Explicit compactly supported smooth cutoffs

Define $a(t)=e^{-1\mathrm{/}t}$ for $t>0$ and $a(t)=0$ for $t\le 0$. On $t>0$, put $s=1\mathrm{/}t$. The quotient rule [F5] applied to $1\mathrm{/}t$ gives $s^{\mathrm{\prime }}=-1\mathrm{/}{t}^2=-s^2$, while [F2] and [F4] give $(e^{-s}{)}^{\mathrm{\prime }}=-e^{-s}$ as a function of $s$. Thus the product and chain rules give $\frac{d}{dt}(P_k(s)e^{-s})=(P_k^{\mathrm{\prime }}(s)-P_k(s))e^{-s}(-s^2)=s^2(P_k(s)-P_k^{\mathrm{\prime }}(s))e^{-s}\mathrm{.}$ Starting at $P_0=1$, this proves $a^{(k)}(t)=P_k(s)e^{-s}$ with the polynomial recurrence $P_{k+1}(s)=s^2(P_k(s)-P_k^{\mathrm{\prime }}(s))$. 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. Write $g(t)=a(t)+a(1-t)$, which is smooth by step 1.1 and the sum and chain rules. Since $g>0$, [F6] makes $q=1\mathrm{/}g$ continuous, and [F5] gives $q^{\mathrm{\prime }}=-g^{\mathrm{\prime }}{q}^2$. For every $k\ge 0$, finite iteration of the product rule [F5] and continuity of products [F6] shows that products of $C^k$ functions are $C^k$. If $q\in C^k$, the displayed derivative is therefore $C^k$ because $g^{\mathrm{\prime }}$ is smooth; hence $q\in C^{k+1}$. Starting with $C^0$, induction proves $q\in C^{\mathrm{\infty }}$ and then $\sigma =aq\in C^{\mathrm{\infty }}$. Thus $0\le \sigma \le 1$, 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.