3
$\begingroup$

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 $|n|\leq N_0$?

$\endgroup$
0

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.