$(\psi ,\phi ,ϵ,\lambda )$-Contraction theorems in probabilistic metric spaces for single valued case

Abstract

In this article, we prove some fixed-point theorems for $(\psi ,\phi ,ϵ,\lambda )$-contraction in probabilistic metric spaces for single valued case. We will generalize the definition of $(\psi ,\phi ,ϵ,\lambda )$-contraction and present fixed-point theorem in the generalized $(\psi ,\phi ,ϵ,\lambda )$-contraction.

1 Introduction

The probabilistic metric space was introduced by Menger [1]. Mihet presented the class of $(\psi ,\phi ,ϵ,\lambda )$-contraction for a single valued case in fuzzy metric spaces [2, 3]. This class is a generalization of the $(ϵ,\lambda )$-contraction which was introduced in [4]. We defined the class of $(\psi ,\phi ,ϵ,\lambda )$-contraction for the multi-valued case in a probabilistic metric space before [5]. Now, we obtain two fixed-point theorems of $(\psi ,\phi ,ϵ,\lambda )$-contraction for single valued case. Also, we extend the concept of $(\psi ,\phi ,ϵ,\lambda )$-contraction to the generalized $(\psi ,\phi ,ϵ,\lambda )$-contraction.

The structure of this paper is as follows: Section 2 is a review of some concepts in probabilistic metric spaces and probabilistic contractions. In Section 3, we will show two theorems for $(\psi ,\phi ,ϵ,\lambda )$-contraction in the single-valued case and explain the generalized $(\psi ,\phi ,ϵ,\lambda )$-contraction.

2 Preliminary notes

We recall some concepts from probabilistic metric space, convergence and contraction. For more details, we refer the reader to [6–8].

Let $D_{+}$ be the set of all distribution of functions F such that $F(0)=0$ (F is a non-decreasing, left continuous mapping from ℝ into $[0,1]$ such that ${lim}_{x\to \mathrm{\infty }}F(x)=1$).

The ordered pair $(S,F)$ is said to be a probabilistic metric space if S is a nonempty set and $F:S\times S\to D_{+}$ ($F(p,q)$ written by $F_{pq}$ for every $(p,q)\in S\times S$) satisfies the following conditions:

• (1) $F_{uv}(x)=1$ for every $x>0\iff u=v$ ($u,v\in S$),

• (2) $F_{uv}=F_{vu}$ for every $u,v\in S$,

• (3) $F_{uv}(x)=1$ and $F_{vw}(y)=1\Rightarrow F_{u,w}(x+y)=1$ for every $u,v,w\in S$, and every $x,y\in {\mathbb{R}}^{+}$.

A Menger space is a triple $(S,F,T)$ where $(S,F)$ is a probabilistic metric space, T is a triangular norm (abbreviated t-norm) and the following inequality holds:

$$F_{uv}(x+y)\ge T(F_{uw}(x),F_{wv}(y))\text{for every }u,v,w\in S,\text{ and every }x,y\in {\mathbb{R}}^{+}.$$

Recall the mapping $T:[0,1]\times [0,1]\to [0,1]$ is called a triangular norm (a t-norm) if the following conditions are satisfied: $T(a,1)=a$ for every $a\in [0,1]$; $T(a,b)=T(b,a)$ for every $a,b\in [0,1]$; $a\ge b,c\ge d\Rightarrow T(a,c)\ge T(b,d)$, $a,b,c,d\in [0,1]$; $T(T(a,b),c)=T(a,T(b,c))$, $a,b,c\in [0,1]$. Basic examples of t-norms are $T_L$ (Lukasiewicz t-norm), $T_P$ and $T_M$, defined by $T_L(a,b)=max\{a+b-1,0\}$, $T_P(a,b)=ab$ and $T_M(a,b)=min\{a,b\}$. If T is a t-norm and $(x_1,x_2,\dots ,x_n)\in {[0,1]}^n$ ($n\in {\mathbb{N}}^{\ast }$), one can define recurrently ${\mathrm{\top }}_{i=1}^n{x}_i=T({\mathrm{\top }}_{i=1}^{n-1}{x}_i,x_n)$ for all $n\ge 2$. One can also extend T to a countable infinitary operation by defining ${\mathrm{\top }}_{i=1}^{\mathrm{\infty }}{x}_i$ for any sequence ${(x_i)}_{i\in {\mathbb{N}}^{\ast }}$ as ${lim}_{n\to \mathrm{\infty }}{\mathrm{\top }}_{i=1}^n{x}_i$.

