Shilov boundary

In functional analysis, the Shilov boundary is the smallest  closed subset of the structure space of a commutative Banach algebra where an analog of the maximum modulus principle holds. It is named after its discoverer, Georgii Evgen'evich Shilov.

Precise definition and existence
Let $$\mathcal A$$ be a commutative Banach algebra and let $$\Delta \mathcal A$$ be its structure space equipped with the relative weak*-topology of the dual $${\mathcal A}^*$$. A closed (in this topology) subset $$F$$ of $$\Delta {\mathcal A}$$ is called a boundary of $${\mathcal A}$$ if $\max_{f \in \Delta {\mathcal A}} |f(x)|=\max_{f \in F} |f(x)|$ for all $$x \in \mathcal A$$. The set $S = \bigcap\{F:F \text{ is a boundary of } {\mathcal A}\}$ is called the Shilov boundary. It has been proved by Shilov that $$S$$ is a boundary of $${\mathcal A}$$.

Thus one may also say that Shilov boundary is the unique set $$S \subset \Delta \mathcal A$$ which satisfies
 * 1) $$S$$ is a boundary of $$\mathcal A$$, and
 * 2) whenever $$F$$ is a boundary of $$\mathcal A$$, then $$S \subset F$$.

Examples
Let $$\mathbb D=\{z \in \Complex:|z|<1\}$$ be the open unit disc in the complex plane and let $${\mathcal A} = H^\infty(\mathbb D)\cap {\mathcal C}(\bar{\mathbb D})$$ be the disc algebra, i.e. the functions holomorphic in $$\mathbb D$$ and continuous in the closure of $$\mathbb D$$ with supremum norm and usual algebraic operations. Then $$\Delta {\mathcal A} = \bar{\mathbb D}$$ and $$S=\{|z|=1\}$$.