Bourbaki

Exercise 10, § 3

Tag 029A, permanent. Cite it as [Bourbaki, Tag 029A].

Spectral Theories, I, p. 167

Let AA be a commutative Banach algebra and let DD be a continuous derivation of AA into AA, that is to say a continuous linear mapping of AA into AA such that D(ab)=(Da)b+a(Db)D(ab) = (Da)b+a(Db) for a,b∈Aa, b\in A.

a) For every χ∈X′(A)\chi \in \mathsf{X}'(A) and every λ∈C\lambda \in \mathbf{C}, the series

∞n\infty n

φλ(a)=∑λn\varphi_{\lambda}(a) =\sum\lambda n! χ(Dn(a))\chi (D^n(a))

n=0n=0

converges;λ↦φλ(a);\lambda \mapsto \varphi_{\lambda}(a) is an entire function, and a↦φλ(a)a\mapsto \varphi_{\lambda}(a) is a character of A.

b) The number φλ(a)\varphi_{\lambda}(a) is independent of λ\lambda. (Observe that ∣φλ(a)∣⩽∥a∥|\varphi_{\lambda}(a)|\leqslant \|a\| according to a).) Hence χ(D(a))=0\chi (D(a)) = 0.

c) The image of D is contained in the radical of A. Deduce a new proof of assertion f ) of Exercise 31 of I, p. 162.

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Machine translation of the English, by gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.