Common Fixed Point under Nonlinear Contractions on Quasi Metric Spaces

Abstract

We introduce in this article the notion of $(\psi ,\varphi )-$quasi contraction for a pair of functions on a quasi-metric space. We also investigate the existence and uniqueness of the fixed point for a couple functions under that contraction.

1. Introduction and Preliminary

Fixed point has been considered by many researchers since it was established by Banach [1] in 1992. The generalizations of the theory were considered by many researchers on various metric spaces (see, for example, [2,3,4,5,6,7]). Quasi-metric space was one of the interesting examples that were considered since it was introduced by Wilson [8] in 1931. We may suggest the following articles to the reader [8,9,10,11,12,13,14,15,16,17,18,19,20].

Definition 1.

[8] Let χ be a non-empty set and $\rho :\chi \times \chi \to [0,\infty )$ be a given function that satisfies the following conditions:

(1) 

$\rho (\alpha ,\beta )=0$ if and only if $\alpha =\beta$.

(2) 

$\rho (\alpha ,\beta )\le \rho (\alpha ,\gamma )+\rho (\gamma ,\beta )$ for all $\alpha ,\beta ,\gamma \in \chi .$

Then, ρ is called a quasi-metric on χ and the pair $(\chi ,\rho )$ is called a quasi-metric space.

Example 1.

Consider the set $\chi =[0,1]$ and define the function $\rho :\chi \times \chi \to [0,\infty )$ such that

$\rho (\alpha ,\beta )=\left\{\begin{array}{cc}{\alpha }^2-{\beta }^2& if\alpha \ge \beta \\ 1& Otherwise.\end{array}\right.$

Then, $(\chi ,\rho )$ is a quasi-metric space. To prove this, we need to verify the two conditions of Definition 1.

Condition 1.

If $\alpha =\beta$, then it is clear that $\rho (\alpha ,\beta )=0$. On the other hand, if $\rho (\alpha ,\beta )=0$ then we have $0={\alpha }^2-{\beta }^2=(\alpha -\beta )(\alpha +\beta )$. Since $\alpha ,\beta \in [0,1]$, we have $\alpha =\beta$.

Condition 2.

Let $\alpha ,\beta ,\gamma \in \chi$. Then, we have three cases:

Case I

If $\alpha >\beta$ and $\beta >\gamma$, then $\alpha >\gamma$ and hence $\rho (\alpha ,\beta )+\rho (\beta ,\gamma )=({\alpha }^2-{\beta }^2)+({\beta }^2-{\gamma }^2)={\alpha }^2-{\gamma }^2=\rho (\alpha ,\gamma )$.

Case II

If $\alpha >\beta$ and $\gamma >\beta$, then we have $\rho (\alpha ,\beta )+\rho (\beta ,\gamma )=({\alpha }^2-{\beta }^2)+1>\rho (\alpha ,\gamma )$. This is because $\rho (\alpha ,\gamma )\le 1$ for all $\alpha ,\gamma \in [0,1]$.

Case III

If $\beta >\alpha$, then using the same reason as in Case II, we have $\rho (\alpha ,\beta )+\rho (\beta ,\gamma )=1+\rho (\beta ,\gamma )>\rho (\alpha ,\gamma )$.

Therefore, $(\chi ,\rho )$ is a quasi-metric space. It is clear that $(\chi ,\rho )$ is not a metric space since $\rho (\alpha ,\beta )\ne \rho (\beta ,\alpha )$, for all $\alpha \ne \beta$.

Now, we introduce the definitions of convergence and Cauchy of such a sequence in quasi-metric spaces:

Definition 2.

[12,13] Let $(\chi ,\rho )$ be a quasi-metric space. A sequence $\left({\alpha }_n\right)$ in χ converges to the element $\alpha \in \chi$ if and only if

$${\displaystyle {\displaystyle \underset{n\to \infty }{lim}}\rho ({\alpha }_n,\alpha )={\displaystyle \underset{n\to \infty }{lim}}\rho (\alpha ,{\alpha }_n)=0.}$$

Definition 3.

[12,13] Let $(\chi ,\rho )$ be a quasi-metric space. A sequence $\left({\alpha }_n\right)$ in the space χ is said to be a Cauchy sequence if and if, for $ϵ>0$, there exists a positive integer $N=N\left(ϵ\right)$ such that $\rho ({\alpha }_n,{\alpha }_m)<ϵ$ for all $m,n>N.$

Moreover, if every Cauchy sequence in the quasi-metric space χ is convergent, then $(\chi ,\rho )$ is said to be complete.

