Concept

Hamburger moment problem

In mathematics, the Hamburger moment problem, named after Hans Ludwig Hamburger, is formulated as follows: given a sequence (m0, m1, m2, ...), does there exist a positive Borel measure μ (for instance, the measure determined by the cumulative distribution function of a random variable) on the real line such that In other words, an affirmative answer to the problem means that (m0, m1, m2, ...) is the sequence of moments of some positive Borel measure μ. The Stieltjes moment problem, Vorobyev moment problem, and the Hausdorff moment problem are similar but replace the real line by (Stieltjes and Vorobyev; but Vorobyev formulates the problem in the terms of matrix theory), or a bounded interval (Hausdorff). 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 is positive definite, i.e., for every arbitrary sequence (cj)j ≥ 0 of complex numbers that are finitary (i.e. cj = 0 except for finitely many values of j). For the "only if" part of the claims simply note that which is non-negative if is non-negative. We sketch an argument for the converse. Let Z+ be the nonnegative integers and F0(Z+) denote the family of complex valued sequences with finitary 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 [f]. Let en be the element in F0(Z+) defined by en(m) = δnm. One notices that Therefore, the "shift" operator T on , with T[en] = [en + 1], is symmetric. On the other hand, the desired expression suggests that μ is the spectral measure of a self-adjoint operator. (More precisely stated, μ is the spectral measure for an operator defined below and the vector [1], ). 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.

About this result
This page is automatically generated and may contain information that is not correct, complete, up-to-date, or relevant to your search query. The same applies to every other page on this website. Please make sure to verify the information with EPFL's official sources.

Graph Chatbot

Chat with Graph Search

Ask any question about EPFL courses, lectures, exercises, research, news, etc. or try the example questions below.

DISCLAIMER: The Graph Chatbot is not programmed to provide explicit or categorical answers to your questions. Rather, it transforms your questions into API requests that are distributed across the various IT services officially administered by EPFL. Its purpose is solely to collect and recommend relevant references to content that you can explore to help you answer your questions.