We now state and prove the main results of this paper.
Definition 3.1 We say that the mappings T:{X}^{k}\to {X}^{k} and G:{X}^{k}\to {X}^{k} are commutative if TG(Y)=GT(Y) for all Y\in {X}^{k}.
Definition 3.2 Let ({X}^{k},\le ) be a partially ordered set and T:{X}^{k}\to {X}^{k}, G:{X}^{k}\to {X}^{k}. We say that T is a Gisotone mapping if, for any {Y}_{1},{Y}_{2}\in {X}^{k}
G({Y}_{1})\le G({Y}_{2})\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}T({Y}_{1})\le T({Y}_{2}).
Definition 3.3 An element Y\in {X}^{k} is called a coincidence point of the mappings T:{X}^{k}\to {X}^{k} and G:{X}^{k}\to {X}^{k} if T(Y)=G(Y). Furthermore, if T(Y)=G(Y)=Y, then we say that Y is a common fixed point of T and G.
Theorem 3.4 Let (X,\le ) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Let G:{X}^{k}\to {X}^{k} and T:{X}^{k}\to {X}^{k} be a Gisotone mapping for which there exists \phi \in \mathrm{\Phi} such that for all Y\in {X}^{k}, V\in {X}^{k} with G(Y)\ge G(V),
{\rho}_{k}(T(Y),T(V))\le \phi \left({\rho}_{k}(G(Y),G(V))\right),
(3.1)
where {\rho}_{k} is defined via (2.2). Suppose T({X}^{k})\subset G({X}^{k}) and also suppose either

(a)
T is continuous, G is continuous and commutes with T or

(b)
(X,d,\le ) is regular and G({X}^{k}) is closed.
If there exists {Y}_{0}\in {X}^{k} such that G({Y}_{0})\asymp T({Y}_{0}), then T and G have a coincidence point.
Proof Since T({X}^{k})\subset G({X}^{k}), it follows that there exists {Y}_{1}\in {X}^{k} such that G({Y}_{1})=T({Y}_{0}). In general, there exists {Y}_{n}\in {X}^{k} such that G({Y}_{n+1})=T({Y}_{n}), n\ge 0. We denote {Z}_{0}=G({Y}_{0}) and
{Z}_{n+1}=G({Y}_{n+1})=T({Y}_{n}),\phantom{\rule{1em}{0ex}}n\ge 0.
(3.2)
Obviously, if {Z}_{n+1}={Z}_{n} for some n\ge 0, then there is nothing to prove. So, we may assume that {Z}_{n+1}\ne {Z}_{n} for all n\ge 0. Since G({Y}_{0})\asymp T({Y}_{0}), without loss of generality, we assume that G({Y}_{0})\le T({Y}_{0}) (the case G({Y}_{0})\ge T({Y}_{0}) is similar), that is, {Z}_{0}\le {Z}_{1}. Assume that {Z}_{n1}\le {Z}_{n}, that is, G({Y}_{n1})\le G({Y}_{n}). Since T is a Gisotone mapping, we get
{Z}_{n}=T({Y}_{n1})\le T({Y}_{n})={Z}_{n+1},
which shows that {Z}_{n}\le {Z}_{n+1} for all n\ge 0. This actually means that the sequence {\{{Z}_{n}\}}_{n=0}^{\mathrm{\infty}} is nondecreasing. Since G({Y}_{n})={Z}_{n}\ge G({Y}_{n1})={Z}_{n1}, from (3.1) and (i_{
φ
}) we have
\begin{array}{rcl}{\rho}_{k}({Z}_{n+1},{Z}_{n})& =& {\rho}_{k}(T({Y}_{n}),T({Y}_{n1}))\\ \le & \phi \left({\rho}_{k}(G({Y}_{n}),G({Y}_{n1}))\right)\\ =& \phi ({\rho}_{k}({Z}_{n},{Z}_{n1}))\\ <& {\rho}_{k}({Z}_{n},{Z}_{n1})\end{array}
(3.3)
for all n\ge 1. Hence, the sequence {\{{\delta}_{n}\}}_{n=0}^{\mathrm{\infty}} given by {\delta}_{n}={\rho}_{k}({Z}_{n+1},{Z}_{n}) is monotone decreasing and bounded below. Therefore, there exists some \delta \ge 0 such that {lim}_{n\to \mathrm{\infty}}{\delta}_{n}=\delta. We shall prove that \delta =0. Assume that \delta >0. Then by letting n\to \mathrm{\infty} in (3.3) and (ii_{
φ
}) we have
\delta \le \underset{n\to \mathrm{\infty}}{lim}\phi ({\delta}_{n1})=\underset{r\to {\delta}^{+}}{lim}\phi (r)<\delta ,
which is a contradiction. Thus,
\underset{n\to \mathrm{\infty}}{lim}{\delta}_{n}=0.
