Lemma 3.1.2623975478AEBA

Proof. We may assume that $K\neq \varnothing $. Let $\delta >0$. By compactness there are points $\zeta _1,\ldots ,\zeta _ N\in K$ with $K\subset \bigcup _ iB(\zeta _ i,\delta )$. Put $\chi _ i(x)=\rho ((x-\zeta _ i)/(2\delta ))$, where $\rho $ is the mollifier of section 2.5; this function is smooth, positive on $B(\zeta _ i,2\delta )$ and zero outside it. The function $\sigma =\sum _ i\chi _ i$ is positive on the open set $W=\bigcup _ iB(\zeta _ i,2\delta )\supset K$. Choose a smooth function $\eta $ with compact support in $W$ which equals one near $K$ (lemma 2.5.1(4)), and put

extended by zero outside $W$. This is a smooth function with compact support on $\mathbb R^ d$, because it is smooth on $W$, where $\sigma >0$, and vanishes on the open set $\mathbb R^ d\setminus \operatorname{Supp}\eta $; it is real-valued if $f$ is. For $\zeta \in K$ we have $\eta (\zeta )=1$, so $F(\zeta )$ is a convex combination of values $f(\zeta _ i)$ with $|\zeta _ i-\zeta |<2\delta $, and therefore $|F(\zeta )-f(\zeta )|\leq \sup \{ |f(y)-f(y')|:y,y'\in K,\ |y-y'|<2\delta \} $. By the uniform continuity of $f$ on the compact set $K$, this bound tends to zero as $\delta \to 0$. $\square$