Given a finite Borel measure $\mu$ on $\mathbb{S}^1 = \mathbb{R}/\mathbb{Z}$, define its Fourier coefficients by
$$ \hat\mu(n) = \int e^{2i\pi nx} d\mu(x) \qquad\forall n\in \mathbb{Z}.$$
Clearly, $(\hat\mu(n))_n$ is bounded.
- What sufficient conditions on a bounded sequence $(a_n)_n$ are known that ensure that there is a finite measure with $\hat\mu(n)=a_n$ ?
- same with probability measures instead of finite ones;
- same with necessary conditions.
I would guess that characterizations are out of reach, but maybe I am wrong?
Added in Edit: Yemon Choi rightfully asks what kind of sufficient or necessary condition I am after. Any is good for my culture, but I am especially interested in sufficient condition that enable one to construct measure satisfying constraints on Fourier coefficient.
To be honest, one of my goals was to understand why it is not easy to disprove Furstenberg's $\times 2$, $\times 3$ conjecture by simply picking Fourier coefficients $(c_n)_n$ such that $c_{2^p3^qm}=c_m$ (and $c_0=1$) inside the set of Fourier series of probability measures. I think I am starting to get the point.
Best Answer
The most elegant solution exists for problem 2: the necessary and sufficient condition for $a_n=\hat{\mu}(n)$ for a positive measure is that he sequence $(a_n)$ is non-negative semi-definite, which means that all Toplitz forms $$\sum_{i,j=0}^na_{i-j}z_i\overline{z}_j\geq 0$$ for all integers $n$ and complex $z_i$. This is a theorem of Caratheodory and Toplitz (see, for example N. Akhiezer, Classical moment problem, Ch V, section 1).
For probability measures one has to add to this $a_0=1$.
For 1, the best necessary and sufficient condition is that $a_n$ is a difference of two on-negative semi-definite sequences, which is not very effective, of course. Of course these conditions are not always easy to check, but one can derive many simple necessary conditions.
Bochner's theorem is a continuous analog of this (for Fourier transforms of non-periodic measures).