I am working through old qual problems at the University of Minnesota and am trying to find an alternate solution to the following problem.
Determine all continuous functions on $\{ z : 0 < \left| z \right| \leq 1 \}$ which are harmonic on $\{ z : 0 < \left| z \right| < 1 \}$ and are identically zero on $\{ z : \left| z \right| = 1 \}$.
One solution, written by one of my fellow graduate students, uses the Schwartz reflection principle to prove that such functions must assume the form $c \, \log(\left| z \right| )$. Verifying that this function is harmonic is easy; verifying that it is the only such harmonic is hard.
As part of my studying, I am reading Krantz's book on complex analysis. He solves a similar problem by showing that the solutions must be radially symmetric. Under this assumption, the Laplacian in polar coordinates depends only on $r$. This gives a separable ODE that one can solve explicitly to obtain $c \, log(\left| z \right| )$.
Krantz's argument relies on the maximum principle for harmonic functions. To show that a solution $u$ must be radially symmetric, he argues that if $u$ is a solution, then $u(e^{i \theta} z)$ must also be a solution. By the maximum principle, solutions are unique, so $u(z) = u(e^{i \theta} z)$. This implies that $u$ is rotationally symmetric.
Checking the hypotheses of the maximum principle, I think this is a valid argument. But if we can apply uniqueness to conclude radially symmetry, why can't we apply uniqueness to conclude that the only solution is 0? I suspect that I am misapplying a theorem or have missed some hypotheses, but since the work lies in proving uniqueness, not existence, I was surprised to find Krantz's shorter solution.
I am aware of Krantz's solution and this solution, but if anyone can offer alternatives, I would appreciate it. My colleague's argument is a bit difficult to reproduce on an exam.
