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 $|n|\leq N\_0$?



Existence of probability measure on the circle with given Fourier coefficients
Yes, this works. Condition (1) says that $\int |p(e^{ix})|^2\, d\mu(x)\ge 0$ for every polynomial $p(z)=\sum\_{n=0}^N p\_n z^n$. By the [Fejer-Riesz theorem](https://encyclopediaofmath.org/wiki/Fej%C3%A9r-Riesz_theorem), these squares $|p|^2$ range exactly over the trigonometric polynomials $f=\sum\_{|n|\le N} f\_n z^n$ with $f\ge 0$ on $|z|=1$.


So we have a positive linear functional on this vector space $\{ f = \sum\_{|n|\le N} f\_n z^n \}$. This can be extended to a positive linear functional on $C(T)$; see [here](https://en.wikipedia.org/wiki/Positive_linear_functional) for background. This extension gives us the desired measure.


