# Reed's and Simon's spectral theorem: Why $A\psi=\lambda\psi\implies f(A)\psi=f(\lambda)\psi$?

Let $$A$$ be a self-adjoint bounded linear operator on a Hilbert space $$H$$, let $$f$$ be a bounded Borel measurable function on $$\mathbb{R}$$ and suppose that $$\psi\in H$$ such that $$A\psi = \lambda\psi$$ for $$\lambda\in\mathbb{R}$$.

My question is really about an answer already posted to the question: Spectral Decomposition: $A\psi = \lambda \psi \implies f(A)\psi = f(\lambda)\psi$. Namely, consider: https://math.stackexchange.com/a/2464732/820472. Even after re-reading the given argument multiple times I still don't seem to get why in the end we can say that $$A\psi = \lambda\psi\implies f(A)\psi = f(\lambda)\psi$$. The poster seems to use something resembling the Stone's formula from Reed and Simon's book,

Theorem VII.13 (Stone's formula) Let $$A$$ be a bounded self-adjoint operator. Then $$s-\lim_{\varepsilon\downarrow 0}(2\pi i)^{-1}\int_a^b\left[(A - \lambda - i\varepsilon)^{-1} + (A - \lambda + i\varepsilon)^{-1}\right]d\lambda$$

and then deduces that $$(A - \lambda I)^{-1}\psi = \frac{1}{t_0 - \lambda}\psi$$, when now $$A\psi = t_0\psi$$. What I still don't understand is how do we move from this to the functional calculus equality $$f(A)\psi = f(\lambda)\psi$$ in general?

• I know that this does not answer your question, but what stops you from approximating $f$ by polynomials instead? May 30 at 12:40
• @Meowdog Can any (Borel) measurable function be approximated by polynomials? I know that any measurable function is almost everywhere the limit of continuous functions, so is you idea to 1.) Take a sequence of continuous functions converging to our a priori chosen Borel measurable function $f$ and 2.) Approximate each continuous function by a sequence of polynomials? May 30 at 13:00
• May 30 at 13:18