The next notion was given by Khan et al. [21].

Definition 4.

[21] A self function ψ on $[0,\infty )$ is called an altering distance function if the following properties hold:

(1) 

ψ is non-decreasing and continuous.

(2) 

$\psi \left(e\right)=0$ if and only if $e=0$.

2. Main Result

Definition 5.

Let $(\chi ,\rho )$ be a quasi-metric space and $S_1,S_2$ be two self-mappings on χ. Then, the pair $(S_1,S_2)$ is said to be $(\psi ,\varphi )-$quasi contraction if there exist two alternating distance functions ψ and ϕ such that, for all $e,w\in \chi$, we have

$$\psi \left(\rho (S_{1}e,S_{2}w)\right)\le \psi \left(M_1(e,w)\right)-\varphi \left(M_1(e,w)\right)$$

and

$$\psi \left(\rho (S_{2}e,S_{1}w)\right)\le \psi \left(M_2(e,w)\right)-\varphi \left(M_2(e,w)\right)$$

where

$$M_1(e,w)=max\left\{\rho (w,S_{2}w)\frac{1+\rho (e,S_{1}e)}{1+\rho (e,w)},\rho (e,S_{1}e),\rho (w,S_{2}w)\right\}$$

and

$$M_2(e,w)=max\left\{\rho (w,S_{1}w)\frac{1+\rho (e,S_{2}e)}{1+\rho (e,w)},\rho (e,S_{2}e),\rho (w,S_{1}w)\right\}.$$

Now, we prove our first result:

Theorem 1.

Let $(\chi ,\rho )$ be a complete quasi-metric space. Let ψ and ϕ be alternating distance functions and $S_1,S_2$ be two self-mappings on χ such that the pair $(S_1,S_2)$ is $(\psi ,\varphi )-$quasi contraction. Then, $S_1$ and $S_2$ have a unique common fixed point.

Proof. 

We start the proof of the result by taking an element ${\tau }_0\in \chi$. We construct a sequence $\left({\tau }_n\right)$ in $\chi$ in the following way: ${\tau }_{2n+1}=S_1{\tau }_{2n}$ and ${\tau }_{2n+2}=S_2{\tau }_{2n+1}$ for all $n\ge 0$.

It is clear that if there exists $s\in \mathbb{N}$ with ${\tau }_{2s}={\tau }_{2s+1}$, then ${\tau }_{2s}$ is a fixed point of $S_1$. Since the pair $(S_1,S_2)$ is $(\psi ,\varphi )-$quasi contraction, we have

$$\begin{array}{cc}& \psi \left(\rho ({\tau }_{2s+1},{\tau }_{2s+2})\right)\hfill \\ =& \psi \left(\rho (S_1{\tau }_{2s},S_2{\tau }_{2s+1})\right)\hfill \\ \le & \psi \left(M_1({\tau }_{2s},{\tau }_{2s+1})\right)-\varphi \left(M_1({\tau }_{2s},{\tau }_{2s+1})\right)\hfill \\ =& \psi \left(max\{\rho ({\tau }_{2s+1},{\tau }_{2s+2}),\rho ({\tau }_{2s},{\tau }_{2s+1})\right\})\hfill \\ & -\varphi \left(max\left\{\rho ({\tau }_{2s+1},{\tau }_{2s+2}),\rho ({\tau }_{2s},{\tau }_{2s+1})\right\}\right)\hfill \\ =& \psi (\rho ({\tau }_{2s+1},{\tau }_{2s+2})-\varphi \left(\rho ({\tau }_{2s+1},{\tau }_{2s+2})\right).\hfill \end{array}$$

From the above inequality, we deduce that $\psi \left(\rho ({\tau }_{2s+1},{\tau }_{2s+2})\right)=0$. Since $\psi$ is an alternating function, we conclude that ${\tau }_{2s}$ is a fixed point of $S_1$ and $S_2$. Thus, ${\tau }_{2s}$ is a common fixed point of $S_1$ and $S_2$.

