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Research article

A note on three different contractions in partially ordered complex valued Gb-metric spaces

  • Received: 01 March 2022 Revised: 09 April 2022 Accepted: 19 April 2022 Published: 25 April 2022
  • MSC : 47H10, 54H25, 54D99, 54E99

  • In this paper, we introduce the complex valued Cp-class function, a type of Geraghty contraction and a type of JS contraction in complete partially ordered complex valued Gb-metric spaces, prove three fixed point theorems in this space, and also we give some examples to support our results.

    Citation: Yiquan Li, Chuanxi Zhu, Yingying Xiao, Li Zhou. A note on three different contractions in partially ordered complex valued Gb-metric spaces[J]. AIMS Mathematics, 2022, 7(7): 12322-12341. doi: 10.3934/math.2022684

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  • In this paper, we introduce the complex valued Cp-class function, a type of Geraghty contraction and a type of JS contraction in complete partially ordered complex valued Gb-metric spaces, prove three fixed point theorems in this space, and also we give some examples to support our results.



    Fixed point theory in metric spaces occupies an extremely important position in modern mathematics, it has been generalized in various aspects. For example, G-metric spaces [1] were introduced and Gb-metric spaces were reported in [2], which successfully popularized the general metric and promoted the research of various types of fixed point theorems. These theorems are accompanied with different contractive conditions (see [3,4,5,6,7,8,9,10,11,12,21,22,23,24,25,26,27]), especially the new Geraghty contraction was given in [13] and the JS contraction was given in [14].

    Recently, Shoaib et al. [15] introduced the ordered dislocated quasi G-metric spaces, and obtained some new fixed point results for a dominated mapping on a close ball in this space. On the other hand, Ege [16] also proposed the complex valued Gb-metric spaces as a new notion, the Banach contraction principle and Kannan's fixed point theorem were proved for this space. Moreover, there are also other interesting fixed point theorems in this space (see [17,18,19,20]).

    In this work, we study some problems about the common solutions of the operator equations Fnx=ux(u1,nN) in complete partially ordered complex valued Gb-metric spaces, introduce the complex valued Cp-class function and a type of Geraghty contraction to this space respectively, and we obtain the common solutions in a closed ball. Furthermore, we also introduce a type of JS contraction to this space and investigate a new theorem.

    Firstly, we recall some basic concepts, which will be used later. For a real Banach space E, a nonempty closed subset QE is called a cone, if

    (a) for all ζQ and τ0, τζQ;

    (b) for all ζ1,ζ2Q, ζ1+ζ2Q;

    (c) Q(Q)=0.

    For ξ1,ξ2E, given a cone Q, we define a partial order on E, which is induced by Q, i.e., ξ1ξ2 iff ξ2ξ1Q. Furthermore, ξ1,ξ2 are said to be comparable if ξ1ξ2 or ξ2ξ1.

    On the other hand, for all ξ1,ξ2C, the partial order on C is defined as follows:

    ξ1ξ2Re(ξ1)Re(ξ2)andIm(ξ1)Im(ξ2).

    Therefore, ξ1ξ2 if one of the following conditions holds:

    (C1) Re(ξ1)=Re(ξ2) and Im(ξ1)=Im(ξ2);

    (C2) Re(ξ1)=Re(ξ2) and Im(ξ1)<Im(ξ2);

    (C3) Re(ξ1)<Re(ξ2) and Im(ξ1)=Im(ξ2);

    (C4) Re(ξ1)<Re(ξ2) and Im(ξ1)<Im(ξ2).

    Moreover, we denote ξ1ξ2 if only (C4) holds. Obviously, 0ξ1ξ2|ξ1||ξ2|, where |ξi| is the magnitude of ξi, i=1,2. For more details, see [25].

    Definition 1.1. ([16]) Let X be a nonempty set, for a real number s1, if the mapping Gb:X×X×XC satisfies:

    (CGb1) Gb(ζ1,ζ2,ζ3)=0 if ζ1=ζ2=ζ3;

    (CGb2) Gb(ζ1,ζ1,ζ2)0 for all ζ1,ζ2X with ζ1ζ2;

    (CGb3) Gb(ζ1,ζ1,ζ2)Gb(ζ1,ζ2,ζ3) for all ζ1,ζ2,ζ3X with ζ3ζ2;

    (CGb4) Gb(ζ1,ζ2,ζ3)=Gb(R{ζ1,ζ2,ζ3}), where R is an arbitrary permutation of {ζ1,ζ2,ζ3};

    (CGb5) Gb(ζ1,ζ2,ζ3)s[Gb(ζ1,υ,υ)+Gb(υ,ζ2,ζ3)] for all ζ1,ζ2,ζ3,υX.

    Then the function Gb is called a complex valued Gb-metric on X, the pair (X,Gb) is called a complex valued Gb-metric space.

    Proposition 1.1. ([16]) For a complex valued Gb-metric space (X,Gb) and all ζ1,ζ2,ζ3X, we have

    (1) Gb(ζ1,ζ2,ζ3)s[Gb(ζ1,ζ1,ζ2)+Gb(ζ1,ζ1,ζ3)];

    (2) Gb(ζ1,ζ2,ζ2)2s[Gb(ζ1,ζ1,ζ2)].

    Definition 1.2. ([16]) Let {xn} be a sequence in a complex valued Gb-metric space (X,Gb),

    (1) {xn} is called complex valued Gb-convergent to ζX, if for any ϵC with ϵ0, there exists ξN such that Gb(ζ,xn,xm)ϵ for all n,mξ. We write xnζ as n, or limnxn=ζ;

    (2) {xn} is called complex valued Gb-Cauchy, if for any ϵC with ϵ0, there exists ξN such that Gb(xn,xm,xl)ϵ for all n,m,lξ;

    (3) (X,Gb) is said to be complex valued Gb-complete, if any complex valued Gb-Cauchy sequence {xn} is complex valued Gb-convergent.

