# Common Fixed Point Theorems for Six Mappings in Generalized Metric Spaces

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

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## Abstract

By using the weakly commutative and weakly compatible conditions
of self-mapping pairs, we prove some new common fixed point theorems for six self-mappings in the framework of generalized metric spaces. An example is provided to support our result. The results presented in this paper generalize the well-known comparable results in the literature due to Abbas, Nazir, Saadati, Mustafa, and Sims.


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## Body

## 1. Introduction and Preliminaries

The study of fixed points of mappings satisfying certain conditions has been at the center of vigorous research activity. In 2006,  Mustafa and Sims [1] introduced a new structure of generalized metric spaces, which are called G-metric spaces as follows.Definition 1.1 (see [1]).
LetX be a nonempty set and let G:X×X×X→R+ be a function satisfying the following properties:(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) is called a G-metric space.Since then the fixed point theory inG-metric spaces has been studied and developed by authors (see [2–43]). Fixed point problems have also been considered in partially ordered G-metric spaces (see [44–56]).The purpose of this paper is to use the concept of weakly commuting mappings and weakly compatible mappings to discuss some new common fixed point problem for six self-mappings inG-metric spaces. The results presented in this paper extend and improve the corresponding results of Abbas et al. [2] and Mustafa and Sims [3].We now recall some of the basic concepts and results inG-metric spaces.Proposition 1.2 (see [1]).
Let(X,G) be a G-metric space, then the function G(x,y,z) is jointly continuous in three of its variables.Definition 1.3 (see [1]).
Let(X,G) be a G-metric space, and let (xn) be a sequence of points of X. A point x∈X is said to be the limit of the sequence (xn), if limn,m→+∞G(x,xn,xm)=0, and we say that the sequence (xn) is G-convergent to x or  (xn)G-convergent to x, that is, for any ϵ>0, there exists N∈ℕ such that G(x,xn,xm)<ϵ for all m,n≥N(throughout this paper we mean by ℕ the set of all natural numbers).Proposition 1.4 (see [1]).
Let(X,G) be a G-metric space, then the following are equivalent:(i)
(xn) is  G-convergent to  x;(ii)
G(xn,xn,x)→0 as  n→+∞;(iii)
G(xn,x,x)→0 as  n→+∞;(iv)
G(xn,xm,x)→0 as  n,m→+∞.Definition 1.5 (see [1]).
Let(X,G) be a G-metric space. A sequence (xn) is called G-cauchy if for every ϵ>0, there is N∈ℕ such that G(xn,xm,xl)<ϵ for all m,n,l≥N, that is G(xn,xm,xl)→0 as n,m,l→+∞.Proposition 1.6 (see [1]).
Let(X,G) be a G-metric space, then the following are equivalent:(i)
the sequence(xn) is  G-cauchy;(ii)
for everyϵ>0, there is  N∈ℕ such that  G(xn,xm,xm)<ϵ for all  m,n≥N.Definition 1.7 (see [1]).
AG-metric space (X,G) is G-complete if every G-cauchy sequence in (X,G) is G-convergent in X.Definition 1.8 (see [1]).
Let(X,G) and (X',G') be G-metric spaces, and let 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 spaces, then a function f:X→X' is G-continuous at a point 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).Definition 1.10 (see [4]).
Two self-mappingsf and g of a G-metric space (X,G) are said to be weakly commuting if G(fgx,gfx,gfx)≤G(fx,gx,gx) for all x in X.Definition 1.11 (see [4]).
Letf and g be two self-mappings from a G-metric space (X,G) into itself. Then the mappings f and g are said to be weakly compatible if G(fgx,gfx,gfx)=0 whenever G(fx,gx,gx)=0.Proposition 1.12 (see [1]).
Let(X,G) be a G-metric space. Then, for all x,y,z,a in X, it follows that(i)
ifG(x,x,y)=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. Common Fixed Point Theorems

Theorem 2.1.
Let(X,G) be a complete G-metric space, and let f, g, h, A, B, and C be six mappings of X into itself satisfying the following conditions:(i)
f(X)⊂B(X), g(X)⊂C(X), h(X)⊂A(X);(ii)
forallx,y,z∈X,
(2.1)G(fx,gy,hz)≤kmax{G(Ax,gy,gy)+G(By,fx,fx),G(By,hz,hz)+G(Cz,gy,gy),G(Cz,fx,fx)+G(Ax,hz,hz)}
or
(2.2)G(fx,gy,hz)≤kmax{G(Ax,Ax,gy)+G(By,By,fx),G(By,By,hz)+G(Cz,Cz,gy),G(Cz,Cz,fx)+G(Ax,Ax,hz)},
where k∈[0,1/3). Then one of the pairs (f,A), (g,B), and (h,C) has a coincidence point in X. Further, if one of the following conditions is satisfied, then the mappings f, g, h, A, B, and C have a unique common fixed point in X.(a)
Eitherf or A is G-continuous, the pair (f,A) is weakly commutative, the pairs (g,B) and (h,C) are weakly compatible;(b)
Eitherg or B is G-continuous, the pair (g,B) is weakly commutative, the pairs (f,A) and (h,C) are weakly compatible;(c)
Eitherh or C is G-continuous, the pair (h,C) is weakly commutative, the pairs (f,A) and (g,B) are weakly compatible.Proof.
