Let $B\Subset \Omega $ be a coordinate ball. A function $u$ on $B$ has boundary trace $g$, a function on $\partial B$, if $u(z)\to g(\xi )$ as $B\ni z\to \xi $, for every $\xi \in \partial B$.
Theorem 15.21.2.9826C9FA5510 Suppose that $u\in \mathrm{PSH}(X,\theta )$ has minimal ordinary singularities and
\[ \mathrm{MA}_\theta (u)=f\, \mathrm{d}V\quad \text{on }\Omega , \qquad f\in C^\infty (\Omega ),\quad f>0. \]
Assume that $\Omega $ is covered by coordinate Euclidean balls $B_ j\Subset \Omega $ such that $\theta $ has a smooth local potential near $\overline{B_ j}$ and $u$ has a smooth boundary trace on $\partial B_ j$. Then $u\in C^\infty (\Omega )$ and $\theta +\mathrm{dd}^{\mathrm{c}}u>0$ there.
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 $,
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\begin{equation} (\mathrm{dd}^{\mathrm{c}}V)^ n=\frac{n!}{\pi ^ n}\det (V_{j\bar k})\, \mathrm{d}\Lambda . \end{equation}
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.
$\square$
The proof uses minimal singularities only to obtain the local lower bound by $\rho $. Consequently the same conclusion holds for any solution already known to be locally bounded on $\Omega $ and satisfying the same trace condition. Closedness of $\theta $ on $\Omega $ provides the local potentials $h$ once the coordinate balls are small enough; the background may be nonclosed elsewhere.
Example 15.21.4.DF2D43B90C47 For every $n\geq 1$, the trace criterion applies to a bounded continuous solution whose global density belongs to $L^1$ but to no $L^ p$ with $p>1$, and which is not Hölder continuous at the points of $X\setminus \Omega $.
Let $E=\mathbb C/(\mathbb Z+\mathrm{i}\mathbb Z)$, $X=E^ n$, and
\[ \omega _ j=\frac{\mathrm{i}}{2\pi }\, \mathrm{d}z_ j\wedge \mathrm{d}\bar z_ j, \qquad \theta =\sum _{j=1}^ n\omega _ j, \qquad D=\{ 0\} \times E^{n-1}. \]
Take $\mathrm{d}V=\theta ^ n$ and $\rho =0$. Choose a nonnegative smooth cutoff $\chi $ supported in a coordinate disk $|z|<r_0<1$ of $E$ and equal to one near zero. For $\beta >0$, define
\[ \psi (z)=\chi (z)(-\log |z|^2)^{-\beta }\quad (z\ne 0), \qquad \psi (0)=0, \]
and extend by zero. On the smaller punctured disk put $t=|z|^2$ and $L=-\log t$. Differentiation gives
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\begin{equation} \psi _{z\bar z}=\frac{\beta (\beta +1)}{tL^{\beta +2}}, \qquad t\frac{\mathrm{d}}{\mathrm{d}t}L^{-\beta } =\beta L^{-\beta -1}\longrightarrow 0. \end{equation}
The first expression is integrable and the second is the radial flux. Integration by parts on punctured disks, using also $\psi \to 0$, shows that the distributional Hessian has no atom at zero. On the cutoff annulus its negative part is bounded. Hence, for a sufficiently small fixed $\eta >0$, $u=\eta \psi (z_1)$ satisfies $\theta +\mathrm{dd}^{\mathrm{c}}u\geq \theta /2$.
Only one coordinate contributes to $\mathrm{dd}^{\mathrm{c}}u$. Locally approximate the bounded subharmonic potential in that coordinate from above and use theorem 5.27.1; the smooth diagonal determinant identity passes to the limit and yields
\[ \mathrm{MA}_\theta (u)=f\theta ^ n, \qquad f=1+\eta \psi _{z\bar z}(z_1). \]
Give $f$ any positive value on the null set $D$. It is smooth and positive on $\Omega =X\setminus D$, where $u$ is smooth and has smooth traces on all relatively compact coordinate balls. Near $D$ its singular term is comparable to $|z_1|^{-2}(-\log |z_1|^2)^{-\beta -2}$. Its radial $L^1$ integral converges, whereas
\[ \int _0^{r_0}s^{1-2p}(-\log s^2)^{-p(\beta +2)}\, \mathrm{d}s=+\infty \qquad (p>1). \]
Also $(-\log |z|^2)^{-\beta }/|z|^\alpha \to +\infty $ for every $\alpha >0$, so $u$ is not Hölder at $D$. Finally, boundedness gives minimal ordinary singularities, since every global $\theta $-plurisubharmonic competitor is bounded above.
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