## 1. Introduction and Preliminaries

Roughly speaking, a quasi-metric is a distance function that is not symmetry but satisfies both the triangle inequality and self-distance property. The notion of quasi-metric was first introduced by Wilson in 1930s [

1]. This is a subject of intensive research not only in the setting of topology [

2,

3,

4] and functional analysis, but also several qualitative sciences, such as theoretical computer science [

5,

6,

7,

8], biology [

9], and many other qualitative disciplines. In particular, as it is mentioned in [

10], the notion of quasi-metric plays crucial roles in several distinct branches of mathematics, such as in the existence and uniqueness of iterated function systems’ attractor (fractal), in the existence and uniqueness of Hamilton-Jacobi equations, and so on.

Another crucial notion that has no metric counterpart is that of an engaged partial order. Each partial order can be associated with a quasi-metric, and vice versa. Consequently, quasi-metric not only generalizes the concept of the metric, but also partial orders. This is a crucial fact for both the theoretical computer science applications and also has significance in the framework of biology [

9].

For the sake of the completeness, we shall give the formal definition of quasi-metric. Throughout the paper, X is a nonempty set A distance function $q:\chi \times \chi \to \left[0,\infty \right)$ is called a quasi-metric on χ if

- $\left({q}_{1}\right)$
$q(u,v)=0\iff u=v$;

- $\left({q}_{2}\right)$
$q(u,w)\le q(u,v)+q(v,w)$, for all $u,v,w\in \chi $.

In addition, the pair $(\chi ,q)$ is called a quasi-metric space.

In what follows, we indicate the close relation between a standard metric and a quasi-metric. Given

q be a quasi-metric on

X, it is clear that the function

${q}_{*}:\chi \times \chi \to \left[0,\infty \right)$ defined by

${q}_{*}(u,v)=q(v,u)$ forms also a quasi-metric and it is also called the dual (conjugate) of

q. The functions

${d}_{1},{d}_{2}:\chi \times \chi \to \left[0,\infty \right)$, where

form standard metrics on

χ.

We will provide an overview of quasi-metric spaces, presenting the notions of convergence, completeness, and continuity.

Let $\left\{{u}_{n}\right\}$ be a sequence in χ, and $u\in \chi $, where $(\chi ,q)$ a quasi-metric space. We say that:

$\left\{{u}_{n}\right\}$ converges to

u if and only if

$\left\{{u}_{n}\right\}$ is left-Cauchy if and only if for every $\u03f5>0$ there exists a positive integer $k=k\left(\u03f5\right)$ such that $q({u}_{n},{u}_{m})<\u03f5$ for all $n\ge m>k$.

$\left\{{u}_{n}\right\}$ is right-Cauchy if and only if for every $\u03f5>0$ there exists a positive integer $k=k\left(\u03f5\right)$ such that $q({u}_{n},{u}_{m})<\u03f5$ for all $m\ge n>k$.

$\left\{{u}_{n}\right\}$ is Cauchy if and only if it is left-Cauchy and right-Cauchy.

We would remark here that, in a quasi-metric space

$(\chi ,q)$, the limit for a convergent sequence is unique. Indeed, if

${u}_{n}\to u$, for all

$v\in \chi $, we have

A quasi-metric space

$(\chi ,q)$ is said to be: complete (respectively, left-complete or right-complete) if and only if each Cauchy sequence (respectively, left-Cauchy sequence or right-Cauchy sequence) in

X is convergent. Notice, in this context, that “right completeness” is equivalent to “Smyth completeness” [

11]. See also [

12].

A mapping

$T:\chi \to \chi $ is continuous provided that, for any sequence

$\left\{{u}_{n}\right\}$ in

χ such that

${u}_{n}\to u\in \chi $, the sequence

$\left\{T{u}_{n}\right\}$ converges to

$Tu$, that is,

If $T:\chi \to \chi $, then the fixed point set of T is ${\mathcal{F}}_{T}\left(\chi \right):=\{x\in X:Tx=x\}$.

A mapping $\zeta :[0,\infty )\times [0,\infty )\to \mathbb{R}$ is called an extended simulation function if the following axioms are fulfilled:

- $\left({z}_{d}\right)$
$\zeta (t,s)<s-t$ for all $t,s>0$;

- $\left({z}_{0}\right)$
$\zeta (t,0)\le 0$ for every $t\ge 0$ and $\zeta (t,0)=0\iff t=0$.

