Timeline for Existence of unbounded $M \subset \Bbb{R}$ of finite measure s.t. $1_M$ is $L^p$-Fourier multiplier
Current License: CC BY-SA 3.0
6 events
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| Apr 24, 2018 at 20:37 | comment | added | PhoemueX | @fedja: Thanks for the comment. I think I need to refresh my knowledge of Littlewood Paley theory to be sure. Once I have done that I will try to answer my own question :) | |
| Apr 23, 2018 at 0:58 | comment | added | fedja | Looks like by Littlewood-Paley you should be fine with $\cup_{n\ge 1}[2^n,2^n+\delta_n]$ with $\delta_n$ tending to $0$ fast enough, shouldn't you? (there is nothing sacred about partitioning exactly at the powers of $2$, so you can just shift a bit and subtract the standard partition). | |
| Apr 22, 2018 at 18:53 | comment | added | Christian Remling | Ah, ok, that was too naïve, I of course assumed that something like this would be true (as it is for the $L^p$ norm of $\widehat{\chi_I}$). | |
| Apr 22, 2018 at 17:41 | comment | added | PhoemueX | @ChristianRemling: That was my original idea. However, I am not sure at all how we can guarantee that this will be an $L^p$ Fourier multiplier. The problem, if I am not completely mistaken, is that in the operator norm $L^p \to L^p$, each of the Fourier multipliers $f \mapsto \mathcal{F}^{-1}(\widehat{f} \cdot 1_I)$ with an interval $I$ has the same norm (it does not decay if the length of the interval goes to zero), so at least we cannot apply some standard argument to argue that we will get a Fourier multiplier. | |
| Apr 22, 2018 at 17:26 | comment | added | Christian Remling | Can't we just take a union of intervals (say $(n,n+d_n)$) with $d_n\to 0$ sufficiently rapidly? | |
| Apr 22, 2018 at 17:14 | history | asked | PhoemueX | CC BY-SA 3.0 |