Approximate *-derivations and approximate quadratic *-derivations on C*-algebras

Abstract

In this paper, we prove the stability of *-derivations and of quadratic *-derivations on Banach *-algebras. We moreover prove the superstability of *-derivations and of quadratic *-derivations on C*-algebras.

2000 Mathematics Subject Classification: 39B52; 47B47; 46L05; 39B72.

1 Introduction and preliminaries

Suppose that $\mathcal{A}$ is a complex Banach *-algebra. A ℂ-linear mapping $\delta :D\left(\delta \right)\to \mathcal{A}$ is said to be a derivation on $\mathcal{A}$ if δ(ab) = δ(a)+ b + aδ(b) for all $a,b\in \mathcal{A}$, where D(δ) is a domain of δ and D(δ) is dense in $\mathcal{A}$. If δ satisfies the additional condition δ(a*) = δ(a)* 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*-algebra and D(δ) is $\mathcal{A}$, then the derivation δ is bounded.

A C*-dynamical system is a triple ($\mathcal{A}$, G, α) consisting of a C*-algebra $\mathcal{A}$, a locally compact group G, and a pointwise norm continuous homomorphism α of G into the group Aut($\mathcal{A}$) of *-automorphisms of $\mathcal{A}$. Every bounded *-derivation δ arises as an infinitesimal generator of a dynamical system for ℝ. In fact, if δ is a bounded *-derivation of $\mathcal{A}$ on a Hilbert space $\mathcal{H}$, then there exists an element h in the enveloping von Neumann algebra ${\mathcal{A}}^{{}^{″}}$ such that

$$\delta \left(x\right)=a{d}_{ih}\left(x\right)$$

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

If, for each t ∈ ℝ, α _t is defined by α _t (x) = eⁱ th xe^-ith for all $x\in \mathcal{A}$, then α _t is a *-automorphism of $\mathcal{A}$ induced by unitaries U _t = eⁱ th for each t ∈ ℝ. The action $\alpha :ℝ\to Aut\left(\mathcal{A}\right)$, t → α _t , is a strongly continuous one-parameter group of *-automorphisms of $\mathcal{A}$. For several reasons, the theory of bounded derivations of C*-algebras is important in the quantumn mechanics (see [1–3]).

A functional equation is called stable if any function satisfying the 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 (see [4]).

In 1940, Ulam [5] proposed the following question concerning stability of group homomorphisms: under what condition does there exist an additive mapping near an approximately additive mapping? Hyers [6] answered the problem of Ulam for the case where G₁ and G₂ are Banach spaces. A generalized version of the theorem of Hyers for an approximately linear mapping was given by Rassias [7]. Since then, the stability problems of various functional equations have been extensively investigated by a number of authors (see [8–19]). In particular, those of the important functional equations are the following functional equations

$$f\left(x+y\right)=f\left(x\right)+f\left(y\right),$$

(1.1)

$$2f\left(\frac{x+y}{2}\right)=f\left(x\right)+f\left(y\right),$$

(1.2)

which are called the Cauchy functional equation and the Jensen functional equation, respectively. The function f(x) = bx is a solution of these functional equations. Every solution of the functional equations (1.1) and (1.2) is said to be an additive mapping.

In this paper, we introduce functional equations of *-derivations and of quadratic *-derivations. we prove the stability of *-derivations associated with the Cauchy functional equation and the Jensen functional equation and of quadratic *-derivations on Banach *-algebra. We moreover prove the superstability of *-derivations and of quadratic *-derivations on C*-algebras.

2 Stability of *-derivations on Banach *-algebras

In this section, let $\mathcal{A}$ be a Banach *-algebra. We prove the stability of *-derivations on $\mathcal{A}$.

Theorem 2.1 Suppose that$f:\mathcal{A}\to \mathcal{A}$is a mapping with f(0) = 0 for which there exists a function$\phi :{\mathcal{A}}^4\to \left[0,\infty \right)$such that

$$\tilde{\phi }\left(a,b,c,d\right):=\sum _{n=0}^{\infty }\frac{1}{2^{n+1}}\phi \left(2^{n}a,2^{n}b,2^{n}c,2^{n}d\right)<\infty ,$$

(2.1)

$$\parallel f\left(\lambda a+b+cd\right)-\lambda f\left(a\right)-f\left(b\right)-f\left(c\right)d-cf\left(d\right)\parallel \le \phi \left(a,b,c,d\right),$$

(2.2)

$$\parallel f\left(a^{*}\right)-f{\left(a\right)}^{*}\parallel \le \phi \left(a,a,a,a\right)$$

(2.3)

for all $\lambda \in T:=\left\{\lambda \in ℂ:\left|\lambda \right|=1\right\}$ and all $a,b,c,d\in \mathcal{A}$. Then there exists a unique *-derivation δ on $\mathcal{A}$ satisfying

$$\parallel f\left(a\right)-\delta \left(a\right)\parallel \le \tilde{\phi }\left(a,a,0,0\right),$$

(2.4)

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

Proof. Setting a = b, c = d = 0 and λ = 1 in (2.2), we have

$$\parallel f\left(2a\right)-2f\left(a\right)\parallel \le \phi \left(a,a,0,0\right)$$

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

$$\Vert \frac{f{(2}^{n}a)}{2^n}-\frac{f{(2}^{m}a)}{2^m}‱={\displaystyle \sum _{k=m}^{n-1}}\frac{1}{2^{k+1}}\mu {(2}^{k}a{\mathrm{,2}}^{k}a\mathrm{,0,0})$$

(2.5)

for all n > m ≥ 0 and all $a\in \mathcal{A}$. It follows from (2.5) and (2.1) that the sequence $\left\{\frac{f{(2}^{n}a)}{2^n}\right\}$ is Cauchy. Due to the completeness of $\mathcal{A}$, this sequence is convergent. Define

