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15.14 Subsolutions and the Gauduchon trace8C0AAADFFCFC

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The strict barrier also yields subsolutions within a bounded distance of any reference between $\rho $ and zero. The curvature estimate comes from composing a convex function with the difference between the reference and a barrier-based subsolution, which produces a convex combination of positive currents. Pairing the equation with a Gauduchon metric then bounds the normalizing constants through a linear trace.

Lemma 15.14.1.869D3138741D Fix $p>1$ and $L>0$. There is $m>0$ such that, whenever

\[ \rho \leq \eta \leq 0,\quad \eta \in \mathrm{PSH}(X,\theta ), \qquad g\geq 0,\quad \| g\| _{L^ p(\mathrm{d}V)}\leq L, \]

there exists $v\in \mathrm{PSH}(X,\theta )$ with

\[ \eta \leq v\leq \eta +1,\qquad \mathrm{MA}_\theta (v)\geq m g\, \mathrm{d}V. \]

The constant $m$ depends only on the fixed geometry and on $\rho ,p,L$.

Proof. Choose $1<q<p$. By Hölder's inequality and the finiteness of all polynomial moments of the fixed quasi-plurisubharmonic function $\rho $, the functions $h=g(2-\rho )^{2n}$ have uniformly bounded $L^ q$ norms. Lemma 15.3.1, applied in the Hermitian background $\omega $, gives $s\in \mathrm{PSH}(X,\omega )$ with $-1\leq s\leq 0$ and

\[ \omega _ s^ n\geq m h\, \mathrm{d}V \]

for a fixed $m>0$. Put $u_1=\rho +s$. Then $\theta _{u_1}\geq \omega _ s$ and $u_1\leq \rho \leq \eta $. On $\Omega $ define

\[ t=1+\eta -u_1,\qquad v=\eta +t^{-1}, \]

so that $1\leq t\leq 2-\rho $. Lemma 5.28.13, applied on relatively compact subsets of $\Omega $ with $a=\eta $, $h=u_1$ and $\chi (x)=(1-x)^{-1}$, gives

\[ \theta _ v=(1-t^{-2})\theta _\eta +t^{-2}\theta _{u_1} +2t^{-3}\, \mathrm{d}(\eta -u_1)\wedge \mathrm d^{\mathrm c}(\eta -u_1) \geq t^{-2}\theta _{u_1}. \]

Since $\eta \leq v\leq \eta +1$, lemma 5.24.7 extends $v$ across $Z$, and the bounds extend by lemma 5.24.3. On $\Omega $, positivity of the local products gives

\[ \mathrm{MA}_\theta (v)\geq t^{-2n}\theta _{u_1}^ n \geq (2-\rho )^{-2n}\omega _ s^ n \geq m g\, \mathrm{d}V. \]

Neither side charges $Z$, so the inequality holds on $X$. $\square$

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

B100E71E56DE 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. $\square$

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