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 $u(x_1)<u(x_2)=u(x_0)<u(x_3)$. Since I can take infinitely many paths (for $N\ge 2$), the result follows.

Is it ok?

  • $\begingroup$ yes it is correct. Actually it is the same proof as Marco's answer in math.stackexchange.com/questions/2892657/… $\endgroup$ – Gio67 Oct 4 '18 at 10:38
  • $\begingroup$ @Gio67 but his proof uses the mean value property $\endgroup$ – Paprika Oct 4 '18 at 10:42
  • $\begingroup$ 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. $\endgroup$ – Gio67 Oct 4 '18 at 11:10
  • $\begingroup$ 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. $\endgroup$ – zhw. Oct 4 '18 at 17:45
  • $\begingroup$ @Gio67 Marco didn't use infinitely many paths. One path was chosen to reach a contradiction. $\endgroup$ – 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.

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  • $\begingroup$ what is $b$? shouldn't it be $x_0$? $\endgroup$ – Paprika Oct 4 '18 at 23:04
  • $\begingroup$ Also, do you see something wrong with my proof? $\endgroup$ – Paprika Oct 5 '18 at 0:34
  • $\begingroup$ Sorry, about $a,b.$ I started out with those and then thought it better to stick with $x_0,$ etc. I edited. $\endgroup$ – zhw. Oct 5 '18 at 1:29
  • $\begingroup$ Your proof has good ideas. But what does "infinitely many paths" mean? They could all be different and still go through $x_0.$ $\endgroup$ – zhw. Oct 5 '18 at 1:36
  • $\begingroup$ 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. $\endgroup$ – zhw. Oct 5 '18 at 16:34

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