Finite-order analytic cutoff sequence

If $K=\mathrm{\varnothing }$, set ${\chi }_N=0$ for all $N$; the empty set is a neighborhood of $K$, so every assertion holds. Assume henceforth $K\mathrm{\ne }\mathrm{\varnothing }$. Consider all pairs of concentric balls $B_r(x)⋐B_R(x)⋐U$ with $x\in K$ and $0<r<R$. Their smaller balls cover $K$, so compactness supplies finitely many pairs $B_{r_j}(x_j)⋐B_{R_j}(x_j)$, $1\le j\le m$, whose smaller balls cover $K$. Choose their finitely many bumps $b_j$ by [F1], translated to $x_j$, with $0\le b_j\le 1$, $b_j=1$ on $B_{r_j}(x_j)$ and compact support in $B_{R_j}(x_j)$. The smooth function $g=1-{\prod }_{j=1}^m(1-b_j)$ satisfies $0\le g\le 1$, equals one on the open set $O={\bigcup }_{j=1}^m{B}_{r_j}(x_j)$ containing $K$, and has compact support $S⋐U$. Only finitely many bumps are selected.

Choose a nonnegative smooth bump ${\rho }_0$ supported in the unit ball and equal to one on the ball of radius $1\mathrm{/}2$, using [F1]. Its compactly supported integral is finite by [F3] and positive: a closed nondegenerate rectangle inside $B_{1\mathrm{/}2}(0)$ has positive volume and ${\rho }_0=1$ there. Normalize $\rho ={\rho }_0\mathrm{/}\int {\rho }_0$, so $\int \rho =1$, $\rho \ge 0$, and $\mathrm{supp}\rho \subset {\bar{B}}_1(0)$. Since $K⋐O$ and $S⋐U$, choose $\epsilon >0$ smaller than one third of each distance $\mathrm{dist}(K,O^c)$ and $\mathrm{dist}(S,U^c)$, omitting the second restriction when $U={\mathbb{R}}^n$.

For each $N\ge 1$, put $\delta =\epsilon \mathrm{/}N$, ${\rho }_{\delta }(x)={\delta }^{-n}\rho (x\mathrm{/}\delta )$, and recursively ${\eta }_1={\rho }_{\delta }$, ${\eta }_{j+1}={\eta }_j\ast {\rho }_{\delta }$ for $1\le j<N$. Compact-rectangle Fubini [F4] and change of variables give $\int {\eta }_N=1$ and $\mathrm{supp}{\eta }_N\subset {\bar{B}}_{\epsilon }(0)$; [F4] also permits reassociation of these compact convolution integrals. Define ${\chi }_N=g\ast {\eta }_N$. The integral defining ${\chi }_N$ is over a fixed compact rectangle. Differentiating the smooth kernel under this compact integral is justified by uniform convergence of its difference quotients on that rectangle, so ${\chi }_N$ is smooth. Positivity and unit mass give $0\le {\chi }_N\le 1$. If $\mathrm{dist}(x,K)<\epsilon$, then every $y\in \mathrm{supp}{\eta }_N$ satisfies $\mathrm{dist}(x-y,K)<2\epsilon <\mathrm{dist}(K,O^c)$, hence $g(x-y)=1$ and ${\chi }_N(x)=1$. Also $\mathrm{supp}{\chi }_N\subset S+{\bar{B}}_{\epsilon }(0)⋐U$.

For smooth compactly supported $f,h$, write $\mathrm{\parallel }h{\mathrm{\parallel }}_1={\int }_{{\mathbb{R}}^n}\mathrm{\mid }h\mathrm{\mid }$. Fubini on a rectangle containing their supports and the triangle inequality give $\mathrm{\parallel }f\ast h{\mathrm{\parallel }}_1\le \int \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\int \mathrm{\mid }f(x-y)\mathrm{\mid }\mathrm{\mid }h(y)\mathrm{\mid }dydx=\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }h{\mathrm{\parallel }}_1\mathrm{.}$ Also ${\mathrm{\partial }}_j(f\ast h)=({\mathrm{\partial }}_{j}f)\ast h=f\ast ({\mathrm{\partial }}_{j}h)$ by differentiating under the compact integral and integration by parts; there is no boundary term because the factors have compact support. For $k=\mathrm{\mid }\alpha \mathrm{\mid }\le N$, distribute the $k$ coordinate derivatives to $k$ distinct convolution factors in ${\eta }_N$. Scaling gives $\mathrm{\parallel }{\mathrm{\partial }}_j{\rho }_{\delta }{\mathrm{\parallel }}_1={\delta }^{-1}\mathrm{\parallel }{\mathrm{\partial }}_j\rho {\mathrm{\parallel }}_1,$ so repeated use of the preceding inequality yields $\mathrm{\parallel }{\mathrm{\partial }}^{\alpha }{\eta }_N{\mathrm{\parallel }}_1\le (AN{)}^k,A={\epsilon }^{-1}\max (1,{\max }_j\mathrm{\parallel }{\mathrm{\partial }}_j\rho {\mathrm{\parallel }}_1)\mathrm{.}$ For $k=0$ this is $\mathrm{\parallel }{\eta }_N{\mathrm{\parallel }}_1=1$.

Since $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$, $\mathrm{\mid }{\mathrm{\partial }}^{\alpha }{\chi }_N(x)\mathrm{\mid }=\mathrm{\mid }g\ast ({\mathrm{\partial }}^{\alpha }{\eta }_N)(x)\mathrm{\mid }\le \mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }{\mathrm{\partial }}^{\alpha }{\eta }_N{\mathrm{\parallel }}_1\le (AN{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}\mathrm{.}$ Thus the lemma holds with $C=1$ (or any larger fixed constant), uniformly for every $\mathrm{\mid }\alpha \mathrm{\mid }\le N$. The cutoffs are smooth and no analytic bump is used.