# Wardowski Type Contractions and the Fixed-Circle Problem onS-Metric Spaces

Authors: Nabil Mlaiki; Ufuk Çelik; Nihal Taş; Nihal Yilmaz Özgür; Aiman Mukheimer
Journal: Journal of Mathematics
 (2018)

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## Abstract

In this paper, we present new fixed-circle theorems for self-mappings on anS-metric space using some Wardowski type contractions, ψ-contractive, and weakly ψ-contractive self-mappings. The common property in all of the obtained theorems for Wardowski type contractions is that the self-mapping fixes both the circle and the disc with the center x0 and the radius r.


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## Body

## 1. Introduction

Fixed-point theory has many applications in different fields; see [1–10]. Recently, using Wardowski’s technique, some new fixed-point theorems on S-metric spaces [11] and some new fixed-circle theorems on metric spaces [12, 13] have been obtained. Our aim in this paper is to obtain various fixed-circle results using this technique. In Section 2, we recall some necessary background on S-metric spaces and give new examples. In Section 3, we introduce the notion of an FcS-contraction to obtain fixed-circle theorems. By means of this notion, we define new types of an FcS-contraction such as Hardy-Rogers type FcS-contraction and Reich type FcS-contraction and present some fixed-circle results on S-metric spaces. Also, we give an illustrative example of a self-mapping satisfying all of the conditions of the obtained theorems. In Section 4, we prove the existence along with the conditions that give us uniqueness of a fixed circle for ψ-contractive and weakly ψ-contractive self-mappings on S-metric spaces. In Section 5, we give an application of fixed-circle results obtained by Wardowski technique to integral type contractive self-mappings.


