Harmonic function : if $u=0$ on $\Omega$ then $u=0$ on $\partial \Omega$.

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 by the maximum principle for Laplace's equation – Fritz Dec 29 '16 at 14:43
• Sorry, but for 1, why isn't continuity enough? I agree that if $u=0$ on $\partial \Omega$ then the maximum principle implies $u=0$ on $\Omega$, but the reverse direction seems much easier. – Matt Dec 29 '16 at 14:49

• Sorry, but there is something I didn't get. Do you agree that $\int_\Omega \|\nabla u\|=0\implies \nabla u=0$ on $\Omega$ If yes, I still don't see why we need simply connexity. May be you could give a counter example ? By the way, a domain is by definition connected (so simply connexity looks more weird here, since I don't think that we need it) tks a lot – MSE Dec 29 '16 at 16:20