# 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 $$\displaystyle\int_{\mathbb{R^N}_{+}}\vert Dw_m - Du\vert^p dx$$.

Also the (12) is not so clear for me. Any kind of help is well accepted. Thank you.

1. 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

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