Dislocated quasi-b-metric spaces and fixed point theorems for cyclic contractions

Abstract

In this paper, we establish dislocated quasi-b-metric spaces and introduce the notions of Geraghty type dqb-cyclic-Banach contraction and dqb-cyclic-Kannan mapping and derive the existence of fixed point theorems for such spaces. Our main theorem extends and unifies existing results in the recent literature.

1 Introduction and preliminaries

Fixed point theory has been studied extensively, which can be seen from the works of many authors [1–6]. Banach contraction principle was introduced in 1922 by Banach [7] as follows:

• (i) Let $(X,d)$ be a metric space and let $T:X\to X$. Then T is called a Banach contraction mapping if there exists $k\in[0, 1)$ such that

  $$d(Tx,Ty)\leq k d(x,y) $$

  for all $x, y \in X$.

The concept of Kannan mapping was introduced in 1969 by Kannan [8] as follows:

• (ii) T is called a Kannan mapping if there exists $r\in[0, \frac{1}{2})$ such that

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

  for all $x, y \in X$.

Now, we recall the definition of cyclic map. Let A and B be nonempty subsets of a metric space $(X,d)$ and $T: A\cup B\to A\cup B$. T is called a cyclic map iff $T(A)\subseteq B$ and $T(B)\subseteq A$.

In 2003, Kirk et al. [9] introduced cyclic contraction as follows:

• (iii) A cyclic map $T: A\cup B\to A\cup B$ is said to be a cyclic contraction if there exists $a\in[0, 1)$ such that

  $$d(Tx,Ty)\leq a d(x,y) $$

  for all $x\in A$ and $y\in B$.

In 2010, Karapinar and Erhan [10] introduced Kannan type cyclic contraction as follows:

• (iv) A cyclic map $T: A\cup B\to A\cup B$ is called a Kannan type cyclic contraction if there exists $b\in[0, \frac{1}{2})$ such that

  $$d(Tx, Ty) \leq bd(x,Tx) + b d(y, Ty) $$

  for all $x\in A$ and $y\in B$.

If $(X,d)$ is a complete metric space, at least one of (i), (ii), (iii) and (iv) holds, then it has a unique fixed point [7–10]. Next, we discuss the development of spaces. The concept of quasi-metric spaces was introduced by Wilson [11] in 1931 as a generalization of metric spaces, and in 2000 Hitzler and Seda [12] introduced dislocated metric spaces as a generalization of metric spaces, [13] generalized the result of Hitzler, Seda and Wilson and introduced the concept of dislocated quasi-metric space. Włodarczyk et al. (see [14–21]) created uniform spaces as this is the concept of metric spaces. In 1989, Bakhtin [22] introduced b-metric space as a generalization of metric space. Moreover, Czerwik [23] made the results of Bakhtin known more in 1998. Finally, many other generalized b-metric spaces such as quasi-b-metric spaces [24], b-metric-like spaces [25] and quasi-b-metric-like spaces [26] were introduced.

We begin with the following definition as a recall from [11, 12].

Definition 1.1

[7, 11, 12]

Let X be a nonempty set. Suppose that the mapping $d: X\times X\rightarrow [0,\infty)$ satisfies the following conditions:

(d₁):

$d(x,x)=0$ for all $x\in X$;

(d₂):

$d(x,y)=d(y,x)=0$ implies $x=y$ for all $x,y\in X$;

(d₃):

$d(x,y)=d(y,x)$ for all $x,y\in X$;

(d₄):

$d(x,y)\leq[d(x,z)+d(z,y)]$ for all $x,y,z\in X$.

If d satisfies conditions (d₁), (d₂) and (d₄), then d is called a quasi-metric on X. If d satisfies conditions (d₂), (d₃) and (d₄), then d is called a dislocated metric on X. If d satisfies conditions (d₁)-(d₄), then d is called a metric on X.

In 2005 the concept of dislocated quasi-metric spaces [13], which is a new generalization of quasi-b-metric spaces and dislocated b-metric spaces, was introduced. By Definition 1.1, if setting conditions (d₂) and (d₄) hold true, then d is called a dislocated quasi-metric on X.