Notice that the axiom $\left({z}_{d}\right)$ implies that $\zeta (t,t)<0\mathrm{for}\mathrm{all}t0$. Let us denote by $\mathcal{Z}$ the family of all extended simulation functions $\zeta :[0,\infty )\times [0,\infty )\to \mathbb{R}$.

A function

$\phi :\left[0,\infty \right)\to \left[0,\infty \right)$ is called a

comparison function [

13] if:

- $\left({c}_{1}\right)$
$\phi $ is increasing;

- $\left({c}_{2}\right)$
$\underset{n\to \infty}{lim}{\phi}^{n}\left(t\right)=0$, for $t\in \left[0,\infty \right)$.

**Proposition** **1.** If φ is a comparison function, then:

- $\left(i\right)$
each ${\phi}^{k}$ is also a comparison function, for all $k\in \mathbb{N}$;

- $\left(ii\right)$
φ is continuous at 0;

- $\left(iii\right)$
$\phi \left(0\right)=0$ and $\phi \left(t\right)<t$ for all $t>0$.

A function

$\psi :\left[0,\infty \right)\to \left[0,\infty \right)$ is called a

c-comparison function [

13,

14] if:

- $\left(c{c}_{1}\right)$
$\psi $ is increasing;

- $\left(c{c}_{2}\right)$
$\sum _{n=0}^{\infty}{\psi}^{n}\left(t\right)<\infty$, for all $t\in \left(0,\infty \right)$.

We denote by

$\Psi $ the family of

c-comparison functions. In some papers, instead of a

c-comparison function, the term of strong comparison function is used. See [

13].

**Remark** **1.** Any c-comparison function is a comparison function.

Let

$\alpha :\chi \times \chi \to [0,\infty )$ be a function. We say that a mapping

$T:\chi \to \chi $ is

$\alpha $-orbital admissible [

15] if for each

$u\in \chi $ we have

**Lemma** **1.** Let $T:\chi \to \chi $ be an α-orbital admissible function. If there exists ${u}_{0}\in \chi $ such that $\alpha ({u}_{0},T{u}_{0})\ge 1$ and $\alpha (T{u}_{0},{u}_{0})\ge 1$, then the sequence ${\left({u}_{n}\right)}_{n\in \mathbb{N}}$, defined by ${u}_{n}=T{u}_{n-1},n\in \mathbb{N}$ satisfies the following relations: We say that the set

χ is

regular with respect to mapping

$\alpha :\chi \times \chi \to [0,\infty )$ if the following condition is satisfied: if

$\left\{{u}_{n}\right\}$ is a sequence in

χ such that

$\alpha ({u}_{n+1},{u}_{n})\ge 1$ and

$\alpha ({u}_{n},{u}_{n+1})\ge 1$ for any

$n\in \mathbb{N}$ and

${u}_{n}\to u\in \chi $ as

$n\to \infty $, then there exists a subsequence

$\left\{{u}_{n\left(i\right)}\right\}$ of

$\left\{{u}_{n}\right\}$ such that

for each

i.

In this manuscript, we will investigate the existence of fixed points for mappings that satisfy some hybrid type contraction conditions in the setting of quasi-metric spaces. We provide examples to assure the validity of the given results. The results of this paper generalize several known theorems in the recent literature, see [

13,

14,

16,

17,

18,

19,

20,

21,

22,

23,

24,

25].

## 2. Main Results

We start with the formal definition of hybrid almost contraction of type $\mathbb{I}$.

**Definition** **1.** Let $(\chi ,q)$ be a quasi-metric space. We say that the mapping $T:\chi \to \chi $ is a hybrid almost contraction of type$\mathbb{I}$, if there exist $\zeta \in \mathcal{Z}$, $\psi \in \Psi $, $p\ge 0$, $L\ge 0$ and ${a}_{1},{a}_{2},{a}_{3}\in [0,1]$ with ${a}_{1}+{a}_{2}>0$, ${a}_{1}+{a}_{2}+{a}_{3}=1,$ such that, for all distinct $u,v\in \chi $, we havewhereand **Theorem** **1.** Let $(\chi ,q)$ be a complete quasi-metric space and $\alpha :\chi \times \chi \to [0,\infty )$ be a mapping such that:

- (i)
$u=Tu$ implies $\alpha (u,v)>0$ for every $v\in \chi $;

- (ii)
$v=Tv$ implies $\alpha (u,v)>0$ for every $u\in \chi $.

