Trigonometric moment problem

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In mathematics, the trigonometric moment problem is formulated as follows: given a finite sequence [math]\displaystyle{ \{c_0,\dotsc,c_n\} }[/math], does there exist a distribution function [math]\displaystyle{ \mu }[/math] on the interval [math]\displaystyle{ [0,2\pi] }[/math] such that:[1]

[math]\displaystyle{ c_k = \frac{1}{2 \pi}\int_0 ^{2 \pi} e^{-ik\theta}\,d \mu(\theta). }[/math]

In other words, an affirmative answer to the problems means that [math]\displaystyle{ \{c_0,\dotsc,c_n\} }[/math] are the first n + 1 Fourier coefficients of some measure [math]\displaystyle{ \mu }[/math] on [math]\displaystyle{ [0,2\pi] }[/math].

Characterization

The trigonometric moment problem is solvable, that is, [math]\displaystyle{ \{c_k\}_{k=0}^{n} }[/math] is a sequence of Fourier coefficients, if and only if the (n + 1) × (n + 1) Hermitian Toeplitz matrix

[math]\displaystyle{ 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) }[/math] with [math]\displaystyle{ c_{-k}=\overline{c_{k}} }[/math] for [math]\displaystyle{ k \geq 1 }[/math],

is positive semi-definite.[2]

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 [math]\displaystyle{ T }[/math] defines a sesquilinear product on [math]\displaystyle{ \mathbb{C}^{n+1} }[/math], resulting in a Hilbert space

[math]\displaystyle{ (\mathcal{H}, \langle \;,\; \rangle) }[/math]

of dimensional at most n + 1. The Toeplitz structure of [math]\displaystyle{ T }[/math] means that a "truncated" shift is a partial isometry on [math]\displaystyle{ \mathcal{H} }[/math]. More specifically, let [math]\displaystyle{ \{e_0,\dotsc,e_n\} }[/math] be the standard basis of [math]\displaystyle{ \mathbb{C}^{n+1} }[/math]. Let [math]\displaystyle{ \mathcal{E} }[/math] and [math]\displaystyle{ \mathcal{F} }[/math] be subspaces generated by the equivalence classes [math]\displaystyle{ \{[e_0],\dotsc,[e_{n-1}]\} }[/math] respectively [math]\displaystyle{ \{[e_1],\dotsc,[e_{n}]\} }[/math]. Define an operator

[math]\displaystyle{ V: \mathcal{E} \rightarrow \mathcal{F} }[/math]

by

[math]\displaystyle{ V[e_k] = [e_{k+1}] \quad \mbox{for} \quad k = 0 \ldots n-1. }[/math]

Since

[math]\displaystyle{ \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, }[/math]

[math]\displaystyle{ V }[/math] can be extended to a partial isometry acting on all of [math]\displaystyle{ \mathcal{H} }[/math]. Take a minimal unitary extension [math]\displaystyle{ U }[/math] of [math]\displaystyle{ V }[/math], on a possibly larger space (this always exists). According to the spectral theorem, there exists a Borel measure [math]\displaystyle{ m }[/math] on the unit circle [math]\displaystyle{ \mathbb{T} }[/math] such that for all integer k

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

For [math]\displaystyle{ k = 0,\dotsc,n }[/math], the left hand side is

[math]\displaystyle{ \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}. }[/math]

So

[math]\displaystyle{ c_k = \int_{\mathbb{T}} z^{-k} dm = \int_{\mathbb{T}} \bar{z}^k dm }[/math]

which is equivalent to

[math]\displaystyle{ c_k = \frac{1}{2 \pi} \int_0 ^{2 \pi} e^{-ik\theta} d\mu(\theta) }[/math]

for some suitable measure [math]\displaystyle{ \mu }[/math].

Parametrization of solutions

The above discussion shows that the trigonometric moment problem has infinitely many solutions if the Toeplitz matrix [math]\displaystyle{ T }[/math] is invertible. In that case, the solutions to the problem are in bijective correspondence with minimal unitary extensions of the partial isometry [math]\displaystyle{ V }[/math].

See also

Notes

References