If $n\geq 3$ and $f: B_1(0)\setminus\{0\}:\to \mathbb{R}$ is continuous and bounded, then can $f$ be defined at $x=0$ so that $f$ is continuous on $B_1(0)$? I believe this should be true but I am not sure how to show it.
This is just one step in this problem: If $u$ is harmonic in $B_1(0)\setminus\{0\}\subset \mathbb{R}$ and $\lim_{x\to 0} |x|^{n-2}u(x)=0$, then $u\in C^2(B_1(0))$. Here $n\geq 3$.
I was able to show that this condition implies that $u$ is bounded. But I am wondering if it can then be redefined at $x=0$ so that it is continuous? Then I was thinking to use the Mean Value Property to show that $u$ is $C^2$. So if it helps, we can assume that $f$ is harmonic as well.
Answer for the question in the title:
$f(x)=\frac {x_1} {\|x\|}$ is a counterexample. Note that $f$ does not have a limit at the origin along the $x_1$ axis.