In this section, we will show (\psi ,\phi ,\u03f5,\lambda )contraction is continuous. By using this assumption, we will also prove two theorems.
Definition 3.1 Let F be a probabilistic distance on S. A mapping f:S\to S is called continuous if for every \u03f5>0 there exists \delta >0 such that
{F}_{u,v}(\delta )>1\delta \phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}{F}_{fu,fv}(\u03f5)>1\u03f5.
Before we start to present the theorems, we will explain the following lemma.
Lemma 3.1 Every (\psi ,\phi ,\u03f5,\lambda )contraction is continuous.
Proof Suppose that \u03f5>0 be given and \delta \in (0,1) be such that \delta <min\{\u03f5,{\psi}^{1}(\u03f5)\} and since ψ is increasing bijection then \psi (\delta )<\u03f5. If {F}_{x,y}(\delta )>1\delta then, by (\psi ,\phi ,\u03f5,\lambda )contraction we have {F}_{fx,fy}(\psi (\delta ))>1\phi (\delta ), from where we obtain that {F}_{fx,fy}(\u03f5)>{F}_{fx,fy}(\psi (\delta ))>1\phi (\delta )>1\delta >1\u03f5. So f is continuous. □
Theorem 3.1 Let (S,F,T) be a complete Menger space and T a tnorm satisfies in {sup}_{0\le a<1}T(a,a)=1. Also, f:S\to S a (\psi ,\phi ,\u03f5,\lambda )contraction where {lim}_{n\to \mathrm{\infty}}{\psi}^{n}(\delta )=0 for every \delta \in (0,\mathrm{\infty}). If {lim}_{t\to \mathrm{\infty}}{F}_{{x}_{0},{f}^{m}{x}_{0}}(t)=1 for some {x}_{0}\in S and all m\in N, then there exists a unique fixed point x of the mapping f and x={lim}_{n\to \mathrm{\infty}}{f}^{n}({x}_{0}).
Proof Let {x}_{n}={f}^{n}{x}_{0}, n\in N. We shall prove that {({x}_{n})}_{n\in \mathbb{N}} is a Cauchy sequence.
Let n,m\in N, \u03f5>0, \lambda \in (0,1). Since {lim}_{t\to \mathrm{\infty}}{F}_{{x}_{0},{f}^{m}{x}_{0}(t)}(t)=1, it follows that for every \xi \in (0,1) there exists \eta >0 such that {F}_{{x}_{0},{f}^{m}({x}_{0})}(\eta )>1\xi and by induction {F}_{{f}^{m}{x}_{0},{f}^{n+m}{x}_{0}}({\psi}^{n}(\eta ))>1{\phi}^{n}(\xi ) for all n\in \mathbb{N}. By choosing n such that {\psi}^{n}(\eta )<\u03f5 and {\phi}^{n}(\xi )<\lambda, we obtain
{F}_{{x}_{n},{x}_{n+m}}(\u03f5)>1\lambda .
Hence, {({x}_{n})}_{n\in \mathbb{N}} is a Cauchy sequence and since S is complete, it follows the existence of x\in S such that x={lim}_{n\to \mathrm{\infty}}{x}_{n}. By continuity of f and {x}_{n+1}=f{x}_{n} for every n\in \mathbb{N}, when n\to \mathrm{\infty}, we obtain that x=fx. □
Example 3.1 Let (S,F,T) be a complete Menger space where S=\{{x}_{1},{x}_{2},{x}_{3},{x}_{4}\}, T(a,b)=min\{a,b\} and {F}_{xy}(t) is defined as
{F}_{{x}_{1},{x}_{2}}(t)={F}_{{x}_{2},{x}_{1}}(t)=\{\begin{array}{cc}0\hfill & \text{if}t\le 0,\hfill \\ 0.9\hfill & \text{if}0t\le 3,\hfill \\ 1\hfill & \text{if}t3\hfill \end{array}
and
\begin{array}{rcl}{F}_{{x}_{1},{x}_{3}}(t)& =& {F}_{{x}_{3},{x}_{1}}(t)={F}_{{x}_{1},{x}_{4}}(t)={F}_{{x}_{4},{x}_{1}}(t)={F}_{{x}_{2},{x}_{3}}(t)={F}_{{x}_{3},{x}_{2}}(t)={F}_{{x}_{2},{x}_{4}}(t)\\ =& {F}_{{x}_{4},{x}_{2}}(t)={F}_{{x}_{3},{x}_{4}}(t)={F}_{{x}_{4},{x}_{3}}(t)=\{\begin{array}{cc}0\hfill & \text{if}t\le 0,\hfill \\ 0.7\hfill & \text{if}0t6,\hfill \\ 1\hfill & \text{if}6\le t\hfill \end{array}\end{array}
f:S\to S is given by f({x}_{1})=f({x}_{2})={x}_{2} and f({x}_{3})=f({x}_{4})={x}_{1}. If we take \phi (\lambda )=\frac{\lambda}{2}, \psi (\u03f5)=\frac{\u03f5}{2}, then f is a (\psi ,\phi ,\u03f5,\lambda )contraction where {lim}_{n\to \mathrm{\infty}}{\psi}^{n}(\delta )={lim}_{n\to \mathrm{\infty}}\frac{\delta}{{2}^{n}}=0 for every \delta \in (0,\mathrm{\infty}) and if we set {x}_{0}={x}_{2}, then for all m\in N, we have {f}^{m}{x}_{0}={f}^{m}{x}_{2}={x}_{2} and {lim}_{t\to \mathrm{\infty}}{F}_{{x}_{2}{x}_{2}}(t)=1, so {x}_{2} is the unique fixed point for f.
