1. Introduction and Preliminaries
It is well known that some applications of the Banach fixed point theorem and its generalizations have been widely studied in various disciplines of mathematics, engineering, economics and statistics. An interesting application of the Banach fixed point theorem has been obtained in the study of the graph neural network model [
1]. On the other hand, the number of the fixed points of an activation function used in a neural network is important (see [
2] and the references therein). There are some applications of the notion of a fixed point (resp. fixed circle) in neural networks. For example, some activation functions with a fixed circle have been used in complex valued Hopfield neural networks [
3]. Discontinuous activation functions are also extensively used in neural networks. Some applications of fixed points and fixed circles have been obtained in discontinuous activation functions (see [
4,
5,
6,
7] and the references therein). In addition, some of popular activation functions existing in the literature have fixed discs (see [
8,
9]).
A recent approach is to consider the geometric properties of fixed points when the number of fixed points is not unique. In this context, the fixed-circle problem has been investigated in metric spaces via different contractive conditions (see [
4,
5,
10,
11,
12] for more details). Since there exist some examples of an
S-metric which is not generated by any metric, the fixed-circle problem has also been considered in
S-metric spaces and some new fixed-circle results have been obtained (see [
13,
14,
15,
16,
17]). In some of these studies, fixed-disc results have been appeared consequently.
Motivated by these studies, our aim in this paper is to consider the fixed-disc problem as a generalization of the fixed-circle (resp. fixed-point) problem.
The notion of a metric space has been extended and generalized in variant directions. One of these generalizations is made by Branciari [
18] where the triangle inequality was replaced by a rectangular one. Last years, many (common) fixed point results have been established in these spaces. For more details, see [
19,
20,
21,
22,
23,
24,
25,
26,
27,
28,
29,
30]. In the sequel, denote by
the set of all positive integer numbers.
Definition 1. [18] (Rectangular (or Branciari) metric space) Given a nonempty set X. The function satisfying: (R1) if and only if ;
(R2) ;
(R3)
for any and all distinct elements , is called a rectangular metric. Here, the pair is said a rectangular metric space.
An
S-metric space generalizes a metric space [
31].
Definition 2. [31] Given a nonempty set X and . Let be such that - 1.
if and only if ,
- 2.
.
Such is said to be an S-metric on X.
The relationships between an S-metric space and a metric space are as follows:
Lemma 1. [32] Let be a metric space. Then, - 1.
the function given as for all , is an S-metric on X.
- 2.
in if in .
- 3.
is Cauchy in iff is Cauchy in
- 4.
is complete iff is complete.
We write
as an
S-metric generated by
d [
33]. In [
32,
33], there are some examples of
S-metrics which are not generated by any metric. On the other hand, Gupta [
34] claimed that each
S-metric on
X defines a metric
on
X:
for all
. However, since the triangle inequality does not hold for all elements of
X everywhere, the function
defined in Equation (
1) is not always a metric (see [
33] for more details). If the
S-metric is generated by a metric
d on
X, then
is a metric on
X. Indeed,
, while, if the
S-metric is not generated by any metric, then
can or can not be a metric on
X. Such
is called the metric generated by
if it is a metric.
In [
17], the notion of a circle was defined on an
S-metric space as follows:
Definition 3. [17] Let be an S-metric space and , . The circle centered at with radius r is given as In [
14], the investigation of circles on metric and
S-metric spaces has been considered.
Proposition 1. [14] Let be an S-metric generated by a metric d on a nonempty set X. Hence, each circle on corresponds to the circle on . Corollary 1. [14] Let be an S-metric generated by a metric d on a nonempty set X. The circle on corresponds to the circle on . Proposition 2. [14] Let be a metric space such that is generated by an S-metric . Then, any circle on corresponds to the circle on . Corollary 2. [14] The circle on an S-metric space corresponds to the circle on where is the metric generated by . Considering the above literature, the study of new fixed-disc results and fixed-circle results on a rectangular metric space gains an importance because a rectangular metric is a generalization of a metric and there exist some examples of a rectangular metric that is not a metric (see the following two examples).
At first, we define the concepts of a circle and a disc on a rectangular metric space
. Let
and
. The circle
and the closed disc
are
and
Following [
29], we present the following.
Example 1. Let , , and be given as Given the rectangular metric as Note that is not a metric. Indeed, if we take , then we getwhich is a contradiction. In this rectangular metric space, the circle is shown in Figure 1. Following [
35], we state the following example.
