In Dunford & Schwartz Linear Operators Part II, after X.5.3, they explain how to derive the
projection-valued measure form of the spectral theorem for a bounded normal operator from Theorem
X.5.3, which reads "Every Hilbert space admits a spectral representation relative to an arbitrary
bounded normal operator defined in it." X.5.1 gives the relevant definition:
Let $T$ be a normal operator in a Hilbert space $\mathfrak{H}$. Let $\{\mu_\alpha\}$ be a family of
finite positive measures on the Borel sets of the complex plane. A map $U$ of $\mathfrak{H}$ onto
$\sum_\alpha L_2(\mu_\alpha)$ is a spectral representation of $\mathfrak{H}$ onto
$\sum_\alpha L_2(\mu_\alpha)$ relative to $T$ if the following conditions are satisfied:
(a) each measure $\mu_\alpha$ vanishes on the complement of the spectrum of $T$;
(b) the operator $U$ is a linear map of $\mathfrak{H}$ onto all of $\sum_\alpha L_2(\mu_\alpha)$
which preserves inner products;
(c) for every Borel function $f$ which is bounded on the spectrum of $T$, we have, for every
$x$ in $\mathfrak{H}$ and every $\alpha$,
$$(U(f(T)x))_\alpha(\lambda)=f(\lambda)(Ux)_\alpha(\lambda),$$
for $\mu_\alpha$-almost all $\lambda$.
In this definition, $\sum_\alpha L_2(\mu_\alpha)$ is the Hilbert space sum (over an arbitrary
index set) of the Hilbert spaces $L_2(\mu_\alpha)$, and the subscript of $\alpha$ in (c) refers to the $\alpha$-th element of a member of $\sum_\alpha L_2(\mu_\alpha)$. The proof, such as it is,
says that if $UTU^{-1}$ has the desired form … and then goes on to give a definition for $E$,
which they claim is the resolution of the identity for $T$. I can fill in almost all (pun intended)
of the
details of the proof. There is just one loose end I am having trouble with. It is the last step,
in which, having shown that $E$ is a resolution of the identity, one must show that
$$(Tx,y)=\int_{\sigma(T)}\!\lambda\,dE_{x,y}(\lambda),$$
for every $x,y\in\mathfrak{H}$, where $E_{x,y}(e)=(E(e)x,y)$.
I am so close. I have defined the family of complex measures,
$$\nu_\alpha(e)=\int_e\!(Ux)_\alpha\overline{(Uy)_\alpha}\,d\mu_\alpha\qquad(e\in\mathscr{B}_T),$$
where $\mathscr{B}_T$ is the collection of Borel subsets of $\sigma(T)$. The proof comes down
to showing that
\begin{equation}\tag{1}\label{1}
\sum_\alpha\int_{\sigma(T)}\!f_1\,d\nu_\alpha
=\int_{\sigma(T)}\!f_1\,d({\textstyle \sum_\alpha}\nu_\alpha),
\end{equation}
where $f_1(\lambda)=\lambda$ defined on $\sigma(T)$, and where
\begin{equation}\tag{2}\label{2}
\sum_\alpha\nu_\alpha(e)=E_{x,y}(e)\qquad(e\in\mathscr{B}_T).
\end{equation}
I had already proved \eqref{2} in the process of showing that $E_{x,y}$ was a complex measure.
I know that both sides of \eqref{1} exist and that $\alpha$ need only vary over at most a countable
set of values, say $\{\alpha_1, \alpha_2, \dots\}$. I can approximate $f_1$ with simple measurable
functions, for which I know that \eqref{1} holds. Therefore, proving \eqref{1} seems to come down
to interchanging a countably infinite sum with a limit. That's where I'm stuck. I cannot seem to
justify exchanging the limits. I know I need some kind of uniformity to do it, but I don't see
where I can get that.
Best Answer
OK. I got the uniformity I needed. Here's how. In the process of proving that $E$ is a resolution of the identity, I also proved that the sum $$\sum_\alpha\lvert\nu_\alpha\rvert(e)\qquad(e\in\mathscr{B}_T)$$ converges.
Split $f_1$ into four positive parts: real positive, real negative, imaginary positive and imaginary negative. Each part is bounded on $\sigma(T)$ so there is a sequence of simple measurable functions converging to the part uniformly on $\sigma(T)$. The sum of these four sequences is then a sequence $\{s_n\}$ of simple measurable functions converging to $f_1$ uniformly on $\sigma(T)$.
