Harmonic Analysis – Equality of Two Subharmonic Functions

Question

Let $u\leq v$ be two locally bounded subharmonic functions in a domain in $\mathbb{R}^n$. Assume that $u=v$ on a dense subset.

Is it true that $u=v$ everywhere?

Answer 1

This is not the case. Here is a counterexample in the case where $n=2$.

Suppose that $A$ is a countable dense subset of $\mathbb{R}^{2}=\mathbb{C}$. Let $r_{a}>0$ for each $a\in A$, and suppose that $\sum_{a\in A}r_{a}\cdot(1+\max(0,\log(a)))<\infty$.

Let $p(z)=\sum_{a\in A}r_{a}\cdot\log(|z-a|)$. Let $s:\mathbb{C}\rightarrow(0,\infty)$ be any subharmonic function. Let $q(z)=p(z)+s(z)$. Then observe that $p(z)=q(z)=-\infty$ on a dense $G_{\delta}$-set. In particular, $p=q$ on a dense set. The functions $p,q$ are not locally bounded, but we can obtain locally bounded functions from $p,q$.

The function $p$ cannot be $-\infty$ everywhere since $$\frac{1}{2\pi}\int_{0}^{2\pi}p(re^{i\theta}+z_{0})d\theta=(\sum_{a\in A,|z_{0}-a|<r}r_{a})\cdot\log(r)+\sum_{a\in A,|z_{0}-a|>r}r_{a}\cdot\log(|z_{0}-a|)>-\infty.$$

In fact, the set $p^{-1}[\{-\infty\}]$ is a polar set and polar subsets of $\mathbb{C}$ have Hausdorff dimension $0$ and therefore Lebesgue measure zero as well.

Let $u(z)=e^{p(z)},v(z)=e^{q(z)}$. As a consequence of Jensen's inequality, the functions $u(z),v(z)$ are subharmonic, locally bounded, and $u\leq v$. However, $u(z)=v(z)=0$ on a dense $G_{\delta}$-set, and if $p(z)>-\infty$, then $p(z)<q(z)$, so $u(z)<v(z)$.

The fact that $u(z)=v(z)=0$ on a dense $G_{\delta}$-set should be contrasted with the fact that if $f,g$ are subharmonic functions that are equal almost everywhere, then $f=g$.