Let $\Omega $ a bounded domain of $\mathbb R^d$ and $u\in \mathcal C^2(\Omega )\cap\mathcal C(\bar \Omega )$ harmonic in $\Omega $.
1) If $u=0$ on $\Omega $ then $u=0$ on $\partial \Omega $
2) If $\Omega $ is simply connected and $\partial _\nu u=0$ on $\partial \Omega $, then $u$ constant.
3) If $(u_n)_n$ is a sequence of harmonic function that converge uniformly to $u$, does $u$ harmonic ?
My attempt
1) Using divergence theorem, we have that $$\int_{\partial \Omega }\nabla u\cdot \nu=\int_\Omega \Delta u=0,$$ but I can't conclude.
2) I don't really see in what simply connectness is important here. If $\partial _\nu u=0$ on $\partial \Omega $, using divergence theorem $$0=\int_{\Omega } u\underbrace{(\nabla u\cdot \nu)}_{=\partial _\nu u=0}\underset{div}{=}\int_\Omega div(u\nabla u)=\int_\Omega \|\nabla u\|^2+\int_\Omega u\underbrace{\Delta u}_{=0}=\int_\Omega \|\nabla u\|^2,$$ and thus $\nabla u=0$ in $\Omega $, and thus $u$ is constant in $\Omega $. So, why the simply connectness of $\Omega $ ?
3) I think we can construct a sequence $\mathcal C^2$ function s.t. the limit is not $\mathcal C^2$, but I can't find which one. Any idea ?
1.) Follows from continuity.
3.) Every harmonic function satisfies the mean value property. Because on compacta (and sequence uniform convergence) you can interchange limit and integration to obtain that the limit function satisfies the mean value property. Thus, the limit function is harmonic.
2.) Edit: I do not see why one needs simply connected.