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

(2.6)

for all $a\in \mathcal{A}$. Then, we have

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

(2.7)

for each k ∈ ℕ. Putting c = d = 0 and replacing a and b by 2 ⁿa and 2 ⁿb, respectively, in (2.2), we get

$$\Vert \frac{1}{2^n}f{(2}^n(\lambda a+b))-\lambda \frac{1}{2^n}f{(2}^{n}a)-\frac{1}{2^n}f{(2}^{n}b)‱\le \frac{1}{2^n}\mu {(2}^{n}a{\mathrm{,2}}^{n}b\mathrm{,0,0}).$$

Taking the limit as n → ∞, we obtain

$$\delta \left(\lambda a+b\right)=\lambda \delta \left(a\right)+\delta \left(b\right)$$

(2.8)

for all $a,b\in \mathcal{A}$ and all $\lambda \in T$. Putting a = b = 0 and replacing c and d by 2 ⁿc and 2 ⁿd, respectively, in (2.2), we get

$$\begin{array}{l}\Vert \frac{1}{2^{2n}}f{(2}^{2n}cd)-\frac{1}{2^{2n}}f{(2}^{n}c{)(2}^{n}d)-\frac{1}{2^{2n}}{(2}^{n}c)f{(2}^{n}d)‱\\ \le \frac{1}{2^{2n}}\mu ({\mathrm{0,0,2}}^{n}c{\mathrm{,2}}^{n}d)\le \frac{1}{2^n}\mu ({\mathrm{0,0,2}}^{n}c{\mathrm{,2}}^{n}d).\end{array}$$

Taking the limit as n → ∞, we obtain

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

(2.9)

for all $c,d\in \mathcal{A}$.

Next, let λ = λ₁ +iλ₂ ∈ ℂ where λ₁, λ₂, ∈ ℝ. Let γ₁ = λ₁ - [λ₁] and γ₂ = λ₂ - [λ₂], where [λ] denotes the integer part of λ. Then, 0 ≤ γ₁ < 1(1 ≤ i ≤ 2). One can represent γ _i as ${\gamma }_i=\frac{{\lambda }_{i,1}+{\lambda }_{i,2}}{2}$ such that ${\lambda }_{i,j}\in T\left(1\le i,j\le 2\right)$. From (2.7) and (2.8), it follows that

$$\begin{array}{lll}\hfill \delta \left(\lambda a\right)& =\delta \left({\lambda }_{1}a\right)+i\delta \left({\lambda }_{2}a\right)& \hfill \text{(1)}\\ =\left(\left[{\lambda }_1\right]\delta \left(a\right)+\delta \left({\gamma }_{1}a\right)\right)+i\left(\left[{\lambda }_2\right]\delta \left(a\right)+\delta \left({\gamma }_{2}a\right)\right)& \hfill \text{(2)}\\ =\left(\left[{\lambda }_1\right]\delta \left(a\right)+\frac{1}{2}\delta \left({\lambda }_{1,1}a+{\lambda }_{1,2}a\right)\right)+i\left(\left[{\lambda }_2\right]\delta \left(a\right)+\frac{1}{2}\delta \left({\lambda }_{2,1}a+{\lambda }_{2,2}a\right)\right)& \hfill \text{(3)}\\ =\left(\left[{\lambda }_1\right]\delta \left(a\right)+\frac{1}{2}{\lambda }_{1,1}\delta \left(a\right)+\frac{1}{2}{\lambda }_{1,2}\delta \left(a\right)\right)+i\left(\left[{\lambda }_2\right]\delta \left(a\right)+\frac{1}{2}{\lambda }_{2,1}\delta \left(a\right)+\frac{1}{2}{\lambda }_{2,2}\delta \left(a\right)\right)& \hfill \text{(4)}\\ ={\lambda }_1\delta \left(a\right)+i{\lambda }_2\delta \left(a\right)=\lambda \delta \left(a\right)& \hfill \text{(5)}\\ \hfill \text{(6) }\end{array}$$

for all $a\in \mathcal{A}$. Hence, δ is ℂ-linear, and so it is a derivation on $\mathcal{A}$. Moreover, it follows from (2.5) with m = 0 and (2.6) that $\parallel \delta \left(a\right)-f\left(a\right)\parallel \le \tilde{\phi }\left(a,a,0,0\right)$ for all $a\in \mathcal{A}$. It is well known that the additive mapping δ satisfying (2.4) is unique (see [3] or [19]). Replacing a and a* by 2 ⁿa and 2 ⁿa*, respectively, in (2.3), we get

$$\Vert \frac{1}{2^n}f{(2}^n{a}^{*})-\frac{1}{2^n}f{{(2}^{n}a)}^{*}‱\le \frac{1}{2^n}\mu {(2}^{n}a{\mathrm{,2}}^{n}a{\mathrm{,2}}^{n}a{\mathrm{,2}}^{n}a).$$

Passing to the limit as n → ∞, we get the δ(a*) = δ(a)* for all $a\in \mathcal{A}$. So δ is a *-derivation on $\mathcal{A}$, as desired. □

Corollary 2.2 Let ε, p be positive real numbers with p < 1. Suppose that$f:\mathcal{A}\to \mathcal{A}$is a mapping satisfying

$$\parallel f\left(\lambda a+b+cd\right)-\lambda f\left(a\right)-f\left(b\right)-cf\left(d\right)-f\left(c\right)d\parallel \le \epsilon \left(\parallel a{\parallel }^p+\parallel b{\parallel }^p+\parallel c{\parallel }^p+\parallel d{\parallel }^p\right),$$

(2.10)

$$\parallel f\left(a^{*}\right)-f{\left(a\right)}^{*}\parallel \le 4\epsilon \parallel a{\parallel }^p$$

(2.11)

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

$$\parallel f\left(a\right)-\delta \left(a\right)\parallel \le \frac{2\epsilon }{2-2^p}\parallel a{\parallel }^p$$

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

Proof. Putting φ(a, b, c, d) = ε(||a|| ^p + ||b|| ^p + ||c|| ^p + ||d| ^p ) in Theorem 2.1, we get the desired result. □

Similarly, we can obtain the following. We will omit the proof.

