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T-Equivalences: The Metric Behavior Revisited
Open AccessArticle

On Hybrid Contractions in the Context of Quasi-Metric Spaces

by 1,*,†, 2,3,*,† and 4,†
1
Department of Mathematics and Computer Science, Transilvania University of Brasov, 500036 Brasov, Romania
2
Department of Medical Research, China Medical University Hospital, China Medical University, Taichung 40402, Taiwan
3
Department of Mathematics, Çankaya University, Ankara 06790, Turkey
4
Faculty of Business, Babeş-Bolyai University, 400084 Cluj-Napoca, Romania
*
Authors to whom correspondence should be addressed.
These authors contributed equally to this work.
Mathematics 2020, 8(5), 675; https://doi.org/10.3390/math8050675
Received: 1 April 2020 / Revised: 23 April 2020 / Accepted: 27 April 2020 / Published: 29 April 2020

Abstract

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.
Keywords: contractions; hybrid contractions; quasi-metric spaces; metric spaces contractions; hybrid contractions; quasi-metric spaces; metric spaces

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 : χ × χ 0 , is called a quasi-metric on χ if
( q 1 )
q ( u , v ) = 0 u = v ;
( q 2 )
q ( u , w ) q ( u , v ) + q ( v , w ) , for all u , v , w χ .
In addition, the pair ( χ , 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 * : χ × χ 0 , 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 : χ × χ 0 , , where
d 1 ( v , u ) = q ( u , v ) + q * ( u , v ) , d 2 ( v , u ) = max q ( u , v ) , q * ( u , v )
form standard metrics on χ.
We will provide an overview of quasi-metric spaces, presenting the notions of convergence, completeness, and continuity.
Let u n be a sequence in χ, and u χ , where ( χ , q ) a quasi-metric space. We say that:
  • u n converges to u if and only if
    lim n q ( u n , u ) = lim n q ( u , u n ) = 0 .
  • u n is left-Cauchy if and only if for every ϵ > 0 there exists a positive integer k = k ( ϵ ) such that q ( u n , u m ) < ϵ for all n m > k .
  • u n is right-Cauchy if and only if for every ϵ > 0 there exists a positive integer k = k ( ϵ ) such that q ( u n , u m ) < ϵ for all m n > k .
  • u n is Cauchy if and only if it is left-Cauchy and right-Cauchy.
We would remark here that, in a quasi-metric space ( χ , q ) , the limit for a convergent sequence is unique. Indeed, if u n u , for all v χ , we have
lim n q ( u n , v ) = q ( u , v ) and lim n q ( v , u n ) = q ( v , u ) .
A quasi-metric space ( χ , 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 : χ χ is continuous provided that, for any sequence u n in χ such that u n u χ , the sequence T u n converges to T u , that is,
lim n q ( T u n , T u ) = lim n q ( T u , T u n ) = 0
If T : χ χ , then the fixed point set of T is F T ( χ ) : = { x X : T x = x } .
A mapping ζ : [ 0 , ) × [ 0 , ) R is called an extended simulation function if the following axioms are fulfilled:
( z d )
ζ ( t , s ) < s t for all t , s > 0 ;
( z 0 )
ζ ( t , 0 ) 0 for every t 0 and ζ ( t , 0 ) = 0 t = 0 .
Notice that the axiom ( z d ) implies that ζ ( t , t ) < 0   for   all   t > 0 . Let us denote by Z the family of all extended simulation functions ζ : [ 0 , ) × [ 0 , ) R .
A function φ : 0 , 0 , is called a comparison function [13] if:
( c 1 )
φ is increasing;
( c 2 )
lim n φ n ( t ) = 0 , for t 0 , .
Proposition 1.
If φ is a comparison function, then:
( i )
each φ k is also a comparison function, for all k N ;
( i i )
φ is continuous at 0;
( i i i )
φ ( 0 ) = 0 and φ ( t ) < t for all t > 0 .
A function ψ : 0 , 0 , is called a c-comparison function [13,14] if:
( c c 1 )
ψ is increasing;
( c c 2 )
n = 0 ψ n ( t ) < , for all t 0 , .
We denote by Ψ 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 α : χ × χ [ 0 , ) be a function. We say that a mapping T : χ χ is α -orbital admissible [15] if for each u χ we have
α ( u , T u ) 1 α ( T u , T 2 u ) 1 .
Lemma 1.
Let T : χ χ be an α-orbital admissible function. If there exists u 0 χ such that α ( u 0 , T u 0 ) 1 and α ( T u 0 , u 0 ) 1 , then the sequence ( u n ) n N , defined by u n = T u n 1 , n N satisfies the following relations:
α ( u n , u n + 1 ) 1   a n d   α ( u n + 1 , u n ) 1 ,   f o r   a l l   n N 0 .
We say that the set χ is regular with respect to mapping α : χ × χ [ 0 , ) if the following condition is satisfied: if { u n } is a sequence in χ such that α ( u n + 1 , u n ) 1 and α ( u n , u n + 1 ) 1 for any n N and u n u χ as n , then there exists a subsequence u n ( i ) of u n such that
α ( u n ( i ) , u ) 1   and   α ( u , u n ( i ) ) 1 ,
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 I .
Definition 1.
Let ( χ , q ) be a quasi-metric space. We say that the mapping T : χ χ is a hybrid almost contraction of type I , if there exist ζ Z , ψ Ψ , p 0 , L 0 and a 1 , a 2 , a 3 [ 0 , 1 ] with a 1 + a 2 > 0 , a 1 + a 2 + a 3 = 1 , such that, for all distinct u , v χ , we have
1 2 min q ( u , T u ) , q ( v , T v ) q ( T v , v ) q ( u , v )   implies ζ ( α ( u , v ) q ( T u , T v ) , ψ ( I p ( u , v ) + L N ( u , v ) ) ) 0 ,
where
I p ( u , v ) = [ a 1 ( q ( u , v ) ) p + a 2 ( q ( u , T u ) ) p + a 3 ( q ( v , T v ) ) p ] 1 / p , for   p > 0 , ( q ( u , v ) ) a 1 · ( q ( u , T u ) ) a 2 · ( q ( v , T v ) ) a 3 for   p = 0
and
N ( u , v ) = min q ( u , T v ) , q ( v , T u ) .