## 2. Preliminaries

In this section, we recall some necessary notions, relations, and results aboutS-metric spaces.Definition 1 (see [14]).
LetX be a nonempty set and S:X×X×X→[0,∞) be a function satisfying the following conditions for all x,y,z,a∈X:(S1)
S(x,y,z)=0 if and only if x=y=z(S2)
S(x,y,z)≤S(x,x,a)+S(y,y,a)+S(z,z,a)ThenS is called an S-metric on X and the pair (X,S) is called an S-metric space.Example 2 (see [15]).
LetX=R (or C) and the function S:X×X×X→[0,∞) be defined as (1)Sx,y,z=x-z+y-z,for all x,y,z∈R (or C). Then the function S is an S-metric on R (or C). This S-metric is called the usual S-metric on R (or C).Lemma 3 (see [14]).
LetX,S be an S-metric space and x,y∈X. Then we have (2)Sx,x,y=Sy,y,x.The relationships between a metric and anS-metric were studied in different papers (see [16–18] for more details). In [17], a formula of an S-metric space which is generated by a metric d was investigated as follows.Let(X,d) be a metric space. Then the function Sd:X×X×X→0,∞ defined by(3)Sdx,y,z=dx,z+dy,z,for all x,y,z∈X, is an S-metric on X. The S-metric Sd is called the S-metric generated by d [18]. We note that there exists an S-metric which is not generated by any metric d as seen in the following example.Example 4.
LetX be a nonempty set, the function d:X×X→[0,∞) be any metric on X, and the function S:X×X×X→[0,∞) be defined by (4)Sx,y,z=min⁡1,dx,y+min⁡1,dy,z+min⁡1,dx,z,for all x,y,z∈X. Then the function S is an S-metric and X,S is an S-metric space. Indeed,( S 1 ) for x,y,z∈X, we have (5)Sx,y,z=0⇔min⁡1,dx,y+min⁡1,dy,z+min⁡1,dx,z=0⇔dx,y=dy,z=dx,z=0⇔x=y=z( S 2 ) using Table 1, we can easily see that the condition (S2) is satisfied.Table 1 1 1 1 ≤ 2 2 2 1 1 1 ≤ 2 2 2 d ( z , a ) 1 1 1 ≤ 1 1 d ( y , z ) ≤ 2 2 d ( y , a ) 2 d ( z , a ) 1 1 1 ≤ d ( x , y ) d ( y , z ) d ( x , z ) ≤ 2 d ( x , a ) 2 d ( y , a ) 2 d ( z , a ) 1 1 d ( y , z ) ≤ 1 1 1 ≤ 2 2 2 1 1 d ( y , z ) ≤ 1 1 1 ≤ 2 2 2 d ( z , a ) 1 1 d ( y , z ) ≤ 2 d ( y , a ) d ( z , a ) ≤ 2 2 d ( y , a ) 2 d ( z , a ) 1 1 d ( y , z ) ≤ 1 d ( x , z ) 1 ≤ 2 2 d ( y , a ) 2 d ( z , a ) 1 1 d ( y , z ) ≤ d ( x , y ) 1 1 ≤ 2 2 d ( y , a ) 2 d ( z , a ) 1 1 d ( y , z ) ≤ d ( x , y ) d ( x , z ) d ( y , z ) ≤ 2 d ( x , a ) 2 d ( y , a ) 2 d ( z , a ) 1 d ( x , z ) d ( y , z ) ≤ 1 1 1 ≤ 2 2 2 1 d ( x , z ) d ( y , z ) ≤ 1 1 1 ≤ 2 2 2 d ( z , a ) 1 d ( x , z ) d ( y , z ) ≤ 1 1 d ( y , z ) ≤ 2 2 d ( y , a ) 2 d ( z , a ) 1 d ( x , z ) d ( y , z ) ≤ d ( x , y ) 1 1 ≤ 2 d ( x , a ) 2 d ( y , a ) 2 1 d ( x , z ) d ( y , z ) ≤ d ( x , y ) d ( x , z ) d ( y , z ) ≤ 2 d ( x , a ) 2 d ( y , a ) 2 d ( z , a ) d ( x , y ) d ( x , z ) d ( y , z ) ≤ 1 1 1 ≤ 2 2 2 d ( x , y ) d ( x , z ) d ( y , z ) ≤ 1 1 1 ≤ 2 2 2 d ( z , a ) d ( x , y ) d ( x , z ) d ( y , z ) ≤ 1 1 d ( y , z ) ≤ 2 2 d ( y , a ) 2 d ( z , a ) d ( x , y ) d ( x , z ) d ( y , z ) ≤ 2 d ( x , a ) 2 d ( y , a ) 2 d ( z , a ) Other possibilities can be proved like this table.Also theS-metric S is not generated by any metric m. Conversely, suppose that there exists a metric m such that (6)Sx,y,z=mx,z+my,z,for all x,y,z∈X. Then we get (7)Sx,x,z=2mx,zand  somx,z=min⁡1,dx,zand (8)Sy,y,z=2my,zand  somy,z=min1,dy,z.Therefore, we obtain (9)min⁡1,dx,y+min⁡1,dy,z+min⁡1,dx,z=min⁡1,dx,z+min⁡1,dy,z,which is a contradiction. Consequently, S is not generated by any metric m.In [19] and [14], a circle and a disc are defined on an S-metric space as follows, respectively:(10)Cx0,rS=x∈X:Sx,x,x0=rand(11)Dx0,rS=x∈X:Sx,x,x0≤r.We give an example.Example 5.
LetX be a nonempty set, the function d:X×X→[0,∞) be any metric on X, and the S-metric space be defined as Example 4. Let us consider the circle Cx0,rS according to the S-metric:(12)Cx0,rS=x∈X:Sx,x,x0=2min⁡1,dx,x0=r.Then we have the following cases:
Case 1. If r=2 then Cx0,rS=x∈X:dx,x0≥1.
Case 2. If r>2 then Cx0,rS=∅.
Case 3. If r<2 then Cx0,rS=Cx0,r/2, where Cx0,r/2=x∈X:dx,x0=r/2.Definition 6 (see [19]).
LetX,S be an S-metric space, Cx0,rS be a circle, and T:X→X be a self-mapping. If Tx=x for every x∈Cx0,rS then the circle Cx0,rS is called the fixed circle of T.


