We first state the following theorem about the existence and uniqueness of a common fixed point which can be considered as a generalization of Theorem 2.1.
Theorem 3.1 Let (X,G) be a Gmetric space. Let T:X\to X and g:X\to X be two mappings such that
G(Tx,Ty,Tz)\le kG(gx,gy,gz)
(6)
for all x, y, z. Assume that T and g satisfy the following conditions:

(A1)
T(X)\subset g(X),

(A2)
g(X) is Gcomplete,

(A3)
g is Gcontinuous and commutes with T.
If k\in [0,1), then there is a unique x\in X such that gx=Tx=x.
Proof Let {x}_{0}\in X. By assumption (A1), there exists {x}_{1}\in X such that T{x}_{0}=g{x}_{1}. By the same arguments, there exists {x}_{2}\in X such that T{x}_{1}=g{x}_{2}. Inductively, we define a sequence \{{x}_{n}\} in the following way:
g{x}_{n+1}=T{x}_{n},\phantom{\rule{1em}{0ex}}n\in \mathbb{N}.
(7)
Due to (6), we have
\begin{array}{rcl}G(g{x}_{n+2},g{x}_{n+2},g{x}_{n+1})& =& G(T{x}_{n+1},T{x}_{n+1},T{x}_{n})\\ \le & kG(g{x}_{n+1},g{x}_{n+1},g{x}_{n})\end{array}
by taking x=y={x}_{n+1} and z={x}_{n}. Thus, for each natural number n, we have
G(g{x}_{n+2},g{x}_{n+2},g{x}_{n+1})\le {k}^{n+1}G(g{x}_{1},g{x}_{1},g{x}_{0}).
(8)
We will show that \{g{x}_{n}\} is a Cauchy sequence. By the rectangle inequality, we have for m>n
\begin{array}{rcl}G(g{x}_{m},g{x}_{m},g{x}_{n})& \le & G(g{x}_{n+1},g{x}_{n+1},g{x}_{n})+G(g{x}_{n+2},g{x}_{n+2},g{x}_{n+1})\\ +\cdots +G(g{x}_{m1},g{x}_{m1},g{x}_{m2})+G(g{x}_{m},g{x}_{m},g{x}_{m1})\\ \le & {k}^{n}G(g{x}_{1},g{x}_{1},g{x}_{0})+{k}^{n+1}G(g{x}_{1},g{x}_{1},g{x}_{0})\\ +\cdots +{k}^{m2}G(g{x}_{1},g{x}_{1},g{x}_{0})+{k}^{m1}G(g{x}_{1},g{x}_{1},g{x}_{0})\\ \le & \left(\sum _{i=n}^{m1}{k}^{i}\right)G(g{x}_{1},g{x}_{1},g{x}_{0}).\end{array}
(9)
Letting n,m\to \mathrm{\infty} in (9), we get that G(g{x}_{m},g{x}_{m},g{x}_{n})\to 0. Hence, \{g{x}_{n}\} is a GCauchy sequence in g(X). Since (g(X),G) is Gcomplete, then there exists z\in X such that \{g{x}_{n}\}\to z. Since g is Gcontinuous, we have \{gg{x}_{n}\} is Gconvergent to gz. On the other hand, we have gg{x}_{n+1}=gT{x}_{n}=Tg{x}_{n} since g and T commute. Thus,
\begin{array}{rcl}G(g{x}_{n+1},Tz,Tz)& =& G(Tg{x}_{n},Tz,Tz)\\ \le & kG(gg{x}_{n},gz,gz).\end{array}
Letting n\to \mathrm{\infty} and using the fact that the metric G is continuous, we get that
G(gz,Tz,Tz)\le kG(gz,gz,gz).
Hence gz=Tz. The sequence \{g{x}_{n+1}\} is Gconvergent to z since \{g{x}_{n+1}\} is a subsequence of \{g{x}_{n}\}. So, we have
\begin{array}{rcl}G(g{x}_{n+1},gz,gz)& =& G(g{x}_{n+1},Tz,Tz)\\ =& G(T{x}_{n},Tz,Tz)\\ \le & kG(g{x}_{n},z,z).\end{array}
Letting n\to \mathrm{\infty} and using the fact that G is continuous, we obtain that
G(z,gz,gz)\le kG(z,z,z)=0.
Hence we have z=gz=Tz. We will show that z is the unique common fixed point of T and g. Suppose that, contrary to our claim, there exists another common fixed point w\in X with w\ne z. From (6) we have
G(w,w,z)=G(Tw,Tw,Tz)\le kG(w,w,z),
which is a contradiction since k<1. Hence, the common fixed point of T and g is unique. □
Theorem 3.2 Let (X,G) be a Gmetric space. Let T:X\to X and g:X\to X be two mappings such that
G(Tx,Ty,Ty)\le kG(gx,gy,gy)
(10)
for all x, y. Assume that T and g satisfy the following conditions:

(A1)
T(X)\subset g(X),

(A2)
g(X) is Gcomplete,

(A3)
g is Gcontinuous and commutes with T.
If k\in [0,1), then there is a unique x\in X such that gx=Tx=x.
Proof Following the lines of the proof of Theorem 3.1 by taking y=z, one can easily get the result. □
In [16], Ran and Reurings established the following fixed point theorem that extends the Banach contraction principle to the setting of ordered metric spaces.
Theorem 3.3 (Ran and Reurings [16])
Let (X,\u2aaf) be an ordered set endowed with a metric d and T:X\to X be a given mapping. Suppose that the following conditions hold:

