#### Abstract

As the generalization of the fixed-point theory, the fixed-circle problems are interesting and notable geometric constructions. In this paper, we prove that some new necessary conditions are investigated for the existence of a fixed circle of a given self-mapping in -metric spaces. The well-known Braincari and Chatterjea contractive conditions are generalized for proving the uniqueness of obtained theorems. Finally, an application to parametric rectified linear unit activation functions are given to show the importance of studying the fixed-circle problem.

#### 1. Introduction and Preliminaries

Recently, there has been a trend to work fixed-circle problems in both metric spaces and some generalized metric spaces . For some self mappings, when the fixed point is not unique, it is an open question about the geometric shape and in some cases the set of fixed point form a circle. For example, in establishing some applicable areas such as neural networks, besides many others. This approach was initiated in [6, 7] to examine the geometry of the set of fixed-points when the number of the fixed-points of self-mappings is more than one on both metric and -metric spaces. Fixed-circle theorems were proved and extended with various aspects and were applied to discontinuous activation functions (for example, see  and the references therein), to rectified linear units activation functions used in the neural networks .

In this paper, we establish various fixed-circle theorems in -metric spaces. Different examples and application to parametric rectified linear unit activation functions are considered to illustrate the usability of our obtained results.

Firstly, we recall the concept of a -metric space.

Definition 1.1 (see ). Consider the set and such that, for all the following conditions are satisfying:
if and only if
for all with
for all with
(symmetry in all three variables);
for all (rectangle inequality).
Then, the function is called a -metric on

Definition 1.2 (see ). A -metric space is called be symmetric if for all

In , Kaplan and Ta introduced the notion of circle on a -metric space. More precisely, let be a -metric space and , The circle of center and radius is defined as

Example 1.1. Let and be a metric space. Let the function be defined by for all . Then, be a -metric space. Let us consider the function as for all Then, we get the circle of center and radius

They also introduced the notion of fixed circle on a -metric space . Let be a -metric space and be a circle. For a self-mapping if for all then, the circle is said to be a fixed circle of

#### 2. Some New Existence Conditions for Fixed Circles with Auxiliary Functions

Now, we present some new existence theorems for fixed circles of self-mappings.

Theorem 2.1. Let be a -metric space and be any circle on Consider as for all If the self-mapping is a function such that, for all the following conditions are fulfilled(1) for all ,(2) for all with ,(3) for all .Then, the circle is a fixed circle of

Proof. Fix . By hypothesis (1), we have for all We claim that that is, is a fixed point of Now, let us suppose that Firstly, using the condition (2), we obtain Using the condition (3), we have Then, it follows from the inequalities (7) and (8), which is a contradiction. Hence, it should be As a consequence, fixes the circle

Remark 2.1. (1)Note that, in Theorem 2.1, the center of need not to be fixed(2)Theorem 2.1 generalizes Theorem 3 given in .(3)Since the notion of a -metric and an -metric are independent (see,  for more details), then Theorem 2.1 is independent from Theorem 4.1 given in .

Example 2.1. Let be the interval of nonnegative real numbers and let be defined by for all Then, is a -metric on
The circle is obtained as follows: If is defined by for all and then satisfies all the hypotheses of Theorem 2.1 and the circle is fixed by That is, the self-mapping has the unique fixed point Notice that the center of the circle is not fixed by the self-mapping .

Theorem 2.2. Let be a -metric space, be any circle on and let define by for Suppose that the following conditions hold(1),(2),for all such that Then, is a fixed circle of

Proof. Let be any arbitrary point. Together with (1), we obtain From (2), the point should lie on or interior of the circle If which leads to a contradiction by the inequality (2.5). Therefore, it should be If then by the inequality (13) we have and we obtain As a consequence, the circle is fixed circle of

Remark 2.2. Notice that the condition (1) implies that is not inside for . Similarly, (2) guarantees that is not outside of the circle for . Thus, for any and so we get (1)Theorem 2.2 generalizes Theorem 2.2 given in .(2)Theorem 2.2 is independent from Theorem 3.11 given in .

Example 2.2. Let and the mapping be defined by for each . Then, is a -metric space. Let us take the circle If we define by for all then confirms that the conditions (1) and (2) in Theorem 2.2. Hence, the circle is a fixed circle of .

In the following example, we present an example of a self-mapping that satisfies the condition (1) and does not satisfy the condition (2).

Example 2.3. Let and be the -metric space defined in Example 2.2. Let us consider the circle and define the self-mapping by for all Then, the self-mapping satisfies the condition (1) in Theorem 2.2 but does not satisfy the condition (2) in Theorem 2.2. Obviously, does not fix the circle

In the next example, we present an example of a self-mapping that satisfies the condition (2) and does not satisfy the condition (1).

Example 2.4. Let and the mapping be defined by for all . Then, is a -metric space. Let us take the circle If we define by for all then confirms that condition (2) in Theorem 2.2 but does not satisfy the condition (1) in Theorem 2.2. Clearly, does not fix the circle

Now, we present the following theorem.

Theorem 2.3. Let be a -metric space and be any circle on Let the mapping be defined as Theorem 2.1. If the self-mapping is a function such that for all and the following conditions are satisfied: (1),(2),then the circle is a fixed circle of .

Proof. Let Conversely, suppose that Then, take into account the conditions (1) and (2), we conclude that which is a contradiction . As a result, we get and is a fixed circle of

Remark 2.3. Notice that the condition (1) guarantees that is not in the exterior of the circle for . Similarly, the condition (2) guarantees that can lies on or exterior or interior of the circle for . Hence should lies on or interior of the circle . (1)Theorem 2.3 generalizes Theorem 2.3 given in .(2)Theorem 2.3 is independent from Theorem 3.2 given in .