Using similar arguments as above, we may show that, if there exists $s\in \mathbb{N}$ such that ${\tau }_{2s+1}={\tau }_{2s+2}$, then ${\tau }_{2s+1}$ is a common fixed point of $S_1$ and $S_2$.

Now, we may assume that ${\tau }_n\ne {\tau }_{n+1}$ for all $n\in \mathbb{N}.$

In view of $(\psi ,\varphi )-$quasi contraction of the pair $(S_1,S_2)$, we deduce that

$$\begin{array}{cc}& \psi \left(\rho ({\tau }_{2n+1},{\tau }_{2n+2})\right)\hfill \\ =& \psi \left(\rho (S_1{\tau }_{2n},S_2{\tau }_{2n+1})\right)\hfill \\ \le & \psi \left(M_1({\tau }_{2n},{\tau }_{2n+1})\right)-\varphi \left(M_1({\tau }_{2n},{\tau }_{2n+1})\right)\hfill \\ =& \psi \left(max\left\{\rho ({\tau }_{2n+1},{\tau }_{2n+2})\frac{1+\rho ({\tau }_{2n},{\tau }_{2n+1})}{1+\rho ({\tau }_{2n},{\tau }_{2n+1})},\rho ({\tau }_{2n},{\tau }_{2n+1}),\rho ({\tau }_{2n+1},{\tau }_{2n+2})\right\}\right)\hfill \\ & -\varphi \left(max\left\{\rho ({\tau }_{2n+1},{\tau }_{2n+2})\frac{1+\rho ({\tau }_{2n},{\tau }_{2n+1})}{1+\rho ({\tau }_{2n},{\tau }_{2n+1})},\rho ({\tau }_{2n},{\tau }_{2n+1}),\rho ({\tau }_{2n+1},{\tau }_{2n+2})\right\}\right)\hfill \\ =& \psi (max\{\rho ({\tau }_{2n},{\tau }_{2n+1}),\rho ({\tau }_{2n+1},{\tau }_{2n+2})\})-\varphi (max\{\rho ({\tau }_{2n},{\tau }_{2n+1}),\rho ({\tau }_{2n+1},{\tau }_{2n+2})\}).\hfill \end{array}$$

(1)

Assume that

$$max\left\{\psi \left(\rho ({\tau }_{2n},{\tau }_{2n+1})\right),\psi \left(\rho ({\tau }_{2n+1},{\tau }_{2n+2})\right)\right\}=\psi \left(\rho ({\tau }_{2n+1},{\tau }_{2n+2})\right).$$

Then, Equation (1) implies

$$\psi \left(\rho ({\tau }_{2n+1},{\tau }_{2n+2})\right)\le \psi \left(\rho ({\tau }_{2n+1},{\tau }_{2n+2})\right)-\varphi \left(\rho ({\tau }_{2n+1},{\tau }_{2n+2})\right)$$

a contradiction. Thus,

$$max\left\{\psi \left(\rho ({\tau }_{2n},{\tau }_{2n+1})\right),\psi \left(\rho ({\tau }_{2n+1},{\tau }_{2n+2})\right)\right\}=\psi \left(\rho ({\tau }_{2n},{\tau }_{2n+1})\right).$$

Therefore, Equation (1) yields

$$\begin{array}{c}\hfill \psi \left(\rho ({\tau }_{2n+1},{\tau }_{2n+2})\right)\le \psi \left(\rho ({\tau }_{2n},{\tau }_{2n+1})\right)-\varphi \left(\rho ({\tau }_{2n},{\tau }_{2n+1})\right).\end{array}$$

(2)

