I am wondering how badly summable the Fourier transform of the characteristic function of a measurable subset of $S^1$ can be.

Question: Let $\alpha \colon \mathbb N \to [1,\infty)$ be a monotone increasing function with $\lim_{n \to \infty} \alpha(n) = \infty$. Is there a measurable subset $E \subset S^1$, such that $$\sum_{n \in \mathbb Z} | \widehat \chi_E(n)|^2 \cdot \alpha(|n|) = \infty \ ?$$ Here, $\widehat \chi(n)$ are the usual moments $$\widehat \chi_E(n):= \int_E z^n \ dz.$$

The only example I know is the Fourier transform of the characteristic function of an interval, which grows like $1/n$. On the other hand, one can easily see that the growth cannot be better than $1/n$ (something like $1/n^{1 + \varepsilon}$), since $\ell^1 \mathbb Z \subset C(S^1)$.

More concretely:

Question: Can anyone compute the growth of the Fourier transform of the characteristic function of something like a Cantor set of non-zero measure?

Again, more abstractly:

Question: What can be said about the growth of the Fourier transform of the characteristic function of a generic subset of $S^1$?


2 Answers 2


It is a difficut problem to compute the Fourier transform of the characteristic function of a union of open intervals (in the general case) and it is known that such Fourier transform can converge to 0 with a very slow growth rate. I am currently working on that problem. Probably the growth rate may depend on some arithmetical properties of the boundary of the set as it is the case in studying the Fourier dimension of sets.


The Fourier transform of the Cantor set is just zero. (It has zero Lebesgue measure.) For a subset $E \subseteq S^1$, the measure $\mu = \chi_E dz$ is absolutely continuous, so the Fourier coefficients converge to $0$. This is just the Riemann--Lebesgue Lemma.

More interesting is what happens with the Cantor measure $\mu$. Then $\limsup_{n \to \infty} |\hat{\mu}(n)| > 0$. (Easy!).

The main difference between these two cases is the mass of the measure ...

  • 3
    $\begingroup$ The Fourier coefficients of a bounded function can dominate any given $\ell^2$ sequence. Your condition is convex and bounded functions are essentially convex combinations of characteristic functions, so the answer to your main question is "Yes" (well, formally what I said only shows that there is no constant upper bound but it is fairly easy to get $+\infty$ from there by the usual sliding hump method). I do not know if you can dominate every $\ell^2$ sequence by the Fourier coefficients of a characteristic function though. $\endgroup$
    – fedja
    May 27, 2011 at 19:13
  • $\begingroup$ OK. But middle fifth at every stage still gets measure zero in the end. $\endgroup$ May 27, 2011 at 19:13
  • $\begingroup$ Fedja, why don't you put this as an answer. $\endgroup$ May 27, 2011 at 20:37
  • $\begingroup$ Gerald, you are right. The obvious Cantor sets have zero measure. Anyhow, if you take out a fraction depending on the step of the iteration, you can achieve that the limit has positive measure. $\endgroup$ May 27, 2011 at 20:44
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    $\begingroup$ @ Andreas: Nah, I knew this well enough. Once I reached my 10K, I made it a general rule not to try to get any rep. points for things that I just happen to know off hand. $\endgroup$
    – fedja
    May 27, 2011 at 23:07

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