(3.4)
We claim that \{{Z}_{n}\} is a Cauchy sequence. Indeed, if it is false, then there exist \u03f5>0 and the sequences \{{Z}_{m(t)}\} and \{{Z}_{n(t)}\} of \{{Z}_{n}\} such that n(t) is the minimal in the sense that n(t)>m(t)\ge t and {\rho}_{k}({Z}_{m(t)},{Z}_{n(t)})>\u03f5. Therefore, {\rho}_{k}({Z}_{m(t)},{Z}_{n(t)1})\le \u03f5.
Using the triangle inequality, we obtain
\begin{array}{rcl}\u03f5& <& {\rho}_{k}({Z}_{m(t)},{Z}_{n(t)})\le {\rho}_{k}({Z}_{m(t)},{Z}_{n(t)1})+{\rho}_{k}({Z}_{n(t)1},{Z}_{n(t)})\\ \le & \u03f5+{\rho}_{k}({Z}_{n(t)1},{Z}_{n(t)}).\end{array}
Letting t\to \mathrm{\infty} in the above inequality and using (3.4), we get
\underset{t\to \mathrm{\infty}}{lim}{\rho}_{k}({Z}_{m(t)},{Z}_{n(t)})={\u03f5}^{+}.
(3.5)
Since n(t)>m(t), we have {Z}_{m(t)}\le {Z}_{n(t),} and hence G({Y}_{n(t)})\ge G({Y}_{m(t)}). Now, by (3.1), we have
\begin{array}{rcl}{\rho}_{k}({Z}_{n(t)+1},{Z}_{m(t)+1})& =& {\rho}_{k}(T({Y}_{n(t)}),T({Y}_{m(t)}))\\ \le & \phi \left({\rho}_{k}(G({Y}_{n(t)}),G({Y}_{m(t)}))\right)=\phi ({\rho}_{k}({Z}_{n(t)},{Z}_{m(t)})).\end{array}
Observe that
\begin{array}{rcl}{\rho}_{k}({Z}_{m(t)},{Z}_{n(t)})& \le & {\rho}_{k}({Z}_{m(t)},{Z}_{m(t)+1})+{\rho}_{k}({Z}_{m(t)+1},{Z}_{n(t)+1})+{\rho}_{k}({Z}_{n(t)+1},{Z}_{n(t)})\\ \le & {\delta}_{m(t)}+{\delta}_{n(t)}+{\rho}_{k}({Z}_{m(t)+1},{Z}_{n(t)+1})\\ \le & {\delta}_{m(t)}+{\delta}_{n(t)}+\phi ({\rho}_{k}({Z}_{n(t)},{Z}_{m(t)})).\end{array}
Letting t\to \mathrm{\infty} in the above inequality and using (3.4)(3.5), we have
\u03f5\le \underset{t\to \mathrm{\infty}}{lim}\phi ({r}_{t})=\underset{r\to {\u03f5}^{+}}{lim}\phi (r)<\u03f5,
where {r}_{t}={\rho}_{k}({Z}_{n(t)},{Z}_{m(t)}), which is a contradiction. Hence, the sequence {\{{Z}_{n}\}}_{n=0}^{\mathrm{\infty}} is a Cauchy sequence in the metric space ({X}^{k},{\rho}_{k}). On the other hand, since (X,d) is a complete metric space, thus the metric space ({X}^{k},{\rho}_{k}) is complete. Therefore, there exists \overline{Z}\in {X}^{k} such that {lim}_{n\to \mathrm{\infty}}{Z}_{n}=\overline{Z}, that is, {lim}_{n\to \mathrm{\infty}}G({Y}_{n})=\overline{Z}.
