# Boundedness of a finite subharmonic function

Let $$u\colon B^n(0,1)\to \mathbb{R}$$ be a subharmonic function in the open unit ball in $\mathbb{R}^n$. The crucial assumption is that $u$ never equals $-\infty$.

Is it true that $|u|$ is bounded on any compact subset of the unit ball?

Remark. (1) $u$ is bounded above on any compact subset of the unit ball since it is upper semi-continuous.

(2) If the answer has the negative answer in the above generality, then I would like to ask it for $u$ being a plurisubharmonic function in the unit ball in $\mathbb{C}^{n/2}=\mathbb{R}^n$.

No. Take a zequence $x_k\to 0$ and a sequence of positive numbers $\epsilon_k$ such that $\sum_k\epsilon_k|K_n(x_k)|<\infty$. Then the function $$\sum_k\max\{\epsilon_k K_n(x-x_k),-k\},$$ where $K_n$ is the fundamental solution of the Laplace equation ($K_n(x)=-|x|^{-n+2}$ when $n\geq 3$ and $K_2=\log|x|$). This function is finite everywhere, so your polar set and your $U$ are empty, but the function is evidently unbounded from below.
When $n=2$ "subharmonic" and "plurisubharmonic" are the same, so it also gives a counterexample for plurisubharmonic functions (in any dimension, just let it be independent of the rest of the complex variables).
• Did you perhaps mean $$\sum_k \max\{\epsilon_k K_n(x - x_k), -k\} ?$$ – Mateusz Kwaśnicki Feb 21 '18 at 20:09