$u$ harmonic then $u^{-1}\{u(x_0)\}$ is infinite for $N\ge 2$

Question

Let $\Omega\subset\mathbb{R}^N$ be an open, $u$ harmonic in $\Omega$ and $x_0\in\Omega$. Show that if $N\ge 2$ then $u^{-1}\{u(x_0)\}$ is infinite. What about $N=1$?

The $N\ge 2$ makes me think that somehow this problem on the line collapses because of dimension $1$, and for higher dimensions it's like I'm in a plane and can walk in many directions.

I don't see how to connect this with the fact that $$\sum\frac{\partial^2 u}{\partial x_i^2} = 0$$

For the one dimensional case, $$\frac{\partial^2 u}{\partial x^2} = 0$$

for all $x$ in some $\Omega$, so it's an interval. This is the same as saying that the function has second derivative $0$ at an open the interval. If it were just a point I'd say the function might have an inflection point, but in fact it's in an entire open interval.

Let's first think of the most basic example: $u(x) = 1$. The set $u^{-1}(u(x))$ is the entire interval on which the function is harmonic (even more, but let's focus on that). So in this case it is infinite. I guess I should find a counterexample with some non obvious function. Anyone has ideas?

Answer 1

Suppose $u^{-1}(u(x_0))$ is a finite set $F.$ Then, as Marco wrote, there is an open ball $B(x_0,r)$ such that $u\ne u(x_0)$ in $B(x_0,r)\setminus \{x_0\}.$ Since $B(x_0,r), B(x_0,r)\setminus \{x_0\}$ are both connected, their images under $u$ are, respectively, intervals $I,J$ by the continuity of $u.$ But $J=I\setminus \{u(x_0)\}.$ So we've removed $u(x_0)$ from $I$ and still have an interval. The only way that can happen is for $u(x_0)$ to be the largest value of $I$ or the smallest value of $I.$ In either case we have $u\equiv u(x_0)$ in $B(x_0,r)$ by the max/min principle for harmonic functions, contradiction. Therefore $u^{-1}(u(x_0))$ is infinite (and in fact it must be uncountable).

Answer 2

Without loss of generality, by making $\Omega$ smaller, suppose $\Omega$ is a ball centered at $x_0$ and $u^{-1}\{u(x_0)\}=\{x_0\}$ and we derive a contradiction. Since the value of $u(x_0)$ is the average of the values of $u(x)$ on the ball, we conclude that there must exist $x_1,x_2 \in \Omega$ such that $u(x_1)<u(x_0)<u(x_2)$. Connect $x_1$ to $x_2$ via a path that doesn't go through $x_0$. By the intermediate-value theorem, there must exist $x_3$ along the path such that $u(x_3)=u(x_0)$, a contradiction.