In this paper, we prove coupled coincidence and common fixed point theorems for mixed gmonotone mappings satisfying more general contractive conditions in partially ordered metric spaces. We also present results on existence and uniqueness of coupled common fixed points. Our results improve those of Luong and Thuan [29]. Our work generalizes, extends and unifies several well known comparable results in the literature.
Let Φ denote all functions \phi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) which satisfy

(1)
φ is continuous and nondecreasing,

(2)
\phi (t)=0 and only if t=0,

(3)
\phi (t+s)\le \phi (t)+\phi (s), \mathrm{\forall}t,s\in [0,\mathrm{\infty})
and Ψ denote all functions \psi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) which satisfy {lim}_{t\to r}\psi (t)>0 for all r>0 and {lim}_{t\to {0}^{+}}\psi (t)=0.
Theorem 4 Let (X,\le ) be a partially ordered set and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Let (X,\le ) be a partially ordered set and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Let F:X\times X\to X be a mapping having the mixed monotone property on X and there exist two elements {x}_{0},{y}_{0}\in X with {x}_{0}\le F({x}_{0},{y}_{0}) and {y}_{0}\ge F({y}_{0},{x}_{0}). Suppose that F, g satisfy
\phi \left(d(F(x,y),F(u,v))\right)\le \frac{1}{2}\phi (d(gx,gu)+d(gy,gv))\psi \left(\frac{d(gx,gu)+d(gy,gv)}{2}\right)
(2.1)
for all x,y,u,v\in X with gx\le gu and gy\ge gv, F(X\times X)\subseteq g(X), g(X) is complete and g is continuous.
Suppose that either

(1)
F
is continuous or

(2)
X has the following property:

(a)
if a nondecreasing sequence \{{x}_{n}\}\to x, then {x}_{n}\le x for all n\in \mathbb{N},

(b)
if a nonincreasing sequence \{{y}_{n}\}\to y, then y\le {y}_{n} for all n\in \mathbb{N}.
Then there exist
x,y\in X
such that
gx=F(x,y)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}gy=F(y,x),
that is, F and g have a coupled coincidence point in X\times X.
Proof Using Lemma 1, there exists E\subseteq X such that g(E)=g(X) and g:E\to X is onetoone. We define a mapping A:g(E)\times g(E)\to X by
A(gx,gy)=F(x,y),\phantom{\rule{1em}{0ex}}\mathrm{\forall}gx,gy\in g(E).
(2.2)
As g is onetoone on g(E), so A is well defined. Thus, it follows from (2.1) and (2.2) that
\phi (A(x,y),A(u,v))\le \frac{1}{2}\phi (d(gx,gu)+d(gy,gv))\psi \left(\frac{d(gx,gu)+d(gy,gv)}{2}\right)
(2.3)
for all gx,gy,gu,gv\in g(E) with gx\le gu and gy\ge gv. Since F has the mixed gmonotone property, for all x,y\in X, we have
{x}_{1},{x}_{2}\in X,\phantom{\rule{1em}{0ex}}g{x}_{1}\le g{x}_{2}\phantom{\rule{1em}{0ex}}\u27f9\phantom{\rule{1em}{0ex}}F({x}_{1},y)\le F({x}_{2},y)
(2.4)
and
{y}_{1},{y}_{2}\in X,\phantom{\rule{1em}{0ex}}g{y}_{1}\ge g{y}_{2}\phantom{\rule{1em}{0ex}}\u27f9\phantom{\rule{1em}{0ex}}F(x,{y}_{1})\le F(x,{y}_{2}).
(2.5)
Thus, it follows from (2.2), (2.4) and (2.5) that, for all gx,gy\in g(E),
g{x}_{1},g{x}_{2}\in g(X),\phantom{\rule{1em}{0ex}}g{x}_{1}\le g{x}_{2}\phantom{\rule{1em}{0ex}}\u27f9\phantom{\rule{1em}{0ex}}A(g{x}_{1},gy)\le A(g{x}_{2},gy)
and
g{y}_{1},g{y}_{2}\in g(X),\phantom{\rule{1em}{0ex}}g{y}_{1}\ge g{y}_{2}\phantom{\rule{1em}{0ex}}\u27f9\phantom{\rule{1em}{0ex}}A(gx,g{y}_{1})\le A(gx,g{y}_{2}),
which implies that A has the mixed monotone property.
Suppose that assumption (1) holds. Since F is continuous, A is also continuous. Using Theorem 3 with the mapping A, it follows that A has a coupled fixed point (u,v)\in g(E)\times g(E).
Suppose that assumption (2) holds. We can conclude similarly in the proof of Theorem 3 that the mapping A has a coupled fixed point (u,v)\in g(X)\times g(X).
Finally, we prove that F and g have a coupled fixed point in X. Since (u,v) is a coupled fixed point of A, we get
u=A(u,v),\phantom{\rule{2em}{0ex}}v=A(v,u).
(2.6)
Since (u,v)\in g(X)\times g(X), there exists a point ({u}^{\prime},{v}^{\prime})\in X\times X such that
u=g{u}^{\prime},\phantom{\rule{2em}{0ex}}v=g{v}^{\prime}.
(2.7)
Thus, it follows from (2.6) and (2.7) that
g{u}^{\prime}=A(g{u}^{\prime},g{v}^{\prime}),\phantom{\rule{2em}{0ex}}g{v}^{\prime}=A(g{v}^{\prime},g{u}^{\prime}).
(2.8)
Also, from (2.2) and (2.8), we get
g{u}^{\prime}=F({u}^{\prime},{v}^{\prime}),\phantom{\rule{2em}{0ex}}g{v}^{\prime}=F({v}^{\prime},{u}^{\prime}).
Therefore, ({u}^{\prime},{v}^{\prime}) is a coupled coincidence point of F and g. This completes the proof. □
Corollary 1 Let (X,\le ) be a partially ordered set and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Let (X,\le ) be a partially ordered set and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Let F:X\times X\to X be a mapping having the mixed monotone property on X and there exist two elements {x}_{0},{y}_{0}\in X with {x}_{0}\le F({x}_{0},{y}_{0}) and {y}_{0}\ge F({y}_{0},{x}_{0}). Suppose that F, g satisfy
\phi \left(d(F(x,y),F(u,v))\right)\le \frac{k}{2}d(gx,gu)+d(gy,gv)
for all x,y,u,v\in X with gx\le gu and gy\ge gv, F(X\times X)\subseteq g(X), g(X) is complete and g is continuous.
Suppose that either

