show that a certain bounded linear operator does not exist

Question

I'm trying to do problem 8 from section 5.10 on Evans' PDE book.

Basically the problem asks if $U$ is a bounded open subset of $\mathbb R^n$ with $C^1$ boundary and $u \in L^p (U)$, show that there does not exist a bounded linear operator from $L^p(U)$ to $L^p (\partial U)$ such that $Tu = u |_{\partial U } $ whenever $u \in C(\bar U) \cap L^p (U)$.

So my attempt is since the Trace Theorem is for all $u \in W^{1,p } (U)$. I have to find a sequence of function $u$ which is in $L^p (U)$ but $Du $ is not in $L^p $ and trying to make their $L^p ( \partial U)$ value blows up. I tried the function $f(x) = 1/|x|^\alpha$ with $p\alpha < n < p (\alpha + 1)$ and $U = B(0,1)$, but this approach doesn't work. I think I need to find functions with the above properties ($u \in L^p$ but $Du \notin L^p $) and grows rapidly near the boundaries but I don't know these functions. Any help is appreciated!

Answer 1

Assume that there is such an operator $T$.

Take $w\equiv 1$ in $U$. Then approximate $u$ by a sequence of functions $w_k \in C_c^\infty(U)$, $w_k\to w$ in $L^p(U)$, which works due to density.

Then $u:=1-w$ is zero, and $u_k:=1-w_k$ converges to $u$ in $L^p(U)$. Then $Tu=0$ and $Tu_k=1$, hence $Tu_k$ cannot converge to $Tu$. Contradiction.

Answer 2

No, you don't need (directly) a blow-up of the derivative near the boundary.

Rather try to construct functions $f_n \in C(\bar U) \cap L^p(\Omega)$, such that $\|f_n\|_{L^p(U)}$ is small, but with $\|T \, f_n\|_{L^p(\partial U)}$ being large.