Let $H^\infty (\mathbb{B}_n)$ be the space of all bounded holomorphic functions from the open unit ball $\mathbb{B}_n$ of $\mathbb{C}^n$ $(n\geq 1)$ to $\mathbb{C}$. It is a Banach algebra. Let $\mathscr{M}_n$ be the maximal ideal space of $H^\infty (\mathbb{B}_n)$. Let $\Delta = \{\phi_z: z\in \mathbb{B}_n\}$ be the evaluation functionals in $\mathscr{M}_n$.
The Shilov boundary is the smallest closed subset of $\mathscr{M}_n$ on which every element of the Gelfand transform of $H^\infty (\mathbb{B}_n)$ attains its supremum.
My question is if $\mathscr{M}_n - \Delta$ is in the Shilov boundary of the Banach algebra $H^\infty (\mathbb{B}_n)$.
Thanks.