In the proof of the Sobolev extension theorem for $W^{1,p}$ spaces where $p\in [1,\infty )$, in Evans' PDE book, he reduces proving the theorem down to proving it for smooth functions on a open set with flat boundary. Then he uses higher-order reflection to solve this case. I understand this proof quite well, but I have one question: is it necessary to approximate the given function with smooth functions? Can't you just reflect the original function $f\in W^{1,p}$?
If this is not possible, it seems that I have an incorrect intuition of weak derivatives.
Evans left $p=\infty $ as an exercise, and I know how to do this in another way, but I thought that this, reflecting the original function, if possible, should be the intended way.
You can extend the original function by reflection. Problem is, how will you prove that the extended function is in $W^{1,p}$? Computing the derivatives pointwise a.e. is not enough, one must verify the definition of weak derivatives. It can be done but is a chore.
The advantage of temporarily switching to smooth functions is that smoothness is verified pointwise. To show something is $C^1$ we only need to check the derivatives at each point individually. Simply put, classical derivatives are easier to handle than weak derivatives.
But the case $p=\infty$ is different. Here smooth approximation is not available, and we have to extend the original function. Fortunately, in this case there is a pointwise characterization that helps: $W^{1,\infty}$ is precisely the space of Lipschitz functions (see relation between $W^{1,\infty}$ and $C^{0,1}$). The Lipschitz condition of the extended function is easy to verify by checking two cases (two points in one closed halfspace, two points in different halfspaces). But actually, in this there is no need for higher order reflection: one can simply let $$\tilde u(x_1,\dots, x_{n-1} ,x_n) = u(x_1,\dots, x_{n-1} , |x_n|)$$ which is Lipschitz, being the composition of Lipschitz functions. The "straightening the boundary" step works by observing that both $\Phi$ and $\Psi$ are Lipschitz continuous, as they are smooth (see $C^1$ function on compact set is Lipschitz for example).