Research article

Some common fixed-point and fixed-figure results with a function family on Sb-metric spaces

  • Received: 18 February 2023 Revised: 22 March 2023 Accepted: 23 March 2023 Published: 03 April 2023
  • MSC : 47H09, 47H10, 54H25

  • In this paper, we prove a common fixed-point theorem for four self-mappings with a function family on Sb-metric spaces. In addition, we investigate some geometric properties of the fixed-point set of a given self-mapping. In this context, we obtain a fixed-disc (resp. fixed-circle), fixed-ellipse, fixed-hyperbola, fixed-Cassini curve and fixed-Apollonious circle theorems on Sb-metric spaces.

    Citation: Nihal Taş, Irshad Ayoob, Nabil Mlaiki. Some common fixed-point and fixed-figure results with a function family on Sb-metric spaces[J]. AIMS Mathematics, 2023, 8(6): 13050-13065. doi: 10.3934/math.2023657

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  • In this paper, we prove a common fixed-point theorem for four self-mappings with a function family on Sb-metric spaces. In addition, we investigate some geometric properties of the fixed-point set of a given self-mapping. In this context, we obtain a fixed-disc (resp. fixed-circle), fixed-ellipse, fixed-hyperbola, fixed-Cassini curve and fixed-Apollonious circle theorems on Sb-metric spaces.



    Fixed-point theory has been extensively researched in various areas, such as mathematics, engineering, and physics. Of particular importance is metric fixed-point theory, which is used in various branches of mathematics like topology, analysis, and applied mathematics. This theory was initiated with the famous Banach fixed-point theorem [1]. This theorem ensures that a self-mapping will have a unique fixed point. Despite this, there remain instances of self-mapping that have a fixed point but do not meet the criteria set by the Banach fixed point theorem, such as:

    Let X=R, and (X,ζ) be the usual metric space. Consider a self-mapping g:RR defined as

    gμ=2μ4,

    for all μR. Then g has a unique fixed point μ=4, but g does not meet the criteria of Banach contraction principle.

    There are two popular methods used by researchers to generalize the Banach contraction principle. The first entails extending the utilized contractive condition while the second centers around generalizing the underlying metric space. For example, Gb-metric spaces, G-metric spaces [2], complex valued Gb-metric spaces [3,4], S-metric spaces, A-metric spaces [5], Sb-metric spaces, fuzzy cone metric spaces [6], modular metric spaces [7,8,9] et al. were defined for this purpose (for more details, see [10,11,12,13] and the references therein). Especially, we focus on the notion of Sb-metric spaces. To do this, we recall the following basic concepts:

    Definition 1.1. [14] Let X be a nonempty set and s1 a given real number. A function Sb:X×X×X[0,) is said to be Sb -metric if and only if for all μ,τ,,ρX the following conditions are satisfied:

    (Sb1) Sb(μ,τ,)=0 if and only if μ=τ=,

    (Sb2) Sb(μ,τ,)s[Sb(μ,μ,ρ)+Sb(τ,τ,ρ)+Sb(,,ρ)].

    The pair (X,Sb) is called an Sb-metric space.

    As every S-metric is a Sb-metric with s=1, we observe that Sb -metric spaces are extensions of S-metric spaces. However, the converse statement is not always true (see [14] and [15] for more details).

    Definition 1.2. [15] Let (X,Sb) be an Sb-metric space and s>1. An Sb-metric Sb is called symmetric if

    Sb(μ,μ,τ)=Sb(τ,τ,μ),

    for all μ,τX.

    Definition 1.3. [14] Let (X,Sb) be an Sb -metric space, and {n} be a sequence in X.

    1) Then the sequence {n} converges to X if and only if Sb(n,n,)0 as n, that is, for each ε>0 there exists n0N such that for all nn0, Sb(n,n,)<ε. It is denoted by

    limnn=.

    2) Then the sequence {n} is called a Cauchy sequence if for each ε>0 there exists n0N such that Sb(n,n,m)<ε for each n,mn0.