Remark 1.2

It is obvious that metric spaces are quasi-metric spaces and dislocated metric spaces, but the converse is not true.

In 1989, Bakhtin [22] introduced the concept of b-metric spaces and investigated some fixed point theorems in such spaces.

Definition 1.3

[22]

Let X be a nonempty set. Suppose that the mapping $b: X\times X\rightarrow[0,\infty)$ such that the constant $s\geq1$ satisfies the following conditions:

(b₁):

$b(x,y)=b(y,x)=0 \Leftrightarrow x=y$ for all $x,y\in X$;

(b₂):

$b(x,y)=b(y,x)$ for all $x,y\in X$;

(b₃):

$b(x,y)\leq s[b(x,z)+b(z,y)]$ for all $x,y,z\in X$.

The pair $(X, b)$ is then called a b-metric space.

Remark 1.4

It is obvious that metric spaces are b-metric spaces, but conversely this is not true.

In 2012, Shah and Hussain [24] introduced the concept of quasi-b-metric spaces and verified some fixed point theorems in quasi-b-metric spaces.

Definition 1.5

[24]

Let X be a nonempty set. Suppose that the mapping $q: X\times X\rightarrow[0,\infty)$ such that constant $s\geq1$ satisfies the following conditions:

(q₁):

$q(x,y)=q(y,x)=0 \Leftrightarrow x=y$ for all $x,y\in X$;

(q₂):

$q(x,y)\leq s[q(x,z)+q(z,y)]$ for all $x,y,z\in X$.

The pair $(X, q)$ is then called a quasi-b-metric space.

Remark 1.6

It is obvious that b-metric spaces are quasi-b-metric spaces, but conversely this is not true.

Recently, the concept of b-metric-like spaces, which is a new generalization of metric-like spaces, was introduced by Alghamdi et al. [25].

Definition 1.7

[25]

Let X be a nonempty set. Suppose that the mapping $D: X\times X\rightarrow[0,\infty)$ such that constant $s\geq1$ satisfies the following conditions:

(D₁):

$D(x,y)=0 \Rightarrow x=y$ for all $x,y\in X$;

(D₂):

$D(x,y)=D(y,x)$ for all $x,y\in X$;

(D₃):

$D(x,y)\leq s[D(x,z)+D(z,y)]$ for all $x,y,z\in X$.

The pair $(X, D)$ is then called a b-metric-like space (or a dislocated b-metric space).

Remark 1.8

It is obvious that b-metric spaces are b-metric-like spaces, but conversely this is not true.

In this paper we introduce dislocated quasi-b-metric spaces which generalize quasi-b-metric spaces and b-metric-like spaces, and we introduce the notions of Geraghty type dqb-cyclic-Banach contraction and dqb-cyclic-Kannan mapping and derive the existence of fixed point theorems for such spaces. Our main theorems extend and unify existing results in the recent literature.

2 Main results

In this section, we begin with introducing the notion of dislocated quasi-b-metric space.

Definition 2.1

Let X be a nonempty set. Suppose that the mapping $d: X\times X\rightarrow[0,\infty)$ such that constant $s\geq 1$ satisfies the following conditions:

• (d1) $d(x,y)=d(y,x)=0$ implies $x=y$ for all $x,y\in X$;

• (d2) $d(x,y)\leq s[d(x,z)+d(z,y)]$ for all $x,y,z\in X$.

The pair $(X, d)$ is then called a dislocated quasi-b-metric space (or simply dqb-metric). The number s is called the coefficient of $(X, d)$.

Remark 2.2

It is obvious that b-metric spaces, quasi-b-metric spaces and b-metric-like spaces are dislocated quasi-b-metric spaces, but the converse is not true.

Example 2.3

Let $X=\mathbb{R}$ and let

$$d(x,y)=|x-y|^{2}+\frac{|x|}{n}+\frac{|y|}{m}, $$

where $n,m \in\mathbb{N}\setminus\{1\}$ with $n\neq m$.