(1)
F
is continuous or

(2)
X has the following property:

(a)
if a nondecreasing sequence \{{x}_{n}\}\to x, then {x}_{n}\le x for all n\in \mathbb{N},

(b)
if a nonincreasing sequence \{{y}_{n}\}\to y, then y\le {y}_{n} for all n\in \mathbb{N}.
Then there exist
x,y\in X
such that
gx=F(x,y)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}gy=F(y,x),
that is, F and g have a coupled coincidence point in X\times X.
Proof In Theorem 4, taking \phi (t)=t, we get Corollary 1. □
Corollary 2 Let (X,\le ) be a partially ordered set and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Let (X,\le ) be a partially ordered set and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Let F:X\times X\to X be a mapping having the mixed monotone property on X, and there exist two elements {x}_{0},{y}_{0}\in X with {x}_{0}\le F({x}_{0},{y}_{0}) and {y}_{0}\ge F({y}_{0},{x}_{0}). Suppose that F, g satisfy
d(F(x,y),F(u,v))\le \frac{1}{2}(d(gx,gu)+d(gy,gv))\psi \left(\frac{d(gx,gu)+d(gy,gv)}{2}\right)
for all x,y,u,v\in X with gx\le gu and gy\ge gv, F(X\times X)\subseteq g(X), g(X) is complete and g is continuous.
Suppose that either

(1)
F
is continuous or

(2)
X has the following property:

(a)
if a nondecreasing sequence \{{x}_{n}\}\to x, then {x}_{n}\le x for all n\in \mathbb{N},

(b)
if a nonincreasing sequence \{{y}_{n}\}\to y, then y\le {y}_{n} for all n\in \mathbb{N}.
Then there exist
x,y\in X
such that
gx=F(x,y)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}gy=F(y,x),
that is, F and g have a coupled coincidence point in X\times X.
Proof In Corollary 1, taking \psi (t)=\frac{1k}{2}t, we get Corollary 2. □
Theorem 5 Let (X,\le ) be a partially ordered set and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Let (X,\le ) be a partially ordered set and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Let F:X\times X\to X be a mapping having the mixed monotone property on X and there exist two elements {x}_{0},{y}_{0}\in X with {x}_{0}\le F({x}_{0},{y}_{0}) and {y}_{0}\ge F({y}_{0},{x}_{0}). Suppose that F, g satisfy
\phi \left(d(F(x,y),F(u,v))\right)\le \frac{1}{2}\phi (d(gx,gu)+d(gy,gv))\psi \left(\frac{d(gx,gu)+d(gy,gv)}{2}\right)
for all x,y,u,v\in X with gx\le gu and gy\ge gv, F(X\times X)\subseteq g(X), g(X) is complete and g is continuous.
Suppose that either