(i)
(X,d) is complete;

(ii)
T is continuous and nondecreasing (with respect to ⪯);

(iii)
there exists {x}_{0}\in X such that {x}_{0}\u2aafT{x}_{0};

(iv)
there exists a constant k\in (0,1) such that for all x,y\in X with x\u2ab0y,
Then T has a fixed point. Moreover, if for all (x,y)\in X\times X there exists z\in X such that x\u2aafz and y\u2aafz, we obtain uniqueness of the fixed point.
The result of Ran and Reurings [16] can be also proved in the framework of a Gmetric space.
Theorem 3.4 Let (X,\u2aaf) be an ordered set endowed with a Gmetric and T:X\to X be a given mapping. Suppose that the following conditions hold:

(i)
(X,G) is Gcomplete;

(ii)
T is Gcontinuous and nondecreasing (with respect to ⪯);

(iii)
there exists {x}_{0}\in X such that {x}_{0}\u2aafT{x}_{0};

(iv)
there exists a constant k\in (0,1) such that for all x,y,z\in X with x\u2ab0y\u2ab0z,
G(Tx,Ty,Tz)\le kG(x,y,z).
(11)
Then T has a fixed point. Moreover, if for all (x,y)\in X\times X there exists w\in X such that x\u2aafw and y\u2aafw, we obtain uniqueness of the fixed point.
Proof Let {x}_{0}\in X be a point satisfying (iii), that is, {x}_{0}\u2aafT{x}_{0}. We define a sequence \{{x}_{n}\} in X as follows:
{x}_{n}=T{x}_{n1}\phantom{\rule{1em}{0ex}}\text{for}n\ge 1.
(12)
Regarding that T is a nondecreasing mapping together with (12), we have {x}_{0}\u2aafT{x}_{0}={x}_{1} implies {x}_{1}=T{x}_{0}\u2aafT{x}_{1}={x}_{2}. Inductively, we obtain
{x}_{0}\u2aaf{x}_{1}\u2aaf{x}_{2}\u2aaf\cdots \u2aaf{x}_{n1}\u2aaf{x}_{n}\u2aaf{x}_{n+1}\u2aaf\cdots .
(13)
Assume that there exists {n}_{0} such that {x}_{{n}_{0}}={x}_{{n}_{0}+1}. Since {x}_{{n}_{0}}={x}_{{n}_{0}+1}=T{x}_{{n}_{0}}, then {x}_{{n}_{0}} is the fixed point of T, which completes the existence part of the proof. Suppose that {x}_{n}\ne {x}_{n+1} for all n\in \mathbb{N}. Thus, by (13) we have
{x}_{0}\prec {x}_{1}\prec {x}_{2}\prec \cdots \prec {x}_{n1}\prec {x}_{n}\prec {x}_{n+1}\prec \cdots .
(14)
Put x=y={x}_{n} and z={x}_{n1} in (11). Then
\begin{array}{rcl}0& \le & G(T{x}_{n},T{x}_{n},T{x}_{n1})=G({x}_{n+1},{x}_{n+1},{x}_{n})\le kG({x}_{n},{x}_{n},{x}_{n1})\\ \le & {k}^{2}G({x}_{n1},{x}_{n1},{x}_{n2})\\ \cdots \\ \le & {k}^{n}G({x}_{1},{x}_{1},{x}_{0}).\end{array}
(15)
Then we have
0\le G({x}_{n+1},{x}_{n+1},{x}_{n})\le {k}^{n}G({x}_{1},{x}_{1},{x}_{0}),
which, upon letting n\to \mathrm{\infty}, implies
\underset{n\to \mathrm{\infty}}{lim}G({x}_{n+1},{x}_{n+1},{x}_{n})=0.
(16)
On the other hand, by Lemma 2.1 we have
G(y,y,x)\le G(y,x,x)+G(x,y,x)=2G(y,x,x).
(17)
The inequality (17) with x={x}_{n} and y={x}_{n1} becomes
G({x}_{n1},{x}_{n1},{x}_{n})=G({x}_{n},{x}_{n1},{x}_{n1})\le 2G({x}_{n1},{x}_{n},{x}_{n}).
(18)
Letting n\to \mathrm{\infty} in (18), we get
\underset{n\to \mathrm{\infty}}{lim}G({x}_{n},{x}_{n1},{x}_{n1})=0.
(19)
We will show that the sequence \{{x}_{n}\} is a Cauchy sequence in the metric space (X,{d}_{G}) where {d}_{G} is given in (1). For n\ge l we have
\begin{array}{rcl}{d}_{G}({x}_{n},{x}_{l})& \le & {d}_{G}({x}_{n},{x}_{n1})+{d}_{G}({x}_{n1},{x}_{n2})+\cdots +{d}_{G}({x}_{l+1},{x}_{l})\\ =& G({x}_{n},{x}_{n1},{x}_{n1})+G({x}_{n1},{x}_{n},{x}_{n})\\ +G({x}_{n1},{x}_{n2},{x}_{n2})+G({x}_{n2},{x}_{n1},{x}_{n1})+\cdots \\ +G({x}_{l+1},{x}_{l},{x}_{l})+G({x}_{l},{x}_{l+1},{x}_{l+1})\\ =& \sum _{i=l+1}^{n}[G({x}_{i},{x}_{i1},{x}_{i1})+G({x}_{i1},{x}_{i},{x}_{i})],\end{array}
(20)
and making use of (15) and (18), we obtain
\begin{array}{rcl}0& \le & {d}_{G}({x}_{n},{x}_{l})\le \sum _{i=l+1}^{n}3{k}^{i1}G({x}_{1},{x}_{1},{x}_{0})\\ \le & 3G({x}_{1},{x}_{1},{x}_{0})[\sum _{i=0}^{n}{k}^{i1}\sum _{i=0}^{l}{k}^{i1}].\end{array}
(21)
Hence,
{d}_{G}({x}_{n},{x}_{l})\to 0\phantom{\rule{1em}{0ex}}\text{as}n,l\to \mathrm{\infty},
(22)
that is, the sequence \{{x}_{n}\} is Cauchy in (X,{d}_{G}) and hence \{{x}_{n}\} is GCauchy in (X,G) (see Proposition 9 in [7]). Since the space (X,G) is Gcomplete, then (X,{d}_{G}) is complete (see Proposition 10 in [7]). Thus, \{{x}_{n}\} is Gconvergent to a number, say u\in X, that is,
\underset{n\to \mathrm{\infty}}{lim}G({x}_{n},{x}_{n},u)=\underset{n\to \mathrm{\infty}}{lim}G({x}_{n},u,u)=0.
(23)
We show now that u\in X is a fixed point of T, that is, u=Tu. By the Gcontinuity of T, the sequence \{T{x}_{n}\}=\{{x}_{n+1}\} converges to Tu, that is,
\underset{n\to \mathrm{\infty}}{lim}G(T{x}_{n},T{x}_{n},Tu)=\underset{n\to \mathrm{\infty}}{lim}G(T{x}_{n},Tu,Tu)=0.
(24)
The rectangle inequality on the other hand gives
\begin{array}{rcl}G(u,Tu,Tu)& \le & G(u,{x}_{n+1},{x}_{n+1})+G({x}_{n+1},Tu,Tu)\\ \le & G(u,{x}_{n+1},{x}_{n+1})+G(T{x}_{n},Tu,Tu).\end{array}
(25)
Passing to limit as n\to \mathrm{\infty} in (25), we conclude that G(u,Tu,Tu)=0. Hence, u=Tu, that is, u∈ is a fixed point of T.
To prove the uniqueness, we assume that v\in X is another fixed point of T such that v\ne u. We examine two cases. For the first case, assume that either v\u2aafu or u\u2aafv. Then we substitute x=u and y=z=v in (11) which yields G(Tv,Tu,Tu)\le kG(u,v,v). This is true only for k=1, but k\in (0,1) by definition. Thus, the fixed point of T is unique.
For the second case, we suppose that neither v\u2aafu nor u\u2aafv holds. Then by assumption (iv), there exists w\in X such that u\u2aafw and v\u2aafw. Substituting x=y=w and z=u in (11), we get that G(Tw,Tw,Tu)=G(Tw,Tw,u)\le kG(w,w,u). Since T is nondecreasing, Tu\u2aafTw. Substitute now x=y=Tw and z=Tu, which implies G({T}^{2}w,{T}^{2}w,{T}^{2}u)\le kG(Tw,Tw,Tu)\le {k}^{2}G(w,w,u). Continuing in this way, we conclude G({T}^{n}w,{T}^{n}w,u)\le {k}^{n}G(w,w,u). Passing to limit as n\to \mathrm{\infty}, we get
\underset{n\to \mathrm{\infty}}{lim}G({T}^{n}w,{T}^{n}w,u)=0.
(26)
Similarly, if we take x=y=w and z=v in (11), then we obtain
\underset{n\to \mathrm{\infty}}{lim}G({T}^{n}w,{T}^{n}w,v)=0.
(27)
From (26) and (27), we deduce \{{T}^{n}w\}\to u and \{{T}^{n}w\}\to v. The uniqueness of the limit implies that u=v. Hence, the fixed point of T is unique. □
Corollary 3.1 Let (X,\u2aaf) be an ordered set endowed with a Gmetric and T:X\to X be a given mapping. Suppose that the following conditions hold:

