1. Introduction and Preliminaries
Metric fixed point theory is a very natural and well-developed branch of mathematical analysis which deals with the existence and uniqueness results for fixed points of the mappings defined on metric spaces. The source of this theory is the classical result due to Banach [
1] which is known as the Banach contraction principle and states that every contraction mapping defined on a complete metric space
admits a unique fixed point. Several authors extended and generalized this principle in various directions (for instance, see [
2,
3,
4,
5,
6,
7,
8,
9,
10]). In this regard, Jleli and Samet [
11] introduced an attractive class of auxiliary functions
:
satisfying some suitable properties and utilized the same functions to prove some fixed point theorems concerning a new type of contractions called
-contractions (or JS-contractions).
The Banach contraction principle and its generalizations have been applied in various disciplines of mathematics, economics and engineering. One of the interesting applications of Banach contraction principle is to study the graph neural network model [
12]. On the other hand, in the case that the mapping
S defined on a metric space
has more than one fixed point, the number of fixed points of an activation function used in a neural network is important (see [
13] and the references therein). There exist some mappings that fix all points of a circle or a disc. These kind of mappings have some applications to neural networks (for instance, see [
14]). For example, let
be the set of all complex numbers with the usual metric
, for all
and consider
to be defined by
where
is the conjugate of
z. Then,
S fixes all points of the circle
. It is worth pointing out that there exist some mappings which map the circle
to itself but do not fix all points of the circle
. For instance, let
T be a self-mapping on
defined by
Then, , but T does not fix all points of . In fact, the mapping T fixes only two points of the unit circle.
Natural and interesting problems were found through the investigations of the geometric properties of fixed points. In this context, the fixed circle and fixed disc problem have been studied in metric and generalized metric spaces via different contractive conditions (see [
15,
16,
17,
18,
19,
20,
21,
22,
23,
24]). For example, in [
15], some fixed circle results were proved using the Caristi type contraction on a metric space. Using Wardowski’s technique and some classical contractive conditions, some fixed circle theorems were obtained in [
19]. In [
20,
21], the fixed circle problem was studied in the setting of
S-metric spaces. In [
16], some fixed disc theorems were obtained using several types of contractive conditions on rectangular metric space.
Motivated by the above, we aim to present in this paper some fixed circle (disc) results for some types of contraction self-mappings, namely
-contractions, Ćirić type
-contractions, Hardy–Rogers type
-contractions and Khan type
-contractions in the setting of metric spaces by using JS technique. Furthermore, we establish some fixed circle (disc) results of integral type contractive self-mappings. To do this, we use the notion of
-function introduced by Jleli and Samet [
11]. The function
:
is said to be a
-function, if the following conditions are satisfied:
- (JS1)
is nondecreasing;
- (JS2)
for every sequence
,
- (JS3)
there exist and such that .
In the sequel, we adopt the following notations:
the class of all functions which satisfy .
the class of all functions which satisfy –.
Remark 1. Observe that the auxiliary function Θ used in [11] satisfies the conditions –, that is, while the auxiliary function Θ used in the present paper need not to satisfies and , that is, . For a metric space , the two sets and are called circle and disc, respectively, with center and radius r. The circle (disc ) is said to be a fixed circle (fixed disc) of a mapping if , for all ().
Before starting our main results, we would like to bring to the reader’s attention that, in the next three sections, we are working on transformations that not only fix one element but also fix a well-defined set of “fixed points” which is either a circle or a disc.
2. Fixed Circle (Disc) Results for -Contraction and Ćirić Type -Contraction
Let
be a metric space. From now on,
S is a self-mapping defined on a non-empty set
Y and
At first, we introduce the notion of -contraction as follows:
Definition 1. Let be a metric space and . A mapping S is said to be a -contraction if there exist and such that Using the -contraction condition, we present the following fixed circle result.
Theorem 1. Let be a metric space. Assume that S is a -contraction mapping defined on Y with and , for all . Then, is a fixed circle of S.
Proof. Let
and assume on contrary that
, that is,
. From the definition of
r, we have
. As
S is a
-contraction, so in view of the fact that
is nondecreasing, we get
which is a contradiction. Therefore,
, for all
, that is,
is a fixed circle of
S. This concludes the proof. □
Next, we prove the following fixed disc result.
Theorem 2. Let be a metric space. Assume that S is a Ćirić type -contraction mapping with and for all . Then, is a fixed disc of S.
Proof. In view of Theorem 1, the mapping
S fixes the circle
. In order to show that
is a fixed disc of
S, it is sufficient to show that
S fixes any circle
with
. Let
, and, contrarily, let us assume that
for some
. Since
S is a
-contraction, we have
Since
is nondecreasing, we then obtain
which leads to a contradiction as
. Therefore,
for all
. Consequently,
is a fixed disc of
S. □
Next, we generalize -contractions to Ćirić type -contractions and prove the following new results in this section.
