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

I want to prove that $$u^{-1}(u(x_0))$$ is infinite for dimension $$N\ge 2$$, for $$x_0\in \Omega\subset\mathbb{R}^N$$ and $$u$$ harmonic. I've found $$u$$ harmonic then $$u^{-1}\{u(x_0)\}$$ is infinite for $$N\ge 2$$ but I think I have an idea

By the maximum principle, both the maximum and the minimum of $$u$$ are attained at $$\partial \Omega$$. So construct a ball around $$x_0$$. $$x_1$$ and $$x_3$$ are the maximum and minimum points of $$u$$ in the border of the ball. By the intermediate value theorem and using the fact that the ball is connected, the image of $$u$$ around a path that passes through $$x_1$$ and $$x_3$$ is an interval. By the intermediate value theorem, there exists a point $$x_2$$ such that $$u(x_1). Since I can take infinitely many paths (for $$N\ge 2$$), the result follows.

Is it ok?

• yes it is correct. Actually it is the same proof as Marco's answer in math.stackexchange.com/questions/2892657/… – Gio67 Oct 4 '18 at 10:38
• @Gio67 but his proof uses the mean value property – Paprika Oct 4 '18 at 10:42
• and you use the maximum principle. Yes I saw that. I like yours better. But the overall idea of using infinitely many paths is the same. – Gio67 Oct 4 '18 at 11:10
• I deleted my previous comment because I see you did indeed reduce to a ball. But again, why do you say the first line in the second paragraph? $u$ is not defined on $\partial \Omega.$ And then you never mention $\partial \Omega$ again. That's confusing. – zhw. Oct 4 '18 at 17:45
• @Gio67 Marco didn't use infinitely many paths. One path was chosen to reach a contradiction. – zhw. Oct 4 '18 at 17:48

Take a closed ball $$\overline B(x_0,r) \subset \Omega.$$ Let $$S_r=\partial \overline B(x_0,r).$$ Then by the max/min principle, the minimum $$m$$ and maximum $$M$$ of $$u$$ over $$\overline B(x_0,r)$$ occur on $$S_r.$$ Now $$S_r$$ is connected. It follows that $$u(S_r)$$ is connected, hence is an interval, and therefore equals $$[m,M].$$ Since $$m\le u(x_0)\le M,$$ we have $$u(x_0) = u(x)$$ for some $$x\in S_r.$$
Now there are uncountably many $$r$$ such that $$\overline B(x_0,r)\subset \Omega.$$ For each one, the above shows there is $$x_r \in S_r$$ such that $$u(x_r)=u(x_0).$$ It follows that $$u^{-1}(\{u(x_0)\})$$ is not just infinite, but uncountable.
• what is $b$? shouldn't it be $x_0$? – Paprika Oct 4 '18 at 23:04
• Sorry, about $a,b.$ I started out with those and then thought it better to stick with $x_0,$ etc. I edited. – zhw. Oct 5 '18 at 1:29
• Your proof has good ideas. But what does "infinitely many paths" mean? They could all be different and still go through $x_0.$ – zhw. Oct 5 '18 at 1:36
• One way to use your path idea is start like Marco and I did previously: If the result is false, then there is a closed ball $\overline B (x_0,r)\subset \Omega$ such that $u\ne u(x_0)$ in $\overline B (x_0,r)\setminus \{x_0\}.$ Now choose a path from $x_1$ to $x_3$ in $\overline B (x_0,r)\setminus \{x_0\}.$ Then $u$ takes the value $u(x_0)$ somewhere on this path, contradiction. This avoids the mean value property. – zhw. Oct 5 '18 at 16:34