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Prove that the zeroes of a Harmonic function is never isolated. All I can think of is a very rough idea of a proof by contradiction.

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Taken from some old lecture notes (note, I believe this only applies to real valued functions):

Suppose $u$ is harmonic and real valued on $D$ with $a \in D$ and $u(a) = 0$. Take $r \gt 0$ such that $\bar B(a,r) \subset D$. As the mean of $u$ over $\partial B$ equals zero, either $u$ is identically $0$ or $u$ takes both positive and negative values on the boundary. But having both $\pm$ values on $\partial B$ implies $u$ has a zero on $\partial B$ which implies $u$ has a zero on the boundary on every sufficiently small ball centered at $a$. Hence $a$ is not an isolated zero of $u$.

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    $\begingroup$ In response to your statement in parenthesis, the result is false for complex-valued harmonic functions: Take $u(z)=\sin(z)$ in $\mathbb{C}$. In fact any zero of an holomorphic function must be isolated, and holomorphic functions are harmonic. $\endgroup$
    – Jose27
    Mar 26, 2015 at 2:33
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    $\begingroup$ @Jose27 Thanks for that, nice to have the proof completed for real and complex functions. $\endgroup$
    – Matthew Cassell
    Mar 26, 2015 at 7:20

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