In this section, we explicitly prove fixed point theorems for coupled maps on partially ordered Vfuzzy metric spaces.
Theorem 3.1
Let
\((X, V, \ast)\)
be a complete
Vfuzzy metric space, and
\((X,\preceq )\)
be a partially ordered set. Let
\(P: X \times X \rightarrow X\)
and
\(Q : X \times X \rightarrow X\)
be two mappings such that

(T1)
\(P(X \times X) \subseteq Q(X)\);

(T2)
P
has the mixed
Qmonotone property;

(T3)
there exists
\(k \in(0, 1)\)
such that
$$\begin{aligned}& V\bigl(P(x, y), P(x, y),\ldots,P(x, y), P(u, v), kt\bigr) \\& \quad \geq V(Qx, Qx, \ldots, Qx, Qu, t) \\& \qquad {}\ast V\bigl(Qx, Qx,\ldots, Qx, P(x, y), t\bigr) \\& \qquad {} \ast V\bigl(Qu, Qu,\ldots, Qu, P(u,v), t\bigr) \end{aligned}$$
for all
\(x, y, u, v \in X\)
and
\(t > 0\)
for which
\(Q(x) \leq Q(u)\)
and
\(Q(y) \geq Q(v)\)
or
\(Q(x) \geq Q(u)\)
and
\(Q(y) \leq Q(v)\);

(T4)
Q
is continuous, and
P
and
Q
are compatible.
Also suppose that

(a)
P
is continuous or

(b)
X
has the following properties:

(i)
if
\(\{x_{r} \}\)
is a nondecreasing sequence such that
\(x_{r}\rightarrow x\), then
\(x_{r} \leq x\)
for all
\(r \in N\);

(ii)
if
\(\{y_{r} \}\)
is a nonincreasing sequence
\(y_{r}\rightarrow y\), then
\(y_{r} \geq y\)
for all
\(r \in N\).
If there exist
\(x_{0}, y_{0} \in X\)
such that
\(Q(x_{0}) \leq P(x_{0},y_{0})\)
and
\(Q(y_{0}) \geq P(y_{0},x_{0})\), then
P
and
Q
have a coupled coincidence point in
X.
Proof
Let \((x_{0}, y_{0})\) be a given point in \(X \times X\) such that \(Q(x_{0}) \leq P(x_{0}, y_{0})\) and \(Q(y_{0}) \geq P(y_{0}, x_{0})\). Using (T1), choose \(x_{1}\), \(y_{1}\) such that
$$ P(x_{0}, y_{0}) = Q(x_{1}) \quad \mbox{and} \quad P(y_{0}, x_{0}) = Q(y_{1}). $$
(3)
Construct two sequences \(\{x_{r} \}\) and \(\{y_{r} \} \) in X such that \(P(x_{r}, y_{r}) = Q(x_{r+1})\) and
$$ P(y_{r}, x_{r}) = Q(y_{r+1}) \quad \mbox{for all } r \geq0. $$
(4)
Now we shall prove that
$$ Q(x_{r}) \leq Q(x_{r+1})\quad \mbox{and} \quad Q(y_{r}) \geq Q(y_{r+1}). $$
(5)
We use mathematical induction.
Step 1. Let \(r = 0\). Since \(Q(x_{0}) \leq P(x_{0}, y_{0})\) and \(Q(y_{0}) \geq P(y_{0}, x_{0})\), using condition (3), we have
$$ Q(x_{0}) \leq Q(x_{1}) \quad \mbox{and} \quad Q(y_{0}) \geq Q(y_{1}). $$
So inequalities (5) hold for \(r = 0\).
