# Common Fixed Point Theorems for a Class of Twice Power Type Contraction Maps inG-Metric Spaces

Authors: Hongqing Ye; Feng Gu
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We introduce a new twice power type contractive condition for three mappings inG-metric spaces, and several new common fixed point theorems are established in complete G-metric space. An example is provided to support our result. The results obtained in this paper differ from other comparable results already known.


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## Body

## 1. Introduction

The study of fixed points of mappings satisfying certain contractive conditions has been in the center of rigorous research activity. In 2006, a new structure of generalized metric space was introduced by Mustafa and Sims [1] as an appropriate notion of generalized metric space called G-metric space. Abbas and Rhoades [2] initiated the study of common fixed point in generalized metric space. Recently, many fixed point theorems for certain contractive conditions have been established in G-metric spaces, and for more details one can refer to [3–27]. Fixed point problems have also been considered in partially ordered G-metric spaces [28–31], cone metric spaces [32], and generalized cone metric spaces [33].In 2006, Gu and He [34] introduced a class of twice power type contractive condition in metric space, proving some common fixed point theorems for four self-maps with twice power type Φ-contractive condition.In this paper, motivated and inspired by the above results, we introduce a new twice power type contractive condition inG-metric space, and we prove some new common fixed point theorems in complete G-metric spaces. Our results obtained in this paper differ from other comparable results already known.Throughout the paper, we mean byℕ the set of all natural numbers. Consistent with Mustafa and Sims [1], the following definitions and results will be needed in the sequel.Definition 1.1 (see [1]).
LetX be a nonempty set, and let G:X×X×X→R+ be a function satisfying the following axioms:(G1)G(x,y,z=0) if x=y=z;(G2)0<G(x,x,y), for all x,y∈X with x≠y;(G3)  G(x,x,y)≤G(x,y,z), for all x,y,z∈X with z≠ y;(G4)  G(x,y,z)=G(x,z,y)=G(y,z,x)=⋯ (symmetry in all three variables);(G5)G(x,y,z)≤G(x,a,a)+G(a,y,z) for all x,y,z,a∈X, (rectangle inequality);then the function G is called a generalized metric, or, more specifically, a G-metric on X and the pair (X,G) are called a G-metric space.Definition 1.2 (see [1]).
Let(X,G) be a G-metric space, and let {xn} be a sequence of points in X, a point x in X is said to be the limit of the sequence {xn} if limm,n→∞G(x,xn,xm)=0, and one says that sequence {xn} is G-convergent to x.Thus, ifxn→x in a G-metric space (X,G), then for any ϵ>0, there exists N∈ℕ such that G(x,xn,xm)<ϵ, for all n,m≥N.Proposition 1.3 (see [1]).
Let(X,G) be a G-metric space, then the followings are equivalent.(1)
{xn} is G-convergent to x.(2)
G(xn,xn,x)→0 as n→∞.(3)
G(xn,x,x)→0 as n→∞.(4)
G(xn,xm,x)→0 as n,m→∞.Definition 1.4 (see [1]).
Let(X,G) be a G-metric space. A sequence {xn} is called G-Cauchy sequence if for each ϵ>0 there exists a positive integer N∈ℕ such that G(xn,xm,xl)<ϵ for all n,m,l≥N; that is, if G(xn,xm,xl)→0 as n,m,l→∞.Definition 1.5 (see [1]).
AG-metric space (X,G) is said to be G-complete if every G-Cauchy sequence in (X,G) is G-convergent in X.Proposition 1.6 (see [1]).
Let(X,G) be a G-metric space. Then the following are equivalent.(1)
The sequence{xn} is G-Cauchy.(2)
For everyϵ>0, there exists k∈ℕ such that G(xn,xm,xm)<ϵ, for all n,m≥k.Proposition 1.7 (see [1]).
Let(X,G) be a G-metric space. Then the function G(x,y,z) is jointly continuous in all three of its variables.Definition 1.8 (see [1]).
Let(X,G) and (X′,G′) be G-metric space, and f:(X,G)→(X′,G′) be a function. Then f is said to be G-continuous at a point a∈X if and only if for every ε>0, there is δ>0 such that x,y∈X and G(a,x,y)<δ implies G′(f(a),f(x),f(y))<ε. A function f is G-continuous at X if and only if it is G-continuous at all a∈X.Proposition 1.9 (see [1]).
Let(X,G) and (X′,G′) be G-metric space. Then f:X→X′ is G-continuous at x∈X if and only if it is G-sequentially continuous at x; that is, whenever {xn} is G-convergent to x, {f(xn)} is G-convergent to f(x).Proposition 1.10 (see, [1]).
Let(X,G) be a G-metric space. Then, for any x,y,z,a in X it follows that:(i)
ifG(x,y,z)=0, then x=y=z;(ii)
G(x,y,z)≤G(x,x,y)+G(x,x,z);(iii)
G(x,y,y)≤2G(y,x,x);(iv)
G(x,y,z)≤G(x,a,z)+G(a,y,z);(v)
G(x,y,z)≤(2/3)(G(x,y,a)+G(x,a,z)+G(a,y,z));(vi)
G(x,y,z)≤(G(x,a,a)+G(y,a,a)+G(z,a,a)).


