Trace operator counterexample

Question

This is homework so no answers please

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, prove there does not exist a bounded linear operator

\begin{equation} T:L^p(U) \to L^p(\partial U) \end{equation}

such that $Tu = \left. u \right|_{\partial U}$ whenever $u \in C(\overline{U}) \cap L^p(U)$.

This problem has been answered at A typical $L^p$ function does not have a well-defined trace on the boundary

I am just wondering if my attempt is correct:

let $f_{n}(x)=\frac{1}{|x-u|}1_{A_{n}}(x)$, where $A_{n}=\{x\in U:dist(x,\partial U)\geq \frac{1}{n}\}$ and for some $u\in \partial U$.

Therefore, $\int_{U}\frac{1}{|x-u|^{p}}1_{A_{n}}(x)dx\leq n^{p}|A_{n}|\leq n^{p}|U|<\infty$ so $f_{n}\in L^{p}(U)$. So the trace of $f_{n}$ is zero i.e. $Tf_{n}=0$ on $\partial U$.

But $ f_{n}\to \frac{1}{|x-u|}$, which is not $L^{p}(\partial U)$ and more importantly at $x=u$ we get $Tf=\infty$.

Since T is continuous, $limT( f_{n})\neq T(f)$.

Thanks

Answer 1

Since you are just wondering if your attempt is correct.

No.

1. Your limit function $1/|x-u|$ is not in $L^p(U)$ either, if $P$ is larger than the dimension of your space.
2. Furthermore, your $f_n$ are not continuous. While the question does not require that, it forms a good hint.

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But your basic idea is good. Two suggestions:

1. You need to factor in the dimension $d$ of the ambient space $\mathbb{R}^d \supset U$; you can do this by replacing $1/|x-u|$ by $1/|x-u|^\alpha$.
2. Instead of approximating by a cut-off in the domain $U$, approximate by truncating the function, setting $f_n = \min (1/|x-u|^\alpha, n)$.