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?
