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# Comment on ‘Approximate ∗-derivations and approximate quadratic ∗-derivations on ${C}^{\ast }$-algebras’ [Jang, Park, J. Inequal. Appl. 2011 (2011), Article ID 55]

Journal of Inequalities and Applications20122012:183

https://doi.org/10.1186/1029-242X-2012-183

• Received: 1 March 2012
• Accepted: 31 May 2012
• Published:

## Abstract

In (J. Inequal. Appl. 2011:Article ID 55, Section 4, 2011), Jang and Park proved the Hyers-Ulam stability of quadratic -derivations on Banach -algebras. One can easily show that all the quadratic -derivations δ in Section 4 must be trivial. So the results are trivial. In this paper, we correct the statements and prove the corrected results.

MSC:39B52, 47B47, 39B72.

## Keywords

• Banach -algebra
• Hyers-Ulam stability

## 1 Introduction and preliminaries

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

A ${C}^{\ast }$-dynamical system is a triple $\left(\mathcal{A},G,\alpha \right)$ consisting of a ${C}^{\ast }$-algebra $\mathcal{A}$, a locally compact group G, and a pointwise norm continuous homomorphism α of G into the group $Aut\left(\mathcal{A}\right)$ of -automorphisms of $\mathcal{A}$. Every bounded -derivation δ arises as an infinitesimal generator of a dynamical system for $\mathbb{R}$. 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 ${\mathcal{A}}^{\mathrm{\prime }\mathrm{\prime }}$ such that
$\delta \left(x\right)=a{d}_{ih}\left(x\right)$

for all $x\in \mathcal{A}$. The theory of bounded derivations of ${C}^{\ast }$-algebras is important in the quantum mechanics (see ).

A functional equation is called stable if any function satisfying the functional equation ‘approximately’ is near to a true solution of the functional equation.

In 1940, Ulam  proposed the following question concerning the stability of group homomorphisms: Under what condition does there exist an additive mapping near an approximately additive mapping? Hyers  answered 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 Rassias . Since then, the stability problems of various functional equations have been extensively investigated by a number of authors (see ).

Jang and Park [, Section 4] proved the Hyers-Ulam stability of quadratic -derivations on Banach -algebras.

Theorem 1.1 ([, Theorem 4.2])

Suppose that $f:\mathcal{A}\to \mathcal{A}$ is a mapping with $f\left(0\right)=0$ for which there exists a function $\phi :{\mathcal{A}}^{4}\to \left[0,\mathrm{\infty }\right)$ such that
$\begin{array}{r}\stackrel{˜}{\phi }\left(a,b,c,d\right):=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{{4}^{k}}\phi \left({2}^{k}a,{2}^{k}b,{2}^{k}c,{2}^{k}d\right)<\mathrm{\infty },\\ \parallel f\left(\lambda a+\lambda b+cd\right)+f\left(\lambda a-\lambda b+cd\right)-2{\lambda }^{2}f\left(a\right)-2{\lambda }^{2}f\left(b\right)-2f\left(c\right){d}^{2}-2{c}^{2}f\left(d\right)\parallel \\ \phantom{\rule{1em}{0ex}}\le \phi \left(a,b,c,d\right),\\ \parallel f\left({a}^{\ast }\right)-f{\left(a\right)}^{\ast }\parallel \le \phi \left(a,a,a,a\right)\end{array}$
(1.1)
for all $a,b,c,d\in \mathcal{A}$ and all $\lambda \in \mathbb{T}:=\left\{\mu \in \mathbb{C}:|\mu |=1\right\}$. Also, if for each fixed $a\in \mathcal{A}$ the mapping $t\to f\left(ta\right)$ from to $\mathcal{A}$ is continuous, then there exists a unique quadratic -derivation δ on $\mathcal{A}$ satisfying
$\parallel f\left(a\right)-\delta \left(a\right)\parallel \le \frac{1}{4}\stackrel{˜}{\phi }\left(a,a,0,0\right)$

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

Letting $\lambda =1$, $b=0$ and $d=I$ (identity) in (1.1) of Theorem 1.1, we get
$\parallel f\left(a+c\right)+f\left(a+c\right)-2f\left(a\right)-2f\left(c\right)-2{c}^{2}f\left(I\right)\parallel \le \phi \left(a,0,c,I\right)$

for all $a,c\in \mathcal{A}$. Thus $2\delta \left(a+c\right)=2\delta \left(a\right)+2\delta \left(c\right)+2{c}^{2}{d}^{\prime }$ for some ${d}^{\prime }\in \mathcal{A}$. Since δ is quadratic, $2\delta \left(a\right)+2\delta \left(-c\right)+2{\left(-c\right)}^{2}{d}^{\prime }=2\delta \left(a\right)+2\delta \left(c\right)+2{c}^{2}{d}^{\prime }$ and so $2\delta \left(a+c\right)=2\delta \left(a-c\right)$. Letting $c=a$ in the last equality, we get $2\delta \left(2a\right)=2\delta \left(0\right)=0$. So δ must be zero. Thus the results are trivial.

In this paper, we correct the wrong statements in  and prove the corrected results.

