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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 ( $\sigma\ll\mu$)?

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It'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. –  Davide Giraudo Sep 26 '12 at 10:45
    
@Davide You mean $f\in L^1(\mu_2)$, right? What justifies approximation by trigonometric polynomials? –  user31373 Sep 29 '12 at 4:23
    
I'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. –  Davide Giraudo Sep 29 '12 at 9:14

1 Answer 1

Such a condition exists. The details are in Math Overflow. To find it, we need the Radon-Nikodym theorem and approximation by trigonometric polynomials.

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