Theorem 2.3 Suppose that$f:\mathcal{A}\to \mathcal{A}$is a mapping with f (0) = 0 for which there exists a function$\phi :{\mathcal{A}}^4\to \left[0,\infty \right)$satisfying (2.2), (2.3) and

$$\sum _{n=1}^{\infty }{2}^{2n-1}\phi \left(\frac{a}{2^n},\frac{b}{2^n},\frac{c}{2^n},\frac{d}{2^n}\right)<\infty$$

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

$$\parallel f\left(a\right)-\delta \left(a\right)\parallel \le \tilde{\phi }\left(a,a,0,0\right),$$

for all$a\in \mathcal{A}$, where

$$\tilde{\phi }\left(a,b,c,d\right):=\sum _{n=1}^{\infty }{2}^{n-1}\phi \left(\frac{a}{2^n},\frac{b}{2^n},\frac{c}{2^n},\frac{d}{2^n}\right).$$

Corollary 2.4 Let ε, p be positive real numbers with p > 2. Suppose that$f:\mathcal{A}\to \mathcal{A}$is a mapping satisfying (2.10) and (2.11). Then there exists a unique *-derivation δ on$\mathcal{A}$satisfying

$$\parallel f\left(a\right)-\delta \left(a\right)\parallel \le \frac{2\epsilon }{2^p-2}\parallel a{\parallel }^p$$

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

Proof. Putting φ(a, b, c, d) = ε(||a|| ^p + ||b|| ^p + ||c|| ^p + ||d| ^p ) in Theorem 2.3, we get the desired result. □

3 Stability of *-derivations associated with the Jensen functional equation

The stability of the Jensen functional equation has been studied first by Kominek and then by several other mathematicians (see [11, 20]).

In this section, we study the stability of *-derivation associated with the Jensen functional equation in a Banach *-algebra $\mathcal{A}$.

Theorem 3.1 Let$\mathcal{A}$be a Banach *-algebra. Suppose that$f:\mathcal{A}\to \mathcal{A}$is a mapping with f (0) = 0 for which there exists a function$\phi :\mathcal{A}\times \mathcal{A}\to \left[0,\infty \right)$such that

$$\tilde{\phi }\left(a,b\right):=\sum _{n=0}^{\infty }\frac{1}{3^n}\phi \left(3^{n}a,3^{n}b\right)<\infty ,$$

(3.1)

$$∥2f\left(\frac{\lambda a+\lambda b}{2}\right)-\lambda f\left(a\right)-\lambda f\left(b\right)∥\le \phi \left(a,b\right),$$

(3.2)

$$\parallel f\left(a^{*}\right)-f{\left(a\right)}^{*}\parallel \le \phi \left(a,a\right),$$

(3.3)

$$\parallel f\left(ab\right)-af\left(b\right)-f\left(a\right)b\parallel \le \phi \left(a,b\right)$$

(3.4)

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

$$\parallel f\left(a\right)-\delta \left(a\right)\parallel \le \frac{1}{3}\left(\tilde{\phi }\left(a,-a\right)+\tilde{\phi }\left(-a,3a\right)\right)$$

(3.5)

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

Proof. Letting λ = 1 and b = -a in (3.2), we get

$$\parallel -f\left(a\right)-f\left(-a\right)\parallel \le \phi \left(a,-a\right)$$

for all $a\in \mathcal{A}$. Letting λ = 1 and replacing a and b by -a and 3a, respectively, in (3.2), we get

$$\parallel 2f\left(a\right)-f\left(-a\right)-f\left(3a\right)\parallel \le \phi \left(-a,3a\right)$$

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

$$\begin{array}{lll}\hfill ∥f\left(a\right)-\frac{1}{3}f\left(3a\right)∥& \le \frac{1}{3}\left(\parallel f\left(a\right)+f\left(-a\right)\parallel +\parallel 2f\left(a\right)-f\left(-a\right)-f\left(3a\right)\parallel \right)& \hfill \text{(1)}\\ \le \frac{1}{3}\left(\phi \left(a,-a\right)+\phi \left(-a,3a\right)\right)& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$$

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

$$\begin{array}{c}\Vert \frac{1}{3^n}f{(3}^{n}a)-\frac{1}{3^m}f{(3}^{m}a)‱\le {\displaystyle \sum _{j=m}^{n-1}}\Vert \frac{1}{3^j}f{(3}^{j}a)-\frac{1}{3^{j+1}}f{(3}^{j+1}a)‱\\ \le \frac{1}{3}{\displaystyle \sum _{j=m}^{n-1}}\frac{1}{3^j}\left(\mu {(3}^{j}a\mathrm{,}-3^{j}a)+\mu (-3^{j}a{\mathrm{,3}}^{j+1}a)\right)\end{array}$$

(3.6)

for all nonnegative integers n, m with n > m and all $a\in \mathcal{A}$. It follows from (3.6) that the sequence $\left\{\frac{1}{3^n}f\left(3^{n}a\right)\right\}$ is a Cauchy sequence for all $a\in \mathcal{A}$. Since $\mathcal{A}$ is complete, the sequence $\left\{\frac{1}{3^n}f\left(3^{n}a\right)\right\}$ is convergent. So one can define the mapping $\delta :\mathcal{A}\to \mathcal{A}$ by

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

for all $a\in \mathcal{A}$. By (3.2),

$$\begin{array}{c}\Vert 2\delta \left(\frac{a+b}{2}\right)-\delta (a)-\delta (b)‱=\underset{n\to \infty }{{\displaystyle \lim }}\frac{1}{3^n}\Vert 2f\left(3^n\frac{a+b}{2}\right)-f{(3}^{n}a)-f{(3}^{n}b)‱\\ \le \underset{n\to \infty }{{\displaystyle \lim }}\frac{1}{3^n}\mu (3^{n}a{\mathrm{,3}}^{n}b)=0\end{array}$$

for all $a,b\in \mathcal{A}$. Thus

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

(3.7)

for all $a,b\in \mathcal{A}$. Since f(0) = 0, we have δ(0) = 0. Putting b = 0 in (3.7), we get $2\delta \left(\frac{a}{2}\right)=\delta \left(a\right)$ for all $a\in \mathcal{A}$ and therefore $\delta \left(a\right)+\delta \left(b\right)=2\delta \left(\frac{a+b}{2}\right)=\delta \left(a+b\right)$ for all $a,b\in \mathcal{A}$. Moreover, letting m = 0 and passing the limit n → ∞ in (3.6), we get (3.5).

