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# 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:

## 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

• 47H10

## 1 Introduction and preliminaries

Very recently, we studied in  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  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  to self-operators defined in a b-metric space.

We start by recalling the following definition.

### Definition 1.1

()

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 ) 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  and the references therein.

### Definition 1.2

()

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 .

### 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

()

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.

### 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

()

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 $$\psi(t)=kt$$. □

### 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 .

### 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 .

### 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 .

Note that a Berinde mapping need not have a unique fixed point (see , 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 .

### 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 .

### 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 .

### 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  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  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, P.O. Box 2455, Riyadh, 11451, Saudi Arabia

## References

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