Trigonometric moment problem

Remarks

The problem is solved. Such a measure exists if and only if, for every $n \ge 0$, the Hermitian Toeplitz matrix $T_n = (c_{j-k})_{j,k=0}^{n}$ is positive semidefinite. This is the Carathéodory–Toeplitz theorem, also known in the form of Herglotz's theorem on positive-definite sequences. Necessity is a short computation: for complex numbers $\xi_0, \ldots, \xi_n$, $$\sum_{j,k=0}^{n} c_{j-k} \xi_j \overline{\xi_k} = \frac{1}{2\pi} \int_{\mathbb{T}} \Big\lvert \sum_{j=0}^{n} \xi_j e^{-ij\theta} \Big\rvert^2 d\mu(\theta) \ge 0.$$ When the measure exists it is unique, because trigonometric polynomials are dense in the continuous functions on $\mathbb{T}$; this contrasts with the Hamburger moment problem. The truncated problem, with only $c_0, \ldots, c_n$ given, is also solvable exactly when $T_n$ is positive semidefinite.

Example: $c_0 = 1$ and $c_k = 0$ for $k \ne 0$ give $T_n = I$, and $\mu$ is Lebesgue measure $d\theta$. The sequence $c_0 = 1$, $c_{\pm 1} = 1$, $c_k = 0$ otherwise is not a moment sequence: the matrix $T_1 = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ is positive semidefinite, but $T_2$ has determinant $-1$.