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My problem is the following: I want to find a bounded domain $\Omega\subset\mathbb{R}^N$ such that if $u\in W_0^{1,p}(\Omega)$, $p\in (1,\infty$), then the extension by zero of $u$ to $\mathbb{R}^N$ is not in $W_0^{1,p}(\mathbb{R}^N)$.

If such $u$ do exist, then the problem must be a problem of "differentiability" in $\partial\Omega$, but I could not figure out how to construct such $u$.

I would like to note that if $\Omega\in C^{0,1}(\Omega)$, then such $u$ does not exist because we have a extension operator between $W^{1,p}(\Omega)$ and $W^{1,p}(\mathbb{R}^N)$.

Thank you

Tomás
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There is no such domain. By definition of $W^{1,p}_0(\Omega)$, $u$ is a $W^{1,p}$ limit of smooth compactly supported functions $u_n$. Extend each $u_n$ by zero; you now have a sequence of smooth compactly supported functions which converges in $W^{1,p}(\mathbb R^N)$. Its limit is an element of $W^{1,p}_0(\mathbb R^N)$, which is nothing else but the zero extension of $u$.

The problem of smoothness of $\partial \Omega$ comes up when you interpret the vanishing of $u$ on the boundary differently, i.e., in the sense of traces. Then the question becomes: does having zero trace imply $u\in W^{1,p}_0(\Omega)$? If $\partial \Omega$ is Lipschitz, this is true (e.g., Theorem 15.29 in A first course in Sobolev spaces by Leoni). (I see that you know the last part, but it was natural to include it for completeness).

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  • You are right, I think I made a confusion here. Thank you. – Tomás Jul 10 '13 at 15:33
  • @user85506 If a function belongs to $W_0^{1,p}(\Omega)$ then clearly its zero extension belongs to $W^{1,p}(\mathbb R^N)$. Why is the converse true? – Umberto P. Jul 10 '13 at 16:04
  • @UmbertoP. I saw a proof once but forgot it. Hence, it's an exercise for the reader. (But I will post a proof if I remember). – 40 votes Jul 10 '13 at 16:27
  • @UmbertoP., Brezis gives a proof of it, but he ask some regularity on the boundary: $C^1$. – Tomás Jul 10 '13 at 17:24
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    Consider the following: let $\Omega$ be the unit ball (in two dimensions) with the segment $[-1/2,1/2]$ deleted from the $x$-axis. Let $f(x) = 1 - |x|^2$. The zero extension of $f$ belongs to $W^{1,p}(\mathbb R^2)$ but $f \notin W_0^{1,p}(\Omega)$. – Umberto P. Jul 10 '13 at 17:40
  • @UmbertoP. Extending this $f$ by $0$ on $\Omega^c$, we get a function that is discontinuous on every line segment crossing $[-1/2,1/2]$. It is not ACL, hence not in $W^{1,p}$. That said, we are getting into the business of representatives, etc... so I think it's best to just delete my unsupported claim. Which I just did. – 40 votes Jul 10 '13 at 18:11
  • Fair enough, but $f$ is in $W^{1,p}$: redefining $f$ on a set of Lebesgue measure zero does not remove it from the Sobolev space. – Umberto P. Jul 10 '13 at 19:06
  • @UmbertoP. I think that this result is also valid for $C^{0,1}$ boundarys. – Tomás Jul 10 '13 at 23:15