Lemma 10.31.14.E0503BA0806A Let $u\in \mathrm{PSH}(X,\theta )$ satisfy $u\sim V$ and $u\leq V$. Let $\chi :\mathbb R_{\leq 0}\to \mathbb R_{\leq 0}$ be increasing and concave, with $\chi (0)=0$ and $\chi '\geq 1$, and assume that $\chi $ is $C^{1,1}$ on compact intervals. Set
\[ b=V+\chi (u-V),\qquad p=P_\theta (b), \]
where the obstacle $b$ is defined off the pole set of $V$ and is set equal to $-\infty $ on that set. Then $p\sim V$ and
\[ \mu _ p\leq \mathbf1_{\{ p=b\} }\chi '(u-V)^ n\mu _ u. \]
The measures and contact sets in this inequality may be restricted to the complement of the pole set of any fixed strict analytic barrier for $\theta $, that is, of any $\rho \in \mathrm{PSH}(X,\theta )$ with analytic singularities such that $\theta +\mathrm{dd}^{\mathrm{c}}\rho \geq a\omega $ for some $a>0$.
Proof.
The difference $u-V$ is bounded, so $V-C\leq b\leq u\leq V$ for some constant $C$. Hence $p\sim V$. The obstacle is quasi-continuous: away from sets of arbitrarily small capacity, $u$ and $V$ are finite and continuous, and their difference has a fixed bounded range. The general contact theorem theorem 15.8.6 therefore applies.
Choose a strict analytic barrier $\rho $ and translate it so that $\rho \leq V$; put $\Omega =X\setminus \{ \rho =-\infty \} $. On this open set $u$, $V$ and $p$ are locally bounded. Let $\tau =\chi ^{-1}$; then $\tau $ is convex and $0<\tau '\leq 1$. By lemma 5.28.13, the function
\[ w=V+\tau (p-V) \]
is $\theta $-plurisubharmonic on $\Omega $ and satisfies
\[ \mathrm{MA}_\theta (w)\geq \tau '(p-V)^ n\mu _ p. \]
Since $p\leq w\leq V$ on $\Omega $, the removability lemma lemma 5.24.7 extends $w$ across the analytic set $X\setminus \Omega $, preserving its smooth lower curvature bound.
The inequality $p\leq b$ implies $w\leq u$, with equality on $D=\{ p=b\} $. Ordered contact gives
\[ \mathbf1_ D\tau '(p-V)^ n\mu _ p \leq \mathbf1_ D\mathrm{MA}_\theta (w) \leq \mathbf1_ D\mu _ u. \]
On $D$ the derivative $\tau '(p-V)$ equals $1/\chi '(u-V)$. Together with contact support for $p$, this proves the claim on $\Omega $. The analytic set $X\setminus \Omega $ is not charged by the non-pluripolar measures involved.
$\square$
Pome
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