Fixed point results in double controlled quasi metric type spaces

1.   Introduction and preliminaries 

The theory of fixed points takes an important place in the transition from classical analysis to modern analysis. One of the most remarkable works on fixed point of functions was done by Banach ^[9]. Various generalizations of Banach fixed point result were made by numerous mathematicians ^{[1,2,3,6,12,13,14,17,19,20,25,26]}. One of the generalizations of the metric space is the quasi metric space that was introduced by Wilson ^[30]. The commutativity condition does not hold in general in quasi metric spaces. Several authors used these concepts to prove some fixed point theorems, see ^[7,10,15,21]. On the other hand Bakhtin ^[8] and Czerwik ^[11] generalized the triangle inequality by multiplying the right hand side of triangle inequality in metric spaces by a parameter $b\geq 1$ and defined $b$-metric spaces, for more results, see ^[4,23,24]. Fixed point results for multivalued mappings generalizes the results for single-valued mappings. Many interesting results have been proved in the setting of multivalued mappings, for example, see ^[5,26,29]. Kamran et al. ^[12] introduced a new concept of generalized b metric spaces, named as extended $b$-metric spaces, see also ^[22]. They replaced the parameter $b\geq 1$ in triangle inequality by the control function $\theta :X\times X\rightarrow \lbrack 1, \infty).$ Mlaiki et al. ^[16] replaced the triangle inequality in $b$-metric spaces by using control function in a different style and introduced controlled metric type spaces. Abdeljawad et al. ^[1] generalized the idea of controlled metric type spaces and introduced double controlled metric type spaces. They replaced the control function $\xi (x, y)$ in triangle inequality by two functions $\xi (x, y)$ and $\top (x, y)$, see also ^[27,28]$.$ In this paper, the concept of double controlled quasi metric type spaces has been discussed. Fixed point results and several examples are established. First of all, we discuss the previous concepts that will be useful to understand the paper.

Now, we define double controlled metric type space.

Definition 1.1.^[1] Given non-comparable functions $\xi, \top :X\times X\rightarrow \lbrack 1, \infty)$. If $\triangle :X\times X\rightarrow \lbrack 0, \infty)$ satisfies:

(q1) $\triangle (\omega, v) = 0$ if and only if $\omega = v$,

(q2) $\triangle (\omega, v) = \triangle (v, \omega),$

(q3) $\triangle (\omega, v)\leq \xi (\omega, e)\triangle (\omega, e)+\top (e, v)\triangle (e, v),$

for all $\omega, v, e\in X$. Then, $\triangle$ is called a double controlled metric with the functions $\xi$, $\top$ and the pair $(X, \triangle)$ is called double controlled metric type space with the functions $\xi, \top$.

The classical result to obtain fixed point of a mapping in double controlled metric type space is given below.

Theorem 1.2. ^[1] Let $(X, \triangle)$ be a complete double controlled metric type space with the functions $\xi, \top$ $:X\times X$ $\rightarrow \lbrack 1, \infty)$ and let $T:X\rightarrow X$ be a given mapping. Suppose that there exists $k\in (0, 1)$ such that

For $\omega _{0}\in X,$ choose $\omega _{g} = T^{g}\omega _{0}.$ Assume that

In the addition, assume that, for every $\omega \in X$, we have

Then $T$ has a unique fixed point $\omega ^{\ast }\in X$.

Now, we are introducing the concept of double controlled quasi metric type space and controlled quasi metric type space.

Definition 1.3. Given non-comparable functions $\xi, \top :X\times X\rightarrow \lbrack 1, \infty)$. If $\triangle :X\times X\rightarrow \lbrack 0, \infty)$ satisfies

(q1) $\triangle (\omega, v) = 0$ if and only if $\omega = v$,

(q2) $\triangle (\omega, v)\leq$ $\xi (\omega, e)\triangle (\omega, e)+\top (e, v)\triangle (e, v),$

for all $\omega, v, e\in X$. Then, $\triangle$ is called a double controlled quasi metric type with the functions $\xi$, $\top$ and $(X, \triangle)$ is called a double controlled quasi metric type space. If $\top (e, v) = \xi (e, v)$ then $(X, \triangle)$ is called a controlled quasi metric type space.

