Now, let us present one of our main results.

**Theorem 2.1**. *Let* (*X*, *d*) *be a complete metric space. Assume that g* : *X* → *X and F* : *X* × *X* → *X are two mappings satisfying*

*(H1) there exists a non-decreasing function ϕ* : [0,+∞) → [0,+∞)

*such that*
*for each t* > 0,

*and*
*for all x*,

*y*,

*u*,

*v* ∈

*X*,

*where*
*(H2) F*(*X* × *X*) ⊆ *g*(*X*), *and g*(*X*) *is a closed subset of X.*

*Then F and g have a coupled coincidence point in X.*

*Proof*. First, let us present some properties about *ϕ* which will be used in the sequel. We claim that *ϕ*(*t)* <*t* for each *t* > 0. In fact, if *ϕ*(*t*
_{0}) ≥ *t*
_{0} for some *t*
_{0} > 0, then, since *ϕ* is non-decreasing, *ϕ*
^{
n
}(*t*
_{0}) ≥ *t*
_{0} for all *n* ∈ ℕ, which contradicts the condition
.

Moreover, it is easy to see that *ϕ*(0) = 0, and thus *ϕ*(*t*) ≤ *t* for all *t* ≥ 0.

Take

*x*
_{0},

*y*
_{0} ∈

*X*. Since

*F*(

*X* ×

*X*) ⊆

*g*(

*X*), one can construct two sequences {

*x*
_{
n
}}, {

*y*
_{
n
}} in

*X* such that

For any fixed

*n* ∈ ℕ, by (H1), we have

Now, let us prove that for each

*n* ∈ ℕ,

We consider the following three cases:

**Case I**. If *M*
_{
n
}= 0 or *M*
_{
n
}= max{*d*(*gx*
_{
n
}, *gx*
_{
n-1}), *d* (*gy*
_{
n
}, *gy*
_{
n-1})}, then (2.3) obviously holds.

**Case II**. *M*
_{
n
}= *d*(*gx*
_{
n
}, *gx*
_{
n+}
_{1}) > 0.

which is a contradiction.

**Case III**. *M*
_{
n
}= *d*(*gy*
_{
n
}, *gy*
_{
n+1}) > 0.

Similar to Case II, by (2.2), we get a contradiction.

Thus, in all cases, (2.3) holds for each

*n* ∈ ℕ. In addition, combining (2.1) and (2.2), we get that for all

*n* ∈ ℕ:

Let

*ε* > 0 be fixed. Since

, by (2.5), there exists

*N* ∈ ℕ such that for all

*n* >

*N*,

Throughout the rest of this article, we denote

for each *p* ∈ ℕ and each *n* ∈ ℕ.

Let

*n* >

*N* be fixed. Let us show that for all

*p* ∈ ℕ:

By (2.5) and (2.6), we get

Next, let us show that

. By (H1), we have

i.e.,

. Thus,

If

, one can similarly show that

*a*
_{
n
}≤

*ε*. Hence, in all cases,

*a*
_{
n
}≤

*ε*, so that

. Then, by (2.6), we get

In general, in order to prove that
, one can first show that
, and then by the inequality
, the conclusion follows easily.

Now, we have proved that (2.7) holds for all

*p* ∈ ℕ, which means that {

*gx*
_{
n
}} and {

*gy*
_{
n
}} are Cauchy sequences. Then, by the completeness of

*g*(

*X*), there exist

*x*,

*y* ∈

*X* such that

Now, we claim that

*gx* =

*F*(

*x*,

*y*) and

*gy* =

*F*(

*y*,

*x*). In fact, if this is not true, then

which, together with (2.9), yield that

*c*
_{
n
}= max{

*d*(

*gx*,

*F*(

*x*,

*y*)),

*d*(

*gy*,

*F*(

*y*,

*x*))} when

*n* is sufficiently large. Letting

*n* → ∞ in (2.10) and (2.11), it follows that

This is a contradiction. Thus, *gx* = *F*(*x*, *y*) and *gy* = *F*(*y*, *x*), i.e., (*x*, *y*) is a coupled coincidence point of *F* and *g*.

**Example 2.2**. Let *X* = [2,+∞), *d*(*x*, *y*) = |*x*-*y*|, *F*(*x*, *y*) = *x* + *y*, *g*(*x*) = *x*
^{2}, and
. It is easy to verify that all the assumptions of Theorem 2.1 are satisfied. So *F* and *g* have a coupled coincidence point. In fact, we have *F*(2, 2) = *g*(2).

If *F* and *g* are *w*-compatible, we have the following result:

**Theorem 2.3**. *Suppose that all of the assumptions of Theorem* 2.1 *are satisfied, and F and g are w-compatible. Then F and g have a unique common fixed point.*

*Proof*. We give the proof in 3 steps.

then

*gx*
_{1} =

*gx*
_{2} =

*gy*
_{1} =

*gy*
_{2}. In fact, by (H1), we have

where

. Then, it follows that

which gives that *ω* = 0, i.e., *gx*
_{1} = *gx*
_{2} and *gy*
_{1} = *gy*
_{2}.

By a similar argument, in the case of

one can also show that

*gx*
_{1} =

*gy*
_{2} and

*gy*
_{1} =

*gx*
_{2}. Then, it follows that

**Step 2**. By Theorem 2.1, (

*x, y*) is a coupled coincidence point of

*F* and

*g*, i.e.,

*gx* =

*F*(

*x*,

*y*) and

*gy* =

*F*(

*y*,

*x*). Then, by Step 1, we have

*gx* =

*gy*. Let

*u* =

*gx* =

*gy*. Since

*F* and

*g* are

*w*-compatible, we have

Again by Step 1, one obtains *gu* = *gx*. Thus *u* = *gx* = *gu* = *F*(*u*, *u*), i.e., *u* is a common fixed point of *F* and *g*.

**Step 3**. Let *v* = *gv* = *F*(*v*, *v*). By Step 1, one can deduce that *gv* = *gu*. So *u* = *gu* = *gv* = *v*, which means that *u* is the unique common fixed point of *F* and *g*.