Replacing both a and b in (3.2) by 3 ⁿa and then dividing both sides of the obtained inequality by 3 ⁿ , we get

$$\Vert \frac{1}{3^n}f(\lambda 3^{n}a)-\frac{\lambda }{3^n}f{(3}^{n}a)‱\le \frac{1}{3^n}\mu {(3}^{n}a{\mathrm{,3}}^{n}a).$$

Passing the limit as n → ∞, we get δ(λa) = λδ(a) for all $\lambda \in T$. Thus we can get δ(λa) = λδ(a) for all λ ∈ ℂ by the similar discussion in the proof of Theorem 2.1.

Replacing a in (3.3) by 3 ⁿa and then dividing the both sides of the obtained inequality by 3ⁿ, we get

$$\Vert \frac{1}{3^n}f{(3}^n{a}^{*})-\frac{1}{3^n}f{{(3}^{n}a)}^{*}‱\le \frac{1}{3^n}\mu {(3}^{n}a{\mathrm{,3}}^{n}a).$$

Passing the limit as n tends to infinity, we get δ(a*) = δ(a)*.

Similarly, replacing a and b in (3.4) by 3 ⁿa and 3 ⁿb, respectively, we get

$$\Vert \frac{f{(3}^{2n}ab)}{3^{2n}}-\frac{3^{n}af{(3}^{n}b)}{3^{2n}}-\frac{f{(3}^{n}a{)(3}^{n}b)}{3^{2n}}‱\le \frac{1}{3^{2n}}\mu {(3}^{n}a{\mathrm{,3}}^{n}b)\le \frac{1}{3^n}\mu {(3}^{n}a{\mathrm{,3}}^{n}b),$$

which tends to zero, as n tends to ∞. So we get δ(ab) = δ(a)d + aδ(b) for all $a,b\in \mathcal{A}$. Hence, δ is a *-derivation on $\mathcal{A}$.

Corollary 3.2 Let ε, p be positive real numbers with p < 1. Suppose that$f:\mathcal{A}\to \mathcal{A}$is a mapping satisfying

$$∥2f\left(\frac{\lambda a+\lambda b}{2}\right)-\lambda f\left(a\right)-\lambda f\left(b\right)∥\le \epsilon \left(\parallel a{\parallel }^p+\parallel b{\parallel }^p\right),$$

(3.8)

$$\parallel f\left(a^{*}\right)-f{\left(a\right)}^{*}\parallel \le 2\epsilon \parallel a{\parallel }^p,$$

(3.9)

$$\parallel f\left(ab\right)-af\left(b\right)-f\left(a\right)b\parallel \le \epsilon \left(\parallel a{\parallel }^p+\parallel b{\parallel }^p\right)$$

(3.10)

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

$$\parallel f\left(a\right)-\delta \left(a\right)\parallel \le \frac{3+3^p}{3-3^p}\epsilon \parallel a{\parallel }^p$$

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

Proof. Putting φ(a, b) = ε(||a||^p+ ||b||^p) in Theorem 3.1, we get the desired result. □

Similarly, we can obtain the following. We will omit the proof.

Theorem 3.3 Let$\mathcal{A}$be a Banach *-algebra. Suppose that$f:\mathcal{A}\to \mathcal{A}$is a mapping with f(0) = 0 for which there exists a function$\parallel f\left(a\right)-\delta \left(a\right)\parallel \le \frac{2\epsilon }{2^p-2}\parallel a{\parallel }^p$satisfying (3.2), (3.3), (3.4) and

$$\sum _{n=1}^{\infty }{3}^{2n}\phi \left(\frac{a}{3^n},\frac{b}{3^n}\right)<\infty$$

for all$a,b\in \mathcal{A}$. Then there exists a unique *-derivation δ on$\mathcal{A}$satisfying

$$\parallel f\left(a\right)-\delta \left(a\right)\parallel \le \frac{1}{3}\left(\tilde{\phi }\left(a,-a\right)+\tilde{\phi }\left(-a,3a\right)\right)$$

for all$a\in \mathcal{A}$, where

$$\tilde{\phi }\left(a,b\right):=\sum _{n=1}^{\infty }{3}^n\phi \left(\frac{a}{3^n},\frac{b}{3^n}\right).$$

Corollary 3.4 Let ε, p be positive real numbers with p > 2. Suppose that$f:\mathcal{A}\to \mathcal{A}$is a mapping satisfying (3.8), (3.9) and (3.10). Then there exists a unique *-derivation δ on$\mathcal{A}$satisfying

$$\parallel f\left(a\right)-\delta \left(a\right)\parallel \le \frac{3^p+3}{3^p-3}\epsilon \parallel a{\parallel }^p$$

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

Proof. Putting φ(a, b) = ε(||a||^p+ ||b||^p) in Theorem 3.3, we get the desired result. □

4 Stability of quadratic *-derivations on Banach *-algebras

In this section, we prove the stability of quadratic *-derivations on a Banach *-algebra $\mathcal{A}$.

Definition 4.1 Let$\mathcal{A}$be a *-normed algebra. A mapping$\delta :\mathcal{A}\to \mathcal{A}$is a quadratic *-derivation on$\mathcal{A}$if δ satisfies the following properties:

(1) δ is a quadratic mapping,

(2) δ is quadratic homogeneous, that is, δ(λa) = λ²δ(a) for all$a\in \mathcal{A}$and all λ ∈ ℂ,

(3) δ(a b) = δ(a)b² + a²δ(b) for all$a,b\in \mathcal{A}$,

(4) δ(a*) = δ(a)* for all$a\in \mathcal{A}$.