Definition 2. Let be a metric space and . A mapping S is said to be a Ćirić type -contraction if there exist and such thatwhere The following proposition follows from Definition 2.
Proposition 1. Let be a metric space. If S is a Ćirić type -contraction mapping with , then .
Proof. Assume that
. From Definition 2, we have
a contradiction as
. Therefore, we must have
. □
Using Ćirić type -contraction condition, we present the following fixed circle result.
Theorem 3. Let be a metric space. Assume that S is a Ćirić type -contraction mapping with and for all . Then, is a fixed circle of S.
Proof. Let
and assume contrarily that
. From the definition of
r, we have
. As
S is a Ćirić type
-contraction, using Proposition 1, we have
Now, we have the following possibilities:
Case 1: If
, then from Equation (
2), the definition of
r and the fact that the function
is nondecreasing, we have
a contradiction.
Case 2: If
, then we have two possibilities: either
or
. Assuming that
from Equation (
2) and the fact that the function
is nondecreasing, we have
a contradiction. If
, then from Equation (
2), we get
this inequality contradicts with the definition of
(as
).
Case 3: If
, then from Equation (
2) and the fact that the function
is nondecreasing, we have
a contradiction. Therefore,
for all
. Consequently,
is a fixed circle of
S. □
Next, we prove the following fixed disc result.
Theorem 4. Let be a metric space. Assume that S is a Ćirić type -contraction mapping with and for all . Then, is a fixed disc of S.
Proof. The mapping
S fixes the circle
(in view of Theorem 3). Now, in order to show that
is a fixed disc of
S, it is sufficient to show that
S fixes any circle
with
. Let
and, contrarily, let us assume that
for some
. Since
S is a Ćirić type
-contraction, using Proposition 1, we have
Now, we have the following possibilities:
Case 1: If
, then from Equation (
3), the definition of
r and the fact that the function
is nondecreasing, we have
a contradiction.
Case 2: If
, then we have two possibilities: either
or
. Assuming that
from Equation (
3) and the fact that the function
is nondecreasing, we have
a contradiction. If
, then from Equation (
2), we get
This inequality contradicts with the definition of (as ).
Case 3: If
, then from Equation (
3) and the fact that the function
is nondecreasing, we have
a contradiction.
Case 4: If
, then by the definition of
r, the inequality Equation (
3) and the fact that the function
is nondecreasing, we have
a contradiction. Therefore,
for all
. Consequently,
is a fixed disc of
S. □
Corollary 1. Let be a metric space and S a mapping defined on Y. If there exist and with for all such thatwhere then is a fixed circle (or disc) of S. Proof. The result follows from Theorems 3 and 4 by taking
in the contractive condition (
1). □
Example 1. Let be a metric space endowed with the usual metric ξ. Define asThe self-mapping S is a -contraction. In fact, let , and . For all such that , we haveandThe self-mapping S is a Ćirić type -contraction. To show this, let , , and . For all such that , we haveandWe also haveTherefore, all the conditions of Theorems 1–4 are satisfied. Observe that is a fixed circle and is a fixed disc of S. 3. Fixed Circle (Disc) Results for Hardy–Rogers -Contraction and Khan Type -Contraction
We start this section by introducing the concept Hardy–Rogers type -contractions as follows:
Definition 3. Let be a metric space and . A mapping S is called a Hardy–Rogers type -contraction if there exist and such thatwhere with and . The following proposition follows from Definition 3.
Proposition 2. Let be a metric space. If S is a Hardy–Rogers type -contraction mapping with , then .
Proof. Assume that
. From Definition 3 and the fact that
is nondecreasing, we obtain
a contradiction. Therefore, we must have
. □
Using a Hardy–Rogers type -contraction, we present the following fixed circle results.
Theorem 5. Let be a metric space. Assume that S is a Hardy–Rogers type -contraction with and for all . Then, is a fixed circle of S.
Proof. Let
and assume contrarily that
. By the definition of
r, we have
. Since
S is a Hardy–Rogers type
-contraction, using Proposition 2 and the fact that
is nondecreasing, we have
a contradiction. Therefore,
for all
. Consequently,
is a fixed circle of
S. □
Next, we prove the following fixed disc result.
Theorem 6. Let be a metric space. Assume that S is a Hardy–Rogers type -contraction with and , for all . Then, is a fixed disc of S.
Proof. In view of Theorem 5,
S fixes the circle
. Now, in order to show that
is a fixed disc of the mapping
S, it is sufficient to show that
S fixes any circle
with
. Let
and assume that
. Since
S is a Hardy–Rogers type
-contraction, using Proposition 2 and the fact that
is nondecreasing, we have
a contradiction. Thus, we obtain
, and
is a fixed disc of
S. □
From Theorems 5 and 6, we deduce the following corollaries.
