Understanding the regular representation of an LCA group as a 'direct integral'

Question

The reference for what I'm asking is page $107$ from Folland's harmonic analysis.

$G$ is a locally compact abelian group with dual $\hat{G}$. Let $H$ denote the Hilbert space $L^2(G)$. I'm trying to understand the 'direct integral' interpretation (and it's utility) for the regular representation $$L : G \to U(H);\ L_x(f)(\cdot) = f(x^{-1}\cdot).$$

By abstract Gelfand theory, we know there is a projection-valued measure $P$ on $\hat{G}$ such that \begin{equation}\label{pvm}\tag{1} L_x = \int \Phi(x)(\xi)dP(\xi),\ L_f = \int \xi(f)dP(\xi). \end{equation} Here $\Phi(x)(\xi) :=\xi(x)$ is just the map from $G$ to its double-dual, the $L$ on the right denotes the induced $*$-representation of $L^1(G)$ (which is actually just convolution), and $\xi(f)$ is from the fourier transform - $\hat{f}(\xi^{-1})$.

A calculation shows that if we write $F$ to be the composition $\Phi^* \circ \hat{(\cdot)}:L^2(\hat{G}) \to L^2(\hat{\hat{G}}) \to L^2(G)$, then we get \begin{equation}\label{conjugation}\tag{2} F^{-1}\circ L_x = m_{\Phi(x)}\circ F^{-1}\ \text{ or equivalently }\ L_x = F\circ m_{\Phi(x)}\circ F^{-1} \end{equation} where $m_{\Phi(x)}$ denotes a multiplication operator on $L^2(\hat{G})$.

Question 1: What kind of convergence would I need to use to go from \eqref{conjugation} to the equation \begin{equation}\label{E-conjugation}\tag{3} P(E) = F\circ m_{1_E} \circ F^{-1}? \end{equation} Attempt: Combining equations \eqref{pvm} and \eqref{conjugation} we see that for $f_1, f_2 \in L^2(G)$, we have \begin{equation} (L_x (f_1), f_2) = \int \Phi(x)(\xi) dP_{f_1, f_2}(\xi) = (F\circ m_{\Phi(x)}\circ F^{-1} (f_1), f_2). \end{equation} I know the $\mathbb{C}$-algebra generated by $\mathcal{A} := \{\Phi(x): x\in G \} \subset C(\hat{G})$ will be dense on compact subsets. Also, $P_{f_1, f_2}$ is a complex measure so that might help me when applying convergence theorems for integrals. But in general, taking $E\subset \hat{G}$ compact, I'm not sure how to ensure pointwise bounded convergence by taking functions in $\mathcal{A}$ (unboundedness could happen outside $E$, right?). Is this the right track?

Question 2: Assuming that we have equation \eqref{E-conjugation}, I'm not sure how to concretely interpret this, even in the case where $G =\mathbb{R}$. In this case, how do I see (Stone's theorem) that $L_x = e^{2i\pi x A}$? What exactly is $A$ in this case?

Attempt: Supposedly $A$ could be an 'unbounded operator' on $H$. Unfortunately, I have read everything in Rudin's functional analysis except chapter 13 where Stone's theorem is proved (theorem 13.38). Would I really need to know 'unbounded operators', 'semi-groups', and 'infinitesimal generators' to understand Stone's theorem in this concrete case?

Folland says $A= \int \xi dP(\xi)$ which I can't make sense of. Is the symbol $\int \xi dP_{f_1, f_2}(\xi)$ standing for some complex number?

I'd be very happy if someone could help me with this.

Answer 1

I have an old edition of the book.

Let $h_1, h_2 \in L^2(G)$, let $l_i := F^{-1}h_i \in L^2(\hat{G})$ and let $E\subset \hat{G}$ be Borel. We know the Gelfand (Fourier) transform sends $L^1(G)$ into a dense subset of $C_0(\hat{G})$. Choose a sequence $f_n$ such that we have pointwise (bounded) convergence $\xi(f_n)\to 1_E(\xi)$. \begin{equation} \begin{split} (P(E)h_1,h_2) &= \int 1_E(\xi)dP_{h_1,h_2}(\xi)\\ &= \lim_{n\to \infty} \int\xi(f_n)dP_{h_1,h_2}(\xi) \\ &= \lim_{n\to \infty} (L(f_n)h_1,h_2) \\ &= \lim_{n\to \infty} \int_x f_n(x)(L(x)h_1,h_2)dx \\ &= \lim_{n\to \infty} \int_x f_n(x)\left(F\circ m_{\Phi(x)} \circ F^{-1}(h_1),h_2\right)dx \\ &= \lim_{n\to \infty} \int_x f_n(x)\left( m_{\Phi(x)} \circ F^{-1}(h_1),F^{-1}h_2\right)_{L^2(\hat{G})}dx\\ &= \lim_{n\to \infty} \int_x f_n(x)\left( m_{\Phi(x)} \circ l_1,l_2\right)_{L^2(\hat{G})}dx\\ &= \lim_{n\to \infty} \int_x f_n(x)\left(\int_\xi \xi(x)l_1(\xi)\overline{l_2(\xi)}d\xi\right) dx\\ &= \lim_{n\to \infty} \int_\xi \left(\int_xf_n(x)\xi(x)dx\right) l_1(\xi)\overline{l_2(\xi)} d\xi \\ &= \lim_{n\to \infty} \int_\xi \xi(f_n)l_1(\xi)\overline{l_2(\xi)} d\xi \\ &= \int 1_E(\xi)l_1(\xi)\overline{l_2(\xi)} d\xi \\ &= (m_{1_E}\circ F^{-1}(h_1), F^{-1}h_2)\\ &= (F\circ m_{1_E}\circ F^{-1}(h_1),h_2)? \end{split} \end{equation}

For the second question, I completely forgot that we are in $\mathbb{R}$. The unbounded operator (whatever that means) is $$ \int_{x\in \mathbb{R}} x dP(x). $$ Note, $x\to x$ is an unbounded function.