If $q\in (0,1)$ is given, we say that the t-norm T is q-convergent if ${lim}_{n\to \mathrm{\infty }}{\mathrm{\top }}_{i=n}^{\mathrm{\infty }}(1-q^i)=1$. We remark that if T is q-convergent, then

$$\mathrm{\forall }\lambda \in (0,1)\mathrm{\exists }s=s(\lambda )\in \mathrm{N}{\mathrm{\top }}_{i=1}^n(1-q^{s+i})>1-\lambda ,\mathrm{\forall }n\in \mathrm{N}.$$

Also, note that if the t-norm T is q-convergent, then ${sup}_{0\le t<1}T(t,t)=1$.

Proposition 2.1 Let $(S,F,T)$ be a Menger space. If ${sup}_{0\le t<1}T(t,t)=1$, then the family ${\{U_{ϵ}\}}_{ϵ>0}$, where

$$U_{ϵ}=\{(x,y)\in S\times S\mid F_{x,y}(ϵ)>1-ϵ\}$$

is a base for a metrizable uniformity on S, called the F-uniformity [6–8]. The F-uniformity naturally determines a metrizable topology on S, called the strong topology or F-topology [9], a subset O of S is F-open if for every $p\in O$ there exists $t>0$ such that $N_p=\{q\in S|F_{pq}(t)>1-t\}\subset O$.

Definition 2.1 [6]

A sequence ${(x_n)}_{n\in \mathrm{N}}$ is called an F-convergent sequence to $x\in S$ if for every $ϵ>0$ and $\lambda \in (0,1)$ there exists $N=N(ϵ,\lambda )\in \mathrm{N}$ such that $F_{x_{n}x}(ϵ)>1-\lambda$, $\mathrm{\forall }n\ge N$.

Definition 2.2 [6]

Let $\phi :(0,1)\to (0,1)$ be a mapping, we say that the t-norm T is φ-convergent if

$$\mathrm{\forall }\delta \in (0,1),\mathrm{\forall }\lambda \in (0,1)\mathrm{\exists }s=s(\delta ,\lambda )\in \mathrm{N}{\mathrm{\top }}_{i=1}^n(1-{\phi }^{s+i}(\delta ))>1-\lambda ,\mathrm{\forall }n\ge 1.$$

Definition 2.3 [6]

A sequence ${(x_n)}_{n\in \mathrm{N}}$ is called a convergent sequence to $x\in S$ if for every $ϵ>0$ and $\lambda \in (0,1)$ there exists $N=N(ϵ,\lambda )\in \mathrm{N}$ such that $F_{x_{n}x}(ϵ)>1-\lambda$, $\mathrm{\forall }n\ge N$.

Definition 2.4 [6]

A sequence ${(x_n)}_{n\in \mathrm{N}}$ is called a Cauchy sequence if for every $ϵ>0$ and $\lambda \in (0,1)$ there exists $N=N(ϵ,\lambda )\in \mathrm{N}$ such that $F_{x_n{x}_{n+m}}(ϵ)>1-\lambda$, $\mathrm{\forall }n\ge N$, $\mathrm{\forall }m\in \mathbb{N}$.

We also have

$$x_n{\to }^{F}x\iff F_{x_{n}x}(t)\to 1\mathrm{\forall }t>0.$$

A probabilistic metric space $(S,F,T)$ is called sequentially complete if every Cauchy sequence is convergent.

The concept of $(\psi ,\phi ,ϵ,\lambda )$-contraction has been introduced by Mihet [3].

We will consider comparison functions from the class ϕ of all mapping $\phi :(0,1)\to (0,1)$ with the properties:

• (1) φ is an increasing bijection;

• (2) $\phi (\lambda )<\lambda$ $\mathrm{\forall }\lambda \in (0,1)$.

Since every such a comparison mapping is continuous, if $\phi \in \varphi$, then ${lim}_{n\to \mathrm{\infty }}{\phi }^n(\lambda )=0$ $\mathrm{\forall }\lambda \in (0,1)$.

