# Harmonic function with bounded preimage

I recently saw a question here about bounded/unbounded preimages of a set under a harmonic function. The question asked did not seem to make sense as it was talking about harmonic functions on $\mathbb{R}$, but got me thinking about a corresponding problem for a harmonic function on $\mathbb{R}^n$, and I would appreciate it if anyone could tell me if my argument below is correct, or if there is a neater way to do it.

Suppose that $u:\mathbb{R}^n\to\mathbb{R}$ is harmonic and suppose that the set $C = u^{-1}(\{c\})$ is bounded. Since adding a constant to $u$ leaves it harmonic, we may assume that $c=0$. Take any point $x\in\mathbb{R}^n$ and consider a ball $B(x,r)$ with $r$ sufficiently large that all points of $C$ are in the interior of $B$. Now either $u>0$ or $u<0$ on $\partial B$ since $u$ is continuous and non-zero on $\partial B$. Take the case $u>0$ on $\partial B$ (the other case is similar), and recall that $u(x)$ is the average of the values of $u$ on the surface $\partial B$, so that $u(x)>0$. Since $x$ is arbitrary, it follows that $u$ is bounded below on $\mathbb{R}$, and harmonic, and hence constant.

Rephrasing: "If $u$ is a non-constant harmonic function on $\mathbb{R}$, then $u^{−1}(c)$ is unbounded."

Your reasoning is perfectly valid. A slight rephrasing: Suppose there exists $R$ such that $u$ does not vanish in $\mathbb R^n\setminus B(0,R)$. Since the set $\mathbb R^n\setminus B(0,R)$ is connected (this is where we need $n\ge 2$), it follows that either $u>0$ or $u<0$ on this set. Also, $u$ is bounded on $B(0,R)$. We conclude that $u$ is either bounded from above or bounded from below on $\mathbb R^n$, hence constant.
• Thanks for checking my reasoning. My first attempt at a proof in fact involved arguing that $u$ was bounded on the closed ball (as a continuous image of a compact set), but didn't manage to express it as elegantly and concisely as you have! – Old John Jun 25 '12 at 16:01