1

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 $u\in C(\bar{U})\cap L^p(U)$

Cookie
  • 13,532
  • 1
    There is another answer at this website, the link is: http://math.stackexchange.com/questions/332599/pde-question-in-evans –  May 06 '13 at 08:46

1 Answers1

1

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.

Christopher A. Wong
  • 22,445
  • 3
  • 51
  • 82
  • Another option is to build such functions out of $\operatorname{dist } (x,\partial U)$. That would work for rough domains too. –  Dec 28 '12 at 16:09