Trigonometric moment problem

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

${\displaystyle A=\left({\begin{matrix}\alpha _{0}&\alpha _{1}&\cdots &\alpha _{n}\\{\bar {\alpha _{1}}}&\alpha _{0}&\cdots &\alpha _{n-1}\\\vdots &\vdots &\ddots &\vdots \\{\bar {\alpha _{n}}}&{\bar {\alpha _{n-1}}}&\cdots &\alpha _{0}\\\end{matrix}}\right)}$

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

${\displaystyle ({\mathcal {H}},\langle \;,\;\rangle )}$

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

${\displaystyle V:{\mathcal {E}}\rightarrow {\mathcal {F}}}$

by

${\displaystyle V[e_{k}]=[e_{k+1}]\quad {\mbox{for}}\quad k=0\ldots n-1.}$

Since

${\displaystyle \langle V[e_{j}],V[e_{k}]\rangle =\langle [e_{j+1}],[e_{k+1}]\rangle =A_{j+1,k+1}=A_{j,k}=\langle [e_{j}],[e_{k}]\rangle ,}$

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

${\displaystyle \langle (U^{*})^{k}[e_{n+1}],[e_{n+1}]\rangle =\int _{\mathbf {T} }z^{k}dm.}$

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

${\displaystyle \langle (U^{*})^{k}[e_{n+1}],[e_{n+1}]\rangle =\langle (V^{*})^{k}[e_{n+1}],[e_{n+1}]\rangle =\langle [e_{n+1-k}],[e_{n+1}]\rangle =A_{n+1,n+1-k}={\bar {\alpha _{k}}}.}$

So

${\displaystyle \int _{\mathbf {T} }z^{-k}dm=\int _{\mathbf {T} }{\bar {z}}^{k}dm=\alpha _{k}.}$

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

${\displaystyle {\frac {1}{2\pi }}\int _{0}^{2\pi }e^{-ikt}d\mu (t)=\alpha _{k}}$

for some suitable measure μ.

• 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.