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Let $A$ be a complex Banach algebra and $M_A$ be the space of all non-zero multiplicative linear functionals on $A$ equipped with the weak$^*$-topology. Let $\widehat A$ be the image of $A$ under the Gelfand transform. Then a set $E \subseteq M_A$ is said to be a boundary for the Banach algebra $A$ if every $\widehat f \in \widehat A$ admits its maximum on $E$ i.e. for any $\psi \in M_A$ and $f \in A$ we have $$\left \lvert \psi (f) \right \rvert \leq \sup\limits_{\phi \in E} \left \lvert \phi (f) \right \rvert$$

Shilov showed that intersection of all the closed boundaries of $A$ is again a closed boundary of $A$ which is known as the Shilov boundary of $A.$

But I don't know what is meant by saying the Shilov boundary of the polydisc $\mathbb D^n \subseteq \mathbb C^n$ is $\mathbb T^n$ because $\mathbb D^n$ is not even a closed subset of $\mathbb C^n.$ Could anyone kindly give me some suggestion regarding this?

Thanks in advance.

EDIT $:$ I think the Shilov boundary for $\mathbb D^n$ is defined as follows $:$

Let $\mathcal C$ be the collection of all complex valued functions which are defined and holomorphic on a neighbourhood of $\overline {\mathbb D^n}\ (= \overline {\mathbb D}^n).$ Then a set $F \subseteq \overline {\mathbb D^n}$ is said to be a boundary of $\mathbb D^n$ if for every $f \in \mathcal C$ we have $$\max\limits_{z \in \overline {\mathbb D^n}} \left \lvert f(z) \right \rvert = \sup\limits_{z \in F} \left \lvert f(z) \right \rvert.$$ Then the Shilov boundary is again defined to be the smallest closed boundary of $\mathbb D^n.$ By applying maximum modulus principle $n$-times it is easy to see that for every $f \in \mathcal C$ $$\max\limits_{z \in \overline {\mathbb D^n}} \left \lvert f(z) \right \rvert = \max\limits_{z \in \mathbb T^n} \left \lvert f(z) \right \rvert.$$ So in order to show that $\mathbb T^n$ is the Shilov boundary of $\mathbb D^n$ all we need to do is to show that for any closed subset $A \subseteq \mathbb T^n$ there exists $f \in \mathcal C$ such that $f$ attains its maximum on $\mathbb T^n \setminus A.$ But how do I show that?

Any help in this regard would be greatly appreciated.

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  • $\begingroup$ For the last step, for $z \in T^n$ and $w \in \Bbb C^n$, define $f_z(w) = (z_1+w_1) \cdots (z_n+w_n)$. The maximum modulus this takes on the polydisc is $2^n$ at $w=z$. Hence if $A \subsetneq T^n$ we can consider $f_z$ for $z \not\in A$. $\endgroup$
    – mme
    Commented Dec 27, 2023 at 15:09

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