## 2. Main Results

Theorem 2.1.
Let(X,G) be a complete G-metric space. Suppose the three self-mappings T,S,R:X→X satisfy the following condition:
(2.1)G2(Tx,Sy,Rz)≤αG(x,Tx,Tx)G(y,Sy,Sy)+βG(y,Sy,Sy)G(z,Rz,Rz)+γG(x,Tx,Tx)G(z,Rz,Rz),
for all x,y,z∈X, where α,β,γ are nonnegative real numbers and α+β+γ<1. Then T,S, and R have a unique common fixed point (say u) and T,S,R are all G-continuous at u.Proof .
We will proceed in two steps.
Step 1.We prove any fixed point of T is a fixed point of S and R and conversely. Assume that p∈X is such that Tp=p. However, by (2.1), we have
(2.2)G2(Tp,Sp,Rp)≤αG(p,Tp,Tp)G(p,Sp,Sp)+βG(p,Sp,Sp)G(p,Rp,Rp)+γG(p,Tp,Tp)G(p,Rp,Rp)=αG(p,p,p)G(p,Sp,Sp)+βG(p,Sp,Sp)G(p,Rp,Rp)+γG(p,p,p)G(p,Rp,Rp)=βG(p,Sp,Sp)G(p,Rp,Rp).
Now we discuss the above inequality in three cases.
Case (i). If p≠Sp and p≠Rp, then, by (G3), we have
(2.3)G(p,Sp,Sp)≤G(p,Sp,Rp),G(p,Rp,Rp)≤G(p,Sp,Rp).
So, the above inequality becomes
(2.4)G2(p,Sp,Rp)=G2(Tp,Sp,Rp)≤βG2(p,Sp,Rp).
Since G2(p,Sp,Rp)>0, hence we have β≥1; however, it contradicts with 0≤β≤α+β+γ<1, so we get p=Sp=Rp.
Case (ii). If p=Rp, then we have
(2.5)G2(p,Sp,Rp)=G2(Tp,Sp,Rp)≤βG(p,Sp,Sp)G(p,Rp,Rp)=0.
Hence we have G2(p,Sp,Rp)=0 and so p=Sp=Rp.
Case (iii). If p=Sp, we can also get G2(p,Sp,Rp)=0. Hence we have p=Sp=Rp. Therefore p is a common fixed point of T,S and R.
The same conclusion holds ifp=Sp or p=Rp.
Step 2.We prove that T, S, and R have a unique common fixed point.
Letx0∈X be an arbitrary point, and define the sequence {xn} by x3n+1=Tx3n,x3n+2=Sx3n+1,x3n+3=Rx3n+2, n∈ℕ. If xn=xn+1, for some n, with n=3m, then p=x3m is a fixed point of T and, by the first step, p is a common fixed point of S, T, and R. The same holds if n=3m+1 or n=3m+2. Without loss of generality, we can assume that xn≠xn+1, for all n∈ℕ.
Next, we prove sequence{xn} is a G-Cauchy sequence. In fact, by (2.1) and (G3), we have
(2.6)G2(x3n+1,x3n+2,x3n+3)=G2(Tx3n,Sx3n+1,Rx3n+2)≤αG(x3n,Tx3n,Tx3n)G(x3n+1,Sx3n+1,Sx3n+1)+βG(x3n+1,Sx3n+1,Sx3n+1)G(x3n+2,Rx3n+2,Rx3n+2)+γG(x3n,Tx3n,Tx3n)G(x3n+2,Rx3n+2,Rx3n+2)=αG(x3n,x3n+1,x3n+1)G(x3n+1,x3n+2,x3n+2)+βG(x3n+1,x3n+2,x3n+2)G(x3n+2,x3n+3,x3n+3)+γG(x3n,x3n+1,x3n+1)G(x3n+2,x3n+3,x3n+3)≤αG(x3n,x3n+1,x3n+2)G(x3n+1,x3n+2,x3n+3)+βG(x3n+1,x3n+2,x3n+3)G(x3n+2,x3n+3,x3n+1)+γG(x3n,x3n+1,x3n+2)G(x3n+2,x3n+3,x3n+1).
Which gives that
(2.7)G(x3n+1,x3n+2,x3n+3)≤(α+γ)G(x3n,x3n+1,x3n+2)+βG(x3n+1,x3n+2,x3n+3).
It follows that
(2.8)(1-β)G(x3n+1,x3n+2,x3n+3)≤(α+γ)G(x3n,x3n+1,x3n+2).
From 0≤β<1 we know that 1-β>0. Then, we have
(2.9)G(x3n+1,x3n+2,x3n+3)≤α+γ1-βG(x3n,x3n+1,x3n+2).