Suppose that mappingsf, g, h, A, B, and C satisfy condition (2.1).
Letx0 in X be arbitrary point, since (X)⊂B(X), g(X)⊂C(X), h(X)⊂A(X), there exist the sequences {xn} and {yn} in X such that
(2.3)y3n=fx3n=Bx3n+1,y3n+1=gx3n+1=Cx3n+2,y3n+2=hx3n+2=Ax3n+3,
for n=0,1,2,….
Ifyn=yn+1 for some n, with n=3m, then p=x3m+1 is a coincidence point of the pair (g,B); if yn+1=yn+2 for some n, with n=3m, then p=x3m+2 is a coincidence point of the pair (h,C); if yn+2=yn+3 for some n, with n=3m, then p=x3m+3 is a coincidence point of the pair (f,A). Without loss of generality, we can assume that yn≠yn+1 for all n=0,1,2….
Now we prove that{yn} is a G-cauchy sequence in X.
Actually, using the condition (2.1) and (G3), we have
(2.4)G(y3n-1,y3n,y3n+1)=G(fx3n,gx3n+1,hx3n-1)≤kmax{G(Ax3n,gx3n+1,gx3n+1)+G(Bx3n+1,fx3n,fx3n),G(Bx3n+1,hx3n-1,hx3n-1)+G(Cx3n-1,gx3n+1,gx3n+1),G(Cx3n-1,fx3n,fx3n)+G(A3n,hx3n-1,hx3n-1)}=kmax{G(y3n-1,y3n+1,y3n+1)+G(y3n,y3n,y3n),G(y3n,y3n-1,y3n-1)+G(y3n-2,y3n+1,y3n+1),G(y3n-2,y3n,y3n)+G(y3n-1,y3n-1,y3n-1)}≤kmax{G(y3n-1,y3n,y3n+1),G(y3n-2,y3n-1,y3n),G(y3n-1,y3n,y3n+1)+[G(y3n-2,y3n-1,y3n-1)+G(y3n-1,y3n+1,y3n+1)]}≤kmax{G(y3n-1,y3n,y3n+1),G(y3n-2,y3n-1,y3n),2G(y3n-1,y3n,y3n+1)+G(y3n-2,y3n-1,y3n)}=k[2G(y3n-1,y3n,y3n+1)+G(y3n-2,y3n-1,y3n)],
which further implies that
(2.5)(1-2k)G(y3n-1,y3n,y3n+1)≤kG(y3n-2,y3n-1,y3n).
Thus
(2.6)G(y3n-1,y3n,y3n+1)≤λG(y3n-2,y3n-1,y3n),
where λ=k/(1-2k). Obviously 0≤λ<1.
Similarly it can be shown that(2.7)G(y3n,y3n+1,y3n+2)≤λG(y3n-1,y3n,y3n+1),G(y3n+1,y3n+2,y3n+3)≤λG(y3n,y3n+1,y3n+2).
It follows from (2.6) and (2.7) that, for all n∈ℕ,
(2.8)G(yn,yn+1,yn+2)≤λG(yn-1,yn,yn+1)≤λ2G(yn-2,yn-1,yn)≤⋯≤λnG(y0,y1,y2).
Therefore, for all n,m∈ℕ, n<m, by (G3) and (G5) we have
(2.9)G(yn,ym,ym)≤G(yn,yn+1,yn+1)+G(yn+1,yn+2,yn+2)+G(yn+2,yn+3,yn+3)+⋯+G(ym-1,ym,ym)≤G(yn,yn+1,yn+2)+G(yn+1,yn+2,yn+3)+⋯+G(ym-1,ym,ym+1)≤(λn+λn+1+λn+2+⋯+λm-1)G(y0,y1,y2)≤λn1-λG(y0,y1,y2)→0,asn→∞.
Hence {yn} is a G-cauchy sequence in X, since X is complete G-metric space, there exists a point u∈X such that yn→u(n→∞).
Since the sequences{fx3n}={Bx3n+1}, {gx3n+1}={Cx3n+2}, and {hx3n-1}={Ax3n} are all subsequences of {yn}, then they all converge to u, that is,
(2.10)y3n=fx3n=Bx3n+1→u,y3n+1=gx3n+1=Cx3n+2→u,y3n-1=hx3n-1=Ax3n→u(n→∞).
Now we prove thatu is a common fixed point of f, g, h, A, B, and C under the condition (a).