## 3.FcS-Contraction and Hardy-Rogers Type FcS-Contraction on S-Metric Spaces

At first, we recall the definition of the following family of functions which was introduced by Wardowski in [20].Definition 7 (see [20]).
LetF be the family of all functions F:0,∞→R such that
F 1   F is strictly increasing
F 2 for each sequence αn in 0,∞ the following holds: limαn=0 if and only if limFαn=-∞
F 3 there exists k∈0,1 such that limα→0+αkFα=0.The following is an example of some functions that satisfies conditionsF1, F2, and F3 of Definition 7.Example 8 (see [20]).
( 1 )   F : 0 , ∞ → R defined by Fx=ln⁡x.
( 2 )   F : 0 , ∞ → R defined by Fx=ln⁡x+x.
( 3 )   F : 0 , ∞ → R defined by Fx=-1/x.
( 4 )   F : 0 , ∞ → R defined by Fx=ln⁡x2+x.
Note that these four functions satisfy conditionsF1, F2, and F3 of Definition 7.Now we introduce the following new contraction type using this family of functions.Definition 9.
LetX,S be an S-metric space. A self-mapping T on X is said to be an FcS-contraction if there exist F∈F, t>0, and x0∈X such that for all x∈X the following holds: (13)STx,Tx,x>0⇒t+FSTx,Tx,x≤FSx,x,x0.
Now, we present the following proposition.Proposition 10.
LetX,S be an S-metric space. If a self-mapping T on X is an FcS-contraction with x0∈X, then we have Tx0=x0.Proof.
Assume thatTx0≠x0. From the definition of an FcS-contraction, we get(14)STx0,Tx0,x0>0⇒t+FSTx0,Tx0,x0≤FSx0,x0,x0.Inequality (14) contradicts with the definition of F since F:0,∞→R and Sx0,x0,x0=0. Therefore, it should be Tx0=x0.Using this new type contraction, we give the following fixed-circle theorem.Theorem 11.
LetX,S be an S-metric space, T be an FcS-contractive self-mapping with x0∈X, and r=minSTx,Tx,x:Tx≠x. Then Cx0,rS is a fixed circle of T. T especially fixes every circle Cx0,ρS where ρ<r.Proof.
Letx∈Cx0,rS. If Tx≠x, by the definition of r we have STx,Tx,x≥r. Hence, using the FcS-contractive property and the fact that F is increasing, we obtain(15)Fr≤FSTx,Tx,x≤FSx,x,x0-t<FSx,x,x0=Fr,which also lead to a contradiction. Therefore, STx,Tx,x=0 and that is Tx=x. Consequently, Cx0,rS is a fixed circle of T.
Now we show thatT also fixes any circle Cx0,ρS with ρ<r. Let x∈Cx0,ρS and assume that STx,Tx,x>0. By the FcS-contractive property, we have(16)FSTx,Tx,x≤FSx,x,x0-t<Fρ.
SinceF is increasing, then we find(17)STx,Tx,x<ρ<r.
Butr=minSTx,Tx,x:for  allTx≠x, which leads us to a contradiction. Thus, STx,Tx,x=0 and Tx=x. Hence, Cx0,ρS is a fixed circle of T.Remark 12.
Notice that, in Theorem11, the FcS-contractive self-mapping T fixes the disc with the center x0 and the radius r. Therefore, the center of any fixed circle is also fixed by T.In the following example, we see that the converse statement of Theorem11 is not always true.Example 13.
