Research  Open  Published:
Common fixed point results for three pairs of selfmaps satisfying new contractive condition
Journal of Inequalities and Applicationsvolume 2014, Article number: 366 (2014)
Abstract
In this paper, we introduce a new twice power type contractive condition for six selfmappings in generalized metric spaces. Using the weakly commuting and weakly compatible conditions of selfmapping pairs, the existence and uniqueness of common fixed point in complete generalized metric spaces is discussed, and a new common fixed point theorem is obtained. We also provide illustrative examples in support of our new results. Our results generalize some wellknown comparable results in the literature due to Ye and Gu.
MSC:47H10, 54H25, 54E50.
1 Introduction and preliminaries
In 1976, Jungck [1] proved a common fixed point theorem for commuting maps, generalizing the Banach contraction principle. This theorem has many applications in mathematics. The notion of weakly commuting maps is introduced by Sessa [2]. Jungck [3] generalized the concept of weak commutativity and showed that weakly commuting maps are compatible but the converse is not true. The concept of weakly compatible maps is defined by Jungck [4].
In 2006, Mustafa and Sims [5] introduced a new notion of generalized metric space called Gmetric space. Based on the notion of generalized metric spaces, Mustafa et al. [6, 7] obtained some fixed point results for mappings satisfying different contractive conditions. Aydi [8] obtained a fixed point result for a selfmapping satisfying $(\psi ,\phi )$weakly contractive conditions. Shatanawi [9] proved some fixed point results for selfmaps in a complete Gmetric space under some contractive conditions related to a nondecreasing map $\varphi :{R}^{+}\to {R}^{+}$ with ${lim}_{n\to \mathrm{\infty}}{\varphi}^{n}(t)=0$ for all $t\ge 0$. Chugh et al. [10] obtained some fixed point results for maps satisfying property P.
In 2009, Abbas and Rhoades [11] initiated the study of a common fixed point theory in generalized metric spaces. Kaewcharoen [12] obtained some common fixed point results for contractive mappings satisfying Φmaps. Abbas et al. [13] obtained some periodic point results. Aydi et al. [14] obtained some common fixed point results for generalized weakly Gcontraction mapping. Ye and Gu [15] obtained some common fixed point theorems of three maps for a class of twice power type contraction condition. In [16], Gu and Ye introduce the concept of φweakly commuting selfmapping pairs in Gmetric space, and used this concept, they establish a new common fixed point theorem of Altman integral type mappings. Aydi [17] obtained a common fixed point theorem of integral type contraction in generalized metric spaces. Tahat et al. [18] obtained some common fixed point theorems for singlevalued and multivalued maps satisfying a generalized contraction in Gmetric spaces. Manro et al. [19] obtained some common fixed point theorems for expansion mappings in Gmetric spaces. Abbas et al. [20] and Manro et al. [21] gives some common fixed point theorems for Rweakly commuting maps in Gmetric spaces. In [22, 23], the authors proved some common fixed point theorems of weakly compatible mappings in Gmetric spaces. In [24–30], the authors proved some common fixed point results of three (or four, or six) mappings in Gmetric spaces.
Recently, Abbas et al. [31] and Mustafa et al. [32] obtained some common fixed point results for a pair of mappings satisfying the $(E.A)$ property under certain generalized strict contractive conditions. Long et al. [33] obtained some common fixed points results of two pairs of mappings when only one pair satisfies the $(E.A)$ property. Gu and Yin [34] obtained some common fixed points results of three pairs of mappings for which only two pairs need to satisfy the common $(E.A)$ property in the framework of a generalized metric space. Very recently, Gu and Shatanawi [35] used the concept of the common $(E.A)$ property, proved some common fixed point theorems for three pairs of weakly compatible selfmaps satisfying a generalized weakly Gcontraction condition in generalized metric spaces.
Very recently, Jleli and Samet [36] and Samet et al. [37] noticed that some fixed point theorems in the context of a generalized metric space can be concluded by some existing results in the setting of a (quasi)metric space. In fact, if the contraction condition of the fixed point theorem on a generalized metric space can be reduced to two variables instead of three variables, then one can construct an equivalent fixed point theorem in the setting of a usual metric space. More precisely, in [36, 37], the authors noticed that $d(x,y)=G(x,y,y)$ forms a quasimetric. Therefore, if one can transform the contraction condition of existence results in a generalized metric space in such terms, $G(x,y,y)$, then the related fixed point results become the wellknown fixed point results in the context of a quasimetric space.
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 a class of twice power type contraction maps in Gmetric spaces. The results presented in this paper extend and improve some wellknown corresponding results in the literature due to Ye and Gu [15].
The following definitions and results will be needed in the sequel.
Definition 1.1 [5]
Let X be a nonempty set and let $G:X\times X\times X\to {R}^{+}$ be a function satisfying the following properties:

(G_{1}) $G(x,y,z)=0$ if $x=y=z$;

(G_{2}) $0<G(x,x,y)$ for all $x,y\in X$ with $x\ne y$;

