In this section, we introduce the new notion of a modified αϕfuzzy contractive mapping and αadmissible mapping with respect to η in fuzzy metric spaces.
Denote by Φ the family of all right continuous functions \(\phi: [0, +\infty) \to[0, +\infty)\), with \(\phi(c) < c\) for all \(c >0\).
Remark 3.1
Note that for every function \(\phi\in\Phi\), \(\lim_{n \to+\infty} \phi^{n}(c) = 0\), where \(\phi^{n}(c)\) denotes the nth iterate of ϕ.
According to [24] (see also [25]) we use the concept of αadmissible mapping in the following form.
Definition 3.1
Let \((X, M, *)\) be a fuzzy metric space in the sense of George and Veeramani and let \(\alpha, \eta: X \times X \times(0, +\infty) \to[0, +\infty)\) be two functions. We say that \(T:X \to X\) is αadmissible with respect to η if
$$ x, y \in X, \quad \alpha(x, y, t) \geq\eta(x, y, t) \quad \Rightarrow\quad \alpha(Tx, Ty, t) \geq\eta(Tx, Ty, t)\quad \mbox{for all } t>0. $$
Note that if we take \(\eta(x, y, t) = 1\), then this definition reduces to the definition of αadmissible mapping (Definition 3.4 of Gopal and Vetro [11]). Also, if we take \(\alpha(x, y, t) = 1\), then we say that T is an ηsubadmissible mapping.
In [11], Gopal and Vetro introduced the concept of an αϕfuzzy contractive mapping in fuzzy metric spaces as follows.
Definition 3.2
([11])
Let \((X, M, *)\) be a fuzzy metric space in the sense of George and Veeramani. We say that \(T: X\to X\) is an αϕfuzzy contractive mapping if there exist two functions \(\alpha: X \times X \times(0, +\infty) \to[0, +\infty)\) and \(\phi\in\Phi\) such that, for all \(x, y \in X\) and for all \(t > 0\), we have
$$ \alpha(x, y, t) \biggl(\frac{1}{M(Tx, Ty, t)} 1 \biggr) \leq\phi \biggl( \frac{1}{M(x, y, t)}1 \biggr). $$
Motivated by the above definition we give the following generalization.
Definition 3.3
Let \((X, M, *)\) be a fuzzy metric space in the sense of George and Veeramani. We say that \(T: X\to X\) is a modified αϕfuzzy contractive mapping if there exist three functions \(\alpha, \eta: X \times X \times(0, +\infty) \to[0, +\infty)\) and \(\phi\in\Phi\) such that, for all \(x, y \in X\) and for all \(t > 0\), we have
$$ \alpha(x, y, t) \geq\eta(x, y, t)\quad \Rightarrow\quad \frac{1}{M(Tx, Ty, t)} 1 \leq\phi \biggl(\frac{1}{N(x, y, t)} 1 \biggr), $$
(2)
where \(N(x, y, t) = \min\{M(x, y, t), M(x, Tx, t), M(y, Ty, t) \}\).
Remark 3.2
If \(\eta(x, y,t) = 1\) and \(N(x, y, t) = M(x, y, t)\), then this definition reduces to Definition 3.2 in [11], thus it will imply the definition of the fuzzy contractive mapping given by Gregori and Sapena [6]. It follows that a fuzzy contractive mapping is a modified αϕfuzzy contractive mapping; in general the converse is not true.
Our first main result is the following theorem.
Theorem 3.1
Let
\((X, M, *)\)
be a
Gcomplete fuzzy metric space in the sense of George and Veeramani. Let
\(T: X\to X\)
be a modified
αϕfuzzy contractive mapping satisfying the following conditions:

(i)
T
is
αadmissible with respect to
η;

(ii)
there exists
\(x_{0} \in X\)
such that
\(\alpha(x_{0}, Tx_{0}, t)\geq\eta(x_{0}, Tx_{0}, t)\)
for each
\(t > 0\);

