# Some Fixed-Circle Results with Different Auxiliary Functions

Authors: Elif Kaplan; Nabil Mlaiki; Nihal Taş; Salma Haque; Asma Karoui Souayah
Journal: Journal of Function Spaces
 (2022)

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## 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 inG-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.


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## Body

## 1. Introduction and Preliminaries

Recently, there has been a trend to work fixed-circle problems in both metric spaces and some generalized metric spaces [1–17]. 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 S-metric spaces. Fixed-circle theorems were proved and extended with various aspects and were applied to discontinuous activation functions (for example, see [18–20] and the references therein), to rectified linear units activation functions used in the neural networks [21].In this paper, we establish various fixed-circle theorems inG-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 aG-metric space.Definition 1.1 (see [22]).
Consider the setF≠∅ and G:F×F×F⟶ℝ∪0 such that, for all ξ,ζ,ϖ,η∈F, the following conditions are satisfying:
G1Gξ,ζ,ϖ=0 if and only if ξ=ζ=ϖ;
G20<Gξ,ξ,ζ for all ξ,ζ∈F with ξ≠ζ;
G3Gξ,ξ,ζ≤Gξ,ζ,ϖ for all ξ,ζ,ϖ∈F with η≠ϖ;
G4Gξ,ζ,ϖ=Gξ,ϖ,ζ=Gζ,ϖ,ξ=⋯, (symmetry in all three variables);
G5Gξ,ζ,ϖ≤Gξ,η,η+Gη,ζ,ϖ for all ξ,ζ,ϖ,η∈F, (rectangle inequality).
Then, the functionG is called a G-metric on F.Definition 1.2 (see [22]).
AG-metric space F,G is called be symmetric if
(1)Gξ,ζ,ζ=Gζ,ξ,ξ,for all ξ,ζ∈F.In [23], Kaplan and Tas introduced the notion of circle on a G-metric space. More precisely, let F,G be a G-metric space and ξ0∈F, r∈0,∞. The circle of center ξ0 and radius r>0 is defined as
(2)CGξ0,r=ξ∈F:Gξ0,ξ,ξ=r.Example 1.1.
LetF=ℝ and d be a metric space. Let the function G:F×F×F⟶0,∞ be defined by
(3)Gξ,ζ,ϖ=maxdξ,ζ,dζ,ϖ,dϖ,ξfor all ξ,ζ,ϖ∈F [22]. Then, F,G be a G-metric space. Let us consider the function d:F×F⟶ℝ as
(4)dξ,ζ=eξ−eζfor all ξ,ζ∈F. Then, we get
(5)CGln2,ln4=ln6the circle of center ln2 and radius ln4.They also introduced the notion of fixed circle on aG-metric space [23]. Let F,G be a G-metric space and CGξ0,r be a circle. For a self-mapping T:F⟶F, if Tξ=ξ for all ξ∈CGξ0,r then, the circle CGξ0,r is said to be a fixed circle of T.


