Harman Patil (Editor)

Trigonometric moment problem

Updated on
Edit
Like
Comment
Share on FacebookTweet on TwitterShare on LinkedInShare on Reddit

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

Contents

α k = 1 2 π ∫ 0 2 π e − i k t d μ ( 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 = ( α 0 α 1 ⋯ α n α 1 ¯ α 0 ⋯ α n − 1 ⋮ ⋮ ⋱ ⋮ α n ¯ α n − 1 ¯ ⋯ α 0 )

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

( H , ⟨ , ⟩ )

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 H . More specifically, let { e0, ...en } be the standard basis of Cn + 1. Let E be the subspace generated by { [e0], ... [en - 1] } and F be the subspace generated by { [e1], ... [en] }. Define an operator

V : E → F

by

V [ e k ] = [ e k + 1 ] for k = 0 … n − 1.

Since

⟨ V [ e j ] , V [ e k ] ⟩ = ⟨ [ e j + 1 ] , [ e k + 1 ] ⟩ = A j + 1 , k + 1 = A j , k = ⟨ [ e j ] , [ e k ] ⟩ ,

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, there exists a Borel measure m on the unit circle T such that for all integer k

⟨ ( U ∗ ) k [ e n + 1 ] , [ e n + 1 ] ⟩ = ∫ T z k d m .

For k = 0,...,n, the left hand side is

⟨ ( U ∗ ) k [ e n + 1 ] , [ e n + 1 ] ⟩ = ⟨ ( V ∗ ) k [ e n + 1 ] , [ e n + 1 ] ⟩ = ⟨ [ e n + 1 − k ] , [ e n + 1 ] ⟩ = A n + 1 , n + 1 − k = α k ¯ .

So

∫ T z − k d m = ∫ T z ¯ k d m = α k .

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

1 2 π ∫ 0 2 π e − i k t d μ ( t ) = α 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

Trigonometric moment problem Wikipedia
(Text) CC BY-SA


Similar Topics
×