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 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.

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 prove u 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 that T 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 lim n→∞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 Theorem 2.1 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.
Let T=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.
Let T=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 for p≠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 prove u 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 that T 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 lim n→∞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 Theorem 2.5.

Example 2.6.
Let X=[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 mappings T, 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.
If x,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.
If x∈[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.
If x∈(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.
If x, 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 mappings T,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 prove T,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)limn→∞Tyn=limn→∞Syn=limn→∞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 Corollary 2.7 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.