Theorem 2.1 Let $(X,G)$ be a complete G-
metric space,
and let S,
T,
R,
A,
B,
and C be six mappings of X into itself satisfying the following conditions:
- (i)
$S(X)\subset B(X)$, $T(X)\subset C(X)$, $R(X)\subset A(X)$;
- (ii)
$\mathrm{\forall}x,y,z\in X$,
${G}^{2}(Sx,Ty,Rz)\le kmax\left\{\begin{array}{c}G(Ax,Sx,Sx)G(By,Ty,Ty),\\ G(By,Ty,Ty)G(Cz,Rz,Rz),\\ G(Cz,Rz,Rz)G(Ax,Sx,Sx)\end{array}\right\}$
(2.1)
or
${G}^{2}(Sx,Ty,Rz)\le kmax\left\{\begin{array}{c}G(Ax,Ax,Sx)G(By,By,Ty),\\ G(By,By,Ty)G(Cz,Cz,Rz),\\ G(Cz,Cz,Rz)G(Ax,Ax,Sx)\end{array}\right\},$
(2.2)
where $k\in [0,1)$.
Then one of the pairs $(S,A)$,
$(T,B)$,
and $(R,C)$ has a coincidence point in X.
Moreover,
if one of the following conditions is satisfied:
- (a)
either S or A is G-continuous, the pair $(S,A)$ is weakly commuting, the pairs $(T,B)$ and $(R,C)$ are weakly compatible;
- (b)
either T or B is G-continuous, the pair $(T,B)$ is weakly commuting, the pairs $(S,A)$ and $(R,C)$ are weakly compatible;
- (c)
either F or C is G-continuous, the pair $(R,C)$ is weakly commuting, the pairs $(S,A)$ and $(T,B)$ are weakly compatible.
Then the mappings S, T, R, A, B, and C have a unique common fixed point in X.
Proof First, we suppose that the condition (2.1) holds.
Let
${x}_{0}$ in
X be an arbitrary point, since
$S(X)\subset B(X)$,
$T(X)\subset C(X)$,
$R(X)\subset A(X)$, there exist the sequences
$\{{x}_{n}\}$ and
$\{{y}_{n}\}$ in
X such that
${y}_{3n}=S{x}_{3n}=B{x}_{3n+1},\phantom{\rule{2em}{0ex}}{y}_{3n+1}=T{x}_{3n+1}=C{x}_{3n+2},\phantom{\rule{2em}{0ex}}{y}_{3n+2}=R{x}_{3n+2}=A{x}_{3n+3}$
for $n=0,1,2,\dots $ . If ${y}_{3n+2}={y}_{3n+3}$, then $Sp=Ap$ where $p={x}_{3n+3}$. If ${y}_{3n}={y}_{3n+1}$, then $Tp=Bp$ where $p={x}_{3n+1}$. If ${y}_{3n+1}={y}_{3n+2}$, then $Rp=Cp$ where $p={x}_{3n+2}$. Without loss of generality, we can assume that ${y}_{n}\ne {y}_{n+1}$, for all $n=0,1,2,\dots $ .
Now we prove that $\{{y}_{n}\}$ is a G-Cauchy sequence in X.
Actually, using condition (2.1) and (G
_{3}) we have
$\begin{array}{rcl}{G}^{2}({y}_{3n-1},{y}_{3n},{y}_{3n+1})& =& {G}^{2}(S{x}_{3n},T{x}_{3n+1},R{x}_{3n-1})\\ \le & kmax\left\{\begin{array}{c}G(A{x}_{3n},S{x}_{3n},S{x}_{3n})G(B{x}_{3n+1},T{x}_{3n+1},T{x}_{3n+1}),\\ G(B{x}_{3n+1},T{x}_{3n+1},T{x}_{3n+1})G(C{x}_{3n-1},R{x}_{3n-1},R{x}_{3n-1}),\\ G(C{x}_{3n-1},R{x}_{3n-1},R{x}_{3n-1})G(A{x}_{3n},S{x}_{3n},S{x}_{3n})\end{array}\right\}\\ =& kmax\left\{\begin{array}{c}G({y}_{3n-1},{y}_{3n},{y}_{3n})G({y}_{3n},{y}_{3n+1},{y}_{3n+1}),\\ G({y}_{3n},{y}_{3n+1},{y}_{3n+1})G({y}_{3n-2},{y}_{3n-1},{y}_{3n-1}),\\ G({y}_{3n-2},{y}_{3n-1},{y}_{3n-1})G({y}_{3n-1},{y}_{3n},{y}_{3n})\end{array}\right\}\\ \le & kmax\left\{\begin{array}{c}G({y}_{3n-2},{y}_{3n-1},{y}_{3n})G({y}_{3n-1},{y}_{3n},{y}_{3n+1}),\\ G({y}_{3n-1},{y}_{3n},{y}_{3n+1})G({y}_{3n-2},{y}_{3n-1},{y}_{3n}),\\ G({y}_{3n-2},{y}_{3n-1},{y}_{3n})G({y}_{3n-1},{y}_{3n},{y}_{3n+1})\end{array}\right\}\\ =& kG({y}_{3n-2},{y}_{3n-1},{y}_{3n})G({y}_{3n-1},{y}_{3n},{y}_{3n+1}).\end{array}$
This implies that
$G({y}_{3n-1},{y}_{3n},{y}_{3n+1})\le kG({y}_{3n-2},{y}_{3n-1},{y}_{3n}).$
(2.3)
Again using condition (2.1) and (G
_{3}) we have
$\begin{array}{rcl}{G}^{2}({y}_{3n},{y}_{3n+1},{y}_{3n+2})& =& {G}^{2}(S{x}_{3n},T{x}_{3n+1},R{x}_{3n+2})\\ \le & kmax\left\{\begin{array}{c}G(A{x}_{3n},S{x}_{3n},S{x}_{3n})G(B{x}_{3n+1},T{x}_{3n+1},T{x}_{3n+1}),\\ G(B{x}_{3n+1},T{x}_{3n+1},T{x}_{3n+1})G(C{x}_{3n+2},R{x}_{3n+2},R{x}_{3n+2}),\\ G(C{x}_{3n+2},R{x}_{3n+2},R{x}_{3n+2})G(A{x}_{3n},S{x}_{3n},S{x}_{3n})\end{array}\right\}\\ =& kmax\left\{\begin{array}{c}G({y}_{3n-1},{y}_{3n},{y}_{3n})G({y}_{3n},{y}_{3n+1},{y}_{3n+1}),\\ G({y}_{3n},{y}_{3n+1},{y}_{3n+1})G({y}_{3n+1},{y}_{3n+2},{y}_{3n+2}),\\ G({y}_{3n+1},{y}_{3n+2},{y}_{3n+2})G({y}_{3n-1},{y}_{3n},{y}_{3n})\end{array}\right\}\\ \le & kmax\left\{\begin{array}{c}G({y}_{3n-1},{y}_{3n},{y}_{3n+1})G({y}_{3n},{y}_{3n+1},{y}_{3n+2}),\\ G({y}_{3n-1},{y}_{3n},{y}_{3n+1})G({y}_{3n},{y}_{3n+1},{y}_{3n+2}),\\ G({y}_{3n},{y}_{3n+1},{y}_{3n+2})G({y}_{3n-1},{y}_{3n},{y}_{3n+1})\end{array}\right\}\\ =& kG({y}_{3n-1},{y}_{3n},{y}_{3n+1})G({y}_{3n},{y}_{3n+1},{y}_{3n+2}).\end{array}$
This gives
$G({y}_{3n},{y}_{3n+1},{y}_{3n+2})\le kG({y}_{3n-1},{y}_{3n},{y}_{3n+1}).$
(2.4)
Similarly, using condition (2.1) and (G
_{3}) we have
$\begin{array}{rcl}{G}^{2}({y}_{3n+1},{y}_{3n+2},{y}_{3n+3})& =& {G}^{2}(S{x}_{3n+3},T{x}_{3n+1},R{x}_{3n+2})\\ \le & kmax\left\{\begin{array}{c}G(A{x}_{3n+3},S{x}_{3n+3},S{x}_{3n+3})G(B{x}_{3n+1},T{x}_{3n+1},T{x}_{3n+1}),\\ G(B{x}_{3n+1},T{x}_{3n+1},T{x}_{3n+1})G(C{x}_{3n+2},R{x}_{3n+2},R{x}_{3n+2}),\\ G(C{x}_{3n+2},R{x}_{3n+2},R{x}_{3n+2})G(A{x}_{3n+3},S{x}_{3n+3},S{x}_{3n+3})\end{array}\right\}\\ =& kmax\left\{\begin{array}{c}G({y}_{3n+2},{y}_{3n+3},{y}_{3n+3})G({y}_{3n},{y}_{3n+1},{y}_{3n+1}),\\ G({y}_{3n},{y}_{3n+1},{y}_{3n+1})G({y}_{3n+1},{y}_{3n+2},{y}_{3n+2}),\\ G({y}_{3n+1},{y}_{3n+2},{y}_{3n+2})G({y}_{3n+2},{y}_{3n+3},{y}_{3n+3})\end{array}\right\}\\ \le & kmax\left\{\begin{array}{c}G({y}_{3n+1},{y}_{3n+2},{y}_{3n+3})G({y}_{3n},{y}_{3n+1},{y}_{3n+2}),\\ G({y}_{3n},{y}_{3n+1},{y}_{3n+2})G({y}_{3n+1},{y}_{3n+2},{y}_{3n+3}),\\ G({y}_{3n},{y}_{3n+1},{y}_{3n+2})G({y}_{3n+1},{y}_{3n+2},{y}_{3n+3})\end{array}\right\}\\ =& kG({y}_{3n},{y}_{3n+1},{y}_{3n+2})G({y}_{3n+1},{y}_{3n+2},{y}_{3n+3}).\end{array}$
