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Article

A New Perspective on Parametric Metric Spaces

1
Department of Mathematics, Ege University, Bornova, 35100 Izmir, Turkey
2
Institute of Research and Development of Processes University of the Basque Country, 48940 Leioa, Bizkaia, Spain
*
Author to whom correspondence should be addressed.
Mathematics 2019, 7(11), 1008; https://doi.org/10.3390/math7111008
Submission received: 25 September 2019 / Revised: 18 October 2019 / Accepted: 21 October 2019 / Published: 24 October 2019

Abstract

:
In this study, we develop a new approach on parametric metric spaces using C -class functions. Moreover, we give some examples to support our findings. The obtained results generalize and extend some existing facts in the literature.

1. Introduction and Mathematical Preliminaries

Fixed point theory has extensive applications in several branches of mathematics, economics, engineering, and statistics with various problems in the theory of differential and integral equations, approximation theory, game theory, and others. To show the existence and uniqueness of fixed points and to determine common fixed points are very popular for researchers in this area.
The concept of coupled fixed point has been introduced in [1]. Coupled coincidence and coupled common fixed point results have been obtained in [2]. There are many studies [3,4,5,6,7,8] on coupled fixed point theorems in metric spaces.
The notion of parametric metric space was introduced in 2014. Rao et al. [9] present parametric S-metric spaces and prove common fixed point theorems. In [10], Banach fixed point theorem has been extended to continuous mappings on complete parametric b-metric spaces. Tas and Ozgur [11] introduce parametric N b -metric spaces, obtain some fixed point results and prove a fixed-circle theorem on a parametric N b -metric space as an application. For other significant studies, see [12,13,14,15,16,17].
We now recall some facts for our main results. For a nonempty set X, a mapping p : X × X × ( 0 , ) [ 0 , ) is called a parametric metric [18] if,
(i)
p ( u , v , t ) = 0 for all t > 0 iff u = v ;
(ii)
p ( u , v , t ) = p ( v , u , t ) for all t > 0 ;
(iii)
p ( u , v , t ) p ( u , w , t ) + p ( w , v , t ) for all u , v , w X and all t > 0 .
Example 1
([18]). The function p : X × X × ( 0 , ) [ 0 , ) defined by p ( f 1 , f 2 , t ) = | f 1 ( t ) f 2 ( t ) | for all f 1 , f 2 X and all t > 0 is parametric metric on X which is the set of all functions f : ( 0 , ) R .
Definition 1
([18]). Let ( X , p ) be a parametric metric space.
  • A point x X is called the limit of a sequence ( x n ) , if lim n p ( x , x n , t ) = 0 for all t > 0 , and the sequence ( x n ) is called convergent to x.
  • A sequence ( x n ) is said to be a Cauchy if and only if lim m , n p ( x n , x m , t ) = 0 for all t > 0 .
  • A parametric metric space ( X , p ) is called complete if and only if every Cauchy sequence is convergent to x X .
Definition 2