On the other hand, by using (2.1) and (G3), we have
(2.10)G2(x3n+2,x3n+3,x3n+4)=G2(Tx3n+3,Sx3n+1,Rx3n+2)≤αG(x3n+3,Tx3n+3,Tx3n+3)G(x3n+1,Sx3n+1,Sx3n+1)+βG(x3n+1,Sx3n+1,Sx3n+1)G(x3n+2,Rx3n+2,Rx3n+2)+γG(x3n+3,Tx3n+3,Tx3n+3)G(x3n+2,Rx3n+2,Rx3n+2)=αG(x3n+3,x3n+4,x3n+4)G(x3n+1,x3n+2,x3n+2)+βG(x3n+1,x3n+2,x3n+2)G(x3n+2,x3n+3,x3n+3)+γG(x3n+3,x3n+4,x3n+4)G(x3n+2,x3n+3,x3n+3)≤αG(x3n+2,x3n+3,x3n+4)G(x3n+1,x3n+2,x3n+3)+βG(x3n+1,x3n+2,x3n+3)G(x3n+2,x3n+3,x3n+4)+γG(x3n+2,x3n+3,x3n+4)G(x3n+2,x3n+3,x3n+4).
Which implies that
(2.11)G(x3n+2,x3n+3,x3n+4)≤(α+β)G(x3n+1,x3n+2,x3n+3)+γG(x3n+2,x3n+3,x3n+4).
It follows that
(2.12)(1-γ)G(x3n+2,x3n+3,x3n+4)≤(α+β)G(x3n+1,x3n+2,x3n+3).
Form the condition 0≤γ≤α+β+γ<1, we know that 1-γ>0. Therefore, we have
(2.13)G(x3n+2,x3n+3,x3n+4)≤α+β1-γG(x3n+1,x3n+2,x3n+3).
Again, using (2.1) and (G3), we can get
(2.14)G2(x3n+3,x3n+4,x3n+5)=G2(Tx3n+3,Sx3n+4,Rx3n+2)≤αG(x3n+3,Tx3n+3,Tx3n+3)G(x3n+4,Sx3n+4,Sx3n+4)+βG(x3n+4,Sx3n+4,Sx3n+4)G(x3n+2,Rx3n+2,Rx3n+2)+γG(x3n+3,Tx3n+3,Tx3n+3)G(x3n+2,Rx3n+2,Rx3n+2)=αG(x3n+3,x3n+4,x3n+4)G(x3n+4,x3n+5,x3n+5)+βG(x3n+4,x3n+5,x3n+5)G(x3n+2,x3n+3,x3n+3)+γG(x3n+3,x3n+4,x3n+4)G(x3n+2,x3n+3,x3n+3)≤αG(x3n+3,x3n+4,x3n+5)G(x3n+3,x3n+4,x3n+5)+βG(x3n+3,x3n+4,x3n+5)G(x3n+2,x3n+3,x3n+4)+γG(x3n+5,x3n+3,x3n+4)G(x3n+2,x3n+3,x3n+4).
Which implies that
(2.15)G(x3n+3,x3n+4,x3n+5)≤αG(x3n+3,x3n+4,x3n+5)+(β+γ)G(x3n+2,x3n+3,x3n+4).
It follows that
(2.16)(1-α)G(x3n+3,x3n+4,x3n+5)≤(β+γ)G(x3n+2,x3n+3,x3n+4).
By the condition 0≤α≤α+β+γ<1, we know that 1-α>0. Hence, we have
(2.17)G(x3n+3,x3n+4,x3n+5)≤β+γ1-αG(x3n+2,x3n+3,x3n+4).
Let q=max{(α+γ)/(1-β),(α+β)/(1-γ),(β+γ)/(1-α)}, then from 0≤α+β+γ<1 we know that 0≤q<1. Combining (2.9), (2.13), and (2.17), we have
(2.18)G(xn,xn+1,xn+2)≤qG(xn-1,xn,xn+1)≤⋯≤qnG(x0,x1,x2).
Thus, by (G3) and (G5), for every m,n∈ℕ,m>n, noting that 0≤q<1, we have
(2.19)G(xn,xm,xm)≤G(xn,xn+1,xn+1)+G(xn+1,xn+2,xn+2)+⋯+G(xm-1,xm,xm),≤G(xn,xn+1,xn+2)+G(xn+1,xn+2,xn+3)+⋯+G(xm-1,xm,xm+1)≤(qn+qn+1+⋯+qm-1)G(x0,x1,x2)≤qn1-qG(x0,x1,x2).
Which implies that G(xn,xm,xm)→0, as n,m→∞. Thus {xn} is a G-Cauchy sequence. Due to the completeness of (X,G), there exists u∈X, such that {xn} is G-convergent to u.
Next we proveu is a common fixed point of T,S, and R. By using (2.1), we have
(2.20)G2(Tu,x3n+2,x3n+3)=G2(Tu,Sx3n+1,Rx3n+2)≤αG(u,Tu,Tu)G(x3n+1,Sx3n+1,Sx3n+1)+βG(x3n+1,Sx3n+1,Sx3n+1)G(x3n+2,Rx3n+2,Rx3n+2)+γG(u,Tu,Tu)G(x3n+2,Rx3n+2,Rx3n+2)=αG(u,Tu,Tu)G(x3n+1,x3n+2,x3n+2)+βG(x3n+1,x3n+2,x3n+2)G(x3n+2,x3n+3,x3n+3)+γG(u,Tu,Tu)G(x3n+2,x3n+3,x3n+3).
Letting n→∞, and using the fact that G is continuous on its variables, we can get
(2.21)G2(Tu,u,u)=0.