This implies that
$G({y}_{3n+1},{y}_{3n+2},{y}_{3n+3})\le kG({y}_{3n},{y}_{3n+1},{y}_{3n+2}).$
(2.5)
Combining (2.3), (2.4), and (2.5) we have
$G({y}_{n},{y}_{n+1},{y}_{n+2})\le kG({y}_{n-1},{y}_{n},{y}_{n+1})\le {k}^{2}G({y}_{n-2},{y}_{n-1},{y}_{n})\le \cdots \le {k}^{n}G({y}_{0},{y}_{1},{y}_{2}).$
Therefore, for all
$n,m\in \mathbb{N}$,
$n<m$, by (G
_{5}) and (G
_{3}) we have
$\begin{array}{rcl}G({y}_{n},{y}_{m},{y}_{m})& \le & G({y}_{n},{y}_{n+1},{y}_{n+1})+G({y}_{n+1},{y}_{n+2},{y}_{n+2})+G({y}_{n+2},{y}_{n+3},{y}_{n+3})+\cdots \\ +G({y}_{m-1},{y}_{m},{y}_{m})\\ \le & G({y}_{n},{y}_{n+1},{y}_{n+2})+G({y}_{n+1},{y}_{n+2},{y}_{n+3})+\cdots +G({y}_{m-1},{y}_{m},{y}_{m+1})\\ \le & ({k}^{n}+{k}^{n+1}+{k}^{n+2}+\cdots +{k}^{m-1})G({y}_{0},{y}_{1},{y}_{2})\\ \le & \frac{{k}^{n}}{1-k}G({y}_{0},{y}_{1},{y}_{2})\to 0,\phantom{\rule{1em}{0ex}}\text{as}n\to \mathrm{\infty}.\end{array}$
Hence $\{{y}_{n}\}$ is a G-Cauchy sequence in X. Since X is a complete G-metric space, there exists a point $u\in X$ such that ${y}_{n}\to u$ ($n\to \mathrm{\infty}$).
Since the sequences
$\{S{x}_{3n}\}=\{B{x}_{3n+1}\}$,
$\{T{x}_{3n+1}\}=\{C{x}_{3n+2}\}$ and
$\{R{x}_{3n-1}\}=\{A{x}_{3n}\}$ are all subsequences of
$\{{y}_{n}\}$, they all converge to
u. We have
$\begin{array}{r}{y}_{3n}=S{x}_{3n}=B{x}_{3n+1}\to u,\phantom{\rule{2em}{0ex}}{y}_{3n+1}=T{x}_{3n+1}=C{x}_{3n+2}\to u,\\ {y}_{3n-1}=R{x}_{3n-1}=A{x}_{3n}\to u\phantom{\rule{1em}{0ex}}(n\to \mathrm{\infty}).\end{array}$
(2.6)
Now we prove that u is a common fixed point of S, T, R, A, B, and C under condition (a).
First, we suppose that A is continuous, the pair $(S,A)$ is weakly commuting, the pairs $(T,B)$ and $(R,C)$ are weakly compatible.
Step 1. We prove that $u=Su=Au$.
By (2.6) and the weakly commuting of the mapping pair
$(S,A)$ we have
$G(SA{x}_{3n},AS{x}_{3n},AS{x}_{3n})\le G(S{x}_{3n},A{x}_{3n},A{x}_{3n})\to 0\phantom{\rule{1em}{0ex}}(n\to \mathrm{\infty}).$
(2.7)
Since A is continuous, ${A}^{2}{x}_{3n}\to Au$ ($n\to \mathrm{\infty}$), $AS{x}_{3n}\to Au$ ($n\to \mathrm{\infty}$). By (2.7) we know $SA{x}_{3n}\to Au$ ($n\to \mathrm{\infty}$).
From condition (2.1) we know
$\begin{array}{r}{G}^{2}(SA{x}_{3n},T{x}_{3n+1},R{x}_{3n+2})\\ \phantom{\rule{1em}{0ex}}\le kmax\left\{\begin{array}{c}G({A}^{2}{x}_{3n},SA{x}_{3n},SA{x}_{3n})G(B{x}_{3n+1},T{x}_{3n+1},T{x}_{3n+1}),\\ G(B{x}_{3n+1},T{x}_{3n+1},T{x}_{3n+1})G(C{x}_{3n+2},R{x}_{3n+2},R{x}_{3n+2}),\\ G(C{x}_{3n+2},R{x}_{3n+2},R{x}_{3n+2})G({A}^{2}{x}_{3n},SA{x}_{3n},SA{x}_{3n})\end{array}\right\}.\end{array}$
Letting
$n\to \mathrm{\infty}$ we have
${G}^{2}(Au,u,u)\le kmax\left\{\begin{array}{c}G(Au,Au,Au)G(u,u,u),\\ G(u,u,u)G(u,u,u),\\ G(u,u,u)G(Au,Au,Au)\end{array}\right\}=0.$
This implies that $G(Au,u,u)$=0, and so $Au=u$.
Again by use of condition (2.1) we have
${G}^{2}(Su,T{x}_{3n+1},R{x}_{3n+2})\le kmax\left\{\begin{array}{c}G(Au,Su,Su)G(B{x}_{3n+1},T{x}_{3n+1},T{x}_{3n+1}),\\ G(B{x}_{3n+1},T{x}_{3n+1},T{x}_{3n+1})G(C{x}_{3n+2},R{x}_{3n+2},R{x}_{3n+2}),\\ G(C{x}_{3n+2},R{x}_{3n+2},R{x}_{3n+2})G(Au,Su,Su)\end{array}\right\}.$
Letting
$n\to \mathrm{\infty}$ we have
${G}^{2}(Su,u,u)\le kmax\left\{\begin{array}{c}G(Au,Su,Su)G(u,u,u),\\ G(u,u,u)G(u,u,u),\\ G(u,u,u)G(Au,Su,Su)\end{array}\right\}=0.$
This implies that $G(Su,u,u)=0$, and so $Su=u$.
So we have $u=Au=Su$.
Step 2. We prove that $u=Tu=Bu$.
Since
$S(X)\subset B(X)$ and
$u=Su\in S(X)$, there is a point
$v\in X$ such that
$u=Su=Bv$. Again, by use of condition (ii), we have
${G}^{2}(Su,Tv,R{x}_{3n+2})\le kmax\left\{\begin{array}{c}G(Au,Su,Su)G(Bv,Tv,Tv),\\ G(Bv,Tv,Tv)G(C{x}_{3n+2},R{x}_{3n+2},R{x}_{3n+2}),\\ G(C{x}_{3n+2},R{x}_{3n+2},R{x}_{3n+2})G(Au,Su,Su)\end{array}\right\}.$
Letting
$n\to \mathrm{\infty}$ and using
$u=Au=Su$ we have
${G}^{2}(u,Tv,u)\le kmax\left\{\begin{array}{c}G(u,u,u)G(u,Tv,Tv),\\ G(u,Tv,Tv)G(u,u,u),\\ G(u,u,u)G(u,u,u)\end{array}\right\}=0.$
This implies that $G(u,Tv,u)=0$, and so $Tv=u$.
Since the pair
$(T,B)$ is weakly compatible, we have
Again, by use of condition (2.1), we have
${G}^{2}(Su,Tu,R{x}_{3n+2})\le kmax\left\{\begin{array}{c}G(Au,Su,Su)G(Bu,Tu,Tu),\\ G(Bu,Tu,Tu)G(C{x}_{3n+2},R{x}_{3n+2},R{x}_{3n+2}),\\ G(C{x}_{3n+2},R{x}_{3n+2},R{x}_{3n+2})G(Au,Su,Su)\end{array}\right\}.$
Letting
$n\to \mathrm{\infty}$ and using
$u=Au=Su$ and
$Tu=Bu$ we have
${G}^{2}(u,Tu,u)\le kmax\left\{\begin{array}{c}G(u,u,u)G(Tu,Tu,Tu),\\ G(Tu,Tu,Tu)G(u,u,u),\\ G(u,u,u)G(u,u,u)\end{array}\right\}=0.$
This implies that $G(u,Tu,u)=0$, and so $Tu=u$.
