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.



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