    3) The Sb-metric space (X,Sb) is called complete if every Cauchy sequence is convergent.

    On the other hand, in the literature, besides various fixed point theorems there are also common fixed point theorems on metric and some extended metric spaces (for example, see [14,16,17,18,19] and the references therein).

    Recently, as a geometric approach to the fixed-point theory, the fixed-circle problem (see [20]) and the fixed-figure problem (see [21]) have been introduced. When there are more than one fixed points, it is interesting to investigate for the possible solutions as follows:

    Let us define a self-mapping, g:RR, where R is with the usual metric

    gμ=μ212,

    for all μR. Then g has two fixed points

    μ1=1 and μ2=1. We consider these fixed points as a unit circle C0,1={1,1}.

    For this reason, there exist some studies related to these recent aspects (for example, see [22,23,24,25,26,27,28] and the references therein).

    By the above motivation, in this paper, we prove a common fixed-point theorem and some fixed-figure results on Sb-metric spaces.

    In this section, we prove a common fixed-point result on Sb-metric spaces. To do this, we are inspired by the function family F6 introduced in [29] and the function family M defined in [30]. We modify these families as follows:

    Let Ψ be the family of all lower semi-continuous functions ψ:R6+R that satisfy the following condition:

    (ψ). For all μ,τ,0 and s1, there exists a k[0,1) such that

    μψ(μ,τ,τ,μ,0,)

    with 2sμ+sτ then

    μkτ.

    If we define the function ψ:R6+R such as

    ψ(t1,t2,t3,t4,t5,t6)=kmax{t1,t2},

    with k[0,1). Then ψΨ.

    Now, we give the following theorem.

    Theorem 2.1. Let (X,Sb) be a complete continuous Sb -metric space with the symmetric metric Sb. Let g,h,G,H:XX be four self-mappings, where g,G and H are continuous and satisfying the following conditions:

    (i) g(X)H(X) and h(X)G(X),

    (ii) For all μ,τX and ψΨ,

    Sb(gμ,gμ,hτ)ψ(Sb(gμ,gμ,hτ),Sb(μ,μ,τ),Sb(μ,μ,gμ),Sb(τ,τ,hτ),Sb(τ,τ,gμ),Sb(μ,μ,hτ)),

    (iii) For all μ,τX and ψΨ,

    Sb(Gμ,Gμ,Hτ)ψ(Sb(Gμ,Gμ,Hτ),Sb(μ,μ,τ),Sb(μ,μ,Gμ),Sb(τ,τ,Hτ),Sb(τ,τ,Gμ),Sb(μ,μ,Hτ)),

    holds, then g, h, G and H have a common fixed point in X.

    Proof. Let 0X, 1=g0 and 2=h1. Using the condition (ii), we get

    Sb(g0,g0,h1)=Sb(1,1,2)ψ(Sb(g0,g0,h1),Sb(0,0,1),Sb(0,0,g0),Sb(1,1,h1),Sb(1,1,g0),Sb(0,0,h1))=ψ(Sb(1,1,2),Sb(0,0,1),Sb(0,0,1),Sb(1,1,2),Sb(1,1,1),Sb(0,0,2))=(Sb(1,1,2),Sb(0,0,1),Sb(0,0,1),Sb(1,1,2),0,Sb(0,0,2)). (2.1)

    By (Sb2) and the symmetry property of Sb, we have

    Sb(0,0,2)=Sb(2,2,0)s[2Sb(2,2,1)+Sb(0,0,1)]=2sSb(1,1,2)+sSb(0,0,1). (2.2)

    Using (2.1), (2.2) and (ψ), there exists a k[0,1) such that

    Sb(1,1,2)kSb(0,0,1).

    Continuing this process with induction with the condition (i), we can define the sequence {n} as follows:

    2n+1=g2n=H2n

    and

    2n=h2n1=G2n1.