Theorem 1.
Let ( χ , q ) be a complete quasi-metric space and α : χ × χ [ 0 , ) be a mapping such that:
(i) 
u = T u implies α ( u , v ) > 0 for every v χ ;
(ii) 
v = T v implies α ( u , v ) > 0 for every u χ .
Suppose that T : χ χ is an hybrid almost contraction of type I and
(C1
T is α-orbital admissible;
(C2
there exists u 0 χ such that α ( u 0 , T u 0 ) 1 and α ( T u 0 , u 0 ) 1 ;
(C3
T is continuous.
Then, T has a fixed point.
Proof. 
Let the sequence u n in χ be defined by
u 1 = T u 0 , u 2 = T u 1 , , u n = T u n 1 = T n 1 u 0
where u 0 χ is the point such that, from ( C 2 ) , α ( u 0 , T u 0 ) 1 and α ( T u 0 , u 0 ) 1 . Indubitably, for all n N , we have u n + 1 u n . As a matter of fact, if we suppose that there is N 0 N such that u N 0 = u N 0 + 1 , from the manner in which the sequence u n was defined, we get
u N 0 = T u N 0 = u N 0 + 1
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 1 2 min q ( u n 1 , T u n 1 ) , q ( u n , T u n ) , q ( T u n , u n ) 1 2 q ( u n 1 , T u n 1 ) < q ( u n 1 , u n ) holds for any n N , by (3), we get
ζ ( α ( u n 1 , u n ) q ( T u n 1 , T u n ) , ψ ( I p ( u n 1 , u n ) + L N ( u n 1 , u n ) ) ) 0 .
In other words, taking into account ( z d ) ,
0 ψ ( I p ( u n 1 , u n ) + L N ( u n 1 , u n ) ) α ( u n 1 , u n ) q ( T u n 1 , T u n ) .
However, T is an α -orbital admissible and, on the strength of Lemma 1, the above inequality yields
q ( T u n 1 , T u n ) α ( u n 1 , u n ) q ( T u n 1 , T u n ) ψ ( I p ( u n 1 , u n ) + L N ( u n 1 , u n ) ) .
Since
N ( u n 1 , u n ) = min q ( u n 1 , T u n ) , q ( u n , T u n 1 ) = min q ( u n 1 , u n ) , q ( u n , u n ) = 0 ,
the inequality (6) becomes
q ( T u n 1 , T u n ) ψ ( I p ( u n 1 , u n ) ) .
In addition, by taking u = u n , respectively, v = u n 1 , we have
1 2 min q ( u n , T u n ) , q ( u n 1 , T u n 1 ) , q ( T u n 1 , u n 1 ) 1 2 min q ( u n , u n + 1 ) , q ( u n 1 , u n ) , q ( u n , u n 1 ) < q ( u n , u n 1 ) .
As a consequence, (3) becomes
ζ ( α ( u n , u n 1 ) q ( T u n , T u n 1 ) , ψ ( I p ( u n , u n 1 ) + L N ( u n , u n 1 ) ) ) 0 ,
or, taking into account ( z d ) ,
0 ψ ( I p ( u n , u n 1 ) + L N ( u n , u n 1 ) ) α ( u n , u n 1 ) q ( T u n , T u n 1 ) .
By Lemma 1, the above inequality yields
q ( u n + 1 , u n ) = q ( T u n , T u n 1 ) α ( u n , u n 1 ) q ( T u n , T u n 1 ) ψ ( I p ( u n , u n 1 ) + L N ( u n , u n 1 ) ) .
However,
N ( u n , u n 1 ) = min q ( u n , T u n 1 ) , q ( u n 1 , T u n ) = min q ( u n , u n ) , q ( u n 1 , u n + 1 ) = 0 ,
and then we get
q ( T u n , T u n 1 ) ψ ( I p ( u n , u n 1 ) ) .
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 ,
I p ( u n 1 , u n ) = [ a 1 ( q ( u n 1 , u n ) ) p + a 2 ( q ( u n 1 , T u n 1 ) ) p + a 3 ( q ( u n , T u n ) ) p ] 1 / p = [ a 1 ( q ( u n 1 , u n ) ) p + a 2 ( q ( u n 1 , u n ) ) p + a 3 ( q ( u n , u n + 1 ) ) p ] 1 / p = [ ( a 1 + a 2 ) ( q ( u n 1 , u n ) ) p + a 3 ( q ( u n , u n + 1 ) ) p ] 1 / p
and the inequality (6) becomes
q ( u n , u n + 1 ) = q ( T u n 1 , T u n ) ψ ( [ ( a 1 + a 2 ) ( q ( u n 1 , u n ) ) p + a 3 ( q ( u n , u n + 1 ) ) p ] 1 / p ) .
Onward, being a c-comparison function, ψ satisfies ( i i i ) by Proposition 1 that is ψ ( t ) < t for any t > 0 , we obtain
q ( u n , u n + 1 ) ψ ( [ ( a 1 + a 2 ) ( q ( u n 1 , u n ) ) p + a 3 ( q ( u n , u n + 1 ) ) p ] 1 / p ) < [ ( a 1 + a 2 ) ( q ( u n 1 , u n ) ) p + ( 1 a 1 a 2 ) ( q ( u n , u n + 1 ) ) p ] 1 / p ,
which is equivalent with
( a 1 + a 2 ) [ q ( u n , u n + 1 ) ] p < ( a 1 + a 2 ) [ q ( u n 1 , u n ) ] p ,
or (since a 1 + a 2 > 0 )
q ( u n , u n + 1 ) < q ( u n 1 , u n ) .
Using the fact that ψ Ψ is increasing, by (13), we have
q ( u n , u n + 1 ) < ψ ( q ( u n 1 , u n ) ) < ψ 2 ( q ( u n 2 , u n 1 ) ) < < ψ n ( q ( u 0 , u 1 ) )
Let now l 1 . By using (15) and the triangle inequality, we get
q ( u n , u n + l ) q ( u n , u n + 1 ) + + q ( u n + l 1 , u n + l ) j = n n + l 1 ψ j ( q ( u 0 , u 1 ) ) j = n ψ j ( q ( u 0 , u 1 ) ) .
