Hamburger Moment Problem - Characterization

Characterization

The Hamburger moment problem is solvable (that is, {mn} is a sequence of moments) if and only if the corresponding Hankel kernel on the nonnegative integers


A =
\left(\begin{matrix}
m_0 & m_1 & m_2 & \cdots \\
m_1 & m_2 & m_3 & \cdots \\
m_2 & m_3 & m_4 & \cdots \\
\vdots & \vdots & \vdots & \ddots \\
\end{matrix}\right)

is positive definite, i.e.,

 \sum_{j,k\ge0}m_{j+k}c_j\bar c_k\ge0

for an arbitrary sequence {cj}j ≥ 0 of complex numbers with finite support (i.e. cj = 0 except for finitely many values of j).

The "only if" part of the claims can be verified by a direct calculation.

We sketch an argument for the converse. Let Z+ be the nonnegative integers and F0(Z+) denote the family of complex valued sequences with finite support. The positive Hankel kernel A induces a (possibly degenerate) sesquilinear product on the family of complex valued sequences with finite support. This in turn gives a Hilbert space

whose typical element is an equivalence class denoted by .

Let en be the element in F0(Z+) defined by en(m) = δnm. One notices that

Therefore the "shift" operator T on, with T =, is symmetric.

On the other hand, the desired expression

suggests that μ is the spectral measure of a self-adjoint operator. If we can find a "function model" such that the symmetric operator T is multiplication by x, then the spectral resolution of a self-adjoint extension of T proves the claim.

A function model is given by the natural isomorphism from F0(Z+) to the family of polynomials, in one single real variable and complex coefficients: for n ≥ 0, identify en with xn. In the model, the operator T is multiplication by x and a densely defined symmetric operator. It can be shown that T always has self-adjoint extensions. Let

be one of them and μ be its spectral measure. So

On the other hand,

Read more about this topic:  Hamburger Moment Problem