Decomposition of a positive semidefinite self-adjoint operator?

Question

If I have a positive semi-definite self-adjoint operator $H:D(H) \rightarrow L^2$, is it true that there is always a decomposition $H=A^* A$ available? If this is true, what can we say about the domains of $A$ and $A^*$?

Answer 1

If you use unbounded operators, then $$ H = \int_{0}^{\infty}\lambda dE(\lambda). $$ The spectrum theorem for unbounded selfadjoint operators has the excellent provision that $$ \mathcal{D}(H) = \left\{ x \in X : \int_{0}^{\infty}\lambda^{2}d\|E(\lambda)x\|^{2} < \infty\right\}. $$ You have $H=A^{2}$, where $Ax=\int_{0}^{\infty}\sqrt{\lambda}dE(\lambda)x$ defines a positive selfadjoint operator with $$ \mathcal{D}(A) = \left\{ x \in X : \int_{0}^{\infty}\lambda\|E(\lambda)x\|^{2} < \infty\right\}. $$ You're guaranteed that $A\mathcal{D}(H)\subseteq\mathcal{D}(A)$ and $A^{2}=H$.

For example, if $H=-\frac{d^{2}}{dx^{2}}$ on the space of twice absolutely continuous functions $f$ with $f,f'' \in L^{2}(\mathbb{R})$, then the operator can be written in terms of the Fourier transform: $$ A=\left(|s|f^{\wedge}(s)\right)^{\vee}. $$ Here $\mathcal{D}(A)=\{ f : f, sf^{\wedge}(s)\in L^{2}\}$.

Answer 2

If $H:X \to X$ is a bounded operator, there is indeed always such a decomposition available. We may construct one such decomposition as follows:

By the spectral theorem, we have $H = UTU^*$ Where the operator $T$ is given by $$ [T(\phi)](x) = f(x)\phi(x) $$ For some $f:\Bbb R \to \sigma(H) \subset [0,\infty)$. We can simply define $\sqrt T$ by $$ [\sqrt T(\phi)](x) = \sqrt{f(x)}\phi(x) $$ and from there, we can set $A = \sqrt{T} U^*$ and find $H = A^*A$.

In general, $A:X \to Y$ and $A^*:Y \to X$, where $Y$ is some space with dimension greater than or equal to the rank of $H$.