In Partial Differential Equations 2ed of Lawrence C.Evans, there is a detail in the proof of trace theorem of Sobolev space where I have some doubt.
I think the underlined formula use Green formula or integration by part. This need the integrand continuously differentiable. If $p>1$, it is right, but if $p=1$, I don't know why it is right.
Best Answer
Since $u\in C^1(\overline U)$, $|u|$ is Lipschitz, hence absolutely continuous. The fundamental theorem of calculus applies to such functions: the value at $x_n=0$ is related to the integral of $x_n$-derivative in the usual way (the other boundary term is zero since $\zeta$ is compactly supported).
(From a comment by John Ma)