Corollary 2. Let be a metric space and S a mapping. If there exists and with , for all such thatwhere and then is a fixed circle (or disc) of S. Proof. The claim follows from Theorems 5 and 6 if we take
and
in the contractive condition (
4). □
Corollary 3. Let be a metric space and S a mapping. If there exists and with , for all such thatwhere then is a fixed circle (or disc) of S. Proof. The claim follows from Theorems 5 and 6 if we put
in the contractive condition (
4). □
The following example exhibits the utility of Theorems 5 and 6.
Example 2. Let be endowed with the usual metric ξ. Define asLet , , , and . Then, S is a Hardy–Rogers type -contraction. Indeed, for andWe also haveHence, all the conditions of Theorems 5 and 6 are satisfied. Observe that S fixes the circle and the disc . We close this section by introducing the concept of the Khan type -contraction followed by the related fixed circle (disc) results.
Definition 4. Let be a metric space and . A mapping S is called a Khan type -contraction if there exist and such that for all , if , then andwhere and if , then . The following proposition is a direct consequence of Definition 4.
Proposition 3. Let be a metric space. If S is a Khan type -contraction with , then .
Proof. Assume that
; then,
. As
S is a Khan type
-contraction, we have
a contradiction. Therefore, we must have
. □
Now, utilizing the definition of the Khan type -contraction, we prove the following fixed circle and fixed disc results.
Theorem 7. Let be a metric space and let be a Khan type -contraction with . Then, is a fixed circle of S.
Proof. Let
and assume contrarily that
; then,
. From the definition of
r, we have
. As
S is a Khan type
-contraction, using Proposition 3 and the fact that
is nondecreasing, we have
a contradiction. Therefore,
for all
. Consequently,
is a fixed circle of
S. □
Theorem 8. Let be a metric space and let be a Khan type -contraction with . Then, is a fixed disc of S.
Proof. In view of Theorem 7,
S fixes the circle
. Now, in order to show that
is a fixed disc of the mapping
S, it is sufficient to show that
S fixes any circle
with
. Let
and assume that
. By the Khan type
-contractive condition, we have
As
is a nondecreasing function, we get
a contradiction (as
). Thus, we obtain
for all
with
. Therefore,
is a fixed disc of
S. □
The following example shows the utility of Theorems 7 and 8.
Example 3. Let be endowed with the usual metric ξ. Define byThen, S is a Khan type Θ-contraction with , , and . In fact,for all such that . Now, we haveandWe also haveTherefore, all the conditions of Theorems 7 and 8 are satisfied. Observe that S fixes the circle and the disc . 4. Fixed Circle (Disc) Results of Integral Type
In this section, we establish some fixed circle and disc results of the integral type.
Let
be a locally integrable function such that
Definition 5. Let and . The mapping S is called an integral type -contraction if there exist and such that for all where is a function defined as in Equation (5). The following proposition is useful in the proof of the main results of this section.
Proposition 4. Let be a metric space and . If S is an integral type -contraction with , then .
Proof. Assume that
. From Definition 5, we have
which contradicts the definition of
, as
:
and
. Hence, we must have
. □
In the following theorem, we present a fixed circle result for integral type -contraction.
Theorem 9. Let be a metric space and . If S is an integral type -contraction with , then is a fixed circle of S.
Proof. Assume that
for some
. Making use of the definition of
r, we have
Since
is a nondecreasing function, we have
and
As
S is an integral type
-contraction, using Equation (
6), we obtain
a contradiction. Therefore, we have
. Consequently,
is a fixed circle of
S. □
Next, we prove the following fixed disc results.
Theorem 10. Let be a metric space, let be defined as in Equation (5) and let S be an integral type -contraction with . Then, is a fixed disc of S. Proof. In view of Theorem 9, the mapping
S fixes the circle
. Now, in order to show that
is a fixed disc of the mapping
S, it is sufficient to show that
S fixes any circle
with
. Let
and assume that
. Making use of the definition of
r, we have
Since
is a nondecreasing function, we have
and
As
S is an integral type
-contraction, using Equation (
7), we obtain
a contradiction. Therefore, we must have
,
. Consequently,
is a fixed disc of
S. □
Remark 2. Using similar arguments as in Definition 5, we can define the notions of an integral type Ćirić type -contraction, an integral Hardy–Rogers type -contraction and an integral Khan type -contraction, and we can obtain corresponding fixed circle and fixed disc theorems.
5. Conclusions
We have obtained some new generalized fixed circle and fixed disc results using our newly introduced contractive conditions in the setting of metric spaces. Precisely, by means of some known techniques which usually are used to obtain some fixed point theorems, we have generated some useful fixed circle and fixed disc results.