(1)
F
is continuous or

(2)
X has the following property:

(a)
if a nondecreasing sequence \{{x}_{n}\}\to x, then {x}_{n}\le x for all n\in \mathbb{N},

(b)
if a nonincreasing sequence \{{y}_{n}\}\to y, then y\le {y}_{n} for all n\in \mathbb{N}.
Then there exist
x,y\in X
such that
gx=F(x,y),\phantom{\rule{2em}{0ex}}gy=F(y,x)
and
x=gx=F(x,y),\phantom{\rule{2em}{0ex}}y=gy=F(y,x),
that is, F and g have a coupled common fixed point (x,y)\in X\times X.
Proof Following the proof of Theorem 4, F and g have a coupled coincidence point. We only have to show that x=gx and y=gy.
Now, {x}_{0} and {y}_{0} are two points in the statement of Theorem 4. Since F(X\times X)\subseteq g(X), we can choose {x}_{1},{y}_{1}\in X such that g{x}_{1}=F({x}_{0},{y}_{0}) and g{y}_{1}=F({y}_{0},{x}_{0}). In the same way, we construct g{x}_{2}=F({x}_{1},{y}_{1}) and g{y}_{2}=F({y}_{1},{x}_{1}). Continuing in this way, we can construct two sequences \{{x}_{n}\} and \{{y}_{n}\} in X such that
g{x}_{n+1}=F({x}_{n},{y}_{n})\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g{y}_{n+1}=F({y}_{n},{x}_{n}),\phantom{\rule{1em}{0ex}}\mathrm{\forall}n\ge 0.
(2.9)
Since gx\ge g{x}_{n+1} and gy\le g{y}_{n+1}, from (2.1) and (2.9), we have
\begin{array}{rcl}\phi (d(g{x}_{n+1},gx))& =& \phi \left(d(F({x}_{n},{y}_{n}),F(x,y))\right)\\ \le & \frac{1}{2}\phi (d(g{x}_{n},gx)+d(g{y}_{n},gy))\psi \left(\frac{d(g{x}_{n},gx)+d(g{y}_{n},gy)}{2}\right).\end{array}
(2.10)
Similarly, since g{y}_{n+1}\ge gy and g{x}_{n+1}\le gx, from (2.1) and (2.9), we have
\begin{array}{rcl}\phi (d(gy,g{y}_{n+1}))& =& \phi \left(d(F(y,x),F({y}_{n},{x}_{n}))\right)\\ \le & \frac{1}{2}\phi (d(gy,g{y}_{n})+d(gx,g{x}_{n}))\psi \left(\frac{d(gy,g{y}_{n})+d(gx,g{x}_{n})}{2}\right).\end{array}
(2.11)
From (2.10) and (2.11), we have
\begin{array}{c}\phi (d(g{x}_{n+1},gx))+\phi (d(gy,g{y}_{n+1}))\hfill \\ \phantom{\rule{1em}{0ex}}\le \phi (d(g{x}_{n},gx)+d(g{y}_{n},gy))2\psi \left(\frac{d(g{x}_{n},gx)+d(g{y}_{n},gy)}{2}\right).\hfill \end{array}
(2.12)
By property (3) of φ, we have
\phi (d(g{x}_{n+1},gx)+d(gy,g{y}_{n+1}))\le \phi (d(g{x}_{n+1},gx))+\phi (d(gy,g{y}_{n+1})).
(2.13)
From (2.12) and (2.13), we have
\phi (d(g{x}_{n+1},gx)+d(gy,g{y}_{n+1}))\le \phi (d(g{x}_{n},gx)+d(g{y}_{n},gy))2\psi \left(\frac{d(g{x}_{n},gx)+d(g{y}_{n},gy)}{2}\right),
which implies
\phi (d(g{x}_{n+1},gx)+d(gy,g{y}_{n+1}))\le \phi (d(g{x}_{n},gx)+d(g{y}_{n},gy)).
Using the fact that φ is nondecreasing, we get
d(g{x}_{n+1},gx)+d(gy,g{y}_{n+1})\le d(g{x}_{n},gx)+d(g{y}_{n},gy).
(2.14)
Set {\delta}_{n}=d(g{x}_{n+1},gx)+d(g{y}_{n+1},gy), then sequence \{{\delta}_{n}\} is decreasing. Therefore, there is some \delta \ge 0 such that
\underset{n\to \mathrm{\infty}}{lim}{\delta}_{n}=\underset{n\to \mathrm{\infty}}{lim}[d(g{x}_{n+1},gx)+d(g{y}_{n+1},gy)]=\delta .
