Trace Theorem: a question about Evans' proof

Question

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.

3. 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

Answer 1

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).