Exercise 27, § 3

Let A be a commutative unital Banach algebra over $\mathbf{C}$. A closed set F of $\mathsf{X}(A)$ is called a boundary of A if one has

sup$_{\chi\in\mathsf{X}(A)}|\chi (x)|=$ sup$_{\chi\in F}|\chi (x)|$

for all $x\in A$.

a) Order the set of boundaries of A by inclusion. There exists a minimal boundary.

b) Let S be a minimal boundary of A. Every boundary of A contains S, hence S is the intersection of all the boundaries of A. (Let F be a boundary of A; now prove that for all $x\in S$, every neighbourhood U of $x$ meets F; use for this the fact that S - U is not a boundary of A.)

We say that S is the Shilov boundary of A, and denote it by $\partial A$.

c) If the image of the Gelfand transform of A is dense in $\mathscr{C}(\mathsf{X}(A))$, the Shilov boundary of A is equal to $\mathsf{X}(A)$.

d) Let A be the algebra of continuous functions on the unit disk Δ of the $z\in \mathbf{C}$ such that $|z|\leqslant 1$ which are holomorphic in the interior of Δ (example 9 of I, p. 20). Then $\mathsf{X}(A)$ identifies with Δ (exercise 6 of I, p. 193) and the Shilov boundary of A identifies with the unit circle $\mathbf{U}$.