Now suppose that the assumption (a) holds. By the continuity of G, we have {lim}_{n\to \mathrm{\infty}}G(G({Y}_{n+1}))=G(\overline{Z}). On the other hand, by the commutativity of T and G, we have
G(G({Y}_{n+1}))=G(T({Y}_{n}))=T(G({Y}_{n})).
(3.6)
By (3.6) and the continuity of T, we have
G(\overline{Z})=\underset{n\to \mathrm{\infty}}{lim}G(G({Y}_{n+1}))=\underset{n\to \mathrm{\infty}}{lim}T(G({Y}_{n}))=T(\overline{Z}),
which shows that \overline{Z} is a coincidence point of T and G.
Suppose that the assumption (b) holds. Using Lemma 2.7, we have ({X}^{k},\le ,{\rho}_{k}) is regular. Since {\{{Z}_{n}\}}_{n=0}^{\mathrm{\infty}} is nondecreasing sequence that converges to \overline{Z}, in view of Definition 2.1, we have {Z}_{n}\le \overline{Z} for all n. Since G({X}^{k}) is closed and by (3.2), we obtain that there exists \overline{Y}\in {X}^{k} for which
\underset{n\to \mathrm{\infty}}{lim}G({Y}_{n})=\underset{n\to \mathrm{\infty}}{lim}T({Y}_{n})=\overline{Z}=G(\overline{Y}).
Then from (3.1), we have
{\rho}_{k}(T({Y}_{n}),T(\overline{Y}))\le \phi \left({\rho}_{k}(G({Y}_{n}),G(\overline{Y}))\right)
for all n\ge 0. Letting n\to \mathrm{\infty} in the above inequality, we have {\rho}_{k}(G(\overline{Y}),T(\overline{Y}))=0, which implies that G(\overline{Y})=T(\overline{Y}). Therefore, \overline{Y} is a coincidence point of T and G. □
Remark 3.5 Different kinds of contractive conditions are studied and we use a distinct methodology to prove Theorem 3.4. The authors proved that any number of sequences are simultaneous Cauchy sequence in [29]. However, we only need to proof that one sequence is a Cauchy sequence.
Taking k=1 in Theorem 3.4, we can obtain the following result immediately.
Corollary 3.6 Let (X,\le ) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Let G:X\to X and T:X\to X be a Gisotone mapping for which there exists \phi \in \mathrm{\Phi} such that for all Y\in X, V\in X with G(Y)\ge G(V),
d(T(Y),T(V))\le \phi \left(d(G(Y),G(V))\right).
Suppose
T(X)\subset G(X)
and also suppose either

(a)
T is continuous, G is continuous and commutes with T or

(b)
(X,d,\le ) is regular and G(X) is closed.
If there exists {Y}_{0}\in X such that G({Y}_{0})\asymp T({Y}_{0}), then T and G have a coincidence point.
Now, we will show that Theorem 3.4 allow us to derive coupled, tripled and quadruple fixedpoint theorems for mixed monotone mappings in partially ordered metric space.
Taking k=2, T(Y)=(F(x,y),F(y,x)) and G(Y)=(g(x),g(y)) for Y=(x,y)\in {X}^{2} in Theorem 3.4, we can obtain the following result.
Corollary 3.7 Let (X,\le ) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Let g:X\to X and F:{X}^{2}\to X be a mixed gmonotone mapping for which there exists \phi \in \mathrm{\Phi} such that for all x,y,u,v\in X with g(x)\ge g(u), g(y)\le g(v),
d(F(x,y),F(u,v))+d(F(y,x),F(v,u))\le 2\phi \left(\frac{d(g(x),g(u))+d(g(y),g(v))}{2}\right).
