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I am trying to solve a problem related to harmonic functions, but I do not see how to begin. The problem is

Let $Ω \subset \mathbb{R}^{n}$ be an open and connected set. Let $u : \Omega \to \mathbb{R}$ be a harmonic function. Suppose that for a pair of points $(x_0, y_0) \in \Omega \times \Omega$, $u$ satisfies $u(x_0) + u(y_0) = M$. Then there are infinitely many pairs $(x, y) \in \Omega \times \Omega$ such that $u(x) + u(y) = M$.

I think that it has something to do with mean value properties. Any tip to approach this problem?.

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Suppose on the contrary, that there are only finitely many such pairs, then we can find $r>0$ such that the ball $B_r(x_0,y_0)\subset \Omega\times \Omega$ contains no other such point, i.e. $\forall(x,y)\in B_r(x_0,y_0)\setminus \{(x_0,y_0)\}: u(x)+u(y)\neq M$.

Now we define the harmonic function $\tilde{u}:B_r(x_0,y_0)\to \mathbb{R}$ by $\tilde{u}(x,y)=u(x)+u(y)$. If $M=\max_{(x,y)\in B_r(x_0,y_0)} \tilde{u}(x,y)$, then we can conclude by the maximum principle that $\tilde{u}$ is constant on $B_r(x_0,y_0)$ and this contradicts the assumption that $(x_0,y_0)$ is the only point in the ball that attains the value $M$. Similarly, we can use the minimum principle to conclude that $M\neq \min_{(x,y)\in B_r(x_0,y_0)} \tilde{u}(x,y)$.

Therefore we have shown that there exists points $(x_1,y_1), (x_2,y_2)\in B_r(x_0,y_0)$ such that \begin{align} \tilde{u}(x_1,y_1)<M<\tilde{u}(x_2,y_2). \end{align} Now we can choose a contiuous curve $c:[0,1]\to B_r(x_0,y_0)$ with $c(0)=(x_1,y_1)$ and $c(1)=(x_2,y_2)$ such that $c(t)\neq (x_0,y_0)$ for all $t\in[0,1]$. If we denote the composition by $f(t)=\tilde{u}(c(t))$, then by construction we have $f(0)<M<f(1)$ and by the intermediate value theorem we can conclude that there has to be a point $t'$ such that $\tilde{u}(c(t'))=f(t')=M$.

Since $c(t')\neq (x_0,y_0)$ we have found another point in $B_r(x_0,y_0)$ that attains the value $M$, which contradicts our assumption that there are only finitely many such points in $\Omega\times\Omega. $

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