Let $u$ be a subharmonic function on some open set $G\subset \mathbb{C}$, that is, $u$ is uppersemicontinuous and $\vartriangle u\geq 0$ in the sense of distribution. Let $P=\{z\in G: u(z)=-\infty\}$. I want to know what can one say about $P$ in general. For instance, would it be discrete? This is the case if $u=\log |f|$ for some holomorphic function $f$.



1 Answer 1


One can say a lot, but this is a difficult subject, and I would refer you to books about potential theory. However, general polar sets need not be discrete. In particular, any countable set is polar, or more generally, any countable union of polar sets is polar. So if $(z_k)$ is any sequence in the complex plane converging to $z_0=0$, then $P = \{z_k\}$ is a polar set with a non-isolated point. (An explicit subharmonic function which is $-\infty$ on $P$ is $u(z) = \sum_k 2^{-k} \log |z-z_k|$.)

On the other hand, all polar sets in the plane are totally disconnected, so they can not contain a continuum like an arc or something.

  • $\begingroup$ Dear Lukas, thanks for your answer! In fact, the question I have in mind is a little different, can you take a look at this related question? math.stackexchange.com/questions/3131561/… Thanks again for your help! $\endgroup$
    – Chun Gan
    Mar 1, 2019 at 16:30
  • $\begingroup$ There is something to fix in the answer. If the sequence is constantly $=0$, you have $u(z)=log \frac{1}{|z|}=-log|z|$, so $u(0)=\infty$ and not $u(0)=-\infty$ $\endgroup$ Jun 16, 2020 at 17:08
  • $\begingroup$ @ClaudioRea: Thanks, it seems I got the signs wrong, I'll fix that. $\endgroup$ Jun 16, 2020 at 23:33
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    $\begingroup$ I also get always mad with this sign. Half of the scientific community use a minus there. Indeed charges do repel thus the opposite sign for the energy would be more natural $\endgroup$ Jun 18, 2020 at 0:28

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