Then $(X, d)$ is a dislocated quasi-b-metric space with the coefficient $s =2$, but since $d(1, 1) \neq 0$, we have $(X, b)$ is not a quasi-b-metric space, and since $d(1, 2) \neq d(2, 1)$, we have $(X, b)$ is not a b-metric-like space. And $(X, b)$ is not a dislocated quasi-metric space. Indeed, let $x,y,z\in X$. Suppose that $d(x,y)=0$.

Then

$$|x-y|^{2}+\frac{|x|}{n}+\frac{|y|}{m}=0. $$

It implies that $|x-y|^{2}=0$, and so $x=y$.

Next, consider

$$\begin{aligned} d(x,y)&=|x-y|^{2}+\frac{|x|}{n}+\frac{|y|}{m} \\ &\leq\bigl(\vert x-z\vert +|z-y|\bigr)^{2}+\frac{|x|}{n}+ \frac{|y|}{m} \\ &\leq |x-z|^{2}+2|x-z|\cdot|z-y|+|z-y|^{2}+ \frac{|x|}{n}+\frac {|y|}{m} \\ &\leq 2\bigl(|x-z|^{2}+|z-y|^{2}\bigr)+\frac{|x|}{n}+ \frac{|z|}{m}+\frac {|z|}{n}+\frac{|y|}{m} \\ &\leq s \bigl[d(x,z)+d(z,y)\bigr], \end{aligned}$$

where $s=2$,

$$\begin{aligned} d\biggl(\frac{1}{2}, \frac{1}{4}\biggr)&=\biggl\vert \frac{1}{2}- \frac {1}{4}\biggr\vert ^{2}+ \frac{|\frac{1}{2}|}{n}+\frac{|\frac{1}{4}|}{m} = \frac{1}{16}+\frac{1}{2n}+\frac{1}{4m} \\ &= \frac{324}{5\text{,}184}+\frac{3}{6n}+\frac{4}{12m} >\frac{180}{5\text{,}184}+\frac{5}{6n}+\frac{7}{12m} \\ &= \frac{1}{36}+\frac{1}{2n}+\frac{1}{3m} +\frac{1}{144}+ \frac {1}{3n}+\frac{1}{4m} \\ &=\biggl\vert \frac{1}{2}- \frac{1}{3}\biggr\vert ^{2}+\frac{|\frac{1}{2}|}{n}+\frac{|\frac {1}{3}|}{m}+\biggl|\frac{1}{3}- \frac{1}{4}\biggr|^{2}+\frac{|\frac{1}{3}|}{n}+\frac {|\frac{1}{4}|}{m} \\ &=d\biggl(\frac{1}{2}, \frac{1}{3}\biggr)+d\biggl( \frac{1}{3}, \frac{1}{4}\biggr), \end{aligned}$$

where $n,m>42$.

Example 2.4

[26]

Let $X = \{0, 1, 2\}$, and let $d: X\times X\to\mathbb{R}^{+}$ be defined by

$$ d(x,y)= \begin{cases} 2 ;& x=y=0, \\ \frac{1}{2};& x=0, y=1, \\ 2; &x=1, y=0, \\ \frac{1}{2}; &\mbox{otherwise}. \end{cases} $$

Then $(X, d)$ is a dislocated quasi-b-metric space with the coefficient $s =2$, but since $d(1, 1) \neq 0$, we have $(X, b)$ is not a quasi-b-metric space, and since $d(1, 2) \neq d(2, 1)$, we have $(X, b)$ is not a b-metric-like space. It is obvious that $(X, b)$ is not a dislocated quasi-metric space.

Example 2.5

Let $X=\mathbb{R}$ and let

$$d(x,y)=|x-y|^{2}+3|x|^{2}+2|y|^{2}. $$

Then $(X, d)$ is a dislocated quasi-b-metric space with the coefficient $s =2$, but since $d(0, 1) \neq d(1, 0)$, we have $(X, b)$ is not a b-metric-like space, since $d(1, 1) \neq 0$, we have $(X, b)$ is not a quasi-b-metric space. It is obvious that $(X, b)$ is not a dislocated quasi-metric space.

