Approximate ∗-derivations on fuzzy Banach ∗-algebras

Abstract

In this paper, we establish functional equations of ∗-derivations and prove the stability of ∗-derivations on fuzzy Banach ∗-algebras. We also prove the superstability of ∗-derivations on fuzzy Banach ∗-algebras.

MSC:39B52, 47B47, 46L05, 39B72.

1 Introduction

Let $\mathcal{A}$ be a Banach ∗-algebra. A linear mapping $\delta :D(\delta )\to \mathcal{A}$ is said to be a derivation on $\mathcal{A}$ if $\delta (ab)=\delta (a)b+a\delta (b)$ for all $a,b\in \mathcal{A}$, where $D(\delta )$ is a domain of δ and $D(\delta )$ is dense in $\mathcal{A}$. If δ satisfies the additional condition $\delta (a^{\ast })=\delta {(a)}^{\ast }$ for all $a\in \mathcal{A}$, then δ is called a ∗-derivation on $\mathcal{A}$. It is well known that if $\mathcal{A}$ is a $C^{\ast }$-algebra and $D(\delta )$ is A, then the ∗-derivation δ is bounded. For several reasons, the theory of bounded derivations of $C^{\ast }$-algebras is very important in the theory of quantum mechanics and operator algebras [3, 4].

A functional equation is called stable if any function satisfying a functional equation “approximately” is near to a true solution of the functional equation. We say that a functional equation is superstable if every approximate solution is an exact solution of it.

In 1940, Ulam [24] proposed the following question concerning stability of group homomorphisms: Under what condition is there an additive mapping near an approximately additive mapping? Hyers [8] answered positively the problem of Ulam for the case where $G_1$ and $G_2$ are Banach spaces. A generalized version of the theorem of Hyers for an approximately linear mapping was given by ThM Rassias [20]. Since then, the stability problems of various functional equations have been extensively investigated by a number of authors (for instances, [1, 2, 9, 10, 19, 20]). In particular, those of the important functional equations are the following functional equations:

(1.1)

(1.2)

which are called the Cauchy equation and the Jensen equation, respectively. Every solution of the functional equations (1.1) and (1.2) is said to be an additive mapping.

Since Katsaras [14] introduced the idea of fuzzy norm on a linear space, several definitions for a fuzzy norm on a linear space have been introduced and discussed from different points of view [5–7]. We use the definition of fuzzy normed spaces given in [5, 17] to investigate the stability of derivation in the fuzzy Banach ∗-algebra setting. The stability of functional equations in fuzzy normed spaces was begun by [17], after then lots of results of fuzzy stability were investigated [11, 13, 16, 18].

Definition 1.1 [5, 17, 21]

Let X be a real vector space. A function $N:X\times \mathbb{R}\to [0,1]$ is called a fuzzy norm on X if for all $x,y\in X$ and all $s,t\in \mathbb{R}$,

($N_1$) $N(x,t)=0$ for $t\le 0$;

($N_2$) $x=0$ if and only if $N(x,t)=1$ for all $t>0$;

($N_3$) $N(cx,t)=N(x,\frac{t}{|c|})$ if $c\ne 0$;

($N_4$) $N(x+y,s+t)\ge min\{N(x,s),N(y,t)\}$;

($N_5$) $N(x,\cdot )$ is a non-decreasing function of $\mathbb{R}$ and ${lim}_{t\to \mathrm{\infty }}N(x,t)=1$;

($N_6$) for $x\ne 0$, $N(x,\cdot )$ is continuous on $\mathbb{R}$.

The pair $(X,N)$ is called a fuzzy normed vector space.

Furthermore, we can make $(X,N)$ a fuzzy normed ∗-algebra if we add ($N_7$) and ($N_8$) as follows:

($N_7$) $N(xy,st)\ge min\{N(x,s),N(y,t)\}$;

($N_8$) $N(x,t)=N(x^{\ast },t)$.

The properties and examples of fuzzy normed vector spaces, fuzzy algebras, and fuzzy norms are given in [17, 18, 22, 23].

Definition 1.2 [5, 17, 21]

Let $(X,N)$ be a fuzzy normed vector space. A sequence $\{x_n\}$ in X is said to be convergent or converge if there exists an $x\in X$ such that ${lim}_{n\to \mathrm{\infty }}N(x_n-x,t)=1$ for all $t>0$. In this case, x is called the limit of the sequence $\{x_n\}$ and we denote it by N-${lim}_{n\to \mathrm{\infty }}{x}_n=x$.

Definition 1.3 [5, 17, 21]

Let $(X,N)$ be a fuzzy normed vector space. A sequence $\{x_n\}$ in X is called Cauchy if for each $\epsilon >0$ and each $t>0$ there exists an $n_0\in \mathbb{N}$ such that for all $n\ge n_0$ and all $p>0$, we have $N(x_{n+p}-x_n,t)>1-\epsilon$.

It is well known that every convergent sequence in a fuzzy normed vector space is Cauchy. If each Cauchy sequence is convergent, then the fuzzy norm is said to be complete and the fuzzy normed vector space is called a fuzzy Banach space.

We say that a mapping $f:X\to Y$ between fuzzy normed vector spaces X and Y is continuous at a point $x_0\in X$ if for each sequence $\{x_n\}$ converging to $x_0$ in X, then the sequence $\{f(x_n)\}$ converges to $f(x_0)$. If $f:X\to Y$ is continuous at each $x\in X$, then $f:X\to Y$ is said to be continuous on X.

In this paper, using the functional equation of ∗-derivations

$$f(\lambda a+b+cd)=\lambda f(a)+f(b)+f(c)d+cf(d)$$

introduced in [12] we prove fuzzy version of the stability of ∗-derivations associated to the Cauchy functional equation and the Jensen functional equation. We also prove the superstability of ∗-derivations on fuzzy Banach ∗-algebras.

