Proposition 10.9.16A3F29E1887A0

Proof. Steps 1–4 show that $\omega _ u^ n$ vanishes on $X\setminus K$: on small balls outside $K$ we compare $u$ with a homogeneous balayage computed with the coefficients of $\omega $ frozen. Step 5 proves the mass bound.

Step 1. Increasing competitors. The competitors for $T_ K$ are bounded above by $\sup _ Xu$, and $u\in \mathrm{PSH}(X,\omega )$ by lemma 10.9.3(1). Cover $X$ by finitely many coordinate charts. Applying Choquet's lemma (lemma 3.4.22) in each chart, we obtain for each chart a countable family of competitors whose supremum has, on that chart, the same upper semicontinuous regularization as $T_ K$. Let $v_1,v_2,\ldots $ be an enumeration of the union of these families. On each chart, $\sup _ kv_ k$ lies between the supremum of the family chosen for that chart and $T_ K$; since the regularization is computed locally, $(\sup _ kv_ k)^*=u$. The functions $u_ j=\max \{ 0,v_1,\ldots ,v_ j\} $ belong to $\mathrm{PSH}(X,\omega )$ by lemma 10.9.3(1), increase with $j$, vanish on $K$, and are competitors; hence $0\leq u_ j\leq T_ K\leq u$, and $(\sup _ ju_ j)^*=u$, because $\sup _ kv_ k\leq \sup _ ju_ j\leq T_ K$.

Step 2. Frozen balayage. Let $G\subset X\setminus K$ be the domain of a coordinate chart, identified with an open subset of $\mathbb C^ n$, and let $L\subset G$ be compact. Since $X\setminus K$ is open in the compact manifold $X$, it is a countable union of such sets $L$, so it suffices to show that $\omega _ u^ n(L)=0$. Choose open sets $G_2\Subset G_1\Subset G_0\Subset G$ with $L\subset G_2$. For $a\in G_2$ let $h_ a$ be the real quadratic polynomial with $\mathrm{dd}^{\mathrm{c}}h_ a=\omega (a)$ described at the beginning of this section. Choose $r_0>0$ with $2r_0<\operatorname {dist}(G_2,\mathbb C^ n\setminus G_1)$. The coefficients of $\omega $ have bounded first derivatives on $G_1$, so by the mean value inequality on balls there is $C_1$ such that, for $a\in G_2$ and $0<r\leq r_0$, the ball $B(a,2r)$ lies in $G_1$ and

indeed a Hermitian matrix whose entries are bounded by $\tau $ lies between $-n\tau $ and $n\tau $ times the identity.

Fix such $a$ and $r$, put $q=\pi |z-a|^2$, so that $\mathrm{dd}^{\mathrm{c}}q=\beta $ by (5.2.9), and let $B=B(a,r)$. The functions

are bounded and plurisubharmonic on $B(a,2r)$, since they are quasi-plurisubharmonic and $\mathrm{dd}^{\mathrm{c}}U_ j=\omega _{u_ j}+(\omega (a)-\omega +\delta \beta )\geq 0$. Let $V_ j$ be the homogeneous balayage of $U_ j$ in $B$ given by corollary 7.7.5, applied on $B(a,2r)$: it equals $U_ j$ outside $B$, satisfies $V_ j\geq U_ j$ and $(\mathrm{dd}^{\mathrm{c}}V_ j)^ n=0$ in $B$, and its restriction to $B$, being a member of the family described there, has upper boundary limits at most $U_ j$ on $\partial B$.

We compare $V_ j$ with $u$. Put $\rho =q-\pi r^2$, which is nonpositive on $B$ and vanishes on $\partial B$, and let $Z_ j=V_ j-h_ a-\delta q+2\delta \rho $ on $B$. Then $Z_ j$ is quasi-plurisubharmonic and

so $Z_ j\in \mathrm{PSH}(B,\omega )$, and for $x\in \partial B$ we have $\varlimsup _{B\ni y\to x}Z_ j(y)\leq U_ j(x)-h_ a(x)-\delta q(x)=u_ j(x)$. By lemma 10.9.3(1), the function equal to $\max \{ u_ j,Z_ j\} $ on $B$ and to $u_ j$ on $X\setminus B$ belongs to $\mathrm{PSH}(X,\omega )$. It vanishes on $K$, because $B\cap K=\varnothing $, so it is a competitor for $T_ K$; hence $\max \{ u_ j,Z_ j\} \leq T_ K\leq u$ on $B$. Therefore

The balayage is increasing in the data, so $V_ j$ increases with $j$. Let $V$ be the upper semicontinuous regularization of $\lim _ jV_ j$ on $B$. Since $U_1\leq V_ j\leq u+h_ a+\delta q+2\pi \delta r^2$, the $V_ j$ are uniformly bounded, $V$ is bounded and plurisubharmonic, and $V_ j\to V$ almost everywhere by theorem 5.10.2; hence $(\mathrm{dd}^{\mathrm{c}}V)^ n=0$ on $B$ by theorem 7.6.1. Put $U=u+h_ a+\delta q$. Since $\sup _ jU_ j=\sup _ ju_ j+h_ a+\delta q$, $h_ a+\delta q$ is continuous and $(\sup _ ju_ j)^*=u$, upper regularization gives $V\geq (\sup _ jU_ j)^*=U$. The upper bound for $V_ j$ is upper semicontinuous, so it also holds for $V$. As $-2\delta \rho \leq 2\pi \delta r^2$ on $B$, we conclude

