PomeProposition 13.15.2.57A94BD59310 There is a constant $C_\omega $ such that, for every $A\geq 0$,
\[ \int _ X(\omega +\mathrm{dd}^{\mathrm{c}}v)^ n\leq C_\omega (1+A)^{r_ n} \quad \text{if }v\in \mathrm{PSH}(X,\omega ),\ 0\leq v\leq A. \]
Proposition 13.15.257A94BD59310
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. $\square$
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