Trigonometric moment problem

In mathematics, the trigonometric moment problem is formulated as follows: given a finite sequence {α0, ... αn }, does there exist a positive Borel measure μ on the interval [0, 2π] such that

\alpha_k = \frac{1}{2 \pi}\int_0 ^{2 \pi} e^{-ikt}\,d \mu(t).

In other words, an affirmative answer to the problems means that {α0, ... αn } are the first n + 1 Fourier coefficients of some positive Borel measure μ on [0, 2π].

Characterization

The trigonometric moment problem is solvable, that is, {αk} is a sequence of Fourier coefficients, if and only if the (n + 1) × (n + 1) Toeplitz matrix


A =
\left(\begin{matrix}
\alpha_0       & \alpha_1           & \cdots   & \alpha_n     \\
\bar{\alpha_1} & \alpha_0           & \cdots   & \alpha_{n-1} \\
\vdots         & \vdots             & \ddots   & \vdots       \\
\bar{\alpha_n} & \bar{\alpha_{n-1}} & \cdots   & \alpha_0     \\
\end{matrix}\right)

is positive semidefinite.

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 A defines a sesquilinear product on Cn + 1, resulting in a Hilbert space

(\mathcal{H}, \langle \;,\; \rangle)

of dimensional at most n + 1, a typical element of which is an equivalence class denoted by [f]. The Toeplitz structure of A means that a "truncated" shift is a partial isometry on \mathcal{H}. More specifically, let { e0, ...en } be the standard basis of Cn + 1. Let \mathcal{E} be the subspace generated by { [e0], ... [en - 1] } and \mathcal{F} be the subspace generated by { [e1], ... [en] }. 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 = A_{j+1, k+1} = A_{j, k} = \langle [e_{j+1}], [e_{k+1}] \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 T such that for all integer k

\langle (U^*)^k [ e_ {n+1} ], [ e_ {n+1} ] \rangle = \int_{\mathbf{T}} z^{k} dm .

For k = 0,...,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 
= A_{n+1, n+1-k} 
= \bar{\alpha_k}.

So


\int_{\mathbf{T}} z^{-k} dm
= \int_{\mathbf{T}} \bar{z}^k dm
= \alpha_k.

Finally, parametrize the unit circle T by eit on [0, 2π] gives

\frac{1}{2 \pi} \int_0 ^{2 \pi} e^{-ikt} d\mu(t) = \alpha_k

for some suitable measure μ.

Parametrization of solutions

The above discussion shows that the trigonometric moment problem has infinitely many solutions if the Toeplitz matrix A is invertible. In that case, the solutions to the problem are in bijective correspondence with minimal unitary extensions of the partial isometry V.

References

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