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Trigonometric moment problem

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In mathematics, the trigonometric moment problem is formulated as follows: given a sequence , does there exist a distribution function on the interval such that:[1][2] with for . An affirmative answer to the problem means that are the Fourier-Stieltjes coefficients for some (consequently positive) unique Radon measure on as distribution function.[3][4][5][6]

In case the sequence is finite, i.e., , it is referred to as the truncated trigonometric moment problem.[7]

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Characterization

The trigonometric moment problem is solvable, that is, is a sequence of Fourier coefficients, if and only if the (n + 1) × (n + 1) Hermitian Toeplitz matrix with for , is positive semi-definite.[6]

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 defines a sesquilinear product on , resulting in a Hilbert space of dimensional at most n + 1. The Toeplitz structure of means that a "truncated" shift is a partial isometry on . More specifically, let be the standard basis of . Let and be subspaces generated by the equivalence classes respectively . Define an operator by Since can be extended to a partial isometry acting on all of . Take a minimal unitary extension of , on a possibly larger space (this always exists). According to the spectral theorem,[8][9] there exists a Borel measure on the unit circle such that for all integer k For , the left hand side is As such, there is a -atomic measure on , with (i.e. the set is finite), such that[10] which is equivalent to

for some suitable measure .

Parametrization of solutions

The above discussion shows that the truncated trigonometric moment problem has infinitely many solutions if the Toeplitz matrix is invertible.[11][12] In that case, the solutions to the problem are in bijective correspondence with minimal unitary extensions of the partial isometry .

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See also

Notes

References

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