Trace Theorem: a question about Evans' proof This is a part of the proof of the Thoerem "Trace-zero functions in $W^{1,p}(\Omega)$ in the book of Evans. I don't understand the inequality involving $\displaystyle\int_{\mathbb{R^N}_{+}}\vert Dw_m - Du\vert^p dx$.
Could anyone please help me to understand why it holds true?
Also the (12) is not so clear for me. Any kind of help is well accepted. Thank you.


  
*Next let $\zeta \in C^{\infty}(\mathbb{R})$ satisfy
  $$
\zeta \equiv 1 \text { on }[0,1], \zeta \equiv 0 \text { on } \mathbb{R}-[0,2], \quad 0 \leq \zeta \leq 1
$$
  and write
  $$
\left\{\begin{array}{l}
\zeta_{m}(x):=\zeta\left(m x_{n}\right) \quad\left(x \in \mathbb{R}_{+}^{n}\right) \\
w_{m}:=u(x)\left(1-\zeta_{m}\right)
\end{array}\right.
$$
  Then
  $$
\left\{\begin{array}{l}
w_{m, x_{n}}=u_{x_{n}}\left(1-\zeta_{m}\right)-m u \zeta^{\prime} \\
D_{x^{\prime}} w_{m}=D_{x^{\prime}} u\left(1-\zeta_{m}\right)
\end{array}\right.
$$
  Consequently
  $$
\begin{aligned}
\int_{\mathbb{R}_{+}^{n}}\left|D w_{m}-D u\right|^{p} d x \leq & C \int_{\mathbb{R}_{+}^{n}}\left|\zeta_{m}\right|^{p}|D u|^{p} d x \\
&+C m^{p} \int_{0}^{2 / m} \int_{\mathbb{R}^{n-1}}|u|^{p} d x^{\prime} d t\\
=:A+B.
\end{aligned}
$$
  Now
  $$
A \rightarrow 0 \quad \text { as } m \rightarrow \infty, \tag{11}
$$
  since $\zeta_{m} \neq 0$ only if $0 \leq x_{n} \leq 2 / m .$ To estimate the term $B$, we utilize inequality (9)
  $$
B \leq C m^{p}\left(\int_{0}^{2 / m} t^{p-1} d t\right)\left(\int_{0}^{2 / m} \int_{\mathbb{R}^{n-1}}|D u|^{p} d x^{\prime} d x_{n}\right) \tag{12}
$$

screenshot direct from book: https://i.stack.imgur.com/dZUOW.png
 A: Note that this section deals with the case $1\le p<\infty$. $\newcommand{\dd}{\mathop{}\!\mathrm{d}}$ First we compute
$$ Dw_n (x)= D\big(u (x)(1-\zeta_m(x))\big) = Du(x) (1-\zeta(mx_n)) - mu(x) \zeta'(mx_n)$$
therefore
\begin{align}I_n:= \int_{\mathbb R_+^n}|Dw_n-Du|^p \dd x  
&= \int_{\mathbb R_+^n} |Du(x) \zeta(mx_n) - mu(x) \zeta'(mx_n)|^p \dd x
\\
&\overset \star\le C \int_{\mathbb R_+^n}|\zeta_m|^p|Du|^p + m^p|\zeta'|^p |u|^p \dd x
\\
&\overset {\star\!\star}\le C \int_{\mathbb R^n_+} |\zeta_m|^p|Du|^p \dd x + C\int_0^{2/m}\int_{\mathbb R^{n-1}} m^p|u|^p \dd x' \dd t
\\
&=: A + B
\end{align}
The line marked $\star$ is by convexity of $\phi:[0,\infty)\to[0,\infty), \phi(t) = t^p$:
$$ (a+b)^p = 2^p\left(\frac{a+b}2\right)^p \le 2^{p-1} (a^p + b^p).$$
The line marked $\star\!\!\star$ is by using that $\zeta'\in C^\infty_c\subset L^\infty$ (know that the constant $C$ changed from line to line), and also $\int_{\mathbb R^n_+} = \int_0^\infty \int_{\mathbb R^{n-1}}$, together with the fact that $\zeta'$ is supported in $[0,2/m]$. Actually, its only nonzero when  $x_n\in [1/m,2/m]$ but this stronger inequality is not important for the proof.
The application of (9) to obtain (12) is easier: first I recall (9),

$$\int_{\mathbb{R}^{n-1}}|u(x', x_{n})|^{p} \dd x^{\prime} \leq C x_{n}^{p-1} \int_{0}^{x_{n}} \int_{\mathbb{R}^{n-1}}|D u|^{p} \dd x' \dd t \tag{9}$$ 

plugging into $B$ gives
$$ B=C\int_0^{2/m}\int_{\mathbb R^{n-1}} m^p|u|^p \dd x' \dd t\le Cm^p\int_0^{2/m} t^{p-1} \int_{0}^{t} \int_{\mathbb{R}^{n-1}}|D u(x',x_n)|^{p} \dd x' \dd x_n \dd t $$
now since the integrands are positive, just use $t<2/m$ to replace $\int_0^t$ with $\int_0^{2/m}$, and then pull $\int_{0}^{2/m} \int_{\mathbb{R}^{n-1}}|D u(x',x_n)|^{p} \dd x' \dd x_n$ out of the $t$ integral. This yields (12).