    Using (ii), for μ=2n and τ=2n+1, we find

    Sb(g2n,g2n,h2n+1)=Sb(2n+1,2n+1,2n+2)ψ(Sb(g2n,g2n,h2n+1),Sb(2n,2n,2n+1),Sb(2n,2n,g2n),Sb(2n+1,2n+1,h2n+1),Sb(2n+1,2n+1,g2n),Sb(2n,2n,h2n+1))=ψ(Sb(2n+1,2n+1,2n+2),Sb(2n,2n,2n+1),Sb(2n,2n,2n+1),Sb(2n+1,2n+1,2n+2),Sb(2n+1,2n+1,2n+1),Sb(2n,2n,2n+2))=ψ(Sb(2n+1,2n+1,2n+2),Sb(2n,2n,2n+1),Sb(2n,2n,2n+1),Sb(2n+1,2n+1,2n+2),0,Sb(2n,2n,2n+2)) (2.3)

    By (Sb2) and the symmetry property of Sb, we have

    Sb(2n,2n,2n+2)=Sb(2n+2,2n+2,2n)s[2Sb(2n+2,2n+2,2n+1)+Sb(2n,2n,2n+1)]=2sSb(2n+1,2n+1,2n+2)+sSb(2n,2n,2n+1). (2.4)

    Using (2.3), (2.4) and (ψ), there exists a k[0,1) such that

    Sb(2n+1,2n+1,2n+2)kSb(2n,2n,2n+1). (2.5)

    Using (iii), for μ=2n1 and τ=2n, we get

    Sb(G2n1,G2n1,H2n)=Sb(2n,2n,2n+1)ψ(Sb(G2n1,G2n1,H2n),Sb(2n1,2n1,2n),Sb(2n1,2n1,G2n1),Sb(2n,2n,H2n),Sb(2n,2n,G2n1),Sb(2n1,2n1,H2n))=ψ(Sb(2n,2n,2n+1),Sb(2n1,2n1,2n),Sb(2n1,2n1,2n),Sb(2n,2n,2n+1),Sb(2n,2n,2n),Sb(2n1,2n1,2n+1))=ψ(Sb(2n,2n,2n+1),Sb(2n1,2n1,2n),Sb(2n1,2n1,2n),Sb(2n,2n,2n+1),0,Sb(2n1,2n1,2n+1)) (2.6)

    By (Sb2) and the symmetry property of Sb, we find

    Sb(2n1,2n1,2n+1)=Sb(2n+1,2n+1,2n1)s[2Sb(2n+1,2n+1,2n)+Sb(2n1,2n1,2n)]=2sSb(2n,2n,2n+1)+sSb(2n1,2n1,2n). (2.7)

    Using (2.6), (2.7) and (ψ), there exists a k[0,1) such that

    Sb(2n,2n,2n+1)kSb(2n1,2n1,2n). (2.8)

    Using the inequalities (2.5) and (2.8), we get

    Sb(2n+1,2n+1,2n+2)kSb(2n,2n,2n+1)k2Sb(2n1,2n1,2n)

    and so, using similar arguments, we have

    Sb(n,n,n+1)knSb(0,0,1). (2.9)

    Now we show that the sequence {n} is Cauchy. Using (Sb2), (2.9) and the symmetry property of Sb, for all n,mN with m>n, we get

    Sb(n,n,m)s[2Sb(n,n,n+1)+Sb(m,m,n+1)]=s[2Sb(n,n,n+1)+Sb(n+1,n+1,m)]2sSb(n,n,n+1)+s2[2Sb(n+1,n+1,n+2)+Sb(m,m,n+2)]=2sSb(n,n,n+1)+s2[2Sb(n+1,n+1,n+2)+Sb(n+2,n+2,m)]2sSb(n,n,n+1)+2s2Sb(n+1,n+1,n+2)+2sknSb(0,0,1)+2s2kn+1Sb(0,0,1)+2skn1skSb(0,0,1).

