Which moments identify an absolutely continuous measure on the unit circle?

Question

Suppose you have a sequence $(s_k)_{k\in\mathbb N}$ of complex values such that there exists a finite measure $\mu$ on the complex unit circle with $s_k$ as its moments $$ s_k = \int_{\mathbb T} z^kd\mu. $$ I want to know some conditions (better if necessary and sufficient) on $s_k$ so that $\mu$ is absolutely continuous wrt the 1-dimensional Lebesgue measure on $\mathbb T$.

This can be translated into a question on a real segment. In fact, in this case, there exists a measure $\nu$ on $[-\pi,\pi]$ such that $$ s_k = \int_{-\pi}^\pi e^{ikx} d\nu. $$ Being absolutely continuous translates into the existence of a real valued measurable function $f(x)$ that has $\overline s_{k}$ as complex Fourier coefficients. Now, if the series of $|s_k|$ converges, we know that the Fourier series converges, so the limit must be the density, but the limit function is absolutely continuous, and this is not always the case.

So, what are sufficient and necessary conditions on $(s_k)$ for a density function $f(x)$ to exist?

Answer 1

There is a famous theorem of F. Riesz and M. Riesz:

Define $$\widehat{\tau}(n):= \int_{\mathbb{T}} \exp(-\rm{i} n \theta) \mathop{d\tau(\theta)}$$ for all $n \in \mathbb{Z}$. Then $\tau$ is absolutely continuous with respect to the Lebesgue-measure, if $\widehat{\tau}(n)=0$ for all $n <0$.

Thus, a sufficient condition is that all $s_k =\widehat{\nu}(-k)$ vanishes. Note that we need also the positive values in order to determine $\tau$ from the Fourier-coefficients.

However, I am not aware of any necessary conditions. The mentioned theorem states also that $\lambda \ll \tau$. Thus, it is easy to construct measures on the unit-circle $\mathbb{T}$ violating the above condition.