Theorem 4.2 Suppose that$f:\mathcal{A}\to \mathcal{A}$is a mapping with f(0) = 0 for which there exists a function$\phi :{\mathcal{A}}^4\to \left[0,\infty \right)$such that

$$\tilde{\phi }\left(a,b,c,d\right):=\sum _{k=0}^{\infty }\frac{1}{4^k}\phi \left(2^{k}a,2^{k}b,2^{k}c,2^{k}d\right)<\infty ,$$

$$\begin{array}{c}\parallel f\left(\lambda a+\lambda b+cd\right)+f\left(\lambda a-\lambda b+cd\right)-2{\lambda }^{2}f\left(a\right)-2{\lambda }^{2}f\left(b\right)-2f\left(c\right)d^2-2c^{2}f\left(d\right)\parallel \\ \le \phi \left(a,b,c,d\right),\end{array}$$

(4.1)

$$\parallel f\left(a^{*}\right)-f{\left(a\right)}^{*}\parallel \le \phi \left(a,a,a,a\right)$$

(4.2)

for all$a,b,c,d\in \mathcal{A}$and all$\lambda \in T$. Also, if for each fixed$a\in \mathcal{A}$the mapping t → f(ta) from ℝ to$\mathcal{A}$is continuous, then there exists a unique quadratic *-derivation δ on$\mathcal{A}$satisfying

$$\parallel f\left(a\right)-\delta \left(a\right)\parallel \le \frac{1}{4}\tilde{\phi }\left(a,a,0,0\right)$$

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

Proof. Putting a = b, c = d = 0, , and λ = 1 in (4.1), we have

$$\parallel f\left(2a\right)-4f\left(a\right)\parallel \le \phi \left(a,a,0,0\right)$$

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

$$\Vert \frac{f{(2}^{n}a)}{4^n}-\frac{f{(2}^{m}a)}{4^m}‱\le \frac{1}{4}{\displaystyle \sum _{k=m}^{n-1}}\frac{\mu {(2}^{k}a{\mathrm{,2}}^{k}a\mathrm{,0,0})}{4^k}$$

(4.3)

for all n > m ≥ 0 and all $a\in \mathcal{A}$. It follows from (4.3) that the sequence $\left\{\frac{f{(2}^{n}a)}{4^n}\right\}$ is Cauchy. Since $\mathcal{A}$ is complete, this sequence is convergent. Define

$$\delta (a):=\underset{n\to \infty }{{\displaystyle \lim }}\frac{f{(2}^{n}a)}{4^n}\mathrm{.}$$

Since f(0) = 0, we have δ(0) = 0. Replacing a and b by 2 ⁿa and 2 ⁿb, c = d = 0, respectively, in (4.1), we get

$$\Vert \frac{f{(2}^n(\lambda a+\lambda b))}{4^n}+\frac{f{(2}^n(\lambda a-\lambda b))}{4^n}-2{\lambda }^2\frac{f{(2}^{n}a)}{4^n}-2{\lambda }^2\frac{f{(2}^{n}b)}{4^n}‱\le \frac{\mu {(2}^{n}a{\mathrm{,2}}^{n}b\mathrm{,0,0})}{4^n}\mathrm{.}$$

Taking the limit as n → ∞, we obtain

$$\delta \left(\lambda a+\lambda b\right)+\delta \left(\lambda a-\lambda b\right)=2{\lambda }^2\delta \left(a\right)+2{\lambda }^2\delta \left(b\right)$$

(4.4)

for all $a,b\in \mathcal{A}$ and all $\lambda \in T$. Putting λ = 1 in (4.4), we obtain that δ is a quadratic mapping. Setting b: = a in (4.4), we get

$$\delta \left(2\lambda a\right)=4{\lambda }^2\delta \left(a\right)$$

for all $a\in \mathcal{A}$ and all $\lambda \in T$. Hence,

$$\delta \left(\lambda a\right)={\lambda }^2\delta \left(a\right)$$

for all $a\in \mathcal{A}$ and all $\lambda \in T$. Under the assumption that f(ta) is continuous in t ∈ ℝ for each fixed $a\in \mathcal{A}$, by the same reasoning as in the proof of [10], we obtain that δ(λa) = λ²δ(a) for all $a\in \mathcal{A}$ and all λ ∈ ℝ. Hence,

$$\delta \left(\lambda a\right)=\delta \left(\frac{\lambda }{\left|\lambda \right|}\left|\lambda \right|a\right)=\frac{{\lambda }^2}{|\lambda {|}^2}\delta \left(\left|\lambda \right|a\right)=\frac{{\lambda }^2}{|\lambda {|}^2}|\lambda {|}^2\delta \left(a\right)={\lambda }^2\delta \left(a\right)$$

for all $a\in \mathcal{A}$ and all λ ∈ ℂ (λ ≠ 0). This means that δ is quadratic homogeneous.

Replacing c and d by 2 ⁿc and 2 ⁿd, respectively, and putting a = b = 0 in (4.1), we get

$$\begin{array}{l}\Vert \frac{f{(2}^{n}c\cdot 2^{n}d)}{4^{2n}}+\frac{f{(2}^{n}c\cdot 2^{n}d)}{4^{2n}}-2\frac{2^{2n}{c}^{2}f{(2}^{n}d)}{4^{2n}}-2\frac{f{(2}^{n}c{)2}^{2n}{d}^2}{4^{2n}}‱\\ =\Vert \frac{f{(2}^{2n}cd)}{4^{2n}}+\frac{f{(2}^{2n}cd)}{4^{2n}}-2\frac{2^{2n}{c}^2}{2^{2n}}\frac{f{(2}^{n}d)}{4^n}-2\frac{f{(2}^{n}c)}{4^n}\frac{2^{2n}{d}^2}{2^{2n}}‱\\ \le \frac{\mu {(\mathrm{0,0,2}}^{n}c{\mathrm{,2}}^{n}d)}{4^{2n}}\le \frac{\mu {(\mathrm{0,0,2}}^{n}c{\mathrm{,2}}^{n}d)}{4^n}\end{array}$$

for all $c,d\in \mathcal{A}$.