Remark 1.4. Any quasi metric space or any double controlled metric type space is also a double controlled quasi metric type space but, the converse is not true in general, see examples (1.5, 2.4, 2.12 and 2.15).

Example 1.5. Let $X = \left \{ 0, 1, 2\right \}.$ Define $\triangle :X\times X\rightarrow \lbrack 0$, $\infty)$ by $\triangle (0, 1) = 4,$ $\triangle (0, 2) = 1,$ $\triangle (1, 0) = 3 = \triangle \left(1, 2\right),$ $\triangle (2, 0) = 0,$ $\triangle (2, 1) = 2,$ $\triangle (0, 0) = \triangle (1, 1) = \triangle (2, 2) = 0.$

Define $\xi, \top :G\times G\rightarrow \lbrack 1$, $\infty)$ as $\xi (0, 1) =$ $\xi (1, 0) =$ $\xi \left(1, 2\right) = 1,$ $\xi (0, 2) = \frac{5}{4},$ $\xi (2, 0) = \frac{10}{9},$ $\xi (2, 1) = \frac{20}{19},$ $\xi (0, 0) =$ $\xi (1, 1) =$ $\xi (2, 2) = 1,$ $\top (0, 1) = \top (1, 0) = \top (0, 2) = \top \left(1, 2\right) = 1,$ $\top (2, 0) = \frac{3}{2},$ $\top (2, 1) = \frac{11}{8},$ $\top (0, 0) = \top (1, 1) = \top (2, 2) = 1.$ It is clear that $\triangle$ is double controlled quasi metric type with the functions $\xi$, $\top.$ Let $w = 0,$ $e = 2,$ $v = 1,$ we have

So $\triangle$ is not a quasi metric. Also, it is not a controlled quasi metric type. Indeed,

Moreover, it is not double controlled metric type space because, we have

The convergence of a sequence in double controlled quasi metric type space is defined as:

Definition 1.6. Let $(X, \triangle)$ be a double controlled quasi metric type space with two functions. A sequence $\{u_{t}\}$ is convergent to some $u$ in $X$ if and only if $\lim_{t\rightarrow \infty }\triangle (u_{t}, u) = \lim_{t\rightarrow \infty }\triangle (u, u_{t}) = 0$.

Now, we discuss left Cauchy, right Cauchy and dual Cauchy sequences in double controlled quasi metric type space.

Definition 1.7. Let $(X, \triangle)$ be a double controlled quasi metric type space with two functions.

(i) The sequence $\{u_{t}\}$ is a left Cauchy if and only if for every $\varepsilon > 0$ such that $\triangle (u_{m}, u_{t}) < \varepsilon$, for all $t > m > t_{\varepsilon }$, where $t_{\varepsilon }$ is some integer or $\lim_{t, m\rightarrow \infty }\triangle (u_{m}, u_{t}) = 0$.

(ii) The sequence $\{u_{t}\}$ is a right Cauchy if and only if for every $\varepsilon > 0$ such that $\triangle (u_{m}, u_{t}) < \varepsilon$, for all $m > t > t_{\varepsilon }$, where $t_{\varepsilon }$ is some integer.

(iii) The sequence $\{u_{t}\}$ is a dual Cauchy if and only if it is both left as well as right Cauchy.

Now, we define left complete, right complete and dual complete double controlled quasi metric type spaces.

Definition 1.8. Let $(X, \triangle)$ be a double controlled quasi metric type space. Then $(X, \triangle)$ is left complete, right complete and dual complete if and only if each left-Cauchy, right Cauchy and dual Cauchy sequence in $X$ is convergent respectively.