Letting n in the above inequality, we derive that j = n ψ j ( q ( u 0 , u 1 ) ) 0 . Hence, q ( u n , u n + l ) 0 as n . Thus, { u n } is a right-Cauchy sequence in ( χ , q ) .
Similarly, since
I p ( u n , u n 1 ) = [ a 1 ( q ( u n , u n 1 ) ) p + a 2 ( q ( u n , T u n ) ) p + a 3 ( q ( u n 1 , T u n 1 ) ) p ] 1 / p = [ a 1 ( q ( u n , u n 1 ) ) p + a 2 ( q ( u n , u n + 1 ) ) p + a 3 ( q ( u n 1 , u n ) ) p ] 1 / p ,
the inequality (12) becomes
q ( u n + 1 , u n ) ψ ( I p ( u n , u n 1 ) ) < I p ( u n , u n 1 ) = [ a 1 ( q ( u n , u n 1 ) ) p + a 2 ( q ( u n , u n + 1 ) ) p + a 3 ( q ( u n 1 , u n ) ) p ] 1 / p .
Taking into account (14), we get
( q ( u n + 1 , u n ) ) p < a 1 ( q ( u n , u n 1 ) ) p + a 2 ( q ( u n , u n + 1 ) ) p + a 3 ( q ( u n 1 , u n ) ) p = a 1 ( q ( u n , u n 1 ) ) p + a 2 ( q ( u n , u n + 1 ) ) p + ( 1 a 1 a 2 ) ( q ( u n 1 , u n ) ) p < a 1 ( q ( u n , u n 1 ) ) p + ( 1 a 1 ) ( q ( u n 1 , u n ) ) p ,   for   any   n N .
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 N , the above inequality becomes
( q ( u n + 1 , u n ) ) p < ( q ( u n 1 , u n ) ) p ,
and then, together with (15),
q ( u n + 1 , u n ) < q ( u n 1 , u n ) < ψ n 1 ( u 0 , u 1 ) , n 1 .
From the triangle inequality and (18), for all l 1 , we get that
q ( u n + l , u n ) q ( u n + l , u n + l 1 ) + + q ( u n + 1 , u n ) j = n n + l 1 ψ j ( q ( u 0 , u 1 ) ) j = n ψ j ( q ( u 0 , u 1 ) ) 0 as n .
b.
If, for any n N , q ( u n , u n 1 ) q ( u n 1 , u n ) , we have
q ( u n + 1 , u n ) < q ( u n , u n 1 )
and, from (17), regarding ψ Ψ , we get that
q ( u n + 1 , u n ) < ψ ( q ( u n , u n 1 ) ) < < ψ n ( q ( u 1 , u 0 ) ) .
Again, by triangle inequality,
q ( u n + l , u n ) q ( u n + l , u n + l 1 ) + + q ( u n + 1 , u n ) j = n n + l 1 ψ j ( q ( u 1 , u 0 ) ) j = n ψ j ( q ( u 1 , u 0 ) ) 0 as n .
c.
If q ( u i , u i 1 ) q ( u i 1 , u i ) for some i N and q ( u k , u k 1 ) > q ( u k 1 , u k ) for some k N , then we have for l N
q ( u n + l , u n ) q ( u n + l , u n + l 1 ) + + q ( u n + 1 , u n ) j = n ψ j ( q ( u 1 , u 0 ) ) + j = n ψ j ( q ( u 0 , u 1 ) ) 0 as n .
Therefore, we proved that { u n } is a left-Cauchy in ( χ , q ) .
Thus, being left and right Cauchy, the sequence { u n } is a Cauchy in complete quasi-metric space ( χ , q ) , which implies that there is u * χ such that
lim n q ( u n , u * ) = lim n q ( u * , u n ) = 0 .
Using the continuity of T and ( q 1 ) , we have
lim n q ( u n , T u * ) = lim n q ( T u n 1 , T u * ) = 0 ,
lim n q ( T u * , u n ) = lim n q ( T u * , T u n 1 ) = 0
and so
lim n q ( u n , T u * ) = lim n q ( T u * , u n ) = 0 .
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
I p ( u n 1 , u n ) = ( q ( u n 1 , u n ) ) a 1 · ( q ( u n 1 , T u n 1 ) ) a 2 · ( q ( u n , T u n ) ) a 3 = ( q ( u n 1 , u n ) ) a 1 · ( q ( u n 1 , u n ) ) a 2 · ( q ( u n , u n + 1 ) ) a 3 .
Replacing in (6) and taking into account (7), we get
q ( u n , u n + 1 ) = q ( T u n 1 , T u n ) α ( u n 1 , u n ) q ( T u n 1 , T u n ) ψ ( I p ( u n 1 , u n ) ) < I p ( u n 1 , u n ) = ( q ( u n 1 , u n ) ) a 1 + a 2 · ( q ( u n , u n + 1 ) ) a 3
and we deduce that
( q ( u n , u n + 1 ) ) a 1 + a 2 < ( q ( u n 1 , u n ) ) a 1 + a 2 .
Thus, taking into account a 1 + a 2 > 0 , we have
q ( u n , u n + 1 ) < q ( u n 1 , u n )
and, from (22), since ψ Ψ we are able to say that, for any n N ,
q ( u n , u n + 1 ) ψ ( q ( u n 1 , u n ) ) < < ψ n ( q ( u 0 , u 1 ) ) .