(3.7)
Suppose
F({X}^{2})\subset g(X)
and also suppose either

(a)
F is continuous, g is continuous and commutes with F or

(b)
(X,d,\le ) is regular and g(X) is closed.
If there exist
{x}_{0},{y}_{0}\in X
such that
g({x}_{0})\le F({x}_{0},{y}_{0})\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}g({y}_{0})\ge F({y}_{0},{x}_{0}),
(3.8)
or
g({x}_{0})\ge F({x}_{0},{y}_{0})\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}g({y}_{0})\le F({y}_{0},{x}_{0}),
(3.9)
then there exist \overline{x},\overline{y}\in X such that g(\overline{x})=F(\overline{x},\overline{y}) and g(\overline{y})=F(\overline{y},\overline{x}), that is, F and g have a couple coincidence point.
Proof For simplicity, we denote Y=(x,y), V=(u,v) and {Y}_{n}=({x}_{n},{y}_{n}) for all n\ge 0. We endow the product space {X}^{2} with the following partial order:
\text{for}Y,V\in {X}^{2},\phantom{\rule{1em}{0ex}}Y\le V\phantom{\rule{1em}{0ex}}\iff \phantom{\rule{1em}{0ex}}x\le u,\phantom{\rule{2em}{0ex}}y\ge v.
(3.10)
Consider the function {\rho}_{2}:{X}^{2}\times {X}^{2}\to [0,+\mathrm{\infty}) defined by
{\rho}_{2}(Y,V)=\frac{1}{2}[d(x,u)+d(y,v)],\phantom{\rule{1em}{0ex}}\mathrm{\forall}Y,V\in {X}^{2}.
(3.11)
Obviously, ({X}^{2},\le ) and {\rho}_{2} are two particular cases of ({X}^{k},\le ) and {\rho}_{k} defined by (2.1) and (2.2), respectively. Now consider the operators T:{X}^{2}\to {X}^{2} and G:{X}^{2}\to {X}^{2} defined by
T(Y)=(F(x,y),F(y,x))
(3.12)
and
G(Y)=(g(x),g(y)),\phantom{\rule{1em}{0ex}}\mathrm{\forall}Y\in {X}^{2}.
(3.13)
Since F({X}^{2})\subset g(X), we have T({X}^{2})\subset G({X}^{2}).
We claim that T is a Gisotone mapping. Indeed, suppose that G({Y}_{1})\le G({Y}_{2}), {Y}_{1},{Y}_{2}\in {X}^{2}. By (3.10) and (3.13), we have g({x}_{1})\le g({x}_{2}) and g({y}_{1})\ge g({y}_{2}). Since F is gmixed monotone, we have
F({x}_{1},{y}_{1})\le F({x}_{2},{y}_{2})\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}F({y}_{1},{x}_{1})\ge F({y}_{2},{x}_{2}),\phantom{\rule{1em}{0ex}}\mathrm{\forall}G({Y}_{1})\le G({Y}_{2}).
(3.14)
From (3.10), (3.12) and (3.14), we have
T({Y}_{1})=(F({x}_{1},{y}_{1}),F({y}_{1},{x}_{1}))\le (F({x}_{2},{y}_{2}),F({y}_{2},{x}_{2}))=T({Y}_{2}),\phantom{\rule{1em}{0ex}}\mathrm{\forall}G({Y}_{1})\le G({Y}_{2}).
Similarly, we can obtain that for any {Y}_{1},{Y}_{2}\in {X}^{2}, G({Y}_{1})\ge G({Y}_{2})\Rightarrow T({Y}_{1})\ge T({Y}_{2}). By (3.8)(3.10), we have there exists {Y}_{0}\in {X}^{2} such that G({Y}_{0})\asymp T({Y}_{0}).