    Theorem 1.1. ([16]) Let {xn} be a sequence in a complex valued Gb-metric space (X,Gb), and ζX, the following are equivalent:

    (1) {xn} is complex valued Gb-convergent to ζ;

    (2) |Gb(xn,xm,ζ)|0 as n,m;

    (3) |Gb(xn,ζ,ζ)|0 as n;

    (4) |Gb(xn,xn,ζ)|0 as n.

    Theorem 1.2. ([16]) A sequence {xn} is complex valued Gb-Cauchy sequence is equivalent to |Gb(xn,xm,xl)|0 as n,m,l.

    Definition 1.3. ([28]) Let QRm be a cone, a mapping S:QRm is said to be dominated if Sxx for all xQ.

    Theorem 1.3. ([15]) Let (X,,G) be an ordered complete dislocated quasi G-metric space, S:XX be a mapping and x0 be an arbitrary point in X. Suppose there exists k[0,12) with

    G(Sx,Sy,Sz)k(G(x,Sx,Sx)+G(y,Sy,Sy)+G(z,Sz,Sz))

    for all comparable elements x,y,z¯B(x0,r), and

    G(x0,Sx0,Sx0)(1θ)r,

    where θ=k12k. If for nonincreasing sequence {xn}u implies that uxn. Then there exists a point x in ¯B(x0,r) such that x=Sx and G(x,x,x)=0. Moreover, if for any three points x,y,z¯B(x0,r), there exists a point v in ¯B(x0,r) such that vx and vy, vz, where

    G(x0,Sx0,Sx0)+G(v,Sv,Sv)+G(v,Sv,Sv)G(x0,v,v)+G(Sx0,Sv,Sv)+G(Sx0,Sv,Sv),

    then the point x is unique.

    In this section, let X=RN, Ω1={Z1C:0Z1}, Ω2={Z2C:0Z2}, Ω3={Z3C:1Z3}. (X,Gb,) is called a partially ordered complex valued Gb-metric space, which shows (X,Gb) is a complex valued Gb-metric space and (X,) is a partially ordered set.

    Let (X,Gb) be a complex valued Gb-metric space, for any x0X, rC and r0, the Gb-ball with ball center x0 is ¯B(x0,r)={xX|Gb(x0,x,x)r}. Moreover, for all nN and x1,x2,...,xnC, the function max{x1,x2,...,xn}xj, j=1,2,...,n.

    Definition 2.1. A continuous mapping P:Ω31C is called complex valued Cp-class function, if it satisfies rP(r,s,t) for all r,s,tΩ1.

    Example 2.1. Some examples of complex valued Cp-class function are given as follows:

    (1) P(r,s,t)=r+s+t, where r,s,tΩ1;

    (2) P(r,s,t)=mr, where m[1,) and r,s,tΩ1;

    (3) P(r,s,t)=η(r)r, where η:Ω1[1,) and r,s,tΩ1.

    Theorem 2.1. Let (X,Gb,) be a complete partially ordered complex valued Gb-metric space with s1, QX be a cone, x0 be an arbitrary element in Q, {Sn:XX,nN} be a dominated mapping sequence. If there exist rΩ2, and nonnegative numbers α,β,γ satisfy α2sγ0,βα2γ[0,δ],δ<1s, such that

    P[ψ(αGb(Six,Sjy,Sjy)),φ(αGb(Six,Sjy,Sjy)),φ(αGb(Six,Sjy,Sjy))]ψ[βGb(x,Six,Six)+γGb(y,Sjy,Sjz)+γGb(z,Sjz,Sjy)] (2.1)

    for any comparable elements x,y,z in ¯B(x0,r), where ¯B(x0,r)Q,i,jN, P is a complex valued Cp-class function, ψ:Ω1Ω1 is a nondecreasing function, φ:Ω1C is a continuous function. And

    Gb(x0,S1x0,S1x0)1sδsr. (2.2)

    Define the operator equations Fnx=ux by Fn=uSn, u1. If a nonincreasing sequence {xn}κ such that κxn, then the operator equations have at least a common solution x in ¯B(x0,r). Moreover, if there exists an element v in ¯B(x0,r) such that vx, and

    βGb(x0,S1x0,S1x0)+2γGb(v,Sjv,Sjv)βGb(x0,v,v)+2γGb(S1x0,Sjv,Sjv), (2.3)

    then the operator equations have an unique solution.

    Proof. By selecting the ball centre x0 in ¯B(x0,r), we construct a sequence {xn}, where xn+1=Sn+1xnxn,nN. From (2.2), we obtain x1¯B(x0,r). Using (2.1), we have

    ψ(αGb(S1x0,S2x1,S2x1))P[ψ(αGb(S1x0,S2x1,S2x1)),φ(αGb(S1x0,S2x1,S2x1)),φ(αGb(S1x0,S2x1,S2x1))]ψ[βGb(x0,S1x0,S1x0)+2γGb(x1,S2x1,S2x1)].

    Since the function ψ is nondecreasing, we can easily get

    Gb(x1,x2,x2)βα2γGb(x0,x1,x1)δGb(x0,x1,x1).

    Hence, Gb(x0,x2,x2)s[Gb(x0,x1,x1)+Gb(x1,x2,x2)]s(1+δ)Gb(x0,x1,x1). Using (2.2), we get Gb(x0,x2,x2)(1δ2)rr, that is x2¯B(x0,r).

    Now we prove {xn}¯B(x0,r). Suppose that x3,x4,...,xk¯B(x0,r), according to (2.1), we have

    ψ(αGb(Skxk1,Sk+1xk,Sk+1xk))P[ψ(αGb(Skxk1,Sk+1xk,Sk+1xk)),φ(αGb(Skxk1,Sk+1xk,Sk+1xk)),φ(αGb(Skxk1,Sk+1xk,Sk+1xk))]ψ[βGb(xk1,Skxk1,Skxk1)+2γGb(xk,Sk+1xk,Sk+1xk)].