Following the above lines and using the triangle inequality, we obtain that the sequence u n is right Cauchy. Likewise, because
I p ( u n , u n 1 ) = ( q ( u n , u n 1 ) ) a 1 · ( q ( u n , T u n ) ) a 2 · ( q ( u n 1 , T u n 1 ) ) a 3 ] = ( q ( u n , u n 1 ) ) a 1 · ( q ( u n , u n + 1 ) ) a 2 · ( q ( u n 1 , u n ) ) a 3 ,
taking into account (11) and (23), we have
q ( u n + 1 , u n ) = q ( T u n , T u n 1 ) α ( u n , u n 1 ) q ( T u n , T u n 1 ) ψ ( I p ( u n , u n 1 ) ) < I p ( u n , u n 1 ) = ( q ( u n , u n 1 ) ) a 1 · ( q ( u n , u n + 1 ) ) a 2 · ( q ( u n 1 , u n ) ) a 3 < ( q ( u n , u n 1 ) ) a 1 · ( q ( u n 1 , u n ) ) a 2 + a 3 ( max q ( u n , u n 1 ) , q ( u n 1 , u n ) ) a 1 + a 2 + a 3 = max q ( u n , u n 1 ) , q ( u n 1 , u n ) .
We must examine two cases.
If max q ( u n , u n 1 ) , q ( u n 1 , u n ) = q ( u n 1 , u n ) , then since q ( u n 1 , u n ) > 0 , we get that
q ( u n + 1 , u n ) ψ ( q ( u n 1 , u n ) ) ,
and recursively
q ( u n + 1 , u n ) ψ n ( q ( u 0 , u 1 ) ) .
If max q ( u n , u n 1 ) , q ( u n 1 , u n ) = q ( u n , u n 1 ) , we have
q ( u n + 1 , u n ) ψ ( q ( u n , u n 1 ) ) < < ψ n ( q ( u 1 , u 0 ) ) .
Therefore, by combining (25) with (26), we derive (due to ( c 2 ) ) that
lim n q ( u n + 1 , u n ) = lim n max ψ n ( q ( u 0 , u 1 ) ) , ψ n ( q ( u 1 , u 0 ) ) = 0 .
Again, using the triangle inequality, and the above inequalities for all l 1 , we get
q ( u n + l , u n ) q ( u n + l , u n + l 1 ) + + q ( u n + 1 , u n ) j = n ψ j ( q ( u 1 , u 0 ) ) + j = n ψ j ( q ( u 0 , u 1 ) ) 0 as n ,
that is, the sequence u n is left Cauchy, so that is a Cauchy sequence in a complete quasi-metric space ( χ , q ) . Thus, there is u * χ such that
lim n q ( u * , u n ) = 0 = lim n q ( u * , u n ) .
Of course, using ( q 1 ) and the continuity of T, we have T u * = u * . □
Corollary 1.
Let ( χ , q ) be a complete quasi-metric space, a function α : X × X [ 0 , ) and a mapping T : χ χ such that there exist ζ Z and ψ Ψ such that, for p 0 , L 0 and a 1 , a 2 , a 3 [ 0 , 1 ) with a 1 + a 2 > 0 and a 1 + a 2 + a 3 = 1 , we have
ζ ( α ( u , v ) q ( T u , T v ) , ψ ( I p ( u , v ) + L N ( u , v ) ) ) 0 , f o r   a l l   d i s t i n c t   u , v χ .
Suppose also that the following assumptions hold:
(i) 
u = T u implies α ( u , v ) > 0 for every v χ ;
(ii) 
v = T v implies α ( u , v ) > 0 for every u χ ;
(i) 
T is α-orbital admissible;
(ii) 
there exists u 0 χ such that α ( u 0 , T u 0 ) 1 and α ( T u 0 , u 0 ) 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 ( χ , q ) be a complete quasi-metric space and a mapping T : χ χ such that there exist ζ Z and ψ Ψ such that, for p 0 , L 0 and a 1 , a 2 , a 3 [ 0 , 1 ) with a 1 + a 2 > 0 and a 1 + a 2 + a 3 = 1 , we have
ζ ( q ( T u , T v ) , ψ ( I p ( u , v ) + L N ( u , v ) ) ) 0 , f o r   a l l   d i s t i n c t   u , v χ .
Then, T has a fixed point.
Proof. 
Let α ( u , v ) = 1 in Corollary 1. □
Corollary 3.
Let ( χ , q ) be a complete quasi-metric space, a function α : χ × χ [ 0 , ) and a continuous mapping T : χ χ such that there exist ψ Ψ such that, for p 0 and a 1 , a 2 , a 3 [ 0 , 1 ) with a 1 + a 2 > 0 and a 1 + a 2 + a 3 = 1 , we have
α ( u , v ) q ( T u , T v ) ψ ( I p ( u , v ) ) , f o r   a l l   d i s t i n c t   u , v χ .
Suppose that there exists u 0 χ such that α ( u 0 , T u 0 ) 1 and α ( T u 0 , u 0 ) 1 . Then, T has a fixed point.
Proof. 
Let ζ ( t , s ) = ψ ( s ) 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 ( χ , q ) be a complete quasi-metric space, a function α : χ × χ [ 0 , ) , and a continuous mapping T : χ χ such that there exists ψ Ψ such that, for a 1 , a 2 , a 3 [ 0 , 1 ) with a 1 + a 2 > 0 and a 1 + a 2 + a 3 = 1 , we have
α ( u , v ) q ( T u , T v ) ψ ( ( q ( u , v ) ) a 1 · ( q ( u , T u ) ) a 2 · ( q ( v , T v ) ) a 3 ) , f o r   a l l   d i s t i n c t   u , v χ .
Suppose that there exists u 0 χ such that α ( u 0 , T u 0 ) 1 and α ( T u 0 , u 0 ) 1 . Then, T has a fixed point.
Inspired by the example of [10], we consider the following:
Example 1.
Let the set χ = [ 1 , ) and the quasi-metric q : χ × χ [ 0 , ) given by
q ( u , v ) = ln v ln u , if   u v 1 3 ( ln u ln v ) , if   u > v .
(see Example 4.1 in [10].) Let the mapping T : χ χ , be defined by
T u = 1 , if   u [ 1 , 2 ] e u 2 , if   u ( 2 , )
and the function α : χ × χ [ 0 , ) be defined by
α ( u , v ) = 2 , if   u , v [ 1 , 2 ) 3 , if   u = 1 , v = 2   or   u = 2 , v = 1 6 , if   u = 3 , v = 2 0 , otherwise .
