$\require{AMScd}$I'm struggling to understand the uniqueness part of Theorem 1.54 in Folland's book on Harmonic analysis.
Theorem 1.54: Let $\mathcal{A}$ be a commutative Banach $*$-algebra (non necessarily unital) and let $\phi$ be a nondegenerate $*$-representation $\phi:\mathcal{A}\to \mathcal{L}(H)$ (the codomain is the set of bounded operators on a Hilbert space). There is a unique projection-valued measure $P$ on $\sigma(\mathcal{A})$ such that $\phi(x) = \int G_{\mathcal{A}}(x)dP
$ for all $x\in\mathcal{A}$. Here $G_{\mathcal{A}}(x)$ is the continuous function on the spectrum $\sigma(\mathcal{A})$ which is given by evaluation on $x$.
Attempt: (Uniqueness) Given two $P, Q$ satisfying the condition, we would have
$$
\int G_{\mathcal{A}}(x) dP_{u,v} = \int G_{\mathcal{A}}(x) dQ_{u,v}
$$
for all $u, v \in H$. In the case of (non-unital) $C^*$-algebras, one could use the fact that the $G_A(x)$ exhaust $C_0(\sigma(\mathcal{A}))$ and conclude equality of $P, Q$ by Riesz representation. However, this case is different since the set $\{G_A(x) : x\in \mathcal{A}\}$ might not be big. So how do I go about proving uniqueness?
I include the existence part of the theorem here, in case it helps. Let $\mathcal{B}$ denote the norm closure of $\phi(\mathcal{A})$, and let $\mathcal{\tilde{A}},\ \mathcal{\tilde{B}}$ denote the algebras with units attached. $\mathcal{\tilde{B}}$ is then $\mathcal{B}\oplus \mathbb{C}I$ and is a commutative unital $C^*$-algebra.
Moreover, the $*$-representatiom $\phi$ extends to give the diagram
$$
\begin{CD}
\mathcal{A} @>{\phi}>> \mathcal{B} \\
@V{}VV @V{}VV \\
\mathcal{\tilde{A}} @>{\phi}>> \mathcal{\tilde{B}}
\end{CD} \qquad\quad (1)
$$
On identifying the spectrum of $\mathcal{\tilde{A}}$ with a subset of bounded linear functionals on $\mathcal{A}$, we can write $\sigma(\mathcal{\tilde{A}}) = \sigma(\mathcal{A})\cup \{0\}$ which, if $\sigma(\mathcal{A})$ is noncompact, is the one-point compactification of $\sigma(\mathcal{A})$. Similarly, this holds for $\mathcal{B}$. Thus $(1)$ gives a dual diagram (the vertical arrows are inclusions and not the dual)
$$
\begin{CD}
\sigma(\mathcal{A}) @<{\phi^*}<< \sigma(\mathcal{B}) \\
@V{}VV @V{}VV \\
\sigma(\mathcal{A})\cup \{0\} @<{\phi^*}<< \sigma(\mathcal{B})\cup \{0\}
\end{CD} \qquad\quad (2)
$$
It follows from the definitions that the arrows are continuous injections. The bottom row is a homeomorphism onto the image. Taking duals again, we have the following commutative diagram of Gelfand transforms
$$
\begin{CD}
\mathcal{A} @>{\phi}>> \mathcal{B} \\
@V{}VV @V{}VV \\
\mathcal{\tilde{A}} @>{\phi}>> \mathcal{\tilde{B}}
\end{CD} \qquad \overset{G_\cdot}{\longrightarrow}
\begin{CD}
C_0(\sigma(\mathcal{A})) @>{(\phi^*)\check{}}>> C_0(\sigma(\mathcal{B})) \\
@V{}VV @V{}VV \\
C(\sigma(\mathcal{A})\cup \{0\}) @>{(\phi^*)\check{}}>> C(\sigma(\mathcal{B})\cup \{0\})
\end{CD}
$$
Now the spectral theorem applies to $\mathcal{\tilde{B}}$ to give a unique projection-valued measure on Borel sets $Q:\text{Borel}(\sigma(\mathcal{B})\cup\{0\}) \to \mathcal{L}(H)$ such that for every $T\in \mathcal{\tilde{B}}$,
$$
T = \int G_\mathcal{\tilde{B}}(T) dQ.
$$
Now, we claim $Q(\{0\})=0$. Put $E=\{0\}$. If $u \in H$ is in the image of $Q(E)$, then we have, for any $S \in \mathcal{B}$ and $v\in H$, by the spectral theorem
$$(Su,v) = (SQ(E)u,v) = \int_{h\in E} G_{\mathcal{\tilde{B}}}(S)(h) dQ_{u,v}(h)=0.
$$
Since $\phi$ is non-degenerate, $Q(E)$ must be the zero operator.
Thus we have a projection-valued measure $R$ on $\sigma(\mathcal{B})$ uniquely defined by the property that, for every $T\in \mathcal{B}$ we have $$
T = \int G_{\mathcal{B}}(T) dR. \qquad (3)
$$
I guess the uniqueness here follows by the fact that $\{G_\mathcal{B}(T) : T\in \mathcal{B}\}=C_0(\sigma(\mathcal{B}))$.
Finally, we define a projection-valued measure $P$ on $\sigma(\mathcal{A})$ using the diagram
$$
\text{Borel}(\sigma(\mathcal{A})) \overset{(\phi^*)\check{}}{\longrightarrow} \text{Borel}(\sigma(\mathcal{B})) \overset{R}{\longrightarrow} \mathcal{L}(H)
$$
With this definition, it is easy to see that $P$ is the pushforward measure from $R$ by $\phi^*$. That is, $P_{u,v} = (\phi^*)_*R_{u,v}$.
To see that $P$ has the required property, take any $x\in\mathcal{A}, u, v \in H$ and observe that, by $(3)$, we have
\begin{equation}\begin{split}
(\phi(x)u,v) &= \int_{\sigma(\mathcal{B})}G_\mathcal{B}(\phi(x))(h)dR_{u,v}(h)\\
&= \int_{\sigma(\mathcal{B})} h(\phi(x)) dR_{u,v}(h)\\
&= \int_{\sigma(\mathcal{B})} \phi^*(h)(x) dR_{u,v}(h)\\
&= \int_{\sigma(\mathcal{B})} G_\mathcal{A}(x)(\phi^*(h)) dR_{u,v}(h)\\
&= \int_{\sigma(\mathcal{A})} G_\mathcal{A}(x)(l) d\left((\phi^*)_*R_{u,v}\right)(l)\\
&= \int_{\sigma(\mathcal{A})} G_\mathcal{A}(x)(l) dP_{u,v}(l)
\end{split}\end{equation}
Again, I don't see why such a measure should be unique. Please send help.
Best Answer
The set of functions of the form $G_{\mathcal{A}}(x)$ is a $*$-subalgebra of $C_0(\sigma(\mathcal{A})$ that separates points from each other and from $0$. So by Stone-Weierstrass, it is dense in $C_0(\sigma(\mathcal{A}))$. It follows that the bounded functionals of the form $G_{\mathcal{A}}(x)\mapsto \int G_{\mathcal{A}}(x) dP_{u,v}$ extend uniquely to bounded functionals on all of $C_0(\sigma(\mathcal{A}))$, and you can apply the Riesz representation theorem.