Trigonometric moment problem

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In mathematics, the trigonometric moment problem is formulated as follows: given a sequence {ck}k0, does there exist a distribution function σ on the interval [0,2π] such that:[1][2] ck=12π02πeikθdσ(θ), with ck=ck for k1. An affirmative answer to the problem means that {ck}k0 are the Fourier-Stieltjes coefficients for some (consequently positive) unique Radon measure μ on [0,2π] as distribution function.[3][4][5][6]

In case the sequence is finite, i.e., {ck}k=0n<, it is referred to as the truncated trigonometric moment problem.[7]

Characterization

The trigonometric moment problem is solvable, that is, {ck}k=0n is a sequence of Fourier coefficients, if and only if the (n + 1) × (n + 1) Hermitian Toeplitz matrix T=(c0c1cnc1c0cn1cncn+1c0) with ck=ck for k1, 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 T defines a sesquilinear product on n+1, resulting in a Hilbert space (,,) of dimensional at most n + 1. The Toeplitz structure of T means that a "truncated" shift is a partial isometry on . More specifically, let {e0,,en} be the standard basis of n+1. Let and be subspaces generated by the equivalence classes {[e0],,[en1]} respectively {[e1],,[en]}. Define an operator V: by V[ek]=[ek+1]fork=0n1. Since V[ej],V[ek]=[ej+1],[ek+1]=Tj+1,k+1=Tj,k=[ej],[ek], V can be extended to a partial isometry acting on all of . Take a minimal unitary extension U of V, on a possibly larger space (this always exists). According to the spectral theorem,[8][9] there exists a Borel measure m on the unit circle 𝕋 such that for all integer k (U*)k[en+1],[en+1]=𝕋zkdm. For k=0,,n, the left hand side is (U*)k[en+1],[en+1]=(V*)k[en+1],[en+1]=[en+1k],[en+1]=Tn+1,n+1k=ck=ck. As such, there is a j-atomic measure m on 𝕋, with j2n+1< (i.e. the set is finite), such that[10] ck=𝕋zkdm=𝕋z¯kdm, which is equivalent to ck=12π02πeikθdμ(θ).

for some suitable measure μ.

Parametrization of solutions

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

See also

Notes

  1. Geronimus 1946.
  2. Akhiezer 1965, pp. 180–181.
  3. Edwards 1982, pp. 72–73.
  4. Zygmund 2002, p. 11.
  5. Katznelson 2004, p. 40.
  6. 6.0 6.1 Schmüdgen 2017, p. 260.
  7. Schmüdgen 2017, pp. 257, 261.
  8. Simon 2005, pp. 26, 42.
  9. Katznelson 2004, pp. 38–45.
  10. Schmüdgen 2017, p. 261.
  11. Curto & Fialkow 1991, p. 634.
  12. Iokhvidov 1982, pp. 97–98.

References