Which gives that Tu=u, that is u is a fixed point of T. By using (2.1) again, we have
(2.22)G2(x3n+1,Su,x3n+3)=G2(Tx3n,Su,Rx3n+2)≤αG(x3n,x3n+1,x3n+1)G(u,Su,Su)+βG(u,Su,Su)G(x3n+2,x3n+3,x3n+3)+γG(x3n,x3n+1,x3n+1)G(x3n+2,x3n+3,x3n+3).
Letting n→∞ at both sides, for G is continuous on its variables, it follows that
(2.23)G2(u,Su,u)=0.
Therefore, Su=u; that is, u is a fixed point of S. Similarly, by (2.1), we can also get
(2.24)G2(x3n+1,x3n+2,Ru)=G2(Tx3n,Sx3n+1,Ru)≤αG(x3n,x3n+1,x3n+1)G(x3n+1,x3n+2,x3n+2)+βG(x3n+1,x3n+2,x3n+2)G(u,Ru,Ru)+γG(x3n,x3n+1,x3n+1)G(u,Ru,Ru).
On taking n→∞ at both sides, since G is continuous on its variables, we get that
(2.25)G2(u,u,Ru)=0.
Which gives that u=Ru, therefore, u is fixed point of R. Consequently, we have u=Tu=Su=Ru, and u is a common fixed point of T,S and R. Suppose v is another common fixed point of T,S, and R, and we have v=Tv=Sv=Rv, then by (2.1), we have
(2.26)G2(u,u,v)=G2(Tu,Su,Rv)≤αG(u,Tu,Tu)G(u,Su,Su)+βG(u,Su,Su)G(v,Rv,Rv)+γG(u,Tu,Tu)G(v,Rv,Rv)=αG(u,u,u)G(u,u,u)+βG(u,u,u)G(v,v,v)+γG(u,u,u)G(v,v,v)=0.
Which implies that G2(u,u,v)=0, hence, u=v. Then we know the common fixed point of T,S, and R is unique.
To show thatT is G-continuous at u, let {yn} be any sequence in X such that {yn} is G-convergent to u. For n∈ℕ, we have
(2.27)G2(Tyn,u,u)=G2(Tyn,Su,Ru)≤αG(yn,Tyn,Tyn)G(u,Su,Su)+βG(u,Su,Su)G(u,Ru,Ru)+γG(yn,Tyn,Tyn)G(u,Ru,Ru)=αG(yn,Tyn,Tyn)G(u,u,u)+βG(u,u,u)G(u,u,u)+γG(yn,Tyn,Tyn)G(u,u,u)=0.
Which implies that limn→∞G2(Tyn,u,u)=0. Hence {Tyn} is G-convergent to u=Tu. So T is G-continuous at u. Similarly, we can also prove that S,R are G-continuous at u. Therefore, we complete the proof.Corollary 2.2.
Let(X,G) be a complete G-metric space. Suppose the three self-mappings T,S,R:X→X satisfy the condition:
(2.28)G2(Tpx,Ssy,Rrz)≤αG(x,Tpx,Tpx)G(y,Ssy,Ssy)+βG(y,Ssy,Ssy)G(z,Rrz,Rrz)+γG(x,Tpx,Tpx)G(z,Rrz,Rrz),
for all x,y,z∈X, where p,s,r∈ℕ,α,β,γ are nonnegative real numbers and α+β+γ<1. Then T,S, and R have a unique common fixed point (say u) and Tp,Ss,Rr are all G-continuous at u.Proof.
From Theorem2.1 we know that Tp,Ss,Rr have a unique common fixed point (say u); that is, Tpu=u,Ssu=u,Rru=u, and Tp,Ss,Rr are G-continuous at u. Since Tu=TTpu=Tp+1u=TpTu, so Tu is another fixed point of Tp, Su=SSsu=Ss+1u=gsgu, so Su is another fixed point of Ss, and Ru=RRru=Rr+1u=RrRu, so Ru is another fixed point of Rr. By (G3) and the condition (2.28) in Corollary 2.2, we have
(2.29)G2(Tu,SsTu,RrTu)=G2(TpTu,SsTu,RrTu)≤αG(Tu,TpTu,TpTu)G(Tu,SsTu,SsTu)+βG(Tu,SsTu,SsTu)G(Tu,RrTu,RrTu)+γG(Tu,TpTu,TpTu)G(Tu,RrTu,RrTu)=βG(Tu,SsTu,SsTu)G(Tu,RrTu,RrTu)≤βG(Tu,SsTu,RrTu)G(Tu,SsTu,RrTu).
Since 0≤β<1, we can get G2(Tu,SsTu,RrTu)=0. That means Tu=TpTu=SsTu=RrTu, hence Tu is another common fixed point of Tp,Ss-and Rr. Since the common fixed point of Tp,Ss-and Rr is unique, we deduce that u=Tu. By the same argument, we can prove u=Su,u=Ru. Thus, we have u=Tu=Su=Ru. Suppose v is another common fixed point of T,S, and R, then v=Tpv=Ssv=Rrv, and by using the condition (2.28) in Corollary 2.2 again, we have
(2.30)G2(v,u,u)=G2(Tpv,Ssu,Rru)≤αG(v,Tpv,Tpv)G(u,Ssu,Ssu)+βG(u,Ssu,Ssu)G(u,Rru,Rru)+γG(v,Tpv,Tpv)G(u,Rru,Rru)=αG(v,v,v)G(u,u,u)+βG(u,u,u)G(u,u,u)+γG(v,v,v)G(u,u,u)=0.
