Proposition 15.14.2.0A5580561320 Let $\gamma $ be a Gauduchon metric on $X$, write $\gamma ^ n=G\, \mathrm{d}V$, and put
\[ T_\gamma =\int _ X\theta \wedge \gamma ^{n-1}. \]
Then $T_\gamma >0$. If $u\in \mathrm{PSH}(X,\theta )$ satisfies $u\geq \rho -C$ and $\mathrm{MA}_\theta (u)\geq F\, \mathrm{d}V$ with $F\geq 0$, then
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15.14.3
\begin{equation} \int _ X F^{1/n}G^{1-1/n}\, \mathrm{d}V\leq T_\gamma . \end{equation}
Consequently, let $0\leq f\in L^ p(\mathrm{d}V)$, $p>1$, be nonzero, and let $\lambda >0$. Every solution of $\mathrm{MA}_\theta (u)=\mathrm{e}^{\lambda u}f\, \mathrm{d}V$ that is bounded below by $\rho $ minus a constant satisfies $\sup _ Xu\leq C'$, where $C'$ depends only on $\theta $, $\gamma $, $f$ and $\lambda $. Moreover, if $v_ j\in \mathrm{PSH}(X,\theta )$, $j\geq 1$, and constants $C_ j\geq 0$ and $c_ j>0$ satisfy
\[ \sup _ Xv_ j=0,\qquad \mathrm{MA}_\theta (v_ j)=c_ j \mathrm{e}^{v_ j/j}f\, \mathrm{d}V, \qquad v_ j\geq \rho -C_ j, \]
then the constants $c_ j$ are uniformly bounded above.
Proof.
The current $\theta _ u\wedge \gamma ^{n-1}$ is linear in $\theta _ u$, so it is a globally defined positive current, even at $Z$. Since $\mathrm{dd}^{\mathrm{c}}\gamma ^{n-1}=0$, Stokes' theorem in the sense of currents gives
\[ \int _ X\theta _ u\wedge \gamma ^{n-1} =\int _ X\theta \wedge \gamma ^{n-1}. \]
Applied to $\rho $, this identity proves $T_\gamma \geq \int _ X\omega \wedge \gamma ^{n-1}>0$.
On $\Omega $, lemma 5.28.7, applied to $\theta _ u$ and $\gamma $, gives
\[ \theta _ u\wedge \gamma ^{n-1}\geq F^{1/n}G^{1-1/n}\, \mathrm{d}V. \]
Exhausting $\Omega $ and using positivity of the global trace measure, we obtain (15.14.3), since the complement $Z$ has zero volume.
For the exponential equation, write $a=\sup _ Xu$ and $w=u-a$. The estimate becomes
\[ \mathrm{e}^{\lambda a/n}\int _ X\mathrm{e}^{\lambda w/n}f^{1/n}G^{1-1/n}\, \mathrm{d}V \leq T_\gamma . \]
The integral has a positive lower bound over all normalized $w\in \mathrm{PSH}(X,\theta )$. Otherwise, normalized compactness would give a sequence of such $w$ with integrals tending to zero and converging almost everywhere to a $\theta $-plurisubharmonic function; since $w\leq 0$ and $f^{1/n}G^{1-1/n}$ is integrable, dominated convergence would give a strictly positive limit integral, a contradiction. This bounds $a$ from above.
For the last assertion, the same estimate contains the factors $c_ j^{1/n}$ and $\mathrm{e}^{v_ j/(jn)}$. If the integrals $\int _ X\mathrm{e}^{v_ j/(jn)}f^{1/n}G^{1-1/n}\, \mathrm{d}V$ tended to zero along a subsequence with $j\to \infty $, normalized compactness and almost-everywhere convergence would give $v_ j/j\to 0$ almost everywhere along a further subsequence. Dominated convergence would then give the positive limit $\int _ Xf^{1/n}G^{1-1/n}\, \mathrm{d}V$. This contradiction bounds $c_ j$ for all large $j$, and each of the finitely many remaining constants is finite. Neither argument uses the total Monge–Ampère mass.
$\square$
Pome
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