Hence, we have

$$\parallel \delta (cd)-c^2\delta (d)-\delta (c)d^2\parallel \le {\lim }_{n\to \infty }\frac{\mu {(\mathrm{0,0,2}}^{n}c{\mathrm{,2}}^{n}d)}{4^n}=0.$$

Thus, δ is a quadratic *-derivation on $\mathcal{A}$.

The rest of the proof is similar to the proof of Theorem 2.1. □

Corollary 4.3 Let ε, p be positive real numbers with p < 2. Suppose that$f:\mathcal{A}\to \mathcal{A}$is a mapping such that

$$\begin{array}{c}\parallel f\left(\lambda a+\lambda b+cd\right)+f\left(\lambda a-\lambda b+cd\right)-2{\lambda }^{2}f\left(a\right)-2{\lambda }^{2}f\left(b\right)-2c^{2}f\left(d\right)-2f\left(c\right)d^2\parallel \\ \le \epsilon \left(\parallel a{\parallel }^p+\parallel b{\parallel }^p+\parallel c{\parallel }^p+\parallel d{\parallel }^p\right)\end{array}$$

(4.5)

for all$a,b,c,d\in \mathcal{A}$and all$\lambda \in T$. Also, if for each fixed$a\in \mathcal{A}$the mapping t → f(ta) is continuous, then there exists a unique derivation δ on$\mathcal{A}$satisfying

$$\parallel f\left(a\right)-\delta \left(a\right)\parallel \le \frac{2\epsilon }{4-2^p}\parallel a{\parallel }^p$$

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

Proof. Putting φ(a, b, c, d) = ε(||a|| ^p + ||b||^p+ ||c||^p+ ||d||^p) in Theorem 4.2, we get the desired result.

Similarly, we can obtain the following. We will omit the proof.

Theorem 4.4 Suppose that$f:\mathcal{A}\to \mathcal{A}$is a mapping with f(0) = 0 for which there exists a function$\phi :{\mathcal{A}}^4\to \left[0,\infty \right)$satisfying (4.1), (4.2) and

$$\sum _{k=1}^{\infty }{4}^{2k}\phi \left(\frac{a}{2^k},\frac{b}{2^k},\frac{c}{2^k},\frac{d}{2^k}\right)<\infty$$

for all$a,b,c,d\in \mathcal{A}$. Also, if for each fixed$a\in \mathcal{A}$the mapping t → f(ta) from ℝ to$\mathcal{A}$is continuous, then there exists a unique quadratic *-derivation δ on$\mathcal{A}$satisfying

$$\parallel f\left(a\right)-\delta \left(a\right)\parallel \le \frac{1}{4}\tilde{\phi }\left(a,a,0,0\right)$$

for all$a\in \mathcal{A}$, where

$$\tilde{\phi }\left(a,b,c,d\right):=\sum _{k=1}^{\infty }{4}^k\phi \left(\frac{a}{2^k},\frac{b}{2^k},\frac{c}{2^k},\frac{d}{2^k}\right)$$

Corollary 4.5 Let ε, p be positive real numbers with p > 4. Suppose that$f:\mathcal{A}\to \mathcal{A}$is a mapping satisfying (4.5). Also, if for each fixed$a\in \mathcal{A}$the mapping t → f(ta) is continuous, then there exists a unique derivation δ on$\mathcal{A}$satisfying

$$\parallel f\left(a\right)-\delta \left(a\right)\parallel \le \frac{2\epsilon }{2^p-4}\parallel a{\parallel }^p$$

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

Proof. Putting φ(a, b, c, d) = ε(||a||^p+ ||b||^p+ ||c||^p+ ||d||^p) in Theorem 4.4, we get the desired result. □

5 Superstability of *-derivations and of quadratic *-derivations On C*-algebras

We prove the superstability of *-derivations and of quadratic *-derivations on C*-algebras. More precisely, we introduce the concept of (ψ, ε) -approximate *-derivations and of (ψ, ε)-approximate quadratic *-derivations on C*-algebras and show that every (ψ, ε)-approximate *-derivation is a *-derivation and that every (ψ, ε)-approximate quadratic *-derivation is a quadratic *-derivation. Thus, we extend the results of [21].

Definition 5.1 Suppose that$\mathcal{A}$is a *-normed algebra and s ∈{1, -1}. Let$\delta :\mathcal{A}\to \mathcal{A}$be a mapping for which there exist a mapping $\epsilon :\mathcal{A}\to \mathcal{A}$, and a function $\psi :\mathcal{A}\times \mathcal{A}\to ℝ$ satisfying

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

(5.1)

such that

$$\begin{array}{c}\parallel a\delta \left(b\right)-\epsilon \left(a\right)b\parallel \le \psi \left(a,b\right)\\ \parallel \epsilon \left(a\right)cd-a\left(\delta \left(c\right)d-c\delta \left(d\right)\right)\parallel \le \psi \left(a,cd\right)\\ \parallel a\delta {\left(b\right)}^{*}-\epsilon \left(a\right)b^{*}\parallel \le \psi \left(a,b\right)\end{array}$$

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

Theorem 5.2 Let$\mathcal{A}$be a C*-algebra. Then any (ψ, ε)-approximate *-derivation δ on$\mathcal{A}$is a *-derivation.

