Spectral Decomposition: $A\psi = \lambda \psi \implies f(A)\psi = f(\lambda)\psi$

Question

I'm reading Simon & Reed's Functional Analysis, and attempting to write out the proof of the functional calculus form of the Spectral Theorem.

See this link for their construction of $f(A)$ where $f \in \mathbb{B}(\mathbb{R})$ a bounded Borel function on $\mathbb{R}$:

Continuity of the functional calculus form of the Spectral Theorem

Let $A \in L(H)$ be self-adjoint.

My question is this:

How do I show that $A\psi = \lambda \psi \implies f(A)\psi = f(\lambda)\psi$?

This argument is easy when proving the same statement in the continuous calculus because when dealing with a continuous $f$ it may be approximated by a sequence $p_n$ that converges to it uniformly, and hence point-wise too.

To my knowledge, a bounded Borel function, $f$, has the property that $\exists \{f_n\} \subset C(\sigma(A))$ s.t $$\lim_n \int_{\sigma(A)}|f_n-f|du_\psi = 0.$$ Also I found that we have point-wise convergence a.e$[u_\psi]$.

Using the definition of $f(A)$ I can show that $$\exists ~~\lim_n f_n(\lambda) \in \mathbb{C}$$ but not that it necessarily equals $f(\lambda)$.

I wanted this to complete the attempt of:

$(\psi, f(A)\psi) = \int_{\sigma(A)} fdu_\psi = \lim_n \int_{\sigma(A)}f_ndu_\psi = \lim_n(\psi, f_n(A)\psi) = \lim_n(\psi, f_n(\lambda)\psi)$

Answer 1

Suppose $\mu$ is a finite Borel measure on $\mathbb{R}$, and consider $$ F(\lambda) = \int_{\mathbb{R}}\frac{1}{t-\lambda}d\mu(t),\;\; \lambda\in\mathbb{C}\setminus\mathbb{R}. $$ For a fixed $t_0$ and $\epsilon > 0$, \begin{align} F(t_0+i\epsilon)-F(t_0-i\epsilon) &=\int_{\mathbb{R}}\frac{1}{(t-t_0)-i\epsilon}-\frac{1}{(t-t_0)+i\epsilon}d\mu(t) \\ &=\int_{\mathbb{R}}\frac{2i\epsilon}{(t-t_0)^2+\epsilon^2}d\mu(t). \end{align} Therefore, by the bounded convergence theorem, $$ \lim_{\epsilon\downarrow 0}i\epsilon\{F(t_0+i\epsilon)-F(t_0-i\epsilon)\} = \lim_{\epsilon\downarrow 0}\int_{\mathbb{R}}\frac{-2\epsilon^2}{(t-t_0)^2+\epsilon^2}d\mu(t) = -2\mu\{t_0\}. $$ Now you can combine this with the functional calculus to get what you want. To be a bit more explicit, suppose $A\psi = t_0\psi$. Then, for $\lambda\notin\mathbb{R}$, $$ (A-\lambda I)\psi = (t_0-\lambda)\psi \\ \psi = (A-\lambda I)^{-1}(t_0-\lambda)\psi \\ (A-\lambda I)^{-1}\psi = \frac{1}{t_0-\lambda}\psi. $$ See if you can take it from here.

Answer 2

$\newcommand{\d}{\operatorname{d}}$ In fact, maybe a little surprise,$$\mu(\{\lambda\})=\mu\{\sigma(A)\}.(\mu:=\mu_\psi)$$ Proof. Let $$p_n(z)=( 1-\lvert\frac{z-\lambda}{R}\rvert^2)^n,$$ where $R>\lVert A\rVert$, then $p_n\to \chi_{\{\lambda\}}$ pointwise, thus $$\int p_n\d\mu \to \int\chi_{\{\lambda\}}\d\mu =\mu\{\lambda\}$$ by the Dominated Convergence Theorem. Since each $p_n$ can be represented as $p_n(z)=1+p(z)( z-\lambda) $ for some polynomial $p$ and $(z-\lambda)(A) \psi=A\psi-\lambda\psi=0$, $$\int p_n(z)\d\mu=\int 1\d\mu+\int p(z)(z-\lambda)\d\mu=\int 1\d\mu+(\psi,p(A)(A-\lambda I)\psi)=\mu(\sigma(A)).$$ Hence $\mu(\{\lambda\})=\mu\{\sigma(A)\}.$