([18]). Let ( X , p ) be a parametric metric space and let T : X X be a mapping. If for any sequence ( x n ) in X such that x n x as n , T x n T x as n , then T is a continuous mapping at x X .
C-class functions have been presented in [19].
Definition 3
([19]). A continuous mapping F : [ 0 , ) × [ 0 , ) R is said to be C-class function if it satisfies the following:
(1) F ( κ , λ ) κ for all κ , λ [ 0 , ) ;
(2) F ( κ , λ ) = κ implies that either κ = 0 or λ = 0 .
The C-class functions will be denoted by C .
Example 2
([19]). Some elements of C are given in the following for all κ , λ [ 0 , ) :
(1) F ( κ , λ ) = κ ( 1 + λ ) r ; for all r ( 0 , ) ,
(2) F ( κ , λ ) = κ ( 1 + κ 2 + κ ) ( λ 1 + λ ) ,
(3) F ( κ , λ ) = κ Γ ( 1 / 2 ) 0 e x x + λ d x , where Γ is the Euler Gamma function.
Let Ψ denote the set of all continuous and monotone nondecreasing functions ψ : [ 0 , ) [ 0 , ) such that ψ ( λ ) = 0 iff λ = 0 , ψ ( κ + λ ) ψ ( κ ) + ψ ( λ ) for all κ , λ [ 0 , ) .
Let Φ 1 denote the set of all continuous functions φ : [ 0 , ) [ 0 , ) such that φ ( λ ) = 0 iff λ = 0 and Φ u denote the set of all continuous functions φ : [ 0 , ) [ 0 , ) such that φ ( 0 ) 0 , note that Φ 1 Φ u .
Ege and Karaca [20] establish a coupled fixed point theorem and give a homotopy application in parametric metric spaces.
Definition 4
([20]). Let ( X , p ) be a parametric metric space, ( x , y ) be an element in X × X and F : X × X X and g : X X be given two functions.
  • If F ( x , y ) = x and F ( y , x ) = y , then ( x , y ) is said to be a coupled fixed point of F.
  • If F ( x , y ) = g x and F ( y , x ) = g y , then ( x , y ) is called a coupled coincidence point of F and g.
  • F and g are said to be commutative if g F ( x , y ) = F ( g x , g y ) .
The goal of this study is to give some generalizations of the following theorems from the literature using C-class functions.
Theorem 1
([20]). Let ( X , p ) be a parametric metric space. Let F : X × X X and g : X X be two maps such that
p ( F ( x , y ) , F ( a , b ) , t ) α [ p ( g x , g a , t ) + p ( g y , g b , t ) ]
for all x , y , a , b X and t > 0 . If
(i) 
F ( X × X ) is a subset of g ( X ) ,
(ii) 
g ( X ) is a complete parametric metric space,
(iii) 
g is continuous,
(iv) 
g and F are commutative,
and α ( 0 , 1 2 ) , then there is a unique x in X such that g x = F ( x , x ) = x .
Theorem 2
([21]). Let ( X , p ) be a complete parametric metric space. If a continuous mapping T : X X satisfies the following
p ( T x , T y , t ) α [ p ( x , T x , t ) , p ( y , T y , t ) ] + β [ p ( x , T y , t ) , p ( T x , y , t ) ] ,
for all x , y X , and α + β < 1 2 , α , β [ 0 , 1 2 ) , then T has a fixed point in X.
Theorem 3
([21]). Let ( X , p ) be a complete parametric metric space and T a continuous map satisfying
p ( T x , T y , t ) α max { p ( x , y , t ) , p ( x , T x , t ) , p ( y , T y , t ) , p ( x , T y , t ) , p ( T x , y , t ) } ,
for all x , y X , x y , and for all t > 0 , where α [ 0 , 1 ) . Then T has a unique fixed point in X.