Since the mapping T is continuous and for u = 2 , α ( 2 , T 2 ) = α ( 2 , 1 ) = 3 and α ( T 2 , 2 ) = α ( 1 , 2 ) = 3 , we have that the assumptions ( C 2 ) , ( C 3 ) are satisfied. Moreover, for any u [ 1 , 2 ) , we have
α ( u , T u ) = α ( u , 1 ) = 2 α ( T 1 , T 2 1 ) = α ( 1 , 1 ) = 2
and
α ( 2 , T 2 ) = α ( 2 , 1 ) = 3 α ( T 2 , T 2 2 ) = α ( 1 , 1 ) = 2 ,
so that T is α-orbital admissible.
Choosing ψ ( t ) = 1 3 t , p = 2 , a 1 = a 2 = a 3 = 1 3 and L = 24 , we have the following cases:
Case 1.  If u , v [ 1 , 2 ] , then q ( u , v ) = q ( 1 , 1 ) = 0 and (3) holds for every ζ Z .
Case 2. If u = 3 , v = 2 , then we have
q ( 3 , T 3 ) = q ( 3 , e ) = 1 3 ln 3 e , q ( 2 , T 2 ) = q ( 2 , 1 ) = 1 3 ln 2 , q ( T 2 , 2 ) = q ( 1 , 2 ) = ln 2 , q ( 3 , 2 ) = 1 3 ln 3 2 , q ( T 3 , T 2 ) = q ( e , 1 ) = 1 3 , q ( 3 , T 2 ) = q ( 3 , 1 ) = 1 3 ln 3 , q ( 2 , T 3 ) = q ( 2 , e ) = ln e 2 .
Thus, we have
1 2 min q ( 3 , T 3 ) , q ( 2 , T 2 ) , q ( T 2 , 2 ) = 1 6 ln 3 e < 1 3 ln 3 2 = q ( 3 , 2 )
and
α ( 3 , 2 ) q ( T 3 , T 2 ) = 6 3 < 1 3 1 3 ( 1 3 ln 3 2 ) 2 + ( 1 3 ln 3 e ) 2 + ( 1 3 ln 2 ) 2 1 / 2 + 24 ln e 2 = ψ ( I p ( 3 , 2 ) + L N ( 3 , 2 ) ) ,
so that T is a hybrid almost contraction for any ζ Z .
The other cases are not interesting, while α ( u , v ) = 0 . (Consequently, the mapping T has two fixed points, u 1 = 1 and u 2 ( 3 , 4 ) . )
On the other hand, since
α ( 3 , 2 ) q ( T 3 , T 2 ) = 2 > ( 1 3 ln 3 2 ) γ ( 1 3 ln 3 e ) β ( 1 3 ln 2 ) 1 γ β > ψ ( q ( 3 , 2 ) ) γ ( q ( 3 , T 3 ) ) β , ( q ( 2 , T 2 ) ) 1 γ β
for every γ , β ( 0 , 1 ) and ψ Ψ , 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 ( χ , q ) be a complete quasi-metric space, a function α : χ × χ [ 0 , ) and a mapping T : χ χ such that there exist ζ Z , ψ Ψ , L 0 and a 1 , a 2 , a 3 [ 0 , 1 ] with a 1 + a 2 + a 3 = 1 , such that, for all distinct u , v χ , we have
1 2 min q ( u , T u ) , q ( v , T v ) , q ( T v , v ) q ( u , v )   implies ζ ( α ( u , v ) q ( T u , T v ) , ψ ( ( q ( u , v ) ) a 1 · ( q ( u , T u ) ) a 2 · ( q ( v , T v ) ) a 3 + L N ( u , v ) ) ) 0 .
Suppose also that
(i) 
u = T u implies α ( u , v ) > 0 for every v χ ;
(ii) 
v = T v implies α ( u , v ) > 0 for every u χ ;
(C1
T is α-orbital admissible;
(C2
there exists u 0 χ such that α ( u 0 , T u 0 ) 1 and α ( T u 0 , u 0 ) 1 ;
(C3
χ is regular with respect to the mapping α.
Then, T has a fixed point.
Proof. 
From the above theorem, there exists u * χ such that (27) hold. In what follows, we claim that
1 2 min q ( u * , T u * ) , q ( u n ( i ) , T u n ( i ) ) , q ( T u n ( i ) , u n ( i ) ) q ( u * , u n ( i ) ) or 1 2 min q ( u n ( i ) 1 , T u n ( i ) 1 ) , q ( u * , T u * ) , q ( T u * , u * ) q ( u n ( i ) 1 , u * ) .
Indeed, using the method of Reductio ad Absurdum, we assume that that there exists k N such that
1 2 min q ( u * , T u * ) , q ( u k , T u k ) , q ( T u k , u k ) > q ( u * , u k ) and 1 2 min q ( u k 1 , T u k 1 ) , q ( u * , T u * ) , q ( T u * , u * ) > q ( u k 1 , u * )
Therefore, we have
q ( u k 1 , u k ) q ( u k 1 , u * ) + q ( u * , u k ) < 1 2 min q ( u k 1 , T u k 1 ) , q ( u * , T u * ) , q ( T u * , u * ) + 1 2 min q ( u * , T u * ) , q ( u k 1 , T u k 1 ) , q ( T u k 1 , u k 1 ) < 1 2 [ min q ( u k 1 , u k ) , q ( u * , T u * ) , q ( T u * , u * ) + min q ( u * , T u * ) , q ( u k 1 , u k ) , q ( u k , u k 1 ) ] 1 2 [ q ( u k 1 , u k ) + q ( u k 1 , u k ) ] = q ( u k 1 , u k ) ,
which is a contradiction.
In the alternative hypothesis, if the space X is regular with respect to mapping α , we have α ( u * , u n ( i ) ) 1 , where { u n ( i ) } is a sub-sequence of { u n } , for i N . We will suppose by reductio ad absurdum that u * T u * . Then, for u = u * and v = u n ( i ) in (3), we get
ζ ( α ( u * , u n ( i ) ) q ( T u * , T u n i ) ) , ψ ( A p ( u * , u n ( i ) ) ) ) 0 .
