Theorem 15.21.29826C9FA5510

Proof. Since $\rho -\sup _ X\rho \leq V_\theta $, minimal singularities give $u\geq \rho -C$ for a constant $C$. Thus $u$ is locally bounded on $\Omega $. Fix one of the balls, write it as $B=B_ R(z_0)$, let $g$ be the boundary trace of $u$ on $\partial B$, and choose a smooth function $h$ near $\overline B$ with $\mathrm{dd}^{\mathrm{c}}h=\theta $. By proposition 5.28.2, the bounded plurisubharmonic function $v=u+h$ has Bedford–Taylor measure $f\, \mathrm{d}V$ and smooth boundary trace $\varphi =g+h|_{\partial B}$.

For a smooth function $V$ and the Euclidean volume $\mathrm{d}\Lambda $,

The smooth Dirichlet theorem recalled in subsection 15.18.7, with zero background and density $f\, \mathrm{d}V$, gives a smooth strictly plurisubharmonic $V$ on $\overline B$ with $(\mathrm{dd}^{\mathrm{c}}V)^ n=f\, \mathrm{d}V$ and boundary value $\varphi $; a smooth extension $\Phi $ of $\varphi $ gives the required strict subsolution $\Phi +A(|z-z_0|^2-R^2)$ for large $A$. In dimension one this is the Dirichlet problem for the Poisson equation.

The difference $v-V$ tends to zero at every boundary point. Compactness of $\partial B$ makes this convergence uniform near the boundary: otherwise a sequence approaching the boundary with difference bounded away from zero would have a subsequence approaching one fixed boundary point. Uniqueness for the bounded Dirichlet problem, that is, the Bedford–Taylor comparison principle, gives $v=V$. Therefore $u=V-h$ is smooth on $B$, and $\theta +\mathrm{dd}^{\mathrm{c}}u=\mathrm{dd}^{\mathrm{c}}V>0$ there. Since the balls cover $\Omega $, the theorem follows. $\square$