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I would like to know if there is a measurable set $M \subset \Bbb{R}$ such that

  1. $M$ has finite Lebesgue measure $0 < \lambda(M) < \infty$,

  2. $M$ is unbounded in the sense that $\lambda(M \setminus [-r,r]) > 0$ for all $r > 0$.

  3. The indicator function of $M$ is an $L^p$-Fourier multiplier for some $p \neq 2$, i.e., the operator $L^2 \to L^2, f \mapsto \mathcal{F}^{-1}(\widehat{f} \cdot 1_M)$ can be extended to a bounded linear operator on $L^p(\Bbb{R})$ for some $p \neq 2$.

  4. We have $\mathcal{F}^{-1} 1_M \in \bigcap_{1 < q < \infty} L^q (\Bbb{R})$.

As a first step, it would also be fine to only have an example where the first three properties are satisfied.

Some observations:

  1. By a result of Lebedev and Olevskii (https://eudml.org/doc/58174) any set satisfying condition 3 from above needs to be an open set, at least after modifying $M$ on a null-set.

  2. Having finite measure is a necessary condition for condition 4, since if $\mathcal{F}^{-1} 1_M \in L^2$, then $1_M \in L^2$, so that $M$ needs to have finite measure.

  3. As noted in a paper by Mockenhaupt and Ricker , the class of all sets $M$ such that $1_M$ is an $L^p$-Fourier multiplier forms an algebra of sets that

    a) is closed under scalings and translations,

    b) contains all (bounded or unbounded) intervals, since the Hilbert transform is bounded on $L^p$,

    c) contains all unions of dyadic intervals $[2^j, 2^{j+1})$ for $j \in \Bbb{Z}$, essentially by Littlewood-Paley theory.

But I was unable to use these properties to either construct a set $M$ as I would like to have, or to show that no such set can exist.

Any help would be appreciated.

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  • $\begingroup$ Can't we just take a union of intervals (say $(n,n+d_n)$) with $d_n\to 0$ sufficiently rapidly? $\endgroup$
    – Christian Remling
    Commented Apr 22, 2018 at 17:26
  • $\begingroup$ @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. $\endgroup$
    – PhoemueX
    Commented Apr 22, 2018 at 17:41
  • $\begingroup$ 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 $\widehat{\chi_I}$). $\endgroup$
    – Christian Remling
    Commented Apr 22, 2018 at 18:53
  • $\begingroup$ 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). $\endgroup$
    – fedja
    Commented Apr 23, 2018 at 0:58
  • $\begingroup$ @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 :) $\endgroup$
    – PhoemueX
    Commented Apr 24, 2018 at 20:37

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