LetX,S be an S-metric space, x0∈X be any point, and the self-mapping T:X→X be defined as(18)Tx=xifSx,x,x0≤rx0ifSx,x,x0>r,for all x∈X with r>0. Then it can be easily seen that T is not an FcS-contractive self-mapping. Indeed, if Sx,x,x0>r for x∈X, then, using Lemma 3 and the FcS-contractive property, we get (19)STx,Tx,x=Sx0,x0,x>0⇒t+FSx0,x0,x≤FSx,x,x0⇒t≤0,which is a contradiction since t>0. Hence T is not an FcS-contractive self-mapping. But T fixes every circle Cx0,ρS where ρ≤r.Related to the number of the elements of the setX, the number of the fixed circles of an FcS-contractive self-mapping T can be infinite as seen in the following example.Example 14.
LetX=x∈Q:0≤x≤2, the metric d:X×X→0,∞ be defined as (20)dx,y=x1+x-y1+y, for all x,y∈X, and the S-metric be defined as in Example 4. Let us define the self-mapping T:X→X as (21)Tx=18ifx=0xotherwise, for all x∈X. Then the self-mapping T is an FcS-contractive self-mapping with F=ln⁡x+x, t=ln⁡3, and x0=1/2. Indeed, we get (22)STx,Tx,x=29>0⇒STx,Tx,x=29<Sx,x,x0=23⇒ln⁡29<ln⁡23⇒ln⁡29+29<ln⁡23+23⇒ln⁡3+ln⁡29+29≤ln⁡23+23⇒t+FSTx,Tx,x≤FSx,x,x0. Using Theorem 11, we have (23)r=min⁡STx,Tx,x:Tx≠x=29. Therefore, T fixes the circle C1/2,2/9S=2/7,4/5 and the disc D1/2,2/9S=x∈X:Sx,x,1/2≤2/9. Evidently, the number of the fixed circles of T is infinite.In the following definition, we introduce the notion of a Hardy-Rogers typeFcS-contraction.Definition 15.
LetX,S be an S-metric space and T be a self-mapping on X. If there exist F∈F, t>0, and x0∈X such that for all x∈X the following holds:(24)STx,Tx,x>0⟹t+FSTx,Tx,x≤FαSx,x,x0+βSTx,Tx,x+γSTx0,Tx0,x0+δSTx0,Tx0,x+ηSTx,Tx,x0,where (25)α+β+γ+δ+η=1,α,β,γ,δ,η≥0andα≠0,then the self-mapping T is called a Hardy-Rogers type FcS-contraction on X.Proposition 16.
LetX,S be an S-metric space. If a self-mapping T on X is a Hardy-Rogers type FcS-contraction with x0∈X then we have Tx0=x0.Proof.
Suppose thatTx0≠x0. Using the hypothesis, we obtain (26)STx0,Tx0,x0>0⇒t+FSTx0,Tx0,x0≤FαSx0,x0,x0+βSTx0,Tx0,x0+γSTx0,Tx0,x0+δSTx0,Tx0,x0+ηSTx0,Tx0,x0=Fβ+γ+δ+ηSTx0,Tx0,x0<FSTx0,Tx0,x0, which is a contradiction since t>0. Therefore, we get Tx0=x0.Remark 17.
Using Proposition16, a Hardy-Rogers type FcS-contraction condition can be changed as follows:(27)STx,Tx,x>0⇒t+FSTx,Tx,x≤FαSx,x,x0+βSTx,Tx,x+δSTx0,Tx0,x+ηSTx,Tx,x0,where (28)α+β+δ+η≤1,α,β,δ,η≥0andα≠0.Now using the Hardy-Rogers typeFcS-contraction condition, we prove the following fixed-circle theorem.Theorem 18.
LetX,S be an S-metric space, T be a Hardy-Rogers type FcS-contractive self-mapping with x0∈X, and r be defined as in Theorem 11. If STx,Tx,x0=r, then Cx0,rS is a fixed circle of T. T especially fixes every circle Cx0,ρS where ρ<r.Proof.
Letx∈Cx0,rS and Tx≠x. Using the Hardy-Rogers type FcS-contraction property, Proposition 16, Lemma 3, and the fact that F is increasing, we get (29)Fr≤FSTx,Tx,x≤FαSx,x,x0+βSTx,Tx,x+δSTx0,Tx0,x+ηSTx,Tx,x0-t<FαSx,x,x0+βSTx,Tx,x+δSTx0,Tx0,x+ηSTx,Tx,x0=Fα+δ+ηr+βSTx,Tx,x≤Fα+β+δ+ηSTx,Tx,x≤FSTx,Tx,x,which is a contradiction. Hence STx,Tx,x=0 and so Tx=x. Consequently, Cx0,rS is a fixed circle of T. By the similar arguments used in the proof of Theorem 11, T also fixes any circle Cx0,ρS where ρ<r.Corollary 19.
(1) LetX,S be an S-metric space, T be a Hardy-Rogers type FcS-contractive self-mapping with x0∈X, and r be defined as in Theorem 11. If STx,Tx,x0=r for all x∈Cx0,rS then T fixes the disc Dx0,rS.
