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.
20.3.5 Comparison with ordinary restriction95B8C7040FB9
When the restriction $u|_ Y$ is not identically $-\infty $, the restriction theorem for multiplier ideals bounds the trace data by its divisorial Lelong numbers. For analytic singularities, comparison with the approximants gives the reverse inequality.
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 $\nu (u,Y)=0$. $\square$
The inequality can be strict, as the following local example shows. On a small bidisk in $\mathbb C^2$ with coordinates $(z,w)$, write
\[ u(z,w)=\max \{ -\log (-\log |z|^2),\ \log |w|^2\} , \qquad Y=\{ z=0\} . \]
The bidisk is chosen with $|z|<\mathrm{e}^{-1}$, and the first function is extended by $-\infty $ along $z=0$. It is plurisubharmonic, since $s\mapsto -\log (-s)$ is increasing and convex for $s<0$, and hence so is the maximum. For every $t>0$,
\[ \mathrm{e}^{-tu}\leq (-\log |z|^2)^ t, \]
which is locally integrable. Thus all multiplier ideals of $u$ are trivial and the trace data vanishes, whereas $u|_ Y=\log |w|^2$ has Lelong number one at the origin.
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