(iii)
T
is continuous.
Then
T
has a fixed point, that is, there exists
\(x^{*} \in X\)
such that
\(Tx^{*} = x^{*}\).
Proof
Let \(x_{0} \in X\) such that \(\alpha(x_{0}, Tx_{0}, t)\geq \eta(x_{0}, Tx_{0}, t)\) for all \(t > 0\). Define a sequence \(\{x_{n}\}\) in X by \(x_{n} = Tx_{n1} = T^{n}x_{0}\) for all \(n \in \mathbb{N}\). Clearly, if \(x_{n} = x_{n+1}\) for some \(n \in\mathbb{N}\) then \(x = x_{n}\) is a fixed point of T. Hence we suppose that \(x_{n+1}\neq x_{n}\) for all \(n \in\mathbb{N}\). Since T is an αadmissible mapping with respect to η and \(\alpha(x_{0}, Tx_{0}, t)\geq\eta(x_{0}, Tx_{0}, t)\), we deduce that
$$\alpha(x_{1}, x_{2}, t) = \alpha\bigl(Tx_{0}, T^{2}x_{0}, t\bigr) \geq\eta\bigl(Tx_{0}, T^{2}x_{0}, t\bigr) = \eta(x_{1}, x_{2}, t). $$
Continuing this process, we get \(\alpha(x_{n}, x_{n+1}, t) \geq\eta(x_{n}, x_{n+1}, t)\) for all \(n \in\mathbb{N}\). Now by (2) with \(x = x_{n1}\), \(y = x_{n}\), we get
$$\frac{1}{M(Tx_{n1}, Tx_{n}, t)} \leq \phi \biggl(\frac{1}{N(x_{n1}, x_{n},t)} \biggr), $$
where
$$\begin{aligned} N(x_{n1}, x_{n}, t)&= \min\bigl\{ M(x_{n1},x_{n}, t), M(x_{n1}, Tx_{n1},t),M(x_{n}, Tx_{n}, t)\bigr\} \\ &= \min\bigl\{ M(x_{n1},x_{n}, t), M(x_{n1}, x_{n},t), M(x_{n}, x_{n+1}, t)\bigr\} \\ &= \min\bigl\{ M(x_{n1},x_{n}, t), M(x_{n}, x_{n+1}, t)\bigr\} . \end{aligned}$$
It follows
$$\frac{1}{M(x_{n}, x_{n+1}, t)}1 \leq\phi \biggl(\frac{1}{\min\{M(x_{n1},x_{n}, t), M(x_{n}, x_{n+1}, t)\}}1 \biggr). $$
Now, if \(\min\{M(x_{n1},x_{n}, t), M(x_{n}, x_{n+1}, t)\} = M(x_{n}, x_{n+1}, t)\) for some \(n \in\mathbb{N}\), then
$$\frac{1}{M(x_{n}, x_{n+1}, t)}1 \leq\phi \biggl(\frac{1}{M(x_{n}, x_{n+1}, t)}1 \biggr) < \frac{1}{M(x_{n},x_{n+1}, t)}1 , $$
which is a contradiction. Hence for all \(n\in\mathbb{N}\) we have
$$\frac{1}{M(x_{n}, x_{n+1}, t)}1 < \frac{1}{M(x_{n1},x_{n}, t)}1. $$
Consequently, \(M(x_{n}, x_{n+1}, t) > M(x_{n1}, x_{n}, t)\) for all \(n \in\mathbb{N}\), thus \(\{M(x_{n1}, x_{n}, t)\}\) is an increasing sequence of positive reals in \([0, 1]\).
Let \(s(t) = \lim_{n\to+\infty} M(x_{n1}, x_{n}, t)\); we show that \(s(t) = 1\) for all \(t > 0\). Indeed, we assume that there exists \(t_{0} > 0\) such that \(s(t_{0}) < 1\). From
$$\frac{1}{M(x_{n}, x_{n+1}, t_{0})}1 \leq\phi \biggl(\frac{1}{M(x_{n1},x_{n}, t_{0})}1 \biggr), $$
by using the right continuity of the function ϕ, letting \(n \to +\infty\), we obtain the contradiction
$$\frac{1}{s(t_{0})}1 \leq\phi \biggl(\frac{1}{s(t_{0})}1 \biggr)< \frac {1}{s(t_{0})}1 . $$
This implies that \(\lim_{n\to+\infty} M(x_{n1}, x_{n}, t) = 1\) for all \(t > 0\). Then, for a fixed \(p \in\mathbb{N}\), we get
$$\begin{aligned} M(x_{n}, x_{n+p}, t) \geq& M(x_{n},x_{n+1}, t/p)*M(x_{n+1},x_{n+2}, t/p)* \cdots \\ &{} *M(x_{n+p1},x_{n+p}, t/p)\to1 * \cdots * 1 = 1 \end{aligned}$$
as \(n \to+\infty\) and hence the sequence \(\{x_{n}\}\) is GCauchy. Since \((X,M,\ast)\) is Gcomplete, then \(\{x_{n}\}\) converges to some \(x^{*} \in X\). Also, the continuity of T leads to \(Tx_{n} \to Tx^{*}\) and hence
$$\lim_{n\to+\infty} M\bigl(x_{n+1}, Tx^{*}, t\bigr) = \lim _{n\to+\infty} M\bigl(Tx_{n}, Tx^{*}, t\bigr) = 1 $$
for all \(t > 0\), that is, the sequence \(x_{n}\) converges to \(Tx^{*}\). By uniqueness of the limit, we deduce that \(x^{*} = Tx^{*}\); this completes the proof. □
In the second theorem, we replace the continuity hypothesis with another regularity hypothesis.
Theorem 3.2
Let
\((X, M, *)\)
be a
Gcomplete fuzzy metric space in the sense of George and Veeramani with
M
triangular. Let
\(T: X\to X\)
be a modified
αϕfuzzy contractive mapping satisfying the following conditions:

(i)
T
is
αadmissible with respect to
η;

(ii)
there exists
\(x_{0} \in X\)
such that
\(\alpha(x_{0}, Tx_{0}, t)\geq\eta(x_{0}, Tx_{0}, t)\)
for each
\(t > 0\);

(iii)
for any sequence
\(\{x_{n}\}\)
in
X
with
\(\alpha(x_{n}, x_{n+1}, t) \geq\eta(x_{n}, x_{n+1}, t)\)
for all
\(n \in\mathbb{N}\), \(t >0\)
and
\(x_{n} \to x\)
as
\(n \to+\infty\), then
\(\alpha(x_{n}, x, t) \geq\eta(x_{n}, x, t)\), for all
\(n \in \mathbb{N}\)
and
\(t > 0\).
Then
T
has a fixed point, that is, there exists
\(x^{*} \in X\)
such that
\(Tx^{*} = x^{*}\).
Proof
By the proof of Theorem 3.1, we see that \(\{x_{n}\}\) is a GCauchy sequence in the Gcomplete fuzzy metric space \((X, M, *)\) such that \(\alpha(x_{n}, x_{n+1},t) \geq\eta(x_{n}, x_{n+1},t)\) for all \(n \in \mathbb{N}\). Then there exists a point \(x^{*}\) in X such that \(x_{n} \to x^{*}\) as \(n \to+\infty\). Now, hypothesis (iii) of the theorem implies that
$$ \alpha\bigl(x_{n}, x^{*},t\bigr) \geq\eta \bigl(x_{n}, x^{*},t\bigr) $$
(3)
for all \(n \in\mathbb{N}\) and for all \(t >0\).
If \(x^{*} \neq Tx^{*}\), that is, \(M(x^{*},Tx^{*},t) <1\) for some \(t>0\), then, from (1), (3), and (2), respectively, since M is triangular we write
$$\begin{aligned} \frac{1}{M( x^{*},Tx^{*}, t)}1 &\leq \biggl(\frac{1}{M(x^{*},x_{n+1}, t)}1 \biggr) + \biggl( \frac{1}{M(Tx_{n}, Tx^{*}, t)}1 \biggr) \\ &\leq \biggl(\frac{1}{M(x^{*},x_{n+1}, t)}1 \biggr) + \phi \biggl(\frac{1}{N(x_{n},x^{*}, t)}1 \biggr). \end{aligned}$$
Letting \(n \to+\infty\), we get
$$\begin{aligned} N\bigl(x_{n},x^{*}, t\bigr) &= \min\bigl\{ M \bigl(x_{n},x^{*},t\bigr),M(x_{n},Tx_{n},t),M \bigl(x^{*},Tx^{*},t\bigr)\bigr\} \\ &\to\min\bigl\{ 1,1,M\bigl(x^{*},Tx^{*}, t\bigr) \bigr\} \\ & = M\bigl(x^{*},Tx^{*}, t\bigr) \end{aligned}$$
and so, to avoid contradiction with \(\phi(c)< c\) for \(c>0\), we conclude that
$$\frac{1}{M(Tx^{*}, x^{*}, t)}1 = 0. $$
It follows that \(Tx^{*}= x^{*}\); this completes the proof. □
Some of the following corollaries can be deduced from the above results. In particular, by taking \(\eta(x, y, t) = 1\) in Theorem 3.1 and Theorem 3.2 (with M triangular), we have the following corollary.
Corollary 3.1
Let
\((X, M, *)\)
be a
Gcomplete fuzzy metric space in the sense of George and Veeramani (with
M
triangular). Let
\(T: X\to X\)
be an
αadmissible mapping. Assume that there exists
\(\phi\in\Phi\)
such that, for
\(x, y \in X\)
and
\(t>0\),
$$ \alpha(x, y, t) \geq1\quad \Rightarrow\quad \frac{1}{M(Tx, Ty, t)} 1 \leq\phi \biggl(\frac{1}{N(x, y, t)} 1 \biggr), $$
(4)
where
\(N(x, y, t) = \min\{M(x, y, t), M(x, Tx, t), M(y, Ty, t) \}\). Also suppose that the following assertions hold:

(i)
there exists
\(x_{0} \in X\)
such that
\(\alpha(x_{0}, Tx_{0}, t)\geq1\)
for all
\(t > 0\);

