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. $\blacksquare$