(2) If we considerα=1 and β=γ=δ=η=0 in Definition 15, then we obtain the concept of an FcS-contractive mapping.In Definition15, if we get δ=η=0 then we have the following definition.Definition 20.
LetX,S be an S-metric space and T be a self-mapping on X. If there exist F∈F, t>0, and x0∈X such that for all x∈X the following holds:(30)STx,Tx,x>0⇒t+FSTx,Tx,x≤FαSx,x,x0+βSTx,Tx,x+γSTx0,Tx0,x0,where (31)α+β+γ<1andα,β,γ≥0, then the self-mapping T is called a Reich type FcS-contraction on X.Proposition 21.
LetX,S be an S-metric space. If a self-mapping T on X is a Reich type FcS-contraction with x0∈X then we get Tx0=x0.Proof.
The proof follows easily sinceβ+γ<1.Remark 22.
Using Proposition21, a Reich type FcS-contraction condition can be changed as follows:(32)STx,Tx,x>0⇒t+FSTx,Tx,x≤FαSx,x,x0+βSTx,Tx,x,where (33)α+β<1andα,β≥0.Theorem 23.
LetX,S be an S-metric space, T be a Reich type FcS-contractive self-mapping with x0∈X, and r be defined as in Theorem 11. Then Cx0,rS is a fixed circle of T. Also, T fixes every circle Cx0,ρS where ρ<r. In other words, T fixes the disc Dx0,rS.Proof.
The proof follows easily since(34)Fr≤FSTx,Tx,x≤Fα+βSTx,Tx,x<FSTx,Tx,x.In Definition15, if we get α=β=γ=0 and δ=η, then we have the following definition.Definition 24.
LetX,S be an S-metric space and T be a self-mapping on X. If there exist F∈F, t>0, and x0∈X such that for all x∈X the following holds:(35)STx,Tx,x>0⇒t+FSTx,Tx,x≤FηSTx0,Tx0,x+STx,Tx,x0, where(36)η∈0,12, then the self-mapping T is called a Chatterjea type FcS-contraction on X.Proposition 25.
LetX,S be an S-metric space. If a self-mapping T on X is a Chatterjea type FcS-contraction with x0∈X then we get Tx0=x0.Proof.
The proof follows easily.Theorem 26.
LetX,S be an S-metric space, T be a Chatterjea type FcS-contractive self-mapping with x0∈X, and r be defined as in Theorem 11. If STx,Tx,x0=r for all x∈Cx0,rS then Cx0,rS is a fixed circle of T. Also, T fixes every circle Cx0,ρS where ρ<r. In other words, T fixes the disc Dx0,rS.Proof.
The proof follows easily by the similar arguments used in the proofs of Theorems11 and 18.Now we give the following illustrative example.Example 27.
LetC be the set of all complex numbers. Consider the set (37)Xz=0,4,z,z2,z4,z8,z8-2,z8+2,z16,z16-2,z16+2⊂C, where z is any complex number with z=2 and the S-metric is defined as in [18] such that (38)Sx,y,t=x-t+x+t-2y, for all x,y,t∈Xz. Let us define the self-mapping T:Xz→Xz as (39)Tx=zifx=0xotherwise, for all x∈Xz. Then the self-mapping T is an FcS-contractive self-mapping with F=-1/x, t=1/28 and x0=z16. Indeed, we obtain (40)STx,Tx,x=4>0, for x=0, and (41)Sx,x,x0=217. Then we have (42)t+STx,Tx,x=128-12≤-1282.
Also we obtain(43)r=min⁡STx,Tx,x:Tx≠x=4. Therefore, the self-mapping T fixes the circle Cz16,4S=z16-2,z16+2 and the disc Dz16,4S=z16-2,z16,z16+2.Also the self-mappingT is a Hardy-Rogers type FcS-contractive self-mapping (resp., a Reich type FcS-contractive self-mapping and a Chatterjea type FcS-contractive self-mapping) on Xz with α=1, β=δ=η=0 (resp., α=(216-214+28)/217(214-28+1), β=1/4 and η=5/(217+41-215)).


