Zeroes of a harmonic function

Question

This question is from Rudin's "Real and Complex Analysis". Let $u$ be a non-negative harmonic function defined on a region such that $u$ vanishes at some point on that region. Then what can we say about the nature of $u$? I am trying to prove that such a function must be 0 on its entire domain. My idea is to apply the identity theorem for harmonic functions on $u$. However for that I need a non-empty open subset of the domain where $u$ vanishes. I am trying to use the MVP to do that but haven't been able to find any such open set. Can anyone help?

Answer 1

There are two versions of Mean Value property. One version is $u(a,b)=\frac 1 {m(B)}\iint_B u(x,y)d(x,y)$ where $B$ is an open ball with center $(a,b)$ contained in the domain. [The other version is one where you integrate over a circle around $(a,b)$]. From this it is clear that if $u \geq 0$ and $u(a,b)=0$ the $u=0$ throught $B$.

[$u(a,b)=\frac 1 {2\pi} \int_0^{2\pi} u(a+r{\cos \theta},b+r \sin \theta)d\theta$. Multiply by $rdr$ and integrate w.r.t $r$ from $0$ to some $R$ to get the new version of Mean Value Property].