Example 2. Consider and . Given the rectangular metric as Here, is not a metric. Indeed, if we take , then we getwhich is a contradiction. Given and , we haveIn the case that , we have , while, in the case that and , . In this paper, we provide some results on fixed-discs for different contraction mappings in the setting of rectangular metric spaces. The given results are supported by several examples. To derive new fixed-disc results, we modify some known techniques and introduce new contractive conditions such as an --contractive condition, an -contractive condition, a Ćirić type -contractive condition, a Branciari -contraction and a Branciari -rational contraction on a rectangular metric space. Using these new contractive conditions, we prove some fixed-disc (fixed-circle) theorems and discuss some related results.
2. Main Results
Throughout the paper,
T is a self-mapping on a rectangular metric space
. Put
We give new contractive conditions to establish some fixed-disc results. The definition of a fixed-disc is given in the following.
Definition 4. The disc is said the fixed disc of T if for all .
2.1. New Contractions via --Admissible Maps
Definition 5. T is an -contractive mapping if there are and such, that for every , we have Now, we prove that, if T is an -contractive mapping, then it fixes a disc.
Theorem 1. Each -contraction T with fixes the disc
Proof. First of all, assume that . In this case, and using the -contractive hypothesis, we get that
Assume that
We claim that
T fixes the disc
Let
be such that
. By Equation (
2), we have
On the other hand, using the
-contractive property of
T, we obtain
which is a contradiction. Thus,
for every
, that is,
T fixes the disc
. □
Now, we introduce the concept of --contractive self-maps.
Definition 6. T is said to be an α--contractive self-mapping if there are and such thatfor all . Now, we introduce --admissible maps.
Definition 7. and T is called α--admissible if for each Theorem 2. Let T be an α--contractive self mapping. Assume that T is α--admissible, and, if , we have . Then, T fixes the disc
Proof. In the case
, we have
. The
-
-contractive hypothesis yields that
Assume that
Let
such that
. We have
We also have
and
T is
-
-admissible, so the
-
-contractive property of
T implies that
which is a contradiction. Thus,
, that is,
T fixes the disc
. □
In [
36], Wardowski initiated a new class of functions.
Definition 8. [36] Let be the set of all functions such that (F1) F is strictly increasing;
(
F2)
For every positive sequence , we have(F3) There is in order that .
The concept of -contractive mappings is as follows:
Definition 9. If there exist , a function and such that for all , the following holds Then, T is said to be an -contractive self-map on X.
Theorem 3. Let T be an -contractive self-mapping with and T be α--admissible. Suppose that, if we have . Then, T fixes the disc
Proof. If
, then we have
and using the
-contractive property, one can easily deduce that
Thus,
T fixes the disc
Now, we assume that
Let
where
. Therefore, by (
2), we have
Moreover, we have
and
T is
-
-admissible. Thus, using the
-contractive property of
T, we get
It is a contradiction because F is strictly increasing, and . Hence, we deduce that , that is, the disc is fixed by T. □
Definition 10. If there are , and such that, for each ,where Then, T is called a Ćirić type -contraction on X.
Proposition 3. If T is a Ćirić type -contraction self-map with such that , then we have .
Proof. Assume that
. By Equations (
6) and (
7), we have
which is a contradiction because of
. Then, we have
. □
A generalization of Theorem 3 is as follows:
Theorem 4. Let T be a Ćirić type -contraction with .Assume that T is α--admissible and if, for every , we have . Then, T fixes the disc
Proof. If
, clearly
is a fixed-disc (point). Consider
Let
. For Equation (
2), we have
. Thus, using Equations (
6), (
7) and the fact that
T is
-
-admissible and
F is increasing, we get
which leads to a contradiction. Hence,
and so
, i.e.,
T fixes the disc
. □
2.2. Branciari Type -Contractions
Definition 11. T is said to be a Branciari -contraction mapping if there are , and so thatfor all . Theorem 5. Let T be a Branciari -contraction self-mapping with .Then, T fixes the disc .
Proof. Suppose that
. Therefore, we get
and, using the Branciari
-contractive property, we can easily see
. Hence,
T fixes the center of the disc
and the whole disc
. Let
and
with
. By Equation (
2), we have
. Because of the Branciari
-contractive property, there are
,
and
so that
for all
. It is a contradiction with
. Hence,
, that is,
T fixes the disc
. □
Now, we introduce a new rational type contractive condition.
Definition 12. T is said to be a Branciari -rational contraction if there exist , and such thatfor all , where Theorem 6. Let T be a Branciari -rational contraction self-mapping with and . Then, T fixes the disc .