We shall show that \delta =0. Suppose, to the contrary, that \delta >0. Then taking the limit as n\to \mathrm{\infty} (equivalently, {\delta}_{n}\to \delta) of both sides of (2.13) and having in mind that we suppose that {lim}_{t\to r}\psi (t)>0 for all r>0 and φ is continuous, we have
\phi (\delta )=\underset{n\to \mathrm{\infty}}{lim}\phi ({\delta}_{n})\le \underset{n\to \mathrm{\infty}}{lim}[\phi ({\delta}_{n1})2\psi \left(\frac{{\delta}_{n1}}{2}\right)]=\phi (\delta )2\underset{{\delta}_{n1}\to \delta}{lim}\psi \left(\frac{{\delta}_{n1}}{2}\right)<\phi (\delta ),
a contradiction. Thus \delta =0, that is,
\underset{n\to \mathrm{\infty}}{lim}{\delta}_{n}=\underset{n\to \mathrm{\infty}}{lim}[d(g{x}_{n+1},gx)+d(g{y}_{n+1},gy)]=0.
(2.15)
Hence d(g{x}_{n+1},gx)=0 and d(g{y}_{n+1},gy)=0, that is, x=gx and y=gy. □
Theorem 6 In addition to the hypotheses of Theorem 4, suppose that for every (x,y), (z,t) in X\times X, there exists (u,v) in X\times X that is comparable to (x,y) and (z,t), then F and g have a unique coupled fixed point.
Proof From Theorem 4, the set of coupled fixed points of F is nonempty. Suppose that (x,y) and (z,t) are coupled coincidence points of F, that is, gx=F(x,y), gy=F(y,x), gz=F(z,t) and gt=F(t,z). We will prove that
gx=gz\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}gy=gt.
By assumption, there exists (u,v) in X\times X such that (F(u,v),F(v,u)) is comparable with (F(x,y),F(y,x)) and (F(z,t),F(t,z)). Put {u}_{0}=u and {v}_{0}=v and choose {u}_{1},{v}_{1}\in X so that g{u}_{1}=F({u}_{0},{v}_{0}) and g{v}_{1}=F({v}_{0},{u}_{0}). Then, similarly as in the proof of Theorem 3, we can inductively define sequences \{g{u}_{n}\}, \{g{v}_{n}\} with
g{u}_{n+1}=F({u}_{n},{v}_{n})\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g{v}_{n+1}=F({v}_{n},{u}_{n})\phantom{\rule{1em}{0ex}}\text{for all}n.
Further set {x}_{0}=x, {y}_{0}=y, {z}_{0}=z and {t}_{0}=t, in a similar way, define the sequences \{g{x}_{n}\}, \{g{y}_{n}\} and \{g{z}_{n}\}, \{g{t}_{n}\}. Then it is easy to show that
g{x}_{n}\to F(x,y),\phantom{\rule{2em}{0ex}}g{y}_{n}\to F(y,x)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g{z}_{n}\to F(z,t),\phantom{\rule{2em}{0ex}}g{t}_{n}\to F(t,z)
as n\to \mathrm{\infty}. Since
(F(x,y),F(y,x))=(g{x}_{1},g{y}_{1})=(gx,gy)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}(F(u,v),F(v,u))=(g{u}_{1},g{v}_{1})
are comparable, then gx\le g{u}_{1} and gy\ge g{v}_{1}, or vice versa. It is easy to show that, similarly, (gx,gy) and (g{u}_{n},g{v}_{n}) are comparable for all n\ge 1, that is, gx\le g{u}_{n} and gy\ge g{v}_{n}, or vice versa. Thus from (2.1), we have