If $\epsilon>0$ is given, then there exists $N$ such that for all $n\geq N$, and for all $\lambda\in\sigma(T)$, \begin{equation*} \lvert f_1(\lambda)-s_n(\lambda)\rvert <\frac{\epsilon}{\sum_\alpha\lvert\nu_\alpha\rvert(\sigma(T))}. \end{equation*} Then \begin{equation*} \begin{split} \Biggl\lvert\int_{\sigma(T)}\!(f_1-s_n)\, d({\textstyle \sum_{i=1}^k}\nu_{\alpha_i})\Biggr| &\leq\int_{\sigma(T)}\!\lvert f_1-s_n\rvert\, d(\lvert{\textstyle\sum_{i=1}^k}\nu_{\alpha_i}\rvert)\\ &<\epsilon\frac{\lvert\sum_{i=1}^k\nu_{\alpha_i}\rvert(\sigma(T))} {\sum_\alpha\lvert\nu_\alpha\rvert(\sigma(T))} \leq\epsilon\frac{\sum_{i=1}^k\lvert\nu_{\alpha_i}\rvert(\sigma(T))} {\sum_\alpha\lvert\nu_\alpha\rvert(\sigma(T))} \leq\epsilon, \end{split} \end{equation*} for all $n\geq N$ and for all $k=1,2,\dots$.
Therefore, \begin{equation}\tag{3}\label{3} \lim_{n\to\infty}\int_{\sigma(T)}\!s_n\,d({\textstyle\sum_{i=1}^k}\nu_{\alpha_i}) =\int_{\sigma(T)}\!f_1\,d({\textstyle\sum_{i=1}^k}\nu_{\alpha_i}) \end{equation} uniformly for $k=1,2,\dots$. \eqref{3} also holds for $k=\infty$.
Now using \eqref{3}, the left hand side of (1) can be rewritten as \begin{equation*} \sum_\alpha\int_{\sigma(T)}\!f_1\,d\nu_\alpha =\lim_{k\to\infty}\sum_{i=1}^k\int_{\sigma(T)}\!f_1\,d\nu_{\alpha_i} =\lim_{k\to\infty}\int_{\sigma(T)}\!f_1\,d({\textstyle\sum_{i=1}^k}\nu_{\alpha_i}) =\lim_{k\to\infty}\lim_{n\to\infty}\int_{\sigma(T)}\!s_n\, d({\textstyle\sum_{i=1}^k}\nu_{\alpha_i}), \end{equation*} and we can interchange the limits, since the $n$ limit is uniform for $k$ and the iterated limit exists. Therefore, interchanging the limits we get \begin{equation*} \begin{split} \sum_\alpha\int_{\sigma(T)}\!f_1\,d\nu_\alpha &=\lim_{n\to\infty}\lim_{k\to\infty}\int_{\sigma(T)}\!s_n\, d({\textstyle\sum_{i=1}^k}\nu_{\alpha_i}) =\lim_{n\to\infty}\lim_{k\to\infty}\sum_{i=1}^k \int_{\sigma(T)}\!s_n\,d\nu_{\alpha_i}\\ &=\lim_{n\to\infty}\sum_\alpha\int_{\sigma(T)}\!s_n\,d\nu_\alpha =\lim_{n\to\infty}\int_{\sigma(T)}\!s_n\,d({\textstyle\sum_\alpha}\nu_\alpha) =\int_{\sigma(T)}\!f_1\,d({\textstyle\sum_\alpha}\nu_\alpha), \end{split} \end{equation*} which is (1). The last equality is because \eqref{3} holds for $k=\infty$. The justification for the next to last equality can be found in Halmos Introduction to Hilbert Space and the Theory of Spectral Multiplicity as Theorems 7.1 and 7.2, remembering that if $s_n=a_1\chi_{e_1}+\cdots+a_j\chi_{e_j}$, then \begin{equation*} \begin{split} \sum_\alpha\int_{\sigma(T)}\!s_n\,d\nu_\alpha &=\sum_\alpha(a_1\nu_\alpha(e_1)+\cdots+a_j\nu_\alpha(e_j)) =a_1\sum_\alpha\nu_\alpha(e_1)+\cdots+a_j\sum_\alpha\nu_\alpha(e_j)\\ &=a_1({\textstyle\sum_\alpha}\nu_\alpha)(e_1) +\cdots+a_j({\textstyle\sum_\alpha}\nu_\alpha)(e_j) =\int_{\sigma(T)}\!s_n\,d({\textstyle\sum_\alpha}\nu_\alpha). \end{split} \end{equation*}