Let $B_{1}$ be the closed unit ball in ${R}^{n}$ centered at 0, n $\geq$ 2. Suppose u is $C^2(B_{1}\setminus \lbrace0 \rbrace)$ and harmonic on $B_{1}\setminus \lbrace0 \rbrace$. Show that if $u(x)$ = $o(lnr)$, as $r = \vert x \vert$ gose to 0, then u can be extended at 0 so that it's harmonic on the whole ball.
The hint is asking us to think about homogeneous harmonic PDE, but I got no idea for that. Moreover, I don't see the possibility of extending a harmonic function at a singular point while keeping it to be harmonic. Any help would be thankful!
This is more of an extended comment than a full answer as it uses a powerful result from the theory for $n \ge 3$
For $n=2$ an easy solution uses the theory of complex analytic functions and I showed it in this answer
For $n \ge 3$ I don't know an easy way to do it as this is included in Bocher theorem (which gives a simple representation for positive harmonic functions on the punctured ball - for $n=2$ we can easily dispense with positivity as showed in the answer linked above) for which this paper of Axler, Bourdon and Ramey gives a reasonably simple proof.
Note that wlog we can assume $u$ real (as we can do the extension on the real and imaginary parts separately) and then since $n>2$ we know that $|x|^{2-n}$ is harmonic on $\mathbb R^n-0$. Since $u$ is continuous on the unit sphere we have that $u \ge -A, A \ge 0$ there so $u+A+|x|^{2-n}$ is harmonic and positive near $0$ because of the bound on $u$ and is positive on the unit sphere by our choices, hence by the minimum principle it must be positive on the whole punctured ball and Bocher's theorem applies to give $u=v+C|x|^{2-n}$ for $v$ harmonic on the full ball. But again the bound at $0$ implies $C=0$ so $u=v$
Of course, there may be a more direct proof using the fundamental solution for $n \ge 3$.