Let $\varphi: A \to \mathbb C$ be a non-zero homomorphism. How can we extend it to an homomorphism $\psi: \overline A \to \mathbb C$?

Question

Let $X$ be a Hausdorff compact set. We consider $C(X)$ as a complex Banach algebra with the supremum norm $\| f \|_X = \sup_{x \in X} |f(x)|$.

Let $A$ be a Banach function algebra on $X$, i.e $A \subset C(X)$ is a Banach algebra that contains the constants and separetes points of $X$ with the norm $\| \cdot \|$, that is not $\|\cdot \|_X$. However, $\| f \|_X \leq \|f\|, \, \forall f \in A$.

We can also consider the algebra $\overline A$, $f \in \overline A$ iff exists a sequence $(f_n)$ in $A$ such that $\|f_n - f \|_X \to 0$, which is a Banach algebra with the supremum norm.

Question: Let $\varphi: A \to \mathbb C$ be a non-zero homomorphism. How can we extend it to an homomorphism $\psi: \overline A \to \mathbb C$?

My attempt: Given $f \in \overline A, \, \exists (f_n)$ sequence in $A$ such that $\|f_n - f \|_X \to 0$. I want to define $\psi (f) = \lim_{n \to \infty} \varphi(f_n)$, however I'm not sure that $\varphi(f_n)$ has a limit in $\mathbb C$.

Help?

Answer 1

A homomorphism on a Banach Algebra is by definition continuous, so your limit exists.