Trace of $L^p$ function

Question

For $U$ a bounded domain in $\mathbb{R}^n$, why is it that, in general, an $L^p$, $1\leq p<+\infty$, function does not have a trace on the boundary of $U$?

Thanks in advance.

Answer 1

Here is a counterexample for $n=1$, which can me easily modified to hold in higher dimensions:

Consider the sequence $f_n(x):=(1-nx)_+$ for $x\in (0,1)$. $(\cdot)_+$ denotes the positive part. By dominated convergence $f_n\to 0$ in $L^p(0,1)$. Assume there exists a contiunous and linear trace operator $T$ on $L^p(0,1)$ that evaluates the functions in $x=0$ (and of course provides the correct result for continuous functions). Then we must have $1=T(f_n)\to T(f)=0$ which is a contradiction. Note that the convergence of $f_n\to 0$ does not hold in $H^{1,p}(0,1)$.

For completeness: $p\in[1,\infty)$.