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),Qℓ=2−ℓ,χℓ(q)=χ(q/Qℓ).\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 χℓ\chi_\ell, ℓ∈Z\ell\in\mathbb Z, form a smooth dyadic scale partition:

χℓ≥0,∑ℓ∈Zχℓ(q)=1,supp⁡χℓ⊂[Qℓ/2,2Qℓ].\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 ℓ=−M\ell=-M to NN telescopes to ρ(2−Mq)−ρ(2N+1q)\rho(2^{-M}q)-\rho(2^{N+1}q), which equals one for sufficiently large M,NM,N. Differentiating gives ∣∂qkχℓ∣≤CkQℓ−k|\partial_q^k\chi_\ell|\le C_k Q_\ell^{-k}; on its support this is at most Ck′q−kC'_k q^{-k}. The constants are independent of ℓ\ell. Hence every fixed power of q∂qq\partial_q is uniformly bounded as well.