(i)
(X,G) is Gcomplete;

(ii)
T is Gcontinuous and nondecreasing (with respect to ⪯);

(iii)
there exists {x}_{0}\in X such that {x}_{0}\u2aafT{x}_{0};

(iv)
there exists a constant k\in (0,1) such that for all x,y\in X with x\u2ab0y,
G(Tx,Ty,Ty)\le kG(x,y,y).
(28)
Then T has a fixed point. Moreover, if for all (x,y)\in X\times X there exists w\in X such that x\u2aafw and y\u2aafw, we obtain uniqueness of the fixed point.
Proof It is sufficient to take z=y in the proof of Theorem 3.4. □
Nieto and López [49] extended the result of Ran and Reurings [16] for a mapping T not necessarily continuous by assuming an additional hypothesis on (X,\u2aaf,d).
Theorem 3.5 (Nieto and López [49])
Let (X,\u2aaf) be an ordered set endowed with a metric d and T:X\to X be a given mapping. Suppose that the following conditions hold:

(i)
(X,d) is complete;

(ii)
X is ordered complete;

(iii)
T is nondecreasing;

(iv)
there exists {x}_{0}\in X such that {x}_{0}\u2aafT{x}_{0};

(v)
there exists a constant k\in (0,1) such that for all x,y\in X with x\u2ab0y,
Then T has a fixed point. Moreover, if for all (x,y)\in X\times X there exists w\in X such that x\u2aafw and y\u2aafw, we obtain uniqueness of the fixed point.
The result of Nieto and López [49] can also be proved in the framework of Gmetric space.
Theorem 3.6 Let (X,\u2aaf) be an ordered set endowed with a Gmetric and T:X\to X be a given mapping. Suppose that the following conditions hold:

(i)
(X,G) is Gcomplete;

(ii)
X is ordered complete;

(iii)
T is nondecreasing;

(iv)
there exists {x}_{0}\in X such that {x}_{0}\u2aafT{x}_{0};

(v)
there exists a constant k\in (0,1) such that for all x,y,z\in X with x\u2ab0y\u2ab0z,
G(Tx,Ty,Tz)\le kG(x,y,z).
(29)
Then T has a fixed point. Moreover, if for all (x,y)\in X\times X there exists w\in X such that x\u2aafw and y\u2aafw, we obtain uniqueness of the fixed point.
Proof Following the lines in the proof of Theorem 3.4, we have a sequence \{{x}_{n}\} which is Gconvergent to u\in X. Due to (ii), we have {x}_{n}\u2aafu for all n. We will show that u is a fixed point of T. Suppose on the contrary that u\ne Tu, that is, {d}_{G}(u,Tu)>0. Regarding (1) and (29) with x={x}_{n}, y=z=Tu, we have
\begin{array}{rcl}0& \le & {d}_{G}({x}_{n},Tu)=G({x}_{n},Tu,Tu)+G(Tu,{x}_{n},{x}_{n})\\ =& G(T{x}_{n1},Tu,Tu)+G(Tu,T{x}_{n1},T{x}_{n1})\\ \le & k[G({x}_{n1},u,u)+G(u,{x}_{n1},{x}_{n1})].\end{array}
(30)
Passing to limit as n\to \mathrm{\infty}, we get {d}_{G}(u,Tu)=0, which is a contradiction. Hence, Tu=u. Uniqueness of u can be observed as in the proof of Theorem 3.4. □
Corollary 3.2 Let (X,\u2aaf) be an ordered set endowed with a Gmetric and T:X\to X be a given mapping. Suppose that the following conditions hold:

(i)
(X,G) is Gcomplete;

(ii)
X is ordered complete;

(iii)
T is nondecreasing;

(iv)
there exists {x}_{0}\in X such that {x}_{0}\u2aafT{x}_{0};

(v)
there exists a constant k\in (0,1) such that for all x,y\in X with x\u2ab0y,
G(Tx,Ty,Ty)\le kG(x,y,y).
(31)
Then T has a fixed point. Moreover, if for all (x,y)\in X\times X there exists w\in X such that x\u2aafw and y\u2aafw, we obtain uniqueness of the fixed point.
Proof It is sufficient to take z=y in the proof of Theorem 3.6. □
Denote by Ψ the set of functions \psi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) satisfying the following conditions:
({\mathrm{\Psi}}_{0}) {\psi}^{1}(\{0\})=0,
({\mathrm{\Psi}}_{1}) \psi (t)<t for all t>0;
({\mathrm{\Psi}}_{2}) {lim}_{r\to {t}^{+}}\psi (r)<t.
Following the work of Ćirić et al. [50], we generalize the abovementioned results by means of introducing a function g. More specifically, we modify the definitions and theorems according to the presence of the function g.
Definition 3.1 (See [50])
Let (X,\u2aaf) be an ordered set and T:X\to X and g:X\to X be given mappings. The mapping T is called gnondecreasing if for every x,y\in X,
gx\u2aafgy\phantom{\rule{1em}{0ex}}\text{implies}\phantom{\rule{1em}{0ex}}Tx\u2aafTy.
Theorem 3.7 Let (X,\u2aaf) be an ordered set endowed with a Gmetric and T:X\to X and g:X\to X be given mappings. Suppose that the following conditions hold:

(i)
(X,G) is Gcomplete;

(ii)
T is Gcontinuous;

(iii)
T is gnondecreasing;

(iv)
there exists {x}_{0}\in X such that g{x}_{0}\u2aafT{x}_{0};

(v)
T(X)\subset g(X) and g is Gcontinuous and commutes with T;

