Core idea

Choose a smooth nonincreasing ρ:R[0,1]\rho:\mathbb R\to[0,1] with ρ=1\rho=1 on (,1]( -\infty,1] and ρ=0\rho=0 on [2,)[2,\infty). Put

χ(s)=ρ(s)ρ(2s),Q=2,χ(q)=χ(q/Q).\chi(s)=\rho(s)-\rho(2s),\qquad Q_\ell=2^{-\ell},\qquad \chi_\ell(q)=\chi(q/Q_\ell).

For q>0q>0, the functions χ\chi_\ell, Z\ell\in\mathbb Z, form a smooth dyadic scale partition:

χ0,Zχ(q)=1,suppχ[Q/2,2Q].\chi_\ell\ge0,\qquad \sum_{\ell\in\mathbb Z}\chi_\ell(q)=1, \qquad \operatorname{supp}\chi_\ell\subset[Q_\ell/2,2Q_\ell].

The family is on (0,)(0,\infty); at most three closed support intervals contain a given point.

Proof and derivative bounds

The finite sum from =M\ell=-M to NN telescopes to ρ(2Mq)ρ(2N+1q)\rho(2^{-M}q)-\rho(2^{N+1}q), which equals one for sufficiently large M,NM,N. Differentiating gives qkχCkQk|\partial_q^k\chi_\ell|\le C_k Q_\ell^{-k}; on its support this is at most CkqkC'_k q^{-k}. The constants are independent of \ell. Hence every fixed power of qqq\partial_q is uniformly bounded as well.