Fixed Point Results for $(\phi ,\psi )$-Weak Contraction in Fuzzy Metric Spaces

Abstract

In the present work, a fixed point result for generalized weakly contractive mapping in fuzzy metric space has been established. An example is cited to illustrate the obtained result.

1 Introduction and Preliminaries

The concept of fuzzy metric spaces have been introduced in different ways by many authors. Among which, KM-fuzzy metric space, introduced by Kramosil and Michalek [2] and GV-fuzzy metric space, introduced by George and Veeramani [3], are two most widely used fuzzy metric spaces. KM-fuzzy metric space is similar to generalized Menger space [4]. George and Veeramani imposed a strong condition on the definition of Kramosil and Michalek for topological reasons. Several fixed point results in these fuzzy metric spaces can be found in [5, 7, 8, 10, 11].

Alber et al. extended the concept of Banach contraction to the weak contraction and established a fixed point result in Hilbert space [1]. There after B.E. Rhoades investigated this result in metric space [6]. Fixed point problem for weak contraction mapping have been investigated by many authors [12,13,14,15, 17,18,19,20,21,22,23,24]. In [9] Dutta et al. extended the results of Rhoades. Motivated by the works of [9, 16, 25], in the present work, a fixed point result in fuzzy metric space, introduced by George and Veeramani, is obtained and an example is added in the support of main result.

Definition 1.1

[4]. A continuous t-norm $*$ is a binary operation on $\left[ 0,1\right]$, which satisfies the following conditions:

• (i) $*$ is associative and commutative,

• (ii) $x*1=x,$ for all $x\in \left[ 0,1\right] ,$

• (iii) $x*y\le u*v,$ whenever $x\le u$ and $y\le v$, for all $x,y,u,v\in \left[ 0,1\right] ,$

• (iv) $*$ is continuous.

For example: (a) The minimum t-norm, $*_{M}$, defined by $x*_{M}y=\min \left\{ x,y\right\}$; (b) The product t-norm, $*_{P}$, defined by $x*_{P}y=x.y$, are two basic t-norms.

Definition 1.2

[3]. The triplet $\left( X,M,*\right)$ is called fuzzy metric space if X is a non-empty set, $*$ is continuous t-norm and M is a fuzzy set on $X^{2}\times \left( 0,\infty \right)$ satisfying the following conditions:  

(i):

$M\left( x,y,t\right) >0,$

(ii):

$M\left( x,y,t\right) =1$ if and only if $x=y,$

(iii):

$M\left( x,y,t\right) =M\left( y,x,t\right) ,$

(iv):

$M\left( x,z,t+s\right) \ge M(x,y,t)*M(y,z,s),$

(v):

$M\left( x,y,.\right) :\left( 0,\infty \right) \rightarrow \left[ 0,1\right]$ is continuous,

 

for all $t,s\in \left( 0,\infty \right)$ and $x,y,z\in X.$

In this paper, we use the notion of fuzzy metric space introduced by George and Veeramani.

Definition 1.3

[3]. Let $\left( X,M,*\right)$ be a fuzzy metric space. Then  

(i):

A sequence $\{x_{n}\}\subseteq X$ is said to converge to a point $x\in X$ if , for all $t>0.$

(ii):

A sequence $\{x_{n}\}\subseteq X$ is called a Cauchy sequence if for each $0<\varepsilon <1$ and $t>0,$ there exists an $N\in \mathbb {N}$ such that $M\left( x_{n},x_{m},t\right) >1-\varepsilon ,$ for each $m,n\ge N.$

(iii):

A fuzzy metric space is called complete if every Cauchy sequence in this space is convergent.

