0

$B=B(0,1)\subset\mathbb{R}^N$ and $\Omega=B\setminus\{0\}$.

(i) Assume $N=1$ and prove $H^1_0(\Omega)\neq H^1_0(B)$.

(ii) Take $N\ge 2$. Does $H^1_0(\Omega)=H^1_0(B)$?

I don't even know where to start with. I think one can probably use the fact that $H_0^1(\Omega)=\overline{C_0^\infty(\Omega)}^{H^1(\Omega)}$ to construct sequences in order to give a counterexample or prove the statement. But the detail is really beyond me. Any insight would be helpful. Thank you very much!

borealopithecus
  • 1
  • question looks like this one: (also not answered yet but you might want to check for updates there) http://math.stackexchange.com/questions/1756751/prove-for-an-open-omega-subset-mathbbrn-with-x-in-omega-that-u-in – Max Apr 27 '16 at 07:18

1 Answers1

0

I think the reason that $H^1_0 (\Omega ) \neq H^1_0 (B) $ for $N=1$, is that $\Omega $ is no longer connected if you remove $ \{ 0 \}$. Thus you can consider $\overline{C_0^\infty((0,1)}^{H^1(\Omega)} \cup \overline{C_0^\infty((-1,0)}^{H^1(\Omega)} \neq \overline{C_0^\infty((-1,1)}^{H^1(B)}$ .

matmat
  • 21
  • 4