I am trying to understand the proof of the Borel measurable functional calculus, especially the construction of the map $\tilde{\Phi}:B_b(\sigma(T)) \rightarrow B(H)$. For that reason I'll try to write it in my own words and give some comments/thoughts about it. I do not understand everything and there are things that slightly differ in my version. (I will come back to that.)
I will write the proof and Theorems it uses in quotations and my thoughts in normal text below.
Notation: Let $B_b(X)$ denote the bounded Borel functions on a set $X$.
Borel measurable functional calculus:
Let $T$ be a bounded linear and self-adjoint operator. Then there exists a unique map $\tilde{\Phi}: B_b(\sigma(T)) \rightarrow B(H)$, that extends the continuous functional calculus $\Phi:C(\sigma(T)) \rightarrow B(H)$ to the bounded Borel functions, such that
(i) $\tilde{\Phi}$ is $\mathbb{C}$-linear
(ii) $\forall f,g \in B_b(\sigma(T)): \tilde{\Phi}(fg)=\tilde{\Phi}(f)\tilde{\Phi}(g)$
(iii) $\tilde{\Phi}(\overline{f})=\tilde{\Phi}(f)^*$.
The proof relies on the Riesz Representation Theorem for the space$ (C(X), \lVert . \rVert_{\infty})$.
Riesz-Representation Theorem: Let X be a compact metric space and let $(C(X), \lVert . \rVert_{\infty})$ be the space of continuous functions on $X$. Let $C(X)'$ denote it's dual and let $M(X)$ denote the space of real/complex regular Borel measures on $X$.
Then the map $R: M(X) \rightarrow C(X)'$ defined by $(R\mu)(f):= \int_X f d \mu$ is an isometric isomorphism.
(I am not very familiar with this Riez-Representation, so sorry if I am just rephrasing the obvious.)
So the statement is that every bounded linear functional on $C(X)$ can be represented in terms of an integral, i.e. there exists a Boreal measure $\mu$ (depending on $f$) such that
$f(g)=\int_X g d \mu \text{ } \forall g \in C(X)$.
Proof:
Let $f \in C(\sigma(T))$ and $x,y \in H$, then we can define $l_{x,y}(f):= \langle \Phi(f)x,y \rangle$. One can show that $l_{x,y}$ is linear and $l_{x,y} \in C(X)'$.
I do assume that linear is meant in the way that $l_{x,y}(\lambda f+g)=\lambda l_{x,y}(f) +l_{x,y}(g)$.
It seems that we use the Riesz-Representation for Hilbert spaces in the backgroud:
Riesz-Representation on Hilbert spaces: Let $H_1,H_2$ be Hilbert spaces and $h:H_1 \times H_2 \rightarrow \mathbb{F}$ a bounded sesquilinear form. Then h has a representation $h(x,y):= \langle Sx,y \rangle$ where $S: H_1 \rightarrow H_2$ is a >bounded linear operator. $S$ is uniquely determined by $h$
So given an operator $\Phi(T)$ we can use it to define a sesquilinear form $l$ on $H$ (depending on $x$ and $y$), i.e. $l_{x,y}:=l(x,y)= \langle \cdot x, y \rangle $, $H \times H \rightarrow \mathbb{F}$.
Since the exact sesquilinear form is dependent on the operator $S$, we can use $\Phi: C(\sigma(T)) \rightarrow B(H)$ to define a specific sesquilinear form $l_{x,y}$ for a fixed $f \in C(\sigma(T))$. We do denote this by $l_{x,y}(f):= \langle \Phi(T)x,y \rangle$.
So $l_{x,y}:B(H) \rightarrow \mathbb{F}$.
(The proof I am reading uses $l_{x,y}(f)$, where I would say that $l_{x,y}(\Phi(T))$ would seem more plausible. Probably just a abuse of notation thing.)
Thus given a continuous function $f$ on $\sigma(T)$. We have $l_{x,y}(f)=l_{x,y} \circ \Phi(f)$, $C(\sigma(T)) \rightarrow \mathbb{F}$. Thus a linear bounded functional on $C(\sigma(T))$.
As $l_{x,y}(f)$ is now an element of $C(\sigma)'$, we can now turn to the Riesz Representation Theorem between $M(X)$ and $C(X)'$ stated at the beginning.
By the Riesz-Representation Theorem there is a complex Borel measure $\mu_{x,y}$ on $\sigma(T) $with $\lVert \mu_{x,y} \rVert=\lVert l_{x,y} \rVert$ such that $\langle \Phi(T)x,y \rangle=\int_{\sigma(T)} f d \mu_{x,y}$ for every $f \in C(\sigma(T))$.
Using said Representation Theorem for the functional $l_{x,y}(f)$ there exists a measure $\mu_{x,y}$ such that $\lVert \mu_{x,y} \rVert=\lVert l_{x,y}(f) \rVert$.
Note in the proof I am reading it is stated $\lVert \mu_{x,y} \rVert=\lVert l_{x,y} \rVert$ whereas I would get $\lVert \mu_{x,y} \rVert=\lVert l_{x,y}(f) \rVert$. (Question 1: Did I make a mistake somewhere or is it just the "abuse of natation"?)
Now the integral also makes sense for $f \in B_b{\sigma(T)}$.
I do not see why that should be true. The bounded Borel functions are a way bigger class of functions. I do assume that I am missing some connection between $C(X)$ and $B_b(X)$ but I can't find out what it is. Especially this should be even more problematic if we consider that we used the Riesz-Representation Theorem for $C'(X)$. I was not able to find a Representation Theorem for $B_b(X)$.\ Question 2: Why can we just go from continuous functions to bounded Borel functions? (And how does this still work with the Representation Theorem?)
Considering the dependence of the integral on $(x,y)$, this defines a continuous sesquilinear form $b_f:H \times H \rightarrow \mathbb{C}$, $(x,y) \mapsto \mu_{x,y}$.
Then the proof shows that $b_f(x,y)$ is bounded. I am quite unsure what exaclty is happening here Since I do not understand how we go from continuoius functions to bounded Borel functions, but I assume that $b_f(x,y)$ is an analogue of $l_{x,y}(f)$ but for $f$ being bounded Borel.
We can use Lax-Milgram to see that $b_f $defines an operator $f(T) \in B(H)$ such that $\langle f(T)x,y \rangle=b_f(x,y) \forall x,y \in H$.
If I did understand it correctly, the Lax-Milgram Theorem is a Representation Theorem for sesquilinear forms on Hilbert spaces, so this step makes sense.
The proof continuous to show that the map $b_f$: $B_b(\sigma(T)) \rightarrow B(H)$ does statisfy the desired properties.
This leads me to the last question. Question 3: Do my comments/thoughts sound plausible? (Just as a check if I did not made any major mistake in my thinking process.)