Example 2.6

Let $X=\mathbb{R}$ and let

$$d(x,y)=|2x-y|^{2}+|2x+y|^{2}. $$

Then $(X, d)$ is a dislocated quasi-b-metric space with the coefficient $s =2$, but since $d(1, 1) \neq 0$, we have $(X, b)$ is not a quasi-b-metric space. It is obvious that $(X, b)$ is not a dislocated quasi-metric space.

We will introduce a dislocated quasi-b-convergent sequence, a Cauchy sequence and a complete space according to Zoto et al. [27].

Definition 2.7

• (1) A sequence $(\{x_{n}\})$ in a dqb-metric space $(X,d)$ dislocated quasi-b-converges (for short, dqb-converges) to $x\in X$ if

  $$\lim_{n\to\infty}d(x_{n},x)=0=\lim_{n\to\infty}d(x,x_{n}). $$

  In this case x is called a dqb-limit of $(\{x_{n}\})$, and we write $(x_{n}\to x)$.

• (2) A sequence $(\{x_{n}\})$ in a dqb-metric space $(X,d)$ is called Cauchy if

  $$\lim_{n,m\to\infty}d(x_{n},x_{m})=0=\lim _{n,m\to\infty}d(x_{m},x_{n}). $$

• (3) A dqb-metric space $(X,d)$ is complete if every Cauchy sequence in it is dqb-convergent in X.

Next, we begin with introducing the concept of a dqb-cyclic-Banach contraction.

Definition 2.8

Let A and B be nonempty subsets of a dislocated quasi-b-metric space $(X,d)$. A cyclic map $T: A\cup B\to A\cup B$ is said to be a dqb-cyclic-Banach contraction if there exists $k\in [0, 1)$ such that

$$ d(Tx,Ty)\leq kd(x,y) $$

(2.1)

for all $x\in A$, $y\in B$ and $s\geq1$ and $sk \leq1$.

Now we prove our main results.

Theorem 2.9

Let A and B be nonempty subsets of a complete dislocated quasi-b-metric space $(X,d)$. Let T be a cyclic mapping that satisfies the condition of a dqb-cyclic-Banach contraction. Then T has a unique fixed point in $A\cap B$.

Proof

Let $x\in A (\mathrm{fix})$ and, using the contractive condition of the theorem, we have

$$\begin{aligned} d\bigl(T^{2}x,Tx\bigr)&=d\bigl(T(Tx),Tx\bigr) \\ &\leq kd(Tx,x) \end{aligned}$$

and

$$\begin{aligned} d\bigl(Tx,T^{2}x\bigr)&=d\bigl(Tx,T(Tx)\bigr) \\ &\leq kd(x,Tx). \end{aligned}$$

So,

$$ d\bigl(T^{2}x,Tx\bigr)\leq k\alpha $$

(2.2)

and

$$ d\bigl(Tx,T^{2}x\bigr)\leq k\alpha, $$

(2.3)

where $\alpha=\max\{d(Tx,x),d(x,Tx)\}$.

By using (2.2) and (2.3), we have $d(T^{3}x,T^{2}x)\leq k^{2}\alpha$, and $d(T^{2}x,T^{3}x)\leq k^{2}\alpha$.

For all $n\in\mathbb{N}$, we get

$$d\bigl(T^{n+1}x,T^{n}x\bigr)\leq k^{n}\alpha $$

and

$$d\bigl(T^{n}x,T^{n+1}x\bigr)\leq k^{n}\alpha. $$

Let $n,m\in\mathbb{N}$ with $m>n$, by using the triangular inequality, we have

$$\begin{aligned} d\bigl(T^{m}x,T^{n}x\bigr) \leq& s^{m-n}d \bigl(T^{m}x,T^{m-1}x\bigr)+s^{m-n-1}d \bigl(T^{m-1}x,T^{m-2}x\bigr)+\cdots+sd\bigl(T^{n+1}x,T^{n}x \bigr) \\ \leq& \bigl(s^{m-n}k^{m-1}+s^{m-n-1}k^{m-2}+s^{m-n-2}k^{m-3}+ \cdots+s^{2}k^{n+1}+sk^{n}\bigr)\alpha \\ =& \bigl((sk)^{m-n}k^{n-1}+(sk)^{m-n-1}k^{m-2}+(sk)^{m-n-2}k^{n-1}+ \cdots \\ &{}+(sk)^{2}k^{n-1}+(sk)k^{n-1}\bigr)\alpha \\ \leq& \bigl(k^{n-1}+k^{n-1}+k^{n-1}+ \cdots+k^{n-1}+k^{n-1}\bigr)\alpha \\ = &\bigl(k^{n-1}\bigr) (m-n+1)\alpha \\ \leq& \bigl(k^{n-1}\bigr)\xi\alpha \end{aligned}$$

for some $\xi>m-n+1$.