(ii)
either
T
is continuous or for any sequence
\(\{x_{n}\}\)
in
X
with
\(\alpha(x_{n}, x_{n+1}, t) \geq1\)
for all
\(n \in\mathbb{N}\), \(t >0\), and
\(x_{n} \to x\)
as
\(n \to+\infty\), then
\(\alpha(x_{n}, x, t) \geq1\), for all
\(n \in\mathbb{N}\)
and
\(t > 0\).
Then
T
has a fixed point.
Also, by taking \(\alpha(x, y, t) = 1\) in Theorem 3.1 and Theorem 3.2 (with M triangular), we have the following corollary.
Corollary 3.2
Let
\((X, M, *)\)
be a
Gcomplete fuzzy metric space in the sense of George and Veeramani (with
M
triangular). Let
\(T: X\to X\)
be an
ηsubadmissible mapping. Assume that there exists
\(\phi\in\Phi\)
such that, for
\(x, y \in X\)
and
\(t>0\),
$$\eta(x, y, t) \leq1\quad \Rightarrow \quad \frac{1}{M(Tx, Ty, t)} 1 \leq\phi \biggl( \frac{1}{N(x, y, t)} 1 \biggr), $$
where
\(N(x, y, t) = \min\{M(x, y, t), M(x, Tx, t), M(y, Ty, t) \}\). Also suppose that the following assertions hold:

(i)
there exists
\(x_{0} \in X\)
such that
\(\eta(x_{0}, Tx_{0}, t)\leq1\)
for all
\(t > 0\);

(ii)
either
T
is continuous or for any sequence
\(\{x_{n}\}\)
in
X
with
\(\alpha(x_{n}, x_{n+1}, t) \geq1\)
for all
\(n \in\mathbb{N}\), \(t >0\)
and
\(x_{n} \to x\)
as
\(n \to+\infty\), then
\(\alpha(x_{n}, x, t) \geq1\), for all
\(n \in\mathbb{N}\)
and
\(t > 0\).
Then
T
has a fixed point.
Now, we give two simple illustrative examples.
Example 3.1
Let \(X = [0, +\infty)\) endowed with the fuzzy metric \(M(x, y, t) = \frac {t}{t+xy}\) for all \(x,y \in X\) and \(t>0\), and the tnorm ∗_{3}. Clearly M is triangular. Define \(T: X \to X\) by
$$Tx= \textstyle\begin{cases} \frac{x^{2}}{4} & \mbox{if } x\in [0,1], \\ 2 & \mbox{if } x\in (1,+\infty), \end{cases} $$
and \(\alpha,\eta: X\times X\times(0, +\infty) \to[0, +\infty)\) by
$$\alpha(x,y, t) = \textstyle\begin{cases} 1 & \mbox{if } x, y\in[0,1], t>0, \\ 0 &\mbox{otherwise}, \end{cases} $$
and \(\eta(x,y,t)= 1/3\) for all \(x,y \in X\) and \(t>0\). Then the existence of a fixed point for T follows by an application of Theorem 3.2 with \(\phi(c) = \frac{c}{2}\) for all \(c \geq0\).
Proof
By completeness of \((X, d)\), where \(d(x,y)=xy\) for all \(x,y \in X\), one deduces easily that \((X,M,\ast_{3})\) is a Gcomplete metric space with M triangular. We show that T is an αadmissible mapping with respect to η. Indeed, let \(x, y \in X\); if \(\alpha(x, y,t) \geq\eta(x, y,t)\) for all \(t>0\), then \(x, y \in[0,1]\). On the other hand, for all \(x,y \in[0,1]\), we have \(Tx,Ty \in[0,1]\), which implies that \(\alpha(Tx, Ty,t) \geq\eta (Tx, Ty,t)\) for all \(t>0\). Trivially, for every \(x_{0} \in[0,1]\) we get \(\alpha(x_{0},Tx_{0},t) \geq\eta(x_{0}, Tx_{0},t)\) for all \(t>0\). Now, if \(\{x_{n}\}\) is a sequence in X such that \(\alpha(x_{n}, x_{n+1},t) \geq\eta(x_{n}, x_{n+1},t)\) for all \(n \in\mathbb{N}\) and \(x_{n} \to x\) as \(n \to+ \infty\), then \(\{x_{n}\} \subset[0, 1]\) and hence \(x \in[0, 1]\). This implies that \(\alpha(x_{n}, x,t) \geq\eta(x_{n}, x,t)\) for all \(n \in\mathbb{N}\) and \(t>0\). Next, let \(\alpha(x,y,t) \geq\eta(x,y,t)\) for all \(t>0\), that is, \(x,y \in[0,1]\). Then the contractive condition in Theorem 3.2, that is,
$$\begin{aligned} \frac{1}{M(Tx,Ty,t)}1 &= \frac{1}{4t} \biglx^{2}y^{2}\bigr \leq \frac{1}{2t} xy \\ & \leq\frac{1}{2t} \max\bigl\{ xy,xTx,yTy\bigr\} \\ &= \frac{1}{2} \biggl(\frac{t+\max\{xy,xTx,yTy\}}{t}1 \biggr) \\ &= \frac{1}{2} \biggl(\frac{1}{\frac{t}{t+\max\{xy,xTx,yTy\} }}1 \biggr) \\ &= \phi \biggl(\frac{1}{N(x, y, t)} 1 \biggr) \end{aligned}$$
is always true since \(x+y \leq2\). Thus, all the hypotheses of Theorem 3.2 are satisfied and T has at least a fixed point; here 0 and 2 are two fixed points of T.
On the other hand, T is not a fuzzy contractive mapping of Gregori and Sapena [6]. In fact, for \(x=1\) and \(y=2\), there does not exist \(k \in(0,1)\) such that
$$\frac{1}{M(Tx,Ty,t)}1 =\frac{7}{4t} \leq\frac{k}{t}=k \biggl( \frac {1}{M(x,y,t)}1 \biggr). $$
□
By a slight modification of Example 3.1, one can obtain, for instance, an example covered by Corollary 3.2.
Example 3.2
Let \(X= \{\frac{1}{n}, n \in\mathbb{N} \} \cup\{0,2\}\) endowed with the fuzzy metric \(M(x, y, t) = \frac{t}{t+xy}\) for all \(x,y \in X\) and \(t>0\), and the tnorm ∗_{3}. Define \(T: X \to X\) by
$$Tx= \textstyle\begin{cases} \frac{x^{2}}{4} & \mbox{if } x \in X \setminus\{2\}, \\ 2 & \mbox{if } x=2, \end{cases} $$
and \(\eta: X\times X\times(0, +\infty) \to[0, +\infty)\) by
$$\eta(x,y, t) = \textstyle\begin{cases} 1 & \mbox{if } x, y \in X \setminus\{2\}, t>0, \\ 3 &\mbox{otherwise}. \end{cases} $$
Then the existence of a fixed point of T follows by an application of Corollary 3.2 with \(\phi(c) = \frac{c}{2}\) for all \(c \geq0\).
Proof
Clearly, T is an ηsubadmissible continuous mapping. Shortly, let \(x,y\in X\) such that \(\eta(x,y,t) \leq1\) for all \(t>0\), this implies that \(x,y \in X \setminus\{2\}\) and, by the definitions of T and η, we have \(Tx, Ty \in X \setminus\{2\} \) and \(\eta(Tx,Ty,t) =1 \) for all \(t>0\), that is, T is ηsubadmissible. Further, for every \(x_{0} \in X \setminus\{2\}\) we get \(\eta(x_{0},Tx_{0},t) = 1\) for all \(t>0\).
Next, let \(\eta(x,y,t) \leq1\) for all \(t>0\), that is, \(x,y \in X \setminus\{2\}\). By using the same argument as in Example 3.1, the contractive condition in Corollary 3.2 is always true since \(x+y \leq2\). Thus, all the hypotheses of Corollary 3.2 are satisfied; again 0 and 2 are two fixed points of T. □
Next, for establishing the uniqueness of fixed point of modified αϕfuzzy contractive mapping, we will consider the following hypothesis:

(H)
for all \(x, y \in X\) and for all \(t>0\) there exists \(z \in X\) such that \(\alpha(x, z, t) \geq\eta(x, z, t)\), \(\alpha(y, z, t) \geq\eta(y, z, t)\) and \(\lim_{n \to+\infty}M(T^{n1}z,T^{n}z,t)=1\).
Theorem 3.3
Adding hypothesis (H) in Theorem
3.1 (Theorem
3.2), one obtains that
\(x^{*}\)
is the unique fixed point of
T
provided that
\(\phi\in\Phi \)
is nondecreasing.
Proof
Assume that \(x^{*}\) and \(y^{*}\) are two fixed points of T. If \(\alpha (x^{*}, y^{*}, t) \geq\eta(x^{*}, y^{*}, t)\), then by condition (2), we get \(x^{*} = y^{*}\). Suppose \(\alpha (x^{*}, y^{*}, t) < \eta(x^{*}, y^{*}, t)\), then from hypothesis (H) there exists \(z \in X\) such that
$$ \alpha\bigl(x^{*}, z, t\bigr) \geq\eta \bigl(x^{*},z, t\bigr) \quad \mbox{and}\quad \alpha\bigl(y^{*}, z, t\bigr) \geq\eta\bigl(y^{*}, z, t\bigr). $$
(5)
Since T is αadmissible with respect to η, then we have
$$ \alpha\bigl(x^{*}, T^{n}z, t\bigr) \geq\eta \bigl(x^{*},T^{n}z, t\bigr) $$
for all \(n \in\mathbb{N}\) and for all \(t>0\). Now we prove that \(M(x^{*},T^{n}z,t) \to1\), as \(n \to+ \infty\), for all \(t>0\). Reasoning by absurd, we suppose that there exists \(t>0\) such that \(\lim_{n \to+\infty}M(x^{*},T^{n}z,t)<1\). Then, from (2) and (5), we get
$$\begin{aligned} \frac{1}{M(x^{*}, T^{n}z, t)}1 &= \frac{1}{M(Tx^{*}, T(T^{n1}z), t)}1 \\ &\leq \phi \biggl(\frac{1}{N(x^{*}, T^{n1}z, t)}1 \biggr), \end{aligned}$$
where
$$\begin{aligned} N\bigl(x^{*}, T^{n1}z, t\bigr) &= \min\bigl\{ M \bigl(x^{*}, T^{n1}z, t\bigr), M\bigl(x^{*},Tx^{*}, t\bigr), M\bigl(T^{n1}z, T\bigl(T^{n1}\bigr)z, t\bigr) \bigr\} \\ &= \min\bigl\{ M\bigl(x^{*}, T^{n1}z, t\bigr), M \bigl(x^{*},Tx^{*}, t\bigr), M\bigl(T^{n1}z, T^{n}z, t\bigr)\bigr\} \\ &= \min\bigl\{ M\bigl(x^{*}, T^{n1}z, t\bigr), 1, M \bigl(T^{n1}z, T^{n}z, t\bigr)\bigr\} . \end{aligned}$$
Let \(n_{0} \in\mathbb{N}\) such that \(M(x^{*},T^{n}z,t) \leq M(T^{n}z,T^{n+1}z,t)\) for all \(n \geq n_{0}\). Therefore, for all \(n > n_{0}\), we obtain
$$\begin{aligned} \frac{1}{M(x^{*}, T^{n}z, t)}1 &= \frac{1}{M(Tx^{*}, T(T^{n1}z), t)}1 \\ &\leq\phi \biggl(\frac{1}{\min\{M(x^{*}, T^{n1}z, t), 1, M(T^{n1}z, T^{n}z, t)\}}1 \biggr) \\ & =\phi \biggl(\frac {1}{M(x^{*}, T^{n1}z, t)}1 \biggr) . \end{aligned}$$
This implies that
$$\begin{aligned} \frac{1}{M(x^{*}, T^{n}z, t)}1 &= \frac{1}{M(Tx^{*}, T(T^{n1}z), t)}1 \\ &\leq\phi^{nn_{0}} \biggl(\frac{1}{M(x^{*}, T^{n_{0}}z, t)} 1 \biggr). \end{aligned}$$
Then, letting \(n \to+\infty\), we get \(T^{n}z \to x^{*}\). Similarly one can obtain \(T^{n}z \to y^{*}\), as \(n \to+\infty\). Consequently, we deduce that \(x^{*} = y^{*}\); the uniqueness is proved. □
The following classes of functions are used in our subsequent results. Let
$$\Psi= \bigl\{ \psi: [0, +\infty) \to[0, +\infty) \mbox{ such that } \psi\mbox{ is nondecreasing and continuous}\bigr\} $$
and
$$\Phi_{1} = \bigl\{ \phi: [0, +\infty) \to[0, +\infty) \mbox{ such that } \phi \mbox{ is lower semicontinuous}\bigr\} , $$
where \(\psi(r) = \phi(r) = 0\) if and only if \(r = 0\).
Theorem 3.4
Let
\((X, M, *)\)
be a
Gcomplete fuzzy metric space and let
\(T:X \to X\)
be an
αadmissible mapping with respect to
η. Assume that there exist
\(\psi\in\Psi\)
and
\(\phi\in\Phi_{1}\)
such that, for
\(x, y \in X\)
and
\(t>0\),
$$\begin{aligned}& \alpha(x, Tx, t)\alpha(y, Ty, t) \geq\eta(x, Tx, t)\eta(y, Ty, t) \\& \quad \Rightarrow\quad \psi \biggl(\frac{1}{M(Tx, Ty, t)} 1 \biggr)\leq\psi \biggl( \frac{1}{M(x, y, t)} 1 \biggr) \phi \biggl(\frac{1}{M(x, y, t)} 1 \biggr). \end{aligned}$$
(6)
Also suppose that the following conditions hold:

(i)
there exists
\(x_{0} \in X\)
such that
\(\alpha(x_{0}, Tx_{0}, t)\geq\eta(x_{0}, Tx_{0}, t)\)
for all
\(t > 0\);

(ii)
either
T
is continuous or for any sequence
\(\{x_{n}\}\)
in
X
with
\(\alpha(x_{n}, x_{n+1}, t) \geq\eta(x_{n}, x_{n+1}, t)\)
for all
\(n \in\mathbb{N}\), \(t >0\), and
\(x_{n} \to x\)
as
\(n \to+\infty\), then
\(\alpha(x_{n}, x, t) \geq\eta(x_{n}, x, t)\)
and
\(\alpha(x, Tx, t) \geq\eta(x, Tx, t)\), for all
\(n \in\mathbb{N}\)
and
\(t > 0\).
Then
T
has a fixed point.
Proof
Let \(x_{0}\) in X such that \(\alpha(x_{0}, Tx_{0}, t)\geq \eta(x_{0}, Tx_{0}, t)\). Define a sequence \(\{x_{n}\}\) in X such that \(x_{n} = Tx_{n1} = T^{n}x_{0}\) for all \(n \in\mathbb{N}\). If \(x_{n+1} = x_{n}\) for some \(n \in X\), then \(x = x_{n}\) is a fixed point for T and the result is proved in this case. Now, we assume that \(x_{n} \neq x_{n+1}\) for all \(n \in \mathbb{N}\). Since T is αadmissible with respect to η and \(\alpha(x_{0}, Tx_{0}, t) \geq\eta(x_{0}, Tx_{0}, t)\), we deduce that
$$\alpha(x_{1},x_{2}, t) = \alpha\bigl(Tx_{0}, T^{2}x_{0}, t\bigr) \geq\eta\bigl(Tx_{0}, T^{2}x_{0}, t\bigr) = \eta(x_{1},x_{2}, t). $$
Continuing this process we get
$$\alpha(x_{n}, x_{n+1}, t) \geq\eta(x_{n}, x_{n+1}, t) $$
for all \(n \in\mathbb{N}\). Clearly,
$$\alpha(x_{n1}, Tx_{n1}, t)\alpha(x_{n}, Tx_{n}, t) \geq\eta(x_{n1}, Tx_{n1}, t) \eta(x_{n}, Tx_{n}, t). $$
Now by (6) with \(x=x_{n1}\), \(y = x_{n}\), we have
$$\psi \biggl(\frac{1}{M(Tx_{n1}, Tx_{n}, t)} 1 \biggr)\leq \psi \biggl(\frac{1}{M(x_{n1}, x_{n}, t)} 1 \biggr) \phi \biggl(\frac{1}{M(x_{n1}, x_{n}, t)}1 \biggr), $$
which implies that
$$\psi \biggl(\frac{1}{M(x_{n}, x_{n+1}, t)} 1 \biggr)\leq \psi \biggl(\frac{1}{M(x_{n1}, x_{n}, t)} 1 \biggr) \phi \biggl(\frac{1}{M(x_{n1}, x_{n}, t)}1 \biggr). $$
If \(M(x_{n1},x_{n},t)=1\), then \(M(x_{n},x_{n+1},t)=1\) too. Otherwise, if \(M(x_{n1},x_{n},t)<1\), then
$$\psi \biggl(\frac{1}{M(x_{n}, x_{n+1}, t)} 1 \biggr) < \psi \biggl(\frac {1}{M(x_{n1}, x_{n}, t)} 1 \biggr) $$
and, since ψ is nondecreasing, we get \(M(x_{n}, x_{n+1}, t) \geq M(x_{n1}, x_{n}, t)\), for all \(n \in\mathbb{N}\). Thus, \(\{M(x_{n1}, x_{n}, t)\}\) is a nondecreasing sequence of positive reals in \([0, 1]\). Let \(s(t) = \lim_{n\to+\infty} M(x_{n1}, x_{n}, t)\). Now we show that \(s(t) = 1\) for all \(t > 0\). If not, then there exists \(t > 0\) such that \(s(t) <1\). Therefore, from the above inequality, on taking \(n \to+\infty\), we obtain
$$\psi \biggl(\frac{1}{s(t)}1 \biggr) \leq\psi \biggl(\frac{1}{s(t)}1 \biggr)  \phi \biggl(\frac{1}{s(t)}1 \biggr), $$