Proof. We assume that (5.1) holds. Let $a,b\in \mathcal{A}$ and λ ∈ ℂ. We have

$$\begin{array}{lll}\hfill \parallel b\left(\delta \left(\lambda a\right)-\lambda \delta \left(a\right)\right)\parallel & \le n^{-s}\parallel n^{s}b\delta \left(\lambda a\right)-\lambda n^{s}b\delta \left(a\right)\parallel & \hfill \text{(1)}\\ \le n^{-s}\parallel n^{s}b\delta \left(\lambda a\right)-\epsilon \left(n^{s}b\right)\lambda a\parallel +n^{-s}\parallel \epsilon \left(n^{s}b\right)\lambda a-\lambda n^{s}b\delta \left(a\right)\parallel & \hfill \text{(2)}\\ \le n^{-s}\psi \left(n^{s}b,\lambda a\right)+n^{-s}\left|\lambda \right|\psi \left(n^{s}b,a\right),& \hfill \text{(3)}\\ \hfill \text{(4) }\end{array}$$

which tends to zero as n → ∞, and so b(δ(λa) - λδ(a)) = 0 for all $b\in \mathcal{A}$. Let {e _i }_i∈Ibe an approximate unit of $\mathcal{A}$. If we replace b with {e _i }, then we have

$$\parallel e_i\left(\delta \left(\lambda a\right)-\lambda \delta \left(a\right)\right)\parallel =0$$

for all i ∈ I. So we conclude that δ(λa) = λδ(a) for all $a\in \mathcal{A}$ and λ ∈ ℂ.

The additivity of δ follows from

$$\begin{array}{l}\parallel c(\delta (a+b)-\delta (a)-\delta (b))\parallel \\ \le n^{-s}\parallel n^{s}c\delta (a+b)-\epsilon (n^{s}c)(a+b)\parallel +n^{-s}\parallel n^{s}c\delta (a)-\epsilon (n^{s}c)a)\parallel +n^{-s}\parallel n^{s}c\delta (b)-\epsilon (n^{s}c)b)\\ \le n^{-s}\psi (n^{s}c\mathrm{,}a+b)+n^{-s}\psi (n^{s}c\mathrm{,}a)+n^{-s}\psi (n^{s}c\mathrm{,}b).\end{array}$$

By the same process, using the approximate unit of $\mathcal{A}$, we have that δ(a + b) - δ(a) -δ(b) for all $a,b\in \mathcal{A}$.

The following computation

$$\begin{array}{c}\parallel z\left(\delta \left(ab\right)-\delta \left(a\right)b-a\delta \left(b\right)\right)\parallel \\ \le n^{-s}\parallel n^{s}z\delta \left(ab\right)-\epsilon \left(n^{s}z\right)\left(ab\right)\parallel +n^{-s}\parallel \epsilon \left(n^{s}z\right)ab-n^{s}z\left(\delta \left(a\right)b+a\delta \left(b\right)\right)\parallel \\ \le n^{-s}\psi \left(n^{s}z,ab\right)+n^{-s}\psi \left(n^{s}z,ab\right)\end{array}$$

yields that δ(ab) = δ(a)b + aδ(b) for all $a,b\in \mathcal{A}$.

Finally, on the involution, we have that

$$\begin{array}{lll}\hfill \parallel z\left(\delta \left(a^{*}\right)-\delta {\left(a\right)}^{*}\right)\parallel & \le n^{-s}\parallel n^{s}z\delta \left(a^{*}\right)-\epsilon \left(n^{s}z\right)a^{*}\parallel & \hfill \text{(1)}\\ +n^{-s}\parallel \epsilon \left(n^{s}z\right)a^{*}-n^{s}z\delta {\left(a\right)}^{*}\parallel & \hfill \text{(2)}\\ \le n^{-s}\psi \left(n^{-s}z,a^{*}\right)+n^{-s}\psi \left(n^{s}z,a\right).& \hfill \text{(3)}\\ \hfill \text{(4)}\end{array}$$

Thus, δ(a)* = δ(a) * for all $a\in \mathcal{A}$. □

Therefore, δ is a *-derivation on $\mathcal{A}$.

Corollary 5.3 Suppose that$\mathcal{A}$is a C*-algebra and that$\delta :\mathcal{A}\to \mathcal{A}$is a mapping for which there exist nonnegative real numbers α, β and positive real numbers p₁, p₂, q₁, q₂with p₁, p₂, q₁, q₂ < 1 such that

$$\begin{array}{c}\parallel a\delta \left(b\right)-\epsilon \left(a\right)b\parallel \le \alpha \left(\parallel a{\parallel }^{p_1}+\parallel b{\parallel }^{p_2}\right)+\beta \parallel a{\parallel }^{q_1}\parallel b{\parallel }^{q_2},\\ \parallel \epsilon \left(a\right)cd-a\left(\delta \left(c\right)d-c\delta \left(d\right)\right)\parallel \le \alpha \left(\parallel a{\parallel }^{p_1}+\parallel cd{\parallel }^{p_2}\right)+\beta \parallel a{\parallel }^{q_1}\parallel cd{\parallel }^{q_2},\\ \parallel a\delta {\left(b\right)}^{*}-\epsilon \left(a\right)b^{*}\parallel \le \alpha \left(\parallel a{\parallel }^{p_1}+\parallel b{\parallel }^{p_2}\right)+\beta \parallel a{\parallel }^{q_1}\parallel b{\parallel }^{q_2}\end{array}$$

for all $a,b,c,d\in \mathcal{A}$. Then δ is a *-derivation of $\mathcal{A}$.

Next, we prove the superstability of quadratic *-derivations on C*-algebras.

Definition 5.4 Suppose that$\mathcal{A}$is a *-normed algebra and s ∈{-1, 1}. Let$\delta :\mathcal{A}\to \mathcal{A}$be a mapping for which there exist a function$\psi :\mathcal{A}\times \mathcal{A}\to \left[0,\infty \right)$and a mapping$\epsilon :\mathcal{A}\to \mathcal{A}$satisfying

$$\underset{n\to \infty }{lim}{n}^{-2s}\psi \left(n^{s}a,b\right)=\underset{n\to \infty }{lim}{n}^{-2s}\psi \left(a,n^{s}b\right)=0\left(a,b\in \mathcal{A}\right)$$

(5.2)

such that

$$\begin{array}{l}\parallel a^2\delta (b)-\epsilon (a)b^2\parallel \le \psi (a\mathrm{,}b)\\ \parallel \epsilon (a)(cd{)}^2-a^2(\delta (c)d^2-c^2\delta (d))\parallel \le \psi (a\mathrm{,}cd)\\ \parallel a^2\delta (b^{*})-\epsilon (a)(b^2{)}^{*}\parallel \le \psi (a\mathrm{,}b)\end{array}$$

for all a,b,c,d ∈ A. Then δ is called a (ψ, ε)-approximate quadratic *-derivation on$\mathcal{A}$.