(vi)
there exists a function \phi \in \mathrm{\Psi} such that for all x,y,z\in X with gx\u2ab0gy\u2ab0gz,
G(Tx,Ty,Tz)\le \phi (G(gx,gy,gz)).
(32)
Then T and g have a coincidence point, that is, there exists w\in X such that gw=Tw.
Proof Let {x}_{0}\in X such that g{x}_{0}\u2aafT{x}_{0}. Since T(X)\subset g(X), we can choose {x}_{1} such that g{x}_{1}=T{x}_{0}. Again, by T(X)\subset g(X), we can choose {x}_{2} such that g{x}_{2}=T{x}_{1}. By repeating the same argument, we construct the sequence \{g{x}_{n}\} in the following way:
g{x}_{n+1}=T{x}_{n},\phantom{\rule{1em}{0ex}}\text{for all}n=0,1,2,\dots .
(33)
Regarding that T is a gnondecreasing mapping together with (33), we observe that
g{x}_{0}\u2aafT{x}_{0}=g{x}_{1}\phantom{\rule{1em}{0ex}}\text{implies}\phantom{\rule{1em}{0ex}}g{x}_{1}=T{x}_{0}\u2aafT{x}_{1}=g{x}_{2}.
Inductively, we obtain
g{x}_{0}\u2aafg{x}_{1}\u2aafg{x}_{2}\u2aaf\cdots \u2aafg{x}_{n1}\u2aafg{x}_{n}\u2aafg{x}_{n+1}\u2aaf\cdots .
(34)
If there exists {n}_{0} such that g{x}_{{n}_{0}}=g{x}_{{n}_{0}+1}, then g{x}_{{n}_{0}}=g{x}_{{n}_{0}+1}=T{x}_{{n}_{0}}, that is, T and g have a coincidence point which completes the proof. Assume that g{x}_{n}\ne g{x}_{n+1} for all n\in \mathbb{N}.
Regarding (34), we set x=y={x}_{n+1} and z={x}_{n} in (32). Then we get
G(T{x}_{n+1},T{x}_{n+1},T{x}_{n})\le \phi (G(g{x}_{n+1},g{x}_{n+1},g{x}_{n})),
which is equivalent to
G(g{x}_{n+2},g{x}_{n+2},g{x}_{n+1})\le \phi (G(g{x}_{n+1},g{x}_{n+1},g{x}_{n}))<G(g{x}_{n+1},g{x}_{n+1},g{x}_{n})
(35)
since \phi (t)<t for all t>0. Let {t}_{n}=G(g{x}_{n+1},g{x}_{n+1},g{x}_{n}). Then \{{t}_{n}\} is a positive nonincreasing sequence. Thus, there exists L\ge 0 such that
\underset{n\to \mathrm{\infty}}{lim}{t}_{n}={L}^{+}.
(36)
We will show that L=0. Suppose that contrary to our claim, L>0. Letting n\to \mathrm{\infty} in (35), we get
L=\underset{n\to \mathrm{\infty}}{lim}{t}_{n+1}\le \underset{n\to \mathrm{\infty}}{lim}\phi ({t}_{n})=\underset{t\to {L}^{+}}{lim}\phi (t)<L,
which is a contradiction. Hence, we have
\underset{n\to \mathrm{\infty}}{lim}G(g{x}_{n+1},g{x}_{n+1},g{x}_{n})=\underset{n\to \mathrm{\infty}}{lim}{t}_{n}=0.
(37)
We will show that \{g{x}_{n}\} is a GCauchy sequence. Suppose on the contrary that the sequence \{g{x}_{n}\} is not GCauchy. Then there exists \epsilon >0 and sequences of natural numbers \{m(k)\}, \{l(k)\} such that for each natural number k,
and we have
{c}_{k}=G(g{x}_{m(k)},g{x}_{m(k)},g{x}_{l(k)})\ge \epsilon .
(38)
Corresponding to l(k), the number m(k) is chosen to be the smallest number for which (38) holds. Hence, we have
G(g{x}_{m(k)1},g{x}_{m(k)1},g{x}_{l(k)})<\epsilon .
(39)
By using (G5), we obtain that
\begin{array}{rcl}\epsilon & \le & G(g{x}_{m(k)},g{x}_{m(k)},g{x}_{l(k)})\\ \le & G(g{x}_{m(k)},g{x}_{m(k)},g{x}_{m(k)1})+G(g{x}_{m(k)1},g{x}_{m(k)1},g{x}_{l(k)})\\ =& {t}_{m(k)1}+G(g{x}_{m(k)1},g{x}_{m(k)1},g{x}_{l(k)})\\ <& {t}_{m(k)1}+\epsilon .\end{array}
Regarding (37) and letting n\to \mathrm{\infty} in the previous inequality, we deduce
\underset{k\to \mathrm{\infty}}{lim}{c}_{k}={\epsilon}^{+}.
(40)
Again by the rectangle inequality (G5), together with (G4) and Lemma 2.1, we get that
\begin{array}{rcl}{c}_{k}& =& G(g{x}_{m(k)},g{x}_{m(k)},g{x}_{l(k)})\\ \le & G(g{x}_{m(k)},g{x}_{m(k)},g{x}_{m(k)+1})+G(g{x}_{m(k)+1},g{x}_{m(k)+1},g{x}_{l(k)+1})+G(g{x}_{l(k)+1},g{x}_{l(k)+1},g{x}_{l(k)})\\ =& {t}_{l(k)}+G(g{x}_{m(k)},g{x}_{m(k)},g{x}_{m(k)+1})+G(g{x}_{m(k)+1},g{x}_{m(k)+1},g{x}_{l(k)+1})\\ \le & {t}_{l(k)}+2G(g{x}_{m(k+1)},g{x}_{m(k)+1},g{x}_{m(k)})+G(g{x}_{m(k)+1},g{x}_{m(k)+1},g{x}_{l(k)+1})\\ \le & {t}_{l(k)}+2{t}_{m(k)}+G(g{x}_{m(k)+1},g{x}_{m(k)+1},g{x}_{l(k)+1}).\end{array}
(41)
Setting x=y={x}_{m(k)} and z={x}_{l(k)}, the inequality (32) implies
\begin{array}{rcl}G(g{x}_{m(k)+1},g{x}_{m(k)+1},g{x}_{l(k)+1})& =& G(T{x}_{m(k)},T{x}_{m(k)},T{x}_{l(k)})\\ \le & \phi (G(g{x}_{m(k)},g{x}_{m(k)},g{x}_{l(k)}))\\ =& \phi ({c}_{k}).\end{array}
(42)
Combining the inequalities (41) and (42), we find
{c}_{k}\le {t}_{l(k)}+2{t}_{m(k)}+\phi ({c}_{k}).
(43)
Taking (37) and (40) into account and letting k\to \mathrm{\infty} in (43), we obtain that
\epsilon \le \underset{k\to \mathrm{\infty}}{lim}\phi ({c}_{k})=\underset{t\to {\epsilon}^{+}}{lim}\phi (t)<\epsilon ,
which is a contradiction. Hence, \{g{x}_{n}\} is a GCauchy sequence in the Gmetric space (X,G). Since (X,G) is Gcomplete, there exists w\in X such that \{g{x}_{n}\} is Gconvergent to w. By Proposition 2.1, we have
\underset{n\to \mathrm{\infty}}{lim}G(g{x}_{n},g{x}_{n},w)=\underset{n\to \mathrm{\infty}}{lim}G(g{x}_{n},w,w)=0.
(44)
The Gcontinuity of g implies that the sequence \{gg{x}_{n}\} is Gconvergent to gw, that is,
\underset{n\to \mathrm{\infty}}{lim}G(gg{x}_{n},gg{x}_{n},gw)=\underset{n\to \mathrm{\infty}}{lim}G(gg{x}_{n},gw,gw)=0.
(45)
On the other hand, due to the commutativity of T and g, we can write
gg{x}_{n+1}=gT{x}_{n}=Tg{x}_{n},
(46)
and the Gcontinuity of T implies that the sequence \{Tg{x}_{n}\}=\{gg{x}_{n+1}\} Gconverges to Tw so that
\underset{n\to \mathrm{\infty}}{lim}G(Tg{x}_{n},Tg{x}_{n},Tw)=\underset{n\to \mathrm{\infty}}{lim}G(Tg{x}_{n},Tw,Tw)=0.
(47)
By the uniqueness of the limit, the expressions (45) and (47) yield that gw=Tw. Indeed, from the rectangle inequality, we get
\begin{array}{rcl}G(gw,Tw,Tw)& \le & G(gz,gg{x}_{n+1},gg{x}_{n+1})+G(gg{x}_{n+1},Tw,Tw)\\ \le & G(gw,gg{x}_{n+1},gg{x}_{n+1})+G(Tg{x}_{n},Tw,Tw),\end{array}
(48)
which implies G(gw,Tw,Tw)=0 upon letting n\to \mathrm{\infty}. Hence, gw=Tw. □
In the next theorem, Gcontinuity of T is no longer required. However, we require the gordered completeness of X.
Theorem 3.8 Let (X,\u2aaf) be an ordered set endowed with a Gmetric and T:X\to X and g:X\to X be given mappings. Suppose that the following conditions hold:

