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?


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

  • $\begingroup$ 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

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