    Thus Gb(xk,xk+1,xk+1)βα2γGb(xk1,xk,xk)δGb(xk1,xk,xk), it can easily get that

    Gb(xk,xk+1,xk+1)δkGb(x0,x1,x1). (2.4)

    By using (CGb5) and (2.4), it follows that

    Gb(x0,xk+1,xk+1)sGb(x0,x1,x1)+s2Gb(x1,x2,x2)+...+sk+1Gb(xk,xk+1,xk+1)(s+s2δ+...+sk+1δk)Gb(x0,x1,x1)s11sδ1sδsr=r,

    i.e., xk+1¯B(x0,r), therefore, {xn}¯B(x0,r).

    Now we show that {xn} is a complex valued Gb-Cauchy sequence, from (2.4), we obtain

    Gb(xn,xn+1,xn+1)δnGb(x0,x1,x1), (2.5)

    thus for all n,mN,n<m, we have

    Gb(xn,xm,xm)sGb(xn,xn+1,xn+1)+s2Gb(xn+1,xn+2,xn+2)+...+smnGb(xm1,xm,xm)(sδn+s2δn+1+...+smnδm1)Gb(x0,x1,x1)sδn11sδGb(x0,x1,x1),

    which implies that

    limn,mGb(xn,xm,xm)=0.

    Therefore, {xn} is a complex valued Gb-Cauchy sequence, and there exists an element x in ¯B(x0,r) such that xnx.

    Next we prove x is the common solution of the operator equations. For any jN, we have

    Gb(x,Sjx,Sjx)s[Gb(x,xn,xn)+Gb(xn,Sjx,Sjx)].

    Furthermore, since Sjxxxnxn1, using (2.1), it can be easily get that

    αGb(xn,Sjx,Sjx)βGb(xn1,xn,xn)+2γGb(x,Sjx,Sjx).

    Hence,

    αGb(x,Sjx,Sjx)sαGb(x,xn,xn)+sαGb(xn,Sjx,Sjx)sαGb(x,xn,xn)+sβGb(xn1,xn,xn)+2sγGb(x,Sjx,Sjx).

    That is,

    Gb(x,Sjx,Sjx)1α2sγ[sαGb(x,xn,xn)+sβGb(xn1,xn,xn)].

    Let n at both sides of the above inequality, we obtain limnGb(x,Sjx,Sjx)=0, i.e. x=Sjx. According to the arbitrariness of j, we get x is a common solution of the operator equations.

    Uniqueness. Assume that y is another solution of the operator equations, yx and y¯B(x0,r).

    Case 1. If x and y are comparable, using (2.1), it follows that

    αGb(x,y,y)=αGb(Six,Sjy,Sjy)βGb(x,Six,Six)+2γGb(y,Sjy,Sjy)=βGb(x,x,x)+2γGb(y,y,y)=0,

    as a result, x=y.

    Case 2. If x and y are not comparable, then there exists an element v¯B(x0,r) such that vx and vy, for any jN, we will show {Snjxn}¯B(x0,r). Owing to (2.1) and (2.3), we have

    αGb(S1x0,Sjv,Sjv)βGb(x0,S1x0,S1x0)+2γGb(v,Sjv,Sjv)βGb(x0,v,v)+2γGb(S1x0,Sjv,Sjv),

    i.e.,

    Gb(S1x0,Sjv,Sjv)βα2γGb(x0,v,v)δr.

    Hence,

    Gb(x0,Sjv,Sjv)s[Gb(x0,x1,x1)+Gb(x1,Sjv,Sjv)]s(1sδsr+δr)=r,

    that is Sjv¯B(x0,r). Suppose that S2jv,S3jv,...,Skjv¯B(x0,r), obviously, SkjvSk1jv...S2jvSjvvxxn...x0. From (2.1), we can immediately obtain

    αGb(Skjv,Sk+1jv,Sk+1jv)βGb(Sk1jv,Skjv,Skjv)+2γGb(Skjv,Sk+1jv,Sk+1jv),

    so we have

    Gb(Skjv,Sk+1jv,Sk+1jv)βα2γGb(Sk1jv,Skjv,Skjv)δGb(Sk1jv,Skjv,Skjv),

    as a result,

    Gb(Skjv,Sk+1jv,Sk+1jv)δGb(Sk1jv,Skjv,Skjv)...δkGb(v,Sjv,Sjv). (2.6)

    In addition, using (2.1), (2.3), (2.5) and (2.6), we can also immediately obtain

    αGb(xk+1,Sk+1jv,Sk+1jv)βGb(xk,xk+1,xk+1)+2γGb(Skjv,Sk+1jv,Sk+1jv)βδkGb(x0,x1,x1)+2γδkGb(v,Sjv,Sjv)βδkGb(x0,v,v)+2γδkGb(S1x0,Sjv,Sjv)βδkGb(x0,v,v)+2γδkβα2γGb(x0,v,v)(βδk+2γδk+1)Gb(x0,v,v),

    i.e.,

    Gb(xk+1,Sk+1jv,Sk+1jv)(βδk+2γδk+1)αGb(x0,v,v)(α2γ)δk+1+2γδk+1αGb(x0,v,v)=δk+1Gb(x0,v,v).

    Thus,

    Gb(x0,Sk+1jv,Sk+1jv)sGb(x0,x1,x1)+...+sk+1Gb(xk,xk+1,xk+1)+sk+1Gb(xk+1,Sk+1jv,Sk+1jv)(s+s2δ+...+sk+1δk)Gb(x0,x1,x1)+sk+1δk+1Gb(x0,v,v)s1(sδ)k+11sδ1sδsr+(sδ)k+1r=[1(sδ)k+1+(sδ)k+1]r=r,

    which implies Sk+1jv¯B(x0,r), so {Snjxn}¯B(x0,r). From (2.6), we obtain

    Gb(Snjv,Sn+1jv,Sn+1jv)δnGb(v,Sjv,Sjv),

    and

    limnGb(Snjv,Sn+1jv,Sn+1jv)=0. (2.7)

    From (2.1), we can easily get

    αGb(x,Snjv,Snjv)=αGb(Six,Snjv,Snjv)βGb(x,Six,Six)+2γGb(Sn1jv,Snjv,Snjv)=2γGb(Sn1jv,Snjv,Snjv).

