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

Fixed Point Theory and Applications20152015:74

https://doi.org/10.1186/s13663-015-0325-2

• Received: 5 February 2015
• Accepted: 11 May 2015
• Published:

## 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.

## Keywords

• fixed points
• dqb-cyclic-Banach contraction
• dqb-cyclic-Kannan mapping
• b-metric spaces
• quasi-b-metric spaces
• b-metric-like spaces
• dislocated quasi-b-metric spaces

• 47H05
• 47H10
• 47J25

## 1 Introduction and preliminaries

Fixed point theory has been studied extensively, which can be seen from the works of many authors . Banach contraction principle was introduced in 1922 by Banach  as follows:
1. (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  as follows:
1. (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.  introduced cyclic contraction as follows:
1. (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  introduced Kannan type cyclic contraction as follows:
1. (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 . Next, we discuss the development of spaces. The concept of quasi-metric spaces was introduced by Wilson  in 1931 as a generalization of metric spaces, and in 2000 Hitzler and Seda  introduced dislocated metric spaces as a generalization of metric spaces,  generalized the result of Hitzler, Seda and Wilson and introduced the concept of dislocated quasi-metric space. Włodarczyk et al. (see ) created uniform spaces as this is the concept of metric spaces. In 1989, Bakhtin  introduced b-metric space as a generalization of metric space. Moreover, Czerwik  made the results of Bakhtin known more in 1998. Finally, many other generalized b-metric spaces such as quasi-b-metric spaces , b-metric-like spaces  and quasi-b-metric-like spaces  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 , 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  introduced the concept of b-metric spaces and investigated some fixed point theorems in such spaces.

### Definition 1.3



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  introduced the concept of quasi-b-metric spaces and verified some fixed point theorems in quasi-b-metric spaces.

### Definition 1.5



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. .

### Definition 1.7



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:
1. (d1)

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

2. (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



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. .

### Definition 2.7

1. (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. (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. (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.

## Declarations

### Acknowledgements

The authors would like to thank Science Achievement Scholarship of Thailand which provides funding for the research. 