2 Stability of ∗-derivations on fuzzy Banach ∗-algebras

In this section, let $\mathcal{A}$ be a fuzzy Banach ∗-algebra.

Theorem 2.1 Let $\phi :{\mathcal{A}}^4\to [0,\mathrm{\infty })$ and $\psi :{\mathcal{A}}^2\to [0,\mathrm{\infty })$ be control functions such that

(2.1)

(2.2)

Suppose that $f:\mathcal{A}\to \mathcal{A}$ is a mapping with $f(0)=0$ satisfying the followings:

$$\underset{t\to \mathrm{\infty }}{lim}N(f(\lambda a+b+cd)-\lambda f(a)-f(b)-f(c)d-cf(d),t\phi (a,b,c,d))=1$$

(2.3)

uniformly on ${\mathcal{A}}^4$ and for all $\lambda \in \mathbb{T}:=\{\lambda \in \mathbb{C}:|\lambda |=1\}$

$$\underset{t\to \mathrm{\infty }}{lim}N(f{(a)}^{\ast }-f\left(a^{\ast }\right),t\psi (a,a^{\ast }))=1$$

(2.4)

uniformly on ${\mathcal{A}}^2$. Then there exists a unique ∗-derivation δ on $\mathcal{A}$ satisfying

$$\underset{t\to \mathrm{\infty }}{lim}N(f(a)-\delta (a),t\tilde{\phi }(a,a,0,0))=1$$

(2.5)

for all $a\in \mathcal{A}$.

Proof Let $0<ϵ<1$ be given. Setting $a=b$, $c=d=0$ and $\lambda =1$ in (2.3), we can find some $t_0>0$ such that

$$N(f(2a)-2f(a),t\phi (a,a,0,0))\ge 1-ϵ$$

for all $a\in \mathcal{A}$ and $t\ge t_0$. One can use induction to show that

$$N(f\left(2^{n}a\right)-2^{n}f(a),t\sum _{k=0}^{n-1}{2}^{n-k-1}\phi (2^{k}a,2^{k}a,0,0))\ge 1-ϵ.$$

(2.6)

Let $t=t_0$ and put $n=p$ then by replacing a with $2^{n}a$ in (2.6), we obtain

$$N(\frac{f(2^{n+p}a)}{2^{n+p}}-\frac{f(2^{n}a)}{2^n},\frac{t_0}{2^{n+p}}\sum _{k=0}^{p-1}{2}^{p-k-1}\phi (2^{n+k}a,2^{n+k}a,0,0))\ge 1-ϵ$$

(2.7)

for all integers $n\ge 0$, $p\ge 0$. By the convergence of (2.1) there is $n_0\in \mathbb{N}$ such that

$$\frac{t_0}{2}\sum _{k=n}^{n+p-1}{2}^{-k}\phi (2^{k}a,2^{k}a,0,0)\le \delta$$

for all $n\ge n_0$ and $p>0$. Since the fuzzy norm $N(x,\cdot )$ is nondecreasing, we can have

(2.8)

It follows from (2.8) and Definition 1.3 that the sequence $\{\frac{f(2^{n}a)}{2^n}\}$ is Cauchy. Due to the completeness of $\mathcal{A}$, this sequence is convergent. Define

$$\delta (a):=N-\underset{n\to \mathrm{\infty }}{lim}\frac{f(2^{n}a)}{2^n}$$

(2.9)

for all $a\in \mathcal{A}$. From the above equation, we have

$$\delta \left(\frac{1}{2^k}a\right)=N-\underset{n\to \mathrm{\infty }}{lim}\frac{1}{2^k}\frac{f(2^{n-k}a)}{2^{n-k}}=\frac{1}{2^k}\delta (a)$$

(2.10)

for each $k\in \mathbb{N}$. Moreover, letting $n=0$ and passing the limit $p\to \mathrm{\infty }$ in (2.8), we get

$$\underset{t\to \mathrm{\infty }}{lim}N(f(a)-\delta (a),t\tilde{\phi }(a,a,0,0))=1$$

(2.11)

for all $a\in \mathcal{A}$. Putting $c=d=0$ and replacing a and b by $2^{n}a$ and $2^{n}b$, respectively, in (2.3), there exists $t_0>0$ such that

$$N(2^{-n}f(2^n(\lambda a+b))-\lambda 2^{-n}f\left(2^{n}a\right)-2^{-n}f\left(2^{n}b\right),t{2}^{-n}\phi (2^{n}a,2^{n}b,0,0))\ge 1-ϵ$$

for all $t\ge t_0$. Let $a,b\in \mathcal{A}$. Temporarily fix $t>0$. Since ${lim}_{n\to \mathrm{\infty }}\frac{1}{2^n}t\phi (2^{n}a,2^{n}b,0,0)=0$, there exists $n_0>0$ such that

$$t\phi (2^{n}a,2^{n}a,0,0)\le \frac{2^{n}t}{4},$$

for all $n\ge n_0$. Hence, we have

$$\begin{array}{c}N(\delta (\lambda a+b)-\lambda \delta (a)-\delta (b),t)\hfill \\ \ge min\{N(\delta (\lambda a+b)-2^{-n}f(2^n(\lambda a+b)),\frac{t}{4}),N(\lambda \delta (a)-\lambda 2^{-n}f\left(2^{n}a\right),\frac{t}{4}),\hfill \\ N(\delta (b)-2^{-n}f\left(2^{n}b\right),\frac{4}{t}),N(f(2^n(\lambda a+b))-\lambda f\left(2^{n}a\right)-f\left(2^{n}b\right),\frac{2^{n}4}{t})\}\hfill \end{array}$$