2. Main Results

In this section, using the C-class functions, we give generalizations of some fixed point theorems from the literature.
Lemma 1.
Let ( X , p ) be a parametric metric space and the mappings g : X X and F : X × X X satisfy the following condition
ψ ( p ( F ( x , y ) , F ( a , b ) , t ) ) 1 2 f ( ψ ( p ( g x , g a , t ) + p ( g y , g b , t ) ) , φ ( p ( g x , g a , t ) + p ( g y , g b , t ) ) ) ,
for all x , y , a , b X and t > 0 , where φ Φ u , ψ Ψ , f C and ( x , y ) is a coupled coincidence point of g and F. Then F ( x , y ) = g x = g y = F ( y , x ) .
Proof. 
Using the definition of a coupled coincidence point, we obtain F ( x , y ) = g x and F ( y , x ) = g y for the mappings g and F. If we assume that g x g y and use the inequality (1), then we have the following statements:
ψ ( p ( g x , g y , t ) ) = ψ ( p ( F ( x , y ) , F ( y , x ) , t ) ) 1 2 f ( ψ ( p ( g x , g y , t ) + p ( g y , g x , t ) ) , φ ( p ( g x , g y , t ) + p ( g y , g x , t ) ) ) = 1 2 f ( ψ ( p ( g x , g y , t ) + p ( g x , g y , t ) ) , φ ( p ( g x , g y , t ) + p ( g x , g y , t ) ) ) = 1 2 f ( ψ ( 2 p ( g x , g y , t ) ) , φ ( 2 p ( g x , g y , t ) ) ) .
There are two cases have to be considered. If f ( ψ ( 2 p ( g x , g y , t ) ) , φ ( 2 p ( g x , g y , t ) ) ) = ψ ( 2 p ( g x , g y , t ) ) , then we have ψ ( 2 p ( g x , g y , t ) ) = 0 or φ ( 2 p ( g x , g y , t ) ) = 0 . Hence p ( g x , g y , t ) = 0 , which is a contradiction. If f ( ψ ( 2 p ( g x , g y , t ) ) , φ ( 2 p ( g x , g y , t ) ) ) ψ ( 2 p ( g x , g y , t ) ) , a contradiction is reached. As a result, we have g x = g y , that is, F ( x , y ) = g x = g y = F ( y , x ) . □
Theorem 4.
Let ( X , p ) be a parametric metric space. Let the mappings g : X X and F : X × X X satisfy (1) for all x , y , a , b X and t > 0 , where φ Φ u , ψ Ψ , f C . Suppose that the following conditions hold:
(I) 
F ( X × X ) is a subset of g ( X ) ,
(II) 
g is continuous,
(III) 
F and g are commutative,
(IV) 
g ( X ) is a complete parametric metric space.
Then there exists a unique element x in X such that g x = F ( x , x ) = x .
Proof. 
Consider two points x 0 and y 0 in X. Using ( I ) , it can be chosen new points x 1 , y 1 X such that g x 1 = F ( x 0 , y 0 ) and g y 1 = F ( y 0 , x 0 ) , and similarly, x 2 , y 2 X such that g x 2 = F ( x 1 , y 1 ) and g y 2 = F ( y 1 , x 1 ) . More generally, it can be constructed two sequences ( x n ) and ( y n ) as follows:
g x n + 1 = F ( x n , y n ) and g y n + 1 = F ( y n , x n ) .
The inequality (1) implies the following:
ψ ( p ( g x n , g x n + 1 , t ) ) = ψ ( p ( F ( x n 1 , y n 1 ) , F ( x n , y n ) , t ) ) 1 2 f ( ψ ( p ( g x n 1 , g x n , t ) + p ( g y n 1 , g y n , t ) ) , φ ( p ( g x n 1 , g x n , t ) + p ( g y n 1 , g y n , t ) ) )
for n N and all t > 0 . By the inequalities
ψ ( p ( g x n 1 , g x n , t ) ) = ψ ( p ( F ( x n 2 , y n 2 ) , F ( x n 1 , y n 1 ) , t ) ) 1 2 f ( ψ ( p ( g x n 2 , g x n 1 , t ) + p ( g y n 2 , g y n 1 , t ) ) , φ ( p ( g x n 2 , g x n 1 , t ) + p ( g y n 2 , g y n 1 , t ) ) )
and
ψ ( p ( g y n 1 , g y n , t ) ) = ψ ( p ( F ( y n 2 , x n 2 ) , F ( y n 1 , x n 1 ) , t ) ) 1 2 f ( ψ ( p ( g y n 2 , g y n 1 , t ) + p ( g x n 2 , g x n 1 , t ) ) , φ ( p ( g y n 2 , g y n 1 , t ) + p ( g x n 2 , g x n 1 , t ) ) ) ,
we see that
ψ ( p ( g x n 1 , g x n , t ) + p ( g y n 1 , g y n , t ) ) ψ ( p ( g x n 1 , g x n , t ) ) + ψ ( p ( g y n 1 , g y n , t ) ) f ( ψ ( p ( g y n 2 , g y n 1 , t ) + p ( g x n 2 , g x n 1 , t ) ) , φ ( p ( g y n 2 , g y n 1 , t ) + p ( g x n 2 , g x n 1 , t ) ) ) ,
that is,
p ( g x n 1 , g x n , t ) + p ( g y n 1 , g y n , t ) p ( g x n 2 , g x n 1 , t ) + p ( g y n 2 , g y n 1 , t ) p ( g x n 3 , g x n 2 , t ) + p ( g y n 3 , g y n 2 , t ) p ( g x 0 , g x 1 , t ) + p ( g y 0 , g y 1 , t )
where n N and t > 0 . So the sequence p ( g x n 1 , g x n , t ) + p ( g y n 1 , g y n , t ) is decreasing. In the limit case, we find
p ( g x n 1 , g x n , t ) + p ( g y n 1 , g y n , t ) r 0
for each t > 0 . As a result, we have
ψ ( r ) f ( ψ ( r ) , φ ( r ) ) r = 0 .
Let m , n N with m > n . From the condition ( i i i ) of the definition of parametric metric space, we obtain
ψ ( p ( g x n , g x m , t ) ) = ψ ( p ( F ( x n 1 , y n 1 ) , F ( x m 1 , y m 1 ) , t ) ) 1 2 f ( ψ ( p ( g x n 1 , g x m 1 , t ) + p ( g y n 1 , g y m 1 , t ) ) , φ ( p ( g x n 1 , g x m 1 , t ) + p ( g y n 1 , g y m 1 , t ) ) )
and
ψ ( p ( g y n , g y m , t ) ) = ψ ( p ( F ( y n 1 , x n 1 ) , F ( y m 1 , x m 1 ) , t ) ) 1 2 f ( ψ ( p ( g y n 1 , g y m 1 , t ) + p ( g x n 1 , g x m 1 , t ) ) , φ ( p ( g y n 1 , g y m 1 , t ) + p ( g x n 1 , g x m 1 , t ) ) ) .
Combining the last two inequalities, we find
ψ ( p ( g x n , g x m , t ) + p ( g y n , g y m , t ) ) ψ ( p ( g x n , g x m , t ) ) + ψ ( p ( g y n , g y m , t ) ) = ψ ( p ( F ( y n 1 , x n 1 ) , F ( y m 1 , x m 1 ) , t ) ) f ( ψ ( p ( g y n 1 , g y m 1 , t ) + p ( g x n 1 , g x m 1 , t ) ) , φ ( p ( g y n 1 , g y m 1 , t ) + p ( g x n 1 , g x m 1 , t ) ) ) .
Taking the limits as m , n , we have
ψ ( ε ) f ( ψ ( ε ) , φ ( ε ) ) ε = 0 .
Then we conclude that ( g x n ) is a Cauchy sequence in g ( X ) . In the same manner, ( g y n ) is also a Cauchy sequence in g ( X ) . From the condition ( I V ) , the sequences ( g x n ) and ( g y n ) are convergent to x X and y X , respectively. The condition ( I I ) shows that ( g g x n ) is convergent to g x and ( g g y n ) is convergent to g y . On the other hand, by ( I I I ) , there are following equalities
g g x n + 1 = g ( F ( x n , y n ) ) = F ( g x n , g y n ) and g g y n + 1 = g ( F ( y n , x n ) ) = F ( g y n , g x n ) .
Using the condition ( I I ) , we conclude that
g x = g ( lim g x n ) = g ( lim ( F ( x n 1 , y n 1 ) ) ) = lim ( F ( g x n 1 , g y n 1 ) ) = F ( lim g x n 1 , lim g y n 1 ) = F ( x , y )
and similarly g y = F ( y , x ) . Hence
ψ ( p ( g g x n + 1 , F ( x , y ) , t ) ) = ψ ( p ( F ( g x n , g y n ) , F ( x , y ) , t ) ) 1 2 f ( ψ ( p ( g g x n , g g x , t ) + p ( g g y n , g y , t ) ) , φ ( p ( g g x n , g g x , t ) + p ( g g y n , g y , t ) ) ) .
Lemma 1 implies that ( x , y ) is a coupled fixed point of F and g. That is, g x = F ( x , y ) = F ( y , x ) = g y . Since ( g x n + 1 ) is a subsequence of ( g x n ) , ( g x n + 1 ) is also convergent to x and
ψ ( p ( g x n + 1 , g x , t ) ) = ψ ( p ( F ( g x n , g y n ) , F ( x , y ) , t ) ) 1 2 f ( ψ ( p ( g x n , g x , t ) + p ( g y n , g y , t ) ) , φ ( p ( g x n , g x , t ) + p ( g y n , g y , t ) ) ) .
In the limit case, we get
ψ ( p ( x , g x , t ) ) 1 2 f ( ψ ( p ( x , g x , t ) + p ( y , g y , t ) ) , φ ( p ( x , g x , t ) + p ( y , g y , t ) ) )
because p is continuous. Similarly,
ψ ( p ( y , g y , t ) ) 1 2 f ( ψ ( p ( x , g x , t ) + p ( y , g y , t ) ) , φ ( p ( x , g x , t ) + p ( y , g y , t ) ) ) .
Combining Equations (2) and (3), we have
ψ ( p ( x , g x , t ) + p ( y , g y , t ) ) f ( ψ ( p ( x , g x , t ) + p ( y , g y , t ) ) , φ ( p ( x , g x , t ) + p ( y , g y , t ) ) ) .
The inequality (4) holds only if p ( x , g x , t ) = 0 and p ( y , g y , t ) = 0 . Moreover, we can write
ψ ( p ( g x , g x n + 1 , t ) ) = ψ ( p ( F ( x , y ) , F ( x n , y n ) , t ) ) 1 2 f ( ψ ( p ( g x , g x n , t ) + p ( g y , g y n , t ) ) , φ ( p ( g x , g x n , t ) + p ( g y , g y n , t ) ) ) .
Letting n and from the fact that the continuity of p, we have
ψ ( p ( g x , x , t ) ) 1 2 f ( ψ ( p ( g x , x , t ) + p ( g y , y , t ) ) , φ ( p ( g x , x , t ) + p ( g y , y , t ) ) ) .
The following inequality can be obtained in a similar way.
ψ ( p ( g y , y , t ) ) 1 2 f ( ψ ( p ( g x , x , t ) + p ( g y , y , t ) ) , φ ( p ( g x , x , t ) + p ( g y , y , t ) ) ) .
Now Equations (5) and (6) imply that
ψ ( p ( g x , x , t ) + p ( g y , y , t ) ) f ( ψ ( p ( g x , x , t ) + p ( g y , y , t ) ) , φ ( p ( g x , x , t ) + p ( g y , y , t ) ) ) .
Since (7) holds only when p ( x , g x , t ) = 0 and p ( y , g y , t ) = 0 , we have g x = F ( x , x ) = x .
All that remains in the proof is to show the uniqueness. Assuming τ X with τ x such that z = g z = F ( z , z ) , we conclude that
ψ ( p ( x , τ , t ) ) = ψ ( p ( F ( x , x ) , F ( τ , τ ) , t ) ) 1 2 f ( ψ ( p ( g x , g τ , t ) + p ( g x , g τ , t ) ) , φ ( p ( g x , g τ , t ) + p ( g x , g τ , t ) ) ) 1 2 f ( ψ ( 2 p ( g x , g τ , t ) ) , φ ( 2 p ( g x , g τ , t ) ) ) ψ ( p ( g x , g τ , t ) ) .
But this is a contradiction. So, there is a unique common fixed point of g and F. □
Example 3.
Let p ( x , y , t ) = t | x y | for all x , y X and all t > 0 be a parametric metric on X = R + { 0 } . It is easy to see that ( X , p ) is a complete parametric metric space. Define
F : X × X X , F ( x , y ) = x + y a n d g : X X , g ( x ) = 5 x .
The map g is continuous, F ( X × X ) = [ 0 , ) g ( X ) and g ( X ) = [ 0 , ) = X is a complete parametric metric space. Since
g F ( x , y ) = g ( x + y ) = 5 x + 5 y = F ( g x , g y ) ,
we have that F and g are commutative. We define the following mappings:
ψ : R + R + , ψ ( t ) = t 2 , φ : R + R + , φ ( t ) = 2 t , f : R + × R + R , f ( u , v ) = u .
Then we obtain
