What can be said about the Fourier transforms of characteristic functions?

What can be said about the Fourier transform of the characteristic function $1_A$, where $A\subset \mathbb{R}^n$ is of finite Lebesgue measure? In particular,

What properties are common to Fourier transforms of all characteristic functions?

Here are a few trivial properties; what other properties are known?

• $\widehat{1_A}$ is a bounded continuous function converging to zero.
• $\widehat{1_A}$ is in $L^2$.

What interesting functional analytic properties does the set $\{\widehat{1_A}: A \in \mathbb{R}^n\}$ of Fourier transforms of characteristic functions have?

At least it is closed in $L^2$ (just apply Plancherel's formula and use the fact that an $L^2$-limit of characteristic functions is a characteristic function). Is it closed in other norms? Is it dense in some interesting spaces (if we are allowed to multiply the functions by a constants)? For which $p$ is the Fourier transform a bounded operator from our set to $L^p$?

How does the regularity (in a vague sense) of $A$ affect on the decay of $\widehat{1_A}$? Are these Fourier transforms always entire?

Here are a few easy remarks:

• If $A$ is a finite union of intervals, then $\hat{1}_A(\xi)$ is a trigonomteric polynomial divided by $\xi$, so it is in every $L^p,p>1$ but not absolutely integrable.
• If $A$ is bounded, the Fourier transform is an entire function in $\mathbb{C}^n$.
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I don't think we expect the regularity of $A$ to affect the REGULARITY of $\widehat{\chi_A}$ (rather, regularity/smoothness <-> decay is the received wisdom). –  Christian Remling Apr 22 '14 at 23:05
By regularity of the Fourier transform, I meant both its decay and smoothness. If these Fourier transforms are always entire, then I just mean decay, but if we for example take a Cantor-type set extending to infinity with positive finite measure, I don't see directly why the Fourier transform should be analytic. –  Joni Teräväinen Apr 22 '14 at 23:22
I think it's hopeless to look for a characterization in terms of norms or asymptotic properties; these would not distinguish small perturbations of a characteristic function. Maybe you could use the fact that they coincide with their square, so that the transform $f$ satisfies $f*f=f$ –  Piero D'Ancona Apr 24 '14 at 11:06

2 Answers

EDIT. I found papers in which the problem of decay and integrability of $\widehat{1_A}$ has been studied and largely answered, at least for domains with some regularity.

This recent paper by Lebedev proves the following:

If $A\subset \mathbb{R}^n (n\geq 2$) has $C^2$-boundary (and finite measure), then the exponent $p=\frac{2n}{n+1}$ is ''critical'' for the integrability of the Fourier transform $\widehat{1_A}$, that is, $\widehat{1_A}\in L^p$ if and only if $p>\frac{2n}{n+1}$. Actually, even the case where $A$ is a ball shows that this cannot be improved. The same conclusion was proved by Herz if $A$ is convex but no regualrity is assumed.

A bit surprisingly, if $\partial A$ has less smoothness, namely it is just $C^1$, then for $n=2$ it is possible that $\widehat{1_A}\in L^p$ for $p>1$.

Also the decay of $\widehat{1_A}$ has been studied; for example in This paper of Svensson and This paper of Brandolini.

According to the first paper, for bounded sets $A$ that are closures of open sets with $C^{\infty}$-boundary, we have $\widehat{1_A}(\xi)=O(|\xi|^{-\frac{n+2}{2}})$ if and only if the Gaussian curvature of $A$ is nonzero everywhere.

Despite these results, it might be that for general measurable sets we cannot say anything; I have not found references yet.

I obtained some partial results for my question.

• By the Hausdorff-Young inequality, $\|\widehat{1_A}\|_q\leq \|1_A\|_p=m(A)^{\frac{1}{p}}$ for every $q\geq 2$, where $p=\frac{q}{q-1}$.
• The Fourier transform $\widehat{1_A}$ does not generally extend to an entire function. Indeed, suppose that $\widehat{1_A}$ is always analytic in $\Omega\subset \mathbb{C}^n$. Then also $\hat{s}$ is analytic in $\Omega$ for every simple function $s$ which is nonzero only in a set of finite measure. Simple functions with bounded support are dense in $L^1(\mathbb{R}^n)$ (for every $k$, choose a function $s_k$ that differs from $f1_{[-k,k]^n}1_{\{|f|\leq k\}}$ by at most $2^{-k}$), so for any $f\in L^1(\mathbb{R}^n)$ we find simple functions $s_k$ with bounded supports such that $\|f-s_k\|_1\to 0$. But then also $\|\hat{f}-\hat{s_k}\|_{\infty}\to 0$, and by Weierstrass' theorem, $\hat{f}$ is analytic in $\Omega$ as a uniform limit of analytic functions. To see the contradiction, it is now enough to take $f(x)=\frac{\sin^2 x}{x^2}$ (or a similar function in higher dimensions) whose Fourier transform is a triangular wave.

However, the first result does not tell whether $\widehat{1_A}$ is in $L^p$ for $1<p<2$ (this would also imply that the set would be closed in $L^p$), and the second one concerns only analyticity instead of weaker forms of smoothness.

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This is by now a rather old question, but I thought it might be worth pointing out that, in fact, $\hat{\mathbb 1}_A$ does extend to an entire function when $A$ is bounded. This is part of the Paley-Wiener family of results. The problem with the uniform approximation argument is that the convergence $\hat{s}_k\to\hat{f}$ takes place in $L^\infty(\mathbb R^n)$ and not in $L^\infty(\mathbb C^n)$. –  Nick Strehlke Sep 12 '14 at 1:49
You are right. I am fairly certain though that there should be an unbounded set of finite measure whose Fourier transform is not analytic. –  Joni Teräväinen Sep 13 '14 at 23:35

I think you will be interested in section 2.2 of Shao's recent paper

http://arxiv.org/pdf/1308.2247v1.pdf

For the case of the characteristic function he improves the optimal constant obtained by Beckner in Hausdorff-Young (optimal over all functions).

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Thanks for the reference, but I think they only consider the case of an even integer $q$. –  Joni Teräväinen Apr 24 '14 at 9:42