Proof. Suppose that
. Thus, we have
. Using the hypothesis
,
T fixes the disc
. Let
and
with
. By Equation (
2), we have
. Because of the Branciari
-rational contractive property, there are
,
and
so that
for all
. Then,
a contradiction. Hence,
. Consequently,
T fixes the disc
. □
2.3. Some Remarks
Let
be any disc on a rectangular metric space
X. We note that all bijective self-mappings
that fix the disc
form a group under composition of functions. That is, the set
is a group under the operation of composition of functions. Besides this main fact, we can give the following remarks considering all of the obtained theorems in the previous sections.
(1) If the given rectangular metric is a metric, then all of the obtained results can be considered in a metric space.
(2) Although the triangle condition
is not used actively in the proofs of the above results. Examples 1 and 2 given in
Section 1, show the importance of studying new fixed-circle (or fixed-disc) theorems in rectangular metric spaces.
(3) If we take the function as for all in Definition 9, then we get Definition 11. In this case, Theorem 3 coincides with Theorem 5.
(4) If the function
is given as
for all
, then every Branciari
-contraction is an
-contraction. Indeed, we get
for all
.
(5) If the function
is given as
for all
, then every
-contraction is a Branciari
-contraction. Indeed, we get
for all
.
(6) Note that the radius r of the fixed-disc is independent from the center in Theorem 3 (resp. Theorem 1, Theorem 2, Theorem 4, Theorem 5 and Theorem 6) (see Example 6 for an example of Theorem 3).
(7) The contractive conditions given in previous subsections have been modified from some classical contractions used to find some fixed-point theorems. For example the notion of an
-contractive mapping, introduced in Definition 5, has been modified using the Banach contraction principle [
37].
(8) All of the obtained fixed-disc results can also be considered as the fixed-circle results.
(9) If the given rectangular metric is a metric, then this metric generate an S-metric as defined in Lemma 1. Then, all of the obtained results can be considered in an S-metric space. In this case, some relationships between circles on a rectangular metric and an S-metric space can be obtained using the similar arguments given in Proposition 1 and Corollary 1.
(10) If an
S-metric generates a metric
, then it generates a rectangular metric space since every metric is a rectangular metric. Then, the obtained fixed-circle results on
S-metric spaces (see [
13,
14,
15,
16,
17]) can be considered in a rectangular metric space. Some relationships between circles on a rectangular metric and an
S-metric space can be obtained using the similar arguments given in Proposition 2 and Corollary 2.
2.4. Illustrative Examples
In this section, we give four illustrative examples for obtained theorems throughout the previous subsections.
Example 3. Consider the rectangular metric space given in Example 2. Given defined byfor all . The-contractive self-mapping The mapping T is an -contraction with and . Indeed, we get the following cases:
Case 1:Let . Then, we have Case 2: Let . Then, we have
Then, T verifies the condition of Theorem 1.
The--contractive and--admissible self-mapping If we take and the function defined as , then T verifies the condition of Theorem 2 similar to the above cases.
The-contractive and--admissible self-mapping If we take , , and such that , then T satisfies the condition of Theorem 3. Indeed, we getfor . Then, we have The Ćirić type-contractive and--admissible self-mapping If we take , , and given as , then T verifies the conditions of Proposition 3 and Theorem 4. Indeed, we getfor . Then, we have The Branciari-contractive self-mapping If we take , and , then T verifies the condition of Theorem 5. Indeed, we getfor . Then, The Branciari-rational contractive self-mapping If we take , and , then T verifies the condition of Theorem 6. Indeed, we getfor . Then, we have Consequently, T fixes the disc In the following, the converse statement of Theorem 1 does not hold everywhere.
Example 4. Let us consider the rectangular metric space given in Example 1. Take asthen we find The mapping T fixes , but T is not an -contractive mapping with any k. Indeed, if then and hencea contradiction. In the following, the converse statements of Theorem 1, Theorem 2, Theorem 3, Theorem 4, Theorem 5 and Theorem 6 are not always true.
Example 5. Let be a rectangular metric space and be any point. If we define asfor each with ; then, T fixes the disc , but T does not satisfy the conditions (3), (4), (5), (6), (8) and (9). In the following example, we see that the radius r of the fixed disc is independent from in Theorem 3.
Example 6. Let be the family of all complex numbers and be defined as for all . Then, is a rectangular metric space. Takefor all . Then, In addition, if we take , , and given as , then T verifies the condition of Theorem 3. Hence T fixes the disc Now, if we take , , and as , again T satisfies the condition of Theorem 3. Hence, T fixes the disc Consequently, the radius r of the fixed disc is independent from the center .