Functional calculus for unitization of an algebra?

Question

I have been stuck on this problem for a week now:

"Let $A$ be a Banach algebra without identity, let $a\in A$, and let $f$ be holomorphic on a neighborhood of $\sigma(a)$, so that $f(a)\in A^\# $ is defined. Show that $f(a)\in A$ if and only if $f(0) = 0$."

Here $A^\#$ is the unitization of $A$.

So I tried approaching this question in two ways - approximating $f$ with rationals using Runge's theorem or via a power series about zero. Intuitively, I get why it makes sense, since the power series for $f$ on a small disk about zero has a zero coefficient for $z^0$, and since the homomorphism

\begin{align*} O(D)&\to A\\ g &\mapsto g(a) \end{align*}

preserves polynomials. (Here $D$ is the domain of $f$ and $O(D)$ is the set of holomorphic functions on $D$). The issue I am facing is of course that the power series only converges on a small disk and not all of $D$.

Any help is much appreciated! (I don't need a full solution - a hint would definitely help.) Thanks!

Answer 1

Hint 1: If $f(0) \ne 0$, consider $$ f(0) = \frac{f(0)-f(z)}{z}z+f(z) $$ Hint 2: If $\lambda \in\rho(a)$ in $A^{\sharp}$, then $a(a-\lambda e)^{-1}\in A$.

Answer 2

Here's how I approached this problem - I used the Stone-Weierstrauss theorem to prove that one can find a sequence of polynomials that converges to $f$ uniformly on compact subsets of $D$. The rest followed easily using the algebra homomorphism, as maintaining $f(0)=0$ requires the constant terms of these polynomials to converge to $0$.