## 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.
LetF,G be a G-metric space and CGξ0,r be any circle on F. Consider Mr:ℝ+∪0⟶ℝ as
(6)Mrη=η−rifη>00ifη=0,for all η∈ℝ+∪0. If the self-mapping T:F⟶F is a function such that, for all ξ∈F, the following conditions are fulfilled:(1)
Gξ0,Tξ,Tξ=r for all ξ∈CGξ0,r,(2)
GTξ,Tξ,Tζ>r for all ξ,ζ∈CGξ0,r with ξ≠ζ,(3)
GTξ,Tξ,Tζ≤Gξ,ξ,ζ−MrGTξ,Tξ,ξ for all ξ,ζ∈CGξ0,r.
Then, the circleCGξ0,r is a fixed circle of T.Proof.
Fixξ∈CGξ0,r. By hypothesis (1), we have Tξ∈CGξ0,r for all ξ∈CGξ0,r. We claim that ξ=Tξ, that is, ξ is a fixed point of T. Now, let us suppose that ξ≠Tξ. Firstly, using the condition (2), we obtain
(7)GT2ξ,T2ξ,Tξ>r.
Using the condition (3), we have(8)GT2ξ,T2ξ,Tξ≤GTξ,Tξ,ξ−MrGTξ,Tξ,ξ=GTξ,Tξ,ξ−GTξ,Tξ,ξ+r=r.
Then, it follows from the inequalities (7) and (8), which is a contradiction. Hence, it should be ξ=Tξ. As a consequence, T fixes the circle CGξ0,r.Remark 2.1.
(1)
Note that, in Theorem2.1, the center of CGξ0,r need not to be fixed(2)
Theorem2.1 generalizes Theorem 3 given in [9].(3)
Since the notion of aG-metric and an S-metric are independent (see, [24] for more details), then Theorem 2.1 is independent from Theorem 4.1 given in [1].Example 2.1.
LetF=0,∞ be the interval of nonnegative real numbers and let G:F×F×F⟶0,∞ be defined by
(9)Gξ,ζ,ϖ=0ifξ=ζ=ϖmaxξ,ζ,ϖotherwisefor all ξ,ζ,ϖ∈F. Then, G is a G-metric on F.
The circleCG1,3 is obtained as follows:
(10)CG1,3=ξ∈F:G1,ξ,ξ=3=3.
IfT1:F⟶F is defined by
(11)T1ξ=κifξ=13ifξ≠1,for all ξ∈F and κ≠1, then T1 satisfies all the hypotheses of Theorem 2.1 and the circle CG1,3 is fixed by T1. That is, the self-mapping T1 has the unique fixed point ξ=3. Notice that the center 1 of the circle CG1,3 is not fixed by the self-mapping T1.Theorem 2.2.
LetF,G be a G-metric space, CGξ0,r be any circle on F and let define φ:F⟶0,∞ by
(12)φξ=Gξ,ξ,ξ0,for ξ∈F. Suppose that the following conditions hold:(1)
Gξ,ξ,Tξ≤φξ+φTξ−2r,(2)
GTξ,Tξ,ξ0≤r,for all ξ∈CGξ0,r such that T:F⟶F. Then, CGξ0,r is a fixed circle of T.Proof.
Letξ0∈CGξ0,r be any arbitrary point. Together with (1), we obtain
(13)Gξ,ξ,Tξ≤φξ+φTξ−2r≤Gξ,ξ,ξ0+GTξ,Tξ,ξ0−2r=GTξ,Tξ,ξ0.
From (2), the pointTξ should lie on or interior of the circle CGξ0,r. If GTξ,Tξ,ξ0<r, which leads to a contradiction by the inequality (2.5). Therefore, it should be GTξ,Tξ,ξ0=r. If GTξ,Tξ,ξ0<r, then by the inequality (13) we have
(14)Gξ,ξ,Tξ≤GTξ,Tξ,ξ0−r=r−r=0and we obtain Tξ=ξ. As a consequence, the circle CGξ0,r is fixed circle of T.Remark 2.2.
Notice that the condition (1) implies thatTξ is not inside CGξ0,r for ξ∈CGξ0,r. Similarly, (2) guarantees that Tξ is not outside of the circle CGξ0,r for ξ∈CGξ0,r. Thus, Tξ∈CGξ0,r for any ξ∈CGξ0,r and so we get TCGξ0,r⊂CGξ0,r.(1)
Theorem2.2 generalizes Theorem 2.2 given in [7].(2)
Theorem2.2 is independent from Theorem 3.11 given in [6].Example 2.2.
LetF=ℝ and the mapping G:F×F×F⟶0,∞ be defined by