Now, we present some examples concerning with self-mappings which have a fixed circle.

Example 2.5. Let and be a -metric space defined in Example 2.4. Let us consider the circle and define the self-mapping by for all . Then, the self-mapping satisfies the condition (1) and (2) in Theorem 2.3. So, is a fixed circle of .

Example 2.6. Let and the function be defined by for all Then, it can be easily checked that is a -metric space. Let us consider the circle and define the self-mapping as for all . So, the self-mapping provides the condition (1) and (2) in Theorem 2.3. Hence, is a fixed circle of .

Next, we give an example of a self-mapping which provides the condition (1) and does not provide the condition (2).

Example 2.7. Let be a -metric space and be a circle on If we take as the self-mapping on then we deduce that the self-mapping satisfies the condition (1) in Theorem 2.3 but does not satisfy the condition (2) in Theorem 2.3. So, it can be easily shown that does not fix a circle

In the next example, we present an example of a self-mapping which satisfies the condition (2) and does not satisfy the condition (1).

Example 2.8. Let and let the function be defined by for all . Let us consider the circle and define the self-mapping as for all Then, the self-mapping provides the condition (2) in Theorem 2.3 but does not provide the condition (1) in Theorem 2.3. It can be easily shown that does not fix the circle

Theorem 2.4. Let be a -metric space and be any circle on Let the mapping be defined as Theorem 2.1. If the self-mapping is a function such that for all and the following conditions are satisfied: (1),(2),then the circle is a fixed circle of .

Proof. Let such that . We show under the following two cases:
Case 1: Let . Then, we get a contradiction. Hence, we get .
Case 2: Let . Then, we obtain a contradiction with . Therefore, we have .

Consequently, the circle is a fixed circle of .

Remark 2.4. Notice that condition (1) guarantees that is not in the interior of the circle for . Similarly, the condition (2) guarantees that is not the exterior of the circle for . Hence for each and so we get (1)Theorem 2.4 is independent from Theorem 4.2 given in .(2)If we consider the self-mapping defined in Example 2.5, then satisfies the conditions (1) and (2) in Theorem 2.4 and so is a fixed circle of .

Notice that the identity mapping defined as for all satisfies conditions (1) and (2) (resp., (1) and (2)) in Theorem 2.2 (resp., Theorem 2.3). Therefore, we need a condition which excludes the identity map in Theorem 2.2 (resp., Theorem 2.3). For this aim, we give in  the following theorem.

Theorem 2.5 (see ). Let be a -metric space, be a self-mapping having a fixed circle and the mapping be defined as 2.2. The self-mapping satisfies the condition for all and some if and only if .

Now we give the another theorem which excludes the identity map using the auxiliary function defined in (6).

Theorem 2.6. Let be a -metric space, be a self-mapping having a fixed circle and the mapping defined in (6). The self-mapping satisfies the condition for all if and only if .

Proof. Let be any point such that . Using the inequality , we get a contradiction. Hence we get and so .

The converse statement is clear.

#### 3. Some New Uniqueness Conditions for Fixed Circles with Integral Type Contractions

In , Braincari gave an integral contractive condition which was a generalization of Banach contraction in a metric space. By the Braincari type contractive condition, we obtain a uniqueness theorem as follows.

Theorem 3.1. Let be a -metric space and be any circle on Let be a self-mapping satisfying the inequalities of Theorem 2.1 (resp., Theorem 2.2, Theorem 2.3 and Theorem 2.4). If the contractive condition is satisfied for all , where and is a Lebesque measurable map which is summable (that is, with a finite integral) on each compact subset of such that for each , then is the unique fixed circle of

Proof. Suppose that the self-mapping has two different fixed circles and Let and be arbitrary points such that We show that and hence By the contractive condition of that is, using the inequality (30), we have which is a contradiction Consequently, is the unique fixed circle of

Taking into consideration that Chatterjea type contraction condition , we prove the following theorem.

Theorem 3.2. Let be a -metric space and be any circle on Let be a self-mapping satisfying the inequalities of Theorem 2.1 (resp., Theorem 2.2, Theorem 2.3 and Theorem 2.4). If the contractive condition is satisfied for all and where is a Lebesque measurable map which is summable (that is, with a finite integral) on each compact subset of such that for each , then the fixed circle of is unique.

Proof. Assume that there exist two different fixed-circles and of the self-mapping Let and be arbitrary points such that Using the inequality (32) and the symmetric property of -metric, we obtain which is a contradiction. Consequently, it should be and thus is the unique fixed circle of .

Remark 3.1. The choice of used contractive condition in uniqueness theorem is not unique. Any contractive condition used to derive the fixed-point theorem can also be selected.

#### 4. An Application to Parametric

In this section, we present a new application to “Parametric Rectified Linear Unit ()” using the obtained fixed-circle results. This activation function was defined to generalize the traditional rectified unit and it adaptively learns the parameters of the rectifiers (see  for more details). This activation function is defined by with parameter . Let us take with the -metric defined as in Example 2.1 and . Then we have for all (see, Figure 1).

If we choose a circle , then satisfies the conditions of Theorem 2.1 (resp., Theorem 2.2, Theorem 2.3 and Theorem 2.4). Thereby, is a fixed circle of . On the other hand, this activation function fixes all circles with , that is, the number of fixed circles of is infinite. In this case, it is important because it increases the learning capacity of the activation function.

#### Data Availability

The data used to support the findings of the study are included within the article.

#### Conflicts of Interest

The authors declare that they have no competing interests.

#### Funding

This research work did not receive any external funding.

#### Acknowledgments

The authors N. Mlaiki and S. Haque would like to thank Prince Sultan University for paying the publication fees for this work through TAS LAB.