# Basic question about Bony decompositions - summation indices

I'm trying to understand some sort of inequality in a larger calculation. I believe my only issue is in counting correctly, so I've also tagged combinatorics. Suppose I have a function $$f$$. Let $$\triangle_q$$, $$q\ge -1$$ be the Littlewood-Paley projection to frequencies $$\sim q$$, and $$S_q=\sum_{-1\le j\le q-1} \triangle_j$$ to be the projection to frequencies $$\lesssim q$$.

By this, I mean that for $$q\ge 0$$, $$\triangle_q f$$ has Fourier support on an annulus $$A[2^{q-1},2^{q+1}]$$, and $$S_qf$$ has Fourier support on a ball of radius $$2^{q}$$. Its also arranged that $$\triangle_q$$ and $$S_q$$ are Fourier multipliers (and so commute). With the standard set-up, we have for example (in a suitable sense) \begin{align} \sum_{q\ge -1} \triangle_q &= \operatorname{Id},\\ \triangle_q \triangle_j &= 0\text{ if }|q-j|\ge 2, \\ S_{q-1}\triangle_q &= 0, \text{ and}\\ S_{q+1}\triangle_q&=\triangle_q \end{align} Also define $$\tilde{\triangle}_q := \triangle_{q-1} + \triangle_q + \triangle_{q+1}.$$I think I've said enough for my problem, for more details you can consult Mathworld, Tao's notes, or the book by Bahouri, Chemin and Danchin "Fourier Analysis and Nonlinear PDEs".

Now I have two functions, lets say $$f,g$$ and I've come across the following sum- $$\sum_{q\ge 1} \triangle_q f\tilde\triangle_q g$$ and I have the following bound, $$\|\triangle_q f\tilde\triangle_q g\|_{L^2} \le Ch_q$$

Why is it that $$\triangle_j \sum_{q\ge 1} \triangle_q f\tilde\triangle_q g \overset{\Huge ?}\le C\sum_{\color{red}{q>j-4}} C h_q$$ My back-of-the-envelope is as follows. The Fourier support of $$\triangle_q f \tilde \triangle_q g$$ should be the sum of the two annuli. Since they can destructively and constructively interfere and everything in between, the Fourier support is now a ball of radius $$2q+3$$. For this ball to intersect the annulus $$A[2^{j-1},2^{j+1}]$$, we need $$2q+3> j-1$$. This leads do $$2q > j-4$$ Did I understand something wrongly?

• I can provide plenty more details, if needed (just ask) – Calvin Khor Jan 9 '20 at 7:20

\begin{align} \DeclareMathOperator{\supp}{supp} &x\in\supp \triangle_qf \implies &c 2^q& \le |x| \le C 2^q, \\&x\in\supp \tilde \triangle_qg\implies &\frac{c}2 2^{q-1} &\le |x| \le 2C 2^{q}, \\&x\in\supp \triangle f \tilde \triangle_qg\implies & &\quad\,|x| \le (2C+C) 2^{q}=3C2^q, \end{align} I believe $$c$$ is taken to be a number in $$(0.5,1)$$ and $$C$$ a number in $$(2,4)$$. In practice one normally fixes $$c=3/4, C=8/3$$, so $$3C2^q = 82^q = 2^{q+3}.$$ So $$\triangle_j$$ of the product is zero if $$j-1>q+3$$. This means we should sum over (as a possible overestimation) $$q$$ such that $$j-1\le q+3$$ (typo in question).
For the range of $$C$$ I indicated above one gets $$3C2^q \in (6\cdot 2^q,\frac{15}2\cdot 2^q)=(0.75\cdot 2^{q+3}, 1.5 2^{q+3})$$.