Thanks to the suggestion by s.harp, I can give an answer to my questions. It is
amazing how long the answer is when written out fully. In order to match my handwritten
notes, I am going to change the notation. So here is the original question, written as a theorem in a different notation:
Theorem. Suppose $T$ is a bounded normal operator on the (not necessarily separable)
Hilbert space $H$, and suppose
$H_\alpha$ is a closed reducing subspace for $T$. That is, $TH_\alpha\subseteq H_\alpha$, and
$T^*H_\alpha\subseteq H_\alpha$ (or equivalently, the second condition can be replaced
with $TH_\alpha^\perp\subseteq H_\alpha^\perp$). Then
- $T_\alpha=T|_{H_\alpha}$ is a bounded normal operator on $H_\alpha$,
- $\sigma(T_\alpha)\subseteq\sigma(T)$
- Let $E$ and $E'$ be the resolutions of the identity for $T$ and $T_\alpha$,
respectively, and let $B_T$ and $B_{T_\alpha}$
be the Borel subsets of $\sigma(T)$ and $\sigma(T_\alpha)$, respectively.
Then for all $e\in B_T$,
\begin{equation*}
e\cap\sigma(T_\alpha)\in B_{T_\alpha}\quad\text{and}\quad
E'(e\cap\sigma(T_\alpha))=E(e)|_{H_\alpha},
\end{equation*}
- If $\mathscr{B}_T$ and $\mathscr{B}_{T_\alpha}$ are the families of bounded complex
Borel functions on $\sigma(T)$ and $\sigma(T_\alpha)$, respectively, and if
$f_\alpha=f|_{\sigma(T_\alpha)}$ then for
all $f\in\mathscr{B}_T$,
\begin{equation*}
f_\alpha\in\mathscr{B}_{T_\alpha}\quad\text{and}\quad
f_\alpha(T_\alpha)=f(T)|_{H_\alpha}
\end{equation*}
Proof. When I mention the spectral theorem for bounded normal operators, I
am referring to Rudin, Functional Analysis, Second Edition, 12.21-12.24.
Since $H_\alpha$ is reducing for $T$, we see that
$T_\alpha$ and $(T^*)_\alpha$ are bounded
operators on the Hilbert space $H_\alpha$. For $x_\alpha,y_\alpha\in H_\alpha$,
\begin{equation*}
(x_\alpha,(T_\alpha)^*y_\alpha)=(T_\alpha x_\alpha,y_\alpha)=(Tx_\alpha,y_\alpha)
=(x_\alpha,T^*y_\alpha)=(x_\alpha,(T^*)_\alpha y_\alpha),
\end{equation*}
so $(T_\alpha)^*=(T^*)_\alpha$ and we will write it as $T_\alpha^*$.
For $x_\alpha\in H_\alpha$,
\begin{equation*}
T_\alpha T_\alpha^*x_\alpha=T_\alpha T^*x_\alpha=TT^*x_\alpha
=T^*Tx_\alpha=T^*T_\alpha x_\alpha=T_\alpha^*T_\alpha x_\alpha,
\end{equation*}
so $T_\alpha$ is normal. This proves #1.
Suppose $\lambda\in\rho(T)$. Let $S=(\lambda I-T)^{-1}\in\mathscr{B}(H)$, the
bounded linear operators on $H$. Let $x\in SH_\alpha$. Let
$x=x_\alpha+x_\alpha^\perp$, where $x_\alpha\in H_\alpha$ and
$x_\alpha^\perp\in H_\alpha^\perp$. Then $(\lambda I-T)x_\alpha\in H_\alpha$,
$(\lambda I-T)x_\alpha^\perp\in H_\alpha^\perp$, and
$$(\lambda I-T)x_\alpha+(\lambda I-T)x_\alpha^\perp=(\lambda I-T)x\in H_\alpha,$$
since $x\in (\lambda I-T)^{-1}H_\alpha$. Hence $(\lambda I- T)x_\alpha^\perp=0$,
so $x_\alpha^\perp=0$ since $\lambda I-T$ is one-to-one. Therefore
$x=x_\alpha\in H_\alpha$, so that $SH_\alpha\subseteq H_\alpha$.
Conversely, if $x_\alpha\in H_\alpha$, then $y=(\lambda I-T)x_\alpha\in H_\alpha$
and $x_\alpha=Sy$ so $SH_\alpha=H_\alpha$. Define $S_\alpha=S|_{H_\alpha}$.
Then $S_\alpha$ is linear, continuous, one-to-one onto $H_\alpha$, so
$S_\alpha\in\mathscr{B}(H_\alpha)$,
and for all $x_\alpha\in H_\alpha$,
\begin{equation*}
S_\alpha(\lambda I_\alpha-T_\alpha)x_\alpha
=S(\lambda I-T)x_\alpha
=x_\alpha
=(\lambda I-T)Sx_\alpha
=(\lambda I_\alpha-T_\alpha)S_\alpha x_\alpha,
\end{equation*}
so $\lambda I_\alpha-T_\alpha$ has an inverse $S_\alpha\in\mathscr{B}(H_\alpha)$,
whence $\lambda\in\rho(T_\alpha)$. This proves #2.
Let $P_\alpha$ be the orthogonal projection on $H_\alpha$. Then for $x\in H$,
write $x=x_\alpha+x_\alpha^\perp$, where $x_\alpha\in H_\alpha$ and
$x_\alpha^\perp\in H_\alpha^\perp$. Then since
$TH_\alpha^\perp\subseteq H_\alpha^\perp$ and $TH_\alpha\subseteq H_\alpha$,
\begin{equation*}
P_\alpha Tx=P_\alpha Tx_\alpha+P_\alpha Tx_\alpha^\perp=P_\alpha Tx_\alpha
=Tx_\alpha=TP_\alpha x,
\end{equation*}
that is, $P_\alpha$ commutes with $T$, so by the spectral theorem, $P_\alpha$
commutes with $E(e)$ for every $e\in B_T$. Therefore
$E(e)H_\alpha\subseteq H_\alpha$.
Let $p(\sigma(T))$ be the algebra of all complex continuous functions on $\sigma(T)$
expressible as $p(\lambda,\bar{\lambda})$, where $p(\nu,\gamma)$ is a polynomial
in two complex variables with complex coefficients. Then by the Stone-Weierstrass
Theorem, $p(\sigma(T))$ is dense in $C(\sigma(T))$. If $f\in C(\sigma(T))$ then
$f_\alpha=f|_{\sigma(T_\alpha)}\in C(\sigma(T_\alpha))$ by #2.
If $f(\lambda)=\sum_{i=1}^kc_i\lambda^{p_i}\bar{\lambda}^{q_i}\in p(\sigma(T))$,
then by the spectral theorem, for $x_\alpha,y_\alpha\in H_\alpha$,
$f(T)=\sum_{i=1}^kc_iT^{p_i}T^{*q_i}$ and
$f_\alpha(T_\alpha)=\sum_{i=1}^kc_iT_\alpha^{p_i}T_\alpha^{*q_i}$ and
\begin{equation*}
\begin{split}
\int_{\sigma(T_\alpha)}\!f_\alpha(\lambda)\,dE'_{x_\alpha,y_\alpha}(\lambda)
&=\Biggl(\biggl(\sum_{i=1}^kc_iT_\alpha^{p_i}T_\alpha^{*q_i}\biggr)x_\alpha,
y_\alpha\Biggr)
=\Biggl(\biggl(\sum_{i=1}^kc_iT^{p_i}T^{*q_i}\biggr)x_\alpha,y_\alpha\Biggr)\\
&=\int_{\sigma(T)}\!f(\lambda)\,dE_{x_\alpha,y_\alpha}(\lambda)
\qquad\qquad\qquad\qquad\qquad(f\in p(\sigma(T))).\quad\text{(1)}
\end{split}
\end{equation*}
If $f\in C(\sigma(T))$, let $\{f_n\}\subseteq p(\sigma(T))$ be such that
$f_n(\lambda)\to f(\lambda)$ uniformly on $\sigma(T)$ (that is, in the norm of
$C(\sigma(T))$). Then $f_{n,\alpha}(\lambda)\to f_\alpha(\lambda)$ uniformly
on $\sigma(T_\alpha)$. By the spectral theorem
\begin{equation*}
\lvert((f(T)-f_n(T))x_\alpha,y_\alpha)\rvert
\leq\lvert\lvert f(T)-f_n(T)\rvert\rvert\,\lvert\lvert x_\alpha\rvert\rvert\,
\lvert\lvert y_\alpha\rvert\rvert\to 0
\end{equation*}
as $n\to\infty$. Similarly,
$\lvert((f_\alpha(T_\alpha)-f_{n,\alpha}(T_\alpha))x_\alpha,y_\alpha)\rvert\to 0$.
Therefore, by (1) and the spectral theorem,
\begin{equation*}
\begin{split}
\int_{\sigma(T_\alpha)}\!f_\alpha(\lambda)\,dE'_{x_\alpha,y_\alpha}(\lambda)
&=\lim_{n\to\infty}\int_{\sigma(T_\alpha)}\!
f_{n,\alpha}(\lambda)\,dE'_{x_\alpha,y_\alpha}(\lambda)
=\lim_{n\to\infty}\int_{\sigma(T)}\!
f_n(\lambda)\,dE_{x_\alpha,y_\alpha}(\lambda)\\
&=\int_{\sigma(T)}\!f(\lambda)\,dE_{x_\alpha,y_\alpha}(\lambda)
\qquad\qquad\qquad\qquad\qquad(f\in C(\sigma(T))).\quad\text{(2)}
\end{split}
\end{equation*}
Let $d(\lambda,S)$ be the distance from $\lambda$ to $S\subseteq\mathbb{C}$:
$d(\lambda,S)=\inf\,\{\lvert\lambda-s\rvert:s\in S\}$. $d$ is continuous as a
function of $\lambda\in\mathbb{C}$. Let $e$ be a closed subset of $\sigma(T)$.
For $n=1,2,\dots$, let $f_n(\lambda)=\max(0,1-nd(\lambda,e))$ for
$\lambda\in\sigma(T)$. Then
$\{f_n\}\subseteq C(\sigma(T))$, $0\leq f_n,f_{n,\alpha}\leq 1$,
$f_n(\lambda)\to\chi_e(\lambda)$ for all $\lambda\in\sigma(T)$, and
$f_{n,\alpha}(\lambda)\to\chi_{e\cap\sigma(T_\alpha)}(\lambda)$ for all
$\lambda\in\sigma(T_\alpha)$. Since
$\chi_{\sigma(T)}\in L^1(\lvert E_{x_\alpha,y_\alpha}\rvert)$,
$\chi_{\sigma(T_\alpha)}\in L^1(\lvert E'_{x_\alpha,y_\alpha}\rvert)$, and
$\lvert f_n\rvert\leq\chi_{\sigma(T)}$ and
$\lvert f_{n,\alpha}\rvert\leq\chi_{\sigma(T_\alpha)}$ for $n=1,2,\dots$, we have
by (2) and the Dominated Convergence Theorem for complex measures and by the spectral
theorem, that
\begin{equation*}
\begin{split}
(E'(e\cap\sigma(T_\alpha))x_\alpha,y_\alpha)
&=\int_{\sigma(T_\alpha)}\!\chi_{e\cap\sigma(T_\alpha)}(\lambda)\,
dE'_{x_\alpha,y_\alpha}(\lambda)
=\lim_{n\to\infty}\int_{\sigma(T_\alpha)}\!f_{n,\alpha}(\lambda)\,
dE'_{x_\alpha,y_\alpha}(\lambda)\\
&=\lim_{n\to\infty}\int_{\sigma(T)}\!f_n(\lambda)\,
dE_{x_\alpha,y_\alpha}(\lambda)
=\int_{\sigma(T)}\!\chi_e(\lambda)\,
dE_{x_\alpha,y_\alpha}(\lambda)\\
&=(E(e)x_\alpha,y_\alpha)\qquad\qquad\qquad\qquad\qquad
\text{($e$ closed $\subseteq\sigma(T)$).}\quad(3)
\end{split}
\end{equation*}
Let
\begin{equation*}
\mathscr{M}=\{e\in B_T: (E'(e\cap\sigma(T_\alpha))x_\alpha,y_\alpha)
=(E(e)x_\alpha,y_\alpha)\text{ for all }x_\alpha,y_\alpha\in H_\alpha\}.