    Owing to (2.7), we have

    limnGb(x,Snjv,Snjv)=0. (2.8)

    Similarly,

    αGb(Snjv,y,y)=αGb(Snjv,Siy,Siy)βGb(Sn1jv,Snjv,Snjv)+2γGb(y,Siy,Siy)=βGb(Sn1jv,Snjv,Snjv).

    According to (2.7), we also have

    limnGb(Snjv,y,y)=0. (2.9)

    Since Gb(x,y,y)s[Gb(x,Snjv,Snjv)+Gb(Snjv,y,y)], using (2.8) and (2.9), we obtain

    Gb(x,y,y)=limnGb(x,y,y)0.

    Therefore, x=y, the proof is completed.

    Following the proof process of Theorem 2.1, we can obtain the following corollary.

    Corollary 2.1. Let (X,Gb,) be a complete partially ordered complex valued Gb-metric space with s1, QX be a cone, {Sn:XQ,nN} be a dominated mapping sequence. If there exist nonnegative numbers α,β,γ satisfy α2sγ0,βα2γ[0,1s), such that

    η(ψ(αGb(Six,Sjy,Sjy)))ψ(αGb(Six,Sjy,Sjy))ψ[βGb(x,Six,Six)+γGb(y,Sjy,Sjz)+γGb(z,Sjz,Sjy)]

    for any comparable elements x,y,z in Q, where i,jN, η:Ω1[1,), ψ:Ω1Ω1 is a nondecreasing function.

    Define the operator equations Fnx=ux by Fn=uSn, u1. If a nonincreasing sequence {xn}κ such that κxn, then the operator equations have at least a common solution x in Q. Moreover, if there exists an element v in Q such that vx, then the operator equations have an unique solution.

    Example 2.2. Let X=R, Q=[0,), α=5,β=γ=1,δ=13, Gb:X×X×XC be defined by Gb(ξ1,ξ2,ξ3)=max{|ξ1ξ2|2,|ξ2ξ3|2,|ξ1ξ3|2}+max{|ξ1ξ2|2,|ξ2ξ3|2,|ξ1ξ3|2}i with s=2, and ψ(r)=η(r)r=r for any r in Ω1.

    For any ξ in X, 0<νn14 and nN, take Snξ=νnξ and Fn=uSn, where u1. The partial order on X is the usual order of R, for any ξ1,ξ2,ξ3 in Q, we have

    αGb(Snξ1,Snξ2,Snξ2)=5ν2n(ξ1ξ2)2+5ν2n(ξ1ξ2)2i,

    and

    β|ξ1νnξ1|2+γ|ξ2νnξ2|2+γ|ξ3νnξ3|2=(1νn)2(ξ21+ξ22+ξ23).

    Hence,

    αGb(Snξ1,Snξ2,Snξ2)β|ξ1νnξ1|2+γ|ξ2νnξ2|2+γ|ξ3νnξ3|2+[β|ξ1νnξ1|2+γ|ξ2νnξ2|2+γ|ξ3νnξ3|2]iβGb(ξ1,Snξ1,Snξ1)+γGb(ξ2,Snξ2,Snξ3)+γGb(ξ3,Snξ3,Snξ2).

    It follows that the operator equations Fnξ=uξ have a common solution ξ=0 in Q, and there exists an element v=0 in Q such that vξ. Therefore, all conditions of Corollary 2.1 are satisfied, the operator equations Fnξ=uξ have an unique solution ξ=0.

    Let B be the set of functions β:Ω1[0,1s), which satisfies if limnβ(xn)=1s, then limnxn=0.

    Theorem 2.2. Let (X,Gb,) be a complete partially ordered complex valued Gb-metric space with s1, QX be a cone, x0 be an arbitrary element in Q, {Sn:XX,nN} be a dominated mapping sequence. Suppose that there exist βB, i,jN and rΩ2, such that

    Gb(Six,Sjy,Sjz)β(M(x,y,z))M(x,y,z) (2.10)

    for any comparable elements x,y,z in ¯B(x0,r), where ¯B(x0,r)Q,

    M(x,y,z)=max{Gb(x,y,z),Gb(x,Six,Six)Gb(y,Sjy,Sjz)1+Gb(x,y,z),Gb(x,Six,Six)Gb(x,Sjy,Sjz)1+s[Gb(x,y,z)+Gb(Six,Sjy,Sjz)]}, (2.11)

    and

    Gb(x0,S1x0,S1x0)1sδsr, (2.12)

    where δ(0,1s).

    Define the operator equations Fnx=ux by Fn=uSn, u1. If a nonincreasing sequence {xn}κ such that κxn, then the operator equations have at least a common solution x in ¯B(x0,r).

    Proof. By selecting the ball centre x0 in ¯B(x0,r), we construct a sequence {xn}, where xn+1=Sn+1xnxn,nN. From (2.12), we know x1¯B(x0,r). Using (2.10), we have

    Gb(x1,x2,x2)=Gb(S1x0,S2x1,S2x1)β(M(x0,x1,x1))M(x0,x1,x1), (2.13)

    where

    M(x0,x1,x1)=max{Gb(x0,x1,x1),Gb(x0,S1x0,S1x0)Gb(x1,S2x1,S2x1)1+Gb(x0,x1,x1),Gb(x0,S1x0,S1x0)Gb(x0,S2x1,S2x1)1+s[Gb(x0,x1,x1)+Gb(S1x0,S2x1,S2x1)]}.

    Since

    Gb(x0,S1x0,S1x0)Gb(x1,S2x1,S2x1)1+Gb(x0,x1,x1)Gb(x1,S2x1,S2x1)=Gb(x1,x2,x2),

    and

    Gb(x0,S1x0,S1x0)Gb(x0,S2x1,S2x1)1+s[Gb(x0,x1,x1)+Gb(S1x0,S2x1,S2x1)]s[Gb(x0,x1,x1)+Gb(x1,S2x1,S2x1)]Gb(x0,S1x0,S1x0)1+s[Gb(x0,x1,x1)+Gb(S1x0,S2x1,S2x1)]Gb(x0,S1x0,S1x0)=Gb(x0,x1,x1),

    thus M(x0,x1,x1)max{Gb(x0,x1,x1),Gb(x1,x2,x2)}.