Theorem 3.2 Let (S,F,T) be a complete Menger space, T be a tnorm such that {sup}_{0\le a<1}T(a,a)=1 and f:S\to S a (\psi ,\phi ,\u03f5,\lambda )contraction where the series {\sum}_{i}{\psi}^{i}(\delta ) is convergent for all \delta >0 and suppose that for some p\in S and j>0
\underset{x>0}{sup}{x}^{j}(1{F}_{p,fp}(x))<\mathrm{\infty}.
If tnorm T is φconvergent, then there exist a unique fixed point z of mapping f and z={lim}_{l\to \mathrm{\infty}}{f}^{l}p.
Proof Choose \u03f5>0 and \lambda \in (0,1). Let {z}_{l}={f}^{l}p, l\in N. We shall prove that {({z}_{l})}_{l\in N} is a Cauchy sequence. It means we prove that there exists {n}_{0}(\u03f5,\lambda )\in N such that
{F}_{{f}^{l}p,{f}^{l+m}p}(\u03f5)>1\lambda \phantom{\rule{1em}{0ex}}\text{for every}l\ge {n}_{0}(\u03f5,\lambda )\text{and every}m\in N.
Suppose that \mu \in (0,1), M>0 are such that
{x}^{j}(1{F}_{p,{f}_{p}}(x))\le M\phantom{\rule{1em}{0ex}}\mathrm{\forall}x>0.
(1)
Let {n}_{1} be such that
1M{\left({\mu}^{j}\right)}^{{n}_{1}}\in [0,1).
From (1), it follows that
{F}_{p,fp}\left(\frac{1}{{\mu}^{n}}\right)>1M{\left({\mu}^{j}\right)}^{n}\phantom{\rule{1em}{0ex}}\mathrm{\forall}n\in N\text{specially for}n={n}_{1}.
Since f is (\psi ,\phi ,\u03f5,\lambda )contraction, we derived by induction {F}_{{f}^{l}p,{f}^{l+1}p}({\psi}^{l}(\frac{1}{{\mu}^{{n}_{1}}}))>1{\phi}^{l}(1M{({\mu}^{j})}^{{n}_{1}}) \mathrm{\forall}l>1. Since the series {\sum}_{i=1}^{\mathrm{\infty}}{\psi}^{i}(\delta ) is convergent, there exists {n}_{2}={n}_{2}(\u03f5)\in N such that {\sum}_{i=l}^{\mathrm{\infty}}{\psi}^{i}(\delta )\le \u03f5 \mathrm{\forall}l\ge {n}_{2}. We know {\sum}_{i=l}^{\mathrm{\infty}}{\psi}^{i}(\frac{1}{{\mu}^{{n}_{1}}})\le \u03f5 for every l>max\{{n}_{1},{n}_{2}\}.