    Since s1 and k[0,1), taking n,m, we get

    Sb(n,n,m)0

    and so {n} is Cauchy. Since (X,Sb) is a complete Sb-metric space, {n} is convergent to a point X, that is,

    limnSb(n,n,)=0.

    Next, we establish that is a common fixed point of g, h, G and H. Using (ii), for μ=2n and τ=, we get

    Sb(g2n,g2n,h)=Sb(2n+1,2n+1,h)ψ(Sb(g2n,g2n,h),Sb(2n,2n,),Sb(2n,2n,g2n),Sb(,,h),Sb(,,g2n),Sb(2n,2n,h))=ψ(Sb(2n+1,2n+1,h),Sb(2n,2n,),Sb(2n,2n,2n+1),Sb(,,h),Sb(,,2n+1),Sb(2n,2n,h))

    and taking n, we obtain

    Sb(,,h)ψ(Sb(,,h),Sb(,,),Sb(,,),Sb(,,h),Sb(,,),Sb(,,h))=ψ(Sb(,,h),0,0,Sb(,,h),0,Sb(,,h)) (2.10)

    and

    Sb(,,h)2sSb(,,h)+s.0. (2.11)

    Using (2.10), (2.11) and (ψ), there exists a k[0,1) such that

    Sb(,,h)k.0=0,

    that is,

    h=.

    Using the continuity hypothesis of g, we have

    limnSb(2n,2n,)=0limnSb(g2n,g2n,g)=0limnSb(2n+1,2n+1,g)=0Sb(,,g)=0g=.

    Hence is a common fixed point g and h. Using (iii), for μ=2n and τ=, we get

    Sb(G2n,G2n,H)=Sb(2n+1,2n+1,H)ψ(Sb(G2n,G2n,H),Sb(2n,2n,),Sb(2n,2n,G2n),Sb(,,H),Sb(,,G2n),Sb(2n,2n,H))=ψ(Sb(2n+1,2n+1,H),Sb(2n,2n,),Sb(2n,2n,2n+1),Sb(,,H),Sb(,,2n+1),Sb(2n,2n,H))

    and taking n, we obtain

    Sb(,,H)ψ(Sb(,,H),Sb(,,),Sb(,,),Sb(,,H),Sb(,,),Sb(,,H))=ψ(Sb(,,H),0,0,Sb(,,H),0,Sb(,,H)) (2.12)

    and

    Sb(,,H)2sSb(,,H)+s.0. (2.13)

    Using (2.12), (2.13) and (ψ), there exists a k[0,1) such that

    Sb(,,H)k.0=0,

    that is,

    H=.

    Using the continuity hypothesis of G, we have

    limnSb(2n,2n,)=0limnSb(G2n,G2n,G)=0limnSb(2n+1,2n+1,G)=0Sb(,,G)=0G=.

    Consequently, we obtain

    =h=g=H=G,

    that is, is a common fixed point of four self-mappings g, h, G and H.

    In this section, we investigate some fixed-figure results on Sb-metric spaces. At first, we recall the following notions:

    Definition 3.1. [22,31] Let (X,Sb) be an Sb-metric space with s1 and 0,1,2X, r[0,).

    The circle is defined by

    CSb0,r={μX:Sb(μ,μ,0)=r}.

    ● The disc is defined by

    DSb0,r={μX:Sb(μ,μ,0)r}.

    ● The ellipse is defined by

    ESbr(1,2)={μX:Sb(μ,μ,1)+Sb(μ,μ,2)=r}.

    ● The hyperbola is defined by

    HSbr(1,2)={μX:|Sb(μ,μ,1)Sb(μ,μ,2)|=r}.

    ● The Cassini curve is defined by

    CSbr(1,2)={μX:Sb(μ,μ,1)Sb(μ,μ,2)=r}.

    ● The Apollonious circle is defined by

    ASbr(1,2)={μX{2}:Sb(μ,μ,1)Sb(μ,μ,2)=r}.

