See here.
Problem 5.14, Evans. 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 $u \in C(\overline{U}) \cap L^p(U)$.
In the link, Sally offers the following solution.
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.
Cookie comments the following in response, which didn't get a response back.
And we shouldn't use absolute values either because if $\text{dist}(x, \partial U) > {1\over n}$, the function $u_n$ would be positive due to the absolute value and not $0$. If we just use the "$+$" like you introduced, then the negative part of the function $u_n$ would just simply be $0$. I hope that it is correct.
Could anyone help resolve this issue?
If you used simply the absolute value, $|\cdot|$, instead of the positive part $(\cdot)^+$, then $u_n(x)$ would be large for $x\in U$ with $\operatorname{dist}(x,U^\mathrm c)>1$. You would not be able to (easily) apply Fatou's lemma (or dominated convergence) to prove $\lim_{n\to\infty} u_n=0$ in $L_p(U)$.