    If max{Gb(x0,x1,x1),Gb(x1,x2,x2)}=Gb(x1,x2,x2), then we have

    Gb(x1,x2,x2)β(M(x0,x1,x1))M(x0,x1,x1)1sGb(x1,x2,x2),

    which is a contradiction, thus

    max{Gb(x0,x1,x1),Gb(x1,x2,x2)}=Gb(x0,x1,x1),

    and

    Gb(x1,x2,x2)β(M(x0,x1,x1))M(x0,x1,x1)δGb(x0,x1,x1).

    So we have

    Gb(x0,x2,x2)s[Gb(x0,x1,x1)+Gb(x1,x2,x2)]s(1+δ)1sδsr(1δ2)rr,

    as a result, x2¯B(x0,r).

    Now we will show {xn}¯B(x0,r). Assume that x3,x4,...,xk¯B(x0,r), owing to (2.10), we get

    Gb(xk,xk+1,xk+1)=Gb(Skxk1,Sk+1xk,Sk+1xk)β(M(xk1,xk,xk))M(xk1,xk,xk).

    Following the above proof process, we can obtain

    M(xk1,xk,xk)max{Gb(xk1,xk,xk),Gb(xk,xk+1,xk+1)}=Gb(xk1,xk,xk). (2.14)

    Thus,

    Gb(xk,xk+1,xk+1)δGb(xk1,xk,xk)δ2Gb(xk2,xk1,xk1)...δkGb(x0,x1,x1). (2.15)

    By using (CGb5) and (2.15), it follows that

    Gb(x0,xk+1,xk+1)sGb(x0,x1,x1)+s2Gb(x1,x2,x2)+...+sk+1Gb(xk,xk+1,xk+1)(s+s2δ+...+sk+1δk)Gb(x0,x1,x1)s1(sδ)k+11sδ1sδsrr.

    Hence, xk+1¯B(x0,r), so {xn}¯B(x0,r). As a result, for all nN,

    Gb(xn,xn+1,xn+1)=Gb(Snxn1,Sn+1xn,Sn+1xn)β(M(xn1,xn,xn))M(xn1,xn,xn), (2.16)

    thus we have Gb(xn,xn+1,xn+1)1sGb(xn1,xn,xn).

    If s>1, then Gb(xn,xn+1,xn+1)(1s)nGb(x0,x1,x1)0 as n.

    If s=1, then Gb(xn,xn+1,xn+1)Gb(xn1,xn,xn), which implies that {Gb(xn,xn+1,xn+1)} is a decreasing sequence.

    Suppose that

    limnGb(xn,xn+1,xn+1)=r0,

    owing to (2.14) and (2.16), we obtain

    r=limnGb(xn,xn+1,xn+1)limnβ(M(xn1,xn,xn))M(xn1,xn,xn)limn1sGb(xn1,xn,xn)r,

    thus limnβ(M(xn1,xn,xn))=1, which implies limnGb(xn1,xn,xn)=0, contradiction. As a result, limnGb(xn,xn+1,xn+1)=0.

    Now we prove {xn} is a complex valued Gb-Cauchy sequence. Suppose that contrary, then there exist ϵ0 and two subsequences xmk and xnk of xn, such that

    Gb(xmk,xnk,xnk)ϵandGb(xmk,xnk1,xnk1)ϵ.

    So we have

    ϵGb(xmk,xnk,xnk)s[Gb(xmk,xmk+1,xmk+1)+Gb(xmk+1,xnk,xnk)].

    Let k, we get

    ϵlimkinfGb(xmk,xnk,xnk)slimkinfGb(xmk+1,xnk,xnk).

    Furthermore, using (2.10) and (2.14),

    limkinfGb(xmk+1,xnk,xnk)limkinfβ(M(xmk,xnk1,xnk1))M(xmk,xnk1,xnk1)limkinfβ(M(xmk,xnk1,xnk1))Gb(xmk,xnk1,xnk1)limkinfβ(M(xmk,xnk1,xnk1))ϵ,

    thus we have

    ϵslimkinfGb(xmk+1,xnk,xnk)limkinfβ(M(xmk,xnk1,xnk1))ϵlimksupβ(M(xmk,xnk1,xnk1))ϵϵs.

    Therefore, limkβ(M(xmk,xnk1,xnk1))=1s, thus limkGb(xmk,xnk1,xnk1)=0. As a result,

    ϵGb(xmk,xnk,xnk)s[Gb(xmk,xnk1,xnk1)+Gb(xnk1,xnk,xnk)]0ask,

    which is a contradiction. Therefore, {xn} is a complex valued Gb-Cauchy sequence, and there exists an element x in ¯B(x0,r) such that xnx.

    Finally, we show that x is a common solution of the operator equations. Let x=xi1,y=z=x in (2.10), we have

    limiGb(Sixi1,Sjx,Sjx)limiβ(M(xi1,x,x))M(xi1,x,x)limi1sM(xi1,x,x),

    where

    M(xi1,x,x)=max{Gb(xi1,x,x),Gb(xi1,Sixi1,Sixi1)Gb(x,Sjx,Sjx)1+Gb(xi1,x,x),Gb(xi1,Sixi1,Sixi1)Gb(xi1,Sjx,Sjx)1+s[Gb(xi1,x,x)+Gb(Sixi1,Sjx,Sjx)]}.

    It can be easily deduced that limiM(xi1,x,x)=0 and limiGb(Sixi1,Sjx,Sjx)=0, thus

    Gb(x,Sjx,Sjx)s[Gb(x,Sixi1,Sixi1)+Gb(Sixi1,Sjx,Sjx)]0asi.

    As a result, x=Sjx, owing to the arbitrariness of j, we obtain that x is a common solution of the operator equations, the proof is completed.