## 4.ψ-Contractive and Weakly ψ-Contractive Self-Mappings on S-Metric Spaces

First, in this section we present this well-known interesting class of functions.Definition 28.
Denote byΨ the family of nondecreasing functions (44)ψ:0,+∞→0,+∞such  that∑n=1+∞ψnt<+∞for  eacht>0, where ψn is the n-th iterate of ψ.Lemma 29.
For every functionψ:[0,+∞)→[0,+∞) the following holds: if ψ is nondecreasing, then, for each t>0, limn→+∞ψn(t)=0 implies that ψ(t)<t.Now, we define theψ-contractive self-mapping in an S-metric space.Definition 30.
LetT be a self-mapping on an S-metric space (X,S). We say that T is ψ-contractive self-mapping if there exist x0∈X and ψ∈Ψ such that for all x,y,z∈X we have (45)STy,Tz,x≤ψSx,x,x0-min⁡ψSTy,Ty,x0,ψSTz,Tz,x0.Theorem 31.
LetT be a ψ-contractive self-mapping with x0∈X on an S-metric space (X,S), and consider the circle Cx0,rS. Thus, for every x∈Cx0,rS, T either fixes x or maps x to the interior of Cx0,rS. Moreover, if for every x∈Cx0,rS we have S(Tx,Tx,x0)=r, then Cx0,rS is a unique fixed circle of T in X.Proof.
Ifx∈Cx0,rS, then since T is ψ-contractive we have (46)STx,Tx,x≤ψSx,x,x0-ψSTx,Tx,x0=ψr-ψSTx,Tx,x0.If S(Tx,Tx,x0)<r, then we are in the case where T maps x to the interior of Cx0,rS. If S(Tx,Tx,x0)≥r, then by using the fact that ψ is a nondecreasing function we have (47)STx,Tx,x≤ψr-ψSTx,Tx,x0.Now, if S(Tx,Tx,x0)>r, then the above inequality implies that S(Tx,Tx,x)<0 which leads to a contradiction. Hence, in this case we must have S(Tx,Tx,x0)=r. Thus, (48)STx,Tx,x≤ψr-ψSTx,Tx,x0=ψr-ψr=0,and that is Tx=x.
Therefore,T either fixes x or maps x to the interior of Cx0,rS as required.
To prove the second part of our theorem, we may assume thatS(Tx,Tx,x0)=r, for all x∈Cx0,rS. Now, we only need to show that if there exists x∈X where Tx=x, then x∈Cx0,rS, and that will prove the uniqueness. So, first let x∈Cx0,rS, and that is Tx=x, and also let y∈X be an arbitrary fixed point of T (i.e., Ty=y) we have two cases.
Case 1. If S(y,y,x0)≥r then by using the fact that ψ is a nondecreasing function we have (49)Sy,y,x=STy,Ty,x≤ψr-ψSTy,Ty,x0. Now, if S(Ty,Ty,x0)>r then the above inequality implies that S(y,y,x)<0 which leads to a contradiction. Hence, in this case we must have S(y,y,x0)=r.(50)Sy,y,x=STy,Ty,x≤ψSx,x,x0-ψSTy,Ty,x0=ψr-ψr=0, and that is x=y.
