# If $u$ is harmonic on $\{0 < |z| < \rho\}$ and $u(|z|) \in o(\log(1/|z|))$, then $u(z)$ is harmonic at $0$.

I am working on problems in complex analysis. I have seen answers on here for a similar question to this one, but the questions involve stronger hypotheses. Instead of assuming that $u(z)$ is bounded, I want to show the same conclusion by only assuming that $u(|z|) \in o(\log(1/|z|)$ (i.e. $\lim_{z \rightarrow 0} \frac{u(|z|)}{\log(1/|z|)} = 0$.

For this exercise, I know that $u(z) = u(re^{i\theta})$ being harmonic on $\{0 < r < 1\}$ has a unique expansion $\sum_{n \in \mathbb{Z}} a_{n}r^{n}\cos(n\theta) + b_{n}r^{n}\sin(n\theta) + c\log(r)$ where $a_{n}r^{n} + a_{-n}r^{-n}$, $b_{n}r^{n} - b_{-n}r^{-n}$, $a_0 + c\log(r)$ are the Fourier coefficients for $u$. Fix $\alpha > 0$ so that $r^{\alpha}u(re^{i\theta}) \rightarrow 0$ as $r \rightarrow 0^{+}$. I already did an exercise where this implies that $a_{n} = 0$ and $b_{n} = 0$ for $n \leq -\alpha$.

Since the limit statement only involves the radius, I think I can use this to conclude that $u$ is harmonic at $0$. But I don't know what I can use to extend harmonicity. I appreciate any help that anyone can give here.