Take $n\to\infty$, we get $d(T^{m}x,T^{n}x)\to0$.

Similarly, let $n,m\in\mathbb{N}$ with $m>n$, by using the triangular inequality, we have

$$ d\bigl(T^{n}x,T^{m}x\bigr)= \bigl(k^{n-1}\bigr)\xi \alpha. $$

Take $n\to\infty$, we get $d(T^{n}x,T^{m}x)\to0$. Thus $T^{n}x$ is a Cauchy sequence.

Since $(X,d)$ is complete, we have $\{T^{n}x\}$ converges to some $z\in X$.

We note that $\{T^{2n}x\}$ is a sequence in A and $\{T^{2n-1}x\}$ is a sequence in B in a way that both sequences tend to the same limit z.

Since A and B are closed, we have $z\in A\cap B$, and then $A\cap B\neq{\emptyset}$.

Now, we will show that $Tz=z$.

By using (2.1), consider

$$\begin{aligned} d\bigl(T^{n}x,Tz\bigr)&=d\bigl(T\bigl(T^{n-1}x\bigr),Tz\bigr) \\ &\leq kd\bigl(T^{n-1}x,z\bigr) \\ &\leq d\bigl(T^{n-1}x,z\bigr). \end{aligned}$$

Taking limit as $n\to\infty$ in the above inequality, we have

$$d(z,Tz)\leq kd(z,Tz)\leq d(z,Tz). $$

And so $d(z,Tz)=kd(z,Tz)$, where $0\leq k<1$. This implies that $d(z,Tz)=0$.

Similarly, considering form (2.1), we get

$$\begin{aligned} d\bigl(Tz,T^{n}x\bigr)&=d\bigl(Tz,T\bigl(T^{n-1}x\bigr)\bigr) \\ &\leq kd\bigl(z,T^{n-1}x\bigr) \\ &\leq d\bigl(z,T^{n-1}x\bigr) . \end{aligned}$$

Taking limit as $n\to\infty$ in the above inequality, we have

$$d(Tz,z)\leq kd(Tz,z)\leq d(Tz,z). $$

And so $d(Tz,z)=kd(Tz,z)$, where $0\leq k<1$. This implies that $d(Tz,z)=0$.

Hence $d(z,Tz)=d(Tz,z)=0$, this implies that $Tz=z$, that is, z is a fixed point of T.

Finally, to prove the uniqueness of a fixed point, let $z^{*}\in X$ be another fixed point of T such that $Tz^{*}=z^{*}$.

Then we have

$$ d\bigl(z,z^{*}\bigr)=d\bigl(Tz,Tz^{*}\bigr)\leq kd\bigl(z,z^{*}\bigr) . $$

(2.4)

On the other hand,

$$ d\bigl(z^{*},z\bigr)=d\bigl(Tz^{*},Tz\bigr)\leq kd\bigl(z^{*},z\bigr) . $$

(2.5)

By forms (2.4) and (2.5), we obtain that $d(z,z^{*})=d(z^{*},z)=0$, this implies that $z^{*}=z$.

Therefore z is a unique fixed point of T. This completes the proof. □

Example 2.10

Let $X=[-1,1]$ and $T: A\cup B\to A\cup B$ defined by $Tx=\frac{-x}{5}$. Suppose that $A=[-1,0]$ and $B=[0,1]$. Define the function $d:X^{2}\to[0,\infty)$ by

$$d(x,y)=|x-y|^{2}+\frac{|x|}{10}+\frac{|y|}{11}. $$

We see that d is a dislocated quasi-b-metric on X.

