PomeProposition 20.3.6.FE94B20EED19 If $v=u|_ Y$ is not identically $-\infty $, then
\[ \tau _{Y,u}(E)\leq \nu (v\circ \pi ,E) \]
for every divisor $E$ over $Y$. If $u$ has analytic singularities, equality holds.
Proposition 20.3.6FE94B20EED19
Proof. The restriction theorem, which is the first inclusion in (5.37.6), and lemma 5.38.9, applied on $Y$, give for every $t>0$
\[ \operatorname {ord}_ E\mathfrak a_ t(u;Y) \leq \operatorname {ord}_ E\mathcal I(tv) \leq t\nu (v\circ \pi ,E). \]
Divide by $t$ and use theorem 20.3.2.
If $u$ has analytic singularities, apply lemma 5.38.12 to the pair $(u,0)$, with $s=u$, $U_ m=u_ m$ and $V_ m=0$. It gives
\[ u_ m\leq (1-\delta _ m)u+O(1),\qquad \delta _ m\downarrow 0. \]
Restricting to $Y$, pulling back by $\pi $ and taking the generic Lelong number along $E$, we obtain, since $u_ m$ has analytic type $(1/m,\mathcal I(mu))$,
\[ \frac1m\operatorname {ord}_ E\mathfrak a_ m(u;Y) \geq (1-\delta _ m)\nu ((u|_ Y)\circ \pi ,E). \]
Letting $m\to \infty $ proves the reverse inequality, hence equality. All the restrictions involved are quasi-plurisubharmonic, because $\nu (u,Y)=0$. $\square$
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