Which implies that G2(v,u,u)=0, hence v=u. So the common fixed of T,S, and R is unique. It is obvious that every fixed point of T is a fixed point of S and R and conversely.Corollary 2.3.
Let(X,G) be a complete G-metric space. Suppose the self-mapping T:X→X satisfies the following condition:
(2.31)G2(Tx,Ty,Tz)≤αG(x,Tx,Tx)G(y,Ty,Ty)+βG(y,Ty,Ty)G(z,Tz,Tz)+γG(x,Tx,Tx)G(z,Tz,Tz),
for all x,y,z∈X, where α,β,γ are nonnegative real numbers and α+β+γ<1. Then T has a unique fixed point (say u) and T is G-continuous at u.Proof.
LetT=S=R in Theorem 2.1, we can get this conclusion holds.Corollary 2.4.
Let(X,G) be a complete G-metric space. Suppose the self-mapping T:X→X satisfies the following condition:
(2.32)G2(Tpx,Tpy,Tpz)≤αG(x,Tpx,Tpx)G(y,Tpy,Tpy)+βG(y,Tpy,Tpy)G(z,Tpz,Tpz)+γG(x,Tpx,Tpx)G(z,Tpz,Tpz).
for all x,y,z∈X, where α,β,γ are nonnegative real numbers and α+β+γ<1. Then T has a unique fixed point (say u) and Tp is G-continuous at u.Proof.
LetT=S=R,p=s=r in Corollary 2.2, we can get this conclusion holds.Theorem 2.5.
Let(X,G) be a complete G-metric space, and let T,S,R:X→X be three self-mappings in X, which satisfy the following condition. 							(2.33)G2(Tx,Sy,Rz)≤αG(x,Tx,Sy)G(y,Sy,Rz)+βG(y,Sy,Rz)G(z,Rz,Tx)+γG(x,Tx,Sy)G(z,Rz,Tx).
for all x,y,z∈X, α,β,γ are nonnegative real numbers and α+β+γ<1. Then T,S and R have a unique common fixed point (say u) and T,S,R are all G-continuous at u.Proof.
We will proceed in two steps.
Step 1. We prove any fixed point of T is a fixed point of S and R and conversely. Assume that p∈X is such that Tp=p. Now we prove that p=Sp and p=Rp. If it is not the case, then for p≠Sp and p≠Rp, by (2.33) and (G3) we have
(2.34)G2(Tp,Sp,Rp)≤αG(p,Tp,Sp)G(p,Sp,Rp)+βG(p,Sp,Rp)G(p,Rp,Tp)+γG(p,Tp,Sp)G(p,Rp,Tp)=αG(p,p,Sp)G(p,Sp,Rp)+βG(p,Sp,Rp)G(p,Rp,p)+γG(p,p,Sp)G(p,Rp,p)≤αG(p,Rp,Sp)G(p,Sp,Rp)+βG(p,Sp,Rp)G(p,Rp,Sp)+γG(p,Rp,Sp)G(p,Rp,Sp)=(α+β+γ)G2(p,Rp,Sp).
It follows that
(2.35)G2(p,Sp,Rp)=G2(Tp,Sp,Rp)≤(α+β+γ)G2(p,Sp,Rp).
Since G2(p,Sp,Rp)>0, hence we have α+β+γ≥1, however it contradicts with the condition 0≤α+β+γ<1, so we can have p=Sp=Rp, hence p is a common fixed point of T,S, and R.
Analogously, following the similar arguments to those given above, we can obtain a contradiction forp≠Sp and p=Rp or p=Sp and p≠Rp. Hence in all the cases, we conclude that p=Sp=Rp. The same conclusion holds if p=Sp or p=Rp.
Step 2. We prove that T, S and R have a unique common fixed point. Let x0∈X be an arbitrary point, and define the sequence {xn} by x3n+1=Tx3n,x3n+2=Sx3n+1,x3n+3=Rx3n+2, n∈ℕ. If xn=xn+1, for some n, with n=3m, then p=x3m is a fixed point of T and, by the first step, p is a common fixed point of S, T, and R. The same holds if n=3m+1 or n=3m+2. Without loss of generality, we can assume that xn≠xn+1, for all n∈ℕ. We first prove the sequence {xn} is a G-Cauchy sequence. In fact, by using (2.33) and (G3), we have
(2.36)G2(x3n+1,x3n+2,x3n+3)=G2(Tx3n,Sx3n+1,Rx3n+2)≤αG(x3n,x3n+1,x3n+2)G(x3n+1,x3n+2,x3n+3)+βG(x3n+1,x3n+2,x3n+3)G(x3n+2,x3n+3,x3n+1)+γG(x3n,x3n+1,x3n+2)G(x3n+2,x3n+3,x3n+1).
