Lemma 6.9.16947654B2D6F

Proof. We work on a bounded polydisc $\Omega $ around $x$ in a chart on which $\varphi $ is bounded above. Subtracting a constant changes neither side of (6.9.2), so we may assume $\varphi \leq 0$ on $\Omega $, and we write $\nu =\nu (\varphi ,x)$. The space $\mathcal H_{m\varphi }(\Omega )$ is not zero: choose $a\in \Omega $ with $\varphi (a)>-\infty $ and apply corollary 6.5.8. By theorem 6.8.9(3), $\mathcal I(m\varphi )_ x$ is generated by germs of elements of $\mathcal H_{m\varphi }(\Omega )$; since $\Omega $ is connected, a nonzero element of $\mathcal H_{m\varphi }(\Omega )$ has a nonzero germ at $x$, by the identity theorem. Hence the order $k$ of $\mathcal I(m\varphi )$ at $x$ is the smallest vanishing order at $x$ of the nonzero elements of $\mathcal H_{m\varphi }(\Omega )$.

For the lower bound, let $F\in \mathcal H_{m\varphi }(\Omega )$ vanish to order exactly $k$ at $x$, and let $P_ k\neq 0$ be its homogeneous term of degree $k$. By (6.7.3), computed in the coordinates of the chart, and since $\varphi \leq 0$, we have $\varphi (z)\leq \nu \log (|z-x|^2/r_0^2)$ for $0<|z-x|<r_0$ and a suitable $r_0>0$; hence $\mathrm e^{-m\varphi (z)}\geq c|z-x|^{-2m\nu }$ there, with $c=r_0^{2m\nu }$. By (6.8.8) and polar coordinates (3.1.2), for small $r>0$

with $c'>0$. The left side is finite, so $2k-2m\nu +2n>0$, that is, $k>m\nu -n$.

For the upper bound, we use that every element of $\mathcal H_{m\varphi }(\Omega )$ vanishes to order at least $k$ at $x$. Choose $R>0$ such that the polydisc $\Delta $ of polyradius $(2R,\ldots ,2R)$ centered at $x$ is relatively compact in $\Omega $. Let $\delta >0$ be half the distance from $\overline\Delta $ to the complement of $\Omega $. Put $A=|B(0,\delta )|^{-1/2}$. For $G\in \mathcal H_{m\varphi }(\Omega )$, (6.8.7) gives $\sup _{\overline\Delta }|G|\leq A\| G\| _{L^2(\Omega )}\leq A\| G\| _{m\varphi }$, and the coefficient bound in (5.2.1) gives $|c_\alpha |\leq (2R)^{-|\alpha |}\sup _{\overline\Delta }|G|$ for the Taylor coefficients $c_\alpha $ of $G$ at $x$. Since $c_\alpha =0$ for $|\alpha |<k$ and $|(a-x)^\alpha |\leq |a-x|^{|\alpha |}$, for $|a-x|<R$ we obtain

For the second inequality, note that there are $\binom {d+n-1}{n-1}$ multi-indices of length $d$; with $\rho =|a-x|/(2R)\leq 1/2$ we get $\sum _{d\geq k}\binom {d+n-1}{n-1}\rho ^ d\leq \rho ^ k\sum _{e\geq 0}\binom {e+k+n-1}{n-1}2^{-e}$, a convergent series. Now let $|a-x|<R$ with $\varphi (a)>-\infty $, and let $G$ be given by corollary 6.5.8, so that $G(a)=1$ and $\| G\| _{m\varphi }^2\leq C_\Omega \mathrm e^{-m\varphi (a)}$, where $C_\Omega $ depends only on $n$ and the diameter of $\Omega $. Then $1\leq C_ k^2C_\Omega |a-x|^{2k}\mathrm e^{-m\varphi (a)}$, that is,

here $C'=m^{-1}\log (C_ k^2C_\Omega )$, and the inequality holds trivially where $\varphi (a)=-\infty $. Since it holds for all $a$ with $|a-x|<R$, the characterization in proposition 6.7.1 gives $k/m\leq \nu $. $\square$