# Square root of unbounded operator

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?

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 $$\sqrt{T}y = \int_{0}^{\infty}\sqrt{\lambda}d(\lambda)y,\\ \mathcal{D}(\sqrt{T})=\left\{ y \in H : \int_{0}^{\infty}\lambda d\|E(\lambda)y\|^{2} < \infty\right\}.$$ 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 $E\psi=H\psi$.

• @TobiasHurth : Or, $A(f) = -if'$ is selfadjoint and $T=A^{2}$. There are lots of roots but only one positive selfadjoint root. The positive one is great for Math, but not so great for Physics. A 'causal' root works better for Quantum, where support of the function is preserved under the action of the operator. – DisintegratingByParts Dec 11 '14 at 12:26
• @TobiasHurth : I'm a little confused by your operator. At first I thought you were defining a linear $A$, but you're not, are you? – DisintegratingByParts Dec 11 '14 at 13:44
• @TobiasHurth : Your previous comment here didn't make sense to me. Now, with more info, I see how you are defining the operators. I'm curious to see what will pop up at MathOverflow. The question is getting +'d which is a good sign. Nice question. – DisintegratingByParts Dec 11 '14 at 19:01
• @TobiasHurth : I see that you added on MathOverflow that your operator is defined on a bounded domain and $V$ is smooth on that domain. Then you need to specify endpoint conditions to make your operator selfadjoint. Part of choosing the domain of $A$ has to involve those conditions. – DisintegratingByParts Dec 11 '14 at 21:04
• @TobiasHurth : When you say $\mathcal{C}^{\infty}$, you should make it clear that this is over the interior of an interval. Without conditions, you can't know the domain of $T$, which means that you can know the domain of $A$. When you write $T=A^{\star}A$, that's an operator equation where $A^{\star}$ is an adjoint. Some conditions may rule out such a relation because $A^{\star}A \ge 0$, and $T \ge 0$ may not be true. – DisintegratingByParts Dec 11 '14 at 22:39

$\text{dom}(A) = \text{dom}(T)$ 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 $\text{dom}(A) = \{f \in L^2(\mu): \sqrt{x} f \in L^2(\mu)\}$.

• thank you. do you have a reference, cause I would like to read a little about this by myself? – user159356 Dec 11 '14 at 0:38
• You might try Reed and Simon, "Methods of Modern Mathematical Physics I: Functional Analysis". – Robert Israel Dec 11 '14 at 5:07
• Hi Robert, I wonder if your answer can be adapted to say something about the following situation. Suppose $T = D^2+f$, where $D$ is a self-adjoint first-order differential operator and $f$ is multiplication by a non-negative function with compact support. Then $T:L^2\rightarrow L^2$ is an unbounded operator with unique positive square root. Then is it true that dom$(D)$ = dom$(\sqrt{D^2+f})$? – ougoah Dec 12 '17 at 23:58