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?

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

It just doesn't add like that :)

we have the following explicit calculations

\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 $3C2^q \in (6\cdot 2^q,\frac{15}2\cdot 2^q)=(0.75\cdot 2^{q+3}, 1.5 2^{q+3})$.

(Thanks Zhu Ning for setting me straight.)


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