I am interested in the following somewhat obscure question:

Is there some $n \in \Bbb{N}$, and a set $A \subset \Bbb{R}^n$ of finite measure such that the Fourier transform $\widehat{1_A}$ of its indicator function satisfies $\widehat{1_A} \in \bigcap_{q > 1} L^q$, but such that there is some $p \in (1,\infty)$ such that $1_A$ is not a Fourier multiplier for $L^p$, i.e., such that the map $L^p \to L^p, g \mapsto g \ast \widehat{1_A}$ is not bounded (and thus not even well-defined).

The motivation for looking at this question is the following: We know that if $f \in L^1$, then (by Young's inequality) the map $g \mapsto g \ast f$ is bounded on every $L^p$ space, $1 \leq p \leq \infty$. Furthermore, even if one is only intersted in boundedness on the $L^p$ spaces with $1<p<\infty$, the assumption $f \in L^1$ cannot be weakened in general to the assumption $f \in \bigcap_{q > 1} L^q$, as can be seen by considering $f(x) = 1_{[1,\infty)}(x) \cdot x^{-1} \cdot \ln x$, $p=2$, and $g(x) = 1_{[e,\infty)} (x) \cdot x^{-1/2} \cdot (\ln x)^{-3/2}$.

However, to provide a counterexample to a certain claim (see below), I need a similar counterexample, but for the case where the convolution $g \mapsto g \ast f$ is a projection, which would be satisfied if $f = \widehat{1_A}$.

Sadly, the simplest example (i.e., the ball multiplier) does not work, since it is an $L^p$ multiplier for all $1 < p < \infty$ in dimension $n=1$, and does not satisfy $\widehat{1_B} \in \bigcap_{q > 1} L^q$ for $n > 1$.

I have found the question What can be said about the Fourier transforms of characteristic functions?, where a paper by Lebedev is cited which shows the following (see Corollary 4 of the paper):

There exists a bounded domain $D \subset \Bbb{R}^2$ with $C^1$ boundary, and with $\widehat{1_D} \in \bigcap_{q>1}L^q$. The boundary of $D$ does not contain line intervals.

I have no idea, however, how I could show (or if it is true) that $1_D$ is not a Fourier multiplier on some $L^p$ space, $1 < p < \infty$.

Also, Corollary 2 of the same paper shows

Let $D \subset \Bbb{R}^n$ ($n \geq 2$) be a bounded domain with $\partial D \in C^{1,1}$. Then $\widehat{1_D} \notin L^p$ for $p \leq 2n/(n+1)$.

Thus, at least for dimension $n \geq 2$, this shows that if a set with the desired properties exist, then it cannot have a $C^{1,1}$ boundary.

Finally, a brief remark on the claim I want to find a counterexample for: A continuous Parseval frame is a family $(\psi_x)_{x \in \Omega}$ in a Hilbert space $\mathcal{H}$, indexed by a measure space $(\Omega, \mu)$, such that $x \mapsto \psi_x$ is weakly measurable, and such that $$ \int_\Omega |\langle f, \psi_x \rangle|^2 d \mu(x) = \| f \|_{\mathcal{H}}^2 \qquad \forall f \in \mathcal{H}. $$ I would like to find such a Parseval frame such that the Gramian kernel $K(x,y) = \langle \psi_y, \psi_x \rangle$ does not yield a bounded operator on some $L^p (\Omega)$ space, $1<p<\infty$, but such that it satisfies $$ \sup_x \int_\Omega |K(x,y)|^q d\mu(y) < \infty $$ for all $1 < q <\infty$. If there exists a set $A$ as in my question above, then I can simply take $\mathcal{H} = L^2(A)$, and $\psi_\xi (x) = e^{2\pi i \xi x}$, $\xi \in \Bbb{R}$.

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    $\begingroup$ 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. $\endgroup$ – Christian Remling Oct 28 '17 at 20:40
  • $\begingroup$ @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. $\endgroup$ – PhoemueX Oct 30 '17 at 7:49
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    $\begingroup$ 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. $\endgroup$ – fedja Oct 31 '17 at 15:57
  • $\begingroup$ @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... $\endgroup$ – PhoemueX Oct 31 '17 at 19:01
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    $\begingroup$ 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 $L^2$. $\endgroup$ – fedja Oct 31 '17 at 20:20

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