D57A1C84047D
We now show that an upper bound for the density controls the normalized potential, following the method of Cherrier. Integrating the equation against $\mathrm{e}^{-pu}$ produces a gradient term and mixed weighted masses; a finite recurrence absorbs the latter together with the torsion error. Moser iteration then gives a set of uniformly positive volume near the minimum, and normalized $L^1$ compactness bounds that minimum. Norms below are computed with respect to the fixed metric $\omega $.
Lemma 14.3.1.E2F31793530D There is a constant $C$ depending on $\omega $ such that, for every positive $(1,1)$-form $\eta $, every smooth real $f$, every $\delta >0$, and $0\leq k\leq n-2$,
\[ \begin{split} |\mathrm d^{\mathrm c}f\wedge \mathrm{d}\omega \wedge \eta ^ k\wedge \omega ^{n-k-2}| \leq {}& \frac C\delta \, \mathrm{d}f\wedge \mathrm d^{\mathrm c}f\wedge \eta ^ k\wedge \omega ^{n-k-1}\\ & +C\delta \, \eta ^ k\wedge \omega ^{n-k}. \end{split} \]
Here the absolute value of a top-degree form is the absolute value of its coefficient relative to the positive volume form $\omega ^ n$.
Proof.
At a given point, choose an $\omega $-unitary frame in which $\eta $ is diagonal, with positive eigenvalues $\lambda _1,\ldots ,\lambda _ n$. Expand $\eta ^ k$ as a sum over $k$-element subsets $I$ of $\{ 1,\ldots ,n\} $, with coefficients that are positive dimensional multiples of $\lambda _ I=\prod _{i\in I}\lambda _ i$. After wedging with $\mathrm d^{\mathrm c}f\wedge \mathrm{d}\omega \wedge \omega ^{n-k-2}$, each surviving term is a bounded coefficient of $\mathrm{d}\omega $ times $\lambda _ I$ times one first derivative $f_ a$ with $a\notin I$. Young's inequality bounds its absolute value by
\[ C\lambda _ I|f_ a|\leq (C/\delta )\lambda _ I|f_ a|^2 +C\delta \lambda _ I. \]
The sum of the first terms is bounded by a fixed multiple of the first form on the right of the assertion, whose diagonal expansion consists of the terms $\lambda _ I|f_ a|^2$ with $a\notin I$. The second form on the right contains all the $\lambda _ I$. Since there are finitely many indices and $\mathrm{d}\omega $ is bounded, the constant $C$ is uniform.
$\square$
Lemma 14.3.2.D570CDA4FA55 Suppose $u$ is smooth, $\eta =\omega +\mathrm{dd}^{\mathrm{c}}u>0$, and $\eta ^ n\leq \Lambda \omega ^ n$ for a fixed $\Lambda >0$. There are constants $p_0\geq 1$ and $C$ depending only on $\omega $ and $\Lambda $ such that
\[ \int _ X\mathrm{d}(\mathrm{e}^{-pu/2})\wedge \mathrm d^{\mathrm c}(\mathrm{e}^{-pu/2})\wedge \omega ^{n-1} \leq Cp\int _ X\mathrm{e}^{-pu}\omega ^ n\quad (p\geq p_0). \]
Proof.
For a fixed real $p\geq 1$, define $M_ k$ for $0\leq k\leq n$ and $J_ k$ for $0\leq k\leq n-1$ by
\[ M_ k=\int _ X\mathrm{e}^{-pu}\eta ^ k\wedge \omega ^{n-k},\qquad J_ k=\int _ X\mathrm{e}^{-pu}\mathrm{d}u\wedge \mathrm d^{\mathrm c}u\wedge \eta ^ k\wedge \omega ^{n-k-1}, \quad J=\sum _{k=0}^{n-1}J_ k. \]
All these numbers are nonnegative. The proof first bounds $pJ$ in terms of the masses $M_ k$, and then bounds these masses by a recurrence. Let $T=\sum _{k=0}^{n-1}\eta ^ k\wedge \omega ^{n-k-1}$, so that $\eta ^ n-\omega ^ n=\mathrm{dd}^{\mathrm{c}}u\wedge T$. Since $\mathrm{d}\eta =\mathrm{d}\omega $, differentiation gives
\[ \mathrm{d}T=n\sum _{k=0}^{n-2}\eta ^ k\wedge \omega ^{n-k-2}\wedge \mathrm{d}\omega . \]
The sum is empty when $n=1$. Stokes' theorem gives the exact identity
\[ \begin{split} \int _ X\mathrm{e}^{-pu}(\eta ^ n-\omega ^ n) & =\int _ X\mathrm{e}^{-pu}\mathrm{dd}^{\mathrm{c}}u\wedge T\\ & =pJ+\int _ X\mathrm{e}^{-pu}\mathrm d^{\mathrm c}u\wedge \mathrm{d}T. \end{split} \]
The density assumption bounds the left-hand side above by $(\Lambda +1)M_0$. Apply lemma 14.3.1 to the last integral, with $\delta =K/p$ for a sufficiently large fixed $K$. The contribution proportional to $J/\delta $ is then at most $pJ/2$, and after absorbing it we obtain
B38AB5EB6BF0
14.3.3
\begin{equation} pJ\leq CM_0+\frac Cp\sum _{k=0}^{n-2}M_ k. \end{equation}
This estimate alone does not suffice, because the mixed weighted masses on the right have not yet been bounded.
