Example of reflecting a function . What is happening here : $\partial U $ is flat and $x^o \in \partial U $ is the center of the sphere, assume $\partial U $ to be flat near $x^0 $ and $x_n=0$
I am looking for a concrete example to appreciate whats happening when i define function the following way . Example can be in 1d or 2d . 
Thanks for ur help. 

$U$ is a open, bounded subset of $\mathbb R^n$. I think flat here means that the $\partial U$ divides the ball symmetrically . I am not pretty sure. 
The text clip is from Partial Differential Equations by Lawrence C. Evans, p. 225, Chapter 5. In the second edition (p. 269, Section 5.4), $C^\infty$ has been replaced by $C^1$. 
 A: It's a constructive way to extend a function $u$ through a flat part of the boundary to get a $C^1$ function $\bar u$. If to use reflection, defining $$\bar u(x_1,\ldots,x_{n-1},-x_n)=u(x_1,\ldots,x_{n-1},x_n),$$ then $\bar u$ for a continuous $u$ will be continuous in $B$. 
It's straightforward to check that $\bar u$ as defined in the book has all continuous derivatives of the first order in $B$. Actually only checking the continuity of $\partial u/\partial x_n$ when $x_n=0$ is needed. Analogously it is possible for $u\in C^m(\bar B_+)$ to obtain a function 
$$
\bar u(x_1,\ldots,x_{n-1},-x_n)=\sum_{k=1}^{m+1}\alpha_k u(x_1,\ldots,x_{n-1},x_n/k)
$$
belonging the class $C^m(\bar B)$ with the proper choice of constants $\alpha_k$.
A: The right kind of example to work through is $x_n^k$ for k=0,1,2... Of course the other variables don't matter here, so $x_n$ can be simply $x$. Plugging this into the reflection formula, you see that for k=0,1 it gives the same monomial. This indicates that the extension is $C^1$ smooth. For k=2 it no longer works; you'd need another term in the reflection formula to achieve $C^2$, as @Andrew wrote.
