Analytic vectors of self-adjoint unbounded operators

Question

I am working with self-adjoint unbounded linear operators on infinite-dimensional separable Hilbert spaces, and I would like to know some references regarding analytic vectors, especially with respect to their properties as subset of the Hilbert space on which they are defined. Moreover I would like to know if there are particular conditions under which the restriction of a self-adjoint, unbounded linear operator to the space of its analytical vector is closed. Thank You.

EDIT

In view of Yurii Savchuk's reply, I understand that the restriction $(A_{r}\,;\mathcal{D}^{\omega}(A))$ of a self-adjoint linear operator $(A\,;\mathcal{D}(A))$ to the set of its analytic vectors, is closed if and only if $A$ is defined over the whole Hilbert space. Therefore, another question arises.

Let $(A\,;\mathcal{D}(A))$ be an unbounded, self-adjoint linear operator. Define:

$$ D^{\infty}(A)=\bigcap_{k=1}^{\infty}(A^{k}) $$ and let $\mathcal{D}(C)$ be a dense subspace of $\mathcal{D}(A)$ which is invariant with respect to $A$. It is true that $\mathcal{D}^{\infty}(A)\subseteq\mathcal{D}(C)$?

Answer 1

Here are some references:

1. Reed, Simon, "Methods of modern mathematical physics. II. Fourier analysis, self-adjointness."

2. Schmüdgen, Konrad "Unbounded self-adjoint operators on Hilbert space."

What kind of properties do you expect from analytical vectors?

If the restriction of a self-adjoint operator $(A,D(A))$ to the space of its analytical vectors $D^\omega(A)$ is closed then $(A,D(A))$ is a self-adjoint extention of a self-adjoint operator $(A,D^\omega(A)).$ Hence $D(A)=D^\omega(A).$ The latter is true only for bounded operators. (Proof uses the spectral theorem.)

Answer 2

A counterexample for your EDIT question is $i\frac{d}{dx}$ on $\mathcal{L}^2(\mathbb R)$ with $\mathcal{D}(C) =$ finite linear combinations of $\mathcal{C^\infty} $ functions of compact support.