(i)
(X,G) is Gcomplete;

(ii)
X is gordered complete;

(iii)
T is gnondecreasing (with respect to ⪯);

(iv)
there exists {x}_{0}\in X such that g{x}_{0}\u2aafT{x}_{0};

(v)
T(X)\subset g(X) and g is Gcontinuous and commutes with T;

(vi)
there exists a function \phi \in \mathrm{\Psi} such that for all x,y,z\in X with gx\u2ab0gy\u2ab0gz,
G(Tx,Ty,Tz)\le \phi (G(gx,gy,gz)).
(49)
Then T and g have a coincidence point, that is, there exists w\in X such that gw=Tw.
Proof Following the lines of the proof of Theorem 3.7, we define a sequence \{g{x}_{n}\} and conclude that it is a GCauchy sequence in the Gcomplete, Gmetric space (X,G). Thus, there exists w\in X such that g{x}_{n} is Gconvergent to gw. Since \{g{x}_{n}\} is nondecreasing and X is gordered complete, we have g{x}_{n}\u2aafgw. If gw=g{x}_{n} for some natural number n, then T and g have a coincidence point. Indeed, gw=g{x}_{n}\u2aafg{x}_{n+1}=T{x}_{n}\u2aafgw and hence g{x}_{n}=T{x}_{n}. Suppose that gw\ne g{x}_{n}. By the rectangle inequality together with the inequality (49) and the property ({\mathrm{\Psi}}_{1}), we have
\begin{array}{rcl}G(Tw,gw,gw)& \le & G(Tw,g{x}_{n+1},g{x}_{n+1})+G(g{x}_{n+1},gw,gw)\\ \le & G(Tw,T{x}_{n},T{x}_{n})+G(g{x}_{n+1},gw,gw)\\ \le & \phi (G(gw,g{x}_{n},g{x}_{n}))+G(g{x}_{n+1},gw,gw)\\ <& G(gw,g{x}_{n},g{x}_{n})+G(g{x}_{n+1},gw,gw).\end{array}
Letting n\to \mathrm{\infty} in the inequality above, we get that G(Tw,gw,gw)=0. Hence, Tw=gw. □
If we take \phi (t)=kt, where k\in [0,1) in Theorem 3.7 and Theorem 3.8, we deduce the following corollaries, respectively.
Corollary 3.3 Let (X,\u2aaf) be an ordered set endowed with a Gmetric and T:X\to X and g:X\to X be given mappings. Suppose that the following conditions hold:

(i)
(X,G) is Gcomplete;

(ii)
T is Gcontinuous;

(iii)
T is gnondecreasing (with respect to ⪯);

(iv)
there exists {x}_{0}\in X such that g{x}_{0}\u2aafT{x}_{0};

(v)
T(X)\subset g(X) and g is Gcontinuous and commutes with T;

(vi)
there exists k\in [0,1) such that for all x,y,z\in X with gx\u2ab0gy\u2ab0gz,
G(Tx,Ty,Tz)\le kG(gx,gy,gz).
(50)
Then T and g have a coincidence point, that is, there exists w\in X such that gw=Tw.
Corollary 3.4 Let (X,\u2aaf) be an ordered set endowed with a Gmetric and T:X\to X and g:X\to X be given mappings. Suppose that the following conditions hold:

(i)
(X,G) is Gcomplete;

(ii)
X is gordered complete;

(iii)
T is gnondecreasing (with respect to ⪯);

(iv)
there exists {x}_{0}\in X such that g{x}_{0}\u2aafT{x}_{0};

(v)
T(X)\subset g(X) and g is Gcontinuous and commutes with T;

(vi)
there exists k\in [0,1) such that for all x,y,z\in X with gx\u2ab0gy\u2ab0gz,
G(Tx,Ty,Tz)\le kG(gx,gy,gz).
(51)
Then T and g have a coincidence point, that is, there exists w\in X such that gw=Tw.
If we take z=y in Theorem 3.7 and Theorem 3.8, we obtain the following particular cases.
Corollary 3.5 Let (X,\u2aaf) be an ordered set endowed with a Gmetric and T:X\to X and g:X\to X be given mappings. Suppose that the following conditions hold:

(i)
(X,G) is Gcomplete;

(ii)
T is Gcontinuous;

(iii)
T is gnondecreasing;

(iv)
there exists {x}_{0}\in X such that g{x}_{0}\u2aafT{x}_{0};

(v)
T(X)\subset g(X) and g is Gcontinuous and commutes with T;

(vi)
there exists a function \phi \in \mathrm{\Psi} such that for all x,y\in X with gx\u2ab0gy,
G(Tx,Ty,Ty)\le \phi (G(gx,gy,gy)).
(52)
Then T and g have a coincidence point, that is, there exists w\in X such that gw=Tw.
Corollary 3.6 Let (X,\u2aaf) be an ordered set endowed with a Gmetric and T:X\to X and g:X\to X be given mappings. Suppose that the following conditions hold:

(i)
(X,G) is Gcomplete;

(ii)
X is gordered complete;

(iii)
T is gnondecreasing;

(iv)
there exists {x}_{0}\in X such that g{x}_{0}\u2aafT{x}_{0};

(v)
T(X)\subset g(X) and g is Gcontinuous and commutes with T;

(vi)
there exists a function \phi \in \mathrm{\Psi} such that for all x,y\in X with gx\u2ab0gy,
G(Tx,Ty,Ty)\le \phi (G(gx,gy,gy)).
(53)
Then T and g have a coincidence point, that is, there exists w\in X such that gw=Tw.
Finally, we let g=i{d}_{X} in the Theorem 3.7 and Theorem 3.8 and conclude the following theorems.
Theorem 3.9 Let (X,\u2aaf) be an ordered set endowed with a Gmetric and T:X\to X be a given mapping. Suppose that the following conditions hold:

(i)
(X,G) is Gcomplete;

(ii)
T is Gcontinuous;

(iii)
T is nondecreasing (with respect to ⪯);

(iv)
there exists {x}_{0}\in X such that {x}_{0}\u2aafT{x}_{0};

(v)
there exists a function \phi \in \mathrm{\Psi} such that for all x,y,z\in X with x\u2ab0y\u2ab0z,
G(Tx,Ty,Tz)\le \phi (G(x,y,z)).
(54)
Then T has a fixed point.
Theorem 3.10 Let (X,\u2aaf) be an ordered set endowed with a Gmetric and T:X\to X be a given mapping. Suppose that the following conditions hold:

(i)
(X,G) is Gcomplete;

(ii)
X is ordered complete;

(iii)
T is nondecreasing;

(iv)
there exists {x}_{0}\in X such that {x}_{0}\u2aafT{x}_{0};

(v)
there exists a function \phi \in \mathrm{\Psi} such that for all x,y,z\in X with x\u2ab0y\u2ab0z,
G(Tx,Ty,Tz)\le \phi (G(x,y,z)).
(55)
Then T has a fixed point.
We next consider some equivalence conditions and their implementation on Gmetric spaces. Let \mathcal{S} denote the set of functions \beta :[0,\mathrm{\infty})\to [0,1) satisfying the condition
\beta ({t}_{n})\u27f61\phantom{\rule{1em}{0ex}}\text{implies}\phantom{\rule{1em}{0ex}}{t}_{n}\u27f60.
In 2007, Jachymski and Jóźwik [51] proved that the classes \mathcal{S} and Ψ are equivalent. Regarding this result, we state the following fixed point theorems on Gmetric spaces.
Theorem 3.11 Let (X,\u2aaf) be an ordered set endowed with a Gmetric and T:X\to X be a given mapping. Suppose that the following conditions hold:

(i)
(X,G) is Gcomplete;

(ii)
T is Gcontinuous;

(iii)
T is nondecreasing (with respect to ⪯);

(iv)
there exists {x}_{0}\in X such that {x}_{0}\u2aafT{x}_{0};

(v)
there exists a function \beta \in \mathcal{S} such that for all x,y,z\in X with x\u2ab0y\u2ab0z,
G(Tx,Ty,Tz)\le \beta (G(x,y,z))G(x,y,z).
(56)
Then T has a fixed point.
Theorem 3.12 Let (X,\u2aaf) be an ordered set endowed with a Gmetric and T:X\to X be a given mapping. Suppose that the following conditions hold:

(i)
(X,G) is Gcomplete;

(ii)
X is ordered complete;

(iii)
T is nondecreasing (with respect to ⪯);