    Definition 3.2. [22] Let g:XX be a self-mapping where (X,Sb) is a Sb-metric space with s1. Let Fix(g) be set of all fixed points of g, then a geometric figure F is said to be a fixed figure of g if F is contained in Fix(g).

    Let us define the number r as

    r=inf{Sb(μ,μ,gμ):μFix(g)}. (3.1)

    Theorem 3.1. Let (X,Sb) be an Sb-metric space with s1, g:XX be a self-mapping, Sb be symmetric and r be defined as in (3.1). If there exist 0X and ψΨ for all μX{0} such that

    μFix(g)Sb(gμ,gμ,μ)<ψ(Sb(gμ,gμ,μ),Sb(μ,μ,0),Sb(μ,μ,g0),Sb(gμ,gμ,μ),Sb(g0,g0,0),Sb(gμ,gμ,0))

    and g0=0, then DSb0,rFix(g). Especially, we have CSb0,rFix(g).

    Proof. Let r=0. Then we have DSb0,r={0}. By the hypothesis g0=0, we obtain

    DSb0,rFix(g).

    Let r>0 and μDSb0,r such that μFix(g). Using the hypothesis, we get

    Sb(gμ,gμ,μ)<ψ(Sb(gμ,gμ,μ),Sb(μ,μ,0),Sb(μ,μ,g0),Sb(gμ,gμ,μ),Sb(g0,g0,0),Sb(gμ,gμ,0))=ψ(Sb(gμ,gμ,μ),Sb(μ,μ,0),Sb(μ,μ,0),Sb(gμ,gμ,μ),Sb(0,0,0),Sb(gμ,gμ,0))=ψ(Sb(gμ,gμ,μ),Sb(μ,μ,0),Sb(μ,μ,0),Sb(gμ,gμ,μ),0,Sb(gμ,gμ,0)). (3.2)

    By (Sb2) and the symmetry property of Sb, we have

    Sb(gμ,gμ,0)s[2Sb(gμ,gμ,μ)+Sb(0,0,μ)]=2sSb(gμ,gμ,μ)+sSb(μ,μ,0). (3.3)

    Using (3.2), (3.3) and (ψ), there exists a k[0,1) such that

    Sb(gμ,gμ,μ)kSb(μ,μ,0)krkSb(gμ,gμ,μ)<Sb(gμ,gμ,μ),

    a contradiction. Hence it should be μFix(g). Consequently, we get

    DSb0,rFix(g).

    Using the similar arguments, it can be easily see that

    CSb0,rFix(g).

    Theorem 3.2. Let (X,Sb) be an Sb-metric space with s1, g:XX be self-mapping, Sb be a symmetric and r be defined as in (3.1). If there exist 1,2X and ψΨ for all μX{1,2} such that

    μFix(g)Sb(gμ,gμ,μ)<ψ(Sb(gμ,gμ,μ),Sb(μ,μ,1)+Sb(μ,μ,2),Sb(μ,μ,g1)+Sb(μ,μ,g2),Sb(gμ,gμ,μ),Sb(g1,g1,1)+Sb(g2,g2,2),Sb(gμ,gμ,1)+Sb(gμ,gμ,2))

    and g1=1, g2=2 with g(μ)ESbr(1,2), then ESbr(1,2)Fix(g).

    Proof. Let r=0. Then we have ESbr(1,2)={1}={2}. By the hypothesis g1=1 and g2=2, we obtain

    ESbr(1,2)Fix(g).