 

Lemma 1.1

[5]. Let $(X,M,*)$ be a fuzzy metric space. Then (X, M, .) is non-decreasing for all $x,y \in X.$

Lemma 1.2

[25]. If $*$ is a continuous t-norm, and $\{\alpha _{n}\}$, $\{\beta _{n}\}$ and $\{\gamma _{n}\}$ are sequences such that $\alpha _{n} \rightarrow \alpha , \gamma _{n}\rightarrow \gamma$ as $n\rightarrow \infty ,$ then $\underset{k\rightarrow \infty }{\overline{\lim }}(\alpha _{k}*\beta _{k}*\gamma _{k})= \alpha *\underset{k\rightarrow \infty }{\overline{\lim }}\beta _{k} *\gamma$ and $\underset{k\rightarrow \infty }{\underline{\lim }}(\alpha _{k}*\beta _{k}*\gamma _{k})= \alpha *\underset{k\rightarrow \infty }{\underline{\lim }}\beta _{k} *\gamma .$

Lemma 1.3

[25]. Let $\{f(k,.):(0,\infty )\rightarrow (0,1], k=0,1,2,........\}$ be a sequence of functions such that f(k, .) is continuous and monotone increasing for each $k\ge 0.$ Then $\underset{k\rightarrow \infty }{\overline{\lim }}f(k,t)$ is a left continuous function in t and $\underset{k\rightarrow \infty }{\underline{\lim }}f(k,t)$ is a right continuous function in t.

2 Main Results

Theorem 2.1

Let $\left( X,M,*\right)$ be a complete fuzzy metric space with an arbitrary continuous t-norm $'*'$ and let $T:X\rightarrow X$ be a self mapping satisfying the following condition:

$$\begin{aligned} \psi (M(Tx,Ty,t))\le \psi (\min (M(x,y,t), M(x,Tx,t), M(y,Ty,t)))\\ \nonumber -\phi (\min (M(x,y,t), M(y,Ty,t))), \end{aligned}$$

(2.1)

where $\psi , \phi :(0,1]\rightarrow [0,\infty )$ are two functions such that:

 

(i):

$\psi$ is continuous and monotone decreasing function with $\psi (t)=0$ if and only if $t=1,$

(ii):

$\phi$ is lower semi continuous function with $\phi (t)=0$ if and only if $t=1.$

 

Then T has a unique fixed point.

Proof: Let $x_{0}\in X.$ We define the sequence $\{x_n\}$ in X such that $x_{n+1}=Tx_{n},$ for each $n\ge 0.$ If there exists a positive integer k such that $x_{k}=x_{k+1},$ then $x_{k}$ is a fixed point of T. Hence, we shall assume that $x_{n}\ne x_{n+1},$ for all $n\ge 0.$ Now, from (2.1)

$$\begin{aligned} \psi (M(x_{n+1},x_{n+2},t))= & {} \psi (M(Tx_{n},Tx_{n+1},t))\nonumber \\\le & {} \psi (\min \{M(x_{n},x_{n+1},t), M(x_{n},x_{n+1},t), M(x_{n+1},x_{n+2},t) \})\nonumber \\&-\phi (\min \{M(x_{n},x_{n+1},t),M(x_{n+1},x_{n+2},t)\}). \end{aligned}$$

(2.2)

Suppose that $M(x_{n},x_{n+1},t) > M(x_{n+1},x_{n+2},t),$ for some positive integer n. Then from (2.2), we have

$\psi (M(x_{n+1},x_{n+2},t))\le \psi (M(x_{n+1},x_{n+2},t))-\phi (M(x_{n+1},x_{n+2},t)),$ that is, $\phi (M(x_{n+1},x_{n+2},t))\le 0,$ which implies that $M(x_{n+1},x_{n+2},t)=1.$ This gives that $x_{n+1}=x_{n+2},$ which is a contradiction.

Therefore, $M(x_{n+1},x_{n+2},t) \le M(x_{n},x_{n+1},t)$ for all $n\ge 0,$ and $\{ M(x_{n},x_{n+1},t)\}$ is a monotone increasing sequence of non-negative real numbers. Hence, there exists an $r>0$ such that .

