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Generally we know that $W^{1,p}(\Omega)$ is bigger then $W^{1,p}_0(\Omega)$ for arbitrary $\Omega\subset \mathbb R^N$ and also we have $W_0^{1,p}(\mathbb R^N)=W^{1,p}(\mathbb R^N)$. Today I found on H. Brezis book a remark states that for domain $\Omega$, if $\mathbb R^N\setminus \Omega$ is "sufficiently thin" and $p<N$, then $W_0^{1,p}(\Omega)= W^{1,p}(\Omega)$.

What does he means by "sufficiently thin"? And how we prove this remark? I don't have an idea to get start...

Also, it gives an example. By defining $\Omega:=\mathbb R^N\setminus\{0\}$ and $N\geq 2$, one can prove that $H_0^1(\Omega)=H^1(\Omega)$.

I tried to prove this example by using approximation. But I think, since $C_c^\infty(\Omega)\neq C^\infty (\Omega)$, why should we have $H_0^1(\Omega)=H^1(\Omega)$?

Any help is really welcome. Thank you!

spatially
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    The necessary and sufficient condition is pretty technical and involves Sobolev capacity. See Theorem 4.8 here. This paper deals with more general metric spaces, but the arguments are adapted after Euclidean spaces; I haven't found an easily available source for the Euclidean case. The proof of Theorem 4.8 will suggest you how to deal with $H_0^1=H^1$: multiply by a suitable smooth cutoff function. –  Jan 05 '15 at 03:23
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    There is also a characterization in terms of polar set. Take a look in Adams-Fournier book, chapter 3. – Tomás Jan 05 '15 at 10:39
  • thanks guys! I'll check the reference you provided. – spatially Jan 05 '15 at 14:33

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