On the other hand, we have

$$\begin{array}{cc}& \psi \left(\rho ({\tau }_{2n},{\tau }_{2n+1})\right)\hfill \\ =& \psi \left(\rho (S_2{\tau }_{2n-1},S_1{\tau }_{2n})\right)\hfill \\ \le & \psi \left(M_2({\tau }_{2n-1},{\tau }_{2n})\right)-\varphi \left(M_2({\tau }_{2n-1},{\tau }_{2n})\right)\hfill \\ =& \psi \left(max\left\{\rho ({\tau }_{2n},{\tau }_{2n+1})\frac{1+\rho ({\tau }_{2n-1},{\tau }_{2n})}{1+\rho ({\tau }_{2n-1},{\tau }_{2n})},\rho ({\tau }_{2n-1},{\tau }_{2n}),\rho ({\tau }_{2n},{\tau }_{2n+1})\right\}\right)\hfill \\ & -\varphi \left(max\left\{\rho ({\tau }_{2n},{\tau }_{2n+1})\frac{1+\rho ({\tau }_{2n-1},{\tau }_{2n})}{1+\rho ({\tau }_{2n-1},{\tau }_{2n})},\rho ({\tau }_{2n-1},{\tau }_{2n}),\rho ({\tau }_{2n},{\tau }_{2n+1})\right\}\right)\hfill \\ =& \psi (max\{\rho ({\tau }_{2n-1},{\tau }_{2n}),\rho ({\tau }_{2n},{\tau }_{2n+1})\})-\varphi (max\{\rho ({\tau }_{2n-1},{\tau }_{2n}),\rho ({\tau }_{2n},{\tau }_{2n+1})\}).\hfill \end{array}$$

(3)

From the last inequality, we get

$$max\{\rho ({\tau }_{2n-1},{\tau }_{2n}),\rho ({\tau }_{2n},{\tau }_{2n+1})\}=\rho ({\tau }_{2n-1},{\tau }_{2n}),$$

and hence

$$\begin{array}{c}\hfill \psi \left(\rho ({\tau }_{2n},{\tau }_{2n+1})\right)\le \psi \left(\rho ({\tau }_{2n-1},{\tau }_{2n})\right)-\varphi \left(\rho ({\tau }_{2n-1},{\tau }_{2n})\right).\end{array}$$

(4)

Combining Equations (2) and (4), we conclude that

$$\begin{array}{c}\hfill \psi \left(\rho ({\tau }_n,{\tau }_{n+1})\right)\le \psi \left(\rho ({\tau }_{n-1},{\tau }_n)\right)-\varphi \left(\rho ({\tau }_{n-1},{\tau }_n)\right)<\psi \left(\rho ({\tau }_{n-1},{\tau }_n)\right)\end{array}$$

(5)

holds for all $n\in \mathbb{N}$.

From Equation (5), we conclude that $\{\rho ({\tau }_{n-1},{\tau }_n):n=1,2,\dots \}$ is a decreasing sequence.

There exists $s\ge 0$ such that

$${\displaystyle \underset{n\to +\infty }{lim}}\rho ({\tau }_{n-1},{\tau }_n)=s.$$

By allowing n tends to $+\infty$ in Equation (5), we conclude that $s=0$ and hence

$$\begin{array}{c}\hfill {\displaystyle \underset{n\to +\infty }{lim}}\rho ({\tau }_{n-1},{\tau }_n)=0.\end{array}$$

(6)

Now, we prove that

$${\displaystyle \underset{n,m\to +\infty }{lim}}\rho ({\tau }_n,{\tau }_m)=0.$$

For two large integer numbers n and m with $m>n$, we discuss the following cases:

Case 1:$n=2l+1$ and $m=2r+2$ for some $l,r\in \mathbb{N}$; that is, n is odd and m is even. By the $(\varphi ,\psi )-$contraction of the pair $(S_1,S_2)$, we have