(15)Gξ,ζ,ϖ=ξ−ζ+ξ−ϖ+ζ−ϖ,for each ξ,ζ,ϖ∈F [25]. Then, F,G is a G-metric space. Let us take the circle CG0,6. If we define T2:F⟶F by
(16)T2ξ=7ξ+933ξ+7,for all ξ∈F, then T2 confirms that the conditions (1) and (2) in Theorem 2.2. Hence, the circle CG0,6 is a fixed circle of T2.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.
LetF=ℝ and F,G be the G-metric space defined in Example 2.2. Let us consider the circle CG−2,4 and define the self-mapping T3:F⟶F by
(17)T3ξ=−5ξ=−45ξ=010otherwise,for all ξ∈F. Then, the self-mapping T3 satisfies the condition (1) in Theorem 2.2 but does not satisfy the condition (2) in Theorem 2.2. Obviously, T3 does not fix the circle CG−2,4.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.
LetF=ℝ and the mapping G:F×F×F⟶0,∞ be defined by
(18)Gξ,ζ,ϖ=maxξ−ζ,ξ−ϖ,ζ−ϖ,for all ξ,ζ,ϖ∈F [25]. Then, F,G is a G-metric space. Let us take the circle CG0,1/2. If we define T4:F⟶F by
(19)T4ξ=−12ifξ=−112ifξ=13otherwise,for all ξ∈F, then T4 confirms that condition (2) in Theorem 2.2 but does not satisfy the condition (1) in Theorem 2.2. Clearly, T4 does not fix the circle CG0,1/2.Now, we present the following theorem.Theorem 2.3.
LetF,G be a G-metric space and CGξ0,r be any circle on F. Let the mapping φ be defined as Theorem 2.1. If the self-mapping T:F⟶F is a function such that for all ξ∈CGξ0,r and k∈0,1, the following conditions are satisfied:
(1)
Gξ,ξ,Tξ≤φξ−φTξ,(2)
kGξ,ξ,Tξ+GTξ,Tξ,ξ0≥r,then the circle CGξ0,r is a fixed circle of T.Proof.
Letξ∈CGξ0,r. Conversely, suppose that ξ≠Tξ. Then, take into account the conditions (1) and (2), we conclude that
(20)Gξ,ξ,Tξ≤φξ−φTξ=Gξ,ξ,ξ0−GTξ,Tξ,ξ0=r−GTξ,Tξ,ξ0≤kGξ,ξ,Tξ+GTξ,Tξ,ξ0−GTξ,Tξ,ξ0=kGξ,ξ,Tξ,which is a contradiction k∈0,1. As a result, we get ξ=Tξ and CGξ0,r is a fixed circle of T.Remark 2.3.
Notice that the condition (1) guarantees thatTξ is not in the exterior of the circle CGξ0,r for ξ∈CGξ0,r. Similarly, the condition (2) guarantees that Tξ can lies on or exterior or interior of the circle CGξ0,r for ξ∈CGξ0,r. Hence Tξ should lies on or interior of the circle CGξ0,r.
(1)
Theorem2.3 generalizes Theorem 2.3 given in [7].(2)
Theorem2.3 is independent from Theorem 3.2 given in [8].Now, we present some examples concerning with self-mappings which have a fixed circle.Example 2.5.
LetF=ℝ and F,G be a G-metric space defined in Example 2.4. Let us consider the circle CG1,3=3 and define the self-mapping T5:F⟶F by
(21)T5ξ=2ξ−3ξ=35otherwise,for all ξ∈F. Then, the self-mapping T5 satisfies the condition (1) and (2) in Theorem 2.3. So, CG1,3 is a fixed circle of T5.Example 2.6.
LetF=ℝ and the function G:F×F×F⟶0,∞ be defined by
(22)Gξ,ζ,ϖ=eξ−eζ+eζ−eϖ+eξ−eϖ,for all ξ,ζ,ϖ∈F. Then, it can be easily checked that F,G is a G-metric space. Let us consider the circle CG0,2=ln2 and define the self-mapping T6:F⟶F as
