Pome

Theorem 14.3.5EEF74F1DEBB8

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 $m=\inf _ Xu$,

0F99936A9B52 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. $\square$

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