Trace theorem for Sobolev functions: what is the significance of continuous extension to the boundary?

Question

Why in the proof of the trace theorem in L.C.Evans' book on PDE, he has considered functions in $W^{1,p}(\Omega)\bigcap C(\overline{\Omega})$ when it is enough to choose functions in $W^{1,p}(\Omega)$, which tackles a bigger issue?.

Answer 1

The trace theorem, as stated by Evans:

Assume $U$ is bounded and $\partial U$ is $C^1$. Then there exists a bounded linear operator $T:W^{1,p}(U)\to L^p(\partial U)$ such that

1. $Tu=u_{|\partial U}$ if $u\in W^{1,p}(U)\cap C(\overline{U})$
2. $\|Tu\|_{L^p(\partial U)}\le C\|u\|_{W^{1,p}(U)}$ for all $u\in W^{1,p}(U)$, with $C$ depending only on $p$ and $U$.

The theorem deals with the entire space $ W^{1,p}(U)$: the trace operator $T$ is defined and studied on this space. Property (1) says that the concept of trace offered by this operator agrees with the restriction to the boundary when the latter is defined. For functions that don't have a continuous extension to the boundary, we don't have $u_{|\partial U}$ to talk about.

To summarize: the theorem defines the trace for every Sobolev function. Part 1 is just a consistency check, which we can do only when there is another concept of boundary values available.