Naive extension by zero from an open set need not be smooth

The set $C$ is closed: its complement is the union of the open intervals $(-\mathrm{\infty },1)$ and $(1,\mathrm{\infty })$. Thus $C\subseteq U$ and the constant smooth function $f$ satisfy the claim's hypotheses. Its prescribed zero extension is $F(x):=1$ for $x>0$ and $F(x):=0$ for $x\le 0$.

Take $\epsilon =1\mathrm{/}2$. For every $\delta >0$, the point $x=\delta \mathrm{/}2$ satisfies $\mathrm{\mid }x\mathrm{\mid }<\delta$ and $\mathrm{\mid }F(x)-F(0)\mathrm{\mid }=1$. Hence $F$ is not continuous at zero by [L2], so it is not smooth by [L1].

Therefore extension by zero is not always smooth, even when the original open set contains the specified closed set. This does not deny the existence of another smooth extension: the constant function one on all of $\mathbb{R}$ already extends this $f$. No assertion that cutoffs are necessary for every extension, and no choice principle, is used.