\end{equation*}
Suppose $e\in\mathscr{M}$. Then
\begin{equation*}
\begin{split}
E'(e\cap\sigma(T_\alpha))+E'(e^c\cap\sigma(T_\alpha))
=E'(\sigma(T_\alpha))&=I_\alpha\quad\text{and}\\
E(e)+E(e^c)=E(\sigma(T))&=I,
\end{split}
\end{equation*}
so
\begin{equation*}
\begin{split}
(E'(e\cap\sigma(T_\alpha))x_\alpha,y_\alpha)
&+(E'(e^c\cap\sigma(T_\alpha))x_\alpha,y_\alpha)
=(I_\alpha x_\alpha,y_\alpha)\\
&=(I x_\alpha,y_\alpha)
=(E(e)x_\alpha,y_\alpha)+(E(e^c)x_\alpha,y_\alpha),
\end{split}
\end{equation*}
hence
$$(E'(e^c\cap\sigma(T_\alpha))x_\alpha,y_\alpha)=(E(e^c)x_\alpha,y_\alpha),$$
so $e^c\in\mathscr{M}$.
If $e,e'\in\mathscr{M}$, then
$E'(e'\cap\sigma(T_\alpha))x_\alpha, E(e)y_\alpha\in H_\alpha$ and
\begin{equation*}
\begin{split}
(E'((e\cap e')\cap\sigma(T_\alpha))x_\alpha,y_\alpha)
&=(E'(e\cap\sigma(T_\alpha))E'(e'\cap\sigma(T_\alpha))x_\alpha,y_\alpha)\\
&=(E(e)E'(e'\cap\sigma(T_\alpha))x_\alpha,y_\alpha)
=(E'(e'\cap\sigma(T_\alpha))x_\alpha,E(e)y_\alpha)\\
&=(E(e')x_\alpha,E(e)y_\alpha)
=(E(e)E(e')x_\alpha,y_\alpha)
=(E(e\cap e')x_\alpha,y_\alpha),
\end{split}
\end{equation*}
so $e\cap e'\in\mathscr{M}$.
Suppose $\{e_1,e_2,\dots\}\subseteq\mathscr{M}$ are disjoint. Then
\begin{equation*}
(E'((\cup_n e_n)\cap\sigma(T_\alpha))x_\alpha,y_\alpha)
=\sum_{n=1}^\infty(E'(e_n\cap\sigma(T_\alpha))x_\alpha,y_\alpha)
=\sum_{n=1}^\infty(E(e_n)x_\alpha,y_\alpha)
=(E(\cup_n e_n)x_\alpha,y_\alpha),
\end{equation*}
so
$$e=\bigcup_{n=1}^\infty e_n\in\mathscr{M}.$$
If $\{e_1,e_2,\dots\}$ are not disjoint, let
$$e_n'=e_n\cap\biggl(\bigcup_{i=1}^{n-1}e_n'\biggr)^c\qquad(n=1,2,\dots).$$
By induction, $\{e_1',e_2',\dots\}$ are disjoint and by induction along with what
has been shown so far, $\{e_1',e_2',\dots\}\subseteq\mathscr{M}$. Since
$$\bigcup_{n=1}^\infty e_n=\bigcup_{n=1}^\infty e_n',$$
we have that $e=\cup_n e_n\in\mathscr{M}$, so $\mathscr{M}$ is a $\sigma$-algebra
containing the closed sets of $\sigma(T)$, so $\mathscr{M}=B_T$. This proves that
\begin{equation*}
(E'(e\cap\sigma(T_\alpha))x_\alpha,y_\alpha)=(E(e)x_\alpha,y_\alpha)
\qquad(x_\alpha,y_\alpha\in H_\alpha,\,e\in B_T).\quad(4)
\end{equation*}
Since (4) holds for all $y_\alpha\in H_\alpha$, we have that
\begin{equation*}
E'(e\cap\sigma(T_\alpha))x_\alpha=E(e)x_\alpha\qquad(x_\alpha\in H_\alpha,\,
e\in B_T)
\end{equation*}
hence $E'(e\cap\sigma(T_\alpha))=E(e)|_{H_\alpha}\quad(e\in B_T)$, which proves #3.