For $1\leq k\leq n-2$, a second integration by parts gives
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14.3.4
\begin{equation} M_ k-M_{k-1}=pJ_{k-1}+R_ k, \qquad R_ k=\int _ X\mathrm{e}^{-pu}\mathrm d^{\mathrm c}u\wedge \mathrm{d}(\eta ^{k-1}\wedge \omega ^{n-k}). \end{equation}
The derivative in $R_ k$ has two terms, with coefficients $k-1$ and $n-k$. Applying the pointwise estimate with $\delta =1/p$ to each of them, and omitting terms with negative indices, we obtain
\[ |R_ k|\leq Cp(J_{k-2}+J_{k-1}) +\frac Cp(M_{k-2}+M_{k-1}), \qquad M_{-1}=J_{-1}=0. \]
Consequently, for $p\geq 1$,
\[ M_ k\leq C(M_{k-1}+M_{k-2}+pJ). \]
Induction over the finitely many values of $k$ proves
\[ \sum _{k=0}^{n-2}M_ k\leq C(M_0+pJ). \]
For $n=2$ this is immediate, and for $n=1$ the sum is empty. Inserting this bound in (14.3.3) gives $pJ\leq CM_0+C'J$. For $p\geq 2C'$ the last term can be absorbed, and $pJ\leq CM_0$.
Finally,
\[ \int _ X\mathrm{d}(\mathrm{e}^{-pu/2})\wedge \mathrm d^{\mathrm c}(\mathrm{e}^{-pu/2})\wedge \omega ^{n-1} =\frac{p^2}{4}J_0\leq \frac{p^2}{4}J\leq CpM_0. \]
This proves the claimed estimate for all $p\geq p_0$.
$\square$
Theorem 14.3.5.EEF74F1DEBB8 For each $\Lambda >0$ there is $C_\Lambda $ such that every smooth $u$ with
\[ \omega +\mathrm{dd}^{\mathrm{c}}u>0,\qquad \sup _ Xu=0,\qquad (\omega +\mathrm{dd}^{\mathrm{c}}u)^ n\leq \Lambda \omega ^ n \]
satisfies $\| u\| _\infty \leq C_\Lambda $.
Proof.
Let $\mathrm{d}\mu =\omega ^ n/\int _ X\omega ^ n$. The Sobolev inequality for the Riemannian metric of $\omega $ gives, for some $\beta >1$ and every smooth real function $w$,
\[ \left(\int _ X |w|^{2\beta }\, \mathrm{d}\mu \right)^{1/\beta } \leq C\left(\int _ X\mathrm{d}w\wedge \mathrm d^{\mathrm c}w\wedge \omega ^{n-1} +\int _ X|w|^2\, \mathrm{d}\mu \right). \]
We may take $\beta =n/(n-1)$ if $n\geq 2$, and for example $\beta =2$ if $n=1$. Choose $p_0$ as in lemma 14.3.2. Applying the Sobolev inequality to $w=\mathrm{e}^{-pu/2}$ and using that lemma, we obtain, with $Z=\mathrm{e}^{-u}$,
\[ \| Z\| _{L^{p\beta }(\mu )}\leq (Cp)^{1/p}\| Z\| _{L^ p(\mu )} \quad (p\geq p_0). \]
Iterating with $p=p_0\beta ^ j$, $j=0,1,2,\ldots $, we obtain a bound for $\sup _ XZ$; the product of the constants is finite because $\sum _ j(1+\log (p_0\beta ^ j))/(p_0\beta ^ j)<\infty $. Hence, for a uniform $C_1\geq 1$ and $m=\inf _ Xu$,
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14.3.6
\begin{equation} \mathrm{e}^{-p_0m}\leq C_1\int _ X\mathrm{e}^{-p_0u}\, \mathrm{d}\mu . \end{equation}
The right-hand side still involves the unknown potential. Divide by $\mathrm{e}^{-p_0m}$ and set $A=p_0^{-1}\log (2C_1)$. The integrand $\mathrm{e}^{-p_0(u-m)}$ is at most one everywhere and at most $(2C_1)^{-1}$ outside $S=\{ u\leq m+A\} $. Therefore (14.3.6) implies
\[ \mu (S)\geq \frac1{2C_1}=:\delta >0. \]
Since $\sup _ Xu=0$, theorem 5.25.2 gives $\int _ X(-u)\, \mathrm{d}\mu \leq C_0$. On $S$ we have $-u\geq -m-A$, so
\[ C_0\geq \int _ X(-u)\, \mathrm{d}\mu \geq \delta (-m-A) \]
when $-m\geq A$; otherwise the desired bound is immediate. Thus $-m\leq A+C_0/\delta $, proving the theorem.
$\square$
Pome
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