Consider a self-adjoint operator $\operatorname{A}$ on a Hilbertspace $\mathcal{A}$ and its spectral decomposition according to the spectral theorem:

$$A = \int_{\mathbb{R}} \lambda \;dP_\lambda$$

Let $f:\mathbb{R}\rightarrow \mathbb{C}$ be a continuous function and consider using functional calculus:

$$f(A) = \int_\mathbb{R}f(\lambda) \; dP_\lambda$$

Is there a "transformation recipe" to another spectral measure $P_\mu^f$, such that one can write:

$$f(A) = \int_\mathbb{R}\mu\; dP_\mu^f,$$

where $P_\mu^f$ depends on $\operatorname{A}$ via its original spectral measure and on the function $f$?


Yes. We can calculate the spectral measure $P^f$ of the operator $f(A).$ Let $M\subseteq\mathbb R$ be a Borel set. Then

$$P^f(M)=\chi_M(f(A))=(\chi_M\circ f)(A)=\int_{\mathbb R}\chi_M(f(\lambda))dP_\lambda=\int_{\mathbb R}\chi_{f^{-1}(M)}(\lambda)dP_\lambda=\chi_{f^{-1}(M)}(A)=P(f^{-1}(M)).$$

You can find a transformation theorem for spectral integrals over general measure spaces in the book by Berezanskii, Us, Sheftel "Functional Analysis".


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