Now
\begin{array}{rcl}{F}_{{f}^{l}p,{f}^{l+m}p}(\u03f5)& \ge & {F}_{{f}^{l}p,{f}^{l+m}p}\left(\sum _{i=l}^{\mathrm{\infty}}{\psi}^{i}\left(\frac{1}{{\mu}^{{n}_{1}}}\right)\right)\ge {F}_{{f}^{l}p,{f}^{l+m}p}\left(\sum _{i=l}^{l+m1}{\psi}^{i}\left(\frac{1}{{\mu}^{{n}_{1}}}\right)\right)\\ \ge & T\left(T\right(\cdots T({F}_{{f}^{l}p,{f}^{l+1}p}\left({\psi}^{l}\left(\frac{1}{{\mu}^{{n}_{1}}}\right)\right),{F}_{{f}^{l+1}p,{f}^{l+2}p}\left({\psi}^{l+1}\left(\frac{1}{{\mu}^{{n}_{1}}}\right)\right)),\dots ,\\ {F}_{{f}^{l+m1}p,{f}^{l+m}p}\left({\psi}^{l+m1}\left(\frac{1}{{\mu}^{{n}_{1}}}\right)\right)\left)\right)\\ \ge & T\left(T\right(\cdots T\left(\right(1{\phi}^{l}(1M{\left({\mu}^{j}\right)}^{{n}_{1}}),(1{\phi}^{l+1}(1M{\left({\mu}^{j}\right)}^{{n}_{1}})),\dots ,\\ (1{\phi}^{l+m1}(1M{\left({\mu}^{j}\right)}^{{n}_{1}}))\left)\right)\\ \ge & {\mathrm{\top}}_{i=l}^{\mathrm{\infty}}(1{\phi}^{i}(1M{\left({\mu}^{j}\right)}^{{n}_{1}})).\end{array}
Since T is φconvergent, we conclude that {({f}^{l}p)}_{l\in \mathbb{N}} is a Cauchy sequence. On the other hand, S is complete, therefore, there is a z\in S such that z={lim}_{l\to \mathrm{\infty}}{f}^{l}p. By the continuity of the mapping f and {z}_{l+1}=f{z}_{l} when l\to +\mathrm{\infty}, it follows that fz=z. □
Example 3.2 Let (S,F,T) and the mappings f, ψ and φ be the same as in Example 3.1. Since {\sum}_{i}{\psi}^{i}(\delta )={\sum}_{i}\frac{\delta}{{2}^{i}}=\delta for all \delta >0 and if we set p={x}_{2}\in S, j>0 then {t}^{j}(1{F}_{{x}_{2}{x}_{2}}(t))=0 for every t>0 or {sup}_{t>0}{t}^{j}(1{F}_{{x}_{2}{x}_{2}}(t))<\mathrm{\infty}, so {x}_{2} is the unique fixed point for f.
Mihet in [3] showed, if f:S\to S is a (\psi ,\phi ,\u03f5,\lambda )contraction and is a complete fuzzy metric space, then f has an unique fixed point. Now we present a generalization of the (\psi ,\phi ,\u03f5,\lambda )contraction. First, we define the class of functions ℵ as follows.
Let ℵ be the family of all the mappings m:\overline{R}\to \overline{R} such that the following conditions are satisfied:

(1)
\mathrm{\forall}t,s\ge 0:m(t+s)\ge m(t)+m(s);

(2)
m(t)=0\iff t=0;

(3)
m is continuous.
Definition 3.2 Let (S,F) be a probabilistic metric space and f:S\to S. The mapping f is a generalized (\psi ,\phi ,\u03f5,\lambda )contraction if there exist a continuous, decreasing function h:[0,1]\to [0,\mathrm{\infty}] such that h(1)=0, {m}_{1},{m}_{2}\in \mathrm{\aleph}, and \lambda \in (0,1) such that the following implication holds for every p,q\in S and for every \u03f5>0:
ho{F}_{p,q}({m}_{2}(\u03f5))<{m}_{1}(\lambda )\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}ho{F}_{f(p),f(q)}\left({m}_{2}(\psi (\u03f5))\right)<{m}_{1}(\phi (\lambda )).
If {m}_{1}(a)={m}_{2}(a)=a, and h(a)=1a for every a\in [0,1], we obtain the Mihet definition.
Theorem 3.3 Let (S,F,T) be a complete Menger space with tnorm T such that {sup}_{0\le a<1}T(a,a)=1 and f:S\to S be a generalized (\psi ,\phi ,\u03f5,\lambda )contraction such that ψ is continuous on (0,\mathrm{\infty}) and {lim}_{n\to \mathrm{\infty}}{\psi}^{n}(\delta )=0 for every \delta \in (0,\mathrm{\infty}). Suppose that there exists \lambda \in (0,1) such that h(0)<{m}_{1}(\lambda ) and φ, ψ satisfy \phi (0)=\psi (0)=0. Then x={lim}_{n\to \mathrm{\infty}}{f}^{n}(p) is the unique fixed point of the mapping f for an arbitrary p\in S.
Proof First we shall prove that f is uniformly continuous. Let \zeta >0 and \eta \in (0,1). We have to prove that there exists N(\overline{\zeta},\overline{\eta})=\{(p,q)(p,q)\in S\times S,{F}_{p,q}(\overline{\zeta})>1\overline{\eta}\} such that
(p,q)\in N(\overline{\zeta},\overline{\eta})\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}{F}_{f(p),f(q)}(\zeta )>1\eta .
Let ϵ be such that {m}_{2}(\psi (\u03f5))<\zeta and \lambda \in (0,1) such that
{m}_{1}(\phi (\lambda ))<h(1\eta ).
