Details in mollifiers proof – Evans’ PDE

Question

I've been reading Evans' PDE book but I stumbled upon the proof of the following theorem (screenshot here):

THEOREM 7 (Properties of mollifiers).

(i) $f^{\epsilon} \in C^{\infty}\left(U_{\epsilon}\right)$.

(ii) $f^{\epsilon} \rightarrow f$ a.e. as $\epsilon \rightarrow 0$.

(iii) If $f \in C(U),$ then $f^{\epsilon} \rightarrow f$ uniformly on
compact subsets of $U$.

(iv) If $1 \leq p<\infty$ and $f \in L_{\text {loc }}^{p}(U),$ then
$f^{\epsilon} \rightarrow f$ in $L_{\text {loc }}^{p}(U)$

And this is part of the proof (screenshot here):

3. Assume now $f \in C(U)$. Given $V \subset \subset U$, we choose $V \subset \subset W \subset \subset U$ and note that $f$ is uniformly continuous on $W .$ Thus the limit (4) holds uniformly for $x \in V .$ Consequently the calculation above implies $f^{\epsilon} \rightarrow f$ uniformly on $V$.

The proof itself is quite clear but the bit that I'm not quite following is how does he choose $W$ satisfying that particular property. I've been trying to use compactness of $\overline{V}$ and openness of $U$ but I'm still not getting it. Some help would be appreciated.

Answer 1

$A\subset\subset B$ is read "$A$ is compactly contained in $B$", sometimes also written with \Subset$\Subset$. According to Evans (2nd ed., pg 698), it means that, for open sets $A,B$, we have $A\subset \overline A\subset B$ and that $\overline A$ is compact:

(vi) $U, V,$ and $W$ usually denote open subsets of $\mathbb{R}^{n} .$ We write $$ V \subset \subset U $$ if $V \subset \bar{V} \subset U$ and $\bar{V}$ is compact, and say $V$ is compactly contained in $U$.

As $V\Subset U \subset \mathbb R^n$, in particular $\overline V\subset U$. I believe $U$ is (as in the above) implicitly an open set; thus $d(\overline V,U^c) \in (0,\infty]$. Set $l = \min(1,d(\overline V,U^c))\in(0,1]$.

For each $x\in \overline V$, let $B_x := \mathbb B(x,l/2) $ be the open ball of radius $l/2$ around $x$. Clearly $\{B_x\}_{x\in \overline V}$ is a cover of $\overline V$, so by compactness there is a finite number of them, $B_1,\dots,B_N$ that cover $\overline V$. Now we can take $W:= B_1\cup \dots\cup B_N$. Since each $\overline{B_i}$ is a subset of $U$ (there is no intersection with $U^c$), $\overline W\subset U$. $\overline W$ is a finite union of closed and bounded balls; hence it is closed and bounded in $\mathbb R^n$, and hence compact. Summary: we created a second open subset $W$ of $U$ such that $V\Subset W \Subset U$. Written using normal subset notation, we have $$ V\subset \overline V \subset W \subset \overline W \subset U.$$

The fact that $\overline W$ is compact immediately implies the uniform continuity of $f$ on $\overline W$, and hence on $W$.

You might ask why we needed $W$ in the first place since $V$ is already precompact; the answer is that we need some space in order to mollify and get a function defined on $V$, for all $\epsilon$ sufficiently small.