Taking into account the properties of function ζ , ψ , and α , the above relation becomes
q ( T u * , u * ) q ( T u * , T u n ) + q ( T u n , u * ) α ( u * , u n ) q ( T u * , T u n ( i ) ) + q ( u n ( i ) + 1 , u * ) ψ ( ( q ( u * , u n ( i ) ) ) a 1 · ( q ( u * , T u * ) ) a 2 · ( q ( u n ( i ) , T u n ( i ) ) ) a 3 + N ( u * , u n ( i ) ) ) + q ( u n ( i ) + 1 , u * ) ,
Letting i , we have
0 < q ( T u * , u * ) < lim i ψ ( ( q ( u * , u n ( i ) ) ) a 1 · ( q ( u * , T u * ) ) a 2 · ( q ( u n ( i ) , T u n ( i ) ) ) a 3 + N ( u * , u n ( i ) ) ) + q ( u n ( i ) + 1 , u * )
and, since ψ is continuous in 0, ψ ( 0 ) = 0 , we get q ( T u * , u * ) = 0 .  □
Corollary 5.
Let ( χ , q ) be a complete quasi-metric space and T : χ χ be a given mapping. Assume that there exist L 0 , ζ Z and ψ Ψ such that, for all distinct u , v χ , we have
1 2 min q ( u , T u ) , q ( v , T v ) q ( T v , v ) q ( u , v )   implies ζ ( q ( T u , T v ) , ψ ( I p ( u , v ) + L N ( u , v ) ) ) 0 ,
for all distinct u , v χ . Then, T has a fixed point.
Proof. 
It is sufficient to take α ( u , v ) = 1 for u , v χ in Theorem 5. □
Corollary 6.
Let ( χ , q ) be a complete quasi-metric space and T : χ χ be a given mapping. Assume that there exist L 0 , ζ Z and ψ Ψ such that, for all distinct u , v χ , we have
1 2 min q ( u , T u ) , q ( v , T v ) q ( T v , v ) q ( u , v )   implies   ( T u , T v ) k I p ( u , v )
for all distinct u , v χ . Then, T has a fixed point.
Proof. 
It is sufficient to take L = 0 , ζ ( t , s ) = k 1 s t , ψ ( u ) = k 2 u with k 1 , k 2 ( 0 , 1 ) and k = k 1 k 2 in Corollary 5. □
Corollary 7.
Let ( χ , q ) be a complete quasi-metric space and T : χ χ a continuous mapping such that
1 2 min q ( u , T u ) , q ( v , T v ) q ( T v , v ) q ( u , v )   implies q ( T u , T v ) k 3 · ( q ( u , v ) ) 2 + ( q ( u , T u ) ) 2 + ( q ( v , T v ) ) 2
for all distinct u , v χ and some k ( 0 , 1 ) . Then, T has a fixed point in χ.
Proof. 
Let p = 2 and a 1 = a 2 = a 3 = 1 3 in Corollary 6. □
In the next theorem, we involve a Jaggi type expression with the hybrid contractions.
Definition 2.
Let ( χ , q ) be a quasi-metric space. A mapping T : χ χ is called a hybrid almost contraction of type J , if there exist ζ Z and ψ Ψ such that, for p 0 , L 0 and a 1 , a 2 > 0 with a 1 + a 2 < 1 , we have
1 2 min q ( u , T u ) , q ( v , T v ) q ( T v , v ) q ( u , v )   implies ζ ( α ( u , v ) q ( T u , T v ) , ψ ( J p ( u , v ) + L N ( u , v ) ) ) 0 ,
for all distinct u , v χ , where
J p ( u , v ) = [ a 1 ( q ( u , v ) ) p + a 2 ( q ( u , T u ) ) · ( q ( v , T v ) q ( u , v ) ) p ] 1 / p , for   p > 0 ( q ( u , v ) ) a 1 · ( q ( u , T u ) ) a 1 · ( q ( v , T v ) ) 1 a 1 a 2 , for   p = 0
and
N ( u , v ) = min q ( u , T v ) , q ( v , T u ) .
Theorem 3.
Let ( χ , q ) be a complete quasi-metric space and α : X × X [ 0 , ) be a mapping such that:
(i) 
u = T u implies α ( u , v ) > 0 for every v χ ;
(ii) 
v = T v implies α ( u , v ) > 0 for every u χ .
Suppose that T : χ χ is a hybrid almost contraction of type J such that the following assumptions hold:
(i) 
T is α-orbital admissible;
(ii) 
there exists u 0 χ such that α ( u 0 , T u 0 ) 1 and α ( T u 0 , u 0 ) 1 ;
(iii) 
there exists Δ > 0 such that, ( a 1 + a 2 Δ 2 p ) 1 / p 1 (where p > 0 ) and
1 Δ q ( u , v ) q ( v , u ) Δ q ( u , v ) , f o r   a l l u , v χ ;
(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 u n χ . Onward, as in the proof of Theorem 1, we suppose that u n + 1 u n for all n N and from (35), we have 1 2 min q ( u n 1 , T u n 1 ) , q ( u n , T u n ) , q ( T u n , u n ) q ( u n 1 , u n ) , which implies
ζ ( α ( u n 1 , u n ) q ( T u n 1 , T u n ) , ψ ( J p ( u n 1 , u n ) + L N ( u n 1 , u n ) ) ) 0 .
By the axiom ( z d ) , Lemma 1 and taking into account (7), this inequality becomes
q ( u n , u n + 1 ) α ( u n 1 , u n ) q ( T u n 1 , T u n ) ψ ( J p ( u n 1 , u n ) + L N ( u n 1 , u n ) ) < J p ( u n 1 , u n ) = [ a 1 ( q ( u n 1 , u n ) ) p + a 2 ( q ( u n 1 , T u n 1 ) · q ( u n , T u n ) q ( u n 1 , u n ) ) p ] 1 / p = [ a 1 ( q ( u n 1 , u n ) ) p + a 2 ( q ( u n 1 , u n ) · ( q u n , u n + 1 ) q ( u n 1 , u n ) ) p ] 1 / p = [ a 1 ( q ( u n 1 , u n ) ) p + a 2 ( q ( u n , u n + 1 ) ) p ] 1 / p .
