Characterspace of algebra of rational functions on compact subset of $\mathbb C$

Question

Let $K\subset \mathbb C$ be compact and denote with $R$ the closure of rational functions on $K$ w.r.t. $\|\cdot \|_\infty$.

Show that the character space of the Banach algebra $R$ is $\{\delta_k:k\in K\}$, where $\delta_k(f) = f(k)$ for $f\in R$.

Any hints for this problem?

I have already shown this for analytic functions on the disc (disc algebra) by using the density of polynomials. Furthermore I also did it for continuous functions, where I used that given $f$ is continuous then also $|f|^2$ is continuous. I cannot use this trick here anymore, so I am stuck there.

Can someone provide any references for this fact?

Answer 1

Sketch of proof:

Fix $\phi\in \Phi_R$ (character space) and let $x\in R$ where $x(z) = z$. Set $\alpha =\phi(x)$, then clearly $\phi(f) = f(\alpha)$ for $f\in R$. Remains to show that $\alpha \in K$. So assume not, then $\alpha - z \ne 0$ for all $z\in K$ and hence $\alpha - x\in R$ is invertible. Therefore, $\alpha\notin \sigma(x)$ contradicting the fact that $\sigma(x)=\{\phi(x) : \phi \in \Phi_R\}$. Hence $\phi = \delta_\alpha$.