Extension Theorem From PDE Evans, 2nd edition, pages 268-270. My question is at the bottom of this post.

THEOREM 1 (Extension Theorem). Assume $U$ is bounded and $\partial U$ is $C^1$. Select a bounded open set $V$ such that $U \subset \subset V$. Then there exists a bounded linear operator $$E : W^{1,p}(U) \rightarrow W^{1,p} (\mathbb{R}^n) \tag{1}$$ such taht for each $u \in \mathbb{W}^{1,p}(U)$:
$\quad$(i) $Eu=u$ a.e. in $U$,
$\quad$(ii) $Eu$ has support within $V$
$\quad$(iii) $$\|Eu\|_{W^{1,p}(\mathbb{R}^n)} \le C \| u \|_{W^{1,p}(U)},$$ the constant $C$ depending only on $p,U$, and $V$.
DEFINITION. We call $Eu$ an extension of $u$ to $\mathbb{R}^n$
Proof. 1. Fix $x^0 \in \partial U$ and suppose first $$\partial u \text{ is flat near $x^0$, lying in the plane } \{x_n=0\}. \tag{2}$$
  Then we may assume there exists an open ball $B$, with center $x^0$ and radius $r$, such that 
  \begin{cases}
B^+ := B \cap \{x_n \ge 0 \} \subset \bar{U} \\
B^- := B \cap \{x_n \le 0 \} \subset \mathbb{R}^n-U.
\end{cases}
$\quad$2. Temporarily suppose also $u \in C^1(\bar{U})$. We define then 
  $$u(x) := \begin{cases}
u(x) & \text{if }x\in B^+ \\
-3u(x_1,\ldots,x_{n-1},-x_n)+4u(x_1,\ldots,x_{n-1},-\frac{x_n}{2}) &\text{if }x \in B^-.
\end{cases} \tag{3}$$
  This is called a higher-order reflection of $u$ from $B^+$ to $B^-$. 
$\quad$3. We claim $$\bar{u} \in C^1(B). \tag{4}$$
  To check this, let us write $u^- := \bar{u}|_{B^-}, u^+ := \bar{u}|_{B^+}$. We demonstrate first $$u_{x_n}^-=u_{x_n}^+ \text{ on } \{x_n=0\}.\tag{5}$$
  Indeed according to $\text{(3)}$, $$u_{x_n}^-(x)=3u_{x_n}(x_1,\ldots,x_{n-1}-x_n)-2u_{x_n}(x_1,\ldots,x_{n-1},-\frac{x_n}{2})$$and so $$u_{x_n}^-|_{\{x_n=0\}}=u_{x_n}^+|_{\{x_n=0\}}.$$ This confirms $\text{(5)}$. Now since $u^+ = u^-$ on $\{x_n=0\}$, we see as well that $$u_{x_i}^-|_{\{x_n=0\}} = u_{x_i}^+|_{\{x_n=0\}}\tag{6}$$ for $i=1,\ldots,n-1$. But then $\text{(5)}$ and $\text{(6)}$ together imply $$D^\alpha u^-|_{\{x_n=0\}} = D^\alpha u^+|_{\{x_n=0\}}\tag{6}$$ for each $|\alpha| \le 1$, and so $\text{(4)}$ follows.
$\quad$4. Using these calculations, we readily check as well $$\|\bar{u} \|_{W^{1,p}(B)} \le C \|u\|_{W^{1,p}(B^+)} \tag{7}$$ for some constant $C$ which does not depend on $u$.

The proof does continue on in the book, but I decide to stop here because my question is: How are the calculations used to derive relation $\text{(7)}$? I do understand the textbook says "readily check", so the proof should be straightforward. 
My work (which is 99.9% inaccurate, thus pointless to read anyway):
If $x \in B^+$, then $\bar{u}(x)=u(x)$. So $$\|\bar{u}\|_{W^{1,p}(B)} =\|u\|_{W^{1,p}(B^+)} \le C\|u\|_{W^{1,p}(B^+)}?$$ 
I don't think that's correct, and I'm also stuck in proving the case of $x \in B^-$.
 A: We want to estimate $\int_B |\nabla\bar u|^p $ in terms of $\int_{B^+} |\nabla u|^p$. To begin with, $$\int_B |\nabla \bar u|^p =  \int_{B^+} |\nabla u|^p +\int_{B^-} |\nabla \bar u|^p$$ 
On $B^-$, the function $\bar u$ is the sum of two functions: 
$$v(x) = -3u(x_1,\ldots,x_{n-1},-x_n)$$ 
and 
$$w(x) =  4u(x_1,\ldots,x_{n-1},-\frac{x_n}{2})$$
Compute the gradients with the chain rule. Get $|\nabla v(x)|=3|\nabla u(x_1,\dots,-x_n)|$. Integration with change of variable: $$\int_{B^-} |\nabla \bar v|^p = 3^p \int_{B^+} |\nabla   u|^p$$ Looks encouraging. 
For the second one: $|\nabla w(x)|\le 4|\nabla u(x_1,\dots,-x_n/2)|$, with inequality sign because there's a factor of $1/2$ attached to one  partial. Integration with change of variable: $$\int_{B^-} |\nabla \bar w|^p \le  4^p  2 \int_{B^+} |\nabla u|^p$$
Here $2$ is the Jacobian determinant. Strictly speaking, the integral on the right could be taken over smaller  domain, kind of flattened half-ball. But that smaller domain is contained in $B^+$, so why not integrate over all of $B^+$. 
Finally, $(A+B)^p\le 2^p (A^p+B^p)$, so 
 $$\int_B |\nabla \bar u|^p \le 2^p(3^p+4^p\cdot 2)\int_{B^+} |\nabla u|^p$$
