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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?

Community
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Student
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  • For the fact that the space of polynomials is dense in $A$, see here. – Daniel Fischer Feb 01 '14 at 21:58
  • @DanielFischer I now also think the statement as I give it in the question is wrong: it should be every unital continuous homomorphism. Shouldn't it? – Student Feb 03 '14 at 10:59
  • That depends on which definition of a (ring/algebra) homomorphism you use. If the one without requiring $\varphi(1) = 1$, then you need to say "unital". If the other one, saying "unital homomorphism" would be pleonastic (specifying continuity is, if I remember correctly, not necessary, I think an algebra homomorphism $A\to\mathbb{C}$ is automatically continuous). The answer is good except for using the wrong argument for the denseness of polynomials. – Daniel Fischer Feb 03 '14 at 12:46
  • For others encountering the issues OP ran into with finding resources, a key idea to look up is "multiplicative linear functional". This will help with hunting for resources. – Cameron Williams Feb 07 '24 at 21:11

1 Answers1

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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 $g\in A$.

Prahlad Vaidyanathan
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  • Thanks. And how to finish the proof? $$ 0 = |\varphi(f-f)| = |\varphi(f-\lim_{n \to \infty}p_n)| = |\varphi(f)-\lim_{n \to \infty}\varphi(p_n))| = |\varphi(f)-\lim_{n \to \infty}\lambda | = |\varphi(f)-\lambda |$$? – Student Jan 07 '14 at 18:49
  • What is $\lambda$? You just need to reprove the fact that if $\varphi$ and $\psi$ are two continuous linear functionals that agree on a dense set, then they agree everywhere. – Prahlad Vaidyanathan Jan 08 '14 at 15:14
  • $\lambda$ is the constant function $f(x) = \lambda$ for all $x$. – Student Jan 10 '14 at 09:34
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    Sorry, could you explain why $\varphi(\overline{z}) = \overline{z_0}$? If this does not hold you can't apply Stone Weierstrass. – Student Jan 29 '14 at 15:27
  • There is a nicer argument for the reasoning. You can prove that the closed subalgebra generated by $1$ and $f$ is equal to $A$, and then the statement follows. – Anton Odina Feb 05 '24 at 20:33