Step 2. Now suppose that (5) hold for some fixed \(s \geq0\). So we get
$$ Q(x_{s}) \leq Q(x_{s+1}) \quad \mbox{and}\quad Q(y_{s}) \geq Q(y_{s+1}). $$
Step 3. Since P has the mixed Qmonotone property, using (4), we have
$$ Q(x_{r+1}) = P(x_{r}, y_{r})\leq P(x_{r+1}, y_{r}) $$
and
$$ Q(y_{r+1}) = P(y_{r}, x_{r})\geq P(y_{r+1}, x_{r}). $$
(6)
Also, \(Q(x_{r+2}) = P(x_{r+1}, y_{r+1})\geq P(x_{r+1}, y_{r})\) and
$$ Q(y_{r+2}) = P(y_{r+1}, x_{r+1})\leq P(y_{r+1}, x_{r}). $$
(7)
From (6) and (7) we get
$$ Q(x_{r}) \leq Q(x_{r+1})\quad \mbox{and}\quad Q(y_{r})\geq Q(y_{r+1}). $$
(8)
From (T3) and (4) we get
$$\begin{aligned}& V\bigl(P(x_{r1}, y_{r1}),P(x_{r1}, y_{r1}), \ldots, P(x_{r1}, y_{r1}), P(x_{r}, y_{r}), kt\bigr) \\& \quad \geq V(Qx_{r1},Qx_{r1},\ldots ,Qx_{r1},Qx_{r},t) \\& \qquad {}\ast V\bigl(Qx_{r1},Qx_{r1},\ldots ,Qx_{r1},P(x_{r1},y_{r1}),t\bigr) \\& \qquad {}\ast V\bigl(Qx_{r},Qx_{r},\ldots, Qx_{r}, P(x_{r}, y_{r}),t\bigr), \\& V(Qx_{r},Qx_{r},\ldots, Qx_{r}, Qx_{r+1}, kt) \\& \quad \geq V(Qx_{r1},Qx_{r1},\ldots,Qx_{r1}, Qx_{r},t) \\& \qquad {}\ast V(Qx_{r},Qx_{r},\ldots, Qx_{r}, Qx_{r+1}, t). \end{aligned}$$
Now, two cases arise.
Case 1. If \(V(Qx_{r1},Qx_{r1},\ldots,Qx_{r1}, Qx_{r},t) < V(Qx_{r},Qx_{r},\ldots, Qx_{r}, Qx_{r+1}, t)\), then
$$\begin{aligned} V (Qx_{r},Qx_{r},\ldots, Qx_{r}, Qx_{r+1}, kt ) \geq& V (Qx_{r1},Qx_{r1}, \ldots,Qx_{r1}, Qx_{r},t ) \\ \geq& V \biggl(Qx_{r2},Qx_{r2},\ldots,Qx_{r2}, Qx_{r1},\frac {t}{k} \biggr) \\ \geq& V \biggl(Qx_{r2},Qx_{r2},\ldots,Qx_{r2}, Qx_{r1},\frac {t}{k^{2}} \biggr) \\ & \cdots \\ \geq& V \biggl(Qx_{0},Qx_{0},\ldots,Qx_{0}, Qx_{1},\frac {t}{k^{r1}} \biggr). \end{aligned}$$
Then by simple induction we have that, for all \(t > 0\) and \(r = 1,2, \ldots,\infty\),
$$ V (Qx_{r},Qx_{r},\ldots, Qx_{r}, Qx_{r+1}, t ) \geq V \biggl(Qx_{0},Qx_{0},\ldots, Qx_{0}, Qx_{1}, \frac{t}{k^{r1}} \biggr). $$
Thus, by condition (VF5) of the definition of a Vfuzzy metric space, for any positive integer p and real number \(t > 0\), we have
$$\begin{aligned} V (Qx_{r},Qx_{r},\ldots, Qx_{r}, Qx_{r+p}, t ) \geq& V \biggl(Qx_{r},Qx_{r}, \ldots, Qx_{r}, Qx_{r+1}, \frac{t}{p} \biggr) \\ &{}\ast V \biggl(Qx_{r+1},Qx_{r+1},\ldots, Qx_{r+1}, Qx_{r+2}, \frac{t}{p} \biggr) \\ &{} \ast\cdots p \mbox{ times} \cdots \\ &{} \ast V \biggl(Qx_{r+p1},Qx_{r+p1},\ldots, Qx_{r+p1}, Qx_{r+p}, \frac {t}{p} \biggr) \\ \geq& V \biggl(Qx_{0},Qx_{0},\ldots, Qx_{0}, Qx_{1}, \frac{t}{pk^{r1}} \biggr) \\ &{} \ast\cdots p \mbox{ times} \cdots \\ &{} \ast V \biggl(Qx_{0},Qx_{0},\ldots, Qx_{0}, Qx_{1}, \frac {t}{pk^{r+p2}} \biggr). \end{aligned}$$
Therefore, taking \(r \rightarrow\infty\), by definition (VF6) we get
$$ V(Qx_{r}, Qx_{r}, \ldots, Qx_{r}, Qx_{r+p}, t) \geq1 \ast\cdots p \mbox{ times} \cdots\ast1, $$
which implies that \(\{Qx_{n} \}\) is a Cauchy sequence in X.