So we have $u=Tu=Bu$.
Step 3. We prove that $u=Ru=Cu$.
Since
$T(X)\subset C(X)$ and
$u=Tu\in T(X)$, there is a point
$w\in X$ such that
$u=Tu=Cw$. Again, by use of condition (2.1), we have
${G}^{2}(Su,Tu,Rw)\le kmax\left\{\begin{array}{c}G(Au,Su,Su)G(Bu,Tu,Tu),\\ G(Bu,Tu,Tu)G(Cw,Rw,Rw),\\ G(Cw,Rw,Rw)G(Au,Su,Su)\end{array}\right\}.$
Using
$u=Au=Su$ and
$u=Tu=Bu=Cw$, we obtain
${G}^{2}(u,u,Rw)\le kmax\left\{\begin{array}{c}G(u,u,u)G(u,u,u),\\ G(u,u,u)G(u,Rw,Rw),\\ G(u,Rw,Rw)G(u,u,u)\end{array}\right\}=0.$
This implies that $G(u,u,Rw)=0$, and so $Rw=u=Cw$.
Since the pair
$(R,C)$ is weakly compatible, we have
Again by use of condition (2.1),
$Su=Au$ and
$Ru=Cu$ we have
${G}^{2}(u,u,Ru)={G}^{2}(Su,Tu,Ru)\le kmax\left\{\begin{array}{c}G(Au,Su,Su)G(Bu,Tu,Tu),\\ G(Bu,Tu,Tu)G(Cu,Ru,Ru),\\ G(Cu,Ru,Ru)G(Au,Su,Su)\end{array}\right\}=0.$
So we have $G(u,u,Ru)=0$, and so $u=Ru=Cu$.
Therefore u is the common fixed point of S, T, R, A, B and C when A is continuous and the pair $(S,A)$ is weakly commuting, the pairs $(T,B)$ and $(F,C)$ are weakly compatible.
Next, we suppose that S is continuous, the pair $(S,A)$ is weakly commuting, the pairs $(T,B)$ and $(R,C)$ are weakly compatible.
Step 1. We prove that $u=Su$.
By (2.6) and the weakly commuting of the mapping pair
$(S,A)$ we have
$G(SA{x}_{3n},AS{x}_{3n},AS{x}_{3n})\le G(S{x}_{3n},A{x}_{3n},A{x}_{3n})\to 0\phantom{\rule{1em}{0ex}}(n\to \mathrm{\infty}).$
(2.8)
Since S is continuous, ${S}^{2}{x}_{3n}\to Su$ ($n\to \mathrm{\infty}$), $SA{x}_{3n}\to Su$ ($n\to \mathrm{\infty}$). By (2.8) we know $AS{x}_{3n}\to Su$ ($n\to \mathrm{\infty}$).
From condition (2.1) we have
$\begin{array}{r}{G}^{2}({S}^{2}{x}_{3n},T{x}_{3n+1},R{x}_{3n+2})\\ \phantom{\rule{1em}{0ex}}\le kmax\left\{\begin{array}{c}G(AS{x}_{3n},{S}^{2}{x}_{3n},{S}^{2}{x}_{3n})G(B{x}_{3n+1},T{x}_{3n+1},T{x}_{3n+1}),\\ G(B{x}_{3n+1},T{x}_{3n+1},T{x}_{3n+1})G(C{x}_{3n+2},R{x}_{3n+2},R{x}_{3n+2}),\\ G(C{x}_{3n+2},R{x}_{3n+2},R{x}_{3n+2})G(AS{x}_{3n},{S}^{2}{x}_{3n},{S}^{2}{x}_{3n})\end{array}\right\}.\end{array}$
Letting
$n\to \mathrm{\infty}$ we have
${G}^{2}(Su,u,u)\le kmax\left\{\begin{array}{c}G(Su,Su,Su)G(u,u,u),\\ G(u,u,u)G(u,u,u),\\ G(u,u,u)G(Su,Su,Su)\end{array}\right\}=0.$
This implies that $G(Su,u,u)=0$, and so $Su=u$.
Step 2. We prove that $u=Tu=Bu$.
Since
$S(X)\subset B(X)$ and
$u=Su\in S(X)$, there is a point
$z\in X$ such that
$u=Su=Bz$. Again by use of condition (2.1), we have
${G}^{2}({S}^{2}{x}_{3n},Tz,R{x}_{3n+2})\le kmax\left\{\begin{array}{c}G(AS{x}_{3n},{S}^{2}{x}_{3n},{S}^{2}{x}_{3n})G(Bz,Tz,Tz),\\ G(Bz,Tz,Tz)G(C{x}_{3n+2},R{x}_{3n+2},R{x}_{3n+2}),\\ G(C{x}_{3n+2},R{x}_{3n+2},R{x}_{3n+2})G(AS{x}_{3n},{S}^{2}{x}_{3n},{S}^{2}{x}_{3n})\end{array}\right\}.$
Letting
$n\to \mathrm{\infty}$ and using
$u=Su$ we have
${G}^{2}(u,Tz,u)\le kmax\left\{\begin{array}{c}G(u,u,u)G(u,Tz,Tz),\\ G(u,Tz,Tz)G(u,u,u),\\ G(u,u,u)G(u,u,u)\end{array}\right\}=0.$
This implies that $G(u,Tz,u)=0$, and so $Tz=u=Bz$.
Since the pair
$(T,B)$ is weakly compatible, we have
Again, by use of condition (2.1), we have
${G}^{2}(S{x}_{3n},Tu,R{x}_{3n+2})\le kmax\left\{\begin{array}{c}G(A{x}_{3n},S{x}_{3n},S{x}_{3n})G(Bu,Tu,Tu),\\ G(Bu,Tu,Tu)G(C{x}_{3n+2},R{x}_{3n+2},R{x}_{3n+2}),\\ G(C{x}_{3n+2},R{x}_{3n+2},R{x}_{3n+2})G(A{x}_{3n},S{x}_{3n},S{x}_{3n})\end{array}\right\}.$
Letting
$n\to \mathrm{\infty}$ and using
$u=Su$ and
$Tu=Bu$ we have
${G}^{2}(u,Tu,u)\le kmax\left\{\begin{array}{c}G(u,u,u)G(Tu,Tu,Tu),\\ G(Tu,Tu,Tu)G(u,u,u),\\ G(u,u,u)G(u,u,u)\end{array}\right\}=0.$
This implies that $G(u,Tu,u)=0$, and so $Tu=u=Bu$.
So we have $u=Tu=Bu$.
Step 3. We prove that $u=Ru=Cu$.
Since
$T(X)\subset C(X)$ and
$u=Tu\in T(X)$, there is a point
$t\in X$ such that
$u=Tu=Ct$. Again by use of condition (2.1), we have
${G}^{2}(S{x}_{3n},Tu,Rt)\le kmax\left\{\begin{array}{c}G(A{x}_{3n},S{x}_{3n},S{x}_{3n})G(Bu,Tu,Tu),\\ G(Bu,Tu,Tu)G(Ct,Rt,Rt),\\ G(Ct,Rt,Rt)G(A{x}_{3n},S{x}_{3n},S{x}_{3n})\end{array}\right\}.$
Letting
$n\to \mathrm{\infty}$ and using
$u=Tu=Bu$, we obtain
${G}^{2}(u,u,Rt)\le kmax\left\{\begin{array}{c}G(u,u,u)G(u,u,u),\\ G(u,u,u)G(u,Rt,Rt),\\ G(u,Rt,Rt)G(u,u,u)\end{array}\right\}=0.$
This implies that $G(u,u,Rt)=0$, and so $Rt=u=Ct$.
Since the pair
$(R,C)$ is weakly compatible, we have
Again, by use of condition (2.1), we have
${G}^{2}(S{x}_{3n},Tu,Ru)\le kmax\left\{\begin{array}{c}G(A{x}_{3n},S{x}_{3n},S{x}_{3n})G(Bu,Tu,Tu),\\ G(Bu,Tu,Tu)G(Cu,Ru,Ru),\\ G(Cu,Ru,Ru)G(A{x}_{3n},S{x}_{3n},S{x}_{3n})\end{array}\right\}.$
Letting
$n\to \mathrm{\infty}$ and using
$u=Tu=Bu$ we have
${G}^{2}(u,u,Ru)\le kmax\left\{\begin{array}{c}G(u,u,u)G(u,u,u),\\ G(u,u,u)G(Ru,Ru,Ru),\\ G(Ru,Ru,Ru)G(u,u,u)\end{array}\right\}=0.$
This implies that $G(u,u,Ru)=0$, and so $Ru=u=Cu$.
Step 4. We prove that $u=Au$.
