Timeline for Existence of $A\subset\Bbb{R}^n$ of finite measure and $\hat{1_A}\in\bigcap_{q>1}L^q$, but s.t. for some $1<p<\infty$, $1_A$ is no $L^p$-Fourier mult
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| Oct 31, 2017 at 20:24 | comment | added | PhoemueX | @fedja: Of course :) Thank you very much! I also just found the following question which is related: mathoverflow.net/questions/3494/lp-multiplier-sets | |
| Oct 31, 2017 at 20:20 | comment | added | fedja | The easiest way is to show that a random set is not a multiplier. Indeed, otherwise we would have any sequence of $\pm 1$ a multiplier but the random choice of signs makes all $L^p$ norms equivalent (Khinchine) and we can easily go outside $L^2$. | |
| Oct 31, 2017 at 19:01 | comment | added | PhoemueX | @fedja: Thanks, that's a great idea! I guess you want to use the "transference principle" between Fourier multipliers on $\Bbb{R}^n$ and $\Bbb{T}^n$ to conclude that we really do not get a Fourier multiplier. When I tried to formalize your idea, I came upon an embarrassing problem: At the moment, I am unable to show that there actually is some $A \subset \Bbb{Z}$ such that $1_A$ is not an $L^p$ Fourier multiplier on the torus :( Do you know any (easy) example of such a set? I might just be too tired at the moment and will have another try tomorrow... | |
| Oct 31, 2017 at 15:57 | comment | added | fedja | Another obvious possibility is to take $n=1$, consider any sequence of integers $\Lambda\subset\mathbb Z$ whose characteristic function is not a multiplier in any $L^p(\mathbb T)$ and to define $A$ as the union of intervals around points in $\Lambda$ whose lengths decay fast enough. | |
| Oct 30, 2017 at 7:49 | comment | added | PhoemueX | @ChristianRemling: Thanks very much for your comment. Up to now, I have not read his article - I only read the statement in books. I will have a look at the article. | |
| Oct 28, 2017 at 20:40 | comment | added | Christian Remling | I would try to take a closer look at the $D$ from your second box. Fefferman's proof that $\chi_B$ is not a multiplier for any $p\not=2$ is fairly robust; he gets rid of $B$ right at the beginning. I think it should work for any set that can approximate half planes with any orientation when rescaled to a large size. | |
| Oct 28, 2017 at 10:05 | history | asked | PhoemueX | CC BY-SA 3.0 |