From (3.11) and (3.12), we have
\frac{d(F(x,y),F(u,v))+d(F(y,x),F(v,u))}{2}={\rho}_{2}(T(Y),T(V))
and
\phi \left(\frac{d(g(x),g(u))+d(g(y),g(v))}{2}\right)=\phi \left({\rho}_{2}(G(Y),G(V))\right)
for any Y\in {X}^{2}, V\in {X}^{2}. It follows from (3.7) that
{\rho}_{2}(T(Y),T(V))\le \phi \left({\rho}_{2}(G(Y),G(V))\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall}G(Y)\ge G(V).
Now suppose that the assumption (a) holds. By the continuity of g, we have G is continuous. From (3.12), (3.13) and using the commutativity of F and g, we have, for any Y\in {X}^{2}
\begin{array}{rcl}TG(Y)& =& T(g(x),g(y))=(F(g(x),g(y)),F(g(y),g(x)))\\ =& (g(F(x,y)),g(F(y,x)))=GT(Y),\end{array}
which implies that G commutes with T. It is easy to see that T is continuous. Indeed, by (3.11), we obtain that {Y}_{n}\to Y (n\to \mathrm{\infty}) if and only if {x}_{n}\to x and {y}_{n}\to y (n\to \mathrm{\infty}). Since F is continuous, we have F({x}_{n},{y}_{n})\to F(x,y) and F({y}_{n},{x}_{n})\to F(y,x) (n\to \mathrm{\infty}), for any {Y}_{n}\to Y (n\to \mathrm{\infty}). Therefore, we have
T({Y}_{n})=(F({x}_{n},{y}_{n}),F({y}_{n},{x}_{n}))\to (F(x,y),F(y,x))=T(Y)\phantom{\rule{1em}{0ex}}(n\to \mathrm{\infty})
for any {Y}_{n}\to Y (n\to \mathrm{\infty}).
Suppose that the assumption (b) holds. It is easy to see that G({X}^{2}) is closed.
All the hypothesis of Theorem 3.4 (k=2) are satisfied, and so we deduce the existence of a coincidence point of T and G. From (3.12) and (3.13), there exists (\overline{x},\overline{y}) such that g(\overline{x})=F(\overline{x},\overline{y}) and g(\overline{y})=F(\overline{y},\overline{x}), that is, (\overline{x},\overline{y}) is a coupled coincidence point of F and g. □
Remark 3.8 Note that in the case of the condition (b) satisfied in Corollary 3.7, we omit the control conditions: g is continuous and commutes with F, which are needed in the proof of Theorem 2.1 in [8] and Theorem 3 in [7].
Taking k=3, T(Y)=(F(x,y,z),F(y,x,y),F(z,y,x)) and G(Y)=(g(x),g(y),g(z)) for Y=(x,y,z)\in {X}^{3} in Theorem 3.4, we can obtain the following result by the similar argument as we did in the proof of Corollary 3.7.
Corollary 3.9 Let (X,\le ) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Let F:{X}^{3}\to X and g:X\to X such that F has the mixed gmonotone property and F({X}^{3})\subset g(X). Assume there is a function \phi \in \mathrm{\Phi} such that
for any x,y,z,u,v,w\in X for which g(x)\ge g(u), g(v)\ge g(y) and g(z)\ge g(w). Suppose either

(a)
F is continuous, g is continuous and commutes with F or

(b)
(X,\le ,d) is regular and g(X) is closed.
If there exist
{x}_{0},{y}_{0},{z}_{0}\in X
such that
g({x}_{0})\le F({x}_{0},{y}_{0},{z}_{0}),\phantom{\rule{2em}{0ex}}g({y}_{0})\ge F({y}_{0},{x}_{0},{y}_{0})\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}g({z}_{0})\le F({z}_{0},{y}_{0},{x}_{0}),
or
g({x}_{0})\ge F({x}_{0},{y}_{0},{z}_{0}),\phantom{\rule{2em}{0ex}}g({y}_{0})\le F({y}_{0},{x}_{0},{y}_{0})\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}g({z}_{0})\ge F({z}_{0},{y}_{0},{x}_{0}),
then there exist
x,y,z\in X
such that
F(x,y,z)=g(x),\phantom{\rule{2em}{0ex}}F(y,x,y)=g(y)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}F(z,y,x)=g(z),
that is, F and g have a tripled coincidence point.
