Question of $u\in L^p(U)$ does not have a trace on $\partial U$.

Question

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

I have seen other questions about this problem. However, the hint I was given is first that the open set is a unit ball.

But I have no idea about how to do it, Can anyone give me some further hint or answer? Thanks!

Answer 1

Hint: Consider $u_n(x) = \max(0, 1 - n\; \text{dist}(x, \partial U))$.