In view of the above facts, from (2.2), we have

$\psi (M(x_{n+1},x_{n+2},t))\le \psi (M(x_{n},x_{n+1},t))-\phi (M(x_{n},x_{n+1},t)),$ for all $n\ge 0,$

Taking the limit as $n\rightarrow \infty$ in the above inequality and using the continuities of $\phi$ and $\psi$ we have $\psi (r)\le \psi (r)-\phi (r),$ which is a contradiction unless $r=1.$ Hence

$$\begin{aligned} M(x_{n}, x_{n+1}, t)\rightarrow 1\,\,\,\,\,\text {as}\,\, n\rightarrow \infty . \end{aligned}$$

(2.3)

Next, we show that $\{x_{n}\}$ is Cauchy sequence. If otherwise, there exist $\lambda ,\, \epsilon >0$ with $\lambda \in (0,1)$ such that for each integer k,  there are two integers l(k) and m(k) such that $m(k)>l(k)\ge k$ and

$$\begin{aligned} M(x_{l(k)}, x_{m(k)}, \epsilon )\le 1-\lambda ,\,\,\,\,\text {for all}\,\,k>0. \end{aligned}$$

(2.4)

By choosing m(k) to be the smallest integer exceeding l(k) for which (2.4) holds, then for all $k>0,$ we have

$$ M(x_{l(k)}, x_{m(k)-1}, \epsilon )> 1-\lambda . $$

Now, by triangle inequality, for any s with $0<s<\frac{\epsilon }{2},$ for all $k>0,$ we have

$$\begin{aligned} 1-\lambda\ge & {} M(x_{l(k)}, x_{m(k)}, \epsilon )\nonumber \\\ge & {} M(x_{l(k)}, x_{l(k)+1}, s)*M(x_{l(k)+1},x_{m(k)+1}, \epsilon -2s)*M(x_{m(k)+1},x_{m(k)}, s).\nonumber \\ \end{aligned}$$

(2.5)

For $t>0,$ we define the function $h_{1}(t)=\underset{n\rightarrow \infty }{\overline{\lim }}M\left( x_{l(k)+1},x_{m(k)+1},t\right) .$

Taking $\limsup$ on both the sides of (2.5), using (2.3) and the continuity property of $*,$ by Lemma (1.2), we conclude that

$$\begin{aligned} 1-\lambda\ge & {} 1*\underset{k\rightarrow \infty }{\overline{\lim }} M(x_{l(k)+1},x_{m(k)+1},\epsilon -2s)*1 \\= & {} \underset{k\rightarrow \infty }{\overline{\lim }} M(x_{l(k)+1},x_{m(k)+1},\epsilon -2s)\\= & {} h_{1}(\epsilon -2s). \end{aligned}$$

By an application of Lemma (1.3), $h_{1}$ is left continuous.

Letting limit as $s\rightarrow 0$ in the above inequality, we obtain

$$\begin{aligned} h_{1}(\epsilon )=\underset{k\rightarrow \infty }{\overline{\lim }}M(x_{l(k)+1},x_{m(k)+1},\epsilon )\le 1-\lambda . \end{aligned}$$

(2.6)

Next, for all $t>0,$ we define the function

$h_{2}(t)=\underset{k\rightarrow \infty }{\underline{\lim }}M\left( x_{l(k)+1},x_{m(k)+1},t\right) .$

In above similar process, we can prove that

$$\begin{aligned} h_{2}(\epsilon )= \underset{k\rightarrow \infty }{\underline{\lim }}M\left( x_{l(k)+1},x_{m(k)+1},\epsilon \right) \ge 1-\lambda . \end{aligned}$$

(2.7)

Combining (2.6) and (2.7), we get

$\underset{k\rightarrow \infty }{\overline{\lim }}M(x_{l(k)+1},x_{m(k)+1},\epsilon )\le 1-\lambda \le \underset{k\rightarrow \infty }{\underline{\lim }}M(x_{l(k)+1},x_{m(k)+1},\epsilon ).$

This implies that

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim }M(x_{l(k)+1},x_{m(k)+1},t)=1-\lambda . \end{aligned}$$

(2.8)

Again by (2.6),

$$ \underset{k\rightarrow \infty }{\overline{\lim }}M(x_{l(k)},x_{m(k)},\epsilon )\le 1-\lambda . $$

For $t>0,$ we define the function

$$\begin{aligned} h_{3}(t)=\underset{k\rightarrow \infty }{\underline{\lim }}M(x_{l(k)},x_{m(k)},\epsilon ). \end{aligned}$$

(2.9)

