Trigonometric moment problem

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In mathematics, the trigonometric moment problem is formulated as follows: given a sequence {ck}k∈ℕ0, does there exist a distribution function σ on the interval [0,2π] such that:Template:SfnTemplate:Sfn ck=12π∫02πe−ikθdσ(θ), with c−k=c‾k for k≥1. An affirmative answer to the problem means that {ck}k∈ℕ0 are the Fourier-Stieltjes coefficients for some (consequently positive) unique Radon measure μ on [0,2π] as distribution function.Template:SfnTemplate:SfnTemplate:SfnTemplate:Sfn

In case the sequence is finite, i.e., {ck}k=0n<∞, it is referred to as the truncated trigonometric moment problem.Template:Sfn

Characterization

The trigonometric moment problem is solvable, that is, {ck}k=0n is a sequence of Fourier coefficients, if and only if the Template:Math Hermitian Toeplitz matrix T=(c0c1⋯cnc−1c0⋯cn−1⋮⋮⋱⋮c−nc−n+1⋯c0) with c−k=ck‾ for k≥1, is positive semi-definite.Template:Sfn

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 ℂn+1, resulting in a Hilbert space (ℋ,⟨,⟩) of dimensional at most Template:Math. The Toeplitz structure of T means that a "truncated" shift is a partial isometry on ℋ. More specifically, let {e0,…,en} be the standard basis of ℂn+1. Let ℰ and ℱ be subspaces generated by the equivalence classes {[e0],…,[en−1]} respectively {[e1],…,[en]}. Define an operator V:ℰ→ℱ by V[ek]=[ek+1]fork=0…n−1. 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 ℋ. Take a minimal unitary extension U of V, on a possibly larger space (this always exists). According to the spectral theorem,Template:SfnTemplate:Sfn there exists a Borel measure m on the unit circle 𝕋 such that for all integer Template:Math ⟨(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+1−k],[en+1]⟩=Tn+1,n+1−k=c−k=ck‾. As such, there is a j-atomic measure m on 𝕋, with j≤2n+1<∞ (i.e. the set is finite), such thatTemplate:Sfn ck=∫𝕋z−kdm=∫𝕋z¯kdm, which is equivalent to ck=12π∫02πe−ikθdμ(θ).

for some suitable measure μ.

Parametrization of solutions

Template:See also The above discussion shows that the truncated trigonometric moment problem has infinitely many solutions if the Toeplitz matrix T is invertible.Template:SfnTemplate:Sfn In that case, the solutions to the problem are in bijective correspondence with minimal unitary extensions of the partial isometry V.

See also

Notes

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References

Template:Probability problems