1
$\begingroup$

I have problems understanding two (for my research important) details in the Proof of Theorem 4 (page 14) in this paper: http://arxiv.org/abs/1209.1518. Notation is very straightforward and is found on page 4. I expect that both issues are in reality the same problem:

  1. In the first line of page 14 Schottdorf claims: $$S_1=\sum_NN^{2s}||P_N \sum_H \sum_{L<<H}I^+(u_L,v_H)||_{U^2_+}^2\color{red}{\lesssim}\sum_HH^{2s}||P_H \sum_{L<<H}I^+(u_L,v_H)||_{U^2_+}^2\lesssim||u||_{Y^s}^2||v||_{Y^s}^2$$

  2. The line below he claims for $S_2$ (I just plugged in the Definition of $X^s$) $$S_2\leq \sum_H\sum_{H\sim H'}(\sum_LL^{2s}||P_LI^+(u_{H'},v_H)||_{U^2_+}^2)^{1/2}\color{red}{\lesssim} \sum_H\sum_{H\sim H'}(\sum_{L\lesssim H}L^{2s}||P_LI^+(u_{H'},v_H)||_{U^2_+}^2)^{1/2}$$

In both problems some Frequencies are just discarded, that should mean that the corresponding terms are 0 and that is probably because the Littlewood Paley Projectors $P_L$ and $P_H$ are on different Frequencies. However if that would always work then the considererd integral term $I^+(u_{L},v_H)$ would always be 0 which wouldn't make sense. Can somebody help me understand the highlighted inequalities?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

To answer your question 1 (and I expect the answer to question 2 is similar):

Note that $I^+(f,g)$ first take the product $fg$ and act on it as a Fourier multiplier. So the frequency support of $I^+(f,g)$ is the same as that of $fg$.

Now, when $L \ll H$ you have that the frequency support of $u_L v_H$ is $\approx H$, using that $2^H \pm 2^L \approx 2^H$ in this case. (You actually have the support being in $[H-3,H+3]$ or something similar depending on how you defined the cut-off functions.) So this tells you that

$$ P_N I^+(u_L, v_H) \neq 0 \iff N \approx H $$

which means in the sum over $N$ and $H$, you only need to sum "diagonally" over the $N\approx H$ terms since all the rest are zero.

Improve this answer
$\endgroup$
8
  • 1
    $\begingroup$ Two points: (1) $L = H \iff (L\leq H \text{ and } H \leq L)$. Not "or". (2) The negation of $L\lesssim H$ is in fact $L \gg H$, not $L \gtrsim H$. $\endgroup$
    – Willie Wong
    Commented Apr 12, 2016 at 15:33
  • 1
    $\begingroup$ A third point, since $H \approx H'$, the relation is $$P_L I^+(u_{H'}, v_H) \neq 0 \iff L \lesssim H $$ Two high frequencies can interact to produce something of much lower frequencies, but not something with very high frequency. $\endgroup$
    – Willie Wong
    Commented Apr 12, 2016 at 15:35
  • 1
    $\begingroup$ You seem to be rusty with your "Littlewood-Paley Trichotomy" computations. I suggest re-reading math.ucla.edu/~tao/254a.1.01w/notes3.ps or the Appendix of Tao's Nonlinear Dispersive Equations (Principle A.5 is what is at work here). $\endgroup$
    – Willie Wong
    Commented Apr 12, 2016 at 15:40
  • 1
    $\begingroup$ You are doing it wrong on multiple levels. The support of $\psi_H$ is something like $ 2^{H-1} \leq |x| \leq 2^{H+1}$. Note first the absolute value sign. And note secondly the dyadic nature of the bounds (2 to the some power and not just a number)! // I am afraid at this point I can only throw my hands up and say that I don't have the time to help you read a paper line-by-line. Go back and really read the references that I pointed you to (Terry's lecture notes very clearly explain this on the bottom of page 2 and top of page 3). If you really want to learn this you have to actually ... $\endgroup$
    – Willie Wong
    Commented Apr 13, 2016 at 14:23
  • 2
    $\begingroup$ ... spend some time sitting down with the definitions and compute a little bit. $\endgroup$
    – Willie Wong
    Commented Apr 13, 2016 at 14:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .