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
-metric spaces, obtain some fixed point results and prove a fixed-circle theorem on a parametric
-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
is called a parametric metric [
18] if,
- (i)
for all iff ;
- (ii)
for all ;
- (iii)
for all and all .
Example 1 ([
18]).
The function defined by for all and all is parametric metric on X which is the set of all functions . Definition 1 ([
18]).
Let be a parametric metric space.A point is called the limit of a sequence , if for all , and the sequence is called convergent to x.
A sequence is said to be a Cauchy if and only if for all .
A parametric metric space is called complete if and only if every Cauchy sequence is convergent to .
Definition 2 ([
18]).
Let be a parametric metric space and let be a mapping. If for any sequence in X such that as , as , then T is a continuous mapping at . C-class functions have been presented in [
19].
Definition 3 ([
19]).
A continuous mapping is said to be C-class function if it satisfies the following:(1) for all ;
(2) implies that either or .
The C-class functions will be denoted by .
Example 2 ([
19]).
Some elements of are given in the following for all :(1) ; for all ,
(2) ,
(3) , where Γ is the Euler Gamma function.
Let denote the set of all continuous and monotone nondecreasing functions such that iff , for all .
Let denote the set of all continuous functions such that iff and denote the set of all continuous functions such that , note that .
Ege and Karaca [
20] establish a coupled fixed point theorem and give a homotopy application in parametric metric spaces.
Definition 4 ([
20]).
Let be a parametric metric space, be an element in and and be given two functions.If and , then is said to be a coupled fixed point of F.
If and , then is called a coupled coincidence point of F and g.
F and g are said to be commutative if .
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 be a parametric metric space. Let and be two maps such thatfor all and . If- (i)
is a subset of ,
- (ii)
is a complete parametric metric space,
- (iii)
g is continuous,
- (iv)
g and F are commutative,
and , then there is a unique x in X such that .
Theorem 2 ([
21])
. Let be a complete parametric metric space. If a continuous mapping satisfies the followingfor all , and , , then T has a fixed point in X. Theorem 3 ([
21]).
Let be a complete parametric metric space and T a continuous map satisfyingfor all , , and for all , where . 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 be a parametric metric space and the mappings and satisfy the following conditionfor all and , where , , and is a coupled coincidence point of g and F. Then . Proof. Using the definition of a coupled coincidence point, we obtain
and
for the mappings
g and
F. If we assume that
and use the inequality (
1), then we have the following statements:
There are two cases have to be considered. If , then we have or . Hence , which is a contradiction. If , a contradiction is reached. As a result, we have , that is, . □
Theorem 4. Let be a parametric metric space. Let the mappings and satisfy (1) for all and , where , , . Suppose that the following conditions hold: - (I)
is a subset of ,
- (II)
g is continuous,
- (III)
F and g are commutative,
- (IV)
is a complete parametric metric space.
Then there exists a unique element x in X such that .
Proof. Consider two points
and
in
X. Using
, it can be chosen new points
such that
and
and similarly,
such that
and
. More generally, it can be constructed two sequences
and
as follows:
The inequality (
1) implies the following:
for
and all
. By the inequalities
and
we see that
that is,
where
and
. So the sequence
is decreasing. In the limit case, we find
for each
. As a result, we have
Let
with
. From the condition
of the definition of parametric metric space, we obtain
and
Combining the last two inequalities, we find
Taking the limits as
, we have
Then we conclude that
is a Cauchy sequence in
. In the same manner,
is also a Cauchy sequence in
. From the condition
, the sequences
and
are convergent to
and
, respectively. The condition
shows that
is convergent to
and
is convergent to
. On the other hand, by
, there are following equalities
Using the condition
, we conclude that
and similarly
. Hence
Lemma 1 implies that
is a coupled fixed point of
F and
g. That is,
. Since
is a subsequence of
,
is also convergent to
x and
In the limit case, we get
because
p is continuous. Similarly,
Combining Equations (
2) and (
3), we have
The inequality (
4) holds only if
and
. Moreover, we can write
Letting
and from the fact that the continuity of
p, we have
The following inequality can be obtained in a similar way.
Now Equations (
5) and (
6) imply that
Since (
7) holds only when
and
, we have
.
All that remains in the proof is to show the uniqueness. Assuming
with
such that
, we conclude that
But this is a contradiction. So, there is a unique common fixed point of g and F. □
Example 3. Let for all and all be a parametric metric on . It is easy to see that is a complete parametric metric space. Define The map g is continuous, and is a complete parametric metric space. Sincewe have that F and g are commutative. We define the following mappings: Then we obtainfor all and all . From Theorem 4, 0 is the unique element in X such that . Theorem 5. Let be a complete parametric metric space and a continuous mapping satisfy the following conditionfor all and all , where , andwhere , . Then T has a fixed point in X. Proof. Let
be an arbitrary element in
X. A sequence
can be defined as
for
. If we take
and
in (
8), we have
and
that is,
If we use (
9) and the definition of parametric metric space for all
with
, we have
where
. Since
, taking the limit
, we obtain
. Hence
is a Cauchy sequence. The completeness of
implies that
is convergent. Let
be the limit of
. Since
T is continuous, we conclude that
As a result, T has a fixed point in X. □
Example 4. Let be a complete parametric metric space withfor all and all . Define the mappings by , by , by , defined by . For , Theorem 5 is satisfied. So is a fixed point of T. Theorem 6. Let T be a continuous self mapping on a complete parametric metric space . If T satisfies the following inequalityfor all distinct and all , where , , andthen T has a unique fixed point in X. Proof. For a point
in
X, we define a sequence
as
for all
. If we use (
10) for
and
, we get
and so
There are two cases:
Case 1: If
, then we have
by induction. Continuing this process for all
with
, we obtain
In the limit case, we find
because
and thus
is a Cauchy sequence in
X. The completeness of
shows that
is convergent. Letting
be the limit of
and using the continuity of
T, we conclude that
Therefore T has a fixed point.
Case 2: In the case
, we have
It can be easily shown that
is a Cauchy sequence by using the above result for all
with
. By the completeness of
,
is convergent. Let
be the limit of
. The continuity of
T implies that
that is,
T has a fixed point.
We will show that this fixed point is unique. Let
T has two different fixed points
, i.e.,
and
. The inequality (
10) implies that
that is,
. Since
, we have
, i.e.,
. This completes the proof. □