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."

Any comments or corrections would be gratefully received.

  • $\begingroup$ Perhaps the down-voter would be good enough to give me a clue as to what is wrong with my question? $\endgroup$ – Old John Dec 5 '13 at 18:49

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.

  • $\begingroup$ 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! $\endgroup$ – Old John Jun 25 '12 at 16:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.