# Quasi-analytic Vectors and Hamburger Moment Problem (Operator Theory)

## Analytic and quasi-analytic vectors

Guided by researches in function theory, operator theorists gave the analogue to quasi-analytic classes. Let \(A\) be an operator in a Banach space \(X\). \(A\) is not necessarily bounded hence the domain \(D(A)\) is not necessarily to be the whole space. We say \(x \in X\) is a \(C^\infty\) vector if \(x \in \bigcap_{n \geq 1}D(A^n)\). This is quite intuitive if we consider the differential operator. A vector is analytic if the series \[ \sum_{n=0}^{\infty}\lVert{A^n x}\rVert\frac{t^n}{n!} \] has a positive radius of convergence. Finally, we say \(x\) is quasi-analytic for \(A\) provided that \[ \sum_{n=0}^{\infty}\left(\frac{1}{\lVert A^n x \rVert}\right)^{1/n} = \infty \] or equivalently its nondecreasing majorant. Interestingly, if \(A\) is symmetric, then \(\lVert{A^nx}\rVert\) is log convex.

Based on the density of quasi-analytic vectors, we have an interesting result.

(Theorem)Let \(A\) be a symmetric operator in a Hilbert space \(\mathscr{H}\). If the set of quasi-analytic vectors spans a dense subset, then \(A\) is essentially self-adjoint.

This theorem can be considered as a corollary to the fundamental theorem of quasi-analytic classes, by applying suitable Banach space techniques in lieu.

## Hamburger moment problem

For a positive sequence \(\{a_n\}\), we see it is the moment of a positive measure \(\mu\), i.e. \(a_n = \int_\mathbb{R}t^n d\mu(t)\) if and only if it is positively definite (proof). But the uniqueness is not guaranteed. Here we have a sufficient condition for this - using the concept of the quasi-analytic vector. This is an old theorem (1922) but we are using operator theory to prove it which appeared decades later.

(Carleman's condition)Suppose \(\{a_n\}\) is the moment sequence of a positive measure \(\mu\) on \(\mathbb{R}\), then \(\mu\) is uniquely determined provided that \(\sum a_{2n}^{-1/2n}=\infty\).

**Proof.** Consider the Hilbert space \[
\mathscr{H}= L^2(\mathbb{R},\gamma)
\] and the operator \[
A:f(t) \mapsto tf(t).
\] It is clear that \(A\) is self-adjoint. We shall work on the constant function \(u(t) \equiv 1 \in \mathscr{H}\). Since \(A^nu = t^n\), we see \(u \in C^\infty\), otherwise, \(a_n\) is not defined. On the other hand, we have \[
(A^n u, u) = a_n \implies (A^{2n} u,u) = \lVert A^n u \rVert^2 = |(A^n u, u)|^2 = a_{2n}.
\] But \(a_{2n}^{-1/2n}=\lVert A^n u \rVert^{-1/n}\) and as a result, we see \(\sum a_{2n}^{-1/2n}= \sum \lVert A^n u \rVert^{-1/n} = \infty\), hence \(u\) is quasi-analytic. In general, \(t^n = A^n u\) is quasi-analytic for all \(n \geq 0\). Consider the space of polynomial \(\mathcal{P}[t]\) with closure \(\mathscr{H}_1\). It follows from the theorem above that \(A_1 = A|_{\mathcal{P}[t]}\) is essentially self-adjoint in \(\mathscr{H}_1\). Hence \(\mathscr{H}_1\) is invariant under the one-parameter group \(e^{iAs}\). Pick \(y \in \mathcal{P}[t]^{\perp}\), we see \[
(y,e^{iAs}u) = \int_\mathbb{R}e^{-ist}y(t)d\gamma(t) = 0,
\] which implies that \(y = 0\) a.e. [\(\gamma\)]. It follows that \(\mathscr{H}_1 = \mathscr{H}\) or equivalently \(\mathcal{P}[t]\) is dense in \(\mathscr{H}\). Suppose now we have another generating measure \(\nu\) of \(\{a_n\}\). With respect to \(\nu\), \(\mathcal{P}[t]\) is still a dense space. But the norm on \(\mathcal{P}[t]\) is fixed by \(\{a_n\}\), hence we obtain an isometry between \(\mathcal{P}[t]_\gamma\) and \(\mathcal{P}[t]_\nu\), which extends to the isometry between \(L^2(\mathbb{R},\gamma)\) and \(L^2(\mathbb{R},\nu)\) which forces \(\gamma\) and \(\nu\) to be equal. \(\blacksquare\)

Quasi-analytic Vectors and Hamburger Moment Problem (Operator Theory)