(iv)
there exists {x}_{0}\in X such that {x}_{0}\u2aafT{x}_{0};

(v)
there exists a function \beta \in \mathcal{S} such that for all x,y,z\in X with x\u2ab0y\u2ab0z,
G(Tx,Ty,Tz)\le \beta (G(x,y,z))G(x,y,z).
(57)
Then T has a fixed point.
The two corollaries below are immediate consequences of Theorem 3.11 and Theorem 3.12.
Corollary 3.7 Let (X,\u2aaf) be an ordered set endowed with a Gmetric and T:X\to X be a given mapping. Suppose that the following conditions hold:

(i)
(X,G) is Gcomplete;

(ii)
T is Gcontinuous;

(iii)
T is nondecreasing (with respect to ⪯);

(iv)
there exists {x}_{0}\in X such that {x}_{0}\u2aafT{x}_{0};

(v)
there exists a function \beta \in \mathcal{S} such that for all x,y\in X with x\u2ab0y,
G(Tx,Ty,Ty)\le \beta (G(x,y,y))G(x,y,y).
(58)
Then T has a fixed point.
Corollary 3.8 Let (X,\u2aaf) be an ordered set endowed with a Gmetric and T:X\to X be a given mapping. Suppose that the following conditions hold:

(i)
(X,G) is Gcomplete;

(ii)
X is ordered complete;

(iii)
T is nondecreasing (with respect to ⪯);

(iv)
there exists {x}_{0}\in X such that {x}_{0}\u2aafT{x}_{0};

(v)
there exists a function \beta \in \mathcal{S} such that for all x,y\in X with x\u2ab0y,
G(Tx,Ty,Ty)\le \beta (G(x,y,y))G(x,y,y).
(59)
Then T has a fixed point.
Denote by Φ the set of functions \phi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) satisfying the conditions ({\mathrm{\Psi}}_{1}) and ({\mathrm{\Psi}}_{2}). Jachymski [52] proved the equivalence of the socalled distance functions (see Lemma 1 in [52]). Inspired by this result, we can state the following theorem.
Theorem 3.13 Let (X,\u2aaf) be an ordered set endowed with a Gmetric and T be a selfmap on a Gcomplete partially ordered Gmetric space (X,G). The following statements are equivalent:

(i)
there exist functions
\psi ,\eta \in \mathrm{\Phi}
such that
\psi (G(Tx,Ty,Tz))\le \psi (G(x,y,z))\eta (G(x,y,z)),
(60)

(ii)
there exist
\alpha \in [0,1)
and a function
\psi \in \mathrm{\Phi}
such that
\psi (G(Tx,Ty,Tz))\le \alpha \psi (G(x,y,z)),
(61)

(iii)
there exists a continuous and nondecreasing function
\alpha :[0,\mathrm{\infty})\to [0,\mathrm{\infty})
such that
\alpha (t)<t
for all
t>0
such that
G(Tx,Ty,Tz)\le \alpha (G(x,y,z)),
(62)

(iv)
there exist a function
\psi \in \mathrm{\Phi}
and a nondecreasing function
\eta :[0,\mathrm{\infty})\to [0,\mathrm{\infty})
with
{\eta}^{1}(0)=0
such that
\psi (G(Tx,Ty,Tz))\le \psi (G(x,y,z))\eta (G(x,y,z)),
(63)

(iv)
there exist a function \psi \in \mathrm{\Phi} and a lower semicontinuous function \eta :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) with {\eta}^{1}(0)=0 and {lim\hspace{0.17em}inf}_{t\to \mathrm{\infty}}\eta (t)>0 such that
\psi (G(Tx,Ty,Tz))\le \psi (G(x,y,z))\eta (G(x,y,z)),
(64)
for any x,y,z\in X with x\u2ab0y\u2ab0z.
As a consequence of Theorem 3.13, we state the next corollary.
Corollary 3.9 Let (X,\u2aaf) be an ordered set endowed with a Gmetric and T be a selfmap on a Gcomplete partially ordered Gmetric space (X,G). The following statements are equivalent:

(i)
there exist functions
\psi ,\eta \in \mathrm{\Phi}
such that
\psi (G(Tx,Ty,Ty))\le \psi (G(x,y,y))\eta (G(x,y,y)),
(65)

(ii)
there exist
\alpha \in [0,1)
and a function
\psi \in \mathrm{\Phi}
such that
\psi (G(Tx,Ty,Ty))\le \alpha \psi (G(x,y,y)),
(66)

(iii)
there exists a continuous and nondecreasing function
\alpha :[0,\mathrm{\infty})\to [0,\mathrm{\infty})
such that
\alpha (t)<t
for all
t>0
such that
G(Tx,Ty,Ty)\le \alpha (G(x,y,y)),
(67)

(iv)
there exist a function
\psi \in \mathrm{\Phi}
and a nondecreasing function
\eta :[0,\mathrm{\infty})\to [0,\mathrm{\infty})
with
{\eta}^{1}(0)=0
such that
\psi (G(Tx,Ty,Ty))\le \psi (G(x,y,y))\eta (G(x,y,y)),
(68)

(iv)
there exist a function \psi \in \mathrm{\Phi} and a lower semicontinuous function \eta :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) with {\eta}^{1}(0)=0 and {lim\hspace{0.17em}inf}_{t\to \mathrm{\infty}}\eta (t)>0 such that
\psi (G(Tx,Ty,Ty))\le \psi (G(x,y,y))\eta (G(x,y,y)),
(69)
for any x,y\in X with x\u2aafy.