Since
$R(X)\subset A(X)$ and
$u=Ru\in R(X)$, there is a point
$p\in X$ such that
$u=Ru=Ap$. Again by use of condition (2.1), we have
${G}^{2}(Sp,Tu,Ru)\le kmax\left\{\begin{array}{c}G(Ap,Sp,Sp)G(Bu,Tu,Tu),\\ G(Bu,Tu,Tu)G(Cu,Ru,Ru),\\ G(Cu,Ru,Ru)G(Ap,Sp,Sp)\end{array}\right\}.$
Using
$u=Tu=Bu$ and
$u=Ru=Cu$, we obtain
${G}^{2}(Sp,u,u)\le kmax\left\{\begin{array}{c}G(u,Sp,Sp)G(u,u,u),\\ G(u,u,u)G(u,u,u),\\ G(u,u,u)G(u,Sp,Sp)\end{array}\right\}=0.$
This implies that $G(Sp,u,u)=0$, and $Sp=u=Ap$.
Since the pair
$(S,A)$ is weakly compatible, we have
Therefore u is the common fixed point of S, T, R, A, B, and C when S is continuous and the pair $(S,A)$ is weakly commuting, the pairs $(T,B)$ and $(F,C)$ are weakly compatible.
Similarly we can prove that u is the unique common fixed point of the maps S, T, R, A, B, and C under the conditions of (b) and (c).
Next we prove the uniqueness of a common fixed point u.
Let
u and
v be two common fixed points of
S,
T,
R,
A,
B, and
C, by use of condition (2.1), we have
${G}^{2}(u,u,v)={G}^{2}(Su,Tu,Rv)\le kmax\left\{\begin{array}{c}G(Au,Su,Su)G(Bu,Tu,Tu),\\ G(Bu,Tu,Tu)G(Cv,Rv,Rv),\\ G(Cv,Rv,Rv)G(Au,Su,Su))\end{array}\right\}=0.$
This shows that $G(u,u,v)=0$, and so $u=v$. Thus the common fixed point is unique.
If condition (2.2) holds, then the argument is similar to that above, so we omit it. □
Theorem 2.2 Let $(X,G)$ be a complete G-
metric space and let S,
T,
R,
A,
B,
and C be six mappings of X into itself satisfying the following conditions:
- (i)
$S(X)\subset B(X)$, $T(X)\subset C(X)$, $R(X)\subset A(X)$;
- (ii)
the pairs $(S,A)$, $(T,B)$, and $(R,C)$ are commuting mappings;
- (iii)
$\mathrm{\forall}x,y,z\in X$,
${G}^{2}({S}^{p}x,{T}^{q}y,{R}^{r}z)\le kmax\left\{\begin{array}{c}G(Ax,{S}^{p}x,{S}^{p}x)G(By,{T}^{q}y,{T}^{q}y),\\ G(By,{T}^{q}y,{T}^{q}y)G(Cz,{R}^{r}z,{R}^{r}z),\\ G(Cz,{R}^{r}z,{R}^{r}z)G(Ax,{S}^{p}x,{S}^{p}x)\end{array}\right\}$
(2.9)
or
${G}^{2}({S}^{p}x,{T}^{q}y,{R}^{r}z)\le kmax\left\{\begin{array}{c}G(Ax,Ax,{S}^{p}x)G(By,By,{T}^{q}y),\\ G(By,By,{T}^{q}y)G(Cz,Cz,{R}^{r}z),\\ G(Cz,Cz,{R}^{r}z)G(Ax,Ax,{S}^{p}x)\end{array}\right\},$
(2.10)
where $k\in [0,1)$, $p,q,r\in \mathbb{N}$, then S, T, R, A, B, and C have a unique common fixed point in X.
Proof Suppose the condition (2.9) holds. Since ${S}^{p}X\subset {S}^{p-1}X\subset \cdots \subset SX$, $SX\subset BX$, so that ${S}^{p}X\subset BX$. Similar, we can show that ${T}^{q}X\subset CX$ and ${R}^{r}X\subset AX$. From Theorem 2.1, we see that ${S}^{p}$, ${T}^{q}$, ${R}^{r}$, A, B, and C have a unique common fixed point u.
Since
$Su=S({S}^{p}u)={S}^{p+1}u={S}^{p}(Su)$, so that
${G}^{2}({S}^{p}Su,{T}^{q}u,{R}^{r}u)\le kmax\left\{\begin{array}{c}G(ASu,{S}^{p}Su,{S}^{p}Su)G(Bu,{T}^{q}u,{T}^{q}u),\\ G(Bu,{T}^{q}u,{T}^{q}u)G(Cu,{R}^{r}u,{R}^{r}u),\\ G(Cu,{R}^{r}u,{R}^{r}u)G(ASu,{S}^{p}Su,{S}^{p}Su)\end{array}\right\}.$
Note that
$u=Au=Bu=Cu={S}^{p}u={T}^{q}u={R}^{r}u$ and
${S}^{p}Su=Su$, we obtain
${G}^{2}(Su,u,u)={G}^{2}({S}^{p}Su,{T}^{q}u,{R}^{r}u)\le kmax\left\{\begin{array}{c}G(ASu,Su,Su)G(u,u,u),\\ G(u,u,u)G(u,u,u),\\ G(u,u,u)G(ASu,Su,Su)\end{array}\right\}=0.$
This implies that $G(Su,u,u)=0$, and so $Su=u$.
By the same argument, we can prove
$Tu=u$ and
$Ru=u$. Thus we have
$u=Su=Tu=Fu=Au=Bu=Cu$, so that
S,
T,
R,
A,
B, and
C have a common fixed point
u in
X. Let
v be any other common fixed point of
S,
T,
R,
A,
B, and
C, then use of condition (2.9), we have
${G}^{2}(u,u,v)={G}^{2}({S}^{p}u,{T}^{q}u,{R}^{r}v)\le kmax\left\{\begin{array}{c}G(Au,{S}^{p}u,{S}^{p}u)G(Bu,{T}^{q}u,{T}^{q}u),\\ G(Bu,{T}^{q}u,{T}^{q}u)G(Cv,{R}^{r}v,{R}^{r}v),\\ G(Cv,{R}^{r}v,{R}^{r}v)G(Au,{S}^{p}u,{S}^{p}u)\end{array}\right\}=0.$
This implies that $G(u,u,v)=0$, and so $u=v$. Thus the common fixed point is unique.
If condition (2.10) holds, then the argument is similar to that above, so we omit it. □
Remark 2.1 Theorems 2.1 and 2.2 improve and extend the corresponding results in Ye and Gu [[15], Theorem 2.1, Corollary 2.2] from three self-mappings to six self-mappings.
Corollary 2.1 Let $(X,G)$ be a complete G-
metric space and let S,
T,
R,
A,
B,
and C be six mappings of X into itself satisfying the following conditions:
- (i)
$S(X)\subset B(X)$, $T(X)\subset C(X)$, $R(X)\subset A(X)$;
- (ii)
$\mathrm{\forall}x,y,z\in X$,
$\begin{array}{rcl}{G}^{2}(Sx,Ty,Rz)& \le & aG(Ax,Sx,Sx)G(By,Ty,Ty)+bG(By,Ty,Ty)G(Cz,Rz,Rz)\\ +cG(Cz,Rz,Rz)G(Ax,Sx,Sx)\end{array}$
(2.11)
or
$\begin{array}{rcl}{G}^{2}(Sx,Ty,Rz)& \le & aG(Ax,Ax,Sx)G(By,By,Ty)+bG(By,By,Ty)G(Cz,Cz,Rz)\\ +cG(Cz,Cz,Rz)G(Ax,Ax,Sx),\end{array}$
(2.12)
where $0\le a+b+c<1$.
Then one of the pairs $(S,A)$,
$(T,B)$,
and $(R,C)$ has a coincidence point in X.
Moreover,
assume one of the following conditions is satisfied:
- (a)
either S or A is G-continuous, the pair $(S,A)$ is weakly commuting, the pairs $(T,B)$ and $(R,C)$ are weakly compatible;
- (b)
either T or B is G-continuous, the pair $(T,B)$ is weakly commuting, the pairs $(S,A)$ and $(R,C)$ are weakly compatible;
- (c)
either F or C is G-continuous, the pair $(R,C)$ is weakly commuting, the pairs $(S,A)$ and $(T,B)$ are weakly compatible.
Then the mappings S, T, R, A, B, and C have a unique common fixed point in X.
Proof Suppose the condition (2.11) holds. For
$x,y,z\in X$, let
$M(x,y,z)=max\left\{\begin{array}{c}G(Ax,Sx,Sx)G(By,Ty,Ty),\\ G(By,Ty,Ty)G(Cz,Rz,Rz),\\ G(Cz,Rz,Rz)G(Ax,Sx,Sx)\end{array}\right\}.$
Then
$\begin{array}{r}aG(Ax,Sx,Sx)G(By,Ty,Ty)+bG(By,Ty,Ty)G(Cz,Rz,Rz)\\ \phantom{\rule{2em}{0ex}}+cG(Cz,Rz,Rz)G(Ax,Sx,Sx)\\ \phantom{\rule{1em}{0ex}}\le (a+b+c)M(x,y,z).\end{array}$
So, if
$\begin{array}{rcl}{G}^{2}(Sx,Ty,Rz)& \le & aG(Ax,Sx,Sx)G(By,Ty,Ty)+bG(By,Ty,Ty)G(Cz,Rz,Rz)\\ +cG(Cz,Rz,Rz)G(Ax,Sx,Sx),\end{array}$
then $G(Sx,Ty,Rz)\le (a+b+c)M(x,y,z)$. Taking $k=a+b+c$ in Theorem 2.1, the conclusion of Corollary 2.1 can be obtained from Theorem 2.1 immediately.