Now, let $x\in A$. Then $-1\leq x\leq0$. So, $0\leq\frac{-x}{5}\leq \frac{1}{5}$. Thus, $Tx\in B$.

On the other hand, let $x\in B$. Then $0\leq x\leq1$. So, $\frac {-1}{5}\leq\frac{-x}{5}\leq0$. Thus, $Tx\in A$.

Hence the map T is cyclic on X because $T(A)\subset B$ and $T(B)\subset A$.

Next, we consider

$$\begin{aligned} d(Tx,Ty)&=|Tx-Ty|^{2}+3|Tx|+2|Ty| \\ &=\biggl\vert \frac{-x}{5}-\frac{-y}{5}\biggr\vert ^{2}+\frac{1}{10}\biggl\vert \frac{-x}{5}\biggr\vert + \frac {1}{11}\biggl\vert \frac{-y}{5}\biggr\vert \\ &=\frac{1}{25}|x-y|^{2}+\frac{1}{50}|x|+ \frac{2}{55}|y| \\ &\leq\frac{1}{5}\biggl[|x-y|^{2}+\frac{1}{10}|x|+ \frac{1}{11}|y|\biggr] \\ &\leq kd(x,y), \end{aligned}$$

so for $\frac{1}{5}\leq k<1$.

Thus T satisfies the dqb-cyclic-Banach contraction of Theorem 2.9 and 0 is the unique fixed point of T.

Finally, we begin with introducing the concept of dqb-cyclic-Kannan mapping.

Definition 2.11

Let A and B be nonempty subsets of a dislocated quasi-b-metric space $(X,d)$. A cyclic map $T: A\cup B\to A\cup B$ is called a dqb-cyclic-Kannan mapping if there exists $r\in[0, \frac{1}{2})$ such that

$$ d(Tx,Ty)\leq r\bigl(d(x,Tx)+d(x,Ty)\bigr) $$

(2.6)

for all $x\in A$, $y\in B$ and $s\geq1$ and $sr \leq\frac{1}{2}$.

In the next theorem, we will prove the fixed point theorem for a cyclic-Kannan mapping in a dislocated quasi-b-metric space.

Theorem 2.12

Let A and B be nonempty subsets of a complete dislocated quasi-b-metric space $(X,d)$. Let T be a cyclic mapping that satisfies the condition of a dqb-cyclic-Kannan mapping. Then T has a unique fixed point in $A\cap B$.

Proof

Let $x\in A (\mathrm{fix})$ and, using the contractive condition of the theorem, we have

$$\begin{aligned} d\bigl(Tx,T^{2}x\bigr)&=d\bigl(Tx,T(Tx)\bigr) \\ &\leq rd(x,Tx)+rd\bigl(Tx,T^{2}x\bigr), \end{aligned}$$

so

$$ d\bigl(Tx,T^{2}x\bigr)\leq\frac{r}{1-r}d(x,Tx). $$

(2.7)

And from (2.7) we have

$$\begin{aligned} d\bigl(T^{2}x,Tx\bigr)&=d\bigl(T(Tx),Tx\bigr) \\ &\leq rd\bigl(Tx,T^{2}x\bigr)+rd(x,Tx) \\ &\leq\frac{r}{1-r}d(x,Tx)+rd(x,Tx) \\ &\leq\biggl(\frac{r}{1-r}+\frac{r}{1-r}\biggr)d(x,Tx) \\ &\leq\frac{r}{1-r}2d(x,Tx), \end{aligned}$$

so

$$ d\bigl(Tx,T^{2}x\bigr)\leq\frac{r}{1-r}\beta, $$

(2.8)

where $\beta=2d(x,Tx)$.

