Why it is not possible to define a continuous functional such that \begin{equation} \begin{split} T\colon L^p(\Omega)& \rightarrow L^p(\partial\Omega)\\ u&\mapsto Tu=u_{|_{\partial \Omega}}\\ \end{split} \end{equation}?
MY ATTEMPT: Since $C_{0}(\Omega)$ is dense in $L^p(\Omega)$. I defined $u_m=\frac{1}{1+m\; \mathscr{d}(x, \partial \Omega)}\in C_o(\Omega).$ That is a sequence in the subset of continuous functions with compact support. It is known $\partial \Omega$ we have $u_m=1$, also $0\leq u_m\leq 1\in \Omega$, that is $u_m\to 0$ a.e. $\in \Omega$.
Then if is considered that this functional exists (as in trace theorem), hence considering dominated convergence theorem, one can write $$\|Tu_n\|_{L^p(\partial \Omega)}\leq C\|u_n\|_{L^p(\Omega)}\to 0$$
However, is a contradiction since $Tu_n=u_n=1\; \forall x\in \partial \Omega$. Therefore this operator does not exist.
MY DOUBT: Do I really need to prove that, or if I say that there is no weak derivatives in the domain (that is, $L^p(\Omega)$) is that enough?