Note that every dual complete double controlled quasi metric type space is left complete but the converse is not true in general, so it is better to prove results in left complete double controlled quasi metric type space instead of dual complete.

Best approximation in a set and proximinal set are defined as:

Definition 1.9. Let $({\Im }, \triangle)$ be a double controlled quasi metric type space. Let $A$ be a non-empty set and $l\in \Im.$ An element $y_{0}\in A$ is called a best approximation in $A$ if

If each $l\in \Im$ has a best approximation in $A,$ then $A$ is know as proximinal set. $P(\Im)$ is equal to the set of all proximinal subsets of $\Im.$

Double controlled Hausdorff quasi metric type space is defined as:

Definition 1.10. The function $H_{_{\triangle }}:P(E)\times P(E)\rightarrow \lbrack 0, \infty),$ defined by

is called double controlled quasi Hausdorff metric type on $P(E).$ Also $(P(E), H_{_{\triangle }})$ is known as double controlled Hausdorff quasi metric type space.

The following lemma plays an important role in the proof of our main result.

Lemma 1.11. Let $\left(X, \triangle \right)$ be a double controlled quasi metric type space. Let $(P(E), H_{_{\triangle }})$ be a double controlled Hausdorff quasi metric type space on $P(E).$ Then, for all $C, F\in P(E)$ and for each $c\in C,$ there exists $f_{c}\in F,$ such that $H_{_{\triangle }}(C, F)\geq \triangle (c, f_{c})$ and $H_{\triangle }(F, C)\geq \triangle (f_{c}, c)$.

2.   Main results 

Let $(X, \triangle)$ be a double controlled quasi metric type space$,$ $u_{0}\in X$ and $T:X\rightarrow P(X)$ be multifunctions on $X$. Let $u_{1}\in Tu_{0}$ be an element such that $\triangle (u_{0}, {T} u_{0}) = \triangle (u_{0}, u_{1}),$ $\triangle ({T}u_{0}, u_{0}) = \triangle (u_{1}, u_{0})$. Let $u_{2}\in Tu_{1}$ be such that $\triangle (u_{1}, Tu_{1}) = \triangle (u_{1}, u_{2}),$ $\triangle (Tu_{1}, u_{1}) = \triangle (u_{2}, u_{1}).$ Let $u_{3}\in {T}u_{2}$ be such that $\triangle (u_{2}, {T} u_{2}) = \triangle (u_{2}, u_{3})$ and so on. Thus, we construct a sequence $u_{\text{t}}\ $of points in $X$ such that $u_{\text{t}+1}\in {T}u_{\text{t}}$ with $\triangle (u_{\text{t}}, {T}u_{\text{t}}) = \triangle (u_{\text{t}}, u_{ \text{t}+1}),$ $\triangle ({T}u_{\text{t}}, u_{\text{t}}) = \triangle (u_{\text{ t}+1}, u_{\text{t}}),$ where $t = 0, 1, 2, \cdots.$ We denote this iterative sequence by $\{XT(u_{\text{t}})\}.$ We say that $\{XT(u_{\text{t}})\}$ is a sequence in $X$ generated by $u_{0}$ under double controlled quasi metric $\triangle.$ If $\triangle$ is quasi $b$-metric then, we say that $\{XT(u_{ \text{t}})\}$ is a sequence in $X$ generated by $u_{0}$ under quasi $b$ -metric $\triangle$. We can define $\{XT(u_{\text{t}})\}$ in other metrics in a similar way.

Now, we define double controlled rational contracion which is a generalization of many other classical contractions.

Definition 2.1. Let $(X, \triangle)$ be a complete double controlled quasi-metric type space with the functions $\alpha, \mu$ $:X\times X$ $\rightarrow \lbrack 1, \infty)$. A multivalued mapping $T:X\rightarrow P\left(X\right)$ is called a double controlled rational contracion if the following conditions are satisfied:

for all $x, y\in X,$ $0 < k < 1$ and

Also$,$ for the terms of the sequence $\{XT(u_{\text{t}})\},$ we have

and for every $u\in X$

Now, we prove that an operator $T$ satisfying certain rational contraction condition has a fixed point in double controlled quasi metric type space.

