Shilov boundary

Remarks

The existence of a smallest closed boundary is a theorem of Shilov: the intersection of all closed boundaries is again a boundary.

Example: the disc algebra $A(\mathbb{D})$ of continuous functions on the closed unit disc $\overline{\mathbb{D}}$ that are holomorphic in the open disc, with the sup norm. Its character space is $\overline{\mathbb{D}}$, through the evaluations at points. By the maximum modulus principle, the unit circle $\mathbb{T}$ is a closed boundary. No proper closed subset $F \subsetneq \mathbb{T}$ is a boundary. Indeed, let $\zeta \in \mathbb{T} \setminus F$; the function $f(z) = (1 + \overline{\zeta} z) / 2$ satisfies $\abs{f} < 1$ on $\overline{\mathbb{D}} \setminus \set{\zeta}$ and $f(\zeta) = 1$. So its supremum on the compact set $F$ is less than $1$. Hence $\partial A(\mathbb{D}) = \mathbb{T}$.

Non-example: for the disc algebra, the circle $\set{\abs{z} = 1/2}$ is closed but not a boundary, since $f(z) = z$ has maximum modulus $1/2$ there and $1$ on $\overline{\mathbb{D}}$.

For $A = C(X)$ with $X$ compact Hausdorff, the Shilov boundary is all of $X$. Indeed, if $F \subsetneq X$ is closed and $x \notin F$, Urysohn's lemma gives $f \in C(X)$ with $f(x) = 1$ and $f = 0$ on $F$, so $F$ is not a boundary.