Imagine $\mu$ and $\nu$ are two Borel probabilty measures in the interval $[0,1]$.

We say that $\mu$ is absolutely continuous with respect to $\nu$, if for every measurable set $A$ such that $\nu(A)=0$, we also have $\mu(A)=0$.

We can define the Fourier coefficients of $\mu$ by $$\hat\mu(n)=\int_{[0,1]} e^{2\pi i n x}d\mu(x),\quad n\in\mathbb Z,$$ and we can define $\hat\nu(n)$ similarly.

Question Is there a way to characterize the absolute continuity of $\mu$ with respect to $\nu$ by looking only at their Fourier coefficients $\hat\mu(n)$ and $\hat\nu(n)$?

Thanks a lot!


1 Answer 1


The condition (C) is stated bellow (but it is only a necessary condition). It does not look easy to check in practical cases.

First, some notations. If $a=(a_n)_{n\in\mathbb Z}$ is a sequence of complex number, then $\tau(a):=(a_{n+1})_{n\in\mathbb Z}$ denotes the shift sequence. If $P=\sum_{j=-N}^Nc_je^{2\pi ijx}$ ($P$ is a trigonometric polynomial), we define the operator $\widetilde P$ define on the sequences of complex numbers by $\widetilde P(a)=\sum_{j=-N}^Nc_j\tau^j(a)$.

Assume that $\mu\ll\nu$. Then there exists a non-negative function $f\in\mathbb L^1(\nu)$ such that for each measurable subset $A$ of $[0,1]$, $\mu(A)=\int_Af(x)\mathrm d\nu(x)$. Fix $\varepsilon\gt 0$. There is some non-negative simple function $g$ such that $\lVert f-g\rVert_{\mathbb L^1(\nu)}\lt\varepsilon$. For each measurable set involved in this simple function, approximate in $\mathbb L^1(\nu)$ its characteristic function by a continuous function. These continuous function may be approximated up to $\varepsilon$ by non-negative trigonometric polynomials. We thus conclude that $$\tag{C}(\widehat{\mu}(n))_{n\in\mathbb Z}\in \overline{\{\widetilde{P}(\widehat{\nu}), P\mbox{ is a non negative trigonometric polynomial}\}}^{\ell^\infty}.$$

  • $\begingroup$ Dear Davide, I am slightly confused by the "only if" part of your statement. Suppose $\mu=\nu=\lambda$ is the Lebesgue measure. The Fourier coefficients are $\hat{\mu}(n) = \delta_{n0}$ hence $P_k = 1$ for any $k\ge 1$ and $S_k = k$ is not a bounded sequence in $L^1(\nu)$? Also if I understood the statement correctly, it implies that $L^1(\lambda)$ can be identified with the measures with Fourier coefficients in $c_0$. However, there exists measures with Fourier coefficients in $c_0$ which are not absolutely continuous with respect to $\lambda$ -- see Rajchman measures. Am I missing something? $\endgroup$ Dec 5, 2015 at 10:41
  • $\begingroup$ Dear Jan, it seems that my argument for this part does not work (thanks for your reading). I will try to think whether the condition (C) is sufficient. $\endgroup$ Dec 6, 2015 at 9:19
  • $\begingroup$ I fear condition (C) cannot be sufficient as it would imply for $\nu$ being the lebesgue measure that every measure with Fourier coefficients in $c_0$ is absolutely continuous; however there exist counterexamples. $\endgroup$ Dec 7, 2015 at 23:36

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