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Let $\Pi\subset\subset U\subset\mathbb{C}^n$ be an analytic polyhedron $$\Pi=\{z\in U:|f_j(z)|<1,1\le j\le m\}$$ where $f_1,\ldots,f_m\in H(U)$, the following equality holds?

$$\overline{\Pi}=\{z\in U:|f_j(z)|\le1,1\le j\le m\}$$

Any hint would be appreciated.

felipeuni
  • 5,080

1 Answers1

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It's clear (by continuity of the $f_j$) that $$\overline{\Pi} \subset \{z\in U:|f_j(z)|\le1,1\le j\le m\}.$$

Conversely, if $|f_j(w)| = 1$, then every neighborhood of $w$ contains a point where $|f_j| < 1$ (otherwise $|f_j| \ge 1$ on $\partial B$ for every small ball centered at $w$, but then $f_j$ is zero-free on $B$, since zero sets can't be compact, so $1/f_j$ is holomorphic on $B$ which would violate the maximum modulus principle). So $\partial \Pi$ contains the union of the hypersurfaces where $|f_j| = 1$, i.e. $$\overline{\Pi} \supset \{z\in U:|f_j(z)|\le1,1\le j\le m\}.$$

mrf
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  • Notice that if $\Pi={(z,w)\in \mathbb{C}^2:|z^2-2|<1,|z-1|<2,|w|<1}$

    ${(z,w)\in \mathbb{C}^2:|z^2-2|\le1,|z-1|\le2,|w|\le1}\not\subset \overline{\Pi}$, consider $(z,w)=(-1,0)$, but if the domain $U$ is reduced, the equality holds.

     – felipeuni Sep 19 '15 at 04:47
  • You are right, I unconsciously assumed that. – mrf Sep 19 '15 at 06:09