15.21 Local density and boundary controlD0D15F32906B

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. $\square$

Proof. The summands defining $h$ are negative on $U$, so their partial sums decrease. Since $h(0)=2\sum _ k\alpha _ k\log a_ k> -\infty $, the limit is subharmonic. On compact subsets of $U\setminus S$, the series and its derivatives converge uniformly, making $h$ harmonic there. Moreover $0\leq b\leq 1$, $b(a_ k)=0$, and $b(0)>0$.

Weighted arithmetic–geometric mean gives, almost everywhere,

For the countable inequality, first use the first $N$ weights and assign the remaining weight to the factor $1$; then let $N$ increase. Since $4A=1/4<2$, the integrals of the terms $|z-a_ k|^{-4A}$ on $U$ are bounded uniformly in $k$. Thus $b^{-1}\in L^2(U)$.

The logarithmic kernel has uniformly bounded $L^2$ norm on the translates of $U$ that occur in the definition of $a$. Cauchy–Schwarz bounds $a$ uniformly. Choose a compactly supported $L^2(\mathbb C)$ kernel agreeing with $\log |z|^2$ on $U-U$. Translation continuity in $L^2$, paired with $b^{-1}\mathbf1_ U\in L^2$, proves continuity of $a$. The distribution identity $\partial _ z\partial _{\bar z}\log |z|^2=\pi \delta _0$ gives $a_{z\bar z}=b^{-1}$. In particular $a$ is subharmonic and, by local elliptic regularity, smooth off $S$.

The second summand of $P$ is $\exp (h(z)+\log |w|^2)$, a convex increasing composition of a plurisubharmonic function, extended by zero at its poles. Thus $P$ is bounded and plurisubharmonic. At smooth points the harmonicity of $h=\log b$ gives $bb_{z\bar z}=|b_ z|^2$, and hence

The set $S\times W$ is contained in the pole set of $h(z)+\log |z|$, a plurisubharmonic function not identically $-\infty $. It is also Lebesgue-null. Pluripolar nullity in theorem 5.27.1 and (15.21.3) prove (15.21.9) throughout the bidisk. Finally,

when $w_0\ne 0$. $\square$

Proof. Set $R=1/8$ and choose $M$ so that $P_0=P+M$ satisfies $0\leq P_0\leq L$ on $B_ R\Subset U\times W$ for some finite $L$. Choose $C>0$ with

and put $Q(Z)=C(|Z|^2-R^2/4)$ for $Z=(z,w)$. Let $\chi $ be the convolution of $|t|$ with an even nonnegative smooth probability kernel supported in $(-1,1)$. Then $\chi $ is smooth and convex, $|\chi '|\leq 1$, $\chi '(0)=0$, and $\chi (t)=|t|$ for $|t|\geq 1$. The function

is bounded and plurisubharmonic, because $m$ is convex and nondecreasing in each variable. The collar inequalities imply

Indeed, $Q\leq -2\leq P_0-2$ on the first ball and $Q>L+2\geq P_0+2$ on the collar. Define

The collar identity makes $u$ identically zero near $\partial B_ R$ and gives a global bounded potential with $\theta +\mathrm{dd}^{\mathrm{c}}u=\mathrm{dd}^{\mathrm{c}}v$ on the ball and $\theta +\mathrm{dd}^{\mathrm{c}}u=\theta $ outside. Thus $u$ is admissible and has minimal ordinary singularities.

Outside the pluripolar set $(S\times W)\cap B_ R$, set $\lambda =(1+\chi '(P_0-Q))/2$. Differentiating the regularized maximum gives

Both $H_ P$ and $CI$ are positive definite there; the final term is positive semidefinite. Hence the classical density is positive. Bounded products charge no pluripolar sets, so there is no further measure on the exceptional set. Their mass on compact $X$ is finite by proposition 5.28.2. This proves $\mathrm{MA}_\theta (u)=F\theta ^2$ for an almost-everywhere positive $F\in L^1$. On $\Omega =B_{R/4}$, (15.21.9) gives the representative $F=C^{-2}$. There $u=P+M-Q$ is discontinuous at $(0,w_0)$ for $0<|w_0|<R/4$.

We prove failure of every global $L^ p$ directly in the transition region. For large $k$, set

Then $w_ k>0$, $P_0(Z_ k)=Q(Z_ k)=m_ k$, and (15.21.13) gives $|Z_ k|^2<9R^2/16$. Near the isolated pole $a_ k$,

where $H_ k$ is smooth and harmonic. Consequently $b$ and $P_0$ are continuous near $Z_ k$. There is a product neighborhood $V_ k\Subset B_ R$ of $Z_ k$, with $|w|$ bounded below by a positive number, on which $1/4\leq \lambda \leq 3/4$.

At smooth points of $V_ k$, positivity and the two-dimensional determinant expansion give

where $c_ k>0$ may depend on $k$ and $C$. Here $P_{z\bar z}=b^{-1}+|w|^2b_{z\bar z}$. With $\zeta =z-a_ k$, shrink the first factor of $V_ k$ so that

for some $c'_ k>0$. Since $F=C^{-2}\det H_ v$, this lower bound implies $F\notin L^ p$ whenever $p(1-\alpha _ k)\geq 1$: the $w$-disk has positive area and the radial integral in $z$ contains

For each $p>1$ such a $k$ exists because $\alpha _ k\downarrow 0$. $\square$