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 $|n|\leq N_0$?


1 Answer 1


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, 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 for background. This extension gives us the desired measure.

  • $\begingroup$ Very nice. The extension of positive linear functional indeed do the job. Thanks a lot! $\endgroup$
    – Goulifet
    Commented Aug 20, 2020 at 2:20
  • $\begingroup$ @Goulifet: By the way, my notation in the first line is a bit sloppy, I really mean what you get if you assume that $\mu$ has the right Fourier coefficients, and then you multiply out $\int |p|^2\, d\mu$ (in other words, this depends on the $f_n$ only and not on a hypothetical $\mu$). But I leave it in this form, it seems clear from the context. $\endgroup$ Commented Aug 20, 2020 at 2:30
  • $\begingroup$ I agree, I got your point so I think the context is indeed clear enough. $\endgroup$
    – Goulifet
    Commented Aug 20, 2020 at 3:11
  • $\begingroup$ @Goulifet: One more afterthought: In the analogous problem for moments (instead of Fourier coefficients), there is a rather explicit description of all such measures $\mu$ ("Nevanlinna parametrization"). There probably must be something similar here. $\endgroup$ Commented Aug 20, 2020 at 14:15

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