Suppose that $T:\chi \to \chi $ is an hybrid almost contraction of type $\mathbb{I}$ and

- (C
_{1}) T is α-orbital admissible;

- (C
_{2}) there exists ${u}_{0}\in \chi $ such that $\alpha ({u}_{0},T{u}_{0})\ge 1$ and $\alpha (T{u}_{0},{u}_{0})\ge 1$;

- (C
_{3}) T is continuous.

Then, T has a fixed point.

**Proof.** Let the sequence

$\left\{{u}_{n}\right\}$ in

χ be defined by

where

${u}_{0}\in \chi $ is the point such that, from

$\left({C}_{2}\right)$,

$\alpha ({u}_{0},T{u}_{0})\ge 1$ and

$\alpha (T{u}_{0},{u}_{0})\ge 1$. Indubitably, for all

$n\in \mathbb{N}$, we have

${u}_{n+1}\ne {u}_{n}$. As a matter of fact, if we suppose that there is

${N}_{0}\in \mathbb{N}$ such that

${u}_{{N}_{0}}={u}_{{N}_{0}+1}$, from the manner in which the sequence

$\left\{{u}_{n}\right\}$ was defined, we get

so that the fixed point of

T is

${u}_{{N}_{0}}$ and the proof is completed. Thus, choosing

$u={u}_{n-1}$ respectively

$v={u}_{n}$ and since

$\frac{1}{2}min\left\{q({u}_{n-1},T{u}_{n-1}),q({u}_{n},T{u}_{n}),q(T{u}_{n},{u}_{n})\right\}\le \frac{1}{2}q({u}_{n-1},T{u}_{n-1})<q({u}_{n-1},{u}_{n})$ holds for any

$n\in \mathbb{N}$, by (

3), we get

In other words, taking into account

$\left({z}_{d}\right),$However,

T is an

$\alpha $-orbital admissible and, on the strength of Lemma 1, the above inequality yields

Since

the inequality (

6) becomes

In addition, by taking

$u={u}_{n}$, respectively,

$v={u}_{n-1}$, we have

As a consequence, (

3) becomes

or, taking into account

$\left({z}_{d}\right),$By Lemma 1, the above inequality yields

From this point of the proof, we will consider the two cases separately: $p>0$ and $p=0$.

**Case 1.** For the case

$p>0$,

and the inequality (

6) becomes

Onward, being a

c-comparison function,

$\psi $ satisfies

$\left(iii\right)$ by Proposition 1 that is

$\psi \left(t\right)<t$ for any

$t>0$, we obtain

which is equivalent with

or (since

${a}_{1}+{a}_{2}>0$)

Using the fact that

$\psi \in \Psi $ is increasing, by (

13), we have

Let now

$l\ge 1$. By using (

15) and the triangle inequality, we get

Letting $n\to \infty $ in the above inequality, we derive that ${\sum}_{j=n}^{\infty}{\psi}^{j}\left(q({u}_{0},{u}_{1})\right)\to 0$. Hence, $q({u}_{n},{u}_{n+l})\to 0$ as $n\to \infty $. Thus, $\left\{{u}_{n}\right\}$ is a right-Cauchy sequence in $(\chi ,q)$.

Similarly, since

the inequality (

12) becomes

Taking into account (

14), we get

We are able to examine it with the following cases.

- a.
If

$q({u}_{n},{u}_{n-1})<q({u}_{n-1},{u}_{n})$ for any

$n\in \mathbb{N}$, the above inequality becomes

and then, together with (

15),

From the triangle inequality and (

18), for all

$l\ge 1$, we get that

- b.
If, for any

$n\in \mathbb{N}$,

$q({u}_{n},{u}_{n-1})\le q({u}_{n-1},{u}_{n})$, we have

and, from (

17), regarding

$\psi \in \Psi $, we get that

Again, by triangle inequality,

- c.
If

$q({u}_{i},{u}_{i-1})\le q({u}_{i-1},{u}_{i})$ for some

$i\in \mathbb{N}$ and

$q({u}_{k},{u}_{k-1})>q({u}_{k-1},{u}_{k})$ for some

$k\in \mathbb{N}$, then we have for

$l\in \mathbb{N}$Therefore, we proved that $\left\{{u}_{n}\right\}$ is a left-Cauchy in $(\chi ,q)$.