Case 2. If \(V(Qx_{r1},Qx_{r1},\ldots,Qx_{r1}, Qx_{r},t) > V(Qx_{r},Qx_{r},\ldots, Qx_{r}, Qx_{r+1}, t)\), then
$$ V (Qx_{r},Qx_{r},\ldots, Qx_{r}, Qx_{r+1}, kt ) \geq V (Qx_{r},Qx_{r}, \ldots,Qx_{r}, Qx_{r+1},t ). $$
By Lemma 2.2 we get \(Qx_{r} = Qx_{r+1}\).
Thus, there exists a positive integer m such that \(r \geq m\) implies \(Qx_{r} = Qx_{m}\), ∀r, which shows that \(\{Qx_{n} \} \) is a convergent sequence and so a Cauchy sequence in X.
Taking \(x = y_{r}\), \(y = x_{r}\), \(u = y_{r1}\), \(v = x_{r1}\) in (T3), we get
$$\begin{aligned}& V\bigl(P(y_{r}, x_{r}), P(y_{r}, x_{r}), \ldots, P(y_{r}, x_{r}), P(y_{r1}, x_{r1}), kt\bigr) \\& \quad \geq V(Qy_{r}, Qy_{r}, \ldots, Qy_{r}, Qu, t) \\& \qquad {}\ast V\bigl(Qy_{r}, Qy_{r}, \ldots, Qy_{r}, P(y_{r}, x_{r}), t\bigr) \\& \qquad {}\ast V\bigl(Qy_{r1}, Qy_{r1}, \ldots, Qy_{r1}, P(y_{r1}, x_{r1}), t\bigr). \end{aligned}$$
So, from equation (4) we have
$$\begin{aligned} V(Qy_{r},Qy_{r}, \ldots, Qy_{r}, Qy_{r+1},kt) \geq& V(Qy_{r1},Qy_{r1}, \ldots, Qy_{r1}, Qy_{r},t) \\ &{}\ast V(Qy_{r},Qy_{r}, \ldots, Qy_{r}, Qy_{r+1},t). \end{aligned}$$
In the same way (discussed before), \(\{Qy_{n} \}\) is a Cauchy sequence in X.
Since X is a complete space, there exist \(x,y \in X\) such that
$$ \lim_{r \rightarrow\infty} P(x_{r}, y_{r}) = \lim _{r \rightarrow\infty } Q(x_{r}) = x,\qquad \lim_{r \rightarrow\infty} P(y_{r}, x_{r}) = \lim_{r \rightarrow\infty} Q(y_{r}) = y. $$
(9)
By considering condition (T4) and \(r \rightarrow\infty\) we have
$$ V\bigl(Q\bigl(P(x_{r}, y_{r})\bigr),Q\bigl(P(x_{r}, y_{r})\bigr), \ldots, Q\bigl(P(x_{r}, y_{r}) \bigr), P\bigl(Q(x_{r}), Q(y_{r}),t\bigr)\bigr)\rightarrow 1 $$
and
$$ V\bigl(Q\bigl(P(y_{r}, x_{r})\bigr),Q \bigl(P(y_{r}, x_{r})\bigr), \ldots,Q\bigl(P(y_{r}, x_{r})\bigr), P\bigl(Q(y_{r}), Q(x_{r}),t\bigr) \bigr)\rightarrow 1 $$
(10)
as \(r \rightarrow\infty\).
