Q:

Spectral Measures: Nelson

Problem
Given a Hilbert space $\mathcal{H}$.
Consider a symmetric operator:
$$T:\mathcal{D}(T)\to\mathcal{H}:\quad T\subseteq\overline{T}\subseteq T^*$$
Denote the convergence radius by:
$$\frac{1}{\rho(\varphi)}:=\limsup_{k\to\infty}\sqrt[k]{\frac{1}{k!}\|T^k\varphi\|}$$
Introduce analytic elements:
$$\mathcal{C}^\omega(T):=\{\varphi\in\mathcal{D}(T):\rho(\varphi)>0\}$$
Then one has as criterion:
$$\overline{T}=T^*\iff\overline{\mathcal{C}^\omega(T)}=\mathcal{H}$$
How to prove this step by step?
Disclaimer
This thread is meant to record. See: Answer own Question

A:

This answer is community wiki.
Construct an extension:
$$A:=\begin{pmatrix}0&\overline{T}\\T^*&0\end{pmatrix}:\quad A=A^*$$
Fix an element:
$$\psi_\pm\in\mathcal{N}(T^*\pm i)$$
One has analytic elements:
$$\varphi_\omega\in\mathcal{C}^\omega(T)\implies\varphi_\omega\oplus\varphi_\omega\in\mathcal{C}^\omega(A)$$
Especially their radii agree:
$$|z|<\rho(\varphi_\omega)=\rho(\varphi_\omega\oplus\varphi_\omega)>0$$
Then one obtains:
$$\langle e^{izA}\varphi_\omega\oplus\varphi_\omega,\psi_\pm\oplus\psi_\pm\rangle=\sum_{k=0}^\infty\frac{1}{k!}(iz)^k\langle A^k\varphi_\omega\oplus\varphi_\omega,\psi_\pm\oplus\psi_\pm\rangle\\
=2\sum_{k=0}^\infty\frac{1}{k!}(iz)^k\langle T^k\varphi_\omega,\psi_\pm\rangle=2\sum_{k=0}^\infty\frac{1}{k!}(iz)^k\langle\varphi_\omega,(T^*)^k\psi_\pm\rangle\\
=2\sum_{k=0}^\infty\frac{1}{k!}(iz)^k\langle\varphi_\omega,(\mp i)^k\psi_\pm\rangle=2\sum_{k=0}^\infty\frac{1}{k!}(\mp z)^k\langle\varphi_\omega,\psi_\pm\rangle=2e^{\mp z}\langle\varphi_\omega,\psi_\pm\rangle$$
By analyticity this equality extends:
$$t\in\mathbb{R}:\quad2e^{\mp t}|\langle\varphi_\omega,\psi_\pm\rangle|\leq\|e^{itA}\varphi_\omega\oplus\varphi_\omega\|\cdot\|\psi_\pm\oplus\psi_\pm\|=2\|\varphi_\omega\|\cdot\|\psi_\pm\|$$
By the uniform bound it must vanish:
$$\overline{\mathcal{C}^\omega(T)}=\mathcal{H}:\quad\langle\varphi_\omega,\psi_\pm\rangle=0\implies\psi_\pm=0$$
So the assertion follows by the basic criterion on selfadjointness!