Case 2. If S(y,y,x0)<r then once again by using the fact that ψ is a nondecreasing function we have (51)Sx,x,y≤ψSy,y,x0-ψSTx,Tx,x0=ψSy,y,x0-ψSx,x,x0=ψSy,y,x0-ψr<ψr-ψr=0, which leads us to a contradiction.
Therefore,Cx0,rS is the unique fixed circle of T in X as desired.Next, we give the definition of a weaklyψ-contractive self-mapping.Definition 32.
LetT be a self-mapping on an S-metric space (X,S). We say that T is a weakly ψ-contractive self-mapping with x0∈X if there exist x0∈X and ψ∈Ψ such that for all x,y,z∈X we have (52)STy,T2z,x≤ψSx,x,x0-min⁡ψSTy,Ty,x0,ψSTz,Tz,x0.Theorem 33.
LetT be a weakly ψ-contractive self-mapping with x0∈X on an S-metric space (X,S) and consider the circle Cx0,rS. Thus, for every x∈Cx0,rST either fixes x or maps x to the interior of Cx0,rS. Moreover, if for every x∈Cx0,rS, we have S(Tx,Tx,x0)=r, then Cx0,rS is a unique fixed circle of T in X.Proof.
Ifx∈Cx0,rS, then since T is weakly ψ-contractive we have (53)STx,T2x,x≤ψSx,x,x0-ψSTx,Tx,x0=ψr-ψSTx,Tx,x0. If S(Tx,Tx,x0)<r, then we are in the case where T maps x to the interior of Cx0,rS. If S(Tx,Tx,x0)≥r, then by using the fact that ψ is a nondecreasing function we have (54)STx,T2x,x≤ψr-ψSTx,Tx,x0. Now, if S(Tx,Tx,x0)>r, then the above inequality implies that S(Tx,T2x,x)<0 which leads to a contradiction. Hence, in this case we must have S(Tx,Tx,x0)=r. Thus, (55)STx,T2x,x≤ψr-ψSTx,Tx,x0=ψr-ψr=0, and that is Tx=x.
Therefore,T either fixes x or maps x to the interior of Cx0,rS as required.
To prove the second part of our theorem, we may assume thatS(Tx,Tx,x0)=r, for all x∈Cx0,rS. Now, we only need to show that if there exists x∈X, where Tx=x, then x∈Cx0,rS, and that will prove the uniqueness. So, first let x∈Cx0,rS, and that is Tx=x, and also let y∈X be an arbitrary fixed point (i.e., Ty=y) we have two cases.
Case 1. If S(y,y,x0)≥r then by using the fact that ψ is a nondecreasing function we have (56)Sy,y,x=STy,T2y,x≤ψr-ψSTy,Ty,x0.
Now, ifS(Ty,Ty,x0)>r, then the above inequality implies that S(y,y,x)<0 which leads to a contradiction. Hence, in this case we must have S(y,y,x0)=r.(57)Sy,y,x=STy,T2y,x≤ψSx,x,x0-ψSTy,Ty,x0=ψr-ψr=0, and that is x=y.
Case 2. If S(y,y,x0)<r then once again by using the fact that ψ is a nondecreasing function we have (58)Sx,x,y=STx,T2x,y≤ψSy,y,x0-ψSTx,Tx,x0=ψSy,y,x0-ψSx,x,x0=ψSy,y,x0-ψr<ψr-ψr=0, which leads us to a contradiction.
Therefore,Cx0,rS is the unique fixed circle of T in X as desired.


