A typical $L^p$ function does not have a well-defined trace on the boundary

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 $L^2$.

For completeness, I'll expand the idea in this comment. The construction does not require $C^1$ boundary, and works in every bounded domain. Let $$ u_n(x) = (1-n\operatorname{dist}(x,\partial U))^+$$ which is a continuous function on $\overline{U}$. Since the sequence $u_n^p$ is decreasing, it is dominated by $u_1^p$, which is integrable. Hence $$ \lim_{n\to\infty}\int_U u_n^p = \int_U \lim_{n\to\infty} u_n^p = 0 $$ On the other hand, $$\int_{\partial U}u_n^p = \int_{\partial U}1\not\to0$$ which yields the claim.

I like your idea of taking a function $\gamma:\mathbb R^+_0\to\mathbb R_0^+$ and composing with the function $d=\operatorname{dist}(\cdot,\partial U)$. This is how your idea will actually work:

You picked $\gamma(t)=1/t$ but that may be too singular to guarantee $L^p$, however $\gamma(t)=[\log 1/t]_+$ should work (possibly using the coarea formula).

To cure the fact that $\gamma(d(\cdot))$ has no trace, you may cut it off by defining the sequence $u_n(x)=n\wedge\gamma(d(x))$, whose trace is constant $n$ and by Lebesgue dominated convergence (or Fatou-Lebesgue theorem) all integrable with integral bounded by that of $\gamma(d(\cdot))$.