    Let r>0 and μESbr(1,2) such that μFix(g). Using the hypothesis, we get

    Sb(gμ,gμ,μ)<ψ(Sb(gμ,gμ,μ),Sb(μ,μ,1)+Sb(μ,μ,2),Sb(μ,μ,g1)+Sb(μ,μ,g2),Sb(gμ,gμ,μ),Sb(g1,g1,1)+Sb(g2,g2,2),Sb(gμ,gμ,1)+Sb(gμ,gμ,2))=ψ(Sb(gμ,gμ,μ),Sb(μ,μ,1)+Sb(μ,μ,2),Sb(μ,μ,1)+Sb(μ,μ,2),Sb(gμ,gμ,μ),Sb(1,1,1)+Sb(2,2,2),Sb(gμ,gμ,1)+Sb(gμ,gμ,2))=ψ(Sb(gμ,gμ,μ),Sb(μ,μ,1)+Sb(μ,μ,2),Sb(μ,μ,1)+Sb(μ,μ,2),Sb(gμ,gμ,μ),0,Sb(gμ,gμ,1)+Sb(gμ,gμ,2))=ψ(Sb(gμ,gμ,μ),r,r,Sb(gμ,gμ,μ),0,r).

    Since

    r2sSb(gμ,gμ,μ)+sr,

    using (ψ), there exists a k[0,1) such that

    Sb(gμ,gμ,μ)krkSb(gμ,gμ,μ)<Sb(gμ,gμ,μ),

    a contradiction. Hence it should be μFix(g). Consequently, we get

    ESbr(1,2)Fix(g).

    Theorem 3.3. Let (X,Sb) be an Sb-metric space with s1, g:XX be self-mapping, Sb be a symmetric and r be defined as in (3.1). If r>0 and there exist 1,2X, ψΨ for all xX{1,2} such that

    μFix(g)Sb(gμ,gμ,μ)<ψ(Sb(gμ,gμ,μ),|Sb(μ,μ,1)Sb(μ,μ,2)|,|Sb(μ,μ,g1)Sb(μ,μ,g2)|,Sb(gμ,gμ,μ),|Sb(g1,g1,1)Sb(g2,g2,2)|,|Sb(gμ,gμ,1)Sb(gμ,gμ,2)|)

    and g1=1, g2=2 with g(μ)HSbr(1,2), then HSbr(1,2)Fix(g).

    Proof. Let r>0 and μHSbr(1,2) such that μFix(g). Using the hypothesis, we get

    Sb(gμ,gμ,μ)<ψ(Sb(gμ,gμ,μ),|Sb(μ,μ,1)Sb(μ,μ,2)|,|Sb(μ,μ,g1)Sb(μ,μ,g2)|,Sb(gμ,gμ,μ),|Sb(g1,g1,1)Sb(g2,g2,2)|,|Sb(gμ,gμ,1)Sb(gμ,gμ,2)|)=ψ(Sb(gμ,gμ,μ),|Sb(μ,μ,1)Sb(μ,μ,2)|,|Sb(μ,μ,1)Sb(μ,μ,2)|,Sb(gμ,gμ,μ),|Sb(1,1,1)Sb(2,2,2)|,|Sb(gμ,gμ,1)Sb(gμ,gμ,2)|)=ψ(Sb(gμ,gμ,μ),|Sb(μ,μ,1)Sb(μ,μ,2)|,|Sb(μ,μ,1)Sb(μ,μ,2)|,Sb(gμ,gμ,μ),0,|Sb(gμ,gμ,1)Sb(gμ,gμ,2)|)=ψ(Sb(gμ,gμ,μ),r,r,Sb(gμ,gμ,μ),0,r).

    Since

    r2sSb(gμ,gμ,μ)+sr,

    using (ψ), there exists a k[0,1) such that

    Sb(gμ,gμ,μ)krkSb(gμ,gμ,μ)<Sb(gμ,gμ,μ),

    a contradiction. Hence it should be μFix(g). Consequently, we get

    HSbr(1,2)Fix(g).

