Extending the functional calculus of bounded self-adjoint linear operators to the Borel functions

Question

I am going through Reed & Simon's textbook on functional analysis and I am trying to understand how the functional calculus is extended from the continuous functions to the Borel functions using the spectral measures.

Let $A \in L(H)$, the set of all bounded linear operator on a Hilbert space $H$ and $\sigma(A)$ to be the spectrum of $A$. Furthermore, let $A$ be self adjoint.

First we define the functional calculus on the continuous functions. Since $A$ is self-adjoint, $\|P(A)\| = \sup_{\lambda \in \sigma(A)} |P(\lambda)|$ for any polynomial $P$. Thus we may use the Weierstrass theorem to obtain a functional calculus defined on $C(\sigma(A))$, the set of continuous functions on $\sigma(A)$. Now for fixed $\varphi \in H$, define the positive linear functional on $C(\sigma(A))$ by $$f \mapsto \langle \varphi, f(A)\varphi\rangle.$$ Then by the Riesz-Markov theorem, to each linear functional of the above form there is an associated measure $\mu_\varphi$ such that $$\langle \varphi, f(A)\varphi\rangle = \int_{\sigma(A)} f(\lambda) d\mu_\varphi$$ which we call a spectral measure.

I've understood everything up untilt his point. To extend the functional calculus to $B(\mathbb{R})$ (the Borel functions on $\mathbb{R}$), Reed & Simon say:

Let $g \in B(\mathbb{R})$. It is natural to define $g(A)$ so that $$\langle \varphi, g(A) \varphi \rangle = \int_{\sigma(A)} g(\lambda)d\mu_\varphi(\lambda).$$ The polarizaiton identity lets us recover $\langle \varphi, g(A) \phi\rangle$ from the proposed $\langle \varphi, g(A) \varphi \rangle$ and then the Riesz lemma lets us construct $g(A)$.

How does

1. the polarization identity let us recover $\langle \varphi, g(A) \phi\rangle$ from $\langle \varphi, g(A) \varphi \rangle$?
2. the Riesz lemma define $g(A)$?

Answer 1

This is only a brief sketch of what i think the authors have in mind:

To the first question: We can define a sesquilinear form (i take them as anti-linear in the first component) $B: H \times H \to \mathbb{K}$ by defining $$ B (\varphi, \varphi) = \int_{\sigma(A)} g(\lambda)d\mu_\varphi(\lambda)$$ and recovering (defining) $B(\varphi, \psi)$ from $B (\varphi, \varphi)$ by using the polarization identity for all $\varphi, \psi \in H$. That this works relies on some properties of the right hand side of course. I guess this is related. This sesqui-linear form is continuous.

To the second question: For every $\psi \in H$ we can define a continuous linear functional $B( \psi,-)$ by $\varphi \mapsto B(\psi, \varphi)$. Now the Riesz lemma shows that there exist a unique $\eta \in H $ so that $B( \psi,-) = \langle \eta , - \rangle$. In fact if $i_H : H \to H^*$ is the anti linear Riesz isomorphism, then $i_H^{-1} B(\psi,-) = \eta$ and since $i_H$ is anti-linear and $B$ is anti-linear in the first argument we see that the map $G: H \to H$ defined by $G(\psi) =i_H^{-1} B(\psi,-)$ is actually a continuous linear map. Now define $T= G^*$ and then we have: $$ B(\psi, \varphi) = \langle G \psi , \varphi \rangle = \langle \psi , T \varphi \rangle. $$ Of course $T= g(A)$. Although i think that this may also require that $g$ is a bounded function (for continuity).