(G_{3}) $G(x,x,y)\le G(x,y,z)$ for all $x,y,z\in X$ with $z\ne y$;

(G_{4}) $G(x,y,z)=G(x,z,y)=G(y,z,x)=\cdots $, symmetry in all three variables;

(G_{5}) $G(x,y,z)\le G(x,a,a)+G(a,y,z)$ for all $x,y,z,a\in X$.
Then the function G is called a generalized metric, or, more specifically, a Gmetric on X, and the pair $(X,G)$ is called a Gmetric space.
It is well known that the function $G(x,y,z)$ on Gmetric space X is jointly continuous in all three of its variables, and $G(x,y,z)=0$ if and only if $x=y=z$ (see [5]).
Definition 1.2 [5]
Let $(X,G)$ be a Gmetric space and let $({x}_{n})$ be a sequence of points of X. A point $x\in X$ is said to be the limit of the sequence $({x}_{n})$ if ${lim}_{n,m\to +\mathrm{\infty}}G(x,{x}_{n},{x}_{m})=0$, and we say that the sequence $({x}_{n})$ is Gconvergent to x or $({x}_{n})$ Gconvergent to x.
Thus, ${x}_{n}\to x$ in a Gmetric space $(X,G)$ if, for any $\u03f5>0$, there exists $k\in N$ such that $G(x,{x}_{n},{x}_{m})<\u03f5$ for all $m,n\ge k$.
Proposition 1.1 [5]
Let $(X,G)$ be a Gmetric space, then the following are equivalent:

1.
$({x}_{n})$ is Gconvergent to x.

2.
$G({x}_{n},{x}_{n},x)\to 0$ as $n\to +\mathrm{\infty}$.

3.
$G({x}_{n},x,x)\to 0$ as $n\to +\mathrm{\infty}$.

4.
$G({x}_{n},{x}_{m},x)\to 0$ as $n,m\to +\mathrm{\infty}$.
Definition 1.3 [5]
Let $(X,G)$ be a Gmetric space. A sequence $({x}_{n})$ is called GCauchy if, for every $\u03f5>0$, there is $k\in N$ such that $G({x}_{n},{x}_{m},{x}_{l})<\u03f5$ for all $m,n,l\ge k$; that is, $G({x}_{n},{x}_{m},{x}_{l})\to 0$ as $n,m,l\to +\mathrm{\infty}$.
Proposition 1.2 [5]
Let $(X,G)$ be a Gmetric space. Then the following are equivalent:

1.
The sequence $({x}_{n})$ is GCauchy.

2.
For every $\u03f5>0$, there is $k\in N$ such that $G({x}_{n},{x}_{m},{x}_{m})<\u03f5$ for all $m,n\ge k$.
Definition 1.4 [5]
Let $(X,G)$ and $({X}^{\prime},{G}^{\prime})$ be Gmetric spaces, and let $f:(X,G)\to ({X}^{\prime},{G}^{\prime})$ be a function. Then f is said to be Gcontinuous at a point $a\in X$ if and only if, for every $\u03f5>0$, there is $\delta >0$ such that $x,y\in X$ and $G(a,x,y)<\delta $ imply ${G}^{\prime}(f(a),f(x),f(y))<\u03f5$. A function f is Gcontinuous at X if only if it is Gcontinuous at $a\in X$.
Definition 1.5 [5]
A Gmetric space $(X,G)$ is Gcomplete if every GCauchy sequence in $(X,G)$ is Gconvergent in X.
Definition 1.6 [19]
Two selfmappings f and g of a Gmetric space $(X,G)$ are said to be weakly commuting if $G(fgx,gfx,gfx)\le G(fx,gx,gx)$ for all x in X.
Definition 1.7 [19]
Let f and g be two selfmappings from a Gmetric 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 1.3 [5]
Let $(X,G)$ be a Gmetric space. Then, for all x, y, z, a in X, it follows that $G(x,y,y)\le 2G(y,x,x)$.
2 Main results
Theorem 2.1 Let $(X,G)$ be a complete Gmetric 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
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 Gcontinuous, the pair $(S,A)$ is weakly commuting, the pairs $(T,B)$ and $(R,C)$ are weakly compatible;

(b)
either T or B is Gcontinuous, the pair $(T,B)$ is weakly commuting, the pairs $(S,A)$ and $(R,C)$ are weakly compatible;