    Similarly, following the proof process of Theorem 2.2, the following corollary will be established.

    Corollary 2.2. Let (X,Gb,) be a complete partially ordered complex valued Gb-metric space with s1, QX be a cone, {Sn:XQ,nN} be a dominated mapping sequence. Suppose that there exist i,jN such that

    Gb(Six,Sjy,Sjz)λM(x,y,z)

    for any comparable elements x,y,z in Q, where λ[0,1s), and

    M(x,y,z)=max{Gb(x,y,z),Gb(x,Six,Six)Gb(y,Sjy,Sjz)1+Gb(x,y,z),Gb(x,Six,Six)Gb(x,Sjy,Sjz)1+s[Gb(x,y,z)+Gb(Six,Sjy,Sjz)]}.

    Define the operator equations Fnx=ux by Fn=uSn, u1. If a nonincreasing sequence {xn}κ such that κxn, then the operator equations have at least a common solution x in Q.

    Example 2.3. Let X=R, Q=[0,), Gb:X×X×XC be defined by Gb(ξ1,ξ2,ξ3)=(|ξ1ξ2|+|ξ2ξ3|+|ξ1ξ3|)2+(|ξ1ξ2|+|ξ2ξ3|+|ξ1ξ3|)2i with s=2, δ=15, x0=1, r=4+4i. For all tΩ1, take

    β(t)={13,t=0;12+|t|2,0<|t|1;152+12+e|t|,|t|>1.

    Obviously, 13β(t)<12, and

    ¯B(1,4+4i)={x|Gb(1,x,x)4+4i}={x|4|1x|2+4|1x|2i4+4i}=[0,2].

    Moreover, for any ξ in X, let Snξ=|ξ|3n,nN and Fn=uSn, where u1. The partial order on X is the usual order of R, for any ξ1,ξ2,ξ3 in ¯B(1,4+4i), we have

    Gb(Snξ1,Snξ2,Snξ3)=13n2[(|ξ1ξ2|+|ξ2ξ3|+|ξ1ξ3|)2+(|ξ1ξ2|+|ξ2ξ3|+|ξ1ξ3|)2i],

    and

    Gb(ξ1,ξ2,ξ3)=(|ξ1ξ2|+|ξ2ξ3|+|ξ1ξ3|)2+(|ξ1ξ2|+|ξ2ξ3|+|ξ1ξ3|)2i.

    It follows that

    Gb(Snξ1,Snξ2,Snξ3)13Gb(ξ1,ξ2,ξ3)13M(ξ1,ξ2,ξ3)β(M(ξ1,ξ2,ξ3))M(ξ1,ξ2,ξ3),

    and

    Gb(1,33,33)=16833+16833i310(4+4i).

    It is clearly that all conditions of Theorem 2.2 are satisfied, as a result, the operator equations Fnξ=uξ have a common solution ξ=0 in ¯B(1,4+4i).

    On the other hand, let Θ be the set of functions θ:Ω2Ω3, which satisfies the following conditions:

    Θ1: θ is continuous;

    Θ2: θ is nondecreasing, i.e. θ(x1)θ(x2) if x1x2;

    Θ3: limnθ(xn)=1limnxn=0+, where {xn}Ω2.

    Theorem 2.3. Let (X,Gb,) be a complete partially ordered complex valued Gb-metric space with s1, QX be a cone, {Sn:XQ,nN} be a dominated mapping sequence. Suppose that there exist θΘ, i,jN,k(0,1),α0 such that

    |θ(Gb(Six,Sjy,Sjz))||θ(1sM(x,y,z)α)|k (2.17)

    for any comparable elements x,y,z in Q, where Gb(Six,Sjy,Sjz)0, and

    M(x,y,z)=max{Gb(x,Six,Six),Gb(y,Sjy,Sjz),Gb(z,Sjz,Sjy),Gb(x,y,z)}. (2.18)

    Define the operator equations Fnx=ux by Fn=uSn, u1. If a nonincreasing sequence {xn}κ such that κxn, then the operator equations have at least a common solution x in Q. Moreover, if there exists an element v in Q such that vx, and

    Gb(Sn1jv,Snjv,Snjv)Gb(x,Sn1jv,Sn1jv), (2.19)

    then the operator equations have an unique solution.

    Proof. By selecting a point x0 in Q, we construct a sequence {xn}, where xn+1=Sn+1xnxn,nN. Let x=xn1,y=z=xn in (2.17), we have

    |θ(1sGb(Snxn1,Sn+1xn,Sn+1xn))||θ(Gb(Snxn1,Sn+1xn,Sn+1xn))||θ(1sM(xn1,xn,xn)α)|k|θ(1sM(xn1,xn,xn))|k,

    where

    M(xn1,xn,xn)=max{Gb(xn1,Snxn1,Snxn1),Gb(xn,Sn+1xn,Sn+1xn),Gb(xn1,xn,xn)}=max{Gb(xn1,xn,xn),Gb(xn,xn+1,xn+1)},

    thus we get

    |θ(1sGb(xn,xn+1,xn+1))||θ(1smax{Gb(xn1,xn,xn),Gb(xn,xn+1,xn+1)})|k.

    If max{Gb(xn1,xn,xn),Gb(xn,xn+1,xn+1)}=Gb(xn,xn+1,xn+1), then

    |θ(1sGb(xn,xn+1,xn+1))||θ(1sGb(xn,xn+1,xn+1))|k,which is contradiction withk(0,1),

    hence,

    |θ(1sGb(xn,xn+1,xn+1))||θ(Gb(xn,xn+1,xn+1))||θ(1sGb(xn1,xn,xn))|k.

    It follows that

    |θ(1sGb(xn,xn+1,xn+1))||θ(1sGb(xn1,xn,xn))|k...|θ(1sGb(x0,x1,x1))|kn,

    and

    limn|θ(1sGb(xn,xn+1,xn+1))|limn|θ(1sGb(x0,x1,x1))|kn=1,

    therefore,

    limnGb(xn,xn+1,xn+1)=0andlimnGb(xn,xn,xn+1)=0.

