There seems to be a removable singularity theorem for harmonic functions that somewhat goes like:
Let $\Omega\subset \Bbb R^n$ be an open region and $x_0\in\Omega$, $u$ is harmonic in $\Omega-\{x_0\}$. Now if we have $u(x)=o(\Phi(x-x_0))$ around $x_0$ then there exists $\tilde u$ harmonic in all of $\Omega$ and $u=\tilde u$ in $\Omega-\{x_0\}$.
Here $\Phi$ is the fundamental solution to $\Delta u=0$ in $\Bbb R^n$, and, since dropping the normalising constant doesn't affect our statement above, you may just take $\Phi=-\log|x-x_0|$ in 2D and $=1/|x-x_0|^{n-2}$ in 3D or higher.
I have a question concerning this theorem:
I found this one, which I am not sure about :
Let $w=u-\tilde{u} .$ Then $w$ is harmonic in $B\left(x_{0}, r\right) \backslash\left\{x_{0}\right\}$ and moreover (check it!)
$$
\lim _{x \rightarrow x_{0}} \frac{w(x)}{r^{2-n}-\left|x-x_{0}\right|^{2-n}}=0 \quad \text { if } n \geq 3 \quad \lim _{x \rightarrow x_{0}} \frac{w(x)}{\log r-\log \left|x-x_{0}\right|}=0 \text { if } n=2
$$
So for every $\varepsilon>0$ there exists $\delta>0$ such that, for all $x$ such that $\left|x-x_{0}\right| \leq \delta$,
$$
|w(x)| \leq \varepsilon\left(r^{2-n}-\left|x-x_{0}\right|^{2-n}\right) \text { if } n \geq 3 \quad|w(x)| \leq \varepsilon\left(\log r-\log \left|x-x_{0}\right|\right) \quad \text { if } n=2
$$
Because I know $u(x)=o(\Phi(x-x_0))$ but I cannot see too easy that $w(x)=o(\Phi(x-x_0)).$ Is that correct or there is a mistake?
Best Answer
Since $\tilde u$ is bounded in $B(x_0,r)$, $w=u-\tilde u=o(\Phi(x-x_0))$ as $x\to x_0$: $$ \lim\limits_{x\to x_0}\frac{w(x)}{\Phi(x-x_0)}=\lim\limits_{x\to x_0}\frac{u(x)-\tilde u(x)}{\Phi(x-x_0)}=0 $$ since $\Phi(x-x_0)\to\infty$ as $x\to x_0$ and $\tilde u(x)$ is bounded.