Note: In the following, $U\subseteq \mathbb R^n$ is an open, simply connected set. The notation $A\subset\subset B$ means $A$ is "compactly contained" in $B$, that is $A\subseteq \bar A\subseteq B^{\circ}$ with $\bar A$ being compact. Here also $k\in\mathbb N$ and $p\in [1,\infty)$.
I am currently reading LC Evans's PDE book. I am reading chapter 5 which establishes the theory of linear PDEs.
In theorem 1, we establish that we can locally approximate functions in Sobolev spaces by smooth functions. That is, if we have a function $u\in W^{k,p}(U)$, then for all $V\subset\subset U$, we can find a sequence of functions $\{u_1,u_2,\dots\}$ with $u_1,u_2,\dots \in C^\infty (V)$ such that $u_m\overset{m\to\infty}{\longrightarrow} u$ in $W^{k,p}(V)$ (i.e $\Vert u_m-u\Vert_{k,p~ (V)}\to 0~\text{as}~m\to\infty$). In fact we can give an explicit formula for these functions in terms of a convolution.
In theorem 2, we are supposed to show that we can not only do this locally, but that we can do this on all of $U$ at once. Evans writes:
THEOREM 2: (Global approximation by smooth functions). Assume $U$ is bounded, and suppose as well that $u\in W^{k,p}(U)$. Then there exist functions $u_1,u_2,\dots \in C^\infty(U)\cap W^{k,p}(U)$ such that $$u_m\overset{m\to\infty}{\longrightarrow}u~~\text{in}~W^{k,p}(U)$$ Note carefully that we do not assert $u_1,u_2,\dots\in C^\infty(\bar U)$, but see Theorem 3 below. Proof. We have $U=\bigcup_{i=1}^\infty U_i$ where $$U_i:=\{x\in U ~:~\operatorname{dist}(x,\partial U)>1/i\}~~(i\in\mathbb N)$$ Write $V_i:=U_{i+3}\setminus \bar U_{i+1}$. Choose also any open set $V_0\subset\subset U$ so that $U=\bigcup_{i=0}^\infty V_i$. Now let $\{\zeta_i\}_{i=0}^\infty$ be a smooth parition of unity subordinate to the open sets $\{V_i\}_{i=0}^\infty$; that is, suppose $$\tag{*}\begin{cases}0\leq \zeta_i\leq 1, & \zeta_i\in C^\infty_{\text{comp}}(V_i) \\ \sum_{i=0}^\infty \zeta_i=1 & \text{on}~U\end{cases}$$ .....
And the proof goes on from there. I have a few questions about it so far.
I don't understand the rationale behind the choice for the sets $V_i$. Using our definition of $U_i$, we can write $V_i:=\{x\in U~:~\frac{1}{i+3}<\operatorname{dist}(x,\partial U)<\frac{1}{i+1}\}$. We can see that these sets are not disjoint (not even a.e!), because a point $x$ satisfying $\operatorname{dist}(x,\partial U)=d$ where $3<d<4$ is contained in the intersection $V_1\cap V_2$. Why bother constructing these sets? Why not just stick with the $U_i$s?
Why does Evans state "choose a set" $V_0\subset\subset U$ such that $V_0$ is open and $U=\bigcup_{i=0}^\infty V_i$. Why can't we just say " let $V_0=\{x\in U~:~\operatorname{dist}(x,\partial U)>1/2\}$ "? Doesn't this meet all of the criteria?
Finally, it's not immediately clear to me we can construct functions $\{\zeta_0,\zeta_1,\dots\}$ that satisfy $(*)$. Can someone provide some justification for this?
That's all for now, thanks.