The mixed density inequality of lemma 5.28.7 passes to unbounded potentials on common finite levels. It gives a lower bound for the extended measure of a convex combination of potentials and, by comparison with a smooth metric, a lower bound for the full trace measure.
Lemma 13.14.10.7051A18CAB21 Let $\theta $ be a smooth semipositive $(1,1)$-form and let $f,g\geq 0$ be measurable functions on $X$. Suppose $u,v\in \mathrm{PSH}(X,\theta )$ satisfy
\[ \mathrm{MA}^{\mathrm{ext}}_\theta (u)\geq f\, \mathrm{d}V, \qquad \mathrm{MA}^{\mathrm{ext}}_\theta (v)\geq g\, \mathrm{d}V. \]
Then, for $0\leq t\leq 1$,
DA998466C80C
13.14.11
\begin{equation} \mathrm{MA}^{\mathrm{ext}}_\theta (tu+(1-t)v) \geq \bigl (tf^{1/n}+(1-t)g^{1/n}\bigr )^ n\, \mathrm{d}V. \end{equation}
Proof.
The cases $t=0$ and $t=1$ are the hypotheses. Let $0<t<1$, translate $u,v$ to be nonpositive, and put $u_ k=\max (u,-k)$ and $v_ k=\max (v,-k)$. Since $\theta $ is semipositive, these are bounded $\theta $-plurisubharmonic functions, and by locality
\[ \mathrm{MA}_\theta (u_ k)\geq \mathbf1_{\{ u>-k\} }f\, \mathrm{d}V, \qquad \mathrm{MA}_\theta (v_ k)\geq \mathbf1_{\{ v>-k\} }g\, \mathrm{d}V. \]
Apply lemma 5.28.7 to each mixed term in the positive expansion of $\theta _{tu_ k+(1-t)v_ k}^{n}$. On $E_ k=\{ u>-k\} \cap \{ v>-k\} $ this gives the density on the right of (13.14.11). On the same set, plurifine locality identifies the left-hand side with the extended measure of $tu+(1-t)v$. As $k\to \infty $, the sets $E_ k$ exhaust the complement of the two pole sets, which have zero smooth volume. Monotone convergence on Borel sets proves the inequality.
$\square$
The same mixed inequality gives a lower bound for the full trace measure of an unbounded potential. Unlike the extended measure, the full current $\omega +\mathrm{dd}^{\mathrm{c}}u$ keeps its mass on the pole set; the bound below concerns only its absolutely continuous part.
Lemma 13.14.12.2CA81432B327 Let $\gamma $ be a Hermitian form. If $u\in \mathrm{PSH}(X,\omega )$ and $0\leq g\in L^1(\mathrm{d}V)$ satisfy $\mathrm{MA}^{\mathrm{ext}}_\omega (u)\geq g\, \mathrm{d}V$, then
\[ (\omega +\mathrm{dd}^{\mathrm{c}}u)\wedge \gamma ^{n-1} \geq (g\, \mathrm{d}V/\gamma ^ n)^{1/n}\gamma ^ n. \]
Proof.
For bounded $u$, lemma 5.28.7, applied in local coordinates with $T_1=\omega +\mathrm{dd}^{\mathrm{c}}u$, $T_2=\gamma $ and $k=1$, gives the asserted inequality.
In general, subtract a constant so that $u\leq 0$, and put $u_ k=\max (u,-k)$. Positivity and locality give
\[ \mathrm{MA}_\omega (u_ k)\geq \mathbf1_{\{ u>-k\} }g\, \mathrm{d}V, \]
so the bounded case applies to $u_ k$ with the density $\mathbf1_{\{ u>-k\} }g$. Since $u_ k\to u$ in $L^1$, the positive measures $(\omega +\mathrm{dd}^{\mathrm{c}}u_ k)\wedge \gamma ^{n-1}$ converge weakly to $(\omega +\mathrm{dd}^{\mathrm{c}}u)\wedge \gamma ^{n-1}$. The densities on the right converge in $L^1$, because $g^{1/n}$ is integrable and the pole set has zero smooth volume. Passing to the limit proves the assertion.
$\square$
Pome
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