This was an exercice in my functional analysis course: Let $\Omega = B(0,1) \subset \mathbb R^N$, $N \ge 3$, and let $u \in C^\infty(\Omega \backslash \{0\})$ be a harmonic function such that $$\lim_{x \to 0} \frac{u(x)}{|x|^{2 - N}} = 0.$$ Under these hypothesis, I must prove that there exists a harmonic function $v \in C^\infty(\Omega)$ such that $v = u$ on $\Omega\backslash \{0\}$, but I really do not see how I should do that. At some point I thought that I could try with Weyl Lemma that states the following:
Let $u \in L^1_{loc}(\mathbb R^n)$ with $\Delta u = 0$ in the sens of distributions. Then there exists $v \in C^2(\mathbb R^n)$ with $u = v$ a.e. and $\Delta v = 0$.
But as nothing ensure me that $u \in L^1_{loc}$ so... Moreover I do not use the fact that $u(x)/|x|^{2 - N}$ goes to zero. Maybe do I have to construct explicitly the function $v$ ? I am really struggling with that, any help is welcome!
Let $v$ solve the Dirichlet problem in $\Omega$ with $v=u$ on its boundary. Let $g(x)=\lvert x \rvert^{2-N}$. Then $g$ is positive and harmonic on $\Omega\setminus\{0\}$. If $w=\left(u-v\right)/g$ we have $w=0$ on $\partial \Omega$ and $\lim_{x \to 0}w(x)=0$. Also $ 0=\Delta\!\left(u-v\right)/g=\Delta w+\left(2/g\right)\sum_{i}g_{x_{i}}w_{x{i}}$ so the maximum principle applies to $w$.
For any $y\in\Omega\setminus\{0\}$ and $\epsilon >0$ there exists $r\le \lvert y\rvert$ such that $\lvert w\rvert \le \epsilon$ on $\partial B(0,r)$ and therefore by the maximum principle applied to $\Omega\setminus B(0,r)$ we have $\lvert w(y)\rvert \le \epsilon$. Therefore $w=0$ throughout its domain.