If condition (2.12) holds, then the argument is similar to that above, so we omit it. This completes the proof of Corollary 2.1. □
Remark 2.2 If $A=B=C=I$ (I is the identity mapping, here and below), Corollary 2.1 is reduced to Theorem 2.1 of Ye and Gu [15].
Corollary 2.2 Let $(X,G)$ be a complete G-
metric space and let S,
T,
R,
A,
B,
and C be six mappings of X into itself satisfying the following conditions:
- (i)
$S(X)\subset B(X)$, $T(X)\subset C(X)$, $R(X)\subset A(X)$;
- (ii)
the pairs $(S,A)$, $(T,B)$, and $(R,C)$ are commuting mappings;
- (iii)
$\mathrm{\forall}x,y,z\in X$,
$\begin{array}{r}{G}^{2}({S}^{p}x,{T}^{q}y,{R}^{r}z)\\ \phantom{\rule{1em}{0ex}}\le aG(Ax,{S}^{p}x,{S}^{p}x)G(By,{T}^{q}y,{T}^{q}y)+bG(By,{T}^{q}y,{T}^{q}y)G(Cz,{R}^{r}z,{R}^{r}z)\\ \phantom{\rule{2em}{0ex}}+cG(Cz,{R}^{r}z,{R}^{r}z)G(Ax,{S}^{p}x,{S}^{p}x)\end{array}$
(2.13)
or
$\begin{array}{r}{G}^{2}({S}^{p}x,{T}^{q}y,{R}^{r}z)\\ \phantom{\rule{1em}{0ex}}\le aG(Ax,Ax,{S}^{p}x)G(By,By,{T}^{q}y)+bG(By,By,{T}^{q}y)G(Cz,Cz,{R}^{r}z)\\ \phantom{\rule{2em}{0ex}}+cG(Cz,Cz,{R}^{r}z)G(Ax,Ax,{S}^{p}x),\end{array}$
(2.14)
where $0\le a+b+c+d<1$, $p,q,r\in \mathbb{N}$, then S, T, R, A, B, and C have a unique common fixed point in X.
Proof The proof follows from Theorem 2.2, and from an argument similar to that used in Corollary 2.1. □
Remark 2.3 If $A=B=C=I$, Corollary 2.2 is reduced to Corollary 2.2 of Ye and Gu [15].
In Theorem 2.1, if we take $A=B=C=I$, then we have the following corollary.
Corollary 2.3 Let $(X,G)$ be a complete G-
metric space and let S,
T,
and R be three mappings of X into itself satisfying the following conditions:
${G}^{2}(Sx,Ty,Rz)\le kmax\left\{\begin{array}{c}G(x,Sx,Sx)G(y,Ty,Ty),\\ G(y,Ty,Ty)G(z,Rz,Rz),\\ G(z,Rz,Rz)G(x,Sx,Sx)\end{array}\right\}$
(2.15)
or
${G}^{2}(Sx,Ty,Rz)\le kmax\left\{\begin{array}{c}G(x,x,Sx)G(y,y,Ty),\\ G(y,y,Ty)G(z,z,Rz),\\ G(z,z,Rz)G(x,x,Sx)\end{array}\right\}$
(2.16)
for all $x,y,z\in X$, where $k\in [0,1)$.
Then the mappings S, T, and R have a unique common fixed point in X.
Remark 2.4 In Theorems 2.1, 2.2, Corollaries 2.1, 2.2 and 2.3, we have taken: (1) $S=T=R$; (2) $A=B=C$; (3) $A=B=C=I$; (4) $T=R$ and $B=C$; (5) $T=R$, $B=C=I$, several new result can be obtain.
Theorem 2.3 Let $(X,G)$ be a complete G-
metric space and let S,
T,
R,
A,
B,
and C be six mappings of X into itself satisfying the following conditions:
- (i)
$S(X)\subset B(X)$, $T(X)\subset C(X)$, $R(X)\subset A(X)$;
- (ii)
$\mathrm{\forall}x,y,z\in X$,
${G}^{2}(Sx,Ty,Rz)\le kmax\left\{\begin{array}{c}G(Ax,Sx,Ty)G(By,Ty,Rz),\\ G(By,Ty,Rz)G(Cz,Rz,Sx),\\ G(Cz,Rz,Sx)G(Ax,Sx,Ty)\end{array}\right\}$
(2.17)
or
${G}^{2}(Sx,Ty,Rz)\le kmax\left\{\begin{array}{c}G(Ax,Ax,Ty)G(By,By,Rz),\\ G(By,By,Rz)G(Cz,Cz,Sx),\\ G(Cz,Cz,Sx)G(Ax,Ax,Ty)\end{array}\right\},$
(2.18)
where $k\in [0,\frac{1}{2})$.
Then one of the pairs $(S,A)$,
$(T,B)$,
and $(R,C)$ has a coincidence point in X.
Moreover,
assume one of the following conditions is satisfied:
- (a)
either S or A is G-continuous, the pair $(S,A)$ is weakly commuting, the pairs $(T,B)$ and $(R,C)$ are weakly compatible;
- (b)
either T or B is G-continuous, the pair $(T,B)$ is weakly commuting, the pairs $(S,A)$ and $(R,C)$ are weakly compatible;
- (c)
either F or C is G-continuous, the pair $(R,C)$ is weakly commuting, the pairs $(S,A)$ and $(T,B)$ are weakly compatible.
Then the mappings S, T, R, A, B, and C have a unique common fixed point in X.
Proof First, we suppose that the condition (2.17) holds.
Let
${x}_{0}$ in
X be an arbitrary point, since
$S(X)\subset B(X)$,
$T(X)\subset C(X)$,
$R(X)\subset A(X)$ there exist the sequences
$\{{x}_{n}\}$ and
$\{{y}_{n}\}$ in
X, such that
${y}_{3n}=S{x}_{3n}=B{x}_{3n+1},\phantom{\rule{2em}{0ex}}{y}_{3n+1}=T{x}_{3n+1}=C{x}_{3n+2},\phantom{\rule{2em}{0ex}}{y}_{3n+2}=R{x}_{3n+2}=A{x}_{3n+3}$
for $n=0,1,2,\dots $ .
If ${y}_{3n+2}={y}_{3n+3}$, then $Sp=Ap$ where $p={x}_{3n+3}$. If ${y}_{3n}={y}_{3n+1}$, then $Tp=Bp$ where $p={x}_{3n+1}$. If ${y}_{3n+1}={y}_{3n+2}$, then $Rp=Cp$ where $p={x}_{3n+2}$. Without loss of generality, we can assume that ${y}_{n}\ne {y}_{n+1}$, for all $n=0,1,2,\dots $ .
Now we prove that $\{{y}_{n}\}$ is a G-Cauchy sequence in X.
