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Let be $u$ a harmonic function defined on an open set $\Omega \setminus \{p\} \subset \mathbb{C}$ of the complex plane. Show that if $u$ is bounded in a neighborhood of $p$ then $u$ admits a harmonic extension to $\Omega$.

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up vote 2 down vote accepted

Take a small disc $D$ around $p$ and solve the Dirichlet problem on it (with datum $u\vert_{bD}$) and call $v(z)$ the solution; then $u_1=u-v$ vanished on $bD$. Wlog we suppose that $p=0$ and $D=\{|z|<1\}$. For any $\epsilon>0$, the function $w_\epsilon=u_1+\epsilon\log|z|-\epsilon$ is negative near $0$ and near $bD$, therefore $w_\epsilon\le0$ on $D\setminus\{0\}$. Letting $\epsilon\to0$, you get $u_1\le 0$ on $D\setminus\{0\}$. In the same way you prove that $-u_1\le0$. Hence $u_1=0$ on $D\setminus\{0\}$, so $u=v$ and, as $v$ is defined on $D$, $u$ extends to $D$ as well.

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