If $u$ is harmonic and bounded in $0<\lvert z \rvert< \rho$, show that $u$ becomes harmonic in $\lvert z \rvert< \rho$ when $u(0)$ is properly defined.
What I was thinking so far is that so long as $u$ is harmonic in an annulus, the arithmetic mean on concentric circles is a linear function of $\log \lvert z \rvert$: $$\frac{1}{2\pi}\int_{C_{r}(z_{0})} u(z)\,dz = \alpha\log r + \beta.$$ But since $u$ is bounded, the left side stays bounded as $r \to 0$, and it must be that $\alpha = 0$. Then $\beta = \lim_{z \rightarrow 0} u(z)$ and $\beta$ is the desired $u(0)$. However, I'm not entirely sure how to show that $\beta$ exists, $u$ is continuous at $0$ with $u(0) = \beta$, or that the resulting $u$ is harmonic.
I was also considering using an argument similar to one used in the characterization of removable singularities, in which Cauchy's integral formula remains valid except at each of finitely many points where the function is not analytic. But the mean value property isn't quite as powerful as Cauchy's integral formula.
(Studying for qualifiers, problem is from Ahlfors)
site:math.stackexchange.com harmonic bounded point– Aug 05 '14 at 04:28