Timeline for What is meant by saying that the Shilov boundary of the polydisc $\mathbb D^n$ is $\mathbb T^n\ $?
Current License: CC BY-SA 4.0
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| Dec 27, 2023 at 15:09 | comment | added | mme | 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 $z \not\in A$. | |
| Dec 27, 2023 at 9:58 | history | edited | Anacardium | CC BY-SA 4.0 |
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| Dec 27, 2023 at 9:01 | history | edited | Anacardium | CC BY-SA 4.0 |
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| Dec 27, 2023 at 7:56 | history | asked | Anacardium | CC BY-SA 4.0 |