## 5. An Application to Integral Type Contractive Self-Mappings

We assume thatφ:0,∞→0,∞ is a Lebesgue-integrable mapping which is summable (that is, with finite integral) on each compact subset of 0,∞, nonnegative, and such that, for each ε>0,(59)∫0εφtdt>0.Now we give the following definition.Definition 34.
LetX,S be an S-metric space and φ:0,∞→0,∞ be defined as in (59). A self-mapping T on X is said to be an integral type FcS-contraction if there exist F∈F, t>0, and x0∈X such that for all x∈X the following holds: (60)STx,Tx,x>0⇒t+∫0FSTx,Tx,xφtdt≤∫0FSx,x,x0φtdt.Proposition 35.
LetX,S be an S-metric space and φ:0,∞→0,∞ be defined as in (59). If a self-mapping T on X is an integral type FcS-contraction with x0∈X then we get Tx0=x0.Proof.
Suppose thatTx0≠x0. From the definition of an integral type FcS-contraction, we have(61)STx0,Tx0,x0>0⇒t+∫0FSTx0,Tx0,x0φtdt≤∫0FSx0,x0,x0φtdt.Inequality (61) contradicts with the definition of F since F:0,∞→R and Sx0,x0,x0=0. Hence, it should be Tx0=x0.Using this new definition, we get the following fixed-circle result.Theorem 36.
LetX,S be an S-metric space, φ:0,∞→0,∞ be defined as in (59), T be an integral type FcS-contraction with x0∈X, and r be defined as in Theorem 11. Then Cx0,rS is a fixed circle of T.Proof.
Letx∈Cx0,rS. Assume that Tx≠x. Then, by the definition of r, we get (62)r≤STx,Tx,x. Using the fact that F is increasing property, we have (63)Fr≤FSTx,Tx,x and(64)∫0Frφtdt≤∫0FSTx,Tx,xφtdt.From inequality (64) and the definition of integral type FcS-contractivity, we obtain (65)∫0Frφtdt≤∫0FSTx,Tx,xφtdt≤∫0FSx,x,x0φtdt-t<∫0FSx,x,x0φtdt=∫0Frφtdt, which is a contradiction. Therefore, we find Tx=x. Consequently, Cx0,rS is a fixed circle of T.Remark 37.
( 1 ) An integral type FcS-contractive self-mapping T fixes also the disc Dx0,rS.
( 2 ) If we set the function φ:0,∞→0,∞ in Theorem 36 as φ(t)=1 for all t∈0,∞, then we get Theorem 11.
( 3 ) By the similar argument used in Definition 34, the notions of an integral Hardy-Rogers type FcS-contractive self-mapping, an integral Reich type FcS-contractive self-mapping, an integral Chatterjea type FcS-contractive self-mapping, and obtained corresponding fixed-circle theorems can be defined.Finally, we give the following example.Example 38.
LetX=e,2e,e+1/2,2e-1/2,2e+1/2⊂R be the S-metric space with the usual S-metric and the function φ:0,∞→0,∞ be defined by (66)φt=2t+1, for all t∈0,∞. Let us define the self-mapping T:X→X as (67)Tx=e+12ifx=exotherwise, for all x∈X. The self-mapping T is an integral type FcS-contractive self-mapping with F=ln⁡x, t=1, and x0=2e. Indeed, we get (68)STx,Tx,x=Se+12,e+12,e=2e+12-e=1>0, for x=e. Then we have (69)STx,Tx,x=1<Sx,x,x0=2e⇒ln⁡1=0<ln⁡2e=ln⁡2+1⇒∫002t+1dt=0<∫0ln⁡2+12t+1dt=ln2⁡2+ln⁡8+2⇒1≤ln2⁡2+ln⁡8+2. Also we obtain (70)r=min⁡STx,Tx,x:Tx≠x=1. Consequently, T fixes the circle C2e,1S=2e-1/2,2e+1/2 and the disc D2e,1S=2e-1/2,2e,2e+1/2.



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