Fourier transforms of characteristic functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T12:13:59Z http://mathoverflow.net/feeds/question/66206 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66206/fourier-transforms-of-characteristic-functions Fourier transforms of characteristic functions Andreas Thom 2011-05-27T17:54:49Z 2012-04-20T14:14:28Z <p>I am wondering how badly summable the Fourier transform of the characteristic function of a measurable subset of $S^1$ can be.</p> <blockquote> <p><strong>Question:</strong> 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 $$\widehat \chi_E(n):= \int_E z^n \ dz.$$</p> </blockquote> <p>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 $\ell^1 \mathbb Z \subset C(S^1)$.</p> <p>More concretely: </p> <blockquote> <p><strong>Question:</strong> Can anyone compute the growth of the Fourier transform of the characteristic function of something like a Cantor set of non-zero measure?</p> </blockquote> <p>Again, more abstractly:</p> <blockquote> <p><strong>Question:</strong> What can be said about the growth of the Fourier transform of the characteristic function of a generic subset of $S^1$?</p> </blockquote> http://mathoverflow.net/questions/66206/fourier-transforms-of-characteristic-functions/66207#66207 Answer by Helge for Fourier transforms of characteristic functions Helge 2011-05-27T18:03:21Z 2011-05-27T18:43:47Z <p>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.</p> <p>More interesting is what happens with the Cantor measure $\mu$. Then $\limsup_{n \to \infty} |\hat{\mu}(n)| > 0$. (Easy!).</p> <p>The main difference between these two cases is the mass of the measure ...</p> http://mathoverflow.net/questions/66206/fourier-transforms-of-characteristic-functions/94643#94643 Answer by Safari for Fourier transforms of characteristic functions Safari 2012-04-20T14:14:28Z 2012-04-20T14:14:28Z <p>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.</p>