Theorem 2.2. Let $(X, \triangle)$ be a left complete double controlled quasi-metric type space with the functions $\alpha, \mu :X\times X\rightarrow \lbrack 1, \infty)$ and $T:X\rightarrow P(X)$ be a double controlled rational contracion. Then, $T$ has a fixed point $u^{\ast }\in X$.

Proof. By Lemma 1.11 and using inequality (2.1), we have

Therefore,

Now,

Therefore,

Using (2.5) in (2.4), we have

Continuing in this way, we obtain

Now, by using (2.6) and by using the technique given in ^[1], it can easily be proved that $\left \{ u_{\text{t}}\right \}$ is a left Cauchy sequence. So, for all natural numbers with $t < m,$ we have

Since $(X, \triangle)$ is a left complete double controlled quasi metric type space, there exists some $u^{\ast }\in X$ such that

By using triangle inequality and then (2.1), we have

Using (2.3), (2.7) and (2.8), we get $\triangle (u^{\ast }, Tu^{\ast })\leq 0$. That is, $u^{\ast }\in Tu^{\ast }.$ Thus $u^{\ast }$ is a fixed point of $T$.

If we take single-valued mapping instead of multivalued mapping, then we obtain the following result.

Theorem 2.3. Let $(X, \triangle)$ be a left complete double controlled quasi-metric type space with the functions $\alpha, \mu :X\times X\rightarrow \lbrack 1, \infty)$ and $T:X\rightarrow X$ be a mapping such that:

for all $x, y\in X,$ $0 < k < 1$ and

Suppose that, for every $u\in X$ and for the Picard sequence $\left \{ u_{ \text{t}}\right \}$

Then, $T$ has a fixed point $u^{\ast }\in X$.

We present the following example to illustrate Theorem 2.3.

Example 2.4. Let $X = \left \{ 0, 1, 2, 3\right \}.$ Define $\triangle :X\times X\rightarrow \lbrack 0$, $\infty)$ by: $\triangle (0, 1) = 1,$ $\triangle (0, 2) = 4,$ $\triangle (0, 3) = 5, \ \triangle (1, 0) = 0,$ $\triangle \left(1, 2\right) = 10,$ $\triangle (1, 3) = 1, \ \triangle (2, 0) = 7,$ $\triangle (2, 1) = 3,$ $\triangle (2, 3) = 5, \ \triangle (3, 0) = 3, \ \triangle (3, 1) = 6, \ \triangle (3, 2) = 2, \ \triangle (0, 0) = \triangle (1, 1) = \triangle (2, 2) = \triangle (3, 3) = 0.$ Define $\alpha, \mu :X\times X\rightarrow \lbrack 1$, $\infty)$ as: $\alpha (0, 1) = 2,$ $\alpha \left(1, 2\right) =$ $\alpha (0, 2) =$ $\alpha (1, 0) = \alpha (2, 0) = \alpha (3, 1) = \alpha (2, 3) = \alpha (0, 3) = 1,$ $\alpha (1, 3) = 2, \ \alpha (2, 1) = \frac{7}{3},$ $\alpha (3, 0) = \frac{4}{3}, \ \alpha (3, 2) = \frac{3}{2}, \ \alpha (0, 0) = \alpha (1, 1) = \alpha (2, 2) = \alpha (3, 3) = 1,$ $\mu (1, 2) = \mu \left(2, 1\right) = \mu (2, 0) = \mu (3, 0) = \mu (0, 3) = 1,$ $\mu (1, 0) = \frac{3}{2},$ $\mu (0, 1) = 2, \ \mu (1, 3) = 3, \ \mu (3, 1) = 1, \ \mu (3, 2) = 4, \ \mu (2, 3) = 1, \ \mu (0, 2) = \frac{5}{2},$ $\mu (0, 0) = \mu (1, 1) = \mu (2, 2) = \mu (3, 3) = 1.$ Clearly $(X, \triangle)$ is a double controlled quasi metric type space$,$ but it is not a controlled quasi metric type space. Indeed,