Now for $s>0,$

$M(x_{l(k)},x_{m(k)},\epsilon +2s) \ge M(x_{l(k)},x_{l(k)+1},s)*M(x_{l(k)+1},x_{m(k)+1},\epsilon )*M(x_{m(k)+1},x_{m(k)},s).$

Taking $\liminf$ both the sides, we have

$\underset{k\rightarrow \infty }{\underline{\lim }}M(x_{l(k)},x_{m(k)},\epsilon +2s)\ge 1*\underset{k\rightarrow \infty }{\underline{\lim }}M(x_{l(k)+1},x_{m(k)+1},\epsilon ) *1$ $=1-\lambda .$

Thus,

$$\begin{aligned} h_{3}(\epsilon +2s)\ge 1-\lambda . \end{aligned}$$

(2.10)

Taking limit as $s\rightarrow 0,$ we get $h_{3}(\epsilon )\ge 1-\lambda .$ Combining (2.9) and (2.10) we obtain

$$ \underset{n\rightarrow \infty }{\lim } M(x_{l(k)},x_{m(k)},\epsilon )=1. $$

Now,

$$\begin{aligned} \psi ({M(x_{l(k)+1},x_{m(k)+1},\epsilon )})\le & {} \psi (\min (M(x_{l(k)},x_{m(k)},\epsilon ),\\&M(x_{l(k)},x_{l(k)+1},\epsilon )),M(x_{m(k)},x_{m(k)+1},\epsilon )) \\&-\phi (\min (M(x_{l(k)},x_{m(k)},\epsilon ), M(x_{m(k)},x_{m(k)+1},\epsilon ))). \end{aligned}$$

Taking limit as $k\rightarrow \infty$, we get

$\psi (1-\lambda )\le \psi (1-\lambda )-\phi (1-\lambda ),$ which is a contradiction.

Thus, $\{x_{n}\}$ is Cauchy sequence. Since X is complete, there exists $p \in X$ such that $x_{n}\rightarrow p$ as $n\rightarrow \infty .$ Now,

$$\begin{aligned} \psi (M(x_{n+1}, Tp, t))= & {} \psi (M(Tx_{n}, Tp, t))\\\le & {} \psi (\min \{M(x_{n}, p, t), M(x_{n}, x_{n+1},t),M(p,Tp,t)\})\\&- \phi (\min \{M(x_{n}, p, t), M(p,Tp,t)\}). \end{aligned}$$

Taking limit as $n \rightarrow \infty ,$ we get

$\psi (M(p,Tp,t))\le \psi (M(p,Tp,t))-\phi (M(p,Tp,t)),$

which implies that $\phi (M(p,Tp,t))=0,$ that is,

$M(p,Tp,t)=1$ or $p=Tp.$

We next establish that fixed point is unique. Let p and q be two fixed points of T.

Putting $x=p$ and $y=q$ in (2.1),

$\psi (M(Tp,Tq,t))\le \psi (\min {M(p,q,t),M(p,Tp,t),M(q,Tq,t)})-\phi (\min {M(p,q,t),M(q,Tq,t)})$ or, $\psi (M(p,q,t))\le \psi (\min {M(p,q,t),M(p,p,t),M(q,q,t)})-\phi (\min {M(p,q,t),M(q,q,t)})$ or, $\psi (M(p,q,t))\le \psi (M(p,q,t))-\phi (M(p,q,t))$ or, $\phi (M(p,q,t))\le 0,$ or, equivalently, $M(p,q,t)=1,$ that is, $p=q.$

The following example is in support of Theorem 2.1.

Example 2.1

Let $X=[0,1].$ Let

$$ M(x,y,t)=e^{-\frac{|x-y|}{t}}, $$

for all $x,y\in X$ and $t>0,$ then $(X,M,*)$ is a complete fuzzy metric space, where $'*'$ is product t-norm. Let $\,\,\psi ,\phi :(0,1]\rightarrow [0,\infty )$ be defined by $\psi (s)=\frac{1}{s}-1$ and $\phi (s)=\frac{1}{s}-\frac{1}{\sqrt{s}}.$ Then $\psi$ and $\phi$ satisfy all the conditions of Theorem (2.1). Let the mapping $T:X\rightarrow X$ be defined by $Tx=\frac{x}{2},$ for all $x\in X.$