By using (2.7) and (2.8), we have

$$d\bigl(T^{3}x,T^{2}x\bigr)\leq\biggl(\frac{r}{1-r} \biggr)^{2}\beta $$

and

$$d\bigl(T^{2}x,T^{3}x\bigr)\leq\biggl(\frac{r}{1-r} \biggr)^{2}\beta. $$

For all $n\in\mathbb{N}$, we get

$$d\bigl(T^{n+1}x,T^{n}x\bigr)\leq\biggl(\frac{r}{1-r} \biggr)^{n}\beta $$

and

$$d\bigl(T^{n}x,T^{n+1}x\bigr)\leq\biggl(\frac{r}{1-r} \biggr)^{n}\beta. $$

Let $n,m\in\mathbb{N}$ with $m>n$, by using the triangular inequality, we have

$$\begin{aligned} d\bigl(T^{m}x,T^{n}x\bigr) \leq& s^{m-n}d \bigl(T^{m}x,T^{m-1}x\bigr)+s^{m-n-1}d \bigl(T^{m-1}x,T^{m-2}x\bigr)+\cdots+sd\bigl(T^{n+1}x,T^{n}x \bigr) \\ \leq& \bigl(s^{m-n}k^{m-1}+s^{m-n-1}k^{m-2}+s^{m-n-2}k^{m-3}+ \cdots+s^{2}k^{n+1}+sk^{n}\bigr)\beta \\ \leq& \biggl(\biggl(\frac{r}{1-r}\biggr)^{n-1}+\biggl( \frac{r}{1-r}\biggr)^{n-1}+\biggl(\frac {r}{1-r} \biggr)^{n-1}+\cdots \\ &{}+\biggl(\frac{r}{1-r}\biggr)^{n-1}+ \biggl(\frac{r}{1-r}\biggr)^{n-1}\biggr)\beta \\ =& \biggl(\frac{r}{1-r}\biggr)^{n-1}(m-n+1)\beta \\ < & \biggl(\frac{r}{1-r}\biggr)^{n-1}\xi\beta \end{aligned}$$

for some $\xi>m-n+1$. Take $n\to\infty$, we get $d(T^{m}x,T^{n}x)\to0$.

Similarly, let $n,m\in\mathbb{N}$ with $m>n$, by using the triangular inequality, we have

$$ d\bigl(T^{n}x,T^{m}x\bigr) < \biggl(\frac{r}{1-r} \biggr)^{n-1}\xi\beta. $$

Take $n\to\infty$, we get $d(T^{n}x,T^{m}x)\to0$. Thus $T^{n}x$ is a Cauchy sequence.

Since $(X,d)$ is complete, we have $\{(T^{n}x)\}$ converges to some $z\in X$.

We note that $\{T^{2n}x\}$ is a sequence in A and $\{T^{2n-1}x\}$ is a sequence in B in a way that both sequences tend to the same limit z.

Since A and B are closed, we have $z\in A\cap B$, and then $A\cap B\neq{\emptyset}$.

Now, we will show that $Tz=z$.

By using (2.6), consider

$$\begin{aligned} d\bigl(T^{n}x,Tz\bigr)&=d\bigl(T\bigl(T^{n-1}x\bigr),Tz\bigr) \\ &\leq rd\bigl(T^{n-1}x,T^{n}x\bigr)+rd(z,Tz). \end{aligned}$$

Taking limit as $n\to\infty$ in the above inequality, we have

$$ d(z,Tz) \leq rd(z,Tz). $$

Since $0\leq r<\frac{1}{2}$, we have $d(z,Tz)=0$.

Similarly, considering form (2.6), we get

$$\begin{aligned} d\bigl(Tz,T^{n}x\bigr)&=d\bigl(Tz,T\bigl(T^{n-1}x\bigr)\bigr) \\ &\leq rd(z,Tz)+rd\bigl(T^{n-1}x,T^{n}x\bigr) . \end{aligned}$$

Taking limit as $n\to\infty$ in the above inequality, we have

$$d(Tz,z)\leq rd(z,Tz). $$

Since $d(z,Tz)=0$, we have $d(z,Tz)=0$.

Hence $d(z,Tz)=d(Tz,z)=0 \Rightarrow Tz=z$ and z is a fixed point of T.

Finally, to prove the uniqueness of a fixed point, let $z^{*}\in X$ be another fixed point of T such that $Tz^{*}=z^{*}$.

