Theorem 6.12.1671AAFAF9C564

Proof. Necessity, even with strict inequality, is proposition 6.12.12. Since the strict condition implies the non-strict one, it suffices to show that the non-strict condition implies $f\in \mathcal I(\varphi )_ x$; we may assume $f\neq 0$. We compare $\varphi $ with a weight $\varphi _ m$ with analytic singularities defined by generators of $\mathcal I(m\varphi )$, apply theorem 6.12.8 to a multiple of $\varphi _ m$, and then return to $\varphi $. Put $\lambda =1+\varepsilon /2$ and choose an integer $m\geq 2$ with $\lambda \geq m/(m-1)$, that is, $(m-1)\lambda \geq m$. On a bounded polydisc $\Omega \subset V$ around $x$ on which $\varphi $ is bounded above, choose $g_1,\ldots ,g_ r\in \mathcal H_{m\varphi }(\Omega )$ generating $\mathcal I(m\varphi )$ near $x$, and put

This is plurisubharmonic by proposition 5.10.3, since $\mathcal I(m\varphi )_ x\neq 0$ by lemma 6.9.1, so that the $g_ i$ do not all vanish identically. By (6.8.11), $\varphi \leq \varphi _ m+C_ m$ on a neighborhood $W\subset V$ of $x$. Pulling this back to a smooth model and using the characterization of Lelong numbers in proposition 6.7.1 at every point, we see that $\nu (\varphi _ m,E)\leq \nu (\varphi ,E)$ for every prime divisor $E$ over an open subset of $W$. Hence, for every prime divisor $E$ over an open neighborhood of $x$ contained in $W$ whose center contains $x$,

and this is strictly larger than $\lambda \nu (\varphi _ m,E)$: when $\nu (\varphi _ m,E)>0$ because $1+\varepsilon >\lambda $, and when $\nu (\varphi _ m,E)=0$ because the left side is at least $A_ X(E)\geq 1$.

Now let $\pi \colon Y\to W'$ be a log resolution of the ideal $(g_1,\ldots ,g_ r)$ over an open neighborhood $W'\subset W$ of $x$, with $(g_1,\ldots ,g_ r)\mathcal O_ Y=\mathcal O_ Y(-D)$ and $D=\sum _ id_ iE_ i$. Let $q\in E_ i$, and choose coordinates $w$ at $q$ as before (6.12.7), so that the components of $D$ through $q$ are coordinate hyperplanes and $E_ i=\{ w_ i=0\} $. By (6.12.7), $\log \sum _ l|g_ l\circ \pi |^2=\sum _ jd_ j\log |w_ j|^2+O(1)\leq d_ i\log |w_ i|^2+C$ near $q$, where $j$ runs over the components through $q$; since $|w_ i|\leq |w-q|$, the characterization in proposition 6.7.1 gives $\nu (\varphi _ m\circ \pi ,q)\geq d_ i/m$. If $q$ lies on no other component of $D$, then $\varphi _ m\circ \pi =(d_ i/m)\log |w_ i|^2+O(1)$ near $q$, and $\nu (\varphi _ m\circ \pi ,q)=d_ i/m$ by proposition 6.7.9. Such points exist, because $E_ i$ is not contained in the union of the other components of the support of $D$ (corollary 9.3.16). Hence $\nu (\varphi _ m,E_ i)=d_ i/m$. Every component $E_ i$ meeting $\pi ^{-1}(x)$ has a center containing $x$, so the strict inequality above reads $\operatorname {ord}_{E_ i}f+A_ X(E_ i)>(\lambda /m)d_ i$ for these components, and theorem 6.12.8, applied on $W'$ to the weight $\lambda \varphi _ m$ with $c=\lambda /m$ and $b=0$, gives $f\in \mathcal I(\lambda \varphi _ m)_ x$.

It remains to pass from $\lambda \varphi _ m$ to $\varphi $. By the definition of the generators,

near $x$. Shrinking the neighborhood and subtracting a constant from $\varphi _ m$, we may assume $\varphi _ m\leq 0$; this multiplies the functions in question by constants and changes none of the integrability statements. We split a small neighborhood of $x$ into the set where $\varphi \geq \lambda \varphi _ m$ and its complement, ignoring the null set formed by the poles of $\varphi $ and the common zeros of the $g_ i$, so that all exponents below are finite. On the first set, $|f|^2\mathrm e^{-\varphi }\leq |f|^2\mathrm e^{-\lambda \varphi _ m}$, which is integrable because $f\in \mathcal I(\lambda \varphi _ m)_ x$. On the complement we have $(m-1)\varphi <(m-1)\lambda \varphi _ m$, since $m\geq 2$; as $(m-1)\lambda \geq m$ and $\varphi _ m\leq 0$, it follows that

and therefore

where $C$ is a local bound for $|f|^2$. By (6.12.18) the right side is integrable near $x$. Hence $|f|^2\mathrm e^{-\varphi }$ is integrable near $x$, that is, $f\in \mathcal I(\varphi )_ x$. $\square$