Lemma: Cutoffs away from Z

Let $Z\subset {\mathbb{R}}^d$ be any set and $\epsilon >0$. There is ${\theta }_{\epsilon }\in C^{\infty }({\mathbb{R}}^d)$ with $0\le {\theta }_{\epsilon }\le 1$, ${\theta }_{\epsilon }=0$ on $\{\mathrm{dist}(x,Z)<\epsilon \}$, ${\theta }_{\epsilon }=1$ on $\{\mathrm{dist}(x,Z)>3\epsilon \}$, and $|{\partial }^{\alpha }{\theta }_{\epsilon }|\le C_{\alpha }{\epsilon }^{-|\alpha |}$ with $C_{\alpha }$ independent of $\epsilon$ and $Z$.

Proof. Let $\rho \in \mathcal{D}({\mathbb{R}}^d)$ be nonnegative, supported in the unit ball, with integral one, and ${\rho }_{\epsilon }(x)={\epsilon }^{-d}\rho (x/\epsilon )$. Put ${\theta }_{\epsilon }=1_{\{\mathrm{dist}(\cdot ,Z)\ge 2\epsilon \}}\ast {\rho }_{\epsilon }$. It vanishes where $\mathrm{dist}<\epsilon$, equals one where $\mathrm{dist}>3\epsilon$, and ${\partial }^{\alpha }{\theta }_{\epsilon }=1\ast {\partial }^{\alpha }{\rho }_{\epsilon }$ is bounded by $\Vert {\partial }^{\alpha }{\rho }_{\epsilon }{\Vert }_{L^1}={\epsilon }^{-|\alpha |}\Vert {\partial }^{\alpha }\rho {\Vert }_{L^1}$.