for all $n\ge n_0$ and $t>0$. The first three terms on the second and third lines of the above inequality tend to 1 as $n\to \mathrm{\infty }$. Furthermore, the last term is greater than

$$N(f(2^n(\lambda a+b))-\lambda f\left(2^{n}a\right)-f\left(2^{n}b\right),t_0\phi (2^{n}a,2^{n}b,0,0)),$$

which is greater than or equal to $1-ϵ$. Therefore,

$$N(\delta (\lambda a+b)-\lambda \delta (a)-\delta (b),t)\ge 1-ϵ$$

for all $t>0$. It follows that $\delta (\lambda a+b)=\lambda \delta (a)+\delta (b)$ by ($N_2$) for all $a,b\in \mathcal{A}$ and all $\lambda \in \mathbb{T}$. Next, let $\lambda ={\lambda }_1+\text{i}{\lambda }_2\in \mathbb{C}$ where ${\lambda }_1,{\lambda }_2\in \mathbb{R}$. Let ${\gamma }_1={\lambda }_1-[{\lambda }_1]$ and ${\gamma }_2={\lambda }_2-[{\lambda }_2]$, where $[\lambda ]$ denotes the integer part of λ. Then $0\le {\gamma }_i<1$ ($1\le i\le 2$). One can represent ${\gamma }_i$ as ${\gamma }_i=\frac{{\lambda }_{i,1}+{\lambda }_{i,2}}{2}$ such that ${\lambda }_{i,j}\in \mathbb{T}$ ($1\le i$, $j\le 2$). From (2.10), we infer that

$$\begin{array}{rcl}\delta (\lambda x)& =& \delta ({\lambda }_{1}x)+\text{i}\delta ({\lambda }_{2}x)\\ =& ([{\lambda }_1]\delta (x)+\delta ({\gamma }_{1}x))+\text{ i}([{\lambda }_2]\delta (x)+\delta ({\gamma }_{2}x))\\ =& ([{\lambda }_1]\delta (x)+\frac{1}{2}\delta ({\lambda }_{1,1}x+{\lambda }_{1,2}x))+\text{ i}([{\lambda }_2]\delta (x)+\frac{1}{2}\delta ({\lambda }_{2,1}x+{\lambda }_{2,2}x))\\ =& ([{\lambda }_1]\delta (x)+\frac{1}{2}{\lambda }_{1,1}\delta (x)+\frac{1}{2}{\lambda }_{1,2}\delta (x))+\text{ i}([{\lambda }_2]\delta (x)+\frac{1}{2}{\lambda }_{2,1}\delta (x)+\frac{1}{2}{\lambda }_{2,2}\delta (x))\\ =& {\lambda }_1\delta (x)+\text{i}{\lambda }_2\delta (x)\\ =& \lambda \delta (x)\end{array}$$

for all $x\in \mathcal{A}$. Hence, δ is $\mathbb{C}$-linear. Putting $a=b=0$ and replacing c and d by $2^{n}c$ and $2^{n}d$, respectively, in (2.3), there exists $t_0>0$ such that

$$N(2^{-2n}f\left(2^{2n}cd\right)-2^{-2n}f\left(2^{n}c\right)\left(2^{n}d\right)-2^{-2n}\left(2^{n}c\right)f\left(2^{n}d\right),t{2}^{-2n}\phi (0,0,2^{n}c,2^{n}d))\ge 1-ϵ$$

for all $t\ge t_0$. Fix $t(>0)$ temporarily. By (2.1) there exists $n_0>0$ such that

$$t\phi (0,0,2^{n}c,2^{n}d)\le \frac{2^{2n}t}{4}$$

for all $n\ge n_0$ and $t>0$. We have

$$\begin{array}{c}N(\delta (cd)-\delta (c)d-c\delta (d),t)\hfill \\ \ge min\{N(\delta (cd)-2^{-2n}f\left(2^{2n}cd\right),\frac{t}{4}),N(\delta (c)d-2^{-2n}f\left(2^{n}c\right)\left(2^{n}d\right),\frac{t}{4}),\hfill \\ N(c\delta (d)-2^{-2n}\left(2^{n}c\right)f\left(2^{n}d\right),\frac{t}{4}),\hfill \\ N(f\left(2^{2n}cd\right)-f\left(2^{n}c\right)\left(2^{n}d\right)-\left(2^{n}c\right)f\left(2^{n}d\right),\frac{2^{2n}4}{t})\}\hfill \\ \ge min\{N(\delta (cd)-2^{-2n}f\left(2^{2n}cd\right),\frac{t}{4}),N(\delta (c)d-2^{-2n}f\left(2^{n}c\right)\left(2^{n}d\right),\frac{t}{4}),\hfill \\ N(c\delta (d)-2^{-2n}\left(2^{n}c\right)f\left(2^{n}d\right),\frac{t}{4}),\hfill \\ N(f\left(2^{2n}cd\right)-f\left(2^{n}c\right)\left(2^{n}d\right)-\left(2^{n}c\right)f\left(2^{n}d\right),t\phi (0,0,2^{n}c,2^{n}d))\}\hfill \end{array}$$

for all $n\ge n_0$ and $t>0$. From the above computation

$$\delta (cd)=\delta (c)d+c\delta (d)$$

(2.12)