(2)
Since {m}_{1} and {m}_{2} are continuous at zero, and {m}_{1}(0)={m}_{2}(0)=0 such numbers ϵ and λ exist. We prove that \overline{\zeta}={m}_{2}(\u03f5), \overline{\eta}=1{h}^{1}({m}_{1}(\lambda )). If (p,q)\in N(\overline{\zeta},\overline{\eta}), we have
\begin{array}{rcl}{F}_{p,q}({m}_{2}(\u03f5))& >& 1(1{h}^{1}({m}_{1}(\lambda )))\\ =& {h}^{1}({m}_{1}(\lambda )).\end{array}
Since h is decreasing, it follows that ho{F}_{p,q}({m}_{2}(\u03f5))<{m}_{1}(\lambda ). Hence,
ho{F}_{f(p),f(q)}\left({m}_{2}(\psi (\u03f5))\right)<{m}_{1}(\phi (\lambda )).
Using (2), we conclude that
ho{F}_{f(p),f(q)}\left({m}_{2}(\psi (\u03f5))\right)<h(1\eta )
and since h is decreasing we have
{F}_{f(p),f(q)}(\zeta )\ge {F}_{f(p),f(q)}\left({m}_{2}(\psi (\u03f5))\right)>1\eta .
Therefore, (f(p),f(q))\in N(\zeta ,\eta ) if (p,q)\in N(\overline{\zeta},\overline{\eta}). We prove that for every \zeta >0 and \eta \in (0,1) there exists {n}_{0}(\zeta ,\eta )\in N such that for every p,q\in S
n>{n}_{0}(\zeta ,\eta )\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}{F}_{{f}^{n}(p),{f}^{n}(q)}(\zeta )>1\eta .
(3)
By assumption, there is a \lambda \in (0,1) such that h(0)<{m}_{1}(\lambda ). From {F}_{p,q}({m}_{2}(\u03f5))\ge 0, it follows that
ho{F}_{p,q}({m}_{2}(\u03f5))\le h(0)<{m}_{1}(\lambda )
which implies that ho{F}_{f(p),f(q)}({m}_{2}(\psi (\u03f5)))<{m}_{1}(\phi (\lambda )), and continuing in this way we obtain that for every n\in N
ho{F}_{{f}^{n}(p),{f}^{n}(q)}\left({m}_{2}({\psi}^{n}(\u03f5))\right)<{m}_{1}({\phi}^{n}(\lambda )).
Let {n}_{0}(\zeta ,\eta ) be a natural number such that {m}_{2}({\psi}^{n}(\u03f5))<\zeta and {m}_{1}({\phi}^{n}(\lambda ))<h(1\eta ), for every n\ge {n}_{0}(\zeta ,\eta ). Then n>{n}_{0}(\zeta ,\eta ) implies that
{F}_{{f}^{n}(p),{f}^{n}(q)}(\zeta )\ge {F}_{{f}^{n}(p),{f}^{n}(q)}\left({m}_{2}({\psi}^{n}(\u03f5))\right)>1\eta .
If q={f}^{m}(p), from (3) we obtain that
{F}_{{f}^{n}(p),{f}^{n+m}(p)}(\zeta )>1\eta \phantom{\rule{1em}{0ex}}\text{for every}n{n}_{0}(\zeta ,\eta )\text{and every}m\in N.
(4)
Relation (4) means that {({f}^{n}(p))}_{n\in N} is a Cauchy sequence, and since S is complete there exists x={lim}_{n\to \mathrm{\infty}}{f}^{n}(p), which is obviously a fixed point of f since f is continuous.
For every p\in S and q\in S such that f(p)=p and f(q)=q we have for every n\in N that {f}^{n}(p)=p, {f}^{n}(q)=q and, therefore, from (3) we have {F}_{p,q}(\zeta )>1\eta for every \eta \in (0,1) and \zeta >0. This implies that {F}_{p,q}(\zeta )=1 for every \zeta >0 and, therefore, p=q. □
Example 3.3 Let (S,F,T) and the mappings f, ψ and φ be the same as in Example 3.1. Set h(a)={e}^{a}{e}^{1} for every a\in [0,1] and {m}_{1}(a)={m}_{2}(a)=a. The mapping f is generalized (\psi ,\phi ,\u03f5,\lambda )contraction and {lim}_{n\to \mathrm{\infty}}{\psi}^{n}(\delta )={lim}_{n\to \mathrm{\infty}}\frac{\delta}{{2}^{n}}=0 for every \delta \in (0,\mathrm{\infty}). On the other hand, there exists \lambda \in (0,1) such that h(0)=1\frac{1}{e}<\lambda and \psi (0)=\phi (0)=0. So {x}_{2} is the unique fixed point for f.