Definition 2.5 [3]

Let $(S,F)$ be a probabilistic space, $\phi \in \varphi$ and ψ be a map from $(0,\mathrm{\infty })$ to $(0,\mathrm{\infty })$. A mapping $f:S\to S$ is called a $(\psi ,\phi ,ϵ,\lambda )$-contraction on S if it satisfies in the following condition:

$$x,y\in S,ϵ>0,\lambda \in (0,1),F_{x,y}(ϵ)>1-\lambda \Rightarrow F_{f(x),f(y)}(\psi (ϵ))>1-\phi (\lambda ).$$

In the rest of paper we suppose that ψ is increasing bijection.

Example 2.1 Let $S=\{0,1,2,\dots \}$ and (for $x\ne y$)

$$F_{x,y}(t)=\{\begin{array}{cc}0\hfill & \text{if }t\le 2^{-min(x,y)},\hfill \\ 1-2^{-min(x,y)}\hfill & \text{if }{2}^{-min(x,y)}<t\le 1,\hfill \\ 1\hfill & \text{if }t>1.\hfill \end{array}$$

Suppose that $f:S\to S$, $f(r)=r+1$.

Then $(S,F,T_L)$ is a probabilistic metric space [10].

Let x, y, ϵ, λ be such that $F_{x,y}(ϵ)>1-\lambda$.

• (i) If $2^{-min(x,y)<ϵ\le 1}$, then $1-2^{-min(x,y)}>1-\lambda$.

  This implies $1-2^{-min(x+1,y+1)}>1-\frac{1}{2}\lambda$, that is,

  $$F_{fx,fy}(ϵ)>1-\frac{1}{2}\lambda .$$

• (ii) If $ϵ>1$ then $F_{fx,fy}(ϵ)=1$, hence again $F_{fx,fy}(ϵ)>1-\frac{1}{2}\lambda$. Thus, the mapping f is a $(\psi ,\phi ,ϵ,\lambda )$-contraction on S with $\psi (ϵ)=ϵ$ and $\phi (\lambda )=\frac{1}{2}\lambda$.

3 Main results

In this section, we will show $(\psi ,\phi ,ϵ,\lambda )$-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 $ϵ>0$ there exists $\delta >0$ such that

$$F_{u,v}(\delta )>1-\delta \Rightarrow F_{fu,fv}(ϵ)>1-ϵ.$$

Before we start to present the theorems, we will explain the following lemma.

Lemma 3.1 Every $(\psi ,\phi ,ϵ,\lambda )$-contraction is continuous.

Proof Suppose that $ϵ>0$ be given and $\delta \in (0,1)$ be such that $\delta <min\{ϵ,{\psi }^{-1}(ϵ)\}$ and since ψ is increasing bijection then $\psi (\delta )<ϵ$. If $F_{x,y}(\delta )>1-\delta$ then, by $(\psi ,\phi ,ϵ,\lambda )$-contraction we have $F_{fx,fy}(\psi (\delta ))>1-\phi (\delta )$, from where we obtain that $F_{fx,fy}(ϵ)>F_{fx,fy}(\psi (\delta ))>1-\phi (\delta )>1-\delta >1-ϵ$. So f is continuous. □

Theorem 3.1 Let $(S,F,T)$ be a complete Menger space and T a t-norm satisfies in ${sup}_{0\le a<1}T(a,a)=1$. Also, $f:S\to S$ a $(\psi ,\phi ,ϵ,\lambda )$-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 $n,m\in N$, $ϵ>0$, $\lambda \in (0,1)$. 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 ${\psi }^n(\eta )<ϵ$ and ${\phi }^n(\xi )<\lambda$, we obtain

$$F_{x_n,x_{n+m}}(ϵ)>1-\lambda .$$

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

$$F_{x_1,x_2}(t)=F_{x_2,x_1}(t)=\{\begin{array}{cc}0\hfill & \text{if }t\le 0,\hfill \\ 0.9\hfill & \text{if }0<t\le 3,\hfill \\ 1\hfill & \text{if }t>3\hfill \end{array}$$

