I am reading the extension theorem in sobolev spaces in the book '' Partial Differential Equation'' by Evan and I get stuck at one point. Let $U\subset\mathbb{R}^n$ is open and bounded, and $\partial U$ is $C^1$. Let $x_0\in\partial U$ and suppose $\partial U$ is flat near $x_0$, lying in the plane $\{x_n=0\}$. Then we may assume that there exists an open ball $B(x_0, r)$ such that $$B^+:=B\cap\{x_n\ge 0\}\subset U$$ $$B^-:=B\cap\{x_n\le 0\}\subset\mathbb{R}- U$$ Let $u\in C^{\infty}(\bar{U})$. We set $\bar{u}(x)=u(x)$ if $x\in B^+$ and $\bar{u}(x)=-3u(x_1,\ldots,x_{n-1},-x_n)+4u(x_1,\ldots,x_{n-1},-\frac{x_n}{2})$ if $x\in B^-$.
The author claims that since $u^+=u^-$ on $\{x_n=0\}$, we have $$\frac{\partial u^-}{\partial x_i}=\frac{\partial u^+}{\partial x_i}$$ on $\{x_n=0\}$.
I do not understand what the author means when he talks about the partial derivative of $u^+,u^-$ on $\{x_n=0\}$ ($\{x_n=0\}$ is not interior points of $B^+, B^-$). How do we get the above equality ?
DISCUSS: In my understanding, to prove that $\bar{u}$ has $\frac{\partial \bar{u}}{\partial x_n}$ at the point $(x_1,\ldots,0)$, we have to prove that the limit:
$$\lim_{h\rightarrow 0} \frac{\bar{u}(x_1,\ldots,h)-\bar{u}(x_1,\ldots,0)}{h}$$ exists. But this is equivalent to prove that $$\lim_{h\rightarrow 0^+} \frac{\bar{u}(x_1,\ldots,h)-u(x_1,\ldots,0)}{h}$$ and $$\lim_{h\rightarrow 0^-} \frac{\bar{u}(x_1,\ldots,h)-u(x_1,\ldots,0)}{h}$$
exist and equal. IS IT RIGHT ?
If it is right, we have
$$\lim_{h\rightarrow 0^+} \frac{\bar{u}(x_1,\ldots,h)-u(x_1,\ldots,0)}{h}=\lim_{h\rightarrow 0^+} \frac{u(x_1,\ldots,h)-u(x_1,\ldots,0)}{h}$$
WHAT IS THE VALUE OF THE SECOND LIMIT ?
Let's first be clear about what he means by $u^\pm$. He's writing this for the restriction of $\bar{u}$ (which is defined on all of $B$ as above) to $B^\pm$, respectively. In other words, $u^+ = u$ in $B^+$ and $u^- = \bar{u}$ in $B^-$. Now he wants to consider these as functions restricted to the set $B \cap \{x_n =0\} \subset \mathbb{R}^{n-1}$, which IS open in $\mathbb{R}^{n-1}$. In fact, it's just an open ball.
Let's write this as $v^\pm(x_1,\dotsc,x_{n-1}) = u^\pm(x_1,\dotsc,x_{n-1},0)$. Then $v^\pm$ is a function defined on an open set in $\mathbb{R}^{n-1}$, so we can compute partials. In fact, we have that $v^+ = v^-$ due to the above construction, so we must have that $$ \frac{\partial v^+ }{\partial x_i} = \frac{\partial v^- }{\partial x_i} $$ for $i=1,\dotsc,n-1$ on $B \cap \{x_n=0\}$. This is then what he means when he writes $$ \frac{\partial u^+ }{\partial x_i} = \frac{\partial u^- }{\partial x_i}. $$
EDIT:
We know by assumption that $u$ is smooth on $B^+$, so the goal is to show that the extended function is smooth. It's clear from the definition that $u^-$ is smooth in the interior of $B^-$, so the only thing to worry about is whether or not the derivatives are continuous across the flat part $B \cap \{x_n=0\}$. The part above shows that all of the partials in the horizontal directions are continuous across the interface. Then you have to go on and use the special form of the extension to show that actually $$ \frac{\partial u^+}{\partial x_n} =\frac{\partial u^-}{\partial x_n} $$ at the interface as well. Once this is established, you then have that all the partial exist and are continuous on $B$, and so $\bar{u}$ is $C^1$.