Open Access

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

Fixed Point Theory and Applications20152015:92

https://doi.org/10.1186/s13663-015-0344-z

Received: 11 March 2015

Accepted: 3 June 2015

Published: 19 June 2015

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.

Keywords

α-ψ-type contraction b-metric space fixed point existence uniqueness

MSC

47H10

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 [25] 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
  1. (i)

    \(d(x,y)=0\) if and only if \(x=y\);

     
  2. (ii)

    \(d(x,y)=d(y,x)\);

     
  3. (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 [68]) 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 [912] 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
  1. (i)

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

     
  2. (ii)

    ψ is continuous at zero;

     
  3. (iii)

    \(\psi(t)< t\) for any \(t >0\);

     
  4. (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
  1. (i)

    ψ is monotone increasing;

     
  2. (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
  1. (i)

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

     
  2. (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
  1. (i)

    \(\operatorname{Fix}(T)\neq\emptyset\);

     
  2. (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
  1. (i)

    T is continuous;

     
  2. (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
  1. (i)

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

     
  2. (ii)

    condition (3.24) holds.

     
Then \(\{T^{n}x_{0}\}\) converges to some \(x^{*}\in X\). Moreover, if
  1. (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
  1. (i)

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

     
  2. (ii)

    \(\operatorname{Fix}(T)\neq\emptyset\);

     
  3. (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 [2628] and others).

Declarations

Acknowledgements

This project was funded by the National Plan for Science, Technology and Innovation (MAARIFAH), King Abdulaziz City for Science and Technology, Kingdom of Saudi Arabia, Award Number (12-MAT 2895-02).

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematics, College of Science, King Saud University

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