By conditions (T4) and (a), since P and Q are continuous, from (10) we have
$$ V\bigl(Q(x), Q(x), \ldots, Q(x), P(x, y), t\bigr) = 1 $$
and
$$ V\bigl(Q(y), Q(y), \ldots, Q(y), P(y, x), t\bigr) = 1. $$
This implies that \(P(x, y) = Q(x)\) and \(P(y, x) = Q(y)\), and thus, we have proved that P and Q have a coupled coincidence point in X.
Now, suppose that conditions (T4) and (b) hold. Since Q is continuous and P, Q are compatible mappings, we have
$$ \lim_{r \rightarrow\infty} P\bigl(Q(x_{r}),Q(y_{r}) \bigr) = \lim_{r \rightarrow \infty} Q\bigl(P(x_{r},y_{r}) \bigr) = \lim_{r \rightarrow\infty} Q(Qx_{r}) = Q(x) $$
and
$$ \lim_{r \rightarrow\infty} P\bigl(Q(y_{r}),Q(x_{r}) \bigr) = \lim_{r \rightarrow \infty} Q\bigl(P(y_{r},x_{r}) \bigr) = \lim_{r \rightarrow\infty} Q(Qy_{r}) = Q(y). $$
(11)
By condition (VF5) of a Vfuzzy metric space, as \(r \rightarrow \infty\), we get
$$\begin{aligned} V\bigl(Qx, Qx, \ldots, Qx, P(x, y),t\bigr) \geq& V\bigl(Qx, Qx, \ldots, Qx, Q(Qx_{r+1}), t  kt\bigr) \\ &{} \ast V\bigl(Q(Qx_{r+1}), Q(Qx_{r+1}), \ldots,Q(Qx_{r+1}), P(x, y), kt\bigr) \\ =& V\bigl(Qx, Qx, \ldots,Qx, Q\bigl(P(x_{r}, y_{r}) \bigr), t  kt\bigr) \\ &{} \ast V\bigl(Q\bigl(P(x_{r}, y_{r})\bigr), Q \bigl(P(x_{r}, y_{r})\bigr), \ldots, \\ & Q\bigl(P(x_{r}, y_{r})\bigr), P(x, y), kt\bigr) \\ \geq& V\bigl(Q\bigl(P(x_{r}, y_{r})\bigr), Q \bigl(P(x_{r}, y_{r})\bigr), \ldots, \\ & Q\bigl(P(x_{r}, y_{r})\bigr), P(x, y), kt\bigr). \end{aligned}$$
We get
$$\begin{aligned} V\bigl(Qx, Qx, \ldots,Qx, P(x, y),t\bigr) \geq& V \bigl(P(Qx_{r}, Qy_{r}),P(Qx_{r}, Qy_{r}),\ldots, \\ &{} P(Qx_{r}, Qy_{r}), P(x, y), kt\bigr). \end{aligned}$$
(12)
Using condition (T3) and equations (11)(12), we get
$$\begin{aligned} V\bigl(Qx, Qx, \ldots,Qx, P(x, y),t\bigr) \geq& V\bigl(Q(Qx_{r}), Q(Qx_{r}),\ldots, Q(Qx_{r}), Qx, t\bigr) \\ &{} \ast V\bigl(Q(Qx_{r}), Q(Qx_{r}),\ldots, P(Qx_{r}, Qy_{r}), t\bigr) \\ &{} \ast V\bigl(Qx, Qx,\ldots, Qx, P(x,y), t\bigr) \\ \geq& V\bigl(Qx, Qx,\ldots, Qx, P(x,y), t\bigr). \end{aligned}$$
By Lemma 2.2 we have \(P(x, y) = Q(x)\). Similarly, we get \(P(y, x) = Q(y)\). Hence, we proved that P and Q have a coupled coincidence point in X. □
Theorem 3.2
Assume that
X
is a totally ordered set in addition to the hypotheses of Theorem
3.1. Then
P
and
Q
have a unique common coupled fixed point.