## 2 Hyers-Ulam stability of quadratic ∗-derivations on Banach ∗-algebras

In this section, we correct the statements of [, Section 4] and prove the Hyers-Ulam stability of the corrected results.

Definition 2.1 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. (1)

δ is a quadratic mapping,

2. (2)

δ is quadratic homogeneous, that is, $\delta \left(\lambda a\right)={\lambda }^{2}\delta \left(a\right)$ for all $a\in \mathcal{A}$ and all ,

3. (3)

$\delta \left(ab\right)=\delta \left(a\right){b}^{2}+{a}^{2}\delta \left(b\right)$ for all $a,b\in \mathcal{A}$,

4. (4)

$\delta \left({a}^{\ast }\right)=\delta {\left(a\right)}^{\ast }$ for all $a\in \mathcal{A}$.

Example 2.2 Let $\mathcal{A}$ be a commutative -normed algebra. For a given self-adjoint element $x\in \mathcal{A}$, let $\delta :\mathcal{A}\to \mathcal{A}$ be given by
$\delta \left(a\right)=i\left(x{a}^{2}-{a}^{2}x\right)$

for all $x\in \mathcal{A}$. Then it is easy to show that $\delta :\mathcal{A}\to \mathcal{A}$ is a quadratic -derivation on $\mathcal{A}$.

Theorem 2.3 Suppose that $f:\mathcal{A}\to \mathcal{A}$ is a mapping with $f\left(0\right)=0$ for which there exists a function $\phi :{\mathcal{A}}^{2}\to \left[0,\mathrm{\infty }\right)$ such that
for all $a,b,c,d\in \mathcal{A}$ and all $\lambda \in \mathbb{T}$. Also, if for each fixed $a\in \mathcal{A}$ the mapping $t\to f\left(ta\right)$ from to $\mathcal{A}$ is continuous, then there exists a unique quadratic -derivation δ on $\mathcal{A}$ satisfying
$\parallel f\left(a\right)-\delta \left(a\right)\parallel \le \frac{1}{4}\stackrel{˜}{\phi }\left(a,a\right)$
(2.4)

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

Proof Putting $a=b$ and $\lambda =1$ in (2.1), we have
$\parallel f\left(2a\right)-4f\left(a\right)\parallel \le \phi \left(a,a\right)$
for all $a\in \mathcal{A}$. One can use induction to show that
$\parallel \frac{f\left({2}^{n}a\right)}{{4}^{n}}-\frac{f\left({2}^{m}a\right)}{{4}^{m}}\parallel \le \frac{1}{4}\sum _{k=m}^{n-1}\frac{\phi \left({2}^{k}a,{2}^{k}a\right)}{{4}^{k}}$
(2.5)
for all $n>m\ge 0$ and all $a\in \mathcal{A}$. It follows from (2.5) that the sequence $\left\{\frac{f\left({2}^{n}a\right)}{{4}^{n}}\right\}$ is Cauchy. Since $\mathcal{A}$ is complete, this sequence is convergent. Define
$\delta \left(a\right):=\underset{n\to \mathrm{\infty }}{lim}\frac{f\left({2}^{n}a\right)}{{4}^{n}}.$
Since $f\left(0\right)=0$, we have $\delta \left(0\right)=0$. Replacing a and b by ${2}^{n}a$ and ${2}^{n}b$, respectively, in (2.1), we get
$\parallel \frac{f\left({2}^{n}\left(\lambda a+\lambda b\right)\right)}{{4}^{n}}+\frac{f\left({2}^{n}\left(\lambda a-\lambda b\right)\right)}{{4}^{n}}-2{\lambda }^{2}\frac{f\left({2}^{n}a\right)}{{4}^{n}}-2{\lambda }^{2}\frac{f\left({2}^{n}b\right)}{{4}^{n}}\parallel \le \frac{\phi \left({2}^{n}a,{2}^{n}b\right)}{{4}^{n}}.$
Taking the limit as $n\to \mathrm{\infty }$, we obtain
$\delta \left(\lambda a+\lambda b\right)+\delta \left(\lambda a-\lambda b\right)=2{\lambda }^{2}\delta \left(a\right)+2{\lambda }^{2}\delta \left(b\right)$
(2.6)

for all $a,b\in \mathcal{A}$ and all . Putting $\lambda =1$ in (2.6), we obtain that δ is a quadratic mapping. It is well known that the quadratic mapping δ satisfying (2.4) is unique (see  or ).

Setting $b:=a$ in (2.6), we get
$\delta \left(2\lambda a\right)=4{\lambda }^{2}\delta \left(a\right)$
for all $a\in \mathcal{A}$ and all . Hence
$\delta \left(\lambda a\right)={\lambda }^{2}\delta \left(a\right)$
for all $a\in \mathcal{A}$ and all . Under the assumption that $f\left(ta\right)$ is continuous in for each fixed $a\in \mathcal{A}$, by the same reasoning as in the proof of , we obtain that $\delta \left(\lambda a\right)={\lambda }^{2}\delta \left(a\right)$ for all $a\in \mathcal{A}$ and all . Hence
$\delta \left(\lambda a\right)=\delta \left(\frac{\lambda }{|\lambda |}|\lambda |a\right)=\frac{{\lambda }^{2}}{{|\lambda |}^{2}}\delta \left(|\lambda |a\right)=\frac{{\lambda }^{2}}{{|\lambda |}^{2}}{|\lambda |}^{2}\delta \left(a\right)={\lambda }^{2}\delta \left(a\right)$

for all $a\in \mathcal{A}$ and all ($\lambda \ne 0$). This means that δ is quadratic homogeneous.