Also, it is not a double controlled metric type space. Take $T0 = T1 = \left \{ 0\right \}$, $T2 = T3 = \left \{ 1\right \}$ and $k = \frac{1}{3}.$ We observe that

Let $u_{0} = 2,$ we have $u_{1} = Tu_{0} = T2 = 1,$ $u_{2} = Tu_{1} = 0,$ $u_{3} = Tu_{2} = 0, \cdots _{\cdot }$

Also, for every $u\in X$, we have

All hypotheses of Theorem 2.3 are satisfied and $u^{\ast } = 0$ is a fixed point.

As every quasi $b$-metric space is double controlled quasi metric type space but the converse is not true in general, so we obtain a new result in quasi $b$-metric space as a corollary of Theorem 2.3.

Theorem 2.5. Let $(X, \triangle)$ be a left complete quasi $b$ -metric space and $T:X\rightarrow X$ be a mapping. Suppose that there exists $k\in \left(0, 1\right)$ such that

whenever,

for all $x, y\in X$. Assume that $0 < bk < 1.$ Then, $T$ has a fixed point $u^{\ast }\in X.$

Remark 2.6. In the Example 2.3, note that $\triangle$ is quasi $b$ -metric with $b = \frac{10}{3},$ but we can not apply Theorem 2.5 for any $b = \frac{10}{3}$ and $k = \frac{1}{3}$, because $bk\nless 1.$

Quasi metric version of Theorem 2.2 is given below:

Theorem 2.7. Let $(X, \triangle)$ be a left complete quasi metric space and $T:X\rightarrow P(X)$ be a multivalued mapping. Suppose that there exists $0 < k < 1$ such that

for all $x, y\in X.$ Then $T$ has a fixed point $u^{\ast }\in X.$

Now, we extend the sequence $\{XT(u_{\text{t}})\}$ for two mappings. Let $({ X}, \triangle)$ be a double controlled quasi metric type space$,$ $u_{0}\in X$ and $S, T:X\rightarrow P(X)$ be the multivalued mappings on $X$. Let $u_{1}\in Su_{0}$ such that $\triangle (u_{0}, Su_{0}) = \triangle (u_{0}, u_{1})$ and $\triangle (Su_{0}, u_{0}) = \triangle (u_{1}, u_{0}).$ Now, for $u_{1}\in X$, there exist $u_{2}\in Tu_{1}$ such that $\triangle (u_{1}, Tu_{1}) = \triangle (u_{1}, u_{2})$ and $\triangle (Tu_{1}, u_{1}) = \triangle (u_{2}, u_{1}).$ Continuing this process, we construct a sequence $u_{\text{t}}\ $of points in $X$ such that $u_{2\text{t} +1}\in Su_{2\text{t}}$, and $u_{2\text{t}+2}\in Tu_{2\text{t}+1}$ with $\triangle (u_{2\text{t}}, Su_{2\text{t}}) = \triangle (u_{2\text{t}}, u_{2\text{t }+1})$, $\triangle (Su_{2\text{t}}, u_{2\text{t}}) = \triangle (u_{2\text{t} +1}, u_{2\text{t}})$ and $\triangle (u_{2\text{t}+1}, Tu_{2\text{t} +1}) = \triangle (u_{2\text{t}+1}, u_{2\text{t}+2})$, $\triangle (Tu_{2\text{t} +1}, u_{2\text{t}+1}) = \triangle (u_{2\text{t}+2}, u_{2\text{t}+1}).$ We denote this iterative sequence by $\{TS(u_{\text{t}})\}$ and say that $\{TS(u_{ \text{t}})\}$ is a sequence in $X$ generated by $u_{0}.$

Now, we introduce double controlled Reich type contraction.

