On functions of a self-adjoint operator of the form $U^{-1} A U$

Question

I found in Reed Simon that, since $\mathcal{F}$ (Fourier Transform) is unitary in $L^2(\mathbb{R}^n)$ and the self-adjoint operator (in a suitable domain) $-\Delta = H_{0}$ can be expressed as $H_{0} = \mathcal{F}^{-1}\lambda^2 \mathcal{F}$, then $f(H_0) = \mathcal{F}^{-1}f(\lambda^2) \mathcal{F}$ for every bounded measurable function $f$. Now, i was wondering how to prove this using functional calculus: I suppose this follows by a more general result like $f(U^{-1}AU) = U^{-1}f(A)U$ but I couldn't find the precise statement of this in the "mare magnum" of Reed Simon. I would greatly appreciate any reference and suggestion. Thank you in advance.

Answer 1

Note that powers of $H_0$ satisfy $H_0^k = (\mathcal{F}^{-1}\lambda^2\mathcal{F})^k = \mathcal{F}^{-1}(\lambda^2)^k\mathcal{F}$. In consequence, if $f$ admits a power expansion (like Taylor or Laurent series), i.e. $f(x) = \sum_k a_kx^k$, then you have : $$ f(H_0) = \sum_k a_k H_0^k = \sum_k a_k \mathcal{F}^{-1}(\lambda^2)^k\mathcal{F} = \mathcal{F}^{-1}f(\lambda^2)\mathcal{F} $$