I'm working on a problem from p. 166 of Lars Ahlfors' Complex Analysis:
If $u$ is harmonic and bounded in $0 < |z| < \rho$, show that the origin is a removable singularity in the sense that $u$ becomes harmonic in $|z|<\rho$ when $u(0)$ is properly defined.
My Solution So Far: This problem is screaming out to use the big Picard theorem...I would like to define a harmonic conjugate function $v:\{0<|z|<\rho\} \to \mathbb{C}$, also bounded, and such that $u,v$ satisfy the Cauchy-Riemann equations. Then we would have that $f(z):=u(x,y) + i v(x,y)$ is analytic and bounded in $\{0<|z|<\rho\}$, so $0$ is a removable singularity, and thus we can fill it in for $u$.
I need to show that there exists such a $v$ and that it is also bounded. If the domain were simply connected, we could define
$$v(x,y) = \int_{(x_0,y_0)}^{(x,y)} -u_2(s,t)ds + u_1(s,t)dt, $$
which would be well-defined because the integrand is closed, owing to the harmonicity of $u$, so the integral is path-independent. But obviously it is not simply connected...how can we overcome this problem?
As to $v$ bounded, even if we could define it as above, I'm not sure how we could bound it on the open punctured disk. (I think we could bound it on any compact subset...)