Definition 2.8. Let $X$ be a non empty set, $(X, \triangle)$ be a left complete double controlled quasi-metric type space with the functions $\alpha, \mu :X\times X\rightarrow \lbrack 1, \infty)$ and $S, T:X\rightarrow P(X)$ be a multivalued mappings. Suppose that the following conditions are satisfied:

for each $x, {y}\in X,$ $0 < c+2k < 1.$ For $u_{0}\in X,$ choose $u_{\text{t} }\in \{TS(u_{\text{t}})\}$, we have

Then the pair $\left(S, T\right)$ is called a double controlled Reich type contraction.

The following results extend the results of Reich ^[18].

Theorem 2.9. Let $S, T:X\rightarrow P(X)$ be the multivalued mappings, $(X, \triangle)$ be a left complete double controlled quasi metric type space and $\left(S, T\right)$ be a pair of double controlled Reich type contraction. Suppose that, for all $u\in X$

Then, $S$ and $T$ have a common fixed point $\dot{z}$ in $X$.

Proof. Consider the sequence $\{TS(u_{\text{t}})\}$. Now, by Lemma 1.11, we have

By using the condition (2.10), we get

where $\eta = \frac{c+k}{1-k}.$ Now, by Lemma 1.11, we have

So, by using the condition (2.9), we have

Using (2.14) in (2.15), we have

Now, by Lemma 1.11 we have

Using the condition (2.10), we have

implies

From (2.16) and (2.17), we have

Using (2.18) in (2.14), we have

Continuing in this way, we get

Similarly, by Lemma 1.11, we have

Now$,$ we can write inequality (2.19) as

Now, by using (2.20) and by using the technique given in ^[1], it can easily be proved that $\left \{ u_{\text{t}}\right \}$ is a left Cauchy sequence. So, for all natural numbers with $t < m,$ we have

Since $(X, \triangle)$ is a left complete double controlled quasi metric type space. So $\{u_{\text{t}}\} \rightarrow \dot{z}\in X,$ that is

Now, we show that $\dot{z}$ is a common fixed point. We claim that $\triangle \left(\dot{z}, T\dot{z}\right) = 0$. On contrary suppose $\triangle (\dot{z}, {T}\dot{z}) > 0.$ Now by Lemma 1.11, we have

Taking $\underset{t\rightarrow \infty }{\lim }$ on both sides of inequality (2.23), we get

By using inequalities (2.21) and (2.22), we get

Now,

Taking $\underset{t\rightarrow \infty }{\lim }$ and by using inequalities (2.12), (2.22) and (2.24), we get

It is a contradiction, therefore

Thus, $\dot{z}\in T\dot{z}.$ Now, suppose $\triangle (\dot{z}, S\dot{z}) > 0.$ By Lemma 1.11, we have

By inequality (2.10), we get

Taking $\underset{t\rightarrow \infty }{\lim }$ on both sides of above inequality, we get

Now,

Taking $\underset{t\rightarrow \infty }{\lim }$ and by using inequality (2.12), (2.22) and (2.25), we get

It is a contradiction. Hence, $\dot{z}\in S\dot{z}.$ Thus, $\dot{z}$ is a common fixed point for $S$ and $T.$

Theorem 2.10 presents a result for single-valued mapping which is a consequence of the previous result.

Theorem 2.10. Let $(X, \triangle)$ be a left complete double controlled quasi-metric type space with the functions $\alpha, \mu :X\times X\rightarrow \lbrack 1, \infty)$ and $S, T:X\rightarrow X$ be the mappings such that:

and

for each $x, {y}\in X,$ $0 < c+2k < 1.$ Suppose that, for every $u\in X$ and for the Picard sequence $\left \{ u_{\text{t}}\right \}$

Then $S$ and $T$ have a common fixed point $u^{\ast }\in X$.