for all $c,d\in \mathcal{A}$. So it is a derivation on $\mathcal{A}$. Moreover, it follows from (2.7) with $n=0$ and (2.9) that ${lim}_{t\to \mathrm{\infty }}N(\delta (a)-f(a),t\tilde{\phi }(a,a,0,0))=1$ for all $a\in \mathcal{A}$. It is well known that the additive mapping δ satisfying (2.5) is unique (see [3] or [20]). Replacing a and $a^{\ast }$ by $2^{n}a$ and $2^n{a}^{\ast }$, respectively, in (2.4) we can find $t_0>0$ such that

$$N(2^{-n}f{\left(2^{n}a\right)}^{\ast }-2^{-n}f\left(2^n{a}^{\ast }\right),t{2}^{-n}\psi (2^{n}a,2^n{a}^{\ast }))\ge 1-ϵ$$

for all $a\in \mathcal{A}$ and all $t>t_0$. Since ${lim}_{n\to \mathrm{\infty }}{2}^{-n}\psi (2^{n}a,2^n{a}^{\ast })=0$, there exists some $n_0>0$ such that $t\psi (2^{n}a,2^n{a}^{\ast })<\frac{t{2}^n}{2}$ for all $n\ge n_0$. Hence,

$$\begin{array}{c}N(\delta {(a)}^{\ast }-\delta \left(a^{\ast }\right),t)\hfill \\ \ge min\{N(\delta {(a)}^{\ast }-2^{-n}f{\left(2^{n}a\right)}^{\ast },\frac{t}{4}),N(\delta \left(a^{\ast }\right)-2^{-n}f\left(2^n{a}^{\ast }\right),\frac{t}{4}),\hfill \\ N(f{\left(2^{n}a\right)}^{\ast }-f\left(2^n{a}^{\ast }\right),\frac{2^{n}t}{2})\}.\hfill \end{array}$$

The first two terms on the right-hand side of the above inequality tend to 1 as $n\to \mathrm{\infty }$. Furthermore, the last term is greater than

$$N(f{\left(2^{n}a\right)}^{\ast }-f\left(2^n{a}^{\ast }\right),t\psi (2^{n}a,2^n{a}^{\ast })),$$

which is greater than or equal to $1-ϵ$. So, we have that $N(\delta {(a)}^{\ast }-\delta (a^{\ast }),t)>1-ϵ$ for all $t>0$. It follows from that $\delta (a^{\ast })=\delta {(a)}^{\ast }$ for all $a\in \mathcal{A}$. So, δ is a *-derivation on $\mathcal{A}$. □

Theorem 2.2 Suppose that $f:\mathcal{A}\to \mathcal{A}$ is a mapping with $f(0)=0$ for which there exist functions $\phi :{\mathcal{A}}^4\to [0,\mathrm{\infty })$ and $\psi :{\mathcal{A}}^2\to [0,\mathrm{\infty })$ such that

$$\begin{array}{c}\tilde{\phi }(a,b,c,d):=\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}{2}^n\phi (2^{-n}a,2^{-n}b,2^{-n}c,2^{-n}d)<\mathrm{\infty },\hfill \\ \underset{n\to \mathrm{\infty }}{lim}{2}^n\psi (2^{-n}a,2^{-n}b)=0,\hfill \\ \underset{t\to \mathrm{\infty }}{lim}N(f(\lambda a+b+cd)-\lambda f(a)-f(b)-f(c)d-cf(d),t\phi (a,b,c,d))=1,\hfill \\ \underset{t\to \mathrm{\infty }}{lim}N(f{(a)}^{\ast }-f{(a)}^{\ast },t\psi (a,a^{\ast }))=1\hfill \end{array}$$

for all $\lambda \in \mathbb{T}$ and all $a,b,c,d\in \mathcal{A}$. Then there exists a unique ∗-derivation δ on $\mathcal{A}$ satisfying

$$\underset{t\to \mathrm{\infty }}{lim}N(f(a)-\delta (a),t\tilde{\phi }(a,a,0,0))=1$$

for all $a\in \mathcal{A}$.

3 Stability of ∗-derivations associated to the Jensen equation

The stability of the Jensen equation has been studied first by Kominek and then by several other mathematicians: ([15]). In this section, we study the stability of ∗-derivation associated to the Jensen equation in a fuzzy Banach ∗-algebra $\mathcal{A}$.

Theorem 3.1 Let $\mathcal{A}$ be a fuzzy Banach ∗-algebra. Suppose that $f:\mathcal{A}\to \mathcal{A}$ is a mapping with $f(0)=0$ for which there exist functions $\phi :{\mathcal{A}}^2\to [0,\mathrm{\infty })$ and ${\psi }_i:{\mathcal{A}}^2\to [0,\mathrm{\infty })$ ($1\le i\le 2$) such that

(3.1)

(3.2)

(3.3)

(3.4)

for all $a,b\in \mathcal{A}$ and all $\lambda \in \mathbb{T}$. Then there exists a unique ∗-derivation δ on $\mathcal{A}$ satisfying

$$\underset{t\to \mathrm{\infty }}{lim}N(f(a)-\delta (a),\frac{t}{3}(\tilde{\phi }(a,-a)+\tilde{\phi }(-a,3a)))=1$$

(3.5)

for all $a\in \mathcal{A}$.

