A smooth bump between concentric Euclidean balls

Put $s=(r+R)\mathrm{/}2$ and $d=s^2-r^2>0$, and define $u(x)=(s^2-{\sum }_{j=1}^n{x}_j^2)\mathrm{/}d$ and $\rho (x)=\sigma (u(x))$. The coordinate derivatives of $u$ are ${\mathrm{\partial }}_{i}u=-2x_i\mathrm{/}d$, ${\mathrm{\partial }}_j{\mathrm{\partial }}_{i}u=-2{\delta }_{ij}\mathrm{/}d$, and zero at every higher order, so $u$ is smooth. Here ${\delta }_{ij}$ is $1$ when $i=j$ and $0$ otherwise.

Every iterated coordinate derivative of $\rho$ is a finite sum of products of derivatives of $u$ and factors ${\sigma }^{(k)}(u)$: induction follows by applying the chain and product rules in [L1] to each factor. These factors are continuous by [F1] and step 1.1, so $\rho$ is smooth.

If $\mathrm{\parallel }x\mathrm{\parallel }\le r$, then $u(x)\ge 1$, so $\rho (x)=1$; if $\mathrm{\parallel }x\mathrm{\parallel }\ge s$, then $u(x)\le 0$, so $\rho (x)=0$. Also $0\le \rho \le 1$ by [F1].

The nonzero locus lies in $B_s(0)$, whose closure is contained in the closed ball ${\bar{B}}_s(0)$. Thus [F2] gives $\mathrm{supp}\rho \subseteq {\bar{B}}_s(0)\subseteq B_R(0)$, since $s<R$. Together with steps 2.1 and 2.2 this proves the claim.