    Theorem 3.4. Let (X,Sb) be an Sb-metric space with s1, g:XX be a self-mapping, Sb be symmetric and r be defined as in (3.1). If there exist 1,2X and ψΨ for all μX{1,2} such that

    μFix(g)Sb(gμ,gμ,μ)<ψ(Sb(gμ,gμ,μ),Sb(μ,μ,1)Sb(μ,μ,2),Sb(μ,μ,g1)Sb(μ,μ,g2),Sb(gμ,gμ,μ),Sb(g1,g1,1)Sb(g2,g2,2),Sb(gμ,gμ,1)Sb(gμ,gμ,2))

    and g1=1, g2=2 with g(μ)CSbr(1,2), then CSbr(1,2)Fix(g).

    Proof. Let r=0. Then we have CSbr(1,2)={1} or {2}. By the hypothesis g1=1 and g2=2, we obtain

    CSbr(1,2)Fix(g).

    Let r>0 and μCSbr(1,2) such that μFix(g). Using the hypothesis, we get

    Sb(gμ,gμ,μ)<ψ(Sb(gμ,gμ,μ),Sb(μ,μ,1)Sb(μ,μ,2),Sb(μ,μ,g1)Sb(μ,μ,g2),Sb(gμ,gμ,μ),Sb(g1,g1,1)Sb(g2,g2,2),Sb(gμ,gμ,1)Sb(gμ,gμ,2))=ψ(Sb(gμ,gμ,μ),Sb(μ,μ,1)Sb(μ,μ,2),Sb(μ,μ,1)Sb(μ,μ,2),Sb(gμ,gμ,μ),Sb(1,1,1)Sb(2,2,2),Sb(gμ,gμ,1)Sb(gμ,gμ,2))=ψ(Sb(gμ,gμ,μ),Sb(μ,μ,1)Sb(μ,μ,2),Sb(μ,μ,1)Sb(μ,μ,2),Sb(gμ,gμ,μ),0,Sb(gμ,gμ,1)Sb(gμ,gμ,2))=ψ(Sb(gμ,gμ,μ),r,r,Sb(gμ,gμ,μ),0,r).

    Since

    r2sSb(gμ,gμ,μ)+sr,

    using (ψ), there exists a k[0,1) such that

    Sb(gμ,gμ,μ)krkSb(gμ,gμ,μ)<Sb(gμ,gμ,μ),

    a contradiction. Hence it should be μFix(g). Consequently, we get

    CSbr(1,2)Fix(g).

    Theorem 3.5. Let (X,Sb) be an Sb-metric space with s1, g:XX be a self-mapping, Sb be symmetric and r be defined as in (3.1). If there exist 1,2X and ψΨ for all μX{1,2} such that

    μFix(g)Sb(gμ,gμ,μ)<ψ(Sb(gμ,gμ,μ),Sb(μ,μ,1)Sb(μ,μ,2),Sb(μ,μ,g1)Sb(μ,μ,g2),Sb(gμ,gμ,μ),0,Sb(gμ,gμ,1)Sb(gμ,gμ,2))

    and g1=1, g2=2 with g(μ)ASbr(1,2), then ASbr(1,2)Fix(g).

    Proof. Let r=0. Then we have ASbr(1,2)={1}. By the hypothesis g1=1, we obtain

    ASbr(1,2)Fix(g).

    Let r>0 and μASbr(1,2) such that μFix(g). Using the hypothesis, we get

    Sb(gμ,gμ,μ)<ψ(Sb(gμ,gμ,μ),Sb(μ,μ,1)Sb(μ,μ,2),Sb(μ,μ,g1)Sb(μ,μ,g2),Sb(gμ,gμ,μ),0,Sb(gμ,gμ,1)Sb(gμ,gμ,2))=ψ(Sb(gμ,gμ,μ),Sb(μ,μ,1)Sb(μ,μ,2),Sb(μ,μ,1)Sb(μ,μ,2),Sb(gμ,gμ,μ),0,Sb(gμ,gμ,1)Sb(gμ,gμ,2))=ψ(Sb(gμ,gμ,μ),r,r,Sb(gμ,gμ,μ),0,r).