Proof Let $0<ϵ<1$ be given. Letting $\lambda =-1$ and $b=-a$ in (3.2), we can find some $t_0>0$ such that

$$N(f(a)+f(-a),t\phi (a,-a))\ge 1-ϵ$$

for all $a\in \mathcal{A}$ and $t\ge t_0$. Letting $\lambda =1$ and replacing a and b by −a and 3a, respectively, in (3.2), we get also $t_1\ge t_0$ such that

$$N(2f(a)-f(-a)-f(3a),t\phi (-a,3a))\ge 1-ϵ$$

for all $a\in \mathcal{A}$ and $t\ge t_1$. Thus,

(3.6)

for all $a\in \mathcal{A}$. Replace a by $3^{n}a$ in (3.6)

$$N(\frac{f(3^{n}a)}{3^n}-\frac{f(3^{n+1}a)}{3^{n+1}},\frac{t}{3^{n+1}}(\phi (3^{n}a,-3^{n}a)+\phi (-3^{n}a,3^{n+1}a)))\ge 1-ϵ.$$

Given $\delta >0$, there exists an integer $n_0>0$ such that

$$\frac{t}{3}\sum _{j=m}^{n-1}{3}^{-j}(\phi (3^{j}a,-3^{j}a)+\phi (-3^{j}a,3^{j+1}a))\le \delta$$

for all $n\ge m\ge n_0$.

So, we have

(3.7)

(3.8)

for all nonnegative integers n, m with $n\ge m\ge n_0$ and all $a\in \mathcal{A}$. It follows from Definition 1.3 that the sequence $\{\frac{1}{3^n}f(3^{n}a)\}$ is a Cauchy sequence for all $a\in \mathcal{A}$. Since $\mathcal{A}$ is complete, the sequence $\{\frac{1}{3^n}f(3^{n}a)\}$ is convergent. So, one can define the mapping $\delta :\mathcal{A}\to \mathcal{A}$ by

$$\delta (a)=N-\underset{n\to \mathrm{\infty }}{lim}\frac{1}{3^n}f\left(3^{n}a\right)$$

(3.9)

for all $a\in \mathcal{A}$. If we put $\lambda =1$ and replace a, b with $3^{n}a$, $3^{n}b$, respectively, in (3.2), we can find some $t_0>0$ such that

$$N(2f\left(3^n\frac{a+b}{2}\right)-f\left(3^{n}a\right)-f\left(3^{n}b\right),3^{-n}t\phi (3^{n}a,3^{n}b))\ge 1-ϵ$$

for all $t\ge t_0$. Fix $t>0$ temporarily. Since ${lim}_{n\to \mathrm{\infty }}{3}^{-n}\phi (3^{n}a,3^{n}b)=0$, there is some $n_0>0$ such that $t\phi (3^{n}a,3^{n}b)<\frac{3^{n}t}{4}$ for all $n\ge n_0$. Then we have

$$\begin{array}{c}N(2\delta \left(\frac{a+b}{2}\right)-\delta (a)-\delta (b),t)\hfill \\ \ge min\{N(2\delta \left(\frac{a+b}{2}\right)-\frac{1}{3^n}2f\left(3^n\frac{a+b}{2}\right),\frac{t}{4}),N(\delta (a)-\frac{f(3^{n}a)}{3^n},\frac{t}{4}),\hfill \\ N(\delta (b)-\frac{f(3^{n}b)}{3^n},\frac{t}{4}),N(2f\left(3^n\frac{a+b}{2}\right)-f\left(3^{n}a\right)-f\left(3^{n}b\right),\frac{3^{n}t}{4})\}\hfill \end{array}$$

for all $a,b\in \mathcal{A}$ and $t>0$. The first three terms on the second and third lines of the above inequality tend to 1 as $n\to \mathrm{\infty }$. Furthermore, the last term is greater than

$$N(2f\left(3^n\frac{a+b}{2}\right)-f\left(3^{n}a\right)-f\left(3^{n}b\right),t\phi (3^{n}a,3^{n}b)),$$

which is greater than or equal to $1-ϵ$.

So, we have

$$N(2\delta \left(\frac{a+b}{2}\right)-\delta (a)-\delta (b),t)\ge 1-ϵ$$

for all $t>0$. By the definition of fuzzy norm, we have

$$2\delta \left(\frac{a+b}{2}\right)=\delta (a)+\delta (b)$$

(3.10)

for all $a,b\in \mathcal{A}$. Since $f(0)=0$, we have $\delta (0)=0$. Putting $b=0$ in (3.10), we get $2\delta (\frac{a}{2})=\delta (a)$ for each $a\in \mathcal{A}$ and, therefore, $\delta (a)+\delta (b)=2\delta (\frac{a+b}{2})=\delta (a+b)$ for all $a,b\in \mathcal{A}$. Moreover, letting $m=0$ and passing the limit $n\to \mathrm{\infty }$ in (3.8), we get

$$N(f(a)-\delta (a),\frac{t}{3}(\tilde{\phi }(a,-a)+\tilde{\phi }(-a,3a)))\ge 1-ϵ$$

for all $a\in \mathcal{A}$. So, we have Eq. (3.5). It is known that such an additive mapping δ is unique. Let $\lambda \in \mathbb{T}$. Replacing both a and b in (3.2) by $3^{n}a$ and dividing the both sides of the obtained inequality by $3^n$, there exists some $t_0>0$ such that

$$N(3^{-n}f\left(\lambda 3^{n}a\right)-\lambda 3^{-n}f\left(3^{n}a\right),3^{-n}t\phi (3^{n}a,3^{n}a))\ge 1-ϵ$$

for all $a\in \mathcal{A}$ and all $t\ge t_0$. Fix $t>0$ temporarily. Since ${lim}_{n\to \mathrm{\infty }}{3}^{-n}\varphi (3^{n}a,3^{n}b)=0$, there exists $n_0>0$ such that $3^{-n}\varphi (3^{n}a,3^{n}b)\le \frac{t}{2}$ for all $n\ge n_0$.

