We say that a Hermitian symmetric (i.e., $f_{-n} = f_n^*$ for any $n \in \mathbb{Z})$ sequence $(f_n)_{n\in \mathbb{Z}}$ is positive-definite if, for any $N \geq 0$ and any $z_0 , \ldots, z_N \in \mathbb{C}$, \begin{equation} \sum_{n,m =0}^N f_{n-m} z_n z_m^* \geq 0. \tag{1} \end{equation}

According to the Herglotz-Bochner theorem, a Hermitian symmetric sequence $(f_n)_{n\in \mathbb{Z}}$ with $f_0 = 1$ is positive-definite if and only if there exists a probability measure $\mu$ in the circle $\mathbb{T} = \mathbb{R} / \mathbb{Z}$ such that \begin{equation} f_n = \hat{\mu}_n := \int_{\mathbb{T}} \mathrm{e}^{2\pi \mathrm{i} n x } \mathrm{d}\mu (x). \end{equation}

Assume now that I am given a vector $(f_{-N_0} , \ldots, f_0 , \ldots , f_{N_0}) \in \mathbb{C}^{2N_0+1}$ such that $f_0 = 1$ and $f_{-n} = f_n^*$ for any $|n|\leq N_0$ and such that (1) holds for any $N \leq N_0$. Is it always possible to complete the vector $(f_n)_{|n|\leq N_0}$ into a positive-definite sequence $(f_n)_{n\in \mathbb{Z}}$, or, equivalently, is there always a probability measure $\mu$ in $\mathbb{T}$ such that $\hat{\mu}_n = f_n$ for $|n|\leq N_0$?