Thereupon,
q ( u n , u n + 1 ) < a 1 1 a 2 1 / p q ( u n 1 , u n ) < q ( u n 1 , u n )
and then, from (36), we have q ( u n , u n + 1 ) < ψ ( q ( u n 1 , u n ) ) . Since ψ Ψ , recursively, we get
q ( u n , u n + 1 ) < ψ ( q ( u n 1 , u n ) ) < < ψ n ( q ( u 0 , u 1 ) ) .
In order to prove that u n is a right-Cauchy sequence, let l N . From (37) and the triangle inequality, we get that
q ( u n , u n + l ) q ( u n , u n + 1 ) + + q ( u n + l 1 , u n + l ) j = n n + l 1 ψ j ( q ( u 0 , u 1 ) ) j = n ψ j ( q ( u 0 , u 1 ) ) 0 , as   n .
We conclude that q n is a right-Cauchy sequence in ( χ , q ) .
Substituting in (35) u = u n and v = u n 1 and since 1 2 min q ( u n , T u n , q ( u n 1 , T u n 1 ) , q ( T u n 1 , u n 1 ) ) q ( u n , u n 1 ) , we have (taking into account (11))
q ( u n + 1 , u n ) α ( u n , u n 1 ) q ( T u n , T u n 1 ) ψ ( J p ( u n , u n 1 ) + L N ( u n , u n 1 ) ) < J p ( u n , u n 1 ) = [ a 1 ( q ( u n , u n 1 ) ) p + a 2 ( q ( u n , u n + 1 ) ) · ( q ( u n 1 , u n ) q ( u n , u n 1 ) p ] 1 / p
i.e.,
( q ( u n + 1 , u n ) ) p < a 1 ( q ( u n , u n 1 ) ) p + a 2 ( q ( u n , u n + 1 ) ) · ( q ( u n 1 , u n ) q ( u n , u n 1 ) p .
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 ( i i i ), there exists a positive constant Δ such that q ( u n 1 , u n ) Δ q ( u n , u n 1 ) for n N . Thus, we have
( q ( u n + 1 , u n ) ) p < a 1 ( q ( u n , u n 1 ) ) p + a 2 ( ( q ( u n 1 , u n ) ) 2 q ( u n , u n 1 ) ) p < a 1 ( q ( u n , u n 1 ) ) p + a 2 ( ( Δ · q ( u n , u n 1 ) ) 2 q ( u n , u n 1 ) ) p = ( a 1 + a 2 Δ 2 p ) · ( q ( u n , u n 1 ) ) p ,
which is equivalent to the next inequality
q ( u n + 1 , u n ) < ( a 1 + a 2 Δ 2 p ) 1 / p q ( u n , u n 1 ) q ( u n , u n 1 ) .
Thus,
q ( u n + 1 , u n ) < ψ ( q ( u n , u n 1 ) ) < < ψ n ( q ( u 1 , u 0 ) )
Again, considering triangle inequality, together with (38), for l N , we have
q ( u n + l , u n ) q ( u n + l , u n + l 1 ) + + q ( u n 1 , u n ) j = n ψ j ( q ( u 0 , u 1 ) ) 0 , as   n .
Analogously, we deduce that u n is left-Cauchy, so that it is a Cauchy sequence in complete quasi-metric space.
Thus, there exists u * χ such that
lim n q ( u n , u * ) = lim n q ( u * , u n ) = 0 .
Under the assumption ( i v ) , from the continuity of T and ( q 1 ) , we have
lim n q ( u n , T u * ) = lim n q ( T u n 1 , T u * ) = 0 ,
lim n q ( T u * , u n ) = lim n q ( T u * , T u n 1 ) = 0
so that
lim n q ( u n , T u * ) = lim n q ( T u * , u n ) = 0 .
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 ( χ , q ) be a complete quasi-metric space, a function α : X × X [ 0 , ) and a mapping T : χ χ such that there exist ζ Z and ψ Ψ such that, for p 0 , L 0 and a 1 , a 2 , [ 0 , 1 ) with a 1 + a 2 < 1 , we have
ζ ( α ( u , v ) q ( T u , T v ) , ψ ( J p ( u , v ) + L N ( u , v ) ) ) 0 , f o r   a l l   d i s t i n c t   u , v χ .
Suppose also that the following assumptions hold:
(i) 
u = T u implies α ( u , v ) > 0 for every v χ ;
(ii) 
v = T v implies α ( u , v ) > 0 for every u χ ;
(i) 
T is α-orbital admissible;
(ii) 
there exists u 0 X such that α ( u 0 , T u 0 ) 1 and α ( T u 0 , u 0 ) 1 ;
(iii) 
there exists Δ > 0 such that, ( a 1 + a 2 Δ 2 p ) 1 / p 1 (where p > 0 ) and
1 Δ q ( u , v ) q ( v , u ) Δ q ( u , v ) , f o r   a l l   u , v χ ;
(iv) 
T is continuous.
Then, T has a fixed point.
Example 2.
Let X = [ 0 , 1 ] and the function
q ( u , v ) = u v , for   u v 2 ( v u ) , for   u < v
It is easy to see that the pair ( χ , q ) forms a quasi-metric space.
Let the map T : χ χ defined by
T u = 1 8 , for   u [ 0 , 1 2 ] u 4 , for   u [ 1 2 , 1 ]
and choose ζ ( u , v ) = 1 2 v u and ψ ( t ) = 1 2 t . For p = 2 , L = 0 , Δ = 2 , a 1 = 1 4 and a 2 = 1 32 because ( a 1 + a 2 · Δ 2 p ) 1 / p = 1 4 + 1 32 · 2 4 = 3 4 1 , the assumption ( i i i ) is satisfied. In this case, (41) becomes
α ( u , v ) q ( u , v ) J p ( u , v ) = 1 4 1 4 ( q ( u , v ) ) 2 + 1 32 ( q ( u , T u ) q ( v , T v ) q ( u , v ) ) 2 .