$$\begin{array}{cc}& \psi \left(\rho ({\tau }_n,{\tau }_m)\right)\hfill \\ =& \psi \left(\rho ({\tau }_{2l+1},{\tau }_{2r+2})\right)\hfill \\ =& \psi \left(\rho (S_1{\tau }_{2l},S_2{\tau }_{2r+1})\right)\hfill \\ \le & \psi \left(M_1({\tau }_{2l},{\tau }_{2r+1})\right)-\varphi \left(M_1({\tau }_{2l},{\tau }_{2r+1})\right)\hfill \\ =& \psi \left(max\left\{\rho ({\tau }_{2r+1},{\tau }_{2r+2})\frac{1+\rho ({\tau }_{2l},{\tau }_{2l+1})}{1+\rho ({\tau }_{2l},{\tau }_{2r+1})},\rho ({\tau }_{2l},{\tau }_{2l+1}),\rho ({\tau }_{2r+1},{\tau }_{2r+2})\right\}\right)\hfill \\ & -\varphi \left(max\left\{\rho ({\tau }_{2r+1},{\tau }_{2r+2})\frac{1+\rho ({\tau }_{2l},{\tau }_{2l+1})}{1+\rho ({\tau }_{2l},{\tau }_{2r+1})},\rho ({\tau }_{2l},{\tau }_{2l+1}),\rho ({\tau }_{2r+1},{\tau }_{2r+2})\right\}\right)\hfill \\ \le & \psi \left(max\left\{\rho ({\tau }_{2r+1},{\tau }_{2r+2})(1+\rho ({\tau }_{2l},{\tau }_{2l+1})),\rho ({\tau }_{2l},{\tau }_{2l+1}),\rho ({\tau }_{2r+1},{\tau }_{2r+2})\right\}\right)\hfill \\ & -\varphi \left(\rho ({\tau }_{2l},{\tau }_{2l+1})\right)\hfill \\ \le & \psi \left(max\left\{\rho ({\tau }_{2r+1},{\tau }_{2r+2})(1+\rho ({\tau }_{2l},{\tau }_{2l+1})),\rho ({\tau }_{2l},{\tau }_{2l+1}),\rho ({\tau }_{2r+1},{\tau }_{2r+2})\right\}\right)\hfill \\ \le & \psi \left(\rho ({\tau }_{2l+1},{\tau }_{2l+2})(1+\rho ({\tau }_{2l},{\tau }_{2l+1}))\right)\hfill \\ =& \psi \left(\rho ({\tau }_n,{\tau }_{n+1})(1+\rho ({\tau }_{n-1},{\tau }_n))\right)\hfill \\ \le & \psi \left(\rho ({\tau }_{n-1},{\tau }_n)(1+\rho ({\tau }_{n-1},{\tau }_n))\right).\hfill \end{array}$$

(7)

In view of Equation (7) and the nondecreasing property of the function $\psi$, we conclude that

$$\rho ({\tau }_n,{\tau }_m)\le \rho ({\tau }_{n-1},{\tau }_n)(1+\rho ({\tau }_{n-1},{\tau }_n))$$

Case 2: $n=2l$ and $m=2r+2$ for some $l,r\in \mathbb{N}$; that is, n and m are both even. Here, we have

$$\begin{array}{ccc}\hfill \rho ({\tau }_n,{\tau }_m)=\rho ({\tau }_{2l},{\tau }_{2r+2})& \le & \rho ({\tau }_{2l},{\tau }_{2l+1})+\rho ({\tau }_{2l+1},{\tau }_{2r+2})\hfill \\ & =& \rho ({\tau }_n,{\tau }_{n+1})+\rho ({\tau }_{n+1},{\tau }_m).\hfill \end{array}$$

From Case 1, we get

$$\begin{array}{ccc}\hfill \rho ({\tau }_n,{\tau }_m)& \le & \rho ({\tau }_n,{\tau }_{n+1})+\rho ({\tau }_n,{\tau }_{n+1})(1+\rho ({\tau }_n,{\tau }_{n+1}))\hfill \\ & \le & \rho ({\tau }_{n-1},{\tau }_n)+\rho ({\tau }_{n-1},{\tau }_n)(1+\rho ({\tau }_{n-1},{\tau }_n)).\hfill \end{array}$$

Case 3: $n=2l$ and $m=2r+3$ for some $l,r\in \mathbb{N}$; that is, n is an even number and m is an odd number. Here, we have

$$\begin{array}{ccc}\hfill \rho ({\tau }_n,{\tau }_m)& =& \rho ({\tau }_{2l},{\tau }_{2r+3})\hfill \\ & \le & \rho ({\tau }_{2l},{\tau }_{2l+1})+\rho ({\tau }_{2l+1},{\tau }_{2r+2})+\rho ({\tau }_{2r+2},{\tau }_{2r+3})\hfill \\ & =& \rho ({\tau }_n,{\tau }_{n+1})+\rho ({\tau }_{n+1},{\tau }_{m-1})+\rho ({\tau }_{m-1},{\tau }_m).\hfill \end{array}$$

From Case 1, we get

$$\begin{array}{ccc}\hfill \rho ({\tau }_n,{\tau }_m)& \le & \rho ({\tau }_n,{\tau }_{n+1})+\rho ({\tau }_{n+1},{\tau }_{m-1})+\rho ({\tau }_{m-1},{\tau }_m)\hfill \\ & \le & \rho ({\tau }_n,{\tau }_{n+1})+\rho ({\tau }_n,{\tau }_{n+1})(1+\rho ({\tau }_n,{\tau }_{n+1}))+\rho ({\tau }_{m-1},{\tau }_m)\hfill \\ & \le & 2\rho ({\tau }_{n-1},{\tau }_n)+\rho ({\tau }_{n-1},{\tau }_n)(1+\rho ({\tau }_{n-1},{\tau }_n)).\hfill \end{array}$$