that is a contradiction and so we get \(\lim_{n\to+\infty} M(x_{n1}, x_{n}, t) = 1\), for all \(t > 0\). Next, for a fixed \(p \in\mathbb{N}\), we have
$$\begin{aligned} M(x_{n}, x_{n+p}, t) \geq& M(x_{n},x_{n+1}, t/p)*M(x_{n+1},x_{n+2}, t/p) \\ &{} * \cdots * M(x_{n+p1},x_{n+p}, t/p) \\ \to& 1 * \cdots * 1 = 1 \end{aligned}$$
as \(n \to+\infty\) and thus \(\{x_{n}\}\) is a GCauchy sequence. Since \((X,M,\ast)\) is Gcomplete, therefore \(\{x_{n}\}\) converges to \(x^{*}\) for some \(x^{*} \in X\). Now, we distinguish two cases.
Case 1. T is continuous. Then this implies that \(Tx_{n} \to Tx^{*}\) and so
$$\lim_{n\to+\infty} M\bigl(x_{n+1}, Tx^{*}, t\bigr) = \lim _{n\to+\infty} M\bigl(Tx_{n}, Tx^{*}, t\bigr) = 1 $$
for all \(t > 0\), that is, \(x_{n} \to Tx^{*}\). By the uniqueness of the limit, we get \(x^{*} = Tx^{*}\), that is, \(x^{*}\) is a fixed point of T.
Case 2. For any sequence \(\{x_{n}\}\) in X with \(\alpha(x_{n}, x_{n+1}, t) \geq\eta(x_{n}, x_{n+1}, t)\) for all \(n \in\mathbb{N}\), \(t >0\), and \(x_{n} \to x^{*}\) as \(n \to+\infty\), then \(\alpha(x_{n}, x^{*}, t) \geq\eta(x_{n}, x^{*}, t)\) and \(\alpha (x^{*}, Tx^{*}, t) \geq\eta(x^{*}, Tx^{*}, t)\), for all \(n \in\mathbb{N}\) and \(t > 0\). In this case, we get easily
$$\alpha(x_{n}, x_{n+1}, t)\alpha\bigl(x^{*}, Tx^{*}, t\bigr) \geq \eta(x_{n}, x_{n+1}, t)\eta\bigl(x^{*}, Tx^{*}, t\bigr). $$
Now by (6), we have
$$\begin{aligned} \psi \biggl(\frac{1}{M(x_{n+1}, Tx^{*}, t)} 1 \biggr)&= \psi \biggl(\frac{1}{M(Tx_{n}, Tx^{*}, t)} 1 \biggr) \\ &\leq\psi \biggl(\frac{1}{M(x_{n}, x^{*}, t)} 1 \biggr) \phi \biggl(\frac {1}{M(x_{n}, x^{*}, t)}1 \biggr). \end{aligned}$$
If \(M(x_{n},x^{*},t)=1\), then \(M(x_{n+1},Tx^{*},t)=1\) too. Otherwise, if \(M(x_{n},x^{*},t)<1\), then
$$\psi \biggl(\frac{1}{M(x_{n+1}, Tx^{*}, t)} 1 \biggr) < \psi \biggl(\frac {1}{M(x_{n}, x^{*}, t)} 1 \biggr). $$
This implies that \(M(x_{n+1}, Tx^{*}, t) \geq M(x_{n}, x^{*}, t) \to1\) as \(n \to+ \infty\) for all \(t>0\) and so \(x^{*} = Tx^{*}\). □
By taking \(\eta(x, y, t) = 1\) in Theorem 3.4, we deduce the following corollary.
Corollary 3.3
Let
\((X, M, *)\)
be a
Gcomplete fuzzy metric space and let
\(T:X \to X\)
be an
αadmissible mapping. Assume that there exist
\(\psi\in\Psi\)
and
\(\phi\in\Phi _{1}\)
such that, for
\(x, y \in X\)
and
\(t>0\),
$$\begin{aligned}& \alpha(x, Tx, t)\alpha(y, Ty, t) \geq1 \\& \quad \Rightarrow\quad \psi \biggl(\frac{1}{M(Tx, Ty, t)} 1 \biggr)\leq\psi \biggl( \frac{1}{M(x, y, t)} 1 \biggr) \phi \biggl(\frac{1}{M(x, y, t)} 1 \biggr). \end{aligned}$$
Also suppose that the following conditions hold:

(i)
there exists
\(x_{0} \in X\)
such that
\(\alpha(x_{0}, Tx_{0}, t)\geq1\)
for all
\(t > 0\);

(ii)
either
T
is continuous or for any sequence
\(\{x_{n}\}\)
in
X
with
\(\alpha(x_{n}, x_{n+1}, t) \geq1\)
for all
\(n \in\mathbb{N}\), \(t >0\)
and
\(x_{n} \to x\)
as
\(n \to+\infty\), then
\(\alpha(x_{n}, x, t) \geq1\)
and
\(\alpha(x, Tx, t) \geq1\), for all
\(n \in\mathbb{N}\)
and
\(t > 0\).
Then
T
has a fixed point.
Also, by taking \(\alpha(x, y, t) = 1\) in Theorem 3.4 we deduce the following corollary.
Corollary 3.4
Let
\((X, M, *)\)
be a
Gcomplete fuzzy metric space and let
\(T: X \to X\)
be an
ηsubadmissible mapping. Assume that there exist
\(\psi\in\Psi\)
and
\(\phi\in\Phi_{1}\)
such that, for
\(x, y \in X\)
and
\(t>0\),
$$\begin{aligned}& \eta(x, Tx, t)\eta(y, Ty, t) \leq1 \\& \quad \Rightarrow\quad \psi \biggl(\frac{1}{M(Tx, Ty, t)} 1 \biggr)\leq\psi \biggl( \frac{1}{M(x, y, t)} 1 \biggr)  \phi \biggl(\frac{1}{M(x, y, t)} 1 \biggr). \end{aligned}$$
Also suppose that the following conditions hold:

(i)
there exists
\(x_{0} \in X\)
such that
\(\eta(x_{0}, Tx_{0}, t)\leq1\)
for all
\(t > 0\);

(ii)
either
T
is continuous or for any sequence
\(\{x_{n}\}\)
in
X
with
\(\eta(x_{n}, x_{n+1}, t) \leq1\)
for all
\(n \in\mathbb{N}\), \(t >0\), and
\(x_{n} \to x\)
as
\(n \to+\infty\), then
\(\eta(x_{n}, x, t) \leq1\)
and
\(\eta (x, Tx, t) \leq1\), for all
\(n \in\mathbb{N}\)
and
\(t > 0\).
Then
T
has a fixed point.