Proof
Suppose that \((x, y)\) and \((l, m)\) are a coupled coincidence point of P and Q, that is,
$$ P(x, y) = Q(x), P(y, x) = Q(y) $$
and
$$ P(l,m) = Q(l), P(m, l) = Q(m). $$
Let us show that \(Q(x) = Q(l)\), \(Q(y) = Q(m)\).
If X is a totally ordered set, then for all \((x,y), (l,m) \in X \times X\), there exists \((\alpha, \beta) \in X \times X\) such that \((P(\alpha, \beta), P(\beta, \alpha))\) is comparable with \((P(x,y),P(y,x))\), \((P(l,m),P(m,l))\).
The sequences \(\{Q(\alpha_{r}) \}\), \(\{Q(\beta_{r}) \} \) and their limits are defined similarly as in Theorem 3.1, so that
$$ P(\alpha_{r},\beta_{r}) = Q(\alpha_{r+1}),\qquad P(\beta_{r},\alpha_{r}) = Q(\beta_{r+1}). $$
By condition (T3) we have
$$\begin{aligned} V(Qx, Qx, \ldots,Qx, Q\alpha_{r+1},kt) =& V\bigl(Qx,Qx, \ldots, Qx, P(\alpha _{r},\beta_{r}), t\bigr) \\ \geq& V(Qx, Qx, \ldots, Qx, Q\alpha_{r}, t) \ast V\bigl(Qx,Qx, \ldots, Qx, P(x, y), t\bigr) \\ &{} \ast V\bigl(Q\alpha_{r}, Q\alpha_{r},\ldots, Q \alpha_{r}, P(\alpha _{r},\beta_{r}), t\bigr) \\ \geq& V(Qx, Qx,\ldots, Qx, Q\alpha_{r}, t) \\ &{} \ast V\bigl(Q\alpha_{r}, Q\alpha_{r}, \ldots,Q \alpha_{r}, P(\alpha _{r},\beta_{r}), t\bigr). \end{aligned}$$
We obtained that
$$ Q(x) = Q(\alpha) \quad \mbox{as } r\rightarrow\infty. $$
(13)
Again by (T3),
$$\begin{aligned} V(Qy, Qy, \ldots,Qy, Q\beta_{r+1},kt) =& V\bigl(Qy,Qy, \ldots, Qy, P( \beta _{r}, \alpha_{r}), t\bigr) \\ \geq& V(Qy, Qy, \ldots, Qy, Q\beta_{r}, t) \\ &{} \ast V\bigl(Qy,Qy, \ldots, Qy, P(y, x), t\bigr) \\ &{} \ast V\bigl(Q\beta_{r}, Q\beta_{r},\ldots, Q \beta_{r}, P(\beta_{r}, \alpha _{r}), t\bigr) \\ \geq& V(Qy, Qy,\ldots, Qy, Q\alpha_{r}, t) \\ &{} \ast V\bigl(Q\alpha_{r}, Q\alpha_{r}, \ldots,Q \alpha_{r}, P(\beta_{r}, \alpha_{r}), t\bigr). \end{aligned}$$
Letting \(r \rightarrow\infty\), we have
$$ \lim_{r \rightarrow\infty}Q(\beta_{r}) = Q(y). $$
(14)
Following the above steps with the help of condition (T3), we obtain
$$ \lim_{r \rightarrow\infty} Q(\alpha_{r}) = Q(l), \qquad \lim_{r \rightarrow \infty} Q(\beta_{r}) = Q(m). $$
(15)
From (13), (14), (15), and the definition of a Vfuzzy metric space we have
$$\begin{aligned} V(Qx,Qx, \ldots, Qx, Ql, t) \geq& V \biggl(Qx,Qx, \ldots, Qx, Q\alpha _{r+1}, \frac{t}{2} \biggr) \\ &{}\ast V \biggl(Q\alpha_{r+1},Q\alpha_{r+1}, \ldots, Q \alpha_{r+1}, Ql, \frac{t}{2} \biggr) \rightarrow1 \quad \mbox{as } r \rightarrow\infty. \end{aligned}$$
This implies that
Now, we can easily prove that
So, we have \(P(x, y) = Q(x)\), \(P(y, x) = Q(y)\), and the compatibility of P and Q implies the wcompatibility of P and Q given by
$$ \begin{aligned} &Q\bigl(Q(x)\bigr) = Q\bigl(P(x,y)\bigr) = P \bigl(Q(x), Q(y)\bigr), \\ &Q\bigl(Q(y)\bigr) = Q\bigl(P(y,x)\bigr) = P\bigl(Q(y), Q(x)\bigr). \end{aligned} $$
(18)
This implies that \((Q(x),Q(y))\) is a coupled coincidence point.
