1
$\begingroup$

So today I obtained Gerard Murphy's $C^*$-Algebras and Operator Theory, and I have been going through the exercises in the earlier chapters to help bring me up to speed. In chapter $1$, exercise $13$ is stated as follows:

Suppose that $d$ is a bounded derivation on a unital Banach algebra $A$ and $\lambda\in\mathbb{C}\setminus\{0\}$ such that $da=\lambda a$. Show that $a$ is nilpotent, that is, that $a^n=0$ for some positive integer $n$ (use the boundedness of $\sigma(d)$).

For completeness, a derivation on an algebra $A$ is a linear map $d:A\to A$ such that $d(ab)=d(a)b+ad(b)$ for all $a,b\in A$.

I came up with the following solution:

By induction, we have $d(a^n)=n\lambda a^n$ for all positive integers $n$. If $a$ is not nilpotent, then $\{n\lambda:n\in\mathbb{N}\}\subset\sigma(d)$. But $r(d)\leq\|d\|$, so $n|\lambda|\leq\|d\|$ for all $n$, a contradiction. Thus $a$ is nilpotent. (Here, $r(d)$ denotes the spectral radius of $d$).

However, my proof does not explicitly use the assumption that $A$ is unital. Am I missing something? Did I make a mistake? Or is the assumption superfluous?

$\endgroup$
1
$\begingroup$

As far as I can tell, the assumption is indeed superfluous and your answer is completely correct.

$\endgroup$
  • $\begingroup$ Thank you. I was fairly certain this was the case, but just wanted input from someone else before I claim this is correct. $\endgroup$ – Aweygan Nov 26 '16 at 5:34
  • $\begingroup$ Fair enough. I made this CW so that passers by would be more likely to either upvote this answer in agreement or correct it as they see fit. Not a lot of traffic on the site right now, so that could take a while. $\endgroup$ – Omnomnomnom Nov 26 '16 at 5:51
  • $\begingroup$ I understand. I just wanted to wait and see if anyone can find something wrong before accepting this answer. $\endgroup$ – Aweygan Nov 26 '16 at 6:04
  • $\begingroup$ @Aweygan yes, and I was trying to say that I understand this $\endgroup$ – Omnomnomnom Nov 26 '16 at 6:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.