Similarly, taking k=4, T(Y)=(F(x,y,z,w),F(y,z,w,x),F(z,w,x,y),F(w,x,y,z)) and G is the identity mapping on {X}^{4} for Y=(x,y,z,w)\in {X}^{4} in Theorem 3.4, we can obtain the following result.
Corollary 3.10 Let (X,\le ) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Let F:{X}^{4}\to X such that F has the mixed monotone property. Assume there is a function \phi \in \mathrm{\Phi} such that
for any x,y,z,w,u,v,r,t\in X for which x\ge u, y\le v, z\ge r and w\le t. Suppose either

(a)
F
is continuous or

(b)
(X,\le ,d) is regular.
If there exist
{x}_{0},{y}_{0},{z}_{0},{w}_{0}\in X
such that
or
then there exist
x,y,z,w\in X
such that
F(x,y,z,w)=x,\phantom{\rule{2em}{0ex}}F(y,z,w,x)=y,\phantom{\rule{2em}{0ex}}F(z,w,x,y)=z\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}F(w,x,y,z)=w,
that is, F have a quadruple fixed point.
Theorem 3.11 In addition to the hypothesis of Theorem 3.4, suppose that for every \overline{Y},{Y}^{\ast}\in {X}^{k} there exists V\in {X}^{k} such that T(V) is comparable to T(\overline{Y}) and to T({Y}^{\ast}). Also, assume that φ is nondecreasing. Let G commute with T if the assumption (b) holds. Then T and G have a unique common fixed point, that is, there exists a unique point \overline{Z}\in {X}^{k} such that \overline{Z}=G(\overline{Z})=T(\overline{Z}).
Proof From Theorem 3.4, the set of coincidence points of T and G is nonempty. Assume that \overline{Y} and {Y}^{\ast}\in {X}^{k} are two coincidence points of T and G. We shall prove that G(\overline{Y})=G({Y}^{\ast}). Put {V}_{0}=V and choose {V}_{1}\in {X}^{k} so that G({V}_{1})=T({V}_{0}). Then, similarly to the proof of Theorem 3.4, we obtain the sequence {\{G({V}_{n})\}}_{n=1}^{\mathrm{\infty}} defined as follows: G({V}_{n+1})=T({V}_{n}), n\ge 0. Since T(\overline{Y})=G(\overline{Y}) and T(V)=G({V}_{1}) are comparable, without loss of generality, we assume that G(\overline{Y})\le G({V}_{1}). Since T is a Gisotone mapping, we have
G(\overline{Y})=T(\overline{Y})\le T({V}_{1})=G({V}_{2}).
Recursively, we get that G(\overline{Y})\le G({V}_{n}), \mathrm{\forall}n\ge 1. Thus, by the contractive condition (3.1), one gets
{\rho}_{k}(G({V}_{n+1}),G(\overline{Y}))={\rho}_{k}(T({V}_{n}),T(\overline{Y}))\le \phi \left({\rho}_{k}(G({V}_{n}),G(\overline{Y}))\right).
Thus, by the above inequality, we get
{\mathrm{\Delta}}_{n+1}\le \phi ({\mathrm{\Delta}}_{n}),\phantom{\rule{1em}{0ex}}n\ge 0,
where {\mathrm{\Delta}}_{n}={\rho}_{k}(G({V}_{n}),G(\overline{Y})). Since φ is nondecreasing, it follows that
{\mathrm{\Delta}}_{n+1}\le {\phi}^{n}({\mathrm{\Delta}}_{1}).