Quasi $b$-metric version of Theorem 2.10 is given below:

Theorem 2.11. Let $(X, \triangle)$ be a left complete quasi $b$-metric type space with the functions $\alpha, \mu :X\times X\rightarrow \lbrack 1, \infty)$ and $S, T:X\rightarrow X$ be the mappings such that:

and

for each $x, {y}\in X,$ $0 < c+2k < 1$ and $b < \frac{1-k}{c+k}$. Then $S$ and $T$ have a common fixed point $u^{\ast }\in X$.

The following example shows that how double controlled quasi metric type spaces can be used where the quasi $b$-metric spaces cannot.

Example 2.12. Let $X = \left \{ 0, \frac{1}{2}, \frac{1}{4}, 1\right \}.$ Define $\triangle :X\times X\rightarrow \lbrack 0$, $\infty)$ by $\triangle (0, \frac{1}{2}) = 1,$ $\triangle (0, \frac{1}{4}) = \frac{1}{3},$ $\triangle (\frac{1}{4}, 0) = \frac{1}{5}, \ \triangle (\frac{1}{2}, 0) = 1,$ $\triangle \left(\frac{1}{4}, \frac{1}{2}\right) = 3,$ $\triangle \left(\frac{ 1}{4}, 1\right) = \frac{1}{2},$ $\triangle \left(1, \frac{1}{4}\right) = \frac{ 1}{3}\ $and $\triangle (x, y) = \left \vert x-y\right \vert,$ if otherwise. Define $\alpha, \mu :X\times X\rightarrow \lbrack 1$, $\infty)$ as follows $\alpha \left(\frac{1}{2}, \frac{1}{4}\right) = \frac{16}{5},$ $\alpha (0, \frac{1}{4}) = \frac{3}{2}$, $\ \alpha (\frac{1}{4}, 1) = 3$, $\ \alpha (1, \frac{1}{ 4}) = \frac{12}{5}$ and $\alpha (x, y) = 1,$ if otherwise. $\mu (0, \frac{1}{2}) = \frac{14}{5}$, $\ \mu (1, \frac{1}{2}) = 3$ and$\ \mu (x, y) = 1,$ if otherwise$.$ Clearly $\triangle$ is double controlled quasi metric type for all $x, y, z\in X.$ Let, $T0 = \left \{ 0\right \},$ $T\frac{1}{2} = \left \{ \frac{1 }{4}\right \}, \ T\frac{1}{4} = \left \{ 0\right \},$ $T1 = \left \{ \frac{1}{4} \right \}$, $\ S0 = S\frac{1}{4} = \left \{ 0\right \}$, $S\frac{1}{2} = \left \{ \frac{1}{4}\right \},$ $S1 = \left \{ 0\right \}$ and $c = \frac{2}{5}$, $k = \frac{1}{4}.$ Now, if we take the case $x = \frac{1}{2}$, $y = \frac{1}{4},$ we have

$H_{\triangle }({S}\frac{1}{2}, T\frac{1}{4}) = H_{\triangle }\left(\left \{ \frac{1}{4}\right \}, \left \{ 0\right \} \right) = \triangle \left(\frac{1}{ 4}, 0\right) = \frac{1}{5}\leq \frac{17}{80} = c(\triangle (x, {y}))+k(\triangle (x, {S}x)+\triangle (y, Ty)).$ Also, satisfied for all cases $x, y\in X.$ That are inequalities (2.9) and (2.10) hold. Take $u_{0} = 1,$ then $u_{1} = Su_{0} = 0,$ $u_{2} = Tu_{1} = 0$, $u_{3} = Su_{2} = 0\cdots.$

which shows that inequality (2.11) holds. Thus the pair $\left(S, T\right)$ is double controlled Reich type contraction. Finally, for every $u\in X,$ we obtain

All hypotheses of Theorem 2.9 are satisfied and $\dot{z} = 0$ is a common fixed point.

