# If a harmonic function is constant in two points then it is constant in infinitely many points

If $u$ is a harmonic function defined on an open set onto $\mathbb{R}$, and we know that there exists $x_0,y_0$ such that $u(x_0)+u(y_0)=M$ for some constant $M$, show that there exist infinitely many points $(x,y)$ such that $u(x)+u(y)=M$.

First I tried to use the mean-value property but as $M$ is not the maximum I cannot conclude anything by using inequalities as to prove the mean-value property itself. I also tried to prove that the set $A=\{(x,y) : u(x)+u(y)=M\}$ is open and closed and as it is nonempty (since $(x_0,y_0) \in A$) it should be the whole set where $u$ is defined (since it is a domain, so it is connected), BUT, I don`t know how to prove it is open, since I would need the mean-value property. If anybody could help me to solve this exercise...Thank you

• Hello. Welcome to MathStack Exchange. I cannot help with your question, but I fixed the math writing using MathJax. Take a look to see how it goes. – Danilo Gregorin Dec 12 '17 at 20:35
• thanks, I don't know how to use it yet. – Irene Gil Dec 12 '17 at 20:38
• There is a lot of LaTeX material on the web, it is worth tô learn as soon as possible – Danilo Gregorin Dec 12 '17 at 20:40
• points $(x,y)$ such that $u(x)+u(y)=M?$ Weird notation. – zhw. Dec 12 '17 at 21:07
• the set of points in the domain of u such that u(x)+u(y)=M – Irene Gil Dec 12 '17 at 21:11

Hint: For all small $r>0,$ there must exist a point $x_r\in \{x: |x-x_0| = r\}$ such that $u(x_r) = u(x_0).$ Otherwise the MVP does not hold at $x_0.$ Same thing at $y_0.$
• @IreneGil For each $r, u(x_0)$ is the average of $u$ over the sphere I mentioned. $u$ is continuous and the sphere is connected. Thus $u$ must take on the value $u(x_0)$ on this sphere. – zhw. Dec 13 '17 at 21:19
• You are making this too difficult. Note my sentence "Same thing at $y_0$" – zhw. Dec 14 '17 at 17:43