Which gives that
(2.37)G(x3n+1,x3n+2,x3n+3)≤(α+γ)G(x3n,x3n+1,x3n+2)+βG(x3n+1,x3n+2,x3n+3).
It follows that
(2.38)(1-β)G(x3n+1,x3n+2,x3n+3)≤(α+γ)G(x3n,x3n+1,x3n+2).
From 0≤β<1, we know that 1-β>0. Then, we have
(2.39)G(x3n+1,x3n+2,x3n+3)≤α+γ1-βG(x3n,x3n+1,x3n+2).
On the other hand, by using (2.33) and (G3), we have
(2.40)G2(x3n+2,x3n+3,x3n+4)=G2(Tx3n+3,Sx3n+1,Rx3n+2)≤αG(x3n+3,x3n+4,x3n+2)G(x3n+1,x3n+2,x3n+3)+βG(x3n+1,x3n+2,x3n+3)G(x3n+2,x3n+3,x3n+4)+γG(x3n+3,x3n+4,x3n+2)G(x3n+2,x3n+3,x3n+4).
Which implies that
(2.41)G(x3n+2,x3n+3,x3n+4)≤(α+β)G(x3n+1,x3n+2,x3n+3)+γG(x3n+2,x3n+3,x3n+4).
It follows that
(2.42)(1-γ)G(x3n+2,x3n+3,x3n+4)≤(α+β)G(x3n+1,x3n+2,x3n+3).
Since 0≤γ<1, we know that 1-γ>0. So, we have
(2.43)G(x3n+2,x3n+3,x3n+4)≤α+β1-γG(x3n+1,x3n+2,x3n+3).
Again, using (2.33) and (G3), we can get
(2.44)G2(x3n+3,x3n+4,x3n+5)=G2(Tx3n+3,Sx3n+4,Rx3n+2)≤αG(x3n+3,x3n+4,x3n+5)G(x3n+4,x3n+5,x3n+3)+βG(x3n+4,x3n+5,x3n+3)G(x3n+2,x3n+3,x3n+4)+γG(x3n+2,x3n+3,x3n+4)G(x3n+3,x3n+4,x3n+5).
Which implies that
(2.45)G(x3n+3,x3n+4,x3n+5)≤αG(x3n+3,x3n+4,x3n+5)+(β+γ)G(x3n+2,x3n+3,x3n+4).
It follows that
(2.46)(1-α)G(x3n+3,x3n+4,x3n+5)≤(β+γ)G(x3n+2,x3n+3,x3n+4).
Since 0≤α≤α+β+γ<1, we know that 1-α>0. So we have
(2.47)G(x3n+3,x3n+4,x3n+5)≤β+γ1-αG(x3n+2,x3n+3,x3n+4).
Let q=max{(α+γ)/(1-β),(α+β)/(1-γ),(β+γ)/(1-α)}, then 0≤q<1, and by combining (2.39), (2.43), and (2.47), we have
(2.48)G(xn,xn+1,xn+2)≤qG(xn-1,xn,xn+1)≤⋯≤qnG(x0,x1,x2).
Thus, by (G3) and (G5), for every m,n∈ℕ, if m>n, noting that 0≤q<1, we have
(2.49)G(xn,xm,xm)≤G(xn,xn+1,xn+1)+G(xn+1,xn+2,xn+2)+⋯+G(xm-1,xm,xm)≤G(xn,xn+1,xn+2)+G(xn+1,xn+2,xn+3)+⋯+G(xm-1,xm,xm+1)≤(qn+qn+1+⋯+qm-1)G(x0,x1,x2)≤qn1-qG(x0,x1,x2).
Which implies that G(xn,xm,xm)→0, as n,m→∞. Thus {xn} is a G-Cauchy sequence. Due to the completeness of (X,G), there exists u∈X, such that {xn} is G-convergent to u.
Now we proveu is a common fixed point of T,S, and R. By using (2.33), we have
(2.50)G2(Tu,x3n+2,x3n+3)=G2(Tu,Sx3n+1,Rx3n+2)≤αG(u,Tu,Sx3n+1)G(x3n+1,Sx3n+1,Rx3n+2)+βG(x3n+1,Sx3n+1,Rx3n+2)G(x3n+2,Rx3n+2,Tu)+γG(u,Tu,Sx3n+1)G(x3n+2,Rx3n+2,Tu)=αG(u,Tu,x3n+2)G(x3n+1,x3n+2,x3n+3)+βG(x3n+1,x3n+2,x3n+3)G(x3n+2,x3n+3,Tu)+γG(u,Tu,x3n+2)G(x3n+2,x3n+3,Tu).
Letting n→∞, and using the fact that G is continuous on its variables and γ<1, we can get
(2.51)G2(Tu,u,u)≤γG2(u,u,Tu).
Which gives that Tu=u, hence u is a fixed point of T. By using (2.33) again, we have
(2.52)G2(x3n+1,Su,x3n+3)=G2(Tx3n,Su,Rx3n+2)≤αG(x3n,x3n+1,Su)G(u,Su,x3n+3)+βG(u,Su,x3n+3)G(x3n+2,x3n+3,x3n+1)+γG(x3n,x3n+1,Su)G(x3n+2,x3n+3,x3n+1).
