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Apr 22, 2012 at 0:09 answer added Piero D'Ancona timeline score: 6
Apr 21, 2012 at 22:40 comment added Analysis Now For my case, $n = 2$.
Apr 21, 2012 at 22:37 comment added Analysis Now @ Prof. Deane Yang : Thank you. If we assume $ p > n $, then is there any condition on $u$ that would make sure that the smooth Sobolev function $u$ on $ U $ can be extended to $Eu \in C^{\infty}(\mathbb{R}^n) \cap W^{1,p}(\mathbb{R}^n)$. Is Holder continuity of $u$ or that $ u $ is Lipchitz enough for this ? Actually, I am looking at this problem as a part of another problem, and there I have $ p > 2, u $ is Lipchitz upto the boundary, because its derivatives are bounded in the interior.
Apr 21, 2012 at 21:58 comment added Deane Yang If $p > n$, then this won't work, because $W^{1,p} \subset C^{\alpha}$ for some $\alpha > 0$. So if $u$ is not Holder continuous on $\overline{U}$, it cannot be extended in $W^{1,p}$ beyond $\overline{U}$.
Apr 21, 2012 at 21:19 answer added Pietro Majer timeline score: 4
Apr 21, 2012 at 20:58 history asked Analysis Now CC BY-SA 3.0