ψ ( p ( F ( x , y ) , F ( a , b ) , t ) ) = ψ ( p ( x + y , a + b , t ) ) = ψ ( t | x + y a b | ) = t 2 | x + y a b | t 2 ( | x a | + | y b | ) 5 4 t ( | x a | + | y b | ) = 1 4 t [ | 5 x 5 a | + | 5 y 5 b | ] = 1 4 [ p ( 5 x , 5 a , t ) + p ( 5 y , 5 b , t ) ] = 1 2 ψ ( p ( 5 x , 5 a , t ) + p ( 5 y , 5 b , t ) ) = 1 2 f ( ψ ( p ( g x , g a , t ) + p ( g y , g b , t ) ) , φ ( p ( g x , g a , t ) + p ( g y , g b , t ) ) )
for all x , y , a , b X and all t > 0 . From Theorem 4, 0 is the unique element in X such that F ( 0 , 0 ) = g ( 0 ) = 0 .
Theorem 5.
Let ( X , p ) be a complete parametric metric space and a continuous mapping T : X X satisfy the following condition
ψ ( p ( T x , T y , t ) ) F ψ ( Δ ( x , y ) ) , φ ( Δ ( x , y ) )
for all x , y X and all t > 0 , where F C , ψ Ψ , φ Φ u , and
Δ ( x , y ) = μ [ p ( x , T x , t ) + p ( y , T y , t ) ] + χ [ p ( x , T y , t ) + p ( T x , y , t ) ] ,
where μ + χ < 1 2 , μ , χ [ 0 , 1 2 ) . Then T has a fixed point in X.
Proof. 
Let x 0 be an arbitrary element in X. A sequence ( x n ) can be defined as T x n = x n + 1 for n N . If we take x = x n and y = x n + 1 in (8), we have
ψ ( p ( x n + 1 , x n + 2 , t ) ) = ψ ( p ( T x n , T x n + 1 , t ) ) F ( ψ ( μ [ p ( x n , T x n , t ) + p ( x n + 1 , T x n + 1 , t ) ] + χ [ p ( x n , T x n + 1 , t ) + p ( T x n , x n + 1 , t ) ] ) , φ ( μ [ p ( x n , T x n , t ) + p ( x n + 1 , T x n + 1 , t ) ] + χ [ p ( x n , T x n + 1 , t ) + p ( T x n , x n + 1 , t ) ] ) ) = F ( ψ ( μ [ p ( x n , x n + 1 , t ) + p ( x n + 1 , x n + 2 , t ) ] + χ [ p ( x n , x n + 2 , t ) + p ( x n + 1 , x n + 1 , t ) ] ) , φ ( μ [ p ( x n , x n + 1 , t ) + p ( x n + 1 , x n + 2 , t ) ] + χ [ p ( x n , x n + 2 , t ) + p ( x n + 1 , x n + 1 , t ) ] ) ) = F ( ψ ( μ [ p ( x n , x n + 1 , t ) + p ( x n + 1 , x n + 2 , t ) ] + χ [ p ( x n , x n + 2 , t ) ] ) , φ ( μ [ p ( x n , x n + 1 , t ) + p ( x n + 1 , x n + 2 , t ) ] + χ [ p ( x n , x n + 2 , t ) ] ) ) ψ ( μ [ p ( x n , x n + 1 , t ) + p ( x n + 1 , x n + 2 , t ) ] + χ [ p ( x n , x n + 2 , t ) ] ) ψ ( μ [ p ( x n , x n + 1 , t ) + p ( x n + 1 , x n + 2 , t ) ] + χ [ p ( x n , x n + 1 , t ) + p ( x n + 1 , x n + 2 , t ) ] ) ψ ( ( μ + χ ) [ p ( x n , x n + 1 , t ) + p ( x n + 1 , x n + 2 , t ) ] )
and
ψ ( p ( x n + 1 , x n + 2 , t ) ) ψ ( ( μ + χ ) [ p ( x n , x n + 1 , t ) + p ( x n + 1 , x n + 2 , t ) ] ) ,
that is,
p ( x n + 1 , x n + 2 , t ) μ + χ 1 ( μ + χ ) p ( x n , x n + 1 , t ) .
Using induction, we get
p ( x n + 1 , x n + 2 , t ) [ μ + χ 1 ( μ + χ ) ] n + 1 p ( x 0 , x 1 , t ) .
If we use (9) and the definition of parametric metric space for all n , m N with n < m , we have
p ( x n , x m , t ) p ( x n , x n + 1 , t ) + p ( x n + 1 , x m , t ) p ( x n , x n + 1 , t ) + p ( x n + 1 , x n + 2 , t ) + p ( x n + 2 , x m , t ) p ( x n , x n + 1 , t ) + p ( x n + 1 , x n + 2 , t ) + + p ( x m 1 , x m , t ) [ Ω n + Ω n + 1 + + Ω m 1 ] p ( x 0 , x 1 , t ) = Ω n [ 1 + Ω + Ω 2 + + Ω m n 1 ] p ( x 0 , x 1 , t ) = Ω n 1 Ω m n 1 Ω p ( x 0 , x 1 , t )
where Ω = μ + χ 1 ( μ + χ ) . Since Ω [ 0 , 1 ) , taking the limit n , m , we obtain p ( x n , x m , t ) 0 . Hence ( x n ) is a Cauchy sequence. The completeness of ( X , p ) implies that ( x n ) is convergent. Let ω X be the limit of ( x n ) . Since T is continuous, we conclude that
