Functional calculus: Does $A$ commute with $e^{iA^2}$?

Question

Let $A$ be an unbounded self-adjoint operator. Is it then true that $A$ commutes with $e^{iA^2}$? This sounds natural to me but I have no clue whether this is true in general. The problem is that we only did such a functional calculus for bounded(!) operators, so I don't even know whether something similarly exists for unbounded ones?

Answer 1

The Spectral Theorem for $A$ is given in terms of a Borel Spectral measure $E$ $$ Ax = \int_{-\infty}^{\infty}\lambda dE(\lambda)x, $$ and $x \in \mathcal{D}(A)$ iff $$ \int_{-\infty}^{\infty}\lambda^2 d\|E(\lambda)x\|^2 < \infty. $$ The operator $e^{iA^2}$ is defined through the functional calculus as $$ e^{iA^2}x = \int_{-\infty}^{\infty}e^{i\lambda^2}dE(\lambda)x. $$ This operator is unitary. For bounded Borel functions $f$, $g$, the operators $f(A)$ and $g(A)$ are bounded operators that satisfy $$ \int f(\lambda)dE(\lambda) \int g(\lambda) dE(\lambda)x = \int (fg)(\lambda)dE(\lambda)x,\;\;\; x\in\mathcal{H}. $$ Therefore, $$ \int_{-R}^{R}\lambda dE(\lambda) e^{iA^2}x = \int_{-R}^{R}\lambda e^{iA^2}dE(\lambda)x =e^{iA^2}\int_{-R}^{R}\lambda dE(\lambda)x \\ \int_{-R}^{R}\lambda^2 d\|E(\lambda)e^{iA^2}x\|^2 = \int_{-R}^{R}\lambda^2 d\|E(\lambda)x\|^2. $$ As $R\rightarrow\infty$, the conclusion from the last integral is that $x\in\mathcal{D}(A)$ iff $e^{iA^2}x\in\mathcal{D}(A)$. Furthermore, using the first vector identity, $$ Ae^{iA^2}x = e^{iA^2}Ax,\;\;\; x\in\mathcal{D}(A). $$