Step 3. A local mass estimate. Put $A=\mathrm{dd}^{\mathrm{c}}U$ and $D=\mathrm{dd}^{\mathrm{c}}V$ on $B$. The function $g=h_ a+\delta q$ satisfies $\mathrm{dd}^{\mathrm{c}}g=\omega (a)+\delta \beta \geq \omega $ on $B(a,2r)$, so lemma 10.9.3(2) gives $A^ n=(\mathrm{dd}^{\mathrm{c}}(u+g))^ n\geq \omega _ u^ n$ on $B$. By (10.9.14), with a fixed function $\eta $ equal to $1$ on $(-\infty ,1/16]$ and to $0$ on $[1/5,\infty )$, there is $\chi \in C^\infty _ c(B(a,r/2))$ with $0\leq \chi \leq 1$, $\chi =1$ on $B(a,r/4)$ and $-C_2r^{-2}\beta \leq \mathrm{dd}^{\mathrm{c}}\chi \leq C_2r^{-2}\beta $, where $C_2$ does not depend on $a$ and $r$. By the definition and the symmetry of Bedford–Taylor products, $A^ n-D^ n=\sum _{j=0}^{n-1}(A-D)\wedge A^ j\wedge D^{n-1-j}=\mathrm{dd}^{\mathrm{c}}\bigl ((U-V)\sum _{j=0}^{n-1}A^ j\wedge D^{n-1-j}\bigr )$. Using $D^ n=0$, integration by parts against $\chi $, the positivity of the currents $A^ j\wedge D^{n-1-j}$ and (10.9.18), we obtain

Apply lemma 10.9.15 with $k=n-1$ and $\kappa =2\pi \delta =2\pi C_1r$; for $r\leq 1/(2\pi C_1)$ we have $\kappa \leq 1$, and the last integral is at most $C_ n\sum _{l=0}^{n-1}\int _{B(a,r)}A^ l\wedge \beta ^{n-l}$.

To sum these estimates we replace $A$ by a current that does not depend on $a$ and $r$. Choose $C_0$ such that $C_0\beta \geq \omega $ on $G_0$ and $C_0\beta \geq \omega (a)+\delta \beta $ for all $a\in G_2$ and $r\leq r_0$. Then $\Gamma =\mathrm{dd}^{\mathrm{c}}u+C_0\beta =\mathrm{dd}^{\mathrm{c}}(u+C_0\pi |z|^2)$, where $u+C_0\pi |z|^2$ is bounded and plurisubharmonic on $G_0$ because $\mathrm{dd}^{\mathrm{c}}(u+C_0\pi |z|^2)=\omega _ u+(C_0\beta -\omega )\geq 0$. On $B$ the difference $\Gamma -A$ is $\mathrm{dd}^{\mathrm{c}}$ of the smooth function $C_0\pi |z|^2-h_ a-\delta q$, and it equals the form $\gamma =C_0\beta -\omega (a)-\delta \beta \geq 0$ with constant coefficients. By (10.9.1), $\Gamma ^ l=\sum _{i=0}^ l\binom liA^ i\wedge \gamma ^{l-i}$, and all terms are positive currents, so $A^ l\wedge \beta ^{n-l}\leq \Gamma ^ l\wedge \beta ^{n-l}$. Altogether, for $a\in G_2$ and $r\leq \min \{ r_0,1/(2\pi C_1)\} $,

with $C_3=2\pi C_1C_2C_ n$ independent of $a$ and $r$.

Step 4. Zero mass off $K$. Let $r$ be as above and also smaller than $4\operatorname {dist}(L,\mathbb C^ n\setminus G_2)$, and consider the cubical grid $h\mathbb Z^{2n}\subset \mathbb C^ n=\mathbb R^{2n}$ of mesh $h=r/(4\sqrt{2n})$. The grid points $a$ within distance $r/4$ of $L$ are finitely many and belong to $G_2$. Every point of $\mathbb C^ n$ lies within distance $\sqrt{2n}\, h/2=r/8$ of a grid point, so the balls $B(a,r/4)$ centered at these grid points cover $L$. The centers of the balls $B(a,r)$ containing a given point $x$ lie in the cube of side $2r$ centered at $x$, which contains at most $(2r/h+1)^{2n}=(8\sqrt{2n}+1)^{2n}=N_ n$ grid points; so each point lies in at most $N_ n$ of these balls. Summing the local estimate over these balls, which lie in $G_1$, gives

The integrals on the right are finite by theorem 7.2.5, applied in $G_0$, and do not depend on $r$. Letting $r\downarrow 0$ gives $\omega _ u^ n(L)=0$. Hence $\omega _ u^ n$ is carried by $K$.

Step 5. The mass bound. Since $T_ K\geq 0$, we have $0\leq u\leq S$. The function $v=u/S$ satisfies $0\leq v\leq 1$ and belongs to $\mathrm{PSH}(X,\omega )$ because $S\geq 1$: indeed $\omega +\mathrm{dd}^{\mathrm{c}}v=S^{-1}\omega _ u+(1-S^{-1})\omega \geq 0$. Since $v=S^{-1}u+(1-S^{-1})\cdot 0$ and the products are affine in each potential, expanding $\omega _ v^ n$ factor by factor gives $\omega _ v^ n=\sum _{k=0}^ n\binom nkS^{-k}(1-S^{-1})^{n-k}\omega _ u^ k\wedge \omega ^{n-k}$, and all terms are positive measures; hence $\omega _ v^ n\geq S^{-n}\omega _ u^ n$. As $\omega _ u^ n$ is carried by $K$,

Finally $\int _ X\omega _ u^ n>0$ by proposition 10.9.10. $\square$