I am working with a book by Axler, Bourdon and Ramey and find the following problem:
Suppose $u$ is a harmonic function on $B \setminus \{0 \}$ such that $$ |x|^{n-2} u(x) \to 0, \qquad x \to 0 $$ Prove that $u$ has a removable singularity at $0$
The solution is trivial if you consider Laurent series. I have been told that the proof is much more elementary just by considering Poisson's kernel. Does anybody know it?
Thank you in advance,
D
I am assuming that the ball $B$ has radius $1$ and $n\ge 3$. Take $r\in (0,1)$ and consider the problem
$$ \left\{ \begin{array}{ccc} -\Delta v=0, &\mbox{ in $B(0,r)$}, \\ v=u, &\mbox{in $\partial B(0,r)$}. \end{array} \right. $$
I - Show that $$\lim_{x\to 0}\frac{u(x)-v(x)}{r^{2-n}-|x|^{2-n}}=0.$$
II - From I, we have that for all $\epsilon>0$, there is $\delta >0$ such that $$|u(x)-v(x)|\le \epsilon |r^{2-n}-|x|^{2-n}|,\ \forall\ x\in \overline{B(0,\delta)}.\tag{1}$$
Use the maximum principle in the domain $B(0,1)\setminus B(0,\delta)$, to conclude an inequality similiar to $(1)$ for this domain.
III - Conclude that $u=v$.