Here is the $H^2$ regularity theorem for the elliptic equations in Evans's Partial Differential Equations (2nd edition):
Here $U\subset\mathbb{R}^n$ is assumed to be open and bounded.
The first big step of the proof gives the following:
All that is left to do for the proof is "replacing" the $H^1(U)$ norm with $L^2(U)$ norm in $(23)$, which one can easily see by comparing (23) with (8). Instead of directly dealing with $\|u\|_{H^1(U)}$, Evans takes one step back and makes the following argument:
Here (9) is given by
I can follow each step in this part logically, but I don't see the point of using of $W$ (and the cutoff function) here.
Evans makes a point at the beginning of the proof:
Here are my questions:
- Would anyone explicitly point out why $W$ is needed in the argument above?
- What would/could be wrong if one erases the cutoff function $\zeta$ and replaces $W$ with $U$ (so that we can use (23) directly) in the red box?
- [Added in light of mcd's answer:] If one does need the cutoff function $\zeta$, then can he use $v=\zeta u$ instead of $v=\zeta^2 u$?
If you transform the problem into its weak formulation you need to apply integration by parts several times. For this you need that your test function $v$ to vanish at the boundary $\partial U$. Evans does impose any condition on $u$ on the boundary, so $u \not \in H^1_0$. He has to use a cut-off so that $B[u,\zeta^2 u] = (f, \zeta^2 u)$ holds. This he uses to deduce the first equation in your red box.
I did not see it immediately, but i strongly suspect that at one point you need that $\partial_{x_j} v = 0$ on the boundary. If you did impose these conditions on $u$ then the problem would no longer be well-defined as you have to many boundary conditions.