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It is well known that if $\Omega\subset\mathbb{R}^{N}$ open, $F\subset\Omega$ closed, such that $\mathcal{H}^{N−1}(F)=0$,where $\mathcal{H}^{N−1}$ denotes (N-1) dimensional Hausdorff measure, then $W^{1,p}(\Omega)=W^{1,p}(\Omega\backslash\Sigma)$ for $p\in (1,+\infty)$.

Question: Assume $F$ and $\Omega$ satisfy the conditions prescribed above. In this case, shall we have the following equality $W^{1,p}_{0}(\Omega)=W^{1,p}_{0}(\Omega\backslash\Sigma)$ ?

Thanks for any suggestion.

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  • $\begingroup$ I do not really see how this is possible, since by removing a set $\Sigma$ from $\Omega$ you could create much more "boudary" where $W_0^{1,p}$ functions would have to vanish, so no. $\endgroup$
    – Sascha
    Oct 1, 2018 at 12:11
  • $\begingroup$ @Sascha Thanks for your comment. I also doubt, but I need proof. $\endgroup$
    – Math777
    Oct 1, 2018 at 12:14
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    $\begingroup$ @Sascha For example, if $\Omega=B(0,1)\subset\mathbb{R}^{2}$ and $\Sigma=\{(0,0)\}$. What we shall obtain? In your example $\mathcal{H}^{0}(\Sigma)=1$. $\endgroup$
    – Math777
    Oct 1, 2018 at 12:18

1 Answer 1

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This is not true and a counterexample is based on the example in the comment of Math777.

Note that $W^{1,p}_0(\Omega\setminus F)\subset W^{1,p}(\mathbb{R}^n)$, because $W^{1,p}_0$ is the closure of $C_0^\infty(\Omega\setminus F)\subset C_0^\infty(\mathbb{R}^n)$. If $p>n$, then Sobolev functions $W^{1,p}_0(\Omega\setminus F)\subset W^{1,p}(\mathbb{R}^n)$ are continuous on $\mathbb{R}^n$. That means, every function in $W^{1,p}_0(\Omega\setminus F)$, when extended by zero outside $\Omega\setminus F$ is continuous.

Let $\Omega=B^n(0,1)$ and let $F=\{0\}$. Then $W^{1,p}_0(\Omega\setminus F)$ functions are continuous at $0$ and vanish at $0$, but functions in $W^{1,p}_0(\Omega)$ need not vanish at $0$.

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  • $\begingroup$ Thanks, Piotr for your answer. But why functions in $W^{1,p}_{0}(\Omega\backslash F)$ vinish at 0, here there is no trace, since $\Omega\backslash F$ does not have $C^{1}$ boundary? $\endgroup$
    – Math777
    Oct 1, 2018 at 13:30
  • $\begingroup$ @Math777 I edited my answer. I hope the argument is clear now. $\endgroup$ Oct 1, 2018 at 14:24

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