Lemma 15.21.85E22269BEE55

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 $w_0\ne 0$. $\square$