This question is from PDE by Evans, 1st edition, Chapter 5, Problem 14. It has been posted here previously, however, I cannot quite put all the information together from the responses there. Hopefully you can help me now. The problem is as follows:
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 $u \in C(\overline{U}) \cap L^p(U)$.
So what we're trying to find is a bounded sequence of functions that go to infinity at the boundary, correct? Because then $Tu = \left. u \right|_{\partial U}$ is undefined.
I like the idea of using the $\mathrm{dist}(x, \partial U)$ function to do this. I was thinking about defining the function
\begin{equation} u(x) = \frac{1}{\mathrm{dist}(x,\partial U)}, \end{equation}
which must be in $L^p(U)$ since $U$ is bounded (hence integral of $\varepsilon$ over $U$ is finite, right?). But this function is not continuous at the boundary, and so $u \notin C(\overline{U})$. Can this be modified somehow?
Another option is to use that the boundary is $C^1$. So for each $x^0 \in \partial U$ there exists an $r > 0$ and a $C^1$ function $\gamma : \mathbb{R}^{n-1} \to \mathbb{R}$ such that $U \cap B(x^0,r) = \{ x \in B(x^0,r) \ | \ x_n > \gamma(x_1,\ldots, x_n) \}$. So around every point on the boundary we can find a ball where the n'th coordinate is greater than the $C^1$ function $\gamma$. How can we use this?
The key is that you can not find a bounded operator. Such an Operator is continuous. So what happen when you plug in a sequence of continuous functions?
Assume $\Omega=(0,1)$ and choose $$f_n (x):=\begin{cases} 1 &\text{ on } [0,1-1/n]\\ n-nx &\text{ on } [1-1/n,1]\end{cases}$$ So $f_n$ falls linearly to zero on $[1-1/n,1]$.
Each $f_n$ is a continuous function and has trace zero in $1$. Now we restrict the trace operator to $x=1$. Now $0=\lim_n T(f_n)\neq T(f)=1$ since $f_n\to 1$ in $L^2$.