In fact, using condition (2.17) we have
$\begin{array}{rcl}{G}^{2}({y}_{3n-1},{y}_{3n},{y}_{3n+1})& =& {G}^{2}(S{x}_{3n},T{x}_{3n+1},R{x}_{3n-1})\\ \le & kmax\left\{\begin{array}{c}G(A{x}_{3n},S{x}_{3n},T{x}_{3n+1})G(B{x}_{3n+1},T{x}_{3n+1},R{x}_{3n-1}),\\ G(B{x}_{3n+1},T{x}_{3n+1},R{x}_{3n-1})G(C{x}_{3n-1},R{x}_{3n-1},S{x}_{3n}),\\ G(C{x}_{3n-1},R{x}_{3n-1},S{x}_{3n})G(A{x}_{3n},S{x}_{3n},T{x}_{3n+1})\end{array}\right\}\\ =& kmax\left\{\begin{array}{c}G({y}_{3n-1},{y}_{3n},{y}_{3n+1})G({y}_{3n},{y}_{3n+1},{y}_{3n-1}),\\ G({y}_{3n},{y}_{3n+1},{y}_{3n-1})G({y}_{3n-2},{y}_{3n-1},{y}_{3n}),\\ G({y}_{3n-2},{y}_{3n-1},{y}_{3n})G({y}_{3n-1},{y}_{3n},{y}_{3n+1})\end{array}\right\}\\ =& kmax\left\{\begin{array}{c}{G}^{2}({y}_{3n-1},{y}_{3n},{y}_{3n+1}),\\ G({y}_{3n-2},{y}_{3n-1},{y}_{3n})G({y}_{3n-1},{y}_{3n},{y}_{3n+1})\end{array}\right\}.\end{array}$
(2.19)
If
$max\{{G}^{2}({y}_{3n-1},{y}_{3n},{y}_{3n+1}),G({y}_{3n-2},{y}_{3n-1},{y}_{3n})G({y}_{3n-1},{y}_{3n},{y}_{3n+1})\}={G}^{2}({y}_{3n-1},{y}_{3n},{y}_{3n+1}),$
then by the inequality (2.19) we obtain
${G}^{2}({y}_{3n-1},{y}_{3n},{y}_{3n+1})\le k{G}^{2}({y}_{3n-1},{y}_{3n},{y}_{3n+1}),$
which is a contradiction since
$0\le k<\frac{1}{2}$, and hence
$\begin{array}{r}max\{{G}^{2}({y}_{3n-1},{y}_{3n},{y}_{3n+1}),G({y}_{3n-2},{y}_{3n-1},{y}_{3n})G({y}_{3n-1},{y}_{3n},{y}_{3n+1})\}\\ \phantom{\rule{1em}{0ex}}=G({y}_{3n-2},{y}_{3n-1},{y}_{3n})G({y}_{3n-1},{y}_{3n},{y}_{3n+1}).\end{array}$
Therefore, the inequality (2.19) implies that
$G({y}_{3n-1},{y}_{3n},{y}_{3n+1})\le kG({y}_{3n-2},{y}_{3n-1},{y}_{3n}).$
(2.20)
Again using the condition (2.17) we have
$\begin{array}{rcl}{G}^{2}({y}_{3n},{y}_{3n+1},{y}_{3n+2})& =& {G}^{2}(S{x}_{3n},T{x}_{3n+1},R{x}_{3n+2})\\ \le & kmax\left\{\begin{array}{c}G(A{x}_{3n},S{x}_{3n},T{x}_{3n+1})G(B{x}_{3n+1},T{x}_{3n+1},R{x}_{3n+2}),\\ G(B{x}_{3n+1},T{x}_{3n+1},R{x}_{3n+2})G(C{x}_{3n+2},R{x}_{3n+2},S{x}_{3n}),\\ G(C{x}_{3n+2},R{x}_{3n+2},S{x}_{3n})G(A{x}_{3n},S{x}_{3n},T{x}_{3n+1})\end{array}\right\}\\ =& kmax\left\{\begin{array}{c}G({y}_{3n-1},{y}_{3n},{y}_{3n+1})G({y}_{3n},{y}_{3n+1},{y}_{3n+2}),\\ G({y}_{3n},{y}_{3n+1},{y}_{3n+2})G({y}_{3n+1},{y}_{3n+2},{y}_{3n}),\\ G({y}_{3n+1},{y}_{3n+2},{y}_{3n})G({y}_{3n-1},{y}_{3n},{y}_{3n+1})\end{array}\right\}\\ =& kmax\left\{\begin{array}{c}G({y}_{3n-1},{y}_{3n},{y}_{3n+1})G({y}_{3n},{y}_{3n+1},{y}_{3n+2}),\\ {G}^{2}({y}_{3n},{y}_{3n+1},{y}_{3n+2})\end{array}\right\}.\end{array}$
(2.21)
If
$max\{G({y}_{3n-1},{y}_{3n},{y}_{3n+1})G({y}_{3n},{y}_{3n+1},{y}_{3n+2}),{G}^{2}({y}_{3n},{y}_{3n+1},{y}_{3n+2})\}={G}^{2}({y}_{3n},{y}_{3n+1},{y}_{3n+2}),$
then the inequality (2.21) implies that
${G}^{2}({y}_{3n},{y}_{3n+1},{y}_{3n+2})\le k{G}^{2}({y}_{3n},{y}_{3n+1},{y}_{3n+2}),$
this is a contradiction, and so
$\begin{array}{r}max\{G({y}_{3n-1},{y}_{3n},{y}_{3n+1})G({y}_{3n},{y}_{3n+1},{y}_{3n+2}),{G}^{2}({y}_{3n},{y}_{3n+1},{y}_{3n+2})\}\\ \phantom{\rule{1em}{0ex}}=G({y}_{3n-1},{y}_{3n},{y}_{3n+1})G({y}_{3n},{y}_{3n+1},{y}_{3n+2}).\end{array}$
Therefore, the inequality (2.21) implies that
$G({y}_{3n},{y}_{3n+1},{y}_{3n+2})\le kG({y}_{3n-1},{y}_{3n},{y}_{3n+1}).$
(2.22)
Similarly, using condition (2.17), we have
$\begin{array}{r}{G}^{2}({y}_{3n+1},{y}_{3n+2},{y}_{3n+3})\\ \phantom{\rule{1em}{0ex}}={G}^{2}(S{x}_{3n+3},T{x}_{3n+1},R{x}_{3n+2})\\ \phantom{\rule{1em}{0ex}}\le kmax\left\{\begin{array}{c}G(A{x}_{3n+3},S{x}_{3n+3},T{x}_{3n+1})G(B{x}_{3n+1},T{x}_{3n+1},R{x}_{3n+2}),\\ G(B{x}_{3n+1},T{x}_{3n+1},R{x}_{3n+2})G(C{x}_{3n+2},R{x}_{3n+2},S{x}_{3n+3}),\\ G(C{x}_{3n+2},R{x}_{3n+2},S{x}_{3n+3})G(A{x}_{3n+3},S{x}_{3n+3},T{x}_{3n+1})\end{array}\right\}\\ \phantom{\rule{1em}{0ex}}=kmax\left\{\begin{array}{c}G({y}_{3n+2},{y}_{3n+3},{y}_{3n+1})G({y}_{3n},{y}_{3n+1},{y}_{3n+2}),\\ G({y}_{3n},{y}_{3n+1},{y}_{3n+2})G({y}_{3n+1},{y}_{3n+2},{y}_{3n+3}),\\ G({y}_{3n+1},{y}_{3n+2},{y}_{3n+3})G({y}_{3n+2},{y}_{3n+3},{y}_{3n+1})\end{array}\right\}\\ \phantom{\rule{1em}{0ex}}=kmax\left\{\begin{array}{c}G({y}_{3n},{y}_{3n+1},{y}_{3n+2})G({y}_{3n+1},{y}_{3n+2},{y}_{3n+3}),\\ {G}^{2}({y}_{3n+1},{y}_{3n+2},{y}_{3n+3})\end{array}\right\}.\end{array}$
(2.23)
If
$\begin{array}{r}max\{G({y}_{3n},{y}_{3n+1},{y}_{3n+2})G({y}_{3n+1},{y}_{3n+2},{y}_{3n+3}),{G}^{2}({y}_{3n+1},{y}_{3n+2},{y}_{3n+3})\}\\ \phantom{\rule{1em}{0ex}}={G}^{2}({y}_{3n+1},{y}_{3n+2},{y}_{3n+3}),\end{array}$
then from the inequality (2.23) we get
${G}^{2}({y}_{3n+1},{y}_{3n+2},{y}_{3n+3})\le k{G}^{2}({y}_{3n+1},{y}_{3n+2},{y}_{3n+3}),$
which is a contradiction, hence we have
$\begin{array}{r}max\{G({y}_{3n},{y}_{3n+1},{y}_{3n+2})G({y}_{3n+1},{y}_{3n+2},{y}_{3n+3}),{G}^{2}({y}_{3n+1},{y}_{3n+2},{y}_{3n+3})\}\\ \phantom{\rule{1em}{0ex}}=G({y}_{3n},{y}_{3n+1},{y}_{3n+2})G({y}_{3n+1},{y}_{3n+2},{y}_{3n+3}).\end{array}$
Therefore, the above inequality (2.23) becomes
${G}^{2}({y}_{3n+1},{y}_{3n+2},{y}_{3n+3})\le kG({y}_{3n},{y}_{3n+1},{y}_{3n+2}).$
(2.24)
By combining (2.20), (2.22), and (2.24),
$\mathrm{\forall}n\in \mathbb{N}$, we have
$G({y}_{n},{y}_{n+1},{y}_{n+2})\le kG({y}_{n-1},{y}_{n},{y}_{n+1})\le \cdots \le {k}^{n}G({y}_{0},{y}_{1},{y}_{2}).$
(2.25)
Therefore, for all
$m,n\in \mathbb{N}$,
$m>n$, by (G
_{3}), (G
_{5}), and (2.25) we have
$\begin{array}{rcl}G({y}_{n},{y}_{m},{y}_{m})& \le & G({y}_{n},{y}_{n+1},{y}_{n+1})+G({y}_{n+1},{y}_{n+2},{y}_{n+2})+\cdots +G({y}_{m-1},{y}_{m},{y}_{m})\\ \le & G({y}_{n},{y}_{n+1},{y}_{n+2})+G({y}_{n+1},{y}_{n+2},{y}_{n+3})+\cdots +G({y}_{m-1},{y}_{m},{y}_{m+1})\\ \le & ({k}^{n}+{k}^{n+1}+\cdots +{k}^{m-1})G({y}_{0},{y}_{1},{y}_{2})\\ <& \frac{{k}^{n}}{1-k}G({y}_{0},{y}_{1},{y}_{2})\to 0\phantom{\rule{1em}{0ex}}(n\to \mathrm{\infty}).\end{array}$
This implies that $G({y}_{n},{y}_{m},{y}_{m})\to 0$, as $n,m\to \mathrm{\infty}$. Thus $\{{y}_{n}\}$ is a G-Cauchy sequence in X. Due to the G-completeness of X, there exists $u\in X$, such that $\{{y}_{n}\}$ is G-convergent to u.
