Trigonometric moment problem
title: "Trigonometric moment problem" type: doc version: 1 created: 2026-02-28 author: "Wikipedia contributors" status: active scope: public tags: ["probability-problems", "measure-theory", "functional-analysis"] topic_path: "general/probability-problems" source: "https://en.wikipedia.org/wiki/Trigonometric_moment_problem" license: "CC BY-SA 4.0" wikipedia_page_id: 0 wikipedia_revision_id: 0
In mathematics, the trigonometric moment problem is formulated as follows: given a sequence {c_k}{k \in \mathbb{N}{0}}, does there exist a distribution function \sigma on the interval [0,2\pi] such that: c_k = \frac{1}{2 \pi}\int_0 ^{2 \pi} e^{-ik\theta},d \sigma(\theta), with c_{-k} = \overline{c}k for k \geq 1. An affirmative answer to the problem means that {c_k}{k \in \mathbb{N}_{0}} are the Fourier-Stieltjes coefficients for some (consequently positive) unique Radon measure \mu on [0,2\pi] as distribution function.
In case the sequence is finite, i.e., {c_k}_{k = 0}^{n , it is referred to as the truncated trigonometric moment problem.
Characterization
The trigonometric moment problem is solvable, that is, {c_k}{k=0}^{n} is a sequence of Fourier coefficients, if and only if the (n + 1) × (n + 1) Hermitian Toeplitz matrix T = \left(\begin{matrix} c_0 & c_1 & \cdots & c_n \ c{-1} & c_0 & \cdots & c_{n-1} \ \vdots & \vdots & \ddots & \vdots \ c_{-n} & c_{-n+1} & \cdots & c_0 \ \end{matrix}\right) with c_{-k}=\overline{c_{k}} for k \geq 1, is positive semi-definite.
The "only if" part of the claims can be verified by a direct calculation. We sketch an argument for the converse. The positive semidefinite matrix T defines a sesquilinear product on \mathbb{C}^{n+1}, resulting in a Hilbert space (\mathcal{H}, \langle ;,; \rangle) of dimensional at most n + 1. The Toeplitz structure of T means that a "truncated" shift is a partial isometry on \mathcal{H}. More specifically, let {e_0,\dotsc,e_n} be the standard basis of \mathbb{C}^{n+1}. Let \mathcal{E} and \mathcal{F} be subspaces generated by the equivalence classes {[e_0],\dotsc,[e_{n-1}]} respectively {[e_1],\dotsc,[e_{n}]}. Define an operator V: \mathcal{E} \rightarrow \mathcal{F} by V[e_k] = [e_{k+1}] \quad \mbox{for} \quad k = 0 \ldots n-1. Since \langle V[e_j], V[e_k] \rangle = \langle [e_{j+1}], [e_{k+1}] \rangle = T_{j+1, k+1} = T_{j, k} = \langle [e_{j}], [e_{k}] \rangle, V can be extended to a partial isometry acting on all of \mathcal{H}. Take a minimal unitary extension U of V, on a possibly larger space (this always exists). According to the spectral theorem, there exists a Borel measure m on the unit circle \mathbb{T} such that for all integer k \langle (U^)^k [ e_ {n+1} ], [ e_ {n+1} ] \rangle = \int_{\mathbb{T}} z^{k} dm . For k = 0,\dotsc,n, the left hand side is \langle (U^)^k [ e_ {n+1} ], [ e_ {n+1} ] \rangle = \langle (V^*)^k [ e_ {n+1} ], [ e_{n+1} ] \rangle = \langle [e_{n+1-k}], [ e_{n+1} ] \rangle = T_{n+1, n+1-k} = c_{-k}=\overline{c_k}. As such, there is a j-atomic measure m on \mathbb{T}, with j \leq 2n + 1 (i.e. the set is finite), such that c_k = \int_{\mathbb{T}} z^{-k} dm = \int_{\mathbb{T}} \bar{z}^k dm, which is equivalent to c_k = \frac{1}{2 \pi} \int_0 ^{2 \pi} e^{-ik\theta} d\mu(\theta).
for some suitable measure \mu.
Parametrization of solutions
The above discussion shows that the truncated trigonometric moment problem has infinitely many solutions if the Toeplitz matrix T is invertible. In that case, the solutions to the problem are in bijective correspondence with minimal unitary extensions of the partial isometry V.
Notes
References
::callout[type=info title="Wikipedia Source"] This article was imported from Wikipedia and is available under the Creative Commons Attribution-ShareAlike 4.0 License. Content has been adapted to SurfDoc format. Original contributors can be found on the article history page. ::