First, we suppose thatA is continuous, the pair (f,A) is weakly commutative, the pairs (g,B) and (h,C) are weakly compatible.
Step  1.We prove that u=fu=Au.
By (2.10) and weakly commutativity of mapping pair (f,A), we have
(2.11)G(fAx3n,Afx3n,Afx3n)≤G(fx3n,Ax3n,Ax3n)→0(n→∞).
Since A is continuous, then A2x3n→Au(n→∞), Afx3n→Au(n→∞). By (2.11), we know that fAx3n→Au(n→∞).
From the condition (2.1), we get
(2.12)G(fAx3n,gx3n+1,hx3n+2)≤kmax{G(A2x3n,gx3n+1,gx3n+1)+G(Bx3n+1,fAx3n,fAx3n),G(Bx3n+1,hx3n+2,hx3n+2)+G(Cx3n+2,gx3n+1,gx3n+1),G(Cx3n+2,fAx3n,fAx3n)+G(A2x3n,hx3n+2,hx3n+2)}.
Letting n→∞ and using the Proposition 1.12(iii), we have
(2.13)G(Au,u,u)≤kmax{G(Au,u,u)+G(u,Au,Au),G(u,u,u)+G(u,u,u),G(u,Au,Au)+G(Au,u,u)}=k[G(Au,u,u)+G(u,Au,Au)]≤3kG(Au,u,u).
Hence, G(Au,u,u)=0 and Au=u, since 0≤k<1/3.
Again by using condition (2.1), we have
(2.14)G(fu,gx3n+1,hx3n+2)≤kmax{G(Au,gx3n+1,gx3n+1)+G(Bx3n+1,fu,fu),G(Bx3n+1,hx3n+2,hx3n+2)+G(Cx3n+2,gx3n+1,gx3n+1),G(Cx3n+2,fu,fu)+G(Au,hx3n+2,hx3n+2)}.
Letting n→∞, we have
(2.15)G(fu,u,u)≤kG(u,fu,fu).
From the Proposition 1.12(iii), we get
(2.16)G(fu,u,u)≤kG(u,fu,fu)≤2kG(fu,u,u).
Hence,G(fu,u,u)=0 and fu=u, since 0≤k<1/3.
So we haveu=Au=fu.
Step  2. We prove that u=gu=Bu.
Sincef(X)⊂B(X) and u=fu∈f(X), there is a point v∈X such that u=fu=Bv. Again by using condition (2.1), we have
(2.17)G(fu,gv,hx3n+2)≤kmax{G(Au,gv,gv)+G(Bv,fu,fu),G(Bv,hx3n+2,hx3n+2)+G(Cx3n+2,gv,gv),G(Cx3n+2,fu,fu)+G(Au,hx3n+2,hx3n+2)}.
Letting n→∞, using u=Au=fu and the Proposition 1.12(iii), we obtain
(2.18)G(u,gv,u)≤kG(u,gv,gv)≤2kG(u,gv,u).
Hence, G(u,gv,u)=0and so gv=u=Bv, since 0≤k<1/3.
Since the pair(g,B) is weakly compatible, we have
(2.19)gu=gBv=Bgv=Bu.
Again by using condition (2.1), we have
(2.20)G(fu,gu,hx3n+2)≤kmax{G(Au,gu,gu)+G(Bu,fu,fu),G(Bu,hx3n+2,hx3n+2)+G(Cx3n+2,gu,gu),G(Cx3n+2,fu,fu)+G(Au,hx3n+2,hx3n+2)}.
Letting n→∞, using u=Au=fu, gu=Bu and the Proposition 1.12(iii), we have
(2.21)G(u,gu,u)≤k[G(u,gu,gu)+G(u,gu,u)]≤3kG(u,gu,u).
Hence, G(u,gu,u)=0  and so u=gu=Bu, since 0≤k<1/3.
So we haveu=gu=Bu.
Step  3. We prove that u=hu=Cu.
Sinceg(X)⊂C(X) and u=gu∈g(X), there is a point w∈X such that u=gu=Cw. Again by using condition (2.1), we have
(2.22)G(fu,gu,hw)≤kmax{G(Au,gu,gu)+G(Bu,fu,fu),G(Bu,hw,hw)+G(Cw,gu,gu),G(Cw,fu,fu)+G(Au,hw,hw)}.
Using u=Au=fu, u=gu=Bu=Cw and the Proposition 1.12(iii), we obtain
(2.23)G(u,u,hw)≤kG(u,hw,hw)≤2kG(u,u,hw).
Hence, G(u,u,hw)=  0 and so hw=u=Cw, since 0≤k<1/3.
Since the pair(h,C) is weakly compatible, we have
(2.24)hu=hCw=Chw=Cu.
Again by using condition (2.1), we have
(2.25)G(fu,gu,hu)≤kmax{G(Au,gu,gu)+G(Bu,fu,fu),G(Bu,hu,hu)+G(Cu,gu,gu),G(Cu,fu,fu)+G(Au,hu,hu)}.
