PomeProposition 13.15.2.57A94BD59310 There is a constant $C_\omega $ such that, for every $A\geq 0$,
\[ \int _ X(\omega +\mathrm{dd}^{\mathrm{c}}v)^ n\leq C_\omega (1+A)^{r_ n} \quad \text{if }v\in \mathrm{PSH}(X,\omega ),\ 0\leq v\leq A. \]
13.15.1 Range and restricted massBD96324E19C7
We first bound the Monge–Ampère mass of a bounded potential polynomially in its range. The first mixed mass $\int _ X(\omega +\mathrm{dd}^{\mathrm{c}}v)\wedge \omega ^{n-1}$ is bounded independently of $v$ by the Gauduchon trace. Local Chern–Levine–Nirenberg estimates turn this into a bound of degree $j-1$ in the range for the $j$th mixed mass, and Stokes' theorem expresses the top mass in terms of the lower ones.
Proof. Choose finitely many coordinate balls whose concentric smaller balls, with compact closure in the larger ones, cover $X$. On each larger ball choose a fixed quadratic function $q$ with $\mathrm{dd}^{\mathrm{c}}q\geq \omega $, shifted so that $p=v+q$ is plurisubharmonic with $0\leq p\leq A+C$. Let $\kappa $ be the Euclidean form. In the identity
\[ \mathrm{dd}^{\mathrm{c}}p=(\omega +\mathrm{dd}^{\mathrm{c}}v)+(\mathrm{dd}^{\mathrm{c}}q-\omega ), \]
the integral of the first term against $\kappa ^{n-1}$ is bounded by a fixed multiple of the global trace mass, which is bounded by corollary 11.5.3, and the second term is smooth and fixed. Thus $\int \mathrm{dd}^{\mathrm{c}}p\wedge \kappa ^{n-1}$ is bounded uniformly in $v$ and $A$; in dimension one, Stokes' theorem replaces the trace bound. By (5.27.3), on each smaller ball,
\[ \int (\mathrm{dd}^{\mathrm{c}}p)^ j\wedge \kappa ^{n-j}\leq C_ j(1+A)^{j-1} \quad (1\leq j\leq n). \]
Expand $(\omega +\mathrm{dd}^{\mathrm{c}}v)^ j=(\mathrm{dd}^{\mathrm{c}}p+\omega -\mathrm{dd}^{\mathrm{c}}q)^ j$ and sum over the finite cover. Since the smooth factors have bounded coefficients, the positive mixed masses satisfy
\[ 0\leq M_ j(v):=\int _ X(\omega +\mathrm{dd}^{\mathrm{c}}v)^ j\wedge \omega ^{n-j} \leq C_ j(1+A)^{\max (j-1,0)} \quad (0\leq j\leq n). \]
For smooth $v$, the exact form $(\mathrm{dd}^{\mathrm{c}}v)^ n$ integrates to zero, as in the proof of theorem 11.1.5; the regularization of theorem 5.32.1 and the continuity of bounded products give the same identity for bounded $v$. Thus
\[ 0=\int _ X(\mathrm{dd}^{\mathrm{c}}v)^ n =\sum _{j=0}^ n(-1)^{n-j}\binom nj M_ j(v). \]
Since the mixed masses are nonnegative, it follows that
\[ M_ n(v)\leq \sum _{j=0}^{n-1}\binom nj M_ j(v) \leq C_\omega (1+A)^{r_ n}. \]
The largest power is $n-2$ when $n\geq 2$; for $n=1$, only the constant term remains. $\square$
For $n\leq 2$ the bound is uniform, so every compact complex curve or surface has the bounded mass property. In higher dimensions the bound grows with the range, and truncation turns it into a bound for the mass of an unbounded potential above a finite depth.
Lemma 13.15.3.357B2DE7779E Let $u\in \mathrm{PSH}(X,\omega )$ be nonpositive. For $t\geq 0$ and every Borel set $E\subset X$,
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13.15.4
\begin{equation} \mu _ u(\{ u\geq -t\} )\leq C_\omega (1+t)^{r_ n}, \qquad \mu _ u(E\cap \{ u\geq -t\} )\leq (1+t)^ n\operatorname {Cap}_\omega (E). \end{equation}
Proof. Put $v_ t=\max (u,-t-1)+t+1$. It is $\omega $-plurisubharmonic and takes values in $[0,t+1]$, and by plurifine locality its Monge–Ampère measure equals $\mu _ u$ on $\{ u\geq -t\} \subset \{ u>-t-1\} $. The first estimate therefore follows from proposition 13.15.2, after enlarging the constant. For the second, $v_ t/(1+t)$ is a test function for the capacity, and positivity gives
\[ (\omega +\mathrm{dd}^{\mathrm{c}}v_ t)^ n \leq (1+t)^ n\bigl (\omega +\mathrm{dd}^{\mathrm{c}}(v_ t/(1+t))\bigr )^ n. \]
Restricting this inequality to $E\cap \{ u\geq -t\} $ proves the capacity bound. $\square$
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