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

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 $$g(A)$$?

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).