If we consider the following inequality

$$\begin{array}{c}N(\delta (\lambda a)-\lambda \delta (a),t)\hfill \\ \ge min\{N(\delta (\lambda a)-3^{-n}f\left(\lambda 3^{n}a\right),\frac{t}{4}),N(\lambda \delta (a)-3^{-n}f\left(\lambda 3^{n}a\right),\frac{t}{4}),\hfill \\ N(3^{-n}f\left(\lambda 3^{n}a\right)-3^{-n}f\left(\lambda 3^{n}a\right),\frac{t}{2})\},\hfill \end{array}$$

then the first two terms on the second line of the above inequality tend to 1 as $n\to \mathrm{\infty }$ and the last term is greater than

$$N(3^{-n}f\left(\lambda 3^{n}a\right)-\lambda 3^{-n}f\left(3^{n}a\right),3^{-n}t\phi (3^{n}a,3^{n}a)),$$

which is greater than or equal to $1-ϵ$. So, we can get $\delta (\lambda a)=\lambda \delta (a)$ for all $\lambda \in \mathbb{C}$ by the similar discussion in the proof Theorem 2.1. Replacing both a and $a^{\ast }$ in (3.3) by $3^{n}a$ and $3^n{a}^{\ast }$, and then dividing the both sides of the obtained inequality by $3^n$, we find some $t_0>0$ such that

$$N(3^{-n}f{\left(3^{n}a\right)}^{\ast }-3^{-n}f\left(3^n{a}^{\ast }\right),t{3}^{-n}{\psi }_1(3^{n}a,3^n{a}^{\ast }))\ge 1-ϵ$$

for all $t\ge t_0$. Fix $t>0$ temporarily. Since ${lim}_{n\to \mathrm{\infty }}{3}^{-n}{\psi }_1(3^{n}a,3^n{a}^{\ast })=0$, there exists $n_0>0$ such that $3^{-n}t{\psi }_1(3^{n}a,3^n{a}^{\ast })\le \frac{t}{2}$ for all $n\ge n_0$. We consider the following inequality:

$$\begin{array}{c}N(\delta \left(a^{\ast }\right)-\delta {(a)}^{\ast },t)\hfill \\ \ge min\{N(\delta \left(a^{\ast }\right)-3^{-n}f\left(3^n{a}^{\ast }\right),\frac{t}{4}),N(\delta {(a)}^{\ast }-3^{-n}f{\left(3^{n}a\right)}^{\ast },\frac{t}{4}),\hfill \\ N(3^{-n}f\left(3^n{a}^{\ast }\right)-3^{-n}f{\left(3^{n}a\right)}^{\ast },\frac{t}{2})\}.\hfill \end{array}$$

Then we get $\delta (a^{\ast })=\delta {(a)}^{\ast }$ for all $a\in \mathcal{A}$. For the derivation property, replacing both a and b in (3.4) by $3^{n}a$ and $3^{n}b$, we can find some $t_0>0$ such that

$$N(\frac{f(3^{2n}ab)}{3^{2n}}-\frac{3^{n}af(3^{n}b)}{3^{2n}}-\frac{f(3^{n}a)(3^{n}b)}{3^{2n}},3^{-n}t{\psi }_2(3^{n}a,3^{n}b))\ge 1-ϵ$$

for all $t\ge t_0$. By (3.4), there exists $n_0\in \mathbf{N}$ such that $3^{-n}t{\psi }_2(3^{n}a,3^{n}b)\le \frac{t}{4}$ for all $n\ge n_0$ and $t>0$. We can get $\delta (ab)=\delta (a)b+a\delta (b)$ for all $a,b\in \mathcal{A}$ from the following computation:

$$\begin{array}{c}N(\delta (ab)-a\delta (b)-\delta (a)b,t)\hfill \\ \ge min\{N(\delta (ab)-\frac{f(3^{2n}ab)}{3^{2n}},\frac{t}{4}),N(a\delta (b)-\frac{3^{n}af(3^{n}b)}{3^{2n}},\frac{t}{4}),\hfill \\ N(\delta (a)b-\frac{f(3^{n}a)(3^{n}b)}{3^{2n}},\frac{t}{4}),N(\frac{f(3^{2n}ab)}{3^{2n}}-\frac{3^{n}af(3^{n}b)}{3^{2n}}-\frac{f(3^{n}a)(3^{n}b)}{3^{2n}},\frac{t}{4})\}.\hfill \end{array}$$

Hence, δ is the ∗-derivation on $\mathcal{A}$ that we want. □

4 Superstability of ∗-derivations

In this section, we prove the superstability of ∗-derivations on a fuzzy Banach ∗-algebras. More precisely, we introduce the concept of $(\psi ,\phi )$-approximate ∗-derivation and show that any $(\psi ,\phi )$-approximate ∗-derivation is just a ∗-derivation.

Definition 4.1 Suppose that $\mathcal{A}$ is a ∗-normed algebra and $s\in \{-1,1\}$. Let $\delta :\mathcal{A}\to \mathcal{A}$ be a mapping for which there exist a function $\phi :\mathcal{A}\to \mathcal{A}$, and functions ${\psi }_i:\mathcal{A}\times \mathcal{A}\to \mathbb{R}$ ($1\le i\le 3$) satisfying

$$\underset{n\to \mathrm{\infty }}{lim}{n}^{-s}{\psi }_i(n^{s}a,b)=\underset{n\to \mathrm{\infty }}{lim}{n}^{-s}{\psi }_i(a,n^{s}b)=0(a,b\in \mathcal{A})$$

(4.1)

such that

(4.2)

(4.3)

(4.4)

for all $a,b,c,d\in \mathcal{A}$. Then δ is called a $(\psi ,\phi )$-approximate ∗-derivation on $\mathcal{A}$.

Theorem 4.2 Let $\mathcal{A}$ be a fuzzy Banach ∗-algebra with approximate unit. Then any $(\psi ,\phi )$-approximate ∗-derivation δ on $\mathcal{A}$ is a ∗-derivation.