    Now we show {xn} is a complex valued Gb-Cauchy sequence. If not, then there exist ϵ0 and two subsequences xmi and xni of xn, where inimi, such that

    Gb(xni,xni,xmi)ϵandGb(xni,xni,xmi1)ϵ.

    Using (CGb5), we have

    ϵGb(xni,xni,xmi)s[Gb(xni,xni,xni+1)+Gb(xni+1,xni+1,xmi)],

    let i at the above inequality, we get

    ϵslimiGb(xmi,xni+1,xni+1). (2.20)

    In addition, owing to (2.17), we obtain

    |θ(Gb(Smixmi1,Sni+1xni,Sni+1xni))||θ(1sM(xmi1,xni,xni)α)|k,

    i.e.,

    |θ(1sGb(xmi,xni+1,xni+1))||θ(Gb(xmi,xni+1,xni+1))||θ(1sM(xmi1,xni,xni)α)|k|θ(1sM(xmi1,xni,xni))|k,

    where

    M(xmi1,xni,xni)=max{Gb(xmi1,xmi,xmi),Gb(xni,xni+1,xni+1),Gb(xni,xni,xmi1)}.

    Since

    limiGb(xmi1,xmi,xmi)=limiGb(xni,xni+1,xni+1)=0,

    obviously, M(xmi1,xni,xni)=Gb(xni,xni,xmi1), it follows that

    |θ(Gb(xmi,xni+1,xni+1))||θ(1sGb(xni,xni,xmi1))|k. (2.21)

    Using (2.20) and (2.21), we have

    |θ(ϵs)|limi|θ(Gb(xmi,xni+1,xni+1))|limi|θ(1sGb(xni,xni,xmi1))|k<|θ(ϵs)|k,

    which is a contradiction with k(0,1). As a result, {xn} is a complex valued Gb-Cauchy sequence, and there exists an element x in Q such that xnx.

    Now we prove that x is a common solution of the operator equations. For all i,jN, we have

    Gb(x,Sjx,Sjx)s[Gb(x,xi,xi)+Gb(xi,Sjx,Sjx)],

    and let i at the above inequality, we get

    Gb(x,Sjx,Sjx)limisGb(xi,Sjx,Sjx). (2.22)

    In addition, since xxi1, according to (2.17), we obtain

    |θ(Gb(Sixi1,Sjx,Sjx))||θ(1sM(xi1,x,x)α)|k|θ(1sM(xi1,x,x))|k,

    where

    M(xi1,x,x)=max{Gb(xi1,xi,xi),Gb(x,Sjx,Sjx),Gb(xi1,x,x)}.

    If M(xi1,x,x)=Gb(x,Sjx,Sjx), using (2.22), it follows that

    limi|θ(Gb(xi,Sjx,Sjx))|limi|θ(1sGb(x,Sjx,Sjx))|klimi|θ(Gb(xi,Sjx,Sjx))|k,

    contradiction, thus we can easily get

    |θ(Gb(xi,Sjx,Sjx))||θ(1sGb(xi1,xi,xi))|k1asi,

    or

    |θ(Gb(xi,Sjx,Sjx))||θ(1sGb(xi1,x,x))|k1asi,

    hence,

    limiGb(xi,Sjx,Sjx)=0.

    From (2.22), we have

    Gb(x,Sjx,Sjx)limisGb(xi,Sjx,Sjx)=0.

    As a result, x=Sjx, owing to the arbitrariness of j, we obtain x is a common solution of the operator equations.

    Uniqueness. If y is another solution of the operator equations, yx, then Gb(x,y,y)0.

    Case 1. x and y are comparable, using (2.17), it follows that

    |θ(Gb(x,y,y))|=|θ(Gb(Six,Sjy,Sjy))||θ(1sM(x,y,y)α)|k|θ(1sM(x,y,y))|k.

    Obviously, M(x,y,y)=Gb(x,y,y), so we have

    |θ(Gb(x,y,y))||θ(1sGb(x,y,y))|k,

    which is a contradiction. As a result, y=x.

    Case 2. x and y are not comparable, then there exists an element vQ such that vx and vy, for any i,jN, we have

    x=Six=S2ix=...=Snix,y=Sjy=S2jy=...=Snjy,

    and

    Snjv...Sjvvx,Snjv...Sjvvy.

    From (2.17), we get

    |θ(Gb(Snix,Snjv,Snjv))||θ(1sM(Sn1ix,Sn1jv,Sn1jv)α)|k|θ(1sM(Sn1ix,Sn1jv,Sn1jv))|k,

    where

    M(Sn1ix,Sn1jv,Sn1jv)=max{Gb(Sn1ix,Snix,Snix),Gb(Sn1jv,Snjv,Snjv),Gb(Sn1ix,Sn1jv,Sn1jv)}.

    According to (2.19), we obtain M(Sn1ix,Sn1jv,Sn1jv)=Gb(Sn1ix,Sn1jv,Sn1jv), and

    |θ(Gb(Snix,Snjv,Snjv))||θ(1sGb(Sn1ix,Sn1jv,Sn1jv))|k|θ(Gb(Sn1ix,Sn1jv,Sn1jv))|k,

    so that we have

    |θ(Gb(Snix,Snjv,Snjv))||θ(Gb(Sn1ix,Sn1jv,Sn1jv))|k...|θ(Gb(x,v,v))|kn.

    It follows that

    limn|θ(Gb(Snix,Snjv,Snjv))|limn|θ(Gb(x,v,v))|kn=1,

    hence,

    limnGb(Snix,Snjv,Snjv)=0. (2.23)

    Similarly, using (2.17) and (2.19), we get

    |θ(Gb(Snjy,Snjv,Snjv))||θ(1sM(Sn1jy,Sn1jv,Sn1jv)α)|k|θ(M(Sn1jy,Sn1jv,Sn1jv))|k,

    where

    M(Sn1jy,Sn1jv,Sn1jv))=max{Gb(Sn1jy,Snjy,Snjy),Gb(Sn1jv,Snjv,Snjv),Gb(Sn1jy,Sn1jv,Sn1jv)}=max{0,Gb(Sn1jv,Snjv,Snjv),Gb(Sn1jy,Sn1jv,Sn1jv)}=Gb(Sn1jy,Sn1jv,Sn1jv).