Thus, being left and right Cauchy, the sequence

$\left\{{u}_{n}\right\}$ is a Cauchy in complete quasi-metric space

$(\chi ,q)$, which implies that there is

${u}^{*}\in \chi $ such that

Using the continuity of

T and

$\left(q1\right)$, we have

and so

It follows from (

20) and (

21) that

$T{u}^{*}={u}^{*}$, that is,

${u}^{*}$ is a fixed point of

T.

**Case 2.** In the case

$p=0$, we have

Replacing in (

6) and taking into account (

7), we get

and we deduce that

Thus, taking into account

${a}_{1}+{a}_{2}>0$, we have

and, from (

22), since

$\psi \in \Psi $ we are able to say that, for any

$n\in \mathbb{N}$,

Following the above lines and using the triangle inequality, we obtain that the sequence

${u}_{n}$ is right Cauchy. Likewise, because

taking into account (

11) and (

23), we have

We must examine two cases.

If

$max\left\{q({u}_{n},{u}_{n-1}),q({u}_{n-1},{u}_{n})\right\}=q({u}_{n-1},{u}_{n})$, then since

$q({u}_{n-1},{u}_{n})>0$, we get that

and recursively

If

$max\left\{q({u}_{n},{u}_{n-1}),q({u}_{n-1},{u}_{n})\right\}=q({u}_{n},{u}_{n-1})$, we have

Therefore, by combining (

25) with (

26), we derive (due to

$\left({c}_{2}\right)$) that

Again, using the triangle inequality, and the above inequalities for all

$l\ge 1$, we get

that is, the sequence

$\left\{{u}_{n}\right\}$ is left Cauchy, so that is a Cauchy sequence in a complete quasi-metric space

$(\chi ,q).$ Thus, there is

${u}^{*}\in \chi $ such that

Of course, using $\left({q}_{1}\right)$ and the continuity of T, we have $T{u}^{*}={u}^{*}$. □

**Corollary** **1.** Let $(\chi ,q)$ be a complete quasi-metric space, a function $\alpha :X\times X\to [0,\infty )$ and a mapping $T:\chi \to \chi $ such that there exist $\zeta \in \mathcal{Z}$ and $\psi \in \Psi $ such that, for $p\ge 0$, $L\ge 0$ and ${a}_{1},{a}_{2},{a}_{3}\in [0,1)$ with ${a}_{1}+{a}_{2}>0$ and ${a}_{1}+{a}_{2}+{a}_{3}=1$, we have Suppose also that the following assumptions hold:

- (i)
$u=Tu$ implies $\alpha (u,v)>0$ for every $v\in \chi $;

- (ii)
$v=Tv$ implies $\alpha (u,v)>0$ for every $u\in \chi $;

- (i)
T is α-orbital admissible;

- (ii)
there exists ${u}_{0}\in \chi $ such that $\alpha ({u}_{0},T{u}_{0})\ge 1$ and $\alpha (T{u}_{0},{u}_{0})\ge 1$;

- (iv)
T is continuous.

Then, T has a fixed point.

**Remark** **2.** Of course, in particular letting $L=0$ in the above Corollary, we find Theorem 2.1. in [16]. **Corollary** **2.** Let $(\chi ,q)$ be a complete quasi-metric space and a mapping $T:\chi \to \chi $ such that there exist $\zeta \in \mathcal{Z}$ and $\psi \in \Psi $ such that, for $p\ge 0$, $L\ge 0$ and ${a}_{1},{a}_{2},{a}_{3}\in [0,1)$ with ${a}_{1}+{a}_{2}>0$ and ${a}_{1}+{a}_{2}+{a}_{3}=1$, we have Then, T has a fixed point.

**Proof.** Let $\alpha (u,v)=1$ in Corollary 1. □

**Corollary** **3.** Let $(\chi ,q)$ be a complete quasi-metric space, a function $\alpha :\chi \times \chi \to [0,\infty )$ and a continuous mapping $T:\chi \to \chi $ such that there exist $\psi \in \Psi $ such that, for $p\ge 0$ and ${a}_{1},{a}_{2},{a}_{3}\in [0,1)$ with ${a}_{1}+{a}_{2}>0$ and ${a}_{1}+{a}_{2}+{a}_{3}=1$, we have Suppose that there exists ${u}_{0}\in \chi $ such that $\alpha ({u}_{0},T{u}_{0})\ge 1$ and $\alpha (T{u}_{0},{u}_{0})\ge 1$. Then, T has a fixed point.