T ω = T ( lim n x n ) = lim n T x n = lim n x n + 1 = ω .
As a result, T has a fixed point in X. □
Example 4.
Let X = R + { 0 } be a complete parametric metric space with
p ( x , y , t ) = t max { x , y } , x y 0 , x = y
for all x , y X and all t > 0 . Define the mappings F : X × X X by F ( x , y ) = x + 2 y , ψ : R + R + by ψ ( t ) = 6 t , φ : R + R + by φ ( t ) = t 2 , T : X X defined by T x = x 3 . For μ = χ = 1 7 , Theorem 5 is satisfied. So x = 0 is a fixed point of T.
Theorem 6.
Let T be a continuous self mapping on a complete parametric metric space ( X , p ) . If T satisfies the following inequality
ψ ( p ( T x , T y , t ) ) θ F ( ψ ( Γ ( x , y ) ) , φ ( Γ ( x , y ) ) )
for all distinct x , y X and all t > 0 , where F C , ψ Ψ , φ Φ u , θ [ 0 , 1 ) , and
Γ ( x , y ) = max { p ( x , y , t ) , p ( x , T x , t ) , p ( y , T y , t ) , p ( x , T y , t ) , p ( T x , y , t ) } ,
then T has a unique fixed point in X.
Proof. 
For a point x 0 in X, we define a sequence ( x n ) as x n + 1 = T x n for all n N . If we use (10) for x = x n and y = x n + 1 , we get
ψ ( p ( x n + 1 , x n + 2 , t ) ) = ψ ( p ( T x n , T x n + 1 , t ) ) θ F ( ψ ( max { p ( x n , x n + 1 , t ) , p ( x n , T x n , t ) , p ( x n + 1 , T x n + 1 , t ) , p ( x n , T x n + 1 , t ) , p ( T x n , x n + 1 , t ) } ) , φ ( max { p ( x n , x n + 1 , t ) , p ( x n , T x n , t ) , p ( x n + 1 , T x n + 1 , t ) , p ( x n , T x n + 1 , t ) , p ( T x n , x n + 1 , t ) } ) ) = θ F ( ψ ( max { p ( x n , x n + 1 , t ) , p ( x n , x n + 1 , t ) , p ( x n + 1 , x n + 2 , t ) , p ( x n , x n + 2 , t ) , p ( x n + 1 , x n + 1 , t ) } ) , φ ( max { p ( x n , x n + 1 , t ) , p ( x n , x n + 1 , t ) , p ( x n + 1 , x n + 2 , t ) , p ( x n , x n + 2 , t ) , p ( x n + 1 , x n + 1 , t ) } ) ) θ ψ ( max { p ( x n , x n + 1 , t ) , p ( x n , x n + 2 , t ) } )
and so
p ( x n + 1 , x n + 2 , t ) θ max { p ( x n , x n + 1 , t ) , p ( x n , x n + 2 , t ) } .
There are two cases:
Case 1: If p ( x n + 1 , x n + 2 , t ) θ p ( x n , x n + 1 , t ) , then we have
p ( x n + 1 , x n + 2 , t ) θ n + 1 p ( x 0 , x 1 , t )
by induction. Continuing this process for all n , m N with n < m , we obtain
p ( x n , x m , t ) p ( x n , x n + 1 , t ) + p ( x n + 1 , x m , t ) p ( x n , x n + 1 , t ) + p ( x n + 1 , x n + 2 , t ) + p ( x n + 2 , x m , t ) p ( x n , x n + 1 , t ) + p ( x n + 1 , x n + 2 , t ) + + p ( x m 1 , x m , t ) [ θ n + θ n + 1 + + θ m 1 ] p ( x 0 , x 1 , t ) = θ n [ 1 + θ + θ 2 + + θ m n 1 ] p ( x 0 , x 1 , t ) = θ n 1 θ m n 1 θ p ( x 0 , x 1 , t ) .
In the limit case, we find p ( x n , x m , t ) 0 because θ [ 0 , 1 ) and thus { x n } is a Cauchy sequence in X. The completeness of ( X , p ) shows that { x n } is convergent. Letting ζ X be the limit of { x n } and using the continuity of T, we conclude that
T ζ = T ( lim n x n ) = lim n T x n = lim n x n + 1 = ζ .
Therefore T has a fixed point.
Case 2: In the case p ( x n + 1 , x n + 2 , t ) θ p ( x n , x n + 2 , t ) , we have