Suppose $f\in\mathscr{B}_T$ and $f\geq 0$. Let $\{s_n\}$ be a sequence of simple
Borel measurable maps, bounded by $f$, converging uniformly on $\sigma(T)$ to $f$.
Then $\{s_{n,\alpha}\}$ converges uniformly on $\sigma(T_\alpha)$ to $f_\alpha$.
Say
$$s_n(\lambda)=\sum_{i=1}^{k_n} c_{n,i}\chi_{e_{n,i}}(\lambda).$$
By the spectral theorem
\begin{equation*}
\lvert((f(T)-s_n(T))x_\alpha,y_\alpha)\rvert
\leq\lvert\lvert f(T)-s_n(T)\rvert\rvert\,\lvert\lvert x_\alpha\rvert\rvert\,
\lvert\lvert y_\alpha\rvert\rvert\to 0
\end{equation*}
as $n\to\infty$. Similarly,
$\lvert((f_\alpha(T_\alpha)-s_{n,\alpha}(T_\alpha))x_\alpha,y_\alpha)\rvert\to 0$.
By the spectral theorem
and by (4),
\begin{equation*}
\begin{split}
(f_\alpha(T_\alpha)x_\alpha,y_\alpha)
&=\lim_{n\to\infty}\,(s_{n,\alpha}(T_\alpha)x_\alpha,y_\alpha)
=\lim_{n\to\infty}\int_{\sigma(T_\alpha)}\!
s_{n,\alpha}(\lambda)\,dE'_{x_\alpha,y_\alpha}(\lambda)\\
&=\lim_{n\to\infty}\sum_{i=1}^{k_n}
c_{n,i}E'_{x_\alpha,y_\alpha}(e_{n,i}\cap\sigma(T_\alpha))
=\lim_{n\to\infty}\sum_{i=1}^{k_n}
c_{n,i}E_{x_\alpha,y_\alpha}(e_{n,i})\\
&=\lim_{n\to\infty}
\int_{\sigma(T)}\!s_n(\lambda)\,dE_{x_\alpha,y_\alpha}(\lambda)
=\lim_{n\to\infty}\,(s_n(T)x_\alpha,y_\alpha)\\
&=(f(T)x_\alpha,y_\alpha)\qquad\qquad\qquad(x_\alpha,y_\alpha\in H_\alpha).
\end{split}
\end{equation*}
Since $P_\alpha$ commutes with $E(e)$ for every $e\in B_T$, by the spectral theorem,
$P_\alpha$ also commutes with $f(T)$ for every $f\in\mathscr{B}_T$. Therefore,
$f(T)H_\alpha\subseteq H_\alpha$, so $f_\alpha(T_\alpha)x_\alpha=f(T)x_\alpha$
for all $x_\alpha\in H_\alpha$, whence $f_\alpha(T_\alpha)=f(T)|_{H_\alpha}$.
If $f$ is not positive, then we can decompose it into its real and imaginary positive and negative parts, all of which are positive, and finally get
$f_\alpha(T_\alpha)=f(T)|_{H_\alpha}$, which proves #4 and completes the proof of
the theorem.