Proof We assume that (4.1) holds. An arbitrary $ϵ>0$ is given. Let $a,b\in \mathcal{A}$ and $\lambda \in \mathbb{C}$. For $n\in \mathbb{N}$ there exists $t_0>0$ by (4.2) such that

$$\begin{array}{c}N(n^{-s}(n^{s}b\delta (\lambda a)-\phi \left(n^{s}b\right)\lambda a),n^{-s}t{\psi }_1(n^{s}b,\lambda a))\ge 1-ϵ,\hfill \\ N(n^{-s}(\phi \left(n^{s}b\right)\lambda a-\lambda n^{s}b\delta (a)),n^{-s}t|\lambda |{\psi }_1(n^{s}b,a))\ge 1-ϵ\hfill \end{array}$$

for all $t\ge t_0$. Fix $t>0$ temporarily. Since ${lim}_{n\to \mathrm{\infty }}{n}^{-s}{\psi }_1(n^{s}a,b)={lim}_{n\to \mathrm{\infty }}{n}^{-s}{\psi }_1(a,n^{s}b)=0$, there exists $n_0>0$ such that $t{n}^{-s}{\psi }_1(n^{s}b,\lambda a)\le \frac{t}{2}$ and $n^{-s}t|\lambda |{\psi }_1(n^{s}b,a)\le \frac{t}{2}$ for all $n\ge n_0$ and $t>0$.

We have

$$\begin{array}{c}N(b(\delta (\lambda a)-\lambda \delta (a)),t)\hfill \\ =N(n^{-s}(n^{s}b\delta (\lambda a)-\phi \left(n^{s}b\right)\lambda a+\phi \left(n^{s}b\right)\lambda a-\lambda n^{s}b\delta (a)),t)\hfill \\ \ge min\{N(n^{-s}(n^{s}b\delta (\lambda a)-\phi \left(n^{s}b\right)\lambda a),\frac{t}{2}),N(n^{-s}(\phi \left(n^{s}b\right)\lambda a-\lambda n^{s}b\delta (a)),\frac{t}{2})\}.\hfill \end{array}$$

Since

$$N(n^{-s}(n^{s}b\delta (\lambda a)-\phi \left(n^{s}b\right)\lambda a),\frac{t}{2})\ge N(n^{-s}(n^{s}b\delta (\lambda a)-\phi \left(n^{s}b\right)\lambda a),t{n}^{-s}{\psi }_1(n^{s}a,b))$$

and

$$N(n^{-s}(\phi \left(n^{s}b\right)\lambda a-\lambda n^{s}b\delta (a)),\frac{t}{2})\ge N(n^{-s}(\phi \left(n^{s}b\right)\lambda a-\lambda n^{s}b\delta (a)),t{n}^{-s}|\lambda |{\psi }_1(n^{s}b,a)),$$

it leads us to have a conclusion that $N(b(\delta (\lambda a)-\lambda \delta (a)),t)\ge 1-ϵ$ for all $t>0$. Therefore, $b(\delta (\lambda a)-\lambda \delta (a))=0$ for all $b\in \mathcal{A}$ by ($N_2$). Let ${\{e_i\}}_{i\in I}$ be an approximate unit of $\mathcal{A}$. If we replace b with ${\{e_i\}}_{i\in I}$, then we have

$$e_i(\delta (\lambda a)-\lambda \delta (a))=0$$

for all $i\in I$. So we conclude that $\delta (\lambda a)=\lambda \delta (a)$ for all $a\in A$ and $\lambda \in \mathbb{C}$. Next, we are going to prove the additivity of δ. By (4.2), there exists $t_0>0$ such that

$$\begin{array}{c}N(n^{-s}(n^{s}c\delta (a+b)-\phi \left(n^{s}c\right)(a+b)),n^{-s}t{\psi }_1(n^{s}c,a+b))\ge 1-ϵ,\hfill \\ N(n^{-s}(n^{s}c\delta (a)-\phi \left(n^{s}c\right)a),n^{-s}t{\psi }_1(n^{s}c,a))\ge 1-ϵ,\hfill \end{array}$$

and

$$N(n^{-s}(n^{s}c\delta (b)-\phi \left(n^{s}c\right)b),n^{-s}t{\psi }_1(n^{s}c,b))\ge 1-ϵ$$

for all $t\ge t_0$. Fix $t>0$ temporarily. By (4.1), we can find $n_0>0$ such that $n^{-s}t{\psi }_1(n^{s}c,a+b)\le \frac{t}{3}$, $n^{-s}t{\psi }_1(n^{s}c,a)\le \frac{t}{3}$, and $n^{-s}t{\psi }_1(n^{s}c,b)\le \frac{t}{3}$ for all $n\ge n_0$.

For the additivity, we can have

$$\begin{array}{c}N(c(\delta (a+b)-\delta (a)-\delta (b)),t)\hfill \\ =N(n^{-s}(n^{s}c\delta (a+b)-\phi \left(n^{s}c\right)(a+b))\hfill \\ +n^{-s}(n^{s}c\delta (a)-\phi \left(n^{s}c\right)a)+n^{-s}(n^{s}c\delta (b)-\phi \left(n^{s}c\right)b),t)\hfill \\ \ge min\{N(n^{-s}(n^{s}c\delta (a+b)-\phi \left(n^{s}c\right)(a+b)),\frac{t}{3}),N(n^{-s}(n^{s}c\delta (a)-\phi \left(n^{s}c\right)a),\frac{t}{3}),\hfill \\ N(n^{-s}(n^{s}c\delta (b)-\phi \left(n^{s}c\right)b),\frac{t}{3})\}\hfill \\ \ge min\{N(n^{-s}(n^{s}c\delta (a+b)-\phi \left(n^{s}c\right)(a+b)),n^{-s}t{\psi }_1(n^{s}c,a+b)),\hfill \\ N(n^{-s}(n^{s}c\delta (a)-\phi \left(n^{s}c\right)a),n^{-s}t{\psi }_1(n^{s}c,a)),\hfill \\ N(n^{-s}(n^{s}c\delta (b)-\phi \left(n^{s}c\right)b),n^{-s}t{\psi }_1(n^{s}c,b))\}.\hfill \end{array}$$