Since the sequences
$\{S{x}_{3n}\}=\{B{x}_{3n+1}\}$,
$\{T{x}_{3n+1}\}=\{C{x}_{3n+2}\}$ and
$\{R{x}_{3n-1}\}=\{A{x}_{3n}\}$ are all subsequences of
$\{{y}_{n}\}$, they all converge to
u. We have
$\begin{array}{r}{y}_{3n}=S{x}_{3n}=B{x}_{3n+1}\to u,\phantom{\rule{2em}{0ex}}{y}_{3n+1}=T{x}_{3n+1}=C{x}_{3n+2}\to u,\\ {y}_{3n-1}=R{x}_{3n-1}=A{x}_{3n}\to u\phantom{\rule{1em}{0ex}}(n\to \mathrm{\infty})\end{array}$
(2.26)
Now we prove that u is a common fixed point of S, T, R, A, B, and C under the condition (a).
First, we suppose that A is continuous, the pair $(S,A)$ is weakly commuting, the pairs $(T,B)$ and $(R,C)$ are weakly compatible.
Step 1. We prove that $u=Su=Au$.
By (2.24) and the weakly commuting of the mapping pair
$(S,A)$ we have
$G(SA{x}_{3n},AS{x}_{3n},AS{x}_{3n})\le G(S{x}_{3n},A{x}_{3n},A{x}_{3n})\to 0\phantom{\rule{1em}{0ex}}(n\to \mathrm{\infty}).$
(2.27)
Since A is continuous, ${A}^{2}{x}_{3n}\to Au$ ($n\to \mathrm{\infty}$), $AS{x}_{3n}\to Au$ ($n\to \mathrm{\infty}$). By (2.27) we know $SA{x}_{3n}\to Au$ ($n\to \mathrm{\infty}$).
From condition (2.17) we know
$\begin{array}{r}{G}^{2}(SA{x}_{3n},T{x}_{3n+1},R{x}_{3n+2})\\ \phantom{\rule{1em}{0ex}}\le kmax\left\{\begin{array}{c}G({A}^{2}{x}_{3n},SA{x}_{3n},T{x}_{3n+1})G(B{x}_{3n+1},T{x}_{3n+1},R{x}_{3n+2}),\\ G(B{x}_{3n+1},T{x}_{3n+1},R{x}_{3n+2})G(C{x}_{3n+2},R{x}_{3n+2},SA{x}_{3n}),\\ G(C{x}_{3n+2},R{x}_{3n+2},SA{x}_{3n})G({A}^{2}{x}_{3n},SA{x}_{3n},T{x}_{3n+1})\end{array}\right\}.\end{array}$
Letting
$n\to \mathrm{\infty}$ we have
${G}^{2}(Au,u,u)\le kmax\left\{\begin{array}{c}G(Au,Au,u)G(u,u,u),\\ G(u,u,u)G(u,u,Au),\\ G(u,u,Au)G(Au,Au,u)\end{array}\right\}=kG(u,u,Au)G(Au,Au,u).$
(2.28)
If
$G(Au,u,u)\ne 0$, then from (2.28) and Proposition 1.3, we obtain
$G(Au,u,u)\le kG(Au,Au,u)\le 2kG(Au,u,u),$
which is a contradiction since $0\le k<\frac{1}{2}$. So $G(Au,u,u)=0$, this is $Au=u$.
Again, by use of condition (2.17) we have
${G}^{2}(Su,T{x}_{3n+1},R{x}_{3n+2})\le kmax\left\{\begin{array}{c}G(Au,Su,T{x}_{3n+1})G(B{x}_{3n+1},T{x}_{3n+1},R{x}_{3n+2}),\\ G(B{x}_{3n+1},T{x}_{3n+1},R{x}_{3n+2})G(C{x}_{3n+2},R{x}_{3n+2},Su),\\ G(C{x}_{3n+2},R{x}_{3n+2},Su)G(Au,Su,T{x}_{3n+1})\end{array}\right\}.$
Letting
$n\to \mathrm{\infty}$ and using
$Au=u$ we have
${G}^{2}(Su,u,u)\le kmax\left\{\begin{array}{c}G(u,Su,u)G(u,u,u),\\ G(u,u,u)G(u,u,Su),\\ G(u,u,Su)G(u,Su,u)\end{array}\right\}=k{G}^{2}(Su,u,u).$
This implies that $G(Su,u,u)=0$, and so $Su=u$. Therefore we have $u=Au=Su$.
Step 2. We prove that $u=Tu=Bu$.
Since
$S(X)\subset B(X)$ and
$u=Su\in S(X)$, there is a point
$v\in X$ such that
$u=Su=Bv$. Again by use of condition (2.15), we have
${G}^{2}(Su,Tv,R{x}_{3n+2})\le kmax\left\{\begin{array}{c}G(Au,Su,Tv)G(Bv,Tv,R{x}_{3n+2}),\\ G(Bv,Tv,R{x}_{3n+2})G(C{x}_{3n+2},R{x}_{3n+2},Su),\\ G(C{x}_{3n+2},R{x}_{3n+2},Su)G(Au,Su,Tv)\end{array}\right\}.$
Letting
$n\to \mathrm{\infty}$ and using
$u=Au=Su=Bv$ we have
${G}^{2}(u,Tv,u)\le kmax\left\{\begin{array}{c}G(u,u,Tv)G(u,Tv,u),\\ G(u,Tv,u)G(u,u,u),\\ G(u,u,u)G(u,u,Tv)\end{array}\right\}=k{G}^{2}(u,Tv,u).$
This implies that $G(u,Tv,u)=0$, and so $Tv=u$.
Since the pair
$(T,B)$ is weakly compatible, we have
Again by use of condition (2.17), we have
${G}^{2}(Su,Tu,R{x}_{3n+2})\le kmax\left\{\begin{array}{c}G(Au,Su,Tu)G(Bu,Tu,R{x}_{3n+2}),\\ G(Bu,Tu,R{x}_{3n+1})G(C{x}_{3n+2},R{x}_{3n+2},Su),\\ G(C{x}_{3n+2},R{x}_{3n+2},Su)G(Au,Su,Tu)\end{array}\right\}.$
Letting
$n\to \mathrm{\infty}$, using
$u=Au=Su$,
$Tu=Bu$, and Proposition 1.3, we have
$\begin{array}{rcl}{G}^{2}(u,Tu,u)& \le & kmax\left\{\begin{array}{c}G(u,u,Tu)G(Tu,Tu,u),\\ G(Tu,Tu,u)G(u,u,u),\\ G(u,u,u)G(u,u,Tu)\end{array}\right\}\\ =& kG(u,u,Tu)G(Tu,Tu,u)\\ \le & 2k{G}^{2}(u,Tu,u).\end{array}$
This implies that $G(u,Tu,u)=0$, and so $Tu=u$.
So we have $u=Tu=Bu$.
Step 3. We prove that $u=Ru=Cu$.
Since
$T(X)\subset C(X)$ and
$u=Tu\in T(X)$, there is a point
$w\in X$ such that
$u=Tu=Cw$. Again by use of condition (2.17), we have
${G}^{2}(Su,Tu,Rw)\le kmax\left\{\begin{array}{c}G(Au,Su,Tu)G(Bu,Tu,Rw),\\ G(Bu,Tu,Rw)G(Cw,Rw,Su),\\ G(Cw,Rw,Su)G(Au,Su,Tu)\end{array}\right\}.$
Using
$u=Au=Su$ and
$u=Tu=Bu=Cw$, we obtain
${G}^{2}(u,u,Rw)\le kmax\left\{\begin{array}{c}G(u,u,u)G(u,u,Rw),\\ G(u,u,Rw)G(u,Rw,u),\\ G(u,Rw,u)G(u,u,u)\end{array}\right\}=k{G}^{2}(u,u,Rw).$
This implies that $G(u,u,Rw)=0$, and so $Rw=u=Cw$.
Since the pair
$(R,C)$ is weakly compatible, we have
Again, by use of condition (2.17),
$u=Su=Au=Tu=Bu$,
$Ru=Cu$, and Proposition 1.3, we have
$\begin{array}{rcl}{G}^{2}(u,u,Ru)& =& {G}^{2}(Su,Tu,Ru)\\ \le & kmax\left\{\begin{array}{c}G(Au,Su,Tu)G(Bu,Tu,Ru),\\ G(Bu,Tu,Ru)G(Cu,Ru,Su),\\ G(Cu,Ru,Su)G(Au,Su,Tu)\end{array}\right\}\\ =& kmax\left\{\begin{array}{c}G(u,u,u)G(u,u,Ru),\\ G(u,u,Ru)G(Ru,Ru,u),\\ G(Ru,Ru,u)G(u,u,u)\end{array}\right\}\\ =& kG(u,u,Ru)G(Ru,Ru,u)\\ \le & 2k{G}^{2}(u,u,Ru).\end{array}$
This implies that $G(u,u,Ru)=0$, and so $Ru=u=Cu$.
