d)$ be a b-metric space with constant $b\geq1$, and let ⪯ be a partial order on X. We denote

$$\Delta=\bigl\{ (x,y)\in X\times X: x\preceq y \text{ or } y\preceq x\bigr\} . $$

Corollary 3.20

Let $T: X\to X$ be a given mapping. Suppose that there exists $\psi\in\Psi_{b}$ such that

$$ d(Tx,Ty)\leq\psi\bigl(d(x,y)\bigr) \quad \textit{for all } (x,y) \in\Delta. $$

(3.23)

Suppose also that

• (i) T is continuous;

• (ii) for some positive integer p, there exists a finite sequence $\{\xi_{i}\}_{i=0}^{p}\subset X$ such that

  $$ \xi_{0}=x_{0},\qquad \xi_{p}=Tx_{0}, \qquad \bigl(T^{n}\xi_{i},T^{n}\xi_{i+1} \bigr)\in\Delta, \quad n\in\mathbb{N}, i=0,\ldots,p-1. $$

  (3.24)

Then $\{T^{n}x_{0}\}$ converges to a fixed point of T.

Proof

Consider the function $\alpha:X\times X\to\mathbb {R}$ defined by

$$ \alpha(x,y)= \textstyle\begin{cases} 1, &\text{if } (x,y)\in\Delta, \\ 0, &\text{otherwise}. \end{cases} $$

(3.25)

From (3.23), we have

$$\alpha(x,y)d(Tx,Ty)\leq\psi\bigl(d(x,y)\bigr) \quad \text{for all } x,y\in X. $$

Then the result follows from Theorem 2.2 with $\sigma=1$. □

Corollary 3.21

Let $T: X\to X$ be a given mapping. Suppose that

• (i) there exists $\psi\in\Psi_{b}$ such that (3.23) holds;

• (ii) condition (3.24) holds.

Then $\{T^{n}x_{0}\}$ converges to some $x^{*}\in X$. Moreover, if

• (iii) there exist a subsequence $\{T^{\gamma(n)}x_{0}\}$ of $\{ T^{n}x_{0}\}$ and $N\in\mathbb{N}$ such that

  $$\bigl(T^{\gamma(n)}x_{0},x^{*}\bigr)\in\Delta, \quad n\geq N, $$

then $x^{*}$ is a fixed point of T.

Proof

We continue to use the same function α defined by (3.25). From the first part of Theorem 2.3, the sequence $\{T^{n}x_{0}\}$ converges to some $x^{*}\in X$. From (iii) and (3.25), we have

$$\alpha\bigl(T^{\gamma(n)}x_{0},x^{*}\bigr)=1, \quad n\geq N. $$

By the second part of Theorem 2.3 (with $\ell=1$), we deduce that $x^{*}$ is a fixed point of T. □

The next result follows from Theorem 2.4 with $\eta=1$.

Corollary 3.22

Let $T: X\to X$ be a given mapping. Suppose that

• (i) there exists $\psi\in\Psi_{b}$ such that (3.23) holds;

• (ii) $\operatorname{Fix}(T)\neq\emptyset$;

• (iii) for every pair $(x,y)\in\operatorname{Fix}(T) \times\operatorname{Fix}(T)$ with $x\neq y$, if $(x,y)\notin\Delta$, there exist a positive integer q and a finite sequence $\{\zeta_{i}(x,y)\}_{i=0}^{q}\subset X$ such that

  $$\zeta_{0}(x,y)=x,\qquad \zeta_{q}(x,y)=y,\qquad \bigl(T^{n}\zeta_{i}(x,y),T^{n} \zeta_{i+1}(x,y)\bigr)\in\Delta $$

  for $n\in\mathbb{N}$ and $i=0,\ldots,q-1$.

Then T has a unique fixed point.

Observe that in our results we do not suppose that T is monotone or T preserves order as it is supposed in many papers (see [26–28] and others).