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 ?

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    $\begingroup$ 1) follows by the maximum principle for Laplace's equation $\endgroup$ – Fritz Dec 29 '16 at 14:43
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    $\begingroup$ 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. $\endgroup$ – Matt Dec 29 '16 at 14:49

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.

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    $\begingroup$ for 2), why what I did is not correct ? $\endgroup$ – MSE Dec 29 '16 at 14:51
  • $\begingroup$ @MSE Sorry for not adressing your question properly. I edited my answer. $\endgroup$ – Severin Schraven Dec 29 '16 at 15:00
  • $\begingroup$ 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 $\endgroup$ – MSE Dec 29 '16 at 16:20

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