Replacing c and d by ${2}^{n}c$ and ${2}^{n}d$, respectively, in (2.2), we get

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

Thus we have
$\parallel \delta \left(cd\right)-{c}^{2}\delta \left(d\right)-\delta \left(c\right){d}^{2}\parallel \le \underset{n\to \mathrm{\infty }}{lim}\frac{\phi \left({2}^{n}c,{2}^{n}d\right)}{{4}^{n}}=0.$
Replacing a and ${a}^{\ast }$ by ${2}^{n}a$ and ${2}^{n}{a}^{\ast }$, respectively, in (2.3), we get
$\parallel \frac{1}{{4}^{n}}f\left({2}^{n}{a}^{\ast }\right)-\frac{1}{{2}^{n}}f{\left({4}^{n}a\right)}^{\ast }\parallel \le \frac{1}{{4}^{n}}\phi \left({2}^{n}a,{2}^{n}a\right).$

Passing to the limit as $n\to \mathrm{\infty }$, we get the $\delta \left({a}^{\ast }\right)=\delta {\left(a\right)}^{\ast }$ for all $a\in \mathcal{A}$. So δ is a quadratic -derivation on $\mathcal{A}$, as desired. □

Corollary 2.4 Let ε, p be positive real numbers with $p<2$. Suppose that $f:\mathcal{A}\to \mathcal{A}$ is a mapping such that
for all $a,b,c,d\in \mathcal{A}$ and all . Also, if for each fixed $a\in \mathcal{A}$ the mapping $t\to f\left(ta\right)$ is continuous, then there exists a unique quadratic -derivation δ on $\mathcal{A}$ satisfying
$\parallel f\left(a\right)-\delta \left(a\right)\parallel \le \frac{2\epsilon }{4-{2}^{p}}{\parallel a\parallel }^{p}$

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

Proof Putting $\phi \left(a,b\right)=\epsilon \left({\parallel a\parallel }^{p}+{\parallel b\parallel }^{p}\right)$ in Theorem 2.3, we get the desired result. □

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

Theorem 2.5 Suppose that $f:\mathcal{A}\to \mathcal{A}$ is a mapping with $f\left(0\right)=0$ for which there exists a function $\phi :{\mathcal{A}}^{2}\to \left[0,\mathrm{\infty }\right)$ satisfying (2.1), (2.2), (2.3) and
$\sum _{k=1}^{\mathrm{\infty }}{4}^{2k}\phi \left(\frac{a}{{2}^{k}},\frac{b}{{2}^{k}}\right)<\mathrm{\infty }$
for all $a,b\in \mathcal{A}$. Also, if for each fixed $a\in \mathcal{A}$ the mapping $t\to f\left(ta\right)$ from to $\mathcal{A}$ is continuous, then there exists a unique quadratic -derivation δ on $\mathcal{A}$ satisfying
$\parallel f\left(a\right)-\delta \left(a\right)\parallel \le \frac{1}{4}\stackrel{˜}{\phi }\left(a,a\right)$
for all $a\in \mathcal{A}$, where
$\stackrel{˜}{\phi }\left(a,b\right):=\sum _{k=1}^{\mathrm{\infty }}{4}^{k}\phi \left(\frac{a}{{2}^{k}},\frac{b}{{2}^{k}}\right).$
Corollary 2.6 Let ε, p be positive real numbers with $p>4$. Suppose that $f:\mathcal{A}\to \mathcal{A}$ is a mapping satisfying (2.7), (2.8) and (2.9). Also, if for each fixed $a\in \mathcal{A}$ the mapping $t\to f\left(ta\right)$ is continuous, then there exists a unique quadratic -derivation δ on $\mathcal{A}$ satisfying
$\parallel f\left(a\right)-\delta \left(a\right)\parallel \le \frac{2\epsilon }{{2}^{p}-4}{\parallel a\parallel }^{p}$

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

Proof Putting $\phi \left(a,b\right)=\epsilon \left({\parallel a\parallel }^{p}+{\parallel b\parallel }^{p}\right)$ in Theorem 2.5, we get the desired result. □

## Declarations

### Acknowledgements

The first author, the second author and the third author were supported by Basic Science Research Program through the National Research Foundation of Korea funded by the Ministry of Education, Science and Technology (NRF-2009-0070788), (NRF-2010-0021792) and (NRF-2010-0009232), respectively.

## Authors’ Affiliations

(1)
Research Institute for Natural Sciences, Hanyang University, Seoul, 133-791, Korea
(2)
Department of Mathematics, University of Seoul, Seoul, 130-743, Korea
(3)
Department of Mathematics, Daejin University, Pocheon, Kyeonggi, 487-711, Korea

## References 