## Harmonic function

Let $e_1,e_2,...,e_n$ be semilines on the plane starting from a common point. Prove that if there doesn’t exist any function $u \not\equiv 0$ harmonic on the whole plane that vanishes on the set $e_1\cup e_2\cup...\cup e_n$, then there exists a pair $(i,j)$ such that there is no function $u\not\equiv 0$ harmonic on the whole plane such that $u$ vanishes on $e_i \cup e_j$.
Miklos Schweitzer 2001

Solution: Use the following Lemma:

If $a,b$ are two semilines starting at the same point $O$ there exists $u \not\equiv 0$ is a harmonic function vanishing on $a,b$ if and only if there exists $q \in \mathbb{Q}$ such that the angle between $a$ and $b$ is $q\pi$.

Proof of the Lemma: Using the following property of harmonic function we can deduce some important and quite interesting properties inherited from holomorphic functions:

1. A harmonic function vanishing on an open set is identically zero.

2. (Schwarz Reflection Principle) If a harmonic function is defined in a neighborhood of a line segment contained in one of the halfplanes determined by that segment and it is continued to zero on that segment, then it can be extended harmonically to the symmetric region by the given line segment.

Suppose $\alpha=\angle(a,b)$ and $\alpha/\pi \notin \mathbb{Q}$. We see that $u$ is defined on a halfplane determined by $a$, containing $b$, and by the Schwarz reflection principle, the restriction of $u$ at that halfplane can be continued harmonically to the other halfplane and the new function $u_0$ is zero on the reflected halfline $b_1$. Since $u$ and $u_0$ agree on an open set (the angle between $a,b$ for example) it follows that $u=u_0$ everywhere, and therefore $u$ vanishes on $b_1$ too. Suppose $b_1$ is rotated by $\alpha$ counterclockwise.

Repeating the above construction, we see that we can construct a sequence of halflines $b_k$ such that $\angle (b_k,b_k+1)=\alpha,\ \forall k$ and the rotation is considered counterclockwise and $u$ vanishes on every $b_k$. Since $\alpha/\pi \notin\mathbb{Q}$ the sequence of halflines is not periodic and therefore is dense in the plane. This means $u$ is identically zero, which is a contradiction.

Conversely, there are the well known solutions of the equation of Laplace in 2 dimensions in polar coordinates
$u(r,\phi)=\left(Ar^m+Br^n\right)(C\cos (m\phi)+D\sin (m\phi))$, which can be chosen to vanish on the two semilines, by taking a rotation and a suitable $m$, when $\alpha/\pi \in \mathbb{Q}$.

For the proof of our theorem, suppose we can’t find such a function $u \not\equiv 0$. If all the angles $\alpha_{ij}=\angle(e_i,e_j)$ are rational multiples of $\pi$, by the above considerations, we can find a harmonic function vanishing on $e_1 \cup e_2\cup ...\cup e_n$, which is a contradiction. Therefore, there exists $i,j$ such that $\alpha_{ij}\pi \notin \mathbb{Q}$, and by the Lemma proved above, there is no harmonic function $u\not\equiv 0$ such that $u$ vanishes on $e_i\cup e_j$.