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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?

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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!

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