# Trigonometric moment problem

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 $\mu$ on the interval $[0,2\pi ]$ such that: $c_{k}={\frac {1}{2\pi }}\int _{0}^{2\pi }e^{-ik\theta }\,d\mu (\theta ),$ with $c_{-k}={\overline {c}}_{k}$ for $k\geq 1$. In case the sequence is finite, i.e., $\{c_{k}\}_{k=0}^{n<\infty }$, it is referred to as the truncated trigonometric moment problem.
An affirmative answer to the problem means that $\{c_{k}\}_{k\in \mathbb {N} _{0}}$ are the Fourier-Stieltjes coefficients for some (consequently positive) Radon measure $\mu$ on $[0,2\pi ]$.

## 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 $j\leq 2n+1<\infty$ (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 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 $V$.