Is there any condition on the Fourier transforms of 2 positive measures $\sigma , \mu$ on the complex unit circle $\mathbb{T}$ that implies absolute continuity ( $\sigma\ll\mu$)?
A condition on Fourier transforms that implies absolute continuity
4
$\begingroup$
measure-theory
fourier-analysis
-
0It's not the answer, but at least, when $\mu_1\ll \mu_2$ we can find a function $f\in L¹$ such that $f\mu_2=\mu_1$.Approximating $f$ by a sequence of trigonometric polynomials, we get that the Fourier transform of $\mu_1$ is in the closure for the uniform norm of elements of the form $\sum_{k\in I}a_k\widehat\mu_2(t+k)$ where $I$ is a finite subset of $\Bbb Z$. Now I will try to see whether this condition is sufficient. ā 2012-09-26
-
0@Davide You mean $f\in L^1(\mu_2)$, right? What justifies approximation by trigonometric polynomials? ā 2012-09-29
-
0I'm not expert in harmonic analysis. At least, what I did works when the continuous functions are dense in $L^1(\mu_2)$ (yes I should precise what measure we are dealing with). I don't know whether it's the case, and under which conditions. I will look in Rudin's book. ā 2012-09-29