Then we have $d(z,z) =d(z^{*},z^{*})=0$, because by assumption

$$\begin{aligned} \begin{aligned}[b] d\bigl(z,z^{*}\bigr)&=d\bigl(Tz,Tz^{*}\bigr) \\ &\leq rd(z,Tz) +rd\bigl(z^{*},Tz^{*}\bigr) \\ &=rd(z,z) +rd\bigl(z^{*},z^{*}\bigr) \\ &=0. \end{aligned} \end{aligned}$$

(2.9)

On the other hand,

$$\begin{aligned} d\bigl(z^{*},z\bigr)&=d\bigl(Tz^{*},Tz\bigr) \\ &\leq r d\bigl(z^{*},Tz^{*}\bigr) +rd(z,Tz) \\ &= rd\bigl(z^{*},z^{*}\bigr) +rd(z,z) \\ &=0. \end{aligned}$$

(2.10)

By forms (2.9) and (2.10), we obtain that $d(z,z^{*})=d(z^{*},z)=0 \Rightarrow z^{*}=z$.

Therefore z is a unique fixed point of T. This completes the proof. □

Example 2.13

Let $X=[-1,1]$ and $T:X\to X$ defined by $Tx=\frac{-x}{7}$. Suppose that $A=[-1,0]$ and $B=[0,1]$. Define the function $d:X^{2}\to[0,\infty)$ by

$$d(x,y)=|x-y|^{2}+3|x|+2|y|. $$

We see that d is a dislocated quasi-b-metric on X.

Now, let $x\in A$. Then $-1\leq x\leq0$. So, $0\leq\frac{-x}{7}\leq\frac{1}{7}$. Thus, $Tx\in B$.

On the other hand, let $x\in B$. Then $0\leq x\leq1$. So, $\frac {-1}{7}\leq\frac{-x}{7}\leq0$. Thus, $Tx\in A$.

Hence the map T is cyclic on X because $T(A)\subset B$ and $T(B)\subset A$.

Next, we consider

$$\begin{aligned} d(Tx,Ty)&=|Tx-Ty|^{2}+3|Tx|+2|Ty| \\ &=\biggl\vert \frac{-x}{7}-\frac{-y}{7}\biggr\vert ^{2}+3\biggl\vert \frac{-x}{7}\biggr\vert +2\biggl\vert \frac{-y}{7}\biggr\vert \\ &=\frac{1}{49}|x-y|^{2}+\frac{3}{7}|x|+ \frac{2}{7}|y| \\ &\leq\frac{1}{49}\bigl(\vert x\vert +|y|\bigr)^{2}+ \frac{3}{7}|x|+\frac{2}{7}|y| \\ &\leq\frac{2}{49}|x|^{2}+\frac{2}{49}|y|^{2}+ \frac{3}{7}|x|+\frac {2}{7}|y| \\ &\leq\frac{2}{23}\biggl(\biggl[\frac{64}{49}|x|^{2}+ \frac{23}{7}|x|\biggr]+\biggl[\frac {64}{49}|y|^{2}+ \frac{23}{7}|y|\biggr] \biggr) \\ &=\frac{2}{23}\biggl(\biggl[\frac{64}{49}|x|^{2}+ \frac{23}{7}|x|\biggr]+\biggl[\frac {64}{49}|y|^{2}+ \frac{23}{7}|y| \biggr]\biggr) \\ &=\frac{2}{23}\biggl(\biggl[\biggl\vert x+\frac{1}{7}x\biggr\vert ^{2}+3|x|+2\biggl\vert \frac{1}{7}x\biggr\vert \biggr]+\biggl[\biggl\vert y+\frac {1}{7}y\biggr\vert ^{2}+3|y|+2\biggl\vert \frac{1}{7}y\biggr\vert \biggr] \biggr) \\ &=\frac{2}{23}\bigl(\bigl[|x-Tx|^{2}+3|x|+2|Tx|\bigr]+ \bigl[|y-Ty|^{2}+3|y|+2|Ty|\bigr]\bigr) \\ &= r\bigl(d(x,Tx)+d(y,Ty)\bigr), \end{aligned}$$

so for $\frac{2}{23}\leq r<\frac{1}{2}$.

Thus T satisfies the dqb-cyclic-Banach contraction of Theorem 2.12 and 0 is the unique fixed point of T.