p ( x n + 1 , x n + 2 , t ) θ [ p ( x n , x n + 1 , t ) + p ( x n + 1 , x n + 2 , t ) ] θ 1 θ p ( x n , x n + 1 , t ) ( θ 1 θ ) 2 p ( x n 1 , x n , t ) ( θ 1 θ ) n + 1 p ( x 0 , x 1 , t ) .
It can be easily shown that ( x n ) is a Cauchy sequence by using the above result for all n , m N with n < m . By the completeness of ( X , p ) , ( x n ) is convergent. Let ρ be the limit of ( x n ) . The continuity of T implies that
T ρ = T ( lim n x n ) = lim n T x n = lim n x n + 1 = ρ ,
that is, T has a fixed point.
We will show that this fixed point is unique. Let T has two different fixed points ξ 1 , ξ 2 , i.e., T ξ 1 = ξ 1 and T ξ 2 = ξ 2 . The inequality (10) implies that
ψ ( p ( ξ 1 , ξ 2 , t ) ) = ψ ( p ( T ξ 1 , T ξ 2 , t ) ) θ F ( ψ ( max { p ( ξ 1 , ξ 2 , t ) , p ( ξ 1 , T ξ 1 , t ) , p ( ξ 2 , T ξ 2 , t ) , p ( ξ 1 , T ξ 2 , t ) , p ( T ξ 1 , ξ 2 , t ) } ) , φ ( max { p ( ξ 1 , ξ 2 , t ) , p ( ξ 1 , T ξ 1 , t ) , p ( ξ 2 , T ξ 2 , t ) , p ( ξ 1 , T ξ 2 , t ) , p ( T ξ 1 , ξ 2 , t ) } ) ) = θ F ( ψ ( max { p ( ξ 1 , ξ 2 , t ) , p ( ξ 1 , ξ 1 , t ) , p ( ξ 2 , ξ 2 , t ) , p ( ξ 1 , ξ 2 , t ) , p ( ξ 1 , ξ 2 , t ) } ) , φ ( max { p ( ξ 1 , ξ 2 , t ) , p ( ξ 1 , ξ 1 , t ) , p ( ξ 2 , ξ 2 , t ) , p ( ξ 1 , ξ 2 , t ) , p ( ξ 1 , ξ 2 , t ) } ) ) = θ F ( ψ ( p ( ξ 1 , ξ 2 , t ) ) , φ ( p ( ξ 1 , ξ 2 , t ) ) ) θ ψ ( p ( ξ 1 , ξ 2 , t ) )
that is, p ( ξ 1 , ξ 2 , t ) θ p ( ξ 1 , ξ 2 , t ) . Since θ [ 0 , 1 ) , we have p ( ξ 1 , ξ 2 , t ) = 0 , i.e., ξ 1 = ξ 2 . This completes the proof. □

Author Contributions

O.E. contributed in conceptualization, investigation, methodology, validation and writing the original draft; M.D.l.S. contributed in funding acquisition, methodology, project administration, supervision, validation, visualization, writing and editing. Both authors agree and approve the final version of this manuscript.

Funding

The authors thank the Basque Government for its support of this work through Grant IT1207-19.

Acknowledgments

The authors thank the Spanish Government and the European Fund of Regional Development FEDER for Grant RTI2018-094336-B-I00 (MCIU/AEI/FEDER, UE) and the Basque Government for Grant IT1207-19. We would like to express our gratitude to the anonymous referees for their helpful suggestions and corrections.

Conflicts of Interest

The authors declare that they have no competing interests concerning the publication of this article.

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Ege, O.; De la Sen, M. A New Perspective on Parametric Metric Spaces. Mathematics 2019, 7, 1008. https://doi.org/10.3390/math7111008

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Ege O, De la Sen M. A New Perspective on Parametric Metric Spaces. Mathematics. 2019; 7(11):1008. https://doi.org/10.3390/math7111008

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Ege, Ozgur, and Manuel De la Sen. 2019. "A New Perspective on Parametric Metric Spaces" Mathematics 7, no. 11: 1008. https://doi.org/10.3390/math7111008

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