Integration by parts in proof of the Trace Theorem in Evans PDE book

Question

After a long hiatus, I have finally resumed my reading of L. C. Evans's PDE book.

I am currently reading chapter 5. With some help (1, 2) from the community, I was able to understand the theorems regarding interior approximation and approximation up to the boundary of functions $u\in W^{k,p}(U)$ with smooth functions, and ultimately, the proof of the Sobolev extension theorem. Now, I am working on the proof of the trace theorem (page 271 in the second edition). I fully understand the motivation of the idea, that is, being able to define boundary values for weak solutions of PDEs. However, I am having difficulty reading the proof. I have included Evans's statement of the theorem and the first few lines of the proof:

THEOREM 1 (Trace Theorem). Assume $U$ is bounded and $\partial U$ is $C^1$. Then there exists a bounded linear operator $$T : W^{1,p}(U) \rightarrow L^p(\partial U)$$ such that $\quad$(i) $$Tu=u|_{\partial U}~~ \text{if}~~u \in W^{1,p}(U) \cap C(\bar{U})$$ and $\quad$(ii) $$\|Tu\|_{L^p(\partial U)} \le C \| U \|_{W^{1,p} (U)},$$ for each $u \in W^{1,p}(U)$, with the constant $C$ depending only on $p$ and $U$.

DEFINITION. We call $Tu$ the trace of $u$ on $\partial U$.

Proof. 1. Assume first $u \in C^1(\bar{U})$. As in the first part of the proof of Theorem 1 in §5.4 let us also intially suppose $x^0 \in \partial U$ and $\partial U$ is flat near $x^0$, lying in the plane $\{x_n=0\}$. Choose an open ball $B$ as in the previous proof and let $\hat{B}$ denote the concentric ball with radius $r/2$. $\quad$Select $\zeta \in C_c^\infty(B)$, with $\zeta \ge 0$ in $B$, $\zeta \equiv 1$ on $\hat{B}$. Denote by $\Gamma$ that portion of $\partial U$ within $\hat{B}$. Set $x'=(x_1,\ldots,x_{n-1}) \in \mathbb{R}^{n-1} = \{x_n=0\}$. Then $$\int_\Gamma |u|^p \, dx'\tag{A}$$ $$\le \int_{\{x_n=0\}} \zeta |u|^p \, dx'\tag{B}$$ $$=-\int_{B^+} (\zeta |u|^p)_{x_n} \, dx\tag{C}$$ $$=-\int_{B^+} |u|^p \zeta_{x_n} + p|u|^{p-1} (\operatorname{sgn} u)u_{x_n} \zeta \, dx\tag{D}$$ $$\le C \int_{B^+} |u|^p + |Du|^p \, dx\tag{E}$$ where we employed Young's inequality, from §B.2.

2. If $x^0\in\partial U$, but $\partial U$ is not flat near $x^0$, we as usual straighten out the boundary near $x^0$ to obtain the setting above. Applying estimate $(1)$ (this is the inequality obtained in the previous step) and changing variable, we obtain the bound $$\int_{\Gamma}|u|^pdS\leq C\int_U |u|^p+|Du|^pdx\tag{F}$$ Where $\Gamma$ is some open subset of $\partial U$ containing $x^0$.

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My questions.

First question: My first problem is with the transformation of the integral on $\Gamma$ in the final section. I am able to understand how he goes from $(\mathrm A)$ to $(\mathrm B)$ - this is because $\zeta\equiv 1$ in $\Gamma\subset \hat B$. I am also able to understand how he goes from $(\mathrm C)$ to $(\mathrm D)$ - this is just the product rule. And, thanks to this question I understand how he goes from $(\mathrm D)$ to $(\mathrm E)$. However, I don't understand how he goes from $(\mathrm B)$ to $(\mathrm C)$. I think it is some kind of integration by parts computation, but I am really not quite seeing the details.

Second question: In the proof of part (ii), I don't understand why the domain of integration on the right hand side of $(\mathrm F)$ is all of $U$. In the previous part, the final inequality obtained only covered a portion of $U$ , namely, $B^+$. So, how are we able to bound the norm on the whole domain $U$?

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symbols used: Here, $U\subset \mathbb R^n$ is an open, bounded, simply connected set with $C^1$ boundary. $B=\mathbb B^n(x^0,r)$ is the open ball centered at $x^0$ with radius $r$ in $\mathbb R^n$. And, $B^+:=B\cap U$ and $B^-:=B\cap (\mathbb R^n\setminus U)$.

Answer 1

Your first question is answered by the fundamental theorem of calculus: First we have $$\int_{\{x_n=0\}} \zeta |u|^p \ dx' = \int_{B \cap \{x_n=0\}}\zeta |u|^p \ dx'$$ by the compact support of $\zeta$ in $B$. Now we artificially add an integral in the $x_n$-direction (FTC): $$\int_{B \cap \{x_n=0\}}\zeta |u|^p \ dx'= -\int_{B \cap \{x_n=0\}} \int_0^{\sqrt{1-|x'|^2}}\left(\zeta |u|^p\right)_{x_n} \ dx_n \ dx'$$ This is a valid equality since the inner integral gives you $$\zeta |u|^p\left(x', \sqrt{1-|x'|^2}\right) - \zeta |u|^p(x',0) = -\zeta |u|^p(x',0)$$ again by the compact support of $\zeta$. Now it only remains to realise how the integration domain looks like to see that $$-\int_{B \cap \{x_n=0\}} \int_0^{\sqrt{1-|x'|^2}}\left(\zeta |u|^p\right)_{x_n} \ dx_n \ dx' = \int_{B^+} \left(\zeta |u|^p\right)_{x_n} \ dx.$$

As to the second question, we have just enlarged the domain of integration from $B^+$ (or the "unstraightened" version of it) to $U$. This is valid because the integrand is non-negative, so we only enlarge the right hand side.