understanding proof of global approximation theorem Evans PDE

Question

I have a couple questions about the following proof of this theorem,

${\bf Global\ Approximation\ Theorem}$(251 page inEvans's PDE book) : Assume $U$ is bounded, then for $u\in W^{k,p}(U)$, there exists $u_m \in C^\infty (U)\cap W^{k,p}(U)$ such that $$u_m \rightarrow u\ in\ W^{k,p}(U)$$

https://i.stack.imgur.com/IdlxA.jpg

My questions are

1. why do Evans choose to work with the partition of unity subordinate? could another partition works?

2.why do Evans choose those $W_i$ sets?

3. why is $v=\sum_{i=0}^{\infty} u^i$ belongs to $C^\infty(U)$, i cannot understand the explanation that Evans gave for that

Thank you in advance.

Answer 1

1. I don't know.

2. To get the estimate in $(3)$ (the inclusion in (3) is used to get the said estimate).

Details: Let us write $u^\varepsilon=\eta_\varepsilon *(\zeta_i u)$. We have $$u^\varepsilon\to \zeta_i u\quad \text{in}\quad W^{k,p}_{\text{loc}}(U).$$ So, given an open set $Y\subset\subset U$, a positive constant $C>0$ and $\delta>0$, we can find $\varepsilon=\varepsilon(Y,C,\delta)$ satisfying $$\|u^\varepsilon-\zeta_i u\|_{W^{k,p}(Y)}<C\delta.$$ But we would like to estimate $\|u^\varepsilon-\zeta_i u\|_{W^{k,p}(U)}$. Of course, if $\text{supp}(\zeta_iu)\subset Y$ and $\text{supp}(u^\varepsilon)\subset Y$ then, from the previous estimate, $$\|u^\varepsilon-\zeta_i u\|_{W^{k,p}(U)}= \|u^\varepsilon-\zeta_i u\|_{W^{k,p}(Y)}<C\delta.$$ Well, we have $\text{supp}(\zeta_iu)\subset V_i$. But we cannot say that $\text{supp}(u^\varepsilon)\subset V_i$ and thus we cannot take $Y=V_i$; we need a little bigger set. So, we take $Y:=W_i$ as defined in the book. Also, by convenience, we take $C=\frac{1}{{2^{i+1}}}$. Then we find $\varepsilon_i=\varepsilon(W_i,\frac{1}{{2^{i+1}}},\delta)$ satisfying $$\|u^{\varepsilon_i}-\zeta_i u\|_{W^{k,p}(U)}= \|u^{\varepsilon_i}-\zeta_i u\|_{W^{k,p}(W_i)}<\frac{\delta}{2^{i+1}}.$$ Note that what we called $u^{\varepsilon_i}$ above is called $u^i$ in the book.

3. Because a finite sum of infinitely differentiable functions is infinitely differentiable.

Details: Take $x\in U$. There exist an open set $V\subset\subset U$ such that $x\in V$. There are a finite number of functions $u^{1},...,u^{k}$ (the number $k$ of functions depends on $V$) such that $$v(y)=\sum_{j=1}^k u^{j}(y),\quad \forall\ y\in V.$$ (In other words, "there are at most finitely many nonzero terms in the sum".) Why? Because $\text{supp}(u^i)\subset W_i$ and $V\cap W_i=\varnothing$ for all sufficiently large values of $i$.

As each $u^{j}$ is infinitely differentiable at $x$, it follows that $v$ (which is a finite summ at $x$) is infinitely differentiable at $x$. Since $x$ is arbitrary, we conclude that that $v\in C^\infty(U)$.

Answer 2

The main idea behind this proof (and similar proofs later in that chapter) is this:

The problem: Remember that if $f$ is defined on $U$, then $f^\epsilon$, the mollified $f$ is defined only on $U_\epsilon$, a proper subset of $U$. And there is no canonical way of extending it smoothly.

The Solution: However, if $f$ happens to be zero close to $\partial U$, then $f^\epsilon$ vanishes close to boundary of $U_\epsilon$ for $\epsilon$ small, and thus we can extend it to all of $U$, all of $R^n$ in fact, by setting it to zero there.

What they do in this proof is that they look at the partitioned functions $f\zeta_k$ that are functions on $V_k$ that vanish near $\partial V_k$. So applying above argument, we know that their mollifiers are now smooth on all of $U$, with support inside $V_k$ in fact. This is the heart of the argument/construction. The rest follows smoothly.