Square root of unbounded operator

Question

Let $T: \operatorname{dom}(T) \subset H \rightarrow H$ be a positive self-adjoint unbounded operator, then I want to define a UNIQUE(!) operator $A$ such that $A^{*}A = T$. Actually, this construction is nothing new, but I am uncertain about the DOMAIN(!) of $A$ in the case of unbounded operators. Therefore, I was wondering if anybody here knows a good reference that treats this "polar decomposition" also in the case of unbounded operators. Alternatively, if somebody wants to comment on this problem, I would highly appreciate this.

I mean, one is somehow tempted to say $\operatorname{dom}(A) = \operatorname{dom}(T)$ and then it is necessary that $\{Ax:x \in \operatorname{dom}(T)\} \subseteq\operatorname{dom}(A^*)$, but how can I see this or is this completely wrong?

Answer 1

The spectral theorem for an unbounded selfadjoint operator allows you to represent $T$ as $$ Tx = \int_{0}^{\infty}\lambda dE(\lambda)x,\\ \mathcal{D}(T) = \left\{ x \in H : \int_{0}^{\infty}\lambda^{2}d\|E(\lambda)x\|^{2} < \infty \right\}. $$ The unique positive square root of $T$ is $$ \sqrt{T}y = \int_{0}^{\infty}\sqrt{\lambda}dE(\lambda)y,\\ \mathcal{D}(\sqrt{T})=\left\{ y \in H : \int_{0}^{\infty}\lambda d\|E(\lambda)y\|^{2} < \infty\right\}. $$ This operator $\sqrt{T}$ is selfadjoint on its domain. In terms of domains $$ x \in \mathcal{D}(T) \iff x \in \mathcal{D}(\sqrt{T}) \mbox{and} \sqrt{T}x\in\mathcal{D}(\sqrt{T}). $$ Note: I apologize for rolling back another user's edit. I understand that some do not like the notation I am using for the Spectral Theorem, but it has been standard since John von Neumann first stated and proved the Spectral Theorem for unbounded selfadjoint operators on a Hilbert Space. John von Neumann was the first to characterize the domain of $T$ as I stated it here; one can view the integral definition for $Tx$ as an improper integral, as one would for the Riemann integral. This notation has been standard for roughly 80 years, including von Neumann's use of $E$ for the spectral measure, motivated by the Schrodinger equation $E\psi=H\psi$.

Answer 2

$\text{dom}(A) = \text{dom}(T)$ is wrong.

By the Spectral Theorem, you can essentially assume $T$ is multiplication by the variable $x$ on $L^2(\mu)$ for some positive measure $\mu$ on $[0,\infty)$, with $\text{dom}(T) = \{f \in L^2(\mu): x f \in L^2(\mu)\}$. Then you want $A$ to be multiplication by $\sqrt{x}$, with $\text{dom}(A) = \{f \in L^2(\mu): \sqrt{x} f \in L^2(\mu)\}$.