Case 4: $n=2l+1$ and $m=2r+3$ for some $l,r\in \mathbb{N}$; that is, n and m are both odd. Here, we have

$$\begin{array}{ccc}\hfill \rho ({\tau }_n,{\tau }_m)& =& \rho ({\tau }_{2l+1},{\tau }_{2r+3})\hfill \\ & \le & \rho ({\tau }_{2l+1},{\tau }_{2r+2})+\rho ({\tau }_{2r+2},{\tau }_{2r+3})\hfill \\ & =& \rho ({\tau }_n,{\tau }_{m-1})+\rho ({\tau }_{m-1},{\tau }_m).\hfill \end{array}$$

Case 1 implies that

$$\begin{array}{ccc}\hfill \rho ({\tau }_n,{\tau }_m)& \le & \rho ({\tau }_{n-1},{\tau }_n)(1+\rho ({\tau }_{n-1},{\tau }_n))+\rho ({\tau }_{m-1},{\tau }_m)\hfill \\ & \le & \rho ({\tau }_{n-1},{\tau }_n)(1+\rho ({\tau }_{n-1},{\tau }_n))+\rho ({\tau }_{n-1},{\tau }_n).\hfill \end{array}$$

By summing all cases together, we conclude that

$$\begin{array}{c}\hfill \rho ({\tau }_n,{\tau }_m)\le 2\rho ({\tau }_{n-1},{\tau }_n)+\rho ({\tau }_n,{\tau }_{n+1})(1+\rho ({\tau }_{n-1},{\tau }_n))\end{array}$$

(8)

holds for all $n,m\in \mathbb{N}$.

Letting $n,m\to +\infty$ in (8), we have

$${\displaystyle \underset{n,m\to +\infty }{lim}}\rho ({\tau }_n,{\tau }_m)=0.$$

Thus, $\left({\tau }_n\right)$ is a Cauchy sequence in $\chi$. In view of the competence of the space $\chi$, we find $a\in \chi$ such that ${\tau }_n\to a$ as n tends to $+\infty$.

For $s\in \mathbb{N}$, we have

$$\begin{array}{cc}& \psi \left(\rho ({\tau }_{2s+1},S_{2}a)\right)\hfill \\ =& \psi \left(\rho (S_1{\tau }_{2s},S_{2}a)\right)\hfill \\ \le & \psi \left(M_1({\tau }_{2s},a)\right)-\varphi \left(M_1({\tau }_{2s},a)\right)\hfill \\ =& \psi \left(max\{\rho (a,S_{2}a)\frac{1+\rho ({\tau }_{2s},{\tau }_{2s+1})}{1+\rho ({\tau }_{2s},a)},\rho ({\tau }_{2s},{\tau }_{2s+1}),\rho (a,S_{2}a)\right\})\hfill \\ & -\varphi \left(max\{\rho (a,S_{2}a)\frac{1+\rho ({\tau }_{2s},{\tau }_{2s+1})}{1+\rho ({\tau }_{2s},a)},\rho ({\tau }_{2s},{\tau }_{2s+1}),\rho (a,S_{2}a)\right\}).\hfill \end{array}$$

Allowing $s\to +\infty$ in above inequality, we get

$$\psi \left(\rho (a,S_{2}a)\right)\le \psi \left(\rho (a,S_{2}a)\right)-\varphi \left(\rho (a,S_{2}a)\right).$$

The above inequality is correct only if $\varphi \left(\rho (a,S_{2}a)\right)=0$ and thus $S_{2}a=a$. Using similar arguments as above, we may figure out $S_{1}a=a$. Thus, a is a common fixed point of $S_1$ and $S_2$.