\begin{array}{rcl}\phi (d(gx,g{u}_{n+1}))& =& \phi (F(x,y),F({u}_{n},{v}_{n}))\\ \le & \frac{1}{2}\phi (d(gx,g{u}_{n})+d(gy,g{v}_{n}))\psi \left(\frac{d(gx,g{u}_{n})+d(gy,g{v}_{n})}{2}\right).\end{array}
(2.16)
Similarly,
\begin{array}{rcl}\phi (d(g{v}_{n+1},gy))& =& \phi (F({v}_{n},{u}_{n}),F(y,x))\\ \le & \frac{1}{2}\phi (d(g{v}_{n},gy)+d(g{u}_{n},gx))\psi \left(\frac{d(g{v}_{n},gy)+d(g{u}_{n},gx)}{2}\right).\end{array}
(2.17)
From (2.16), (2.17) and the property of φ, we have
\begin{array}{c}\phi (d(gx,g{u}_{n+1})+d(g{v}_{n+1},gy))\hfill \\ \phantom{\rule{1em}{0ex}}\le \phi (d(gx,g{u}_{n+1}))+\phi (d(g{v}_{n+1},gy))\hfill \\ \phantom{\rule{1em}{0ex}}\le \phi (d(gx,g{u}_{n})+d(gy,g{v}_{n}))2\psi \left(\frac{d(gx,g{u}_{n})+d(gy,g{v}_{n})}{2}\right),\hfill \end{array}
(2.18)
which implies
\phi (d(gx,g{u}_{n+1})+d(g{v}_{n+1},gy))\le \phi (d(gx,g{u}_{n})+d(gy,g{v}_{n})).
Thus,
d(gx,g{u}_{n+1})+d(g{v}_{n+1},gy)\le d(gx,g{u}_{n})+d(gy,g{v}_{n}).
That is, the sequence \{d(gx,g{u}_{n})+d(gy,g{v}_{n})\} is decreasing. Therefore, there exists \alpha \ge 0 such that
\underset{n\to \mathrm{\infty}}{lim}[d(gx,g{u}_{n})+d(gy,g{v}_{n})]=\alpha .
We shall show that \alpha =0. Suppose, to the contrary, that \alpha >0. Taking the limit as n\to \mathrm{\infty} in (2.18), we have
\phi (\alpha )\le \phi (\alpha )2\underset{n\to \mathrm{\infty}}{lim}\psi \left(\frac{d(gx,g{u}_{n})+d(gy,g{v}_{n})}{2}\right)<\phi (\alpha ),
a contradiction. Thus, \alpha =0, that is,
\underset{n\to \mathrm{\infty}}{lim}[d(gx,g{u}_{n})+d(gy,g{v}_{n})]=0.
It implies
\underset{n\to \mathrm{\infty}}{lim}d(gx,g{u}_{n})=\underset{n\to \mathrm{\infty}}{lim}d(gy,g{v}_{n})=0.
(2.19)
Similarly, we show that
\underset{n\to \mathrm{\infty}}{lim}d(gz,g{u}_{n})=\underset{n\to \mathrm{\infty}}{lim}d(gt,g{v}_{n})=0.
(2.20)
From (2.19), (2.20) and by the uniqueness of the limit, it follows that we have gx=gz and gy=gt. Hence (gx,gy) is the unique coupled point of coincidence of F and g. □
Example 1 Let X=[0,+\mathrm{\infty}) endowed with the standard metric d(x,y)=xy for all x,y\in X. Then (X,d) is a complete metric space. Define the mapping F:X\times X\to X by
F(x,y)=\{\begin{array}{ll}y& \text{if}x\ge y,\\ x& \text{if}xy.\end{array}
Suppose that g:X\to X is such that gx={x}^{2} for all x\in X and \phi (t):[0,+\mathrm{\infty})\to [0,+\mathrm{\infty}) is such that \phi (t)=t. Assume that \psi (t)=\frac{t}{1+t}.
It is easy to show that for all x,y,u,v\in X with gx\le gu and gy\ge gv, we have
\phi \left(d(F(x,y),F(u,v))\right)\le \frac{1}{2}\phi (d(gx,gu)+d(gy,gv))\psi \left(\frac{d(gx,gu)+d(gy,gv)}{2}\right).
Thus, it satisfies all the conditions of Theorem 4. So we deduce that F and g have a coupled coincidence point (x,y)\in X\times X. Here, (0,0) is a coupled coincidence point of F and g.