    Since

    r2sSb(gμ,gμ,μ)+sr,

    using (ψ), there exists a k[0,1) such that

    Sb(gμ,gμ,μ)krkSb(gμ,gμ,μ)<Sb(gμ,gμ,μ),

    a contradiction. Hence it should be μFix(g). Consequently, we get

    ASbr(1,2)Fix(g).

    Now we give the following illustrative example of above proved geometric results.

    Example 3.1. Let us consider Example 2.2 given in [22]. Let X=[1,1]{7,2,2,73,7,8,21} and the S-metric defined as

    S(μ,τ,)=|μ|+|μ+2τ|,

    for all μ,τ,X [32]. Then the function S is also an Sb-metric with s=1. Let us define the function g:XX as

    gμ={7,μ=8μ,μX{8},

    for all μX and the function ψ:R6+R as

    ψ(t1,t2,t3,t4,t5,t6)=kt2,

    with k[0,1). Under these assumptions, we get

    r=inf{S(μ,μ,gμ):μFix(g)}=inf{S(μ,μ,gμ):μ=8}=2.

    If we take 0=0 and k=12, then the function g satisfies the conditions of Theorem 3.1. Therefore, we obtain

    DSb0,2=[1,1]Fix(g)=X{8}

    and

    CSb0,2={1,1}Fix(g)=X{8}.

    If we take 1=12, 2=12 and k=12, then the function g satisfies the conditions of Theorem 3.2. Therefore, we obtain

    ESb2(12,12)=[12,12]Fix(g)=X{8}.

    If we take 1=1, 2=1 and k=34, then the function g satisfies the conditions of Theorem 3.3. Therefore, we obtain

    HSb2(1,1)={12,12}Fix(g)=X{8}.

    If we take 1=1, 2=1 and k=34, then the function g satisfies the conditions of Theorem 3.4. Therefore, we obtain

    CSb2(1,1)={2,0,2}Fix(g)=X{8}.

    If we take 1=7, 2=7 and k=15, then the function g satisfies the conditions of Theorem 3.5. Therefore, we obtain

    ASb2(7,7)={73,21}Fix(g)=X{8}.

    Recently, activation functions have been used in applicable areas. Especially, these functions are used in neural network. For example, for state-of-the-art neural networks, rectified activation units are essential. Therefore, in this section, we focus on the parametric rectified linear unit (PReLU) activation functions [33]. PReLU is defined as

    PReLU(μ)={αμ,μ<0μ,μ0,

    where α is a coefficient.

    Let X=R+{2,1,0}, the S-metric defined as in Example 3.1 and the function ψ:R6+R defined as in Example 3.1. Let us consider α=0.8, then we obtain the function PReLU as

    PReLU(μ)={0.8μ,μ<0μ,μ0,

    for all μR. If we take 0=0 and k=12, then the function PReLU satisfies the conditions of Theorem 3.1. Indeed, for μ(,0), we get

    S(PReLU(μ),PReLU(μ),μ)=2|0.8μμ|=0.4|μ|<|μ|=2k|μ|=kS(μ,μ,0)=kψ(Sb(PReLUμ,PReLUμ,μ),Sb(μ,μ,0),Sb(μ,μ,PReLU0),Sb(PReLUμ,PReLUμ,μ),Sb(PReLU0,PReLU0,0),Sb(PReLUμ,PReLUμ,0)).

    Also, we obtain

    r=inf{S(μ,μ,PReLUμ):μFix(PReLU)}=inf{0.2|μ|:μ<0}=inf{0.2,0.4}=0.2.

    Consequently, we have

    DSb0,0.2=[0,0.1]Fix(PReLU)=R+{0}

    and similarly

    CSb0,0.2={0.1}Fix(PReLU)=R+{0}.

    Finally, we say that the parametric rectified linear unit (PReLU) activation function fixes the disc DSb0,0.2 and CSb0,0.2, that is, PReLU has at least two fixed figure. In this way, the learning capacity of the activation function PReLU increases.

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

    The authors declare no conflicts of interest.



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