Pome89D22AB4CBD8Proposition 20.3.4.89D22AB4CBD8 Let $v\in \mathrm{QPSH}(X)$ also have zero generic Lelong number along $Y$, and recall that $u\preceq _{\mathcal I}v$ means $\mathcal I(ku)\subseteq \mathcal I(kv)$ for every real $k>0$ (section 5.33). Then, as functions on divisors over $Y$,
$u\preceq _{\mathcal I}v$ implies $\tau _{Y,u}\geq \tau _{Y,v}$;
$\tau _{Y,au}=a\tau _{Y,u}$ for every $a>0$;
$\tau _{Y,u+v}=\tau _{Y,u}+\tau _{Y,v}$;
$\tau _{Y,\max (u,v)}\leq \min \{ \tau _{Y,u},\tau _{Y,v}\} $.
Proof. The inclusions of ideals defining $\preceq _{\mathcal I}$ persist after restriction and reverse the vanishing orders, which proves (1). Replacing $t$ by $at$ in (20.3.3) proves (2). Moreover,
Since the order of a sum of ideals is the minimum of their orders, restriction and passage to the limit prove (4); note that $\max (u,v)$ also has zero generic Lelong number along $Y$. Generic Lelong numbers of sums add, so $u+v$ has zero generic Lelong number along $Y$ as well.
For (3), subadditivity of multiplier ideals gives $\tau _{Y,u+v}\geq \tau _{Y,u}+\tau _{Y,v}$. For the reverse inequality, fix a sufficiently large integer $k$ and the analytic approximant $S_ k$ of $u+v$ given by theorem 5.36.1, of analytic type $(1/k,\mathcal I(k(u+v)))$; that theorem gives $u+v\leq S_ k$. Let $U_ m,V_ m$ be the corresponding approximants of $u$ and $v$. Lemma 5.38.12 gives
Here both $\delta _ m$ and the bounded error may depend on the fixed $k$. The restrictions of $U_ m$, $V_ m$ and $S_ k$ to $Y$ are not identically $-\infty $ by corollary 5.43.26. Restricting, pulling back to a fixed modification and taking the generic Lelong number along a divisor $E$ on it, we obtain
Letting first $m\to \infty $, with $k$ fixed, and then $k\to \infty $, the finite limits of theorem 20.3.2 give the reverse inequality. $\square$
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