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Denote by $B = B(0,1) \subset \mathbb{C}$ the open unit disc and by $B' = B \setminus \{ 0 \} \subset \mathbb{C}$ the punctured unit disc. Assume that $u : B' \rightarrow \mathbb{R}$ is a harmonic function that also belongs to the Sobolev space $W^{1,2}(B)$. That is, $ \int_B |u|^2 < \infty \;\;\; \mathrm{and} \;\;\; \int_B |\nabla u|^2 < \infty. $ Does $u$ extend to a harmonic function on $B$?


If $u$ would belong to $W^{1,p}(B)$ for $p > 2$, then $u$ would be necessarily bounded and hence would extend smoothly to a harmonic function. If $p < 2$, then the fundamental solution $u(z) = \log(|z|)$ provides a counterexample. The case $p = 2$ is the limit case, where in general the $W^{1,2}$ norm is not enough to control the $L^\infty$ norm. I suspect this should also hold in the limit case, but I'm not sure how to show it.

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According to your conditions, $\frac{\partial{u}}{\partial z}$ is holomorphic on $B'$ and belongs to $L^2(B')$. By evaluating the $L^2$ norm of $\frac{\partial{u}}{\partial z}$ in terms of its Laurent expansion, you will see the singular part of the Laurent expansion vanishes.

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    @levap: Yes, thank you for pointing it out. I have corrected now.2012-11-02