Let $A$ and $B$ be two self adjoint operators on $L^2(\mathbb{R}, \mu)$ and $L^2(\mathbb{R}, \gamma)$, suppose the spectral measure $\mu, \gamma$ are absolutely continuous. Show that $A$ and $B$ are unitarily equivalent.
I don't see how abs. continuous can help proving the claim. Any hints?
Thanks
Let $\varphi = d\gamma/d\mu$ be the Radon-Nikodym derivative, then for any $f \geq 0$ we have $$ \int fd\gamma = \int f\varphi d\mu \qquad (\ast) $$ In addition to absolute continuity, you also need the fact that $$ \varphi \in L^{\infty}(\mu) $$ The proof is in the following steps : - Since $\mu$ and $\gamma$ are both positive measures, $\varphi \geq 0$ a.e. $[\mu]$ - Define $U : L^2(\gamma) \to L^2(\mu)$ by $$ f \mapsto \sqrt{\varphi}f $$ By $(\ast)$ it follows that $$ \|U(f)\|_2 = \|f\|_2 $$ so $U$ is an isometry. - Now do the same with $$ \psi := d\gamma/d\mu $$ to get an operator $$ V : L^2(\mu) \to L^2(\gamma) $$ - Check that this satisfies both $V = U^{-1}$ and $V = U^{\ast}$ (once again, using $(\ast)$) - Finally check that $U$ conjugates $A$ to $B$ (or rather, the multiplication operators corresponding to $A$ and $B$)