Lemma 21.11.60.B1B557E6282F Let $\alpha ,\beta _1,\ldots ,\beta _ m\in \mathbb {Z}^ n$. Let $\Delta $ be the polytope generated by $\beta _1,\ldots ,\beta _ m$. Then the following are equivalent:
45E8CA3F6710
EF4C4E285442
21.11.61
\begin{equation} |z^{\alpha }|^2\left(\sum _{i=1}^ m |z^{\beta _ i}|^2 \right)^{-1} \end{equation}
is a bounded function on $\mathbb {C}^{*n}$.
B62429A6EB39
$\alpha \in \Delta $.
Proof.
2 $\implies $ 1. Write $\alpha =\sum _ i t_ i\beta _ i$, where $t_ i\in [0,1]$, $\sum _ i t_ i=1$. Then
\[ \begin{aligned} |z^{\alpha }|^2\left(\sum _{i=1}^ m |z^{\beta _ i}|^2 \right)^{-1} & =\prod _ i |z^{\beta _ i}|^{2t_ i}\left(\sum _{i=1}^ m |z^{\beta _ i}|^2 \right)^{-1}\\ & \leq \prod _ i\left(\sum _{j=1}^ m |z^{\beta _ j}|^2\right)^{t_ i} \left(\sum _{i=1}^ m |z^{\beta _ i}|^2 \right)^{-1}=1. \end{aligned} \]
1 $\implies $ 2. Assume that $\alpha \not\in \Delta $. By the strict separation theorem, there is $a=(a_1,\ldots ,a_ n)\in \mathbb {R}^ n$ such that
\[ a\cdot \alpha >\max _ i a\cdot \beta _ i. \]
Set
\[ z(t)\coloneqq (t^{a_1},\ldots ,t^{a_ n}). \]
Then (21.11.61) evaluated at $z(t)$ grows like
\[ \frac{t^{2a\cdot \alpha }}{\sum _ i t^{2a\cdot \beta _ i}} \]
as $t\to \infty $, and hence is not bounded.
$\square$
Pome
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