and

$$\begin{array}{rcl}{F}_{x_1,x_3}(t)& =& F_{x_3,x_1}(t)=F_{x_1,x_4}(t)=F_{x_4,x_1}(t)=F_{x_2,x_3}(t)=F_{x_3,x_2}(t)=F_{x_2,x_4}(t)\\ =& F_{x_4,x_2}(t)=F_{x_3,x_4}(t)=F_{x_4,x_3}(t)=\{\begin{array}{cc}0\hfill & \text{if }t\le 0,\hfill \\ 0.7\hfill & \text{if }0<t<6,\hfill \\ 1\hfill & \text{if }6\le t\hfill \end{array}\end{array}$$

$f:S\to S$ is given by $f(x_1)=f(x_2)=x_2$ and $f(x_3)=f(x_4)=x_1$. If we take $\phi (\lambda )=\frac{\lambda }{2}$, $\psi (ϵ)=\frac{ϵ}{2}$, then f is a $(\psi ,\phi ,ϵ,\lambda )$-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 t-norm such that ${sup}_{0\le a<1}T(a,a)=1$ and $f:S\to S$ a $(\psi ,\phi ,ϵ,\lambda )$-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 t-norm T is φ-convergent, then there exist a unique fixed point z of mapping f and $z={lim}_{l\to \mathrm{\infty }}{f}^{l}p$.

Proof Choose $ϵ>0$ 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 $n_0(ϵ,\lambda )\in N$ such that

$$F_{f^{l}p,f^{l+m}p}(ϵ)>1-\lambda \text{for every }l\ge n_0(ϵ,\lambda )\text{ and every }m\in N.$$

Suppose that $\mu \in (0,1)$, $M>0$ are such that

$$x^j(1-F_{p,f_p}(x))\le M\mathrm{\forall }x>0.$$

(1)

Let $n_1$ be such that

$$1-M{\left({\mu }^j\right)}^{n_1}\in [0,1).$$

From (1), it follows that

$$F_{p,fp}\left(\frac{1}{{\mu }^n}\right)>1-M{\left({\mu }^j\right)}^n\mathrm{\forall }n\in N\text{ specially for }n=n_1.$$

Since f is $(\psi ,\phi ,ϵ,\lambda )$-contraction, we derived by induction $F_{f^{l}p,f^{l+1}p}({\psi }^l(\frac{1}{{\mu }^{n_1}}))>1-{\phi }^l(1-M{({\mu }^j)}^{n_1})$ $\mathrm{\forall }l>1$. Since the series ${\sum }_{i=1}^{\mathrm{\infty }}{\psi }^i(\delta )$ is convergent, there exists $n_2=n_2(ϵ)\in N$ such that ${\sum }_{i=l}^{\mathrm{\infty }}{\psi }^i(\delta )\le ϵ$ $\mathrm{\forall }l\ge n_2$. We know ${\sum }_{i=l}^{\mathrm{\infty }}{\psi }^i(\frac{1}{{\mu }^{n_1}})\le ϵ$ for every $l>max\{n_1,n_2\}$.

Now

$$\begin{array}{rcl}{F}_{f^{l}p,f^{l+m}p}(ϵ)& \ge & F_{f^{l}p,f^{l+m}p}\left(\sum _{i=l}^{\mathrm{\infty }}{\psi }^i\left(\frac{1}{{\mu }^{n_1}}\right)\right)\ge F_{f^{l}p,f^{l+m}p}\left(\sum _{i=l}^{l+m-1}{\psi }^i\left(\frac{1}{{\mu }^{n_1}}\right)\right)\\ \ge & T\left(T\right(\cdots T(F_{f^{l}p,f^{l+1}p}\left({\psi }^l\left(\frac{1}{{\mu }^{n_1}}\right)\right),F_{f^{l+1}p,f^{l+2}p}\left({\psi }^{l+1}\left(\frac{1}{{\mu }^{n_1}}\right)\right)),\dots ,\\ F_{f^{l+m-1}p,f^{l+m}p}\left({\psi }^{l+m-1}\left(\frac{1}{{\mu }^{n_1}}\right)\right)\left)\right)\\ \ge & T\left(T\right(\cdots T\left(\right(1-{\phi }^l(1-M{\left({\mu }^j\right)}^{n_1}),(1-{\phi }^{l+1}(1-M{\left({\mu }^j\right)}^{n_1})),\dots ,\\ (1-{\phi }^{l+m-1}(1-M{\left({\mu }^j\right)}^{n_1}))\left)\right)\\ \ge & {\mathrm{\top }}_{i=l}^{\mathrm{\infty }}(1-{\phi }^i(1-M{\left({\mu }^j\right)}^{n_1})).\end{array}$$

