Proposition 13.18.3.169DBA6BAF3B Let $\pi :Z\to X$ be a smooth compact modification. There is a finite constant $C_\pi $, depending only on $\pi $ and $\theta $, such that, for $u,v,\delta ,D$ as in lemma 13.18.1,
\[ 0\leq \nu (u\circ \pi ,z)-\nu (v\circ \pi ,z) \leq C_\pi \min \left\{ 1,\left(\frac D\delta \right)^{1/n}\right\} \qquad (z\in Z). \]
The same estimate holds for generic Lelong numbers along every prime divisor on $Z$.
Proof.
Let $C_\pi $ be the uniform bound for Lelong numbers of $\pi ^*\theta $-plurisubharmonic functions on $Z$ given by lemma 5.38.8. Since $u\preceq v$, the difference of Lelong numbers is nonnegative, and it is at most $\nu (u\circ \pi ,z)\leq C_\pi $. This proves the assertion when $D\geq \delta $.
If $D<\delta $, then by lemma 13.18.1 every $b>1$ with $b^ nD<\delta $ admits some $h\in \mathcal A_ b(u,v)$. Pulling back the defining convex inequality and taking Lelong numbers gives
\[ \nu (u\circ \pi ,z)-\nu (v\circ \pi ,z) \leq b^{-1}\bigl (\nu (h\circ \pi ,z)-\nu (v\circ \pi ,z)\bigr ) \leq b^{-1}C_\pi . \]
For $D>0$, let $b\uparrow (\delta /D)^{1/n}$; for $D=0$, let $b\to \infty $. Taking generic values along a prime divisor proves the divisorial assertion.
$\square$
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