Skype security

In mathematics, the trigonometric moment problem is formulated as follows: given a finite sequence {α₀, ... α_n }, does there exist a positive Borel measure μ on the interval [0, 2π] such that

${\alpha }_k=\frac{1}{2\pi }{\int }_0^{2\pi }{e}^{-ikt}d\mu (t).$

In other words, an affirmative answer to the problems means that {α₀, ... α_n } 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, {α_k} is a sequence of Fourier coefficients, if and only if the (n + 1) × (n + 1) Toeplitz matrix

$A=\left(\begin{array}{cccc}{\alpha }_0& {\alpha }_1& \cdots & {\alpha }_n\\ \overline{{\alpha }_1}& {\alpha }_0& \cdots & {\alpha }_{n-1}\\ ⋮& ⋮& \ddots & ⋮\\ \overline{{\alpha }_n}& \overline{{\alpha }_{n-1}}& \cdots & {\alpha }_0\end{array}\right)$

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 C^{n + 1}, resulting in a Hilbert space

$(\mathscr{H},⟨,⟩)$

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 $\mathscr{H}$. More specifically, let { e₀, ...e_n } be the standard basis of C^{n + 1}. Let $ℰ$ be the subspace generated by { [e₀], ... [e_{n - 1}] } and $ℱ$ be the subspace generated by { [e₁], ... [e_n] }. Define an operator

$V:ℰ\to ℱ$

by

$V[e_k]=[e_{k+1}]\text{for}k=0\dots n-1.$

Since

$⟨V[e_j],V[e_k]⟩=⟨[e_{j+1}],[e_{k+1}]⟩=A_{j+1,k+1}=A_{j,k}=⟨[e_{j+1}],[e_{k+1}]⟩,$

V can be extended to a partial isometry acting on all of $\mathscr{H}$. 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

$⟨(U^{\ast }{)}^k[e_{n+1}],[e_{n+1}]⟩={\int }_{𝐓}{z}^{k}dm.$

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

$⟨(U^{\ast }{)}^k[e_{n+1}],[e_{n+1}]⟩=⟨(V^{\ast }{)}^k[e_{n+1}],[e_{n+1}]⟩=⟨[e_{n+1-k}],[e_{n+1}]⟩=A_{n+1,n+1-k}=\overline{{\alpha }_k}.$

So

${\int }_{𝐓}{z}^{-k}dm={\int }_{𝐓}{\overline{z}}^{k}dm={\alpha }_k.$

Finally, parametrize the unit circle T by e^it on [0, 2π] gives

$\frac{1}{2\pi }{\int }_0^{2\pi }{e}^{-ikt}d\mu (t)={\alpha }_k$

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

• N.I. Akhiezer, The Classical Moment Problem, Olivier and Boyd, 1965.
• N.I. Akhiezer, M.G. Krein, Some Questions in the Theory of Moments, Amer. Math. Soc., 1962.