Since T is φ-convergent, 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 $(\psi ,\phi ,ϵ,\lambda )$-contraction and is a complete fuzzy metric space, then f has an unique fixed point. Now we present a generalization of the $(\psi ,\phi ,ϵ,\lambda )$-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 $(\psi ,\phi ,ϵ,\lambda )$-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 $ϵ>0$:

$$ho{F}_{p,q}(m_2(ϵ))<m_1(\lambda )\Rightarrow ho{F}_{f(p),f(q)}\left(m_2(\psi (ϵ))\right)<m_1(\phi (\lambda )).$$

If $m_1(a)=m_2(a)=a$, and $h(a)=1-a$ for every $a\in [0,1]$, we obtain the Mihet definition.

Theorem 3.3 Let $(S,F,T)$ be a complete Menger space with t-norm T such that ${sup}_{0\le a<1}T(a,a)=1$ and $f:S\to S$ be a generalized $(\psi ,\phi ,ϵ,\lambda )$-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

$$(p,q)\in N(\overline{\zeta },\overline{\eta })\Rightarrow F_{f(p),f(q)}(\zeta )>1-\eta .$$

Let ϵ be such that $m_2(\psi (ϵ))<\zeta$ 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 $\overline{\zeta }=m_2(ϵ)$, $\overline{\eta }=1-h^{-1}(m_1(\lambda ))$. If $(p,q)\in N(\overline{\zeta },\overline{\eta })$, we have

$$\begin{array}{rcl}{F}_{p,q}(m_2(ϵ))& >& 1-(1-h^{-1}(m_1(\lambda )))\\ =& h^{-1}(m_1(\lambda )).\end{array}$$

Since h is decreasing, it follows that $ho{F}_{p,q}(m_2(ϵ))<m_1(\lambda )$. Hence,

$$ho{F}_{f(p),f(q)}\left(m_2(\psi (ϵ))\right)<m_1(\phi (\lambda )).$$

Using (2), we conclude that

$$ho{F}_{f(p),f(q)}\left(m_2(\psi (ϵ))\right)<h(1-\eta )$$

and since h is decreasing we have

$$F_{f(p),f(q)}(\zeta )\ge F_{f(p),f(q)}\left(m_2(\psi (ϵ))\right)>1-\eta .$$

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 $p,q\in S$

$$n>n_0(\zeta ,\eta )\Rightarrow F_{f^n(p),f^n(q)}(\zeta )>1-\eta .$$

(3)

By assumption, there is a $\lambda \in (0,1)$ such that $h(0)<m_1(\lambda )$. From $F_{p,q}(m_2(ϵ))\ge 0$, it follows that

$$ho{F}_{p,q}(m_2(ϵ))\le h(0)<m_1(\lambda )$$

which implies that $ho{F}_{f(p),f(q)}(m_2(\psi (ϵ)))<m_1(\phi (\lambda ))$, and continuing in this way we obtain that for every $n\in N$

$$ho{F}_{f^n(p),f^n(q)}\left(m_2({\psi }^n(ϵ))\right)<m_1({\phi }^n(\lambda )).$$

Let $n_0(\zeta ,\eta )$ be a natural number such that $m_2({\psi }^n(ϵ))<\zeta$ 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

$$F_{f^n(p),f^n(q)}(\zeta )\ge F_{f^n(p),f^n(q)}\left(m_2({\psi }^n(ϵ))\right)>1-\eta .$$

If $q=f^m(p)$, from (3) we obtain that

$$F_{f^n(p),f^{n+m}(p)}(\zeta )>1-\eta \text{for every }n>n_0(\zeta ,\eta )\text{ and every }m\in N.$$

(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 $(\psi ,\phi ,ϵ,\lambda )$-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.