Letting n→∞ at both sides, for G is continuous in its variables, it follows that
(2.53)G2(u,Su,u)≤αG2(u,Su,u).
For 0≤α<1, Therefore, we can get G2(u,Su,u)=0, hence Su=u, hence u is a fixed point of S. Similarly, by (2.33), we can also get
(2.54)G2(x3n+1,x3n+2,Ru)=G2(Tx3n,Sx3n+1,Ru)≤αG(x3n,x3n+1,x3n+2)G(x3n+1,x3n+2,Ru)+βG(x3n+1,x3n+2,Ru)G(u,Ru,x3n+1)+γG(x3n,x3n+1,x3n+2)G(u,Ru,x3n+1).
On taking n→∞ at both sides, since G is continuous in its variables, we get that
(2.55)G2(u,u,Ru)≤βG2(u,u,Ru).
Since 0≤β<1, so we get G2(u,u,Ru)=0, hence u=Ru, therefore, u is a fixed point of R. Consequently, we have u=Tu=Su=Ru, and u is a common fixed point of T,S, and R. Suppose v≠u is another common fixed point of T,S, and R, and we have v=Tv=Sv=Rv, then by (2.33), we have
(2.56)G2(u,u,v)=G2(Tu,Su,Rv)≤αG(u,Tu,Su)G(u,Su,Rv)+βG(u,Su,Rv)G(v,Rv,Tu)+γG(u,Tu,Su)G(v,Rv,Tu)=αG(u,u,u)G(u,u,v)+βG(u,u,v)G(v,v,u)+γG(u,u,u)G(v,v,u).
Which gives that
(2.57)G2(u,u,v)≤βG(u,u,v)G(v,v,u).
Hence, we can get G(u,u,v)≤βG(v,v,u). By using (2.33) again, we get
(2.58)G2(u,v,v)=G2(Tu,Sv,Rv)≤αG(u,Tu,Sv)G(v,Sv,Rv)+βG(v,Sv,Rv)G(v,Rv,Tu)+γG(u,Tu,Sv)G(v,Rv,Tu)=αG(u,u,v)G(v,v,v)+βG(v,v,v)G(v,v,u)+γG(u,u,v)G(v,v,u).
Which implies that
(2.59)G2(u,v,v)≤γG(u,u,v)G(v,v,u).
Hence, we can get
(2.60)G(u,v,v)≤γG(u,u,v).
By combining G(u,u,v)≤βG(v,v,u), we can have
(2.61)G(u,v,v)≤γG(u,u,v)≤βγG(v,v,u).
Since v≠u,G(u,v,v)>0, so we have that βγ≥1. Since 0≤β,γ<1, we know 0≤βγ<1, so it’s a contradiction. Hence, we get u=v. Then we know the common fixed point of T,S, and R is unique.
To show thatT is G-continuous at u, let {yn} be any sequence in X such that {yn} is G-convergent to u. For n∈ℕ, we have
(2.62)G2(Tyn,u,u)=G2(Tyn,Su,Ru)≤αG(yn,Tyn,Su)G(u,Su,Ru)+βG(u,Su,Ru)G(u,Ru,Tyn)+γG(yn,Tyn,Su)G(u,Ru,Tyn)=αG(yn,Tyn,u)G(u,u,u)+βG(u,u,u)G(u,u,Tyn)+γG(yn,Tyn,u)G(u,u,Tyn)=γG(yn,Tyn,u)G(u,u,Tyn).
Which implies that
(2.63)G(Tyn,u,u)≤γG(yn,Tyn,u).
On taking n→∞ at both sides, considering γ<1, we get limn→∞G(Tyn,u,u)=0. Hence {Tyn} is G-convergent to u=Tu. So T is G-continuous at u. Similarly, we can also prove that S,R are G-continuous at u. Therefore, we complete the proof.Now we introduce an example to support Theorem2.5.Example 2.6.
LetX=[0,1], and let (X,G) be a G-metric space defined by G(x,y,z)=|x-y|+|y-z|+|z-x|, for all x,y,z in X. Let T, S, and R be three self-mappings defined by
(2.64)Tx={1,x∈[0,12]67,x∈(12,1],Sx={78,x∈[0,12]67,x∈(12,1],Rx=67,x∈[0,1].
Next we proof the mappingsT, S, and R are satisfying Condition (2.33) of Theorem 2.5 with α=1/7, β=1/7 and γ=4/7.Case 1.
Ifx,y∈[0,1/2], z∈[0,1], then
(2.65)G2(Tx,Sy,Rz)=G2(1,78,67)=449,G(x,Tx,Sy)=G(x,1,78)=|x-1|+|x-78|+18≥12+38+18=1,G(y,Sy,Rz)=G(y,78,67)=|y-78|+|y-67|+156≥38+514+156=34,G(z,Rz,Tx)=G(z,67,1)=|z-67|+17+|z-1|≥0+17+0=17.
Thus, we have
(2.66)G2(Tx,Sy,Rz)=449≤α⋅1⋅34+β⋅34⋅17+γ⋅1⋅17≤αG(x,Tx,Sy)G(y,Sy,Rz)+βG(y,Sy,Rz)G(z,Rz,Tx)+γG(x,Tx,Sy)G(z,Rz,Tx).Case 2.