Assuming that \(l = Q(x)\), \(m = Q(y)\), by (16)(17) we have
$$ Q\bigl(Q(x)\bigr) = Q(x),\qquad Q\bigl(Q(y)\bigr) = Q(y). $$
(19)
From (18)(19) we obtain
$$ Q(x) = Q\bigl(Q(x)\bigr) = P\bigl(Q(x), Q(y)\bigr) $$
and
$$ Q(y) = Q\bigl(Q(y)\bigr) = P\bigl(Q(y), Q(x)\bigr). $$
So, \((Q(x), Q(y))\) is a common coupled fixed point of P and Q.
We can easily prove the uniqueness of common coupled fixed point under the assumption that \((x^{\ast}, y^{\ast})\) is another common coupled fixed point of P and Q.
From (16)(19) we can show that
$$ x^{\ast} = Q\bigl(x^{\ast}\bigr) = Q\bigl(Q(x)\bigr)= Q(x) $$
and
$$ y^{\ast} = Q\bigl(y^{\ast}\bigr) = Q\bigl(Q(y)\bigr)= Q(y). $$
This implies that P and Q have a unique common coupled fixed point. □
Here, we furnish an example to demonstrate the validity of the hypothesis of the above results.
Example 3.1
Let \((X, \leq)\) be a partially ordered set with \(X = [0,1]\), \(a \ast b = \min \{a, b \}\). Let \(P: X \times X \rightarrow X\) and \(Q: X \rightarrow X\) be two mappings defined as
$$P(x,y) = \textstyle\begin{cases} \frac{xy}{2} &\mbox{if } x \geq y, \\ 0 & \mbox{if } x < y, \end{cases}\displaystyle \qquad Q(x) = x. $$
This implies that P satisfies the definition of the mixed Qmonotone property.
Let
$$ V(x_{1}, x_{2},\ldots, x_{n},t) = \frac{t}{t+A(x_{1}, x_{2},\ldots, x_{n})}, $$
where \(A(x_{1}, x_{2},\ldots, x_{n})\) is the Ametric space defined as
$$ A(x_{1}, x_{2},\ldots,x_{n}) = \vert x_{1}  x_{2}\vert +\vert x_{2}  x_{3}\vert +\cdots+ \vert x_{n1}  x_{n}\vert $$
for all \(x_{1}, x_{2},\ldots,x_{n} \in X\), \(t > 0\).
Then \((X, V, \ast)\) is a complete Vfuzzy metric space.
We take \(k = \frac{1}{2}\) and consider the sequences \(\{x_{r} \}\), \(\{y_{r} \}\) in X defined by \(x_{r} = \frac{1}{2r}\), \(y_{r} = \frac{1}{3r}\).