    Therefore,

    |θ(Gb(Snjy,Snjv,Snjv))||θ(Gb(Sn1jy,Sn1jv,Sn1jv))|k...|θ(Gb(y,v,v))|kn,

    let n, we have

    limn|θ(Gb(Snjy,Snjv,Snjv))|limn|θ(Gb(y,v,v))|kn=1,

    so we obtain

    limnGb(Snjy,Snjv,Snjv)=0,

    and also

    limnGb(Snjv,Snjy,Snjy)=0. (2.24)

    Using (2.23) and (2.24), we also have

    Gb(Snix,Snjy,Snjy)s[Gb(Snix,Snjv,Snjv)+Gb(Snjv,Snjy,Snjy)]0asn.

    Owing to Gb(x,y,y)=Gb(Snix,Snjy,Snjy), as a result, x=y, the proof is completed.

    Example 2.4. Let X=R, Q=[0,), θ(t)=1+t, α=0, k=12, Gb:X×X×XC be defined by Gb(ξ1,ξ2,ξ3)=max{|ξ1ξ2|2,|ξ2ξ3|2,|ξ1ξ3|2}+max{|ξ1ξ2|2,|ξ2ξ3|2,|ξ1ξ3|2}i with s=2. For any ξ in X, take Snξ=|ξ|4n and Fn=uSn, where u1,nN, the partial order on X is the usual order of R.

    Suppose that ξ1ξ2ξ3, if ξ1ξ31 for any ξ1,ξ2,ξ3 in Q, or ξ1,ξ2,ξ3[0,1], we can easily obtain

    1+Gb(Snξ1,Snξ2,Snξ3)=1+(ξ1ξ34n)2+(ξ1ξ34n)2i,

    and

    1+12Gb(ξ1,ξ2,ξ3)=1+12(ξ1ξ3)2+12(ξ1ξ3)2i.

    Hence,

    |1+Gb(Snξ1,Snξ2,Snξ3)|4=[(1+(ξ1ξ34n)2)2+(ξ1ξ34n)4]4=1+4(ξ1ξ34n)2+8(ξ1ξ34n)4+8(ξ1ξ34n)6+4(ξ1ξ34n)81+12(ξ1ξ34n)2+12(ξ1ξ34n)41+(ξ1ξ3)2+12(ξ1ξ3)4,

    and

    |1+12Gb(ξ1,ξ2,ξ3)|2=1+(ξ1ξ3)2+12(ξ1ξ3)4.

    Thus we obtain

    |1+Gb(Snξ1,Snξ2,Snξ3)||1+12Gb(ξ1,ξ2,ξ3)|12|1+12M(ξ1,ξ2,ξ3)|12.

    It follows that the operator equations Fnξ=uξ have a common solution ξ=0 in Q and (2.19) is established with v=0. Therefore, all conditions of Theorem 2.3 are satisfied, the operator equations Fnξ=uξ have an unique solution ξ=0.

    The following two corollaries can be easily obtained, if we let θ(t)=e|t|+t and θ(t)=22πarctan(1|t|γ) in Theorem 2.3 respectively.

    Corollary 2.3. Let (X,Gb,) be a complete partially ordered complex valued Gb-metric space with s1, QX be a cone, {Sn:XQ,nN} be a dominated mapping sequence. Suppose that there exist i,jN,k(0,1),α0 such that

    |e|Gb(Six,Sjy,Sjz)|+Gb(Six,Sjy,Sjz)||e|1sM(x,y,z)α|+1sM(x,y,z)α|k

    for any comparable elements x,y,z in Q, where Gb(Six,Sjy,Sjz)0, and

    M(x,y,z)=max{Gb(x,Six,Six),Gb(y,Sjy,Sjz),Gb(z,Sjz,Sjy),Gb(x,y,z)}.

    Define the operator equations Fnx=ux by Fn=uSn, u1. If a nonincreasing sequence {xn}κ such that κxn, then the operator equations have at least a common solution x in Q. Moreover, if there exists an element v in Q such that vx, and

    Gb(Sn1jv,Snjv,Snjv)Gb(x,Sn1jv,Sn1jv),

    then the operator equations have an unique solution.

    Corollary 2.4. Let (X,Gb,) be a complete partially ordered complex valued Gb-metric space with s1, QX be a cone, {Sn:XQ,nN} be a dominated mapping sequence. Suppose that there exist i,jN,γ,k(0,1),α0 such that

    22πarctan(1|Gb(Six,Sjy,Sjz)|γ)|22πarctan(1|1sM(x,y,z)α|γ)|k

    for any comparable elements x,y,z in Q, where Gb(Six,Sjy,Sjz)0, and

    M(x,y,z)=max{Gb(x,Six,Six),Gb(y,Sjy,Sjz),Gb(z,Sjz,Sjy),Gb(x,y,z)}.

    Define the operator equations Fnx=ux by Fn=uSn, u1. If a nonincreasing sequence {xn}κ such that κxn, then the operator equations have at least a common solution x in Q. Moreover, if there exists an element v in Q such that vx, and

    Gb(Sn1jv,Snjv,Snjv)Gb(x,Sn1jv,Sn1jv),

    then the operator equations have an unique solution.

    In this paper, we have obtained some new theorems for the common solutions of the operator equations Fnx=ux(u1,nN) via complex valued Cp-class function, a type of Geraghty contraction and a type of JS contraction in complete partially ordered complex valued Gb-metric spaces, and some of which are established in a closed ball. These new results generalize many known results in complex valued Gb-metric spaces and Gb-metric spaces, in addition, it would be interesting and worthwhile to further investigate some similar problems in other types of spaces.

    This work was supported by National Natural Science Foundation of China (Grant No. 11771198).

    The authors declare no conflict of interest.



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