Ifx∈[0,1/2],y∈(1/2,1],z∈[0,1], then we can get
(2.67)G2(Tx,Sy,Rz)=G2(1,67,67)=449,G(x,Tx,Sy)=G(x,1,67)=|x-1|+|x-67|+17≥12+514+17=1,G(y,Sy,Rz)=G(y,67,67)=|y-67|+|y-67|≥0+0=0,G(z,Rz,Tx)=G(z,67,1)=|z-67|+17+|z-1|≥0+17+0=17.
Thus, we have
(2.68)G2(Tx,Sy,Rz)=449≤α⋅1⋅0+β⋅0⋅17+γ⋅1⋅17≤αG(x,Tx,Sy)G(y,Sy,Rz)+βG(y,Sy,Rz)G(z,Rz,Tx)+γG(x,Tx,Sy)G(z,Rz,Tx).Case 3.
Ifx∈(1/2,1],y∈[0,1/2],z∈[0,1], then we have
(2.69)G2(Tx,Sy,Rz)=G2(67,78,67)=1784,G(x,Tx,Sy)=G(x,67,78)=|x-67|+|x-78|+156≥0+0+156=156,G(y,Sy,Rz)=G(y,78,67)=|y-78|+|y-67|+156≥38+514+156=34,G(z,Rz,Tx)=G(z,67,67)=|z-67|+|z-67|≥0+0=0.
Thus, we have
(2.70)G2(Tx,Sy,Rz)=1784≤α⋅156⋅34+β⋅34⋅0+γ⋅156⋅0≤αG(x,Tx,Sy)G(y,Sy,Rz)+βG(y,Sy,Rz)G(z,Rz,Tx)+γG(x,Tx,Sy)G(z,Rz,Tx).Case 4.
Ifx,y∈(1/2,1],z∈[0,1], then we have
(2.71)G2(Tx,Sy,Rz)=G2(67,67,67)=0.
Thus, we have
(2.72)G2(Tx,Sy,Rz)=0≤αG(x,Tx,Sy)G(y,Sy,Rz)+βG(y,Sy,Rz)G(z,Rz,Tx)+γG(x,Tx,Sy)G(z,Rz,Tx).Then in all of the above cases, the mappingsT,S, and R satisfy the contractive condition (2.33) of Theorem 2.5 with α=1/7,β=1/7,γ=4/7. So that all the conditions of Theorem 2.5 are satisfied. Moreover, 6/7 is the unique common fixed point for all of the three mappings T,S, and R.At last, we proveT,S, and R are all G-continuous at the common fixed point 6/7. Since 6/7∈(1/2,1], and let the sequence {yn}⊂(0,1] and yn→(6/7)(n→∞), then there exists N∈ℕ such that {yn}⊂(1/2,1], for all n>N. Without loss of generality, we can assume that {yn}⊂(1/2,1], and so Tyn=6/7,Syn=6/7 and Ryn=6/7. Therefore,
(2.73)lim⁡n→∞Tyn=lim⁡n→∞Syn=lim⁡n→∞Ryn=67.
Which implies that T,S, and R are all G-continuous at the common fixed point 6/7.Corollary 2.7.
Let(X,G) be a complete G-metric space. Suppose the three self-mappings T,S,R:X→X satisfy the condition:
(2.74)G2(Tpx,Ssy,Rrz)≤αG(x,Tpx,Ssy)G(y,Ssy,Rrz)+βG(y,Ssy,Rrz)G(z,Rrz,Tpx)+γG(x,Tpx,Ssy)G(z,Rrz,Tpx),
for all x,y,z∈X, where p,s,r∈ℕ, α,β,γ are nonnegative real numbers and α+β+γ<1. Then T,S, and R have a unique common fixed point (say u) and Tp,Ss,Rr are all G-continuous at u.Proof.
Since the proof of Corollary2.7 is very similar to that of Corollary 2.2, so we delete it.Corollary 2.8.
Let(X,G) be a complete G-metric space, and let T:X→X be a self-mapping in X, which satisfies the following condition:
(2.75)G2(Tx,Ty,Tz)≤αG(x,Tx,Ty)G(y,Ty,Tz)+βG(y,Ty,Tz)G(z,Tz,Tx)+γG(x,Tx,Ty)G(z,Tz,Tx).
for all x,y,z∈X, where α,β,γ are nonnegative real numbers and α+β+γ<1. Then T has a unique fixed point (say u) and T is G-continuous at u.Corollary 2.9.
Let(X,G) be a complete G-metric space, and let T:X→X be a self-mapping in X, which satisfies the following condition:
(2.76)G2(Tpx,Tpy,Tpz)≤αG(x,Tpx,Tpy)G(y,Tpy,Tpz)+βG(y,Tpy,Tpz)G(z,Tpz,Tpx)+γG(x,Tpx,Tpy)G(z,Tpz,Tpx).
for all x,y,z∈X, where p∈N,α,β,γ are nonnegative real numbers and α+β+γ<1. Then T has a unique fixed point (say u) and Tp is G-continuous at u.



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*Source: 736214-2012-08-29.xml*