Since
$$\begin{aligned}& \lim_{r\rightarrow\infty}P(x_{r}, y_{r}) = \lim _{r \rightarrow\infty }Q(x_{r}) = 0 = w \quad (\mbox{say}), \\& \lim_{r\rightarrow\infty}P(y_{r}, x_{r}) = \lim _{r \rightarrow\infty }Q(y_{r}) = 0 = w'\quad ( \mbox{say}). \end{aligned}$$
Also, \(P: X \times X \rightarrow X\) and \(Q: X \rightarrow X\) are compatible mappings in X. From Theorem 3.1 we have that \(Q(x) \leq Q(u)\) and \(Q(y) \geq Q(v)\). This implies \(x \leq u\), \(y \geq v\). If we consider \(x \geq y\), \(u \geq v\), then we have
$$\begin{aligned}& V \biggl(P(x, y), P(x, y),\ldots,P(x, y), P(u, v)\frac{t}{2} \biggr) \\& \quad = \frac{\frac{t}{2}}{\frac{t}{2}+2\vert \frac{(xy)(uv)}{2} \vert } \geq\frac{t}{t + \vert u+v \vert } \\& \quad = V\bigl(Q(u),Q(u),\ldots, Q(u), P(u,v),t\bigr) \\& \quad \geq V\bigl(Q(x),Q(x),\ldots, Q(x), Q(u),t\bigr) \\& \qquad {} \ast V\bigl(Q(x),Q(x),\ldots, Q(x), P(x,y),t\bigr) \\& \qquad {} \ast V\bigl(Q(u),Q(u),\ldots, Q(u), P(u,v),t\bigr). \end{aligned}$$
If we consider \(x < y\), \(u \geq v\), then we have
$$\begin{aligned}& V \biggl(P(x, y), P(x, y),\ldots,P(x, y), P(u, v)\frac{t}{2} \biggr) \\ & \quad = \frac{\frac{t}{2}}{\frac{t}{2}+2\vert \frac{(uv)}{2} \vert } \geq \frac{t}{t + 2\vert u  x \vert } \\ & \quad \geq V\bigl(Q(x),Q(x),\ldots, Q(x), Q(u),t\bigr) \\ & \quad \geq V\bigl(Q(x),Q(x),\ldots, Q(x), Q(u),t\bigr) \\ & \qquad {} \ast V\bigl(Q(x),Q(x),\ldots, Q(x), P(x,y),t\bigr) \\ & \qquad {} \ast V\bigl(Q(u),Q(u),\ldots, Q(u), P(u,v),t\bigr). \end{aligned}$$
If we consider \(x < y\), \(u < v\), then we get directly condition (T3) of Theorem 3.1.
Therefore, all hypotheses of Theorem 3.1 hold. So we conclude that \((w, w')\) is a common coupled fixed point of P and Q.
Theorem 3.3
Let
\((X, V, \ast)\)
be a complete
Vfuzzy metric space, and
\((X, \leq )\)
be a partially ordered set. Let
\(P: X \times X \rightarrow X\)
be a mapping such that
P
has the mixed monotone property and there exists
\(k \in(0,1)\)
such that
$$\begin{aligned}& V \bigl(P(x, y), P(x, y),\ldots,P(x, y), P(u, v),kt\bigr) \\& \quad \geq V(x,x,\ldots ,x,u,t) \ast V\bigl(x,x,\ldots,x,P(x,y),t\bigr) \\& \qquad {}\ast V\bigl(u, u, \ldots, u,P(u,v),t\bigr) \end{aligned}$$
for all
\(x, y, u, v \in X\), \(t > 0\)
such that
\(x \leq u\)
and
\(y \geq v\).
Also suppose that

(a)
P
is continuous or

(b)
X
has the following properties:

(i)
if
\(\{x_{n} \}\)
is a nondecreasing sequence
\(x_{r} \rightarrow x\), then
\(x_{r} \leq x\)
for all
\(r \in N\),

(ii)
if
\(\{y_{n} \}\)
is a nondecreasing sequence
\(y_{r} \rightarrow y\), then
\(y_{r} \geq x\)
for all
\(r \in N\).
If there exist
\(x_{0}, y_{0} \in X\)
such that
\(x_{0} \leq P(x_{0},y_{0})\)
and
\(y_{0} \geq P(y_{0},x_{0})\), then
P
has a coupled fixed point in
X.
Proof
By assuming \(Q = I\) (the identity mapping) in Theorem 3.1 we get the result. □