Exercise 20, § 5

Let $u$ be a partial operator on a Hilbert space E. Let dom($u$)$^{an}$ denote the space of vectors $x\in E$ such that $x\in$ dom($u^n$) for every integer $n\in \mathbf{N}$, and such that the power series

(23) $\sum\|u^nn($!$x)\|z^n$

$n\in \mathbf{N}$ has radius of convergence $>0$. The elements of dom($u$)$^{an}$ are called the analytic vectors of $u$.

Let $x$ be an element of E such that $x$ belongs to dom($u^n$) for every $n\in \mathbf{N}$. Let $E_x$ be the closure of the subspace of E generated by the vectors $u^n(x)$ for $n\in \mathbf{N}$, and let $u_x$ be the partial operator on $\overline{E}_x$ with domain $E_x$ deduced from $u$ by passing to subspaces. One says that $x$ is a uniqueness vector of $u$ if $u_x$ is essentially self-adjoint.

a) Suppose that $u$ is symmetric and that E contains a total subset of uniqueness vectors for $u$. Then $u$ is essentially self-adjoint. (Show that the image of $u+i1_E$ and that of $u-i1_E$ are dense in E.)

Suppose henceforth that $u$ is symmetric and that E contains a total subset of analytic vectors.

b) For every analytic vector $x$ of $u$, the partial operator $u_x$ on $\overline{E}_x$ admits a self-adjoint extension.

c) Let $x$ be an analytic vector for $u$ and let $r >0$ be the radius of convergence of the power series (23). Let $v$ be a self-adjoint extension of $u_x$. For every $t\in \mathbf{R}$ such that $|t|< r$, one has

$\langle e^{itv}(x)|x\rangle =\sum(itn)$!$^n\langle u^n(x)|x\rangle$.

$n\in \mathbf{N}$

(Using the spectral measure of $x$ relative to $v$, prove that the left-hand side is the restriction of an analytic function on the set of $t\in \mathbf{C}$ such that $|t|< r$, and calculate the power series expansion of this function at 0.)

d) For every $t\in \mathbf{R}$, the unitary operator $e^{itv}$ depends only on $u_x$ and the operator $u_x$ is essentially self-adjoint. (Use Theorem V, p. 428.) e) Deduce that $u$ is essentially self-adjoint.

f) A symmetric and closed partial operator $u$ on E is self-adjoint if and only if dom($u$)$^{an}$ is dense in E.