The disk algebra is the set of continuous functions $f: D \to \mathbb C$ where $D$ is the closed unit disc in $\mathbb C$ and $f$ is analytic on the interior of $D$. It is endowed with the $\sup$-norm.
Let $A$ denote the disk algebra.
I read that every continuous homomorphism $\varphi : A \to \mathbb C$ is of the form $f \mapsto f(z_0)$ for some $z_0 \in D$. The problem is I tried to look up the proof but I can't remember where I read it and I also can't find an alternative source. I also can't seem to prove it. I'm even starting to doubt the truth of the statement. How to prove this?
Let $\varphi : A\to \mathbb{C}$ be a continuous homomorphism, and let $f \in A$ be the function $z\mapsto z$, and let $z_0 := \varphi(f)$. Then for any polynomial $$ g(z) := \sum_{k=0}^n \alpha_k z^k \Rightarrow g = \sum_{k=0}^n \alpha_k f^k $$ we have $$ \varphi(g) = \sum_{k=0}^n \alpha_k z_0^k = g(z_0) $$ Now the set of such polynomials is dense in $A$ (by Stone-Weierstrass), and hence $\varphi(g) = g(z_0)$ for all $g\in A$.