Using u=Au=fu, u=gu=Bu, Cu=hu and the Proposition 1.12(iii), we have
(2.26)G(u,u,hu)≤k[G(u,hu,hu)+G(hu,u,u)]≤3kG(u,u,hu).
Hence, G(u,u,hu)=  0 and so u=hu=Cu, since 0≤k<1/3.
Therefore,u is the common fixed point of f, g, h, A, B, and C when A is continuous and the pair (f,A) is weakly commutative, the pairs (g,B) and (h,C) are weakly compatible.
Next, we suppose thatf is continuous, the pair (f,A) is weakly commutative, the pairs (g,B) and (h,C) are weakly compatible.
Step  1. We prove that u=fu.
By (2.10) and weak commutativity of mapping pair (f,A), we have
(2.27)G(fAx3n,Afx3n,Afx3n)≤G(fx3n,Ax3n,Ax3n)→0(n→∞).
Since f is continuous, then f2x3n→fu(n→∞), fAx3n→fu(n→∞). By (2.10), we know that Afx3n→fu(n→∞).
From the condition (2.1), we have
(2.28)G(f2x3n,gx3n+1,hx3n+2)≤kmax{G(Afx3n,gx3n+1,gx3n+1)+G(Bx3n+1,f2x3n,f2x3n),G(Bx3n+1,hx3n+2,hx3n+2)+G(Cx3n+2,gx3n+1,gx3n+1),G(Cx3n+2,f2x3n,f2x3n)+G(Afx3n,hx3n+2,hx3n+2)}.
Letting n→∞ and noting the Proposition 1.12(iii), we have
(2.29)G(fu,u,u)≤kmax{G(fu,u,u)+G(u,fu,fu),G(u,u,u)+G(u,u,u),G(u,fu,fu)+G(fu,u,u)}=k[G(fu,u,u)+G(u,fu,fu)]≤3kG(fu,u,u).
Hence, G(fu,u,u)=  0 and so fu=u, since 0≤k<1/3.
Step  2. We prove that u=gu=Bu.
Sincef(X)⊂B(X) and u=fu∈f(X), there is a point z∈X such that u=fu=Bz. Again by using condition (2.1), we have
(2.30)G(f2x3n,gz,hx3n+2)≤kmax{G(Afx3n,gz,gz)+G(Bz,f2x3n,f2x3n),G(Bz,hx3n+2,hx3n+2)+G(Cx3n+2,gz,gz),G(Cx3n+2,f2x3n,f2x3n)+G(Afx3n,hx3n+2,hx3n+2)}.
Letting n→∞and  using u=fu and the Proposition 1.12(iii), we have
(2.31)G(u,gz,u)≤kG(u,gz,gz)≤2kG(u,gz,u).
Hence G(u,gz,u)=  0 and so gz=u=Bz, since 0≤k<1/3.
Since the pair(g,B) is weakly compatible, we have
(2.32)gu=gBz=Bgz=Bu.
Again by using condition (2.1), we have
(2.33)G(fx3n,gu,hx3n+2)≤kmax{G(Ax3n,gu,gu)+G(Bu,fx3n,fx3n),G(Bu,hx3n+2,hx3n+2)+G(Cx3n+2,gu,gu),G(Cx3n+2,fx3n,fx3n)+G(Ax3n,hx3n+2,hx3n+2)}.
Letting n→∞and using u=fu, gu=Bu and the Proposition 1.12(iii), we have
(2.34)G(u,gu,u)≤k[G(u,gu,gu)+G(gu,u,u)]≤3kG(u,gu,u).
Hence, G(u,gu,u)=0 and so gu=u=Bu, since 0≤k<1/3.
So we haveu=gu=Bu.
Step  3.We prove that u=hu=Cu.
Sinceg(X)⊂C(X) and u=gu∈g(X), there is a point t∈X such that u=gu=Ct. Again by using condition (2.1), we have
(2.35)G(fx3n,gu,ht)≤kmax{G(Ax3n,gu,gu)+G(Bu,fx3n,fx3n),G(Bu,ht,ht)+G(Ct,gu,gu),G(Ct,fx3n,fx3n)+G(Ax3n,ht,ht)}.
Letting n→∞and using u=gu=Bu and the Proposition 1.12(iii), we obtain
(2.36)G(u,u,ht)≤kG(u,ht,ht)≤2kG(u,u,ht).
Hence, G(u,u,ht)=0 and so ht=u=Ct, since 0≤k<1/3.
Since the pair(h,C) is weakly compatible, we have
(2.37)hu=hCt=Cht=Cu.
Again by using  condition (2.1), we have
(2.38)G(fx3n,gu,hu)≤kmax{G(Ax3n,gu,gu)+G(Bu,fx3n,fx3n),G(Bu,hu,hu)+G(Cu,gu,gu),G(Cu,fx3n,fx3n)+G(Ax3n,hu,hu)}.
