There is a sentence in Evans I cannot justify.
The claim made is $u=g$ on $\partial U$ in the trace sense. Why? I understand that $u\in H^1$ implies $u\in W^{1,p}(U)$. But we also need the assumption $u\in C(\bar{U})$ to conclude $u=g$ on $\partial U$ in the trace sense, right? But there is no where in the book this is specified.
Hope someone can clarify. This is on p.315.