From the definition of Φ, we get {lim}_{n\to \mathrm{\infty}}{\phi}^{n}(t)=0, for each t>0. Then, we have {lim}_{n\to \mathrm{\infty}}{\mathrm{\Delta}}_{n}=0. Thus,
\underset{n\to \mathrm{\infty}}{lim}{\rho}_{k}(G(\overline{Y}),G({V}_{n}))=0.
(3.15)
Similarly, we obtain that
\underset{n\to \mathrm{\infty}}{lim}{\rho}_{k}(G\left({Y}^{\ast}\right),G({V}_{n}))=0.
(3.16)
Combining (3.15) and (3.16) yields that G(\overline{Y})=G({Y}^{\ast}). Since G(\overline{Y})=T(\overline{Y}), by the commutativity of T and G, we have
G(G(\overline{Y}))=G(T(\overline{Y}))=T(G(\overline{Y})).
(3.17)
Denote G(\overline{Y})=\overline{Z}. By (3.17), we have G(\overline{Z})=T(\overline{Z}), that is \overline{Z} is a coincidence point of T and G. Thus, we have G(\overline{Z})=G(\overline{Y})=\overline{Z}. Therefore, \overline{Z} is a common fixed point of T and G.
To prove the uniqueness, assume {Z}^{\ast} is another common fixed point of T and G. Then we have
{Z}^{\ast}=G\left({Z}^{\ast}\right)=G(\overline{Z})=\overline{Z}.
□
Corollary 3.12 In addition to the hypothesis of Corollary 3.7, suppose that for every (\overline{x},\overline{y}),({x}^{\ast},{y}^{\ast})\in {X}^{2} there exists (u,v)\in X\times X such that (F(u,v),F(v,u)) is comparable to (F({x}^{\ast},{y}^{\ast}),F({y}^{\ast},{x}^{\ast})) and to (F(\overline{x},\overline{y}),F(\overline{y},\overline{x})). Also, assume that φ is nondecreasing. Let g commute with F if the assumption (b) holds. Then F and g have a unique coupled common fixed point, that is, there exists a unique point (\overline{z},\overline{w})\in {X}^{2} such that
\overline{z}=g(\overline{z})=F(\overline{z},\overline{w})\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}\overline{w}=g(\overline{w})=F(\overline{w},\overline{z}).
Proof Similarly to the proof of Corollary 3.7, we can obtain all conditions of Theorem 3.4 (k=2) are satisfied. In addition, by the commutativity of g and F, we have G commutes with T. For simplicity, we denote \overline{Y}=(\overline{x},\overline{y}), {Y}^{\ast}=({x}^{\ast},{y}^{\ast}) and V=(u,v)\in {X}^{2}. By (3.12), we have
By hypothesis, there exists V\in {X}^{2} such that T(V) is comparable to T(\overline{Y}) and to T({Y}^{\ast}). Hence, there is no doubt that all conditions of Theorem 3.11 are satisfied (k=2). Therefore, there exists a unique point \overline{Z}=(\overline{z},\overline{w})\in {X}^{2} such that \overline{Z}=G(\overline{Z})=T(\overline{Z}). That is, \overline{z}=g(\overline{z})=F(\overline{z},\overline{w}) and \overline{w}=g(\overline{w})=F(\overline{w},\overline{z}). □
By the similar argument as we did in the proof of Corollary 3.12, we deduce the following corollary from Theorem 3.11 (k=3).
Corollary 3.13 In addition to the hypothesis of Corollary 3.9, suppose that for all (x,y,z) and (u,v,r) in {X}^{3}, there exists (a,b,c) in {X}^{3} such that (F(a,b,c),F(b,a,b),F(c,b,a)) is comparable to (F(x,y,z),F(y,x,y),F(z,y,x)) and (F(u,v,r),F(v,u,v),F(r,v,u)). Also, assume that φ is nondecreasing. Let g commute with F if the assumption (b) holds. Then F and g have a unique tripled common fixed point (x,y,z), that is,
x=g(x)=F(x,y,z),\phantom{\rule{2em}{0ex}}y=g(y)=F(y,x,y)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}z=g(z)=F(z,y,x).