Letting n→∞and using u=gu=Bu and the Proposition 1.12(iii), we have
(2.39)G(u,u,hu)≤k[G(u,hu,hu)+G(u,u,hu)]≤3kG(u,u,hu).
Hence, G(u,u,hu)=0 and so hu=u=Cu, since 0≤k<1/3.
Step  4. We prove that u=Au.
Sinceh(X)⊂A(X) and u=hu∈h(X), there is a point p∈X such that u=hu=Ap. Again by using condition (2.1), we have
(2.40)G(fp,gu,hu)≤kmax{G(Ap,gu,gu)+G(Bu,fp,fp),G(Bu,hu,hu)+G(Cu,gu,gu),G(Cu,fp,fp)+G(Ap,hu,hu)}.
Using u=gu=Bu, u=hu=Cu, and the Proposition 1.12(iii), we obtain
(2.41)G(fp,u,u)≤kG(u,fp,fp)≤2kG(fp,u,u).
Hence G(fp,u,u)=0 and fp=u=Ap, since 0≤k<1/3.
Since the pair(f,A) is weakly compatible, we have
(2.42)fu=fAp=Afp=Au=u.
Therefore, u is the common fixed point of f, g, h, A, B, and C when f is continuous and the pair (f,A) is weakly commutative, the pairs (g,B) and (h,C) are weakly compatible.
Similarly, we can prove the result thatu is a common fixed point of f, g, h, A, B, and C when under the condition of (b) or (c).
Finally, we prove uniqueness of common fixed pointu.
Letu and q be two common fixed points of f, g, h, A, B, and C, by using condition (2.1), we have
(2.43)G(q,u,u)=G(fq,gu,hu)≤kmax{G(Aq,gu,gu)+G(Bu,fq,fq),G(Bu,hu,hu)+G(Cu,gu,gu),G(Cu,fq,fq)+G(Aq,hu,hu)}=k[G(q,u,u)+G(u,q,q)]≤3kG(q,u,u).
Hence, G(q,u,u)=0 and so q=u, since 0≤k<1/3.Thus common fixed point is unique.
The proof using (2.2) is similar. This completes the proof.Now we introduce an example to support Theorem2.1.Example 2.2.
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,andz in X. Let f, g, h, A, B, and C be self-mappings defined by
(2.44)fx={1,x∈[0,12],78,x∈(12,1],gx={1011,x∈[0,12],78,x∈(12,1],hx={910,x∈[0,12],78,x∈(12,1],Ax=x,Bx={1,x∈[0,12],78,x∈(12,1),0,x=1,Cx={1,x∈[0,12],78,x∈(12,1),1011,x=1.
Note that A is G-continuous in X, and f, g, h, B, and C are not G-continuous in X.Clearly we can getf(X)⊂B(X), g(X)⊂C(X),  and h(X)⊂A(X).Actually, sincefX={7/8,1}, BX={0,7/8,1}, gX={7/8,10/11}, CX={7/8,10/11,1}, hX = {7/8,9/10},  and AX=X=[0,1], so we know f(X)⊂B(X), g(X)⊂C(X), and h(X)⊂A(X).By the definition of the mappings off and A, for all x∈[0,1], G(fAx,Afx,Afx) = G(fx,fx,fx)=0≤G(fx,Ax,Ax), so we can get the pair (f,A) is weakly commuting.By the definition of the mappings ofg and B, only for x∈(1/2,1), gx=Bx=7/8, at this time gBx=g(7/8) = 7/8 = B(7/8) = Bgx, so gBx=Bgx, so we can obtain that the pair (g,B) is weakly compatible. Similarly, we can show that the pair (h,C) is also weakly compatible.Now we proof that the mappingsf, g, h, A, B, and C are satisfying the condition (2.1) of Theorem 2.1 with k=5/16∈[0,1/3). Let
(2.45)M(x,y,z)=max{G(Ax,gy,gy)+G(By,fx,fx),G(By,hz,hz)+G(Cz,gy,gy),G(Cz,fx,fx)+G(Ax,hz,hz)}.Case 1.
Ifx,y,z∈[0,1/2], then
(2.46)G(fx,gy,hz)=G(1,1011,910)=15,G(Ax,gy,gy)+G(By,fx,fx)=G(x,1011,1011)+G(1,1,1)=2|x-1011|≥911.
Thus, we have
(2.47)G(fx,gy,hz)=15<516·911≤k(G(Ax,gy,gy)+G(By,fx,fx))≤kM(x,y,z).Case 2.
Ifx,y∈[0,1/2],z∈(1/2,1], then
(2.48)G(fx,gy,hz)=G(1,1011,78)=14,G(Ax,gy,gy)+G(By,fx,fx)=G(x,1011,1011)+G(1,1,1)=2|x-1011|≥911.
