Asymptotic summation by shrinking cutoffs

Statement

Let $0<q<q_0\le1$ on $\Omega$ , with $q$ smooth, bounded below on compact subsets, and $|\partial^\alpha q|\le C_\alpha q^{1-c|\alpha|}$ . Suppose smooth functions $f_j$ , all on $\Omega$ , satisfy

|f_j|_m\le C_{j,m}q^{g_j-\ell_m}(1+|\log q|)^{P_{j,m}}, \qquad 0<g_1\le g_2\le\cdots\to\infty,

where $\ell_m\ge0$ is nondecreasing and independent of $j$ . There are $a_{j+1}\ge2a_j$ , $a_1^{-1}<q_0$ , such that

f=\sum_{j\ge1}\chi(a_jq)f_j

is smooth and locally finite. With $f^{[J]}=\sum_{j=1}^Jf_j$ , for $J\ge\max(1,m)$ and $q<(2a_J)^{-1}$ ,

|f-f^{[J]}|_m\le2^{-J}q^{g_{J+1}/2-L_m},\qquad L_m=\ell_m+cm.

Here $\chi=1$ on $[0,1/2]$ and $\chi=0$ on $[1,\infty)$ . This is shrinking-cutoff asymptotic summation.

Choice and tail proof

The cutoff derivative bound and Leibniz rule bound the $j$ -th cutoff term through order $m$ by $\widehat C_{j,m}q^{g_j-L_m}(1+|\log q|)^{\widehat P_{j,m}}$ . Choose $a_j$ so that the coefficient times $q^{g_j/2}$ is at most $2^{-j}$ for $m\le j$ throughout $q\le a_j^{-1}$ . This is a finite list of vanishing requirements.

On a compact subset, $q\ge\delta>0$ , so only finitely many cutoff terms survive. For $q<(2a_J)^{-1}$ , the first $J$ cutoffs equal one. The remaining derivative bounds sum to at most $q^{g_{J+1}/2-L_m}\sum_{j>J}2^{-j}$ , proving the tail estimate.

Retaining the original orders

To compare after a fixed truncation $N$ , choose a larger fixed $J$ with $g_{J+1}/2\ge g_{N+1}$ . The finite block $N+1,\ldots,J$ retains its original orders near zero; the tail has the desired order by the estimate. Logs can be retained or absorbed by a small power loss. This realizes an asymptotic expansion without asserting convergence of the uncut series.

References

Melrose, Lectures on Pseudodifferential Operators, Lecture 2 and its addenda.