Now, we will show that

$$\begin{aligned} \psi (M(Tx,Ty,t)) \le \psi (M(x,y))-\phi (N(x,y)), \end{aligned}$$

(2.11)

where $M(x,y)=\min \{M(x,y,t),\,M(x,Tx,t),\,M(y,Ty,t)\}$ and $N(x,y)=\phi (\min \{M(x,y,t),\,M(y,Ty,t)\}).$ Herein;

$$ \max \{|x-y|,\,\frac{x}{2},\,\frac{y}{2}\}= {\left\{ \begin{array}{ll} x-y &{} \,\,\,\,\,0 \le y \le \frac{x}{2}\\ \frac{x}{2} &{}\,\,\,\,\, \frac{x}{2}< y \le x\\ \frac{y}{2} &{}\,\,\,\,\, x< y \le 2x\\ y-x &{}\,\,\,\,\, 2x < y \le 1 \end{array}\right. } $$

and

$$ \max \{|x-y|,\,\frac{y}{2}\}= {\left\{ \begin{array}{ll} x-y &{}\,\,\,\,\, 0 \le y \le \frac{2x}{3}\\ \frac{y}{2} &{}\,\,\,\,\, \frac{2x}{3}< y \le 2x\\ y-x &{}\,\,\,\,\, 2x < y \le 1. \end{array}\right. } $$

Case (1): When $0\le y\le \frac{x}{2}$ or $2x < y\le 1,$ then

$$ \psi (M(Tx,Ty,t))=\psi (e^{-|\frac{x-y}{2t}|})=e^{|\frac{x-y}{2t}|}-1 $$

and

$$ \psi (M(x,y))-\phi (N(x,y))=\psi (e^{-\frac{|{x-y}|}{t}})-\phi (e^{-\frac{|{x-y}|}{t}}) =e^{|\frac{x-y}{2t}|}-1. $$

Obviously, in this case, (2.11) is satisfied.

Case (2): When $\frac{x}{2}< y \le \frac{2x}{3},$ then

$$ \psi (M(Tx,Ty,t))=\psi (e^{-\frac{x-y}{2t}})=e^{\frac{x-y}{2t}}-1 $$

and

$$ \psi (M(x,y))-\phi (N(x,y))=\psi (e^{-\frac{{x}}{2t}})-\phi (e^{-\frac{x-y}{t}}) =e^{\frac{x}{2t}}-1-e^{\frac{x-y}{t}}+e^{\frac{x-y}{2t}}. $$

In this case, $\frac{x}{2}\ge x-y$ and exponential function is an increasing function. Therefore, $e^{\frac{x-y}{2t}} \le e^{\frac{x}{2t}}-e^{\frac{x-y}{t}}+e^{\frac{x-y}{2t}}$ and hence (2.11) is satisfied.

Case (3): When $\frac{2x}{3}< y \le x,$ then

$$ \psi (M(Tx,Ty,t))=\psi (e^{-\frac{x-y}{2t}})=e^{\frac{x-y}{2t}}-1 $$

and

$$ \psi (M(x,y))-\phi (N(x,y))=\psi (e^{-\frac{{x}}{2t}})-\phi (e^{-\frac{y}{2t}}) =e^{\frac{x}{2t}}-1-e^{\frac{y}{2t}}+e^{\frac{y}{4t}}. $$

Since, in this case, $\frac{x-y}{2}\le \frac{y}{4}$ and $\frac{x}{2}\ge \frac{y}{2}$, (2.11) is satisfied.

Case (4): $x<y\le 2x,$ then

$$ \psi (M(Tx,Ty,t))=\psi (e^{-\frac{y-x}{2t}})=e^{\frac{y-x}{2t}}-1 $$

and

$$ \psi (M(x,y))-\phi (N(x,y))=\psi (e^{-\frac{{y}}{2t}})-\phi (e^{-\frac{y}{2t}}) =e^{\frac{x}{4t}}-1. $$

Since, in this case, $\frac{y}{2}\ge y-x,$ (2.11) is satisfied. Hence, all the conditions of Theorem (2.1) are satisfied. Thus, 0 is the unique fixed point of T.