Note that $\triangle$ is quasi $b$-metric with $b = 3,$ but Theorem 2.11 can not be applied because $b\nless \frac{1-k}{c+k},$ for all $b = 3.$ Therefore, this example shows that generalization from a quasi $b$-metric spaces to a double controlled quasi metric type spaces is real.

Taking $c = 0$ in Theorem 2.9, we get the following result of Kannan-type.

Theorem 2.13. Let $(X, \triangle)$ be a left complete double controlled quasi-metric type space with the functions $\alpha, \mu :X\times X\rightarrow \lbrack 1, \infty)$ and $S, T:X\rightarrow P(X)$ be the multivalued mappings such that:

and

for each $x, {y}\in X,$ $0 < c+2k < 1.$ Suppose that, for every $u\in X$ and for the sequence $\{TS(u_{\text{t}})\}$, we have

Then $S$ and $T$ have a common fixed point $u^{\ast }\in X$. Then, $S$ and $T$ have a common fixed point $\dot{z}$ in $X$.

Taking $c = 0$ and $S = T$ in Theorem 2.9, we get the following result.

Theorem 2.14. Let $(X, \triangle)$ be a left complete double controlled quasi-metric type space with the functions $\alpha, \mu :X\times X\rightarrow \lbrack 1, \infty)$ and $T:X\rightarrow P(X)$ be a multivalued mapping such that:

for each $x, {y}\in X,$ $0 < c+2k < 1.$ Suppose that, for every $u\in X$ and for the sequence $\{T(u_{\text{t}})\}$, we have

Then $T$ has a fixed point.

Now, the following example illustrates Theorem 2.14.

Example 2.15. Let $X = [0, 3).$ Define $\triangle :X\times X\rightarrow \lbrack 0$, $\infty)$ as

with

Clearly $\left(X, \triangle \right)$ is double controlled quasi metric type space. Choose $Tx = \left \{ \frac{x}{4}\right \}$ and $k = \frac{2}{5}$. It is clear that $T$ is Kannan type double controlled contraction. Also, for each $u$ $\in X$, we have

Thus, all hypotheses of Theorem 2.14 are satisfied and $\dot{z} = 0$ is the fixed point.

Quasi $b$-metric version of Theorem 2.14 is given below:

Theorem 2.16. Let $(X, \triangle)$ be a left complete quasi $b$ -metric space and $T:X\rightarrow P\left(X\right)$ be a mapping such that:

for all $x, y\in X,$ $k\in \lbrack 0, \frac{1}{2})$ and

Then $T$ has a fixed point $u^{\ast }\in X$.

The following remark compare, distinguish and relate the quasi $b$-metric with the double controlled quasi metric type spaces and illustrate the importance of double controlled quasi metric type spaces.

Remark 2.17. In the Example 2.15, $\triangle (x, y) = \left(x-y\right) ^{2}+x$ is a quasi $b$-metric with $b\geq 2,$ but we can not apply Theorem 2.16 because $T$ is not Kannan type $b$-contraction. Indeed $b\nless \frac{ 1-k}{k},$ for all $b\geq 2.$

3.   Conclusion and future work 

It has been shown that double controlled quasi metric is general and better than other metrics, like controlled quasi metric, controlled metric, extended quasi-b-metric, extended b-metric, quasi-b-metric and quasi metric. Also, left, right and dual completeness has been discussed. Results in dual complete spaces can be obtained as corollaries. These results may be extended to obtain results for other contractions. Double controlled quasi metric like ordered spaces can be introduced to establish new results. These results may be applied to find applications to random impulsive differential equations, dynamical systems, graph theory and integral equations.

Acknowledgments 

The authors would also like to thank the editor and the reviewers for their helpful comments and suggestions. This research was funded by the Center of Excellence in Theoretical and Computational Science (TaCS-CoE), KMUTT.

Conflict of interest 

The authors declare that they do not have any competing interests.