Now, assume that $S_1{w}_1=S_2{w}_1=w_1$ and $S_1{w}_2=S_2{w}_2=w_2$. In view of $(\psi ,\varphi )-$contraction of the pair $(S_1,S_2)$, we have

$$\psi \left(\rho (w_1,w_2)\right)=\psi \left(\rho (S_1{w}_1,S_2{w}_2)\right)\le 0.$$

Thus, $\psi \left(\rho (w_1,w_2)\right)=0$. Therefore, $w_1=w_2$. Thus, the common fixed point of $S_1$ and $S_2$ is unique. □

By taking

$$max\left\{\rho (w,S_{2}w)\frac{1+\rho (e,S_{1}e)}{1+\rho (e,w)},\rho (e,S_{1}e),\rho (w,S_{2}w)\right\}=\rho (w,S_{2}w)\frac{1+\rho (e,S_{1}e)}{1+\rho (e,w)}$$

and

$$max\left\{\rho (w,S_{1}w)\frac{1+\rho (e,S_{2}e)}{1+\rho (e,w)},\rho (e,S_{2}e),\rho (w,S_{1}w)\right\}=\rho (w,S_{1}w)\frac{1+\rho (e,S_{2}e)}{1+\rho (e,w)}$$

in Definition 5. Then, the following result holds:

Corollary 1.

Let $(\chi ,\rho )$ be a complete quasi-metric space and $S_1,S_2:\chi \to \chi$ be two mappings. Let ψ and ϕ be two altering distance functions such that

$$\begin{array}{ccc}\hfill \psi \left(\rho (S_{1}e,S_{2}w)\right)& \le & \psi \left(\rho (w,S_{2}w)\frac{1+\rho (e,S_{1}e)}{1+\rho (e,w)}\right)-\varphi \left(\rho (w,S_{2}w)\frac{1+\rho (e,S_{1}e)}{1+\rho (e,w)}\right),\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \psi \left(\rho (S_{2}e,S_{1}w)\right)& \le & \psi \left(\rho (w,S_{1}w)\frac{1+\rho (e,S_{2}e)}{1+\rho (e,w)}\right)-\varphi \left(\rho (w,S_{1}w)\frac{1+\rho (e,S_{2}e)}{1+\rho (e,w)}\right).\hfill \end{array}$$

Then, $S_1$ and $S_2$ have a unique common fixed point.

If we define $\psi$ and $\varphi$ on the interval $[0,+\infty )$ such that $\psi \left(\tau \right)=\tau$ and $\varphi \left(\tau \right)=(1-a)\tau$ where $a\in [0,1)$ in Theorem 1, we formulate the following result.

Corollary 2.

Let $(\chi ,\rho )$ be a complete quasi-metric space and $S_1,S_2:\chi \to \chi$ be two mappings. Let $a\in [0,1)$ such that

$$\begin{array}{c}\hfill \psi \left(\rho (S_{1}e,S_{2}w)\right)\le amax\left\{\rho (w,S_{2}w)\frac{1+\rho (e,S_{1}e)}{1+\rho (e,w)},\rho (e,S_{1}e),\rho (w,S_{2}w)\right\},\end{array}$$

and

$$\begin{array}{c}\hfill \psi \left(\rho (S_{2}e,S_{1}w)\right)\le amax\left\{\rho (w,S_{1}w)\frac{1+\rho (e,S_{1}2e)}{1+\rho (e,w)},\rho (e,S_{1}2e),\rho (w,S_{1}w)\right\}.\end{array}$$

Then, $S_1$ and $S_2$ have a unique common fixed point.

In addition, if we assume $S_1=S_2$ in Theorem 1, Corollary 1, and Corollary 2, then the following results hold.

Corollary 3.

Let $(\chi ,\rho )$ be a complete quasi-metric space and $S_1$ be a self-mapping on χ. Assume ψ and ϕ are two altering distance functions such that

$$\begin{array}{ccc}\hfill \psi \left(\rho (S_{1}e,S_{1}w)\right)& \le & \psi \left(\left\{\rho (w,S_{1}w)\frac{1+\rho (e,S_{1}e)}{1+\rho (e,w)},\rho (e,S_{1}e),\rho (w,S_{1}w)\right\}\right)\hfill \\ & & -\varphi \left(\left\{\rho (w,S_{1}w)\frac{1+\rho (e,S_{1}e)}{1+\rho (e,w)},\rho (e,S_{1}e),\rho (w,S_{1}w)\right\}\right).\hfill \end{array}$$

Then, $S_1$ has a unique fixed point.

Corollary 4.