**Proof.** Let $\zeta (t,s)=\psi \left(s\right)-t$ in Corollary 1. □

Moreover, it easy to see that Theorem 1 is a generalization of Theorem 2.1 in [

18] in the context of quasi-metric space. Indeed, if we take

$L=0$ and

$p=0$ in Corollary 3, we find:

**Corollary** **4.** Let $(\chi ,q)$ be a complete quasi-metric space, a function $\alpha :\chi \times \chi \to [0,\infty )$, and a continuous mapping $T:\chi \to \chi $ such that there exists $\psi \in \Psi $ such that, for ${a}_{1},{a}_{2},{a}_{3}\in [0,1)$ with ${a}_{1}+{a}_{2}>0$ and ${a}_{1}+{a}_{2}+{a}_{3}=1$, we have Suppose that there exists ${u}_{0}\in \chi $ such that $\alpha ({u}_{0},T{u}_{0})\ge 1$ and $\alpha (T{u}_{0},{u}_{0})\ge 1$. Then, T has a fixed point.

Inspired by the example of [

10], we consider the following:

**Example** **1.** Let the set $\chi =[1,\infty )$ and the quasi-metric $q:\chi \times \chi \to [0,\infty )$ given by(see Example 4.1 in [10].) Let the mapping $T:\chi \to \chi $, be defined byand the function $\alpha :\chi \times \chi \to [0,\infty )$ be defined by Since the mapping T is continuous and for $u=2$, $\alpha (2,T2)=\alpha (2,1)=3$ and $\alpha (T2,2)=\alpha (1,2)=3$, we have that the assumptions $\left({C}_{2}\right),\left({C}_{3}\right)$ are satisfied. Moreover, for any $u\in [1,2)$, we haveandso that T is α-orbital admissible. Choosing $\psi \left(t\right)=\frac{1}{3}t$, $p=2$, ${a}_{1}={a}_{2}={a}_{3}=\frac{1}{3}$ and $L=24$, we have the following cases:

**Case 1. ** If $u,v\in [1,2]$, then $q(u,v)=q(1,1)=0$ and (3) holds for every $\zeta \in \mathcal{Z}$.**Case 2. **If $u=3,v=2$, then we haveThus, we haveandso that T is a hybrid almost contraction for any $\zeta \in \mathcal{Z}$. The other cases are not interesting, while $\alpha (u,v)=0.$ (Consequently, the mapping T has two fixed points, ${u}_{1}=1$ and ${u}_{2}\in (3,4).$)

On the other hand, sincefor every $\gamma ,\beta \in (0,1)$ and $\psi \in \Psi $, Theorem 2.1 in [18] can not be applied. In particular, for the case $p=0$, the continuity condition of T can be replaced with the regularity condition of the space X.

**Theorem** **2.** Let $(\chi ,q)$ be a complete quasi-metric space, a function $\alpha :\chi \times \chi \to [0,\infty )$ and a mapping $T:\chi \to \chi $ such that there exist $\zeta \in \mathcal{Z}$, $\psi \in \Psi $, $L\ge 0$ and ${a}_{1},{a}_{2},{a}_{3}\in [0,1]$ with ${a}_{1}+{a}_{2}+{a}_{3}=1,$ such that, for all distinct $u,v\in \chi $, we have Suppose also that

- (i)
$u=Tu$ implies $\alpha (u,v)>0$ for every $v\in \chi $;

- (ii)
$v=Tv$ implies $\alpha (u,v)>0$ for every $u\in \chi $;

- (C
_{1}) T is α-orbital admissible;

- (C
_{2}) there exists ${u}_{0}\in \chi $ such that $\alpha ({u}_{0},T{u}_{0})\ge 1$ and $\alpha (T{u}_{0},{u}_{0})\ge 1$;

- (C
_{3}) χ is regular with respect to the mapping α.

Then, T has a fixed point.

**Proof.** From the above theorem, there exists

${u}^{*}\in \chi $ such that (

27) hold. In what follows, we claim that

Indeed, using the method of Reductio ad Absurdum, we assume that that there exists

$k\in \mathbb{N}$ such that

Therefore, we have

which is a contradiction.

