Let $\Omega\subset\mathbb{R}^3$ an open bounded domain (without holes) with boundary $\partial\Omega$ and let $\Omega_1\subset\Omega$.
We consider the domain $\Omega\setminus\Omega_1$, that is, a domain with one hole.
My question is:
$\{w\in H^1(\Omega-\Omega_1):w=0\textrm{ on }\partial\Omega\textrm{ and }w=\textrm{ constant on }\partial\Omega_1\}$ a closed set? Why?
I want to know if it is a Hilbert to use Lax Milgram Theorem in other problem.
Thanks in advance !
This is the result of Trace operator. Namely, we have there exists a linear bounded operator $T$: $H^(\Omega)\to L^2(\partial \Omega)$, provided that $\Omega$ has at least Lipschitz boundary. That is, you have for $u\in H^1(\Omega)$, $$\|u\|_{L^1(\partial \Omega)}\leq C\|u\|_{H^1(\Omega)} $$
Let define set $D:=\{u\in H^1(\Omega\setminus\Omega_1):\, u=1\text{ on }\partial\Omega\text{ and }u=a\text{ on }\partial \Omega_1\}$ where $a$ is a constant. Now, assume $(u_n)\subset D$ such that $u_n\to u$ in $H^1(\Omega\setminus\Omega_1)$, then by Trace operator, we have $$\|u_n-u\|_{L^2(\partial (\Omega\setminus\Omega_1))}\leq C\|u_n-u\|_{H^1(\Omega\setminus\Omega_1)} $$
Then we know that $u_n\to u$ a.e. on $\partial(\Omega\setminus\Omega_1)$ with respect $\mathcal{H}^{N-1}$ measure. Hence, we conclude that $u\in D$ and thus $D$ is closed.