Fourier transforms of characteristic functions - MathOverflow

Fourier transforms of characteristic functions

2011-05-27T17:54:49Z

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 $S^1$?

Answer by Helge for Fourier transforms of characteristic functions

2011-05-27T18:03:21Z

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 ...

Answer by Safari for Fourier transforms of characteristic functions

2012-04-20T14:14:28Z

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.