In the alternative hypothesis, if the space

X is regular with respect to mapping

$\alpha $, we have

$\alpha ({u}^{*},{u}_{n\left(i\right)})\ge 1$, where

$\left\{{u}_{n\left(i\right)}\right\}$ is a sub-sequence of

$\left\{{u}_{n}\right\}$, for

$i\in \mathbb{N}.$ We will suppose by

reductio ad absurdum that

${u}^{*}\ne T{u}^{*}$. Then, for

$u={u}^{*}$ and

$v={u}_{n\left(i\right)}$ in (

3), we get

Taking into account the properties of function

$\zeta $,

$\psi $, and

$\alpha $, the above relation becomes

Letting

$i\to \infty $, we have

and, since

$\psi $ is continuous in 0,

$\psi \left(0\right)=0$, we get

$q(T{u}^{*},{u}^{*})=0.$ □

**Corollary** **5.** Let $(\chi ,q)$ be a complete quasi-metric space and $T:\chi \to \chi $ be a given mapping. Assume that there exist $L\ge 0$, $\zeta \in \mathcal{Z}$ and $\psi \in \Psi $ such that, for all distinct $u,v\in \chi $, we havefor all distinct $u,v\in \chi $. Then, T has a fixed point. **Proof.** It is sufficient to take $\alpha (u,v)=1$ for $u,v\in \chi $ in Theorem 5. □

**Corollary** **6.** Let $(\chi ,q)$ be a complete quasi-metric space and $T:\chi \to \chi $ be a given mapping. Assume that there exist $L\ge 0$, $\zeta \in \mathcal{Z}$ and $\psi \in \Psi $ such that, for all distinct $u,v\in \chi $, we havefor all distinct $u,v\in \chi $. Then, T has a fixed point. **Proof.** It is sufficient to take $L=0$, $\zeta (t,s)={k}_{1}s-t$, $\psi \left(u\right)={k}_{2}u$ with ${k}_{1},{k}_{2}\in (0,1)$ and $k={k}_{1}{k}_{2}$ in Corollary 5. □

**Corollary** **7.** Let $(\chi ,q)$ be a complete quasi-metric space and $T:\chi \to \chi $ a continuous mapping such thatfor all distinct $u,v\in \chi $ and some $k\in (0,1)$. Then, T has a fixed point in χ. **Proof.** Let $p=2$ and ${a}_{1}={a}_{2}={a}_{3}=\frac{1}{3}$ in Corollary 6. □

In the next theorem, we involve a Jaggi type expression with the hybrid contractions.

**Definition** **2.** Let $(\chi ,q)$ be a quasi-metric space. A mapping $T:\chi \to \chi $ is called a hybrid almost contraction of type

$\mathbb{J}$, if there exist $\zeta \in \mathcal{Z}$ and $\psi \in \Psi $ such that, for $p\ge 0$, $L\ge 0$ and ${a}_{1},{a}_{2}>0$ with ${a}_{1}+{a}_{2}<1$, we havefor all distinct $u,v\in \chi $, whereand **Theorem** **3.** Let $(\chi ,q)$ be a complete quasi-metric space and $\alpha :X\times X\to [0,\infty )$ be a mapping such that:

- (i)
$u=Tu$ implies $\alpha (u,v)>0$ for every $v\in \chi $;

- (ii)
$v=Tv$ implies $\alpha (u,v)>0$ for every $u\in \chi $.

Suppose that $T:\chi \to \chi $ is a hybrid almost contraction of type $\mathbb{J}$ such that the following assumptions hold:

- (i)
T is α-orbital admissible;

- (ii)
there exists ${u}_{0}\in \chi $ such that $\alpha ({u}_{0},T{u}_{0})\ge 1$ and $\alpha (T{u}_{0},{u}_{0})\ge 1$;

- (iii)
there exists $\Delta >0$ such that, ${({a}_{1}+{a}_{2}{\Delta}^{2p})}^{1/p}\le 1$ (where $p>0$) and - (iv)
T is continuous.

Then, T has a fixed point.

