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Let $U$ be bounded, with a $C^1$ boundary. Show that a ''typical'' function $u\in L^p(U) (1\leq p < \infty)$ does not have a trace on $\partial U$. More precisely there does not exist a bounded linear operator $$T:L^p(U)\rightarrow L^p(\partial U) $$ such that $Tu=u|_{\partial U}$ whenever $u\in C(\bar{U})\cap L^p(U)$

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marked as duplicate by Meta, studiosus, Jack D'Aurizio, Jonas Meyer, graydad Jan 11 '15 at 20:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

There is another answer at this website, the link is: – user76346 May 6 '13 at 8:46

Hint: Construct an $L^p$ bounded sequence of continuous functions that look "nice" in the interior of $U$, but grow fast near the boundary. In order to construct this sequence explicitly, you can use the $C^1$ boundary condition in order to pass to local coordinates.

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Another option is to build such functions out of $\operatorname{dist } (x,\partial U)$. That would work for rough domains too. – user53153 Dec 28 '12 at 16:09