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Let $\Omega$ be an open set of $\Bbb R^d$: consider the following function spaces

  • $H_0^1(\Omega)$, i.e. the closure of $C_c^\infty(\Omega)$ in $H^1(\Omega)$
  • $H_*(\Omega)$, i.e. the closure of $C_c^\infty(\Omega)$ in $H^1(\Bbb R^d)$.
  • $H_{\Omega}(\Omega)=\{u\in H^1(\Bbb R^d):\ u= 0 \text{ a.e on } \Omega^c\}$.

Question: does the above spaces coincide? If not when are they equal?

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    $\begingroup$ The first two should be equivalent, as the $H^1(\Omega)$ norm and $H^1(\mathbb{R}^d)$ norm of $C^\infty_c(\Omega)$ functions are the same. $\endgroup$ Commented Apr 8, 2022 at 16:38

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The first two are equivalent, as the $H^1(\Omega)$ norm and $H^1(\mathbb{R}^d)$ norm coincide for $C^\infty_c(\Omega)$ functions.

The third is in general different:

If you let $d = 1$ and $\Omega = \mathbb{R}\setminus \{0\}$, you see that $H_{\Omega}(\Omega) = H^1(\mathbb{R}) \supsetneq H^1_0(\Omega)$. You can create similar examples in higher dimensions (by omitting a hyperplane instead of a point). On the other hand, if $\Omega$ is a Sobolev extension domain (see e.g. Leoni's First Course in Sobolev Spaces) then $u\in H^1_0(\Omega) \iff$ extending $u$ by zero to the exterior gives an $H^1(\mathbb{R}^d)$ function. And in that case, the third is equivalent to the first two.

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    $\begingroup$ Your next question would be: when is $\Omega$ a Sobolev extension domain? It is sufficient that $\Omega$ has Lipschitz boundary. But I don't think there is a known (useful) characterization result. $\endgroup$ Commented Apr 8, 2022 at 17:44
  • $\begingroup$ please how do you show that the inclusion is strict? $\endgroup$
    – Guy Fsone
    Commented Nov 8, 2022 at 12:06
  • $\begingroup$ @GuyFsone: by Morrey's inequality every element in $H^1_0(\Omega)$ is almost everywhere equal to a continuous function that vanishes at $0$. So a smooth bump function that does not vanish at $0$ is an element of $H^1(\mathbb{R})$ but not in $H^1_0(\Omega)$. $\endgroup$ Commented Nov 8, 2022 at 15:57

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