(23)T6ξ=ξξ∈CG0,2ln5otherwise,for all ξ∈F. So, the self-mapping T6 provides the condition (1) and (2) in Theorem 2.3. Hence, CG0,2 is a fixed circle of T6.Next, we give an example of a self-mapping which provides the condition (1) and does not provide the condition (2).Example 2.7.
LetF,G be a G-metric space and CGξ0,r be a circle on F. If we take T7ξ=ξ0 as the self-mapping on F, then we deduce that the self-mapping T7 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 T7 does not fix a circle CGξ0,r.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.
LetF=ℝ and let the function G:F×F×F⟶0,∞ be defined by
(24)Gξ,ζ,ϖ=maxξ−ζ,ζ−ϖ,ξ−ϖ,for all ξ,ζ,ϖ∈F [25]. Let us consider the circle CG0,5 and define the self-mapping T8:F⟶F as T8ξ=5 for all ξ∈F. Then, the self-mapping T8 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 T8 does not fix the circle CG0,5.Theorem 2.4.
LetF,G be a G-metric space and CGξ0,r be any circle on F. Let the mapping φ be defined as Theorem 2.1. If the self-mapping T:F⟶F is a function such that for all ξ∈CGξ0,r and k∈0,1, the following conditions are satisfied:
(1)
Gξ,ξ,Tξ≤maxφξ,φTξ−r,(2)
GTξ,Tξ,ξ0−kGξ,ξ,Tξ≤r,then the circle CGξ0,r is a fixed circle of T.Proof.
Letξ∈CGξ0,r such that ξ≠Tξ. We show ξ=Tξ under the following two cases:
Case 1: Letmaxφξ,φTξ=φξ. Then, we get
(25)Gξ,ξ,Tξ≤maxφξ,φTξ−r=φξ−r=r−r=0,
a contradiction. Hence, we getξ=Tξ.
Case 2: Letmaxφξ,φTξ=φTξ. Then, we obtain
(26)Gξ,ξ,Tξ≤maxφξ,φTξ−r=φTξ−r=GTξ,Tξ,ξ0−r≤r+kGξ,ξ,Tξ−r=kGξ,ξ,Tξ,
a contradiction withk∈0,1. Therefore, we have ξ=Tξ.Consequently, the circleCGξ0,r is a fixed circle of T.Remark 2.4.
Notice that condition (1) guarantees thatTξ is not in the interior of the circle CGξ0,r for ξ∈CGξ0,r. Similarly, the condition (2) guarantees that Tξ is not the exterior of the circle CGξ0,r for ξ∈CGξ0,r. Hence Tξ∈CGξ0,r for each ξ∈CGξ0,r and so we get TCGξ0,r⊂CGξ0,r.(1)
Theorem2.4 is independent from Theorem 4.2 given in [1].(2)
If we consider the self-mappingT5:F⟶F defined in Example 2.5, then T5 satisfies the conditions (1) and (2) in Theorem 2.4 and so CG1,3 is a fixed circle of T5.Notice that the identity mappingIF defined as IFξ=ξ for all ξ∈F 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 [23] the following theorem.Theorem 2.5 (see [23]).
LetF,G be a G-metric space, T:F⟶F be a self-mapping having a fixed circle CGξ0,r and the mapping φ be defined as 2.2. The self-mapping T satisfies the condition
(27)IGGξ,ξ,Tξ≤hϕξ−ϕTξ,for all ξ∈F and some h∈0,1/4 if and only if T=IF.Now we give the another theorem which excludes the identity map using the auxiliary functionξr defined in (6).Theorem 2.6.
LetF,G be a G-metric space, T:F⟶F be a self-mapping having a fixed circle CGξ0,r and the mapping Mr defined in (6). The self-mapping T satisfies the condition
(28)IG∗Gξ,ξ,Tξ<MrGξ,ξ,Tξ+r,for all ξ∈F if and only if T=IF.Proof.
Letξ∈F be any point such that ξ≠Tξ. Using the inequality IG∗, we get
(29)Gξ,ξ,Tξ<MrGξ,ξ,Tξ+r=Gξ,ξ,Tξ−r+r=Gξ,ξ,Tξ,
a contradiction. Hence we getξ=Tξ and so T=IF.The converse statement is clear.