Define α : χ × χ [ 0 , ) such that
α ( u , v ) = 3 , for   u , v [ 0 , 1 2 ) 1 , for   u = 1 , v = 0 0 , otherwise
It is easy to see that T is α-admissible. Indeed, we have
α ( u , v ) = 3 α ( T u , T v ) = α ( 1 / 8 , 1 / 8 ) = 3 , for   u , v [ 0 , 1 2 )
and
α ( 1 , 0 ) = 1 α ( T 1 , T 0 ) = α ( 1 / 4 , 1 / 8 ) = 3 .
On the other hand, for q 0 = 0 ,
α ( 0 , T 0 ) = α ( T 0 , 0 ) = α ( 0 , 0 ) = 3 ,
so that the presumptions ( i ) , ( i i ) , and ( i v ) are satisfied. Of course, if u , v [ 0 , 1 2 ) , we have q ( T u , T v ) = q ( 1 8 , 1 8 ) = 0 and (41) is verified. For u = 1 and v = 0 , we have q ( T 1 , T 0 ) = 1 4 1 8 = 1 8 , q ( 0 , T 0 ) = q ( 0 , 1 / 8 ) = 2 ( 1 / 8 0 ) = 1 / 4 , q ( 1 , T 1 ) = q ( 1 , 1 / 4 ) = 3 / 4 and
α ( 1 , 0 ) q ( T 1 , T 0 ) = 1 8 1 4 1 4 + 1 32 ( 3 16 ) 2 = 1 4 1 4 ( q ( 1 , 0 ) ) 2 + 1 32 ( q ( 0 , T 0 ) q ( 1 , T 1 ) q ( 1 , 0 ) ) 2
The other cases are not interesting since α ( u , v ) = 0 and the condition (42) is fulfilled trivially. Thus, the presumptions of Theorem 8 are provided and u = 1 8 is the fixed point of T.
Corollary 9.
Let ( χ , q ) be a complete quasi-metric space and T be a continuous self-mapping on χ. Suppose that there exist ζ Z , ψ Ψ such that
ζ ( q ( T u , T v ) , ψ ( J p ( u , v ) ) ) 0 ,
for each distinct u , v χ . If there exists Δ > 0 such that ( a 1 + a 2 · Δ 2 p ) 1 / p 1 for p > 0 , and 1 Δ q ( u , v ) q ( v , u ) Δ q ( u , v ) for all u , v χ , then T has a fixed point.
Proof. 
It is sufficient to take L = 0 and α ( u , v ) = 1 for u , v χ 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 Δ > 0 such that ( a 1 + a 2 · Δ 2 p ) 1 / p 1 for p > 0 , and 1 Δ q ( u , v ) q ( v , u ) Δ q ( u , v ) for all u , v χ . The mapping T has a fixed point provided that
q ( T u , T v ) c · J p ( u , v )
for each distinct u , v χ and some c ( 0 , 1 ) .
Proof. 
We set ζ ( t , s ) = c 1 s t , ψ ( u ) = c 2 u with c 1 , c 2 [ 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 ( χ , q ) be the quasi-metric space, where χ = [ 1 , ) and
q ( u , v ) = u v , for   u v 2 ( v u ) , for   u < v
Let
T u = u 3 8 u 2 + 19 u 9 , for   u [ 1 , 5 ] l n ( u 2 24 ) + u + 6 , for   u ( 5 , ) .
Consider the function ζ be arbitrary in Z , ψ Ψ with ψ ( t ) = t 3 and α : χ × χ [ 0 , ) such that
α ( u , v ) = u 2 + 1 , for   ( u , v ) ( 3 , 3 ) , ( 3 , 4 ) , ( 4 , 3 ) , ( 3 , 1 ) , ( 1 , 3 ) 1 , for   ( u , v ) = ( 2 , 1 ) 0 , otherwise .
It is easily verified that T is α-orbital admissible. Whereas T 1 = T 3 = T 4 = 3 , taking into account the definition of function α, we have that the inequality (41) holds for every pair ( u , v ) χ 2 \ ( 2 , 1 ) . For the case u = 2 and v = 1 , choosing a 1 = 1 2 , a 2 = 1 48 and p = 2 , we find that axiom ( i i i ) holds. On the other hand,
J p ( 2 , 1 ) = 1 2 ( q ( 2 , 1 ) ) 2 + 1 48 q ( 2 , T 2 ) · q ( 1 , T 1 ) q ( 2 , 1 ) 2 1 / 2 = 1 2 + 1 48 · q ( 2 , 5 ) · q ( 1 , 3 ) q ( 2 , 1 ) 2 = 25 2
and
α ( 2 , 1 ) q ( T 2 , T 1 ) = q ( 5 , 3 ) = 2 < 25 6 = ψ ( J p ( 2 , 1 ) ) .
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 ,
q ( T 1 , T ( 4.5 ) ) = q ( 2 , 5.625 ) = 7.25 , q ( 1 , T 1 ) = q ( 1 , 2 ) = 2 , q ( 4.5 , T ( 4.5 ) ) = q ( 4.5 , 5.625 ) = 1.125 ,
so that, since
q ( T 1 , T ( 4.5 ) ) > λ ( q ( 1 , T 1 ) ) α ( q ( 4.5 , T ( 4.5 ) ) ) 1 α
for any λ [ 0 , 1 ) and α ( 0 , 1 ) , Theorem 2.2 in [20] can not be applied.
Corollary 11.
Let ( χ , q ) be a complete quasi-metric space and T : χ χ a continuous mapping. Then, T has a fixed point provided that
q ( T u , T v ) k 1 · q ( u , v ) + k 2 · q ( u , T u ) q ( v , T v ) q ( u , v )
for each u , v X and k 1 , k 2 ( 0 , 1 ) with k 1 + k 2 < 1
Proof. 
Let p = 1 and k i = c · a i , for i { 1 , 2 } in Corollary 10. □

Author Contributions

All authors contributed equally and significantly in writing this article. Writing—original draft, A.F. and E.K.; writing—review and editing, A.F., G.P., and E.K. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Acknowledgments

The authors thank the anonymous referees for their remarkable comments, suggestions, and ideas that helped to improve this paper.

Conflicts of Interest

The authors declare no conflict of interest.

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