I want to proof that $\|\nabla (\cdot)\|_{L^2(\Omega)}$ is a norm on $H^1_0(\Omega)$, $\Omega \subseteq \mathbb{R^2}$, $\Omega$ a polygonal domain.
Does the fact that $\Omega$ is a polygonal domain, changes the usual proof somehow?
For the non-degeneracy I have that
$0=\|\nabla (u)\|_{L^2(\Omega)} \implies 0=\int_\Omega |\nabla u|^2 dx$
$\implies \nabla (u)=0 $ a.e
$\implies u=c \in \mathbb{R} $ a.e
If $u$ is in $H^1_0(\Omega)$, it should be zero at the boundary, but I guess I can't say u=0, because u does not have to be continuous. Can someone clear up this? HOw do I conclude u =0 ?
Additionaly the rest of the proof I did it like this, for instance , for $\alpha \in \mathbb{R}$
$\|\nabla (\alpha u)\|_{L^2(\Omega)}= |\alpha| \|\nabla u\|_{L^2(\Omega)}$
$\|\nabla (u+v)\|_{L^2(\Omega)} = \|\nabla u + \nabla v\|_{L^2(\Omega)} \le \|\nabla u\|_{L^2(\Omega)} + \|\nabla v\|_{L^2(\Omega)}$
But I am not sure, since I guess I have to use the fact that $\Omega$ is polygonal
Normally, you use the definition $$ H_0^1(\Omega)=\overline{C_c^{\infty}(\Omega)} $$ where the closure is taken in the Sobolev norm. Furthermore, the trace operator is continuous, that means: $$ T:H^1_0(\Omega) \to L^2(\partial \Omega) $$ is a continuous linear operator assigning boundary values to a Sobolev function. This is where you need your domain to be polygonal: It has a Lipschitz boundary, which is a requiremnt for the trace operator to be well defined and continuous. Now, you can use the equivalence of defintions $$ u \in H^1_0(\Omega) \iff u \in H^1(\Omega) \; \textrm{and }\; Tu=0. $$ and note that the trace operator agrees on functions continuous up to the boundary $u \in C( \overline{\Omega }) $. Since $u(x)=c \; a.e.$, you can chose the representative $u(x)=c$, dropping the almost everywhere and conclude that $Tu=0 \iff c=0$.