Let $(\chi ,\rho )$ be a complete quasi-metric space and $S_1:\chi \to \chi$ be a mapping. Let ψ and ϕ be two altering distance functions such that

$$\begin{array}{c}\hfill \psi \left(\rho (S_{1}e,S_{1}w)\right)\le \psi \left(max\left\{\rho (w,S_{1}w)\frac{1+\rho (e,S_{1}e)}{1+\rho (e,w)}\right\}\right)-\varphi \left(max\left\{\rho (w,S_{1}w)\frac{1+\rho (e,S_{1}e)}{1+\rho (e,w)}\right\}\right).\end{array}$$

Then, $S_1$ has a unique fixed point.

Corollary 5.

Let $(\chi ,\rho )$ be a complete quasi-metric space and $S_1:\chi \to \chi$ be a mapping. Let $a\in [0,1)$ such that

$$\begin{array}{c}\hfill \psi \left(\rho (S_{1}e,S_{1}w)\right)\le amax\left\{\rho (w,S_{1}w)\frac{1+\rho (e,S_{1}e)}{1+\rho (e,w)},\rho (e,S_{1}e),\rho (w,S_{1}w)\right\}.\end{array}$$

Then, $S_1$ has a unique fixed point.

The following example shows the validate of our results:

Example 2.

On the space $\chi =[0,1]$, define the quasi-metric via

$$\rho (\alpha ,\beta )=\left\{\begin{array}{cc}0,\hfill & \mathrm{if}\alpha =\beta ;\hfill \\ max\{\alpha ,\beta \},\hfill & \mathrm{if}\alpha \ne \beta ,\hfill \end{array}\right.$$

In addition, on $\chi =[0,1]$, define the mappings $S_1$ and $S_2$ via $S_1\tau =\frac{1}{2}{sin}^2\tau$ and $S_2\tau =\frac{1}{2}{\tau }^2$. Take the following altering functions $\psi \left(\alpha \right)=\frac{\alpha }{1+\alpha }$ and $\varphi \left(\alpha \right)=\frac{\alpha }{5+5\alpha }$. Then,

1. 

ρ induces complete quasi-metric on χ.

2. 

$(S_1,S_2)$ is $(\psi ,\varphi )-quasi$ contraction.

Proof. 

The proof of Part (1) is clear. To verify Part (2), given $(e,w)\in [0,1]\times [0,1]$ with $e\ne w$. Without loss of generality, we may assume that $e>w$. Then,

$$\begin{array}{ccc}\hfill M_1(e,w)& =& \left\{\rho \left(w,\frac{1}{2}{w}^2\right)\left(\frac{1+\rho (e,\frac{1}{2}{sin}^{2}e)}{1+\rho (e,w)}\right),\rho (e,\frac{1}{2}{sin}^{2}e),\rho (w,\frac{1}{2}{w}^2)\right\}\hfill \\ & =& e.\hfill \end{array}$$

Thus,

$$\begin{array}{ccc}\hfill \psi \left(\rho (S_{1}e,S_{2}w)\right)& =& \psi \left(\rho \left(\frac{1}{2}{sin}^{2}e,\frac{1}{2}{w}^2\right)\right)\hfill \\ & =& \frac{max\left\{\frac{1}{2}{sin}^{2}e,\frac{1}{2}{w}^2\right\}}{1+max\left\{\frac{1}{2}{sin}^{2}e,\frac{1}{2}{w}^2\right\}}\hfill \\ & \le & \frac{\frac{1}{2}e}{1+\frac{1}{2}e}\hfill \\ & =& \frac{e}{2+e}\hfill \\ & \le & \left(\frac{4}{5}\right)\left(\frac{e}{1+e}\right)\hfill \\ & =& \frac{e}{1+e}-\frac{e}{5+5e}\hfill \\ & =& \psi \left(M_1(e,w)\right)-\varphi \left(M_1(e,w)\right).\hfill \end{array}$$

Using similar arguments as for the above method, we can prove that

$$\psi \left(\rho (S_{2}e,S_{1}w)\right)=\psi \left(M_2(e,w)\right)-\varphi \left(M_2(e,w)\right).$$

Thus, $(S_1,S_2)$ is $(\psi ,\varphi )-$quasi contraction. Thus, by Theorem 1, we deduce that $S_1$ and $S_2$ have a unique common fixed point. □