Therefore, we get
(2.49)G(fx,gy,hz)=14<516·911≤k(G(Ax,gy,gy)+G(By,fx,fx))≤kM(x,y,z).Case 3.
Ifx,z∈[0,1/2],y∈(1/2,1], then
(2.50)G(fx,gy,hz)=G(1,78,910)=14,G(Cz,fx,fx)+G(Ax,hz,hz)=G(1,1,1)+G(x,910,910)=2|x-910|≥45.
Hence, we have
(2.51)G(fx,gy,hz)=14=516·45≤k(G(Cz,fx,fx)+G(Ax,hz,hz))≤kM(x,y,z).Case 4.
Ify,z∈[0,1/2],x∈(1/2,1], then
(2.52)G(fx,gy,hz)=G(78,1011,910)=344,G(By,hz,hz)+G(Cz,gy,gy)=G(1,910,910)+G(1,1011,1011)=2155.
So we get
(2.53)G(fx,gy,hz)=344<516·2155≤k(G(By,hz,hz)+G(Cz,gy,gy))≤kM(x,y,z).Case 5.
Ifx∈[0,1/2],y,z∈(1/2,1], then
(2.54)G(fx,gy,hz)=G(1,78,78)=14.
If y∈(1/2,1), then
(2.55)G(Ax,gy,gy)+G(By,fx,fx)=G(x,78,78)+G(78,1,1)=2|x-78|+14≥34+14=1.
If y=1, then
(2.56)G(Ax,gy,gy)+G(By,fx,fx)=G(x,78,78)+G(0,1,1)=2|x-78|+2≥34+2=114.
And so we have
(2.57)G(Ax,gy,gy)+G(By,fx,fx)≥1
for all y∈(1/2,1]. Hence we have
(2.58)G(fx,gy,hz)=14<516·1≤k(G(Ax,gy,gy)+G(By,fx,fx))≤kM(x,y,z).Case 6.
Ify∈[0,1/2],x,z∈(1/2,1], then
(2.59)G(fx,gy,hz)=G(78,1011,78)=344,G(Ax,gy,gy)+G(By,fx,fx)=G(x,1011,1011)+G(1,78,78)=2|x-1011|+14≥14.
Thus, we have
(2.60)G(fx,gy,hz)=344<516·14≤k(G(Ax,gy,gy)+G(By,fx,fx))≤kM(x,y,z).Case 7.
Ifz∈[0,1/2],x,y∈(1/2,1], then
(2.61)G(fx,gy,hz)=G(78,78,910)=120.
If y∈(1/2,1), then
(2.62)G(By,hz,hz)+G(Cz,gy,gy)=G(78,910,910)+G(1,78,78)=310.
If y=1, then
(2.63)G(By,hz,hz)+G(Cz,gy,gy)=G(0,910,910)+G(1,78,78)=4120.
And so we have
(2.64)G(By,hz,hz)+G(Cz,gy,gy)≥310
for all y∈(1/2,1]. Hence we have
(2.65)G(fx,gy,hz)=120<516·310≤k(G(By,hz,hz)+G(Cz,gy,gy))≤kM(x,y,z).Case 8.
Ifx,y,z∈(1/2,1], then
(2.66)G(fx,gy,hz)=G(78,78,78)=0≤516M(x,y,z)=kM(x,y,z).
Then in all the above cases, the mappings f, g, h, A, B, and C are satisfying the condition (2.1) of  Theorem 2.1 with k=5/16 so that all the conditions of Theorem 2.1 are satisfied. Moreover, 7/8 is the unique common fixed point for all of the mappings f, g, h, A, B, and C.In Theorem2.1, if we take A=B=C=I (I is identity mapping, the same as  below), then we have the following corollary.Corollary 2.3 (see [2, Theorem 2.4]).
Let(X,G) be a complete G-metric space, and let f, g, and h be three mappings of X into itself satisfying the following conditions:
(2.67)G(fx,gy,hz)≤kmax{G(x,gy,gy)+G(y,fx,fx),G(y,hz,hz)+G(z,gy,gy),G(z,fx,fx)+G(x,hz,hz)}
or
(2.68)G(fx,gy,hz)≤kmax{G(x,x,gy)+G(y,y,fx),G(y,y,hz)+G(z,z,gy),G(z,z,fx)+G(x,x,hz)}forallx,y,z∈X, where k∈[0,1/3).Then f, g, and h have a unique common fixed point in X.Also, if we takef=g=h and A=B=C=I in Theorem 2.1, then we get the following.Corollary 2.4 (see [3, Theorem  2.4]).
Let(X,G) be a complete G-metric space, and let f be a mapping of X into itself satisfying the following conditions:
(2.69)G(fx,fy,fz)≤kmax{G(x,fy,fy)+G(y,fx,fx),G(y,fz,fz)+G(z,fy,fy),G(z,fx,fx)+G(x,fz,fz)}
or
(2.70)G(fx,fy,fz)≤kmax{G(x,x,fy)+G(y,y,fx),G(y,y,fz)+G(z,z,fy),G(z,z,fx)+G(x,x,fz)},forallx,y,z∈X, where k∈[0,1/3). Then, f has  a unique fixed point in X.Remark 2.5.