**Proof.** We will consider only the case

$p>0$ because, for

$p=0$, the expression is similar to the one in Theorem 1. By verbatim of the first lines in the proof of Theorem 1, starting from a point

${u}_{0}$, we are able to build a sequence

$\left\{{u}_{n}\right\}\subset \chi .$ Onward, as in the proof of Theorem 1, we suppose that

${u}_{n+1}\ne {u}_{n}$ for all

$n\in \mathbb{N}$ and from (

35), we have

$\frac{1}{2}min\left\{q({u}_{n-1},T{u}_{n-1}),q({u}_{n},T{u}_{n}),q(T{u}_{n},{u}_{n})\right\}\le q({u}_{n-1},{u}_{n})$, which implies

By the axiom

$\left({z}_{d}\right)$, Lemma 1 and taking into account (

7), this inequality becomes

Thereupon,

and then, from (

36), we have

$q({u}_{n},{u}_{n+1})<\psi \left(q({u}_{n-1},{u}_{n})\right)$. Since

$\psi \in \Psi $, recursively, we get

In order to prove that

$\left\{{u}_{n}\right\}$ is a right-Cauchy sequence, let

$l\in \mathbb{N}.$ From (

37) and the triangle inequality, we get that

We conclude that $\left\{{q}_{n}\right\}$ is a right-Cauchy sequence in $(\chi ,q)$.

Substituting in (

35)

$u={u}_{n}$ and

$v={u}_{n-1}$ and since

$\frac{1}{2}min\left\{q({u}_{n},T{u}_{n},q({u}_{n-1},T{u}_{n-1}),q(T{u}_{n-1},{u}_{n-1}))\right\}\le q({u}_{n},{u}_{n-1})$, we have (taking into account (

11))

i.e.,

On one hand, we have already proved that

$q({u}_{n},{u}_{n+1})<q({u}_{n-1},{u}_{n})$. On the other hand, by (

$iii$), there exists a positive constant

$\Delta $ such that

$q({u}_{n-1},{u}_{n})\le \Delta q({u}_{n},{u}_{n-1})$ for

$n\in \mathbb{N}$. Thus, we have

which is equivalent to the next inequality

Again, considering triangle inequality, together with (

38), for

$l\in \mathbb{N}$, we have

Analogously, we deduce that $\left\{{u}_{n}\right\}$ is left-Cauchy, so that it is a Cauchy sequence in complete quasi-metric space.

Thus, there exists

${u}^{*}\in \chi $ such that

Under the assumption

$\left(iv\right)$, from the continuity of

T and

$\left({q}_{1}\right)$, we have

so that

Hence, $T{u}^{*}={u}^{*}$ that is, ${u}^{*}$ is a fixed point of T. □

The following is a special case for $p=0$.

**Corollary** **8.** Let $(\chi ,q)$ be a complete quasi-metric space, a function $\alpha :X\times X\to [0,\infty )$ and a mapping $T:\chi \to \chi $ such that there exist $\zeta \in \mathcal{Z}$ and $\psi \in \Psi $ such that, for $p\ge 0$, $L\ge 0$ and ${a}_{1},{a}_{2},\in [0,1)$ with ${a}_{1}+{a}_{2}<1$, we have Suppose also that the following assumptions hold:

- (i)
$u=Tu$ implies $\alpha (u,v)>0$ for every $v\in \chi $;

- (ii)
$v=Tv$ implies $\alpha (u,v)>0$ for every $u\in \chi $;

- (i)
T is α-orbital admissible;

- (ii)
there exists ${u}_{0}\in X$ such that $\alpha ({u}_{0},T{u}_{0})\ge 1$ and $\alpha (T{u}_{0},{u}_{0})\ge 1$;

- (iii)
there exists $\Delta >0$ such that, ${({a}_{1}+{a}_{2}{\Delta}^{2p})}^{1/p}\le 1$ (where $p>0$) and - (iv)
T is continuous.

Then, T has a fixed point.

**Example** **2.** Let $X=[0,1]$ and the function It is easy to see that the pair $(\chi ,q)$ forms a quasi-metric space.

