Let $A$ be a commutative unital Banach Algebra. Let $F$ be a weak*-closed subset of $\Omega(A)$ with the property that for $A$ if for all $x \in A$, $\sup\{ |\hat{x}(\phi)| : \phi \in F\} = \rho(x)$, where $\Omega(A)$ is the set of characters in $A$, $\hat{x} $ is the Gelfand Transform, and $\rho$ is the spectral radius.
Show that $A$ admits a minimal set with respect to inclusion amongst all sets with properties of $F$ for $A$. If $A$ is the disk algebra, what is the minimal set as above?
Things I know:
- Since $A$ is a commutative unital Banach Algebra, $\rho(x) = \| \hat{x}\|$.
- If $A$ is the disk algebra then the Gelfand transform is the identity map.
Otherwise I have no clue on how to construct the minimal set.
Any help will be appreciated!
To see that a minimal $F$ exists, we can use Zorn's Lemma. We consider the set of all weak$^*$-closed $F\subset \Omega(A)$ (the unit ball of $A^*$) with reverse inclusion as order. If $\{F_j\}$ is a decreasing chain of sets satisfying $$ \rho(x)=\sup\{|\hat x(\varphi)|:\ \varphi\in F_j\},\qquad x\in A, $$ then it has the finite-intersection property. As these sets are weak$^*$-compact, they have the finite intersection property, so $\displaystyle\bigcap_jF_j\ne\emptyset$. So the chain has an upper bound. By Zorn's Lemma, a minimal $F$ exists.
Knowing that the characters are the point evaluations, and since every function in the disk algebra achieves its extremes on the boundary, the obvious candidate for the minimal set is the boundary.
For every $f\in A(\mathbb D)$, we have that $\sigma(f)=f(\overline{\mathbb D})$. So $$ \rho(f)=\max\{|f(z)|:\ z\in\overline{\mathbb D}\}=\max\{|f(z)|:\ |z|=1\}. $$ The second equality shows that the unit circle $\mathbb T$ satisfies the "F-condition". It remains to show that any proper subset of $\mathbb T$ does not satisfy the condition.
Let $S\subsetneq\mathbb T$ be closed. So there exists $z_0$ with $|z_0|=1$ and $z_0\not\in S$. Let $g:\mathbb T\to\mathbb C$ be given by $g(z)=z_0-z$. Then $\rho(g)=2$. Write $z_0=e^{i\gamma}$. For any $s=e^{i\theta}\in S$, $$ |g(s)|=|e^{i\gamma}-e^{i\theta}|=|1-e^{i(\theta-\gamma)}|=2(1-\cos(\theta-\gamma)). $$ Since $z_0\not\in S$, we get that $|g(s)|<2$ for all $s\in S$. As $S$ is compact, $$\sup\{|g(s)|:\ s\in S\}<2,$$ and so $S$ fails the $F$ property. So indeed $\mathbb T$ is the minimal set.