Pome

15.21.6 A bounded discontinuous potentialAD0BD024C9A0

We construct a bounded plurisubharmonic function on a bidisk whose Monge–Ampère measure is a constant multiple of Lebesgue measure, but which is discontinuous. The construction places logarithmic poles with summable weights at points accumulating at the origin; the resulting weight $b$ vanishes at the poles but not at their limit. It is the local model for theorem 15.21.12.

Let $U=W=\{ |z|<1/4\} \subset \mathbb C$, with coordinates $z$ and $w$ on the two factors. Write $\mathrm{d}A$ for planar Lebesgue measure and set

\[ a_ k=\alpha _ k=2^{-k-4},\qquad A=\sum _{k\geq 1}\alpha _ k=\frac1{16},\qquad S=\{ 0\} \cup \{ a_ k:k\geq 1\} . \]

Define on $U$

6321AD313CED 15.21.7
\begin{equation} h(z)=\sum _{k\geq 1}\alpha _ k\log |z-a_ k|^2, \qquad b(z)=\mathrm{e}^{h(z)}, \end{equation}

with $b=0$ where $h=-\infty $.

Lemma 15.21.8.5E22269BEE55 The function

\[ a(z)=\frac1\pi \int _ U\log |z-\zeta |^2\, b(\zeta )^{-1}\, \mathrm{d}A(\zeta ) \]

is bounded and continuous on $U$. The bounded function

\[ P(z,w)=a(z)+b(z)|w|^2 \]

is plurisubharmonic on $U\times W$, is smooth off $S\times W$, and satisfies

69692418B3EE 15.21.9
\begin{equation} (\mathrm{dd}^{\mathrm{c}}P)^2=\frac2{\pi ^2}\, \mathrm{d}A(z)\, \mathrm{d}A(w). \end{equation}

It is discontinuous at every $(0,w_0)$ with $w_0\ne 0$.

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,

331160A2E4B5 15.21.10
\begin{equation} b(z)^{-2}\leq \sum _{k\geq 1}\frac{\alpha _ k}{A}|z-a_ k|^{-4A}. \end{equation}

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

\[ \det (P_{j\bar k}) =ba_{z\bar z}+|w|^2\bigl (bb_{z\bar z}-|b_ z|^2\bigr )=1. \]

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,

\[ P(a_ k,w_0)=a(a_ k)\longrightarrow a(0) <a(0)+b(0)|w_0|^2=P(0,w_0) \]

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

Cite this page

Comments

Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.

Read this page’s discussion on GitHubPost a comment on GitHub