(c)
either F or C is Gcontinuous, 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
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 GCauchy sequence in X.
Actually, using condition (2.1) and (G_{3}) we have
This implies that
Again using condition (2.1) and (G_{3}) we have
This gives
Similarly, using condition (2.1) and (G_{3}) we have
This implies that
Combining (2.3), (2.4), and (2.5) we have
Therefore, for all $n,m\in \mathbb{N}$, $n<m$, by (G_{5}) and (G_{3}) we have
Hence $\{{y}_{n}\}$ is a GCauchy sequence in X. Since X is a complete Gmetric 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 $\{R{x}_{3n1}\}=\{A{x}_{3n}\}$ are all subsequences of $\{{y}_{n}\}$, they all converge to u. We have
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
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
Letting $n\to \mathrm{\infty}$ we have
This implies that $G(Au,u,u)$=0, and so $Au=u$.
Again by use of condition (2.1) we have
Letting $n\to \mathrm{\infty}$ we have
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
Letting $n\to \mathrm{\infty}$ and using $u=Au=Su$ we have
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
Letting $n\to \mathrm{\infty}$ and using $u=Au=Su$ and $Tu=Bu$ we have
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
Using $u=Au=Su$ and $u=Tu=Bu=Cw$, we obtain
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
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
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
Letting $n\to \mathrm{\infty}$ we have
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
Letting $n\to \mathrm{\infty}$ and using $u=Su$ we have
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
Letting $n\to \mathrm{\infty}$ and using $u=Su$ and $Tu=Bu$ we have
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
Letting $n\to \mathrm{\infty}$ and using $u=Tu=Bu$, we obtain
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
Letting $n\to \mathrm{\infty}$ and using $u=Tu=Bu$ we have
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
Using $u=Tu=Bu$ and $u=Ru=Cu$, we obtain
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
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 Gmetric 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
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 ${S}^{p}X\subset {S}^{p1}X\subset \cdots \subset SX$, $SX\subset BX$, 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
Note that $u=Au=Bu=Cu={S}^{p}u={T}^{q}u={R}^{r}u$ and ${S}^{p}Su=Su$, we obtain
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
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 selfmappings to six selfmappings.
Corollary 2.1 Let $(X,G)$ be a complete Gmetric 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
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 Gcontinuous, the pair $(S,A)$ is weakly commuting, the pairs $(T,B)$ and $(R,C)$ are weakly compatible;

(b)
either T or B is Gcontinuous, the pair $(T,B)$ is weakly commuting, the pairs $(S,A)$ and $(R,C)$ are weakly compatible;

(c)
either F or C is Gcontinuous, 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
Then
So, if
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 Gmetric 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
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 Gmetric space and let S, T, and R be three mappings of X into itself satisfying the following conditions:
or
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 Gmetric 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
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 Gcontinuous, the pair $(S,A)$ is weakly commuting, the pairs $(T,B)$ and $(R,C)$ are weakly compatible;

(b)
either T or B is Gcontinuous, the pair $(T,B)$ is weakly commuting, the pairs $(S,A)$ and $(R,C)$ are weakly compatible;

(c)
either F or C is Gcontinuous, 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
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 GCauchy sequence in X.
In fact, using condition (2.17) we have
If
then by the inequality (2.19) we obtain
which is a contradiction since $0\le k<\frac{1}{2}$, and hence
Therefore, the inequality (2.19) implies that
Again using the condition (2.17) we have
If
then the inequality (2.21) implies that
this is a contradiction, and so
Therefore, the inequality (2.21) implies that
Similarly, using condition (2.17), we have
If
then from the inequality (2.23) we get
which is a contradiction, hence we have
Therefore, the above inequality (2.23) becomes
By combining (2.20), (2.22), and (2.24), $\mathrm{\forall}n\in \mathbb{N}$, we have
Therefore, for all $m,n\in \mathbb{N}$, $m>n$, by (G_{3}), (G_{5}), and (2.25) we have
This implies that $G({y}_{n},{y}_{m},{y}_{m})\to 0$, as $n,m\to \mathrm{\infty}$. Thus $\{{y}_{n}\}$ is a GCauchy sequence in X. Due to the Gcompleteness of X, there exists $u\in X$, such that $\{{y}_{n}\}$ is Gconvergent to u.
Since the sequences $\{S{x}_{3n}\}=\{B{x}_{3n+1}\}$, $\{T{x}_{3n+1}\}=\{C{x}_{3n+2}\}$ and $\{R{x}_{3n1}\}=\{A{x}_{3n}\}$ are all subsequences of $\{{y}_{n}\}$, they all converge to u. We have
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
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
Letting $n\to \mathrm{\infty}$ we have
If $G(Au,u,u)\ne 0$, then from (2.28) and Proposition 1.3, we obtain
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
Letting $n\to \mathrm{\infty}$ and using $Au=u$ we have
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
Letting $n\to \mathrm{\infty}$ and using $u=Au=Su=Bv$ we have
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
Letting $n\to \mathrm{\infty}$, using $u=Au=Su$, $Tu=Bu$, and Proposition 1.3, we have
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
Using $u=Au=Su$ and $u=Tu=Bu=Cw$, we obtain
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
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
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
Letting $n\to \mathrm{\infty}$, and using Proposition 1.3, we have
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
Letting $n\to \mathrm{\infty}$ and using $u=Su=Bz$ we have
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
Letting $n\to \mathrm{\infty}$, using $u=Su$, $Tu=Bu$, and Proposition 1.3, we have
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
Letting $n\to \mathrm{\infty}$ and using $u=Tu=Bu=Ct$, we obtain
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
Letting $n\to \mathrm{\infty}$, using $u=Tu=Bu$, $Ru=Cu$, and Proposition 1.3, we have
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