# Subharmonic function in $B(0;1) \setminus \{ 0 \}$ is subharmonic in $B(0;1)$

Let $D = B(0; 1)$ be the unit open ball in $\mathbb{C}$. Let $u: D \rightarrow \mathbb{R}$ be a continuous function such that $u$ is subharmonic in $D \setminus \{ 0 \}$. How to prove that $u$ is subharmonic in $D$?

Thank you!

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I notice that there's no question here ... $\:$ – Ricky Demer Oct 30 '12 at 20:50
Why? It is nontrivial that u is subharmonic in the origin. – rla Oct 30 '12 at 21:45

Let $G$ be a disk compactly contained in $D$, and let $u_G$ be the harmonic function in $G$ with boundary values $u$. For subharmonicity we have to show that $u \le u_G$ in $G$. If $G$ does not contain $0$ in its interior, this follows directly from the assumption. Otherwise, for $\epsilon>0$ small, let $u_{G,\epsilon}$ be the harmonic function in $G \setminus B(0,\epsilon)$ with boundary values $u$. Then by a simple harmonic measure estimate $\lim\limits_{\epsilon \to 0} u_{G,\epsilon}(z) = u_G(z)$ for $z\in G \setminus \{0\}$. For every such fixed $z$ and $0<\epsilon<|z|$ we also have $u(z) \le u_{G,\epsilon}(z)$, so $u(z) \le u_G(z)$ for $z\in G \setminus \{0\}$. By continuity this also holds for $z=0$.