# The image of a derivation on a Banach algebra is contained in the kernel of a character

It is known that if $D$ is a continuous derivation on a commutative Banach algebra $\mathcal{A}$, then for any nonzero character $\theta$ on $\mathcal{A}$ we have $D(\mathcal{A})⊆ker\theta$.

Please help me with these questions or give some references. How this can be restated in the case of a non-commutative Banach algebra?

Does there exist a similar result for a non-commutative Banach algebra?

• You should first clarify whether you are assuming that the derivation is continuous. For continuous derivations the Singer-Wermer theorem has an elegant and (relatively) easy proof, accessible to anyone with a good first course in Banach algebras. For not-necessarily continuous derivations on commutative Banach algebras, the corresponding statement is still true, but the proof is much much much much harder. – Yemon Choi Jan 28 '16 at 0:19
• @YemonChoi : You are right. Sorry it was a typo. I am assuming the derivation $D$ is continuous. I corrected it. – Fermat Jan 28 '16 at 4:52

I shall assume that in your question, you are asking about continuous derivations on a Banach algebra $A$. Results for everywhere-defined-but-not-continuous derivations were studied intensively 30-40 years ago but my understanding is that the remaining open problems are thought to be inaccessible.

So, let $A$ be a Banach algebra and let $D:A\to A$ be a continuous derivation. Without loss of generality, we can assume that $A$ is unital — if not, then we simply adjoin an identity element and extend $D$ by making it $0$ on the adjoined identity.

We then have the following old result of A. M. Sinclair (Proc. AMS 20 (1969), 166–170):

Let $P$ be a primitive ideal of $A$. Then $D(P)\subseteq P$.

(This can also be found as Proposition 2.7.22(ii) of Dales's behemoth Banach Algebras and Automatic Continuity, but the original paper of Sinclair is reasonably self-contained and easy to follow.)

If $A$ is commutative, then primitive ideals are the same thing as maximal ideals, and so Sinclair's theorem implies that $D(M)\subseteq M$ for every maximal ideal $M$. But since $A={\mathbb C}1_A+M$ and $D(1_A)=0$, we get $D(A)\subseteq M$ for all maximal ideals $M$, which is to say $D(A)\subseteq {\rm rad}(A)$ — and that is the result which you quoted in your question.

• Thank you very much @YemonChoi for your detailed answer. – Fermat Jan 28 '16 at 4:57