Explicit compactly supported smooth cutoffs

Define $a(t)=e^{-1\mathrm{/}t}$ for $t>0$ and $a(t)=0$ for $t\le 0$. If $s=1\mathrm{/}t$, induction gives $a^{(k)}(t)=P_k(s)e^{-s}$ on $t>0$, with $P_0=1$ and $P_{k+1}(s)=s^2(P_k(s)-P_k^{\mathrm{\prime }}(s))$. By [F1], both $P_k(s)e^{-s}$ and $s{P}_k(s)e^{-s}$ tend to zero as $s\to \mathrm{\infty }$. Extending each derivative by zero to $t\le 0$ is therefore continuous; its difference quotient at zero also tends to zero. Induction proves $a\in C^{\mathrm{\infty }}(\mathbb{R})$ with every derivative zero at zero.

Set $\sigma (t)=a(t)\mathrm{/}(a(t)+a(1-t))$. For $t\le 0$, the second summand in the denominator is positive; for $t\ge 1$, the first is positive; for $0<t<1$, both are positive. Thus the quotient is smooth, $0\le \sigma \le 1$, and $\sigma =0$ on $t\le 0$, $\sigma =1$ on $t\ge 1$. Define $\chi (x)=\sigma ((4-\mathrm{\mid }x{\mathrm{\mid }}^2)\mathrm{/}3)$. Repeated [F2] proves smoothness, and the two constant regions give the claimed unit and zero regions, including their boundaries. Its closed support lies in the closed radius-two ball and is compact by [F3].

Differentiating $\chi (x\mathrm{/}R)$ once in coordinate $j$ gives $R^{-1}({\mathrm{\partial }}_j\chi )(x\mathrm{/}R)$. Iterating this identity in the prescribed multi-index order gives the asserted factor, including $\beta =0$. No selection was made in any construction.