Theorem2.1 and Corollaries 2.3 and 2.4 generalize and extend the corresponding results of Abbas and Rhoades [5] and Mustafa et al. [6].Remark 2.6.
In Theorem2.1, if we take: (1) f=g=h; (2) A=B=C; (3) g=h and B=C; (4) g=h, B=C=I, several new results can be obtained.Theorem 2.7.
Let(X,G) be a complete G-metric space, and let f, g, h, A, B, and C be six mappings of X into itself satisfying the following conditions:(i)
f(X)⊂B(X), g(X)⊂C(X), h(X)⊂A(X);(ii)
the pairs(f,A), (g,B), and (h,C) are commutative mappings;(iii)
forallx,y,z∈X,
(2.71)G(fmx,gmy,hmz)≤kmax{G(Ax,gmy,gmy)+G(By,fmx,fmx),G(By,hmz,hmz)+G(Cz,gmy,gmy),G(Cz,fmx,fmx)+G(Ax,hmz,hmz)}or
(2.72)G(fmx,gmy,hmz)≤kmax{G(Ax,Ax,gmy)+G(By,By,fmx),G(By,By,hmz)+G(Cz,Cz,gmy),G(Cz,Cz,fmx)+G(Ax,Ax,hmz)},
where k∈[0,1/2),m∈ℕ, then f, g, h, A, B, and C have a unique common fixed point in X.Proof.
Suppose that mappingsf, g, h, A, B, and C satisfy the condition (2.71). Since fmX⊂fm-1X⊂⋯⊂fX, fX⊂BX so that fmX⊂BX. Similarly, we can show that gmX⊂CX and hmX⊂AX. From the the Theorem 2.1, we see that fm, gm, hm, A, B, and C have a unique common fixed point u.
Sincefu=f(fmu)=fm+1u=fm(fu), so that
(2.73)G(fmfu,gmu,hmu)≤kmax{G(Afu,gmu,gmu)+G(Bu,fmfu,fmfu),G(Bu,hmu,hmu)+G(Cu,gmu,gmu),G(Cu,fmfu,fmfu)+G(Afu,hmu,hmu)},
note that Afu=fAu=fu and the Proposition 1.12(iii), we obtain
(2.74)G(fu,u,u)≤kmax{G(fu,u,u)+G(u,fu,fu),G(u,u,u)+G(u,u,u),G(u,fu,fu)+G(fu,u,u)}=k[G(u,fu,fu)+G(fu,u,u)]≤3kG(fu,u,u).
Since k∈[0,1/3), hence G(fu,u,u)=0 and so fu=u.
By the same argument, we can prove thatgu=u and hu=u. Thus, we have u = fu = gu = hu = Au = Bu = Cu so that f, g, h, A, B, and C have a common fixed point u in X. Let v be any other common fixed point of f, g, h, A, B, and C, then by using condition (2.71), we have
(2.75)G(u,u,v)=G(fmu,gmu,hmv)≤kmax{G(Au,gmu,gmu)+G(Bu,fmu,fmu),G(Bu,hmv,hmv)+G(Cv,gmu,gmu),G(Cv,fmu,fmu)+G(Au,hmv,hmv)}=kmax{G(u,u,u)+G(u,u,u),G(u,v,v)+G(v,u,u),G(v,u,u)+G(u,v,v)}=k[G(u,v,v)+G(v,u,u)]≤3kG(u,u,v).
Hence, G(u,u,v)=0 and so u=v, since 0≤k<1/3. Thus, common fixed point is unique.
The proof using (2.72) is similar. This completes the proof.In Theorem2.7, if we take A=B=C=I, then we have the following corollary.Corollary 2.8 (see [2, Corollary  2.5]).
Let(X,G) be a complete G-metric space, and let f, g, and h be three mappings of X into itself satisfying the following conditions:
(2.76)G(fmx,gmy,hmz)≤kmax{G(x,gmy,gmy)+G(y,fmx,fmx),G(y,hmz,hmz)+G(z,gmy,gmy),G(z,fmx,fmx)+G(x,hmz,hmz)}
or
(2.77)G(fmx,gmy,hmz)≤kmax{G(x,x,gmy)+G(y,y,fmx),G(y,y,hmz)+G(z,z,gmy),G(z,z,fmx)+G(x,x,hmz)}forallx,y,z∈X, where k∈[0,1/3), m∈ℕ; then f, g, and h have a unique common fixed point in X.Also, if we takef=g=h and A=B=C=I in Theorem 2.7, then we get the following.Corollary 2.9 (see [3, Corollary  2.5]).