Therefore u is the common fixed point of S, T, R, A, B and C when A is continuous and the pair $(S,A)$ is weakly commuting, the pairs $(T,B)$ and $(F,C)$ are weakly compatible.
Next, we suppose that S is continuous, the pair $(S,A)$ is weakly commuting, the pairs $(T,B)$ and $(R,C)$ are weakly compatible.
Step 1. We prove that $u=Su$.
By (2.24) and the weakly commuting of the mapping pair
$(S,A)$ we have
$G(SA{x}_{3n},AS{x}_{3n},AS{x}_{3n})\le G(S{x}_{3n},A{x}_{3n},A{x}_{3n})\to 0\phantom{\rule{1em}{0ex}}(n\to \mathrm{\infty}).$
(2.29)
Since S is continuous, ${S}^{2}{x}_{3n}\to Su$ ($n\to \mathrm{\infty}$), $SA{x}_{3n}\to Su$ ($n\to \mathrm{\infty}$). By (2.29) we know $AS{x}_{3n}\to Su$ ($n\to \mathrm{\infty}$).
From condition (2.17) we have
$\begin{array}{r}{G}^{2}({S}^{2}{x}_{3n},T{x}_{3n+1},R{x}_{3n+2})\\ \phantom{\rule{1em}{0ex}}\le kmax\left\{\begin{array}{c}G(AS{x}_{3n},{S}^{2}{x}_{3n},T{x}_{3n+1})G(B{x}_{3n+1},T{x}_{3n+1},R{x}_{3n+2}),\\ G(B{x}_{3n+1},T{x}_{3n+1},R{x}_{3n+2})G(C{x}_{3n+2},R{x}_{3n+2},{S}^{2}{x}_{3n}),\\ G(C{x}_{3n+2},R{x}_{3n+2},{S}^{2}{x}_{3n})G(AS{x}_{3n},{S}^{2}{x}_{3n},T{x}_{3n+1})\end{array}\right\}.\end{array}$
Letting
$n\to \mathrm{\infty}$, and using Proposition 1.3, we have
$\begin{array}{rcl}{G}^{2}(Su,u,u)& \le & kmax\left\{\begin{array}{c}G(Su,Su,u)G(u,u,u),\\ G(u,u,u)G(u,u,Su),\\ G(u,u,Su)G(Su,Su,u)\end{array}\right\}\\ =& kG(u,u,Su)G(Su,Su,u)\\ \le & 2k{G}^{2}(Su,u,u).\end{array}$
This implies that $G(Su,u,u)=0$, and so $Su=u$.
Step 2. We prove that $u=Tu=Bu$.
Since
$S(X)\subset B(X)$ and
$u=Su\in S(X)$, there is a point
$z\in X$ such that
$u=Su=Bz$. Again by use of condition (2.17), we have
$\begin{array}{r}{G}^{2}({S}^{2}{x}_{3n},Tz,R{x}_{3n+2})\\ \phantom{\rule{1em}{0ex}}\le kmax\left\{\begin{array}{c}G(AS{x}_{3n},{S}^{2}{x}_{3n},Tz)G(Bz,Tz,R{x}_{3n+2}),\\ G(Bz,Tz,R{x}_{3n+2})G(C{x}_{3n+2},R{x}_{3n+2},{S}^{2}{x}_{3n}),\\ G(C{x}_{3n+2},R{x}_{3n+2},{S}^{2}{x}_{3n})G(AS{x}_{3n},{S}^{2}{x}_{3n},Tz)\end{array}\right\}.\end{array}$
Letting
$n\to \mathrm{\infty}$ and using
$u=Su=Bz$ we have
${G}^{2}(u,Tz,u)\le kmax\left\{\begin{array}{c}G(u,u,Tz)G(u,Tz,u),\\ G(u,Tz,u)G(u,u,u),\\ G(u,u,u)G(u,u,Tz)\end{array}\right\}=k{G}^{2}(u,Tz,u).$
This implies that $G(u,Tz,u)=0$, and so $Tz=u=Bz$.
Since the pair
$(T,B)$ is weakly compatible, we have
Again, by use of condition (2.17), we have
${G}^{2}(S{x}_{3n},Tu,R{x}_{3n+2})\le kmax\left\{\begin{array}{c}G(A{x}_{3n},S{x}_{3n},Tu)G(Bu,Tu,R{x}_{3n+2}),\\ G(Bu,Tu,R{x}_{3n+2})G(C{x}_{3n+2},R{x}_{3n+2},S{x}_{3n}),\\ G(C{x}_{3n+2},R{x}_{3n+2},S{x}_{3n})G(A{x}_{3n},S{x}_{3n},Tu)\end{array}\right\}.$
Letting
$n\to \mathrm{\infty}$, using
$u=Su$,
$Tu=Bu$, and Proposition 1.3, we have
$\begin{array}{rl}{G}^{2}(u,Tu,u)& \le kmax\left\{\begin{array}{c}G(u,u,Tu)G(Tu,Tu,u),\\ G(Tu,Tu,u)G(u,u,u),\\ G(u,u,u)G(u,u,Tu)\end{array}\right\}\\ =kG(u,u,Tu)G(Tu,Tu,u)\\ \le 2k{G}^{2}(u,Tu,u).\end{array}$
This implies that $G(u,Tu,u)=0$, and so $Tu=u=Bu$.
So we have $u=Tu=Bu$.
Step 3. We prove that $u=Ru=Cu$.
Since
$T(X)\subset C(X)$ and
$u=Tu\in T(X)$, there is a point
$t\in X$ such that
$u=Tu=Ct$. Again, by use of condition (2.17), we have
${G}^{2}(S{x}_{3n},Tu,Rt)\le kmax\left\{\begin{array}{c}G(A{x}_{3n},S{x}_{3n},Tu)G(Bu,Tu,Rt),\\ G(Bu,Tu,Rt)G(Ct,Rt,S{x}_{3n}),\\ G(Ct,Rt,S{x}_{3n})G(A{x}_{3n},S{x}_{3n},Tu)\end{array}\right\}.$
Letting
$n\to \mathrm{\infty}$ and using
$u=Tu=Bu=Ct$, we obtain
${G}^{2}(u,u,Rt)\le kmax\left\{\begin{array}{c}G(u,u,u)G(u,u,Rt),\\ G(u,u,Rt)G(u,Rt,u),\\ G(u,Rt,u)G(u,u,u)\end{array}\right\}=k{G}^{2}(u,u,Rt).$
This implies that $G(u,u,Rt)=0$, and so $Rt=u=Ct$.
Since the pair
$(R,C)$ is weakly compatible, we have
Again, by use of condition (2.17), we have
${G}^{2}(S{x}_{3n},Tu,Ru)\le kmax\left\{\begin{array}{c}G(A{x}_{3n},S{x}_{3n},Tu)G(Bu,Tu,Ru),\\ G(Bu,Tu,Ru)G(Cu,Ru,S{x}_{3n}),\\ G(Cu,Ru,S{x}_{3n})G(A{x}_{3n},S{x}_{3n},Tu)\end{array}\right\}.$
Letting
$n\to \mathrm{\infty}$, using
$u=Tu=Bu$,
$Ru=Cu$, and Proposition 1.3, we have
$\begin{array}{rcl}{G}^{2}(u,u,Ru)& \le & kmax\left\{\begin{array}{c}G(u,u,u)G(u,u,Ru),\\ G(u,u,Ru)G(Ru,Ru,u),\\ G(Ru,Ru,u)G(u,u,u)\end{array}\right\}\\ =& kG(u,u,Ru)G(Ru,Ru,u)\\ \le & 2k{G}^{2}(u,u,Ru).\end{array}$
This implies that $G(u,u,Ru)=0$, and so $Ru=u=Cu$.
Step 4. We prove that $u=Au$.
Since
$R(X)\subset A(X)$ and
$u=Ru\in R(X)$, there is a point
$p\in X$ such that
$u=Ru=Ap$. Again, by use of condition (2.17), we have
${G}^{2}(Sp,Tu,Ru)\le kmax\left\{\begin{array}{c}G(Ap,Sp,Tu)G(Bu,Tu,Ru),\\ G(Bu,Tu,Ru)G(Cu,Ru,Sp),\\ G(Cu,Ru,Sp)G(Ap,Sp,Tu)\end{array}\right\}.$
Using
$u=Tu=Bu$ and
$u=Ru=Cu=Ap$, we obtain
${G}^{2}(Sp,u,u)\le kmax\{\begin{array}{c}G(u,Sp,uu)G(u,u,u),\\ G(u,u,u)G(u,u,Sp),\end{array}$