Since all terms of the final inequality of the above inequality are larger than $1-ϵ$, we can have $N(c(\delta (a+b)-\delta (a)-\delta (b)),t)>1-ϵ$ for all $t>0$. We can get $c(\delta (a+b)-\delta (a)-\delta (b))=0$ for all $a,b,c\in \mathcal{A}$ by ($N_2$). By using the approximate unit of $\mathcal{A}$, we have that $\delta (a+b)=\delta (a)+\delta (b)$ for all $a,b\in \mathcal{A}$. Next, we are going to show the derivation property of δ. From (4.2) and (4.1), there exists $t_0>0$ such that

$$\begin{array}{c}N(n^{-s}(n^{s}z\delta (ab)-\phi \left(n^{s}z\right)(ab)),n^{-s}t{\psi }_1(n^{s}z,ab))\ge 1-ϵ,\hfill \\ N(n^{-s}(\phi \left(n^{s}z\right)ab-n^{s}z(\delta (a)b+a\delta (b))),n^{-s}t{\psi }_2(n^{s}z,ab))\ge 1-ϵ\hfill \end{array}$$

for all $t\ge t_0$. By (4.1), we can find $n_0>0$ such that $n^{-s}t{\psi }_1(n^{s}z,ab)\le \frac{t}{2}$ and $n^{-s}t{\psi }_2(n^{s}z,ab)\le \frac{t}{2}$ for all $n\ge n_0$. The following computation

$$\begin{array}{c}N(z(\delta (ab)-\delta (a)b-a\delta (b)),t)\hfill \\ \ge min\{N(n^{-s}(n^{s}z\delta (ab)-\phi \left(n^{s}z\right)(ab)),\frac{t}{2}),\hfill \\ N(n^{-s}(\phi \left(n^{s}z\right)ab-n^{s}z(\delta (a)b+a\delta (b))),\frac{t}{2})\}\hfill \\ \ge min\{N(n^{-s}(n^{s}z\delta (ab)-\phi \left(n^{s}z\right)(ab)),n^{-s}t{\psi }_1(n^{s}z,ab)),\hfill \\ N(n^{-s}(\phi \left(n^{s}z\right)ab-n^{s}z(\delta (a)b+a\delta (b))),n^{-s}t{\psi }_2(n^{s}z,ab))\}\ge 1-ϵ\hfill \end{array}$$

yields that $\delta (ab)=\delta (a)b+a\delta (b)$ for all $a,b\in \mathcal{A}$. By (4.2) and (4.4) there exists $t_0>0$ such that

$$\begin{array}{c}N(n^{-s}(n^{s}z\delta \left(a^{\ast }\right)-\phi \left(n^{s}z\right)a^{\ast }),n^{-s}t{\psi }_1(n^{s}z,a^{\ast }))\ge 1-ϵ,\hfill \\ N(n^{-s}(\phi \left(n^{s}z\right)a^{\ast }-n^{s}z\delta {(a)}^{\ast }),n^{-s}t{\psi }_3(n^{s}z,a))\ge 1-ϵ\hfill \end{array}$$

for all $t\ge t_0$. For fixing $t>0$ temporarily, there exists $n_0>0$ such that $n^{-s}t{\psi }_1(n^{-s}z,a^{\ast })\le \frac{t}{2}$ and $n^{-s}t{\psi }_3(n^{s}z,a)\le \frac{t}{2}$ for $n\ge n_0$. From the following computation

$$\begin{array}{c}N(z(\delta \left(a^{\ast }\right)-\delta {(a)}^{\ast }),t)\hfill \\ =N(n^{-s}(n^{s}z\delta \left(a^{\ast }\right)-\phi \left(n^{s}z\right)a^{\ast })+n^{-s}(\phi \left(n^{s}z\right)a^{\ast }-n^{s}z\delta {(a)}^{\ast }),t)\hfill \\ \ge min\{N(n^{-s}(n^{s}z\delta \left(a^{\ast }\right)-\phi \left(n^{s}z\right)a^{\ast }),\frac{t}{2}),N(n^{-s}(\phi \left(n^{s}z\right)a^{\ast }-n^{s}z\delta {(a)}^{\ast }),\frac{t}{2})\}\hfill \\ \ge min\{N(n^{-s}(n^{s}z\delta \left(a^{\ast }\right)-\phi \left(n^{s}z\right)a^{\ast }),n^{-s}t{\psi }_1(n^{-s}z,a^{\ast })),\hfill \\ N(n^{-s}(\phi \left(n^{s}z\right)a^{\ast }-n^{s}z\delta {(a)}^{\ast }),n^{-s}t{\psi }_3(n^{s}z,a))\}>1-ϵ\hfill \end{array}$$

we can have $N(z(\delta (a^{\ast })-\delta {(a)}^{\ast }),t)>1-ϵ$ for all $t>0$. By ($N_2$) and using approximate unit $\delta (a^{\ast })=\delta {(a)}^{\ast }$ for all $a\in \mathcal{A}$. Thus, δ is a ∗-derivation on $\mathcal{A}$. □