I was reading the book "Partial Differential Equation" written by Lawrence C. Evans, coming up with a question.
On page $271$, Evans wrote,
$\tilde{u} := \begin{cases} u(x) && \text{if } x \in B^+ \\ -3u(x_1,...,x_{n-1},-x_n)+4u(x_1,...,x_{n-1},-\frac{x_n}{2}) && \text{if } x \in B^- \end{cases}$
Where $B^{\pm} := B \ \cap \ \{sgn(\pm)x_n \geq0 \}.$ an $B$ is an openball near a boundary point $x_0$.
Then he uses straighten boundary to proof the inequality between $\tilde{u}$ and $u$ in region near the boundary.
After that he uses partition of unity to generalize the inequality to $\mathbb{R}^n$.
His definition for $\tilde{u}$ becomes $\sum_{i=0}^N\tilde{u_i}\xi_i$, on page $272$, where $\xi_i$ is partition of unity, and $\tilde{u_i} = \begin{cases} u && \text{i = 0} \\ \tilde{u'}_i && \text{i} \neq 0 \end{cases}$
Here $\tilde{u'}_i$ denotes the $\tilde{u'}$ with respect to $i$-th point that we choose on the boundary.
But on page $273$, Evans claims that $Eu:=\tilde{u}$ is linear. I do not see how it works because for specific $\tilde{u}$ on the page $271$, there are two cases. One is the identity transformation but the other is change a little bit of parameter. I am not sure it will make the mapping becomes non-linear or not.
PS: I also did not see why on page $273$, Evans says $\tilde{u}$ is in $W^{2,p}(B).$ I do see it is not in $C^2$ though.