Theorem 5.5 Suppose that$\mathcal{A}$is a C*-algebra and s ∈{-1, 1}. Let$\delta :\mathcal{A}\to \mathcal{A}$be a (ψ, ε)-approximate quadratic *-derivation on$\mathcal{A}$. Then δ is a quadratic *-derivation on$\mathcal{A}$.

Proof. We assume that (5.2) holds. We first show that δ is quadratic homogeneous. To do this, pick λ ∈ ℂ and $a,b\in \mathcal{A}$. Then, we have

$$\begin{array}{c}\parallel b^2(\delta (\lambda a)-{\lambda }^2\delta (a))\parallel =n^{-2s}\parallel n^{2s}{b}^2\delta (\lambda a)-{\lambda }^2{n}^{2s}{b}^2\delta (a)\parallel \\ \le n^{-2s}\parallel n^{2s}{b}^2\delta (\lambda a)-\epsilon (n^{s}b)(\lambda a{)}^2\parallel +n^{-2s}\parallel {\lambda }^2\epsilon (n^{s}b)a^2-{\lambda }^2{n}^{2s}{b}^2\delta (a)\parallel \\ \le n^{-2s}\psi (n^{s}b\mathrm{,}\lambda a)+n^{-2s}|\lambda {|}^2\psi (n^{s}b\mathrm{,}a).\end{array}$$

So

$$\parallel b^2\left(\delta \left(\lambda a\right)-{\lambda }^2\delta \left(a\right)\right)\parallel \le n^{-2s}\psi \left(n^{s}b,\lambda a\right)+|\lambda {|}^2{n}^{-2s}\psi \left(n^{s}b,a\right),$$

which tends to 0 as n → ∞. Let {e _i }_i∈Ibe an approximate unit of $\mathcal{A}$. Then, {f(e _i )|i ∈ I} is also an approximate unit of $\mathcal{A}$ for every polynomial f. Considering e _i instead of b in the above inequality, we conclude that δ(λa) = λ²δ(a) for all λ ∈ ℂ.

The quadraticity of δ follows from

$$\begin{array}{l}\parallel d^2(\delta (a+b)+\delta (a-b)-2\delta (a)-2\delta (b))\parallel \\ =n^{-2s}\parallel n^{2s}{d}^2\delta (a+b)+n^{2s}{d}^2\delta (a-b)-2n^{2s}{d}^2\delta (a)-2n^{2s}{d}^2\delta (b)\parallel \\ \le n^{-2s}[\parallel n^{2s}{d}^2\delta (a+b)-\epsilon (n^{s}d)(a+b{)}^2\parallel \\ +n^{-2s}\parallel n^{2s}{d}^2\delta (a-b)-\epsilon (n^{s}d)(a-b{)}^2\parallel \\ +2n^{-2s}\parallel \delta (n^{s}d)a^2-n^{2s}{d}^2\delta (a)\parallel +2n^{-2s}\parallel \delta (n^{s}d)b^2-n^{2s}{d}^2\delta (b)\parallel ]\\ \le n^{-2s}[\psi (n^{s}d\mathrm{,}a+b)+\psi (n^{s}d\mathrm{,}a-b)+2\psi (a\mathrm{,}{n}^{s}d)+2\psi (b\mathrm{,}{n}^{s}d)]\end{array}$$

for all $a,b,d\in \mathcal{A}$. Thus, we have δ(a + b) + δ(a - b) -2δ(a) - 2δ(b) = 0 for all $a,b\in \mathcal{A}$.

$$\begin{array}{c}\Vert d^2(\delta (ab)-(\delta (a)b^2+a^2\delta (b))‱=n^{-2s}\parallel n^{2s}{d}^2(\delta (ab)-\delta (a)b^2-a^2\delta (b))\parallel \\ \le n^{-2s}[\parallel n^{2s}{d}^2\delta (ab)-\epsilon (n^{s}d)(ab{)}^2\parallel \\ +n^{-2s}\parallel \epsilon (n^{s}d)(ab{)}^2-n^{2s}{d}^2\delta (a)b^2+n^{2s}{d}^2{a}^2\delta (b)\parallel ]\\ \le n^{-2s}[\psi (n^{s}d\mathrm{,}ab)+\psi (n^{s}d\mathrm{,}ab)]\end{array}$$

for all $a,b,d\in \mathcal{A}$. So δ(ab) = δ(a)b² + a²δ(b).

The rest of the proof is similar to the proof of Theorem 5.2.

Therefore, δ is a quadratic *-derivation on $\mathcal{A}$. □

Corollary 5.6 Suppose that$\mathcal{A}$is a C*-algebra and that$\delta :\mathcal{A}\to \mathcal{A}$is a mapping for which there exist a nonnegative real number α and a positive real number p with p < 2 such that

$$\begin{array}{l}\parallel a^2\delta (b)-\delta (a)b^2\parallel \le \alpha \parallel a{\parallel }^p\parallel b{\parallel }^p\mathrm{,}\\ \parallel \epsilon (a)(cd{)}^2-a^2(\delta (c)d^2-c^2\delta (d))\parallel \le \alpha \parallel a{\parallel }^p\parallel cd{\parallel }^p\mathrm{,}\\ \parallel a^2\delta (b^{*})-\epsilon (a)(b^2{)}^{*}\parallel \le \alpha \parallel a{\parallel }^p\parallel b{\parallel }^p\end{array}$$

for all a, b, c, d ∈ A. Then δ is a quadratic *-derivation on $\mathcal{A}$.