Let(X,G) be a complete G-metric space, and let f be a mapping of X into itself satisfying the following conditions:
(2.78)G(fmx,fmy,fmz)≤kmax{G(x,fmy,fmy)+G(y,fmx,fmx),G(y,fmz,fmz)+G(z,fmy,fmy),G(z,fmx,fmx)+G(x,fmz,fmz)}
or
(2.79)G(fmx,fmy,fmz)≤kmax{G(x,x,fmy)+G(y,y,fmx),G(y,y,fmz)+G(z,z,fmy),G(z,z,fmx)+G(x,x,fmz)}forallx,y,z∈X, where k∈[0,1/3),m∈ℕ; then f has  a unique fixed point in X.Remark 2.10.
In Theorem2.7, if we take: (1) f=g=h; (2) g=h and B=C; (3) g=h, B=C=I, several new results can be obtained.Corollary 2.11.
Let(X,G) be a complete G-metric space, and let f, g, h, A, B, and C be six mappings of X into itself satisfying the following conditions:(i)
f(X)⊂B(X), g(X)⊂C(X), h(X)⊂A(X);(ii)
forallx,y,z∈X,
(2.80)G(fx,gy,hz)≤a{G(Ax,gy,gy)+G(By,fx,fx)}+b{G(By,hz,hz)+G(Cz,gy,gy)}+c{G(Cz,fx,fx)+G(Ax,hz,hz)}or(2.81)G(fx,gy,hz)≤a{G(Ax,Ax,gy)+G(By,By,fx)}+b{G(By,By,hz)+G(Cz,Cz,gy)}+c{G(Cz,Cz,fx)+G(Ax,Ax,hz)},where0≤a+b+c<1/3.Then one of the pairs (f,A), (g,B), and (h,C) has a coincidence point in X. Further, if one of the following conditions is satisfied, then the mappings f, g, h, A, B, and C have a unique common fixed point in X.(a)
Eitherf or A is G-continuous, the pair  (f,A) is weakly commutative, the pairs (g,B) and (h,C) are weakly compatible;(b)
Eitherg or B is G-continuous, the pair (g,B) is weakly commutative, the pairs (f,A) and (h,C) are weakly compatible;(c)
Eitherh or C is G-continuous, the pair (h,C) is weakly commutative, the pairs (f,A) and (g,B) are weakly compatible.Proof.
Suppose that mappingsf, g, h, A, B, and C satisfy the condition (2.80). For x,y,z∈X, let
(2.82)M(x,y,z)=max{G(Ax,gy,gy)+G(By,fx,fx),G(By,hz,hz)+G(Cz,gy,gy),G(Cz,fx,fx)+G(Ax,hz,hz)}.
Then
(2.83)a{G(Ax,gy,gy)+G(By,fx,fx)}+b{G(By,hz,hz)+G(Cz,gy,gy)}+c{G(Cz,fx,fx)+G(Ax,hz,hz)}≤(a+b+c)M(x,y,z).
So, if
(2.84)G(fx,gy,hz)≤a{G(Ax,gy,gy)+G(By,fx,fx)}+b{G(By,hz,hz)+G(Cz,gy,gy)}+c{G(Cz,fx,fx)+G(Ax,hz,hz)},
then G(fx,gy,hz)≤(a+b+c)M(x,y,z). Taking k=a+b+c in Theorem 2.1, the conclusion of Corollary 2.11 can be obtained from Theorem 2.1 immediately.
The proof using (2.81) is similar. This completes the proof.Corollary 2.12.
Let(X,G) be a complete G-metric space, and let f, g, h, A, B, and C be six mappings of X into itself satisfying the following conditions:(i)
f(X)⊂B(X),  g(X)⊂C(X),  h(X)⊂A(X);(ii)
the pairs(f,A),  (g,B) and  (h,C) are commutative mappings;(iii)
forallx,y,z∈X,
(2.85)G(fmx,gmy,hmz)≤a{G(Ax,gmy,gmy)+G(By,fmx,fmx)}+b{G(By,hmz,hmz)+G(Cz,gmy,gmy)}+c{G(Cz,fmx,fmx)+G(Ax,hmz,hmz)}
or
(2.86)G(fmx,gmy,hmz)≤a{G(Ax,Ax,gmy)+G(By,By,fmx)}+b{G(By,By,hmz)+G(Cz,Cz,gmy)}+c{G(Cz,Cz,fmx)+G(Ax,Ax,hmz)},
where 0≤a+b+c<1/3, m∈ℕ; then f, g, h, A, B, and C have a unique common fixed point in X.Proof.
The proof follows from Corollary2.11, and from an argument similar to that used in Theorem 2.7.Remark 2.13.
In Corollaries2.11 and 2.12, if we take: (1) A=B=C=I; (2) f=g=h; (3) A=B=C; (4) g=h and B=C; (5) g=h, B=C=I, several new results can be obtained.



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*Source: 379212-2012-12-03.xml*