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

In [26], 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.
LetF,G be a G-metric space and CGξ0,r be any circle on F. Let T:F⟶F 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
(30)∫0GTξ,Tξ,Tζωtdt≤c∫0Gξ,ξ,ζωtdtis satisfied for all ξ∈CGξ0,r, ζ∈F−CGξ0,r where c∈0,1 and ω:0,∞⟶0,∞ is a Lebesque measurable map which is summable (that is, with a finite integral) on each compact subset of 0,∞ such that ∫0εωtdt>0 for each ε>0, then CGξ0,r0 is the unique fixed circle of T.Proof.
Suppose that the self-mappingT has two different fixed circles CGξ0,r0 and CGξ1,r1. Let u∈CGξ0,r0 and v∈CGξ1,r1 be arbitrary points such that u≠v. We show that Gu,u,v=0 and hence u=v. By the contractive condition of T, that is, using the inequality (30), we have
(31)∫0Gu,u,vωtdt=∫0GTu,Tu,Tvωtdt≤c∫0Gu,u,vωtdtwhich is a contradiction c∈0,1. Consequently, CGξ0,r0 is the unique fixed circle of T.Taking into consideration that Chatterjea type contraction condition [27], we prove the following theorem.Theorem 3.2.
LetF,G be a G-metric space and CGξ0,r0 be any circle on F. Let T:F⟶F 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
(32)∫0GTξ,Tξ,Tζωtdt≤η∫0Gξ,ξ,Tζωtdt+∫0Gζ,ζ,Tξωtdtis satisfied for all ξ∈CGξ0,r,ζ∈F−CGξ0,r and η∈0,1/2 where ω:0,∞⟶0,∞ is a Lebesque measurable map which is summable (that is, with a finite integral) on each compact subset of 0,∞ such that ∫0εωtdt>0 for each ε>0, then the fixed circle of T is unique.Proof.
Assume that there exist two different fixed-circlesCGξ0,r0 and CGξ1,r1 of the self-mapping T:F⟶F. Let u∈CGξ0,r0 and v∈CGξ1,r1 be arbitrary points such that u≠v. Using the inequality (32) and the symmetric property of G-metric, we obtain
(33)∫0Gu,u,vωtdt=∫0GTu,Tu,Tvωtdt≤η∫0Gu,u,Tvωtdt+∫0Gv,v,Tuωtdt=η∫0Gu,u,vωtdt+∫0Gv,v,uωtdt=2η∫0Gu,u,vωtdt,which is a contradiction. Consequently, it should be u=v and thus CGξ0,r0 is the unique fixed circle of T.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 ParametricReLU

In this section, we present a new application to “Parametric Rectified Linear Unit (PReLU)” using the obtained fixed-circle results. This activation function PReLU was defined to generalize the traditional rectified unit and it adaptively learns the parameters of the rectifiers (see [28] for more details). This activation function is defined by
(34)PReLUξ=cξifξ<0ξifξ≥0,with parameter c. Let us take F=0,∞ with the G-metric defined as in Example 2.1 and c=5. Then we have
(35)PReLUξ=5ξifξ<0ξifξ≥0,for all ξ∈0,∞ (see, Figure 1).Figure 1
The activation functionPReLU.If we choose a circleCG0,1=1, then PReLU satisfies the conditions of Theorem 2.1 (resp., Theorem 2.2, Theorem 2.3 and Theorem 2.4). Thereby, CG0,1 is a fixed circle of PReLU. On the other hand, this activation function fixes all circles CG0,r with r>0, that is, the number of fixed circles of PReLU is infinite. In this case, it is important because it increases the learning capacity of the activation function.



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