Let the map $T:\chi \to \chi $ defined byand choose $\zeta (u,v)=\frac{1}{2}v-u$ and $\psi \left(t\right)=\frac{1}{2}t.$ For $p=2$, $L=0$, $\Delta =2$, ${a}_{1}=\frac{1}{4}$ and ${a}_{2}=\frac{1}{32}$ because ${({a}_{1}+{a}_{2}\xb7{\Delta}^{2p})}^{1/p}=\frac{1}{4}+\frac{1}{32}\xb7{2}^{4}=\frac{3}{4}\le 1$, the assumption $\left(iii\right)$ is satisfied. In this case, (41) becomes Define $\alpha :\chi \times \chi \to [0,\infty )$ such that It is easy to see that T is α-admissible. Indeed, we haveand On the other hand, for ${q}_{0}=0$,so that the presumptions $\left(i\right)$, $\left(ii\right)$, and $\left(iv\right)$ are satisfied. Of course, if $u,v\in [0,\frac{1}{2})$, we have $q(Tu,Tv)=q(\frac{1}{8},\frac{1}{8})=$ 0 and (41) is verified. For $u=1$ and $v=0$, we have $q(T1,T\mathit{0})=\frac{1}{4}-\frac{1}{8}=\frac{1}{8}$, $q(0,T\mathit{0})=q(\mathit{0},1/8)=2(1/8-0)=1/4$, $q(1,T1)=q(1,1/4)=3/4$ and The other cases are not interesting since $\alpha (u,v)=0$ and the condition (42) is fulfilled trivially. Thus, the presumptions of Theorem 8 are provided and $u=\frac{1}{8}$ is the fixed point of T. **Corollary** **9.** Let $(\chi ,q)$ be a complete quasi-metric space and T be a continuous self-mapping on χ. Suppose that there exist $\zeta \in \mathcal{Z}$, $\psi \in \Psi $ such thatfor each distinct $u,v\in \chi $. If there exists $\Delta >0$ such that ${({a}_{1}+{a}_{2}\xb7{\Delta}^{2p})}^{1/p}\le 1$ for $p>0$, and $\frac{1}{\Delta}q(u,v)\le q(v,u)\le \Delta q(u,v)$ for all $u,v\in \chi $, then T has a fixed point. **Proof.** It is sufficient to take $L=0$ and $\alpha (u,v)=1$ for $u,v\in \chi $ in Corollary 8. □

**Corollary** **10.** Let $(X,q)$ be a complete quasi-metric space and T be a self-mapping on χ. Suppose that there exists $\Delta >0$ such that ${({a}_{1}+{a}_{2}\xb7{\Delta}^{2p})}^{1/p}\le 1$ for $p>0$, and $\frac{1}{\Delta}q(u,v)\le q(v,u)\le \Delta q(u,v)$ for all $u,v\in \chi $. The mapping T has a fixed point provided thatfor each distinct $u,v\in \chi $ and some $c\in (0,1)$. **Proof.** We set $\zeta (t,s)={c}_{1}s-t$, $\psi \left(u\right)={c}_{2}u$ with ${c}_{1},{c}_{2}\in [0,1)$ and $c={c}_{1}+{c}_{2}$ in Corollary 9. □

**Remark** **3.** Letting $p=0$ in Corollary 10, we find Theorem 2.2. in [20]. **Example** **3.** Let $(\chi ,q)$ be the quasi-metric space, where $\chi =[1,\infty )$ and Consider the function ζ be arbitrary in $\mathcal{Z}$, $\psi \in \Psi $ with $\psi \left(t\right)=\frac{t}{\sqrt{3}}$ and $\alpha :\chi \times \chi \to [0,\infty )$ such that It is easily verified that T is α-orbital admissible. Whereas $T1=T3=T4=3$, taking into account the definition of function α, we have that the inequality (41) holds for every pair $(u,v)\in {\chi}^{2}\backslash \left\{(2,1)\right\}$. For the case $u=2$ and $v=1$, choosing ${a}_{1}=\frac{1}{2}$, ${a}_{2}=\frac{1}{48}$ and $p=2$, we find that axiom $\left(iii\right)$ holds. On the other hand,and Consequently, by Theorem 8, we have that the mapping T has a fixed point in χ.

On the other hand, we can observed that, for $u=1$ and $v=5$,so that, sincefor any $\lambda \in [0,1)$ and $\alpha \in (0,1)$, Theorem 2.2 in [20] can not be applied. **Corollary** **11.** Let $(\chi ,q)$ be a complete quasi-metric space and $T:\chi \to \chi $ a continuous mapping. Then, T has a fixed point provided thatfor each $u,v\in X$ and ${k}_{1},{k}_{2}\in (0,1)$ with ${k}_{1}+{k}_{2}<1$ **Proof.** Let $p=1$ and ${k}_{i}=c\xb7{a}_{i}$, for $i\in \{1,2\}$ in Corollary 10. □