Continuity of the functional calculus form of the Spectral Theorem I'm trying to understand the proof of the functional calculus form of the Spectral Theorem.
From what I understand the construction of a definition of $f(A)$ for $f$ measurable and $A$ $\in L(H)$ self-adjoint begins as follows.
Assume we know the proof of the continuous functional calculus so that $f(A) \in L(H)$ is defined for $f$ continuous on the spectrum of $A$.
We can define a function $T_\psi: C(\sigma(A)) \to \mathbb{C}$ by $$T_\psi(f) = (\psi,f(A)\psi)$$ for a set $\psi \in H.$
This gives a positive linear functional on $C(\sigma(A))$, and from the Riesz-Markov theorem there is a measure $\mu_\psi$ on the compact set $\sigma(A)$ s.t
$$T_\psi(f) = (\psi,f(A)\psi)=\int_{\sigma(A)}fdu_\psi.$$
We note that the right hand side is defined for any $f \in \mathbb{B}(\mathbb{R})$, a bounded measurable function, and so we define:
$(\psi,g(A)\psi):=\int_{\sigma(A)}gdu_\psi$ $\forall g \in \mathbb{B}(\mathbb{R})$.
*Now this is a definition, the left hand side is not an inner product because the second term is not yet defined.
We next extend the definition:
$\forall \psi, \phi \in H $, $(\phi, g(A)\psi) := $
$\frac{1}{4}[(\phi + \psi, g(A)(\phi + \psi)) - (\phi - \psi, g(A)(\phi -
 \psi)) + (\phi + i\psi, g(A)(\phi + i\psi)) - (\phi - i\psi, g(A)(\phi -
 i\psi))]$
Lastly we can define a function $T_1: H \to \mathbb{C}$ by $T_1(\phi) = (\phi, g(A)\psi) $ for a set $\psi, g$.
$T_1$ is a continuous linear function, and so by the Riesz rep. theorem we have $\exists!$ $h_{\psi,g}\in H$ s.t $\forall \phi$ $(\phi, g(A)\psi) = (\phi, h_{\psi,g})$.
Now, the functional calculus theorem states that exists, and unique with the following properties, $\Phi: \mathbb{B}(\mathbb{R}) \to L(H)$ s.t


*

*$\Phi$ is a $*$-homomorphism.

*$\|\Phi(f)\| \leq \|f\|_{\infty}$
among other properties I won't list.
I need help showing 1. and 2.
For 2. I thought of trying to show that $\forall f \in \mathbb{B}(\mathbb{R})$ $f(A)$ is self adjoint but kind of got lost in the details of this. Is it true?
If it is, then we have 
$\|\Phi(f)\| = sup_{\|\phi\|=1}[\|f(A)\phi\|] = sup_{\|\phi\|=1}(\phi,f(A)\phi) = \int_{\sigma(A)} f du_\phi \leq \|f\|_{\infty}\mu_\phi(\sigma(A)) = \|f\|_{\infty}$ where the last equality should hold by the Riesz Markov theorem.
For 1. I'm not sure where to begin.
 A: Throughout the post you mean $B(\mathbb{R})$, right?
The proof for 2. seems O.K.
For 1., observe that it is naturally true for polynomials, and so it is true for all continuous functions since every continuous function on a compact set is a limit of polynomials. Now it is true for all of $B(\mathbb{R}$) since $C(\mathbb{R})$ is dense in $B(\mathbb{R})$. 
Let $f,g\in C(\mathbb{R})$. Since $\sigma (A)$ is compact, there exists polynomials $p_n$ s.t. $p_n\longrightarrow f$, and $q_n$ s.t. $q_n\longrightarrow g$  uniformly. Therefore, 
$<\phi,fg(A)\phi>=<\phi,\underset{n,m\longrightarrow \infty}{lim}p_nq_n(A)\phi>=\underset{n,m\longrightarrow \infty}{lim}<\phi,p_nq_n(A)\phi>=$ $\underset{n,m\longrightarrow \infty}{lim}<\phi,p_n(A)q_n(A)\phi>=...=<\phi,f(A)g(A)\phi> $ and from here we conclude that $fg(A)=f(A)g(A)$ for every $f,g\in C(\mathbb{R})$.
Let $f,g\in B(\mathbb{R})$. Since $C(\mathbb{R})$ is dense in $B(\mathbb{R})$ with respect to $L^1$ convergence, there exist $f_n,g_n\in C(\mathbb{R})$ bounded s.t. $f_n,g_n$ converge to $f,g$ respectively in $L^1$. Now similarly, the same equalities as the previous case hold. Note that this time the equality $<\phi,\underset{n,m\longrightarrow \infty}{lim}f_ng_n(A)\phi>=\underset{n,m\longrightarrow \infty}{lim}<\phi,f_ng_n(A)\phi>$ holds because since $f_n,g_n$ are bounded and $\mu_\phi$ is finite, we can use the bounded convergence theorem.
A: There is a lot of problems in your question, so I don't know where to begin my answer. But I believe your question will be answered if you answer my questions step by step.
First, the definition of $g(A)$  has some prblems. Why dose there exist $g(A)$ such that $(\psi,g(A)\psi):=\int_{\sigma(A)}gdu_\psi$?
Second, what's the definition of $f(A)$ for $f\in B(\mathbb{R})$?
Maybe the problem in your question is: Why dose there exist such $h_{\psi, g}$? It is because $T_1$ is bounded. So once $\Phi$ is defined, it is continuous.

For 1, (a) $\Phi$ is linear.
(b)To prove $\Phi$ preserves involution, we only need to verify $\Phi(f)$ is hermitian if $f$ is real because of  (a). Since $(\phi, \Phi(f)\phi)=\int f d\mu_\phi$ is real for all $\phi$, $\Phi(f)$ is hermitian. 
(c)To prove $\Phi$ preservers product, we have to prove:for all $ f, g\in B(\mathbb{R})$,
$$(\phi, \Phi(fg)\phi)=(\Phi(\bar f)\phi, \Phi(g)\phi)=\sum_{k=0}^3 i^k(\phi_k, \Phi(g)\phi_k)~~(polarization),$$
i.e., \begin{align}\label{eq}
\int fg d\mu_\phi=\sum_{k=0}^3 i^k\int gd\mu_{\phi_k}.\tag{1}\end{align}
(c.1)Fixed $f\in C(\mathbb{R})$. Since the above equation holds for all continous function $g$, it also holds for all bounded measurable function $g$.
(c.2)Because of (c.1), for all continuous function $f$ and bounded measurable function $g$  $$\overline{(\phi, \Phi(\bar f\bar g)\phi)}=\overline{(\Phi( f)\phi, \Phi(\bar g)\phi)},$$ 
i.e.
$$(\phi, \Phi(gf)\phi)=(\Phi(\bar g)\phi, \Phi(f)\phi),$$
that is,
\begin{align}
\int gf d\mu_\phi=\sum_{k=0}^3 i^k\int fd\mu_{\phi'_k}.\end{align}
So it also holds for all bounded measurable function $f$. Done!

Details for (c.1):
By Riesz Representation theorm,

Theorem 1. $$C(X)^*\cong M(X),$$
  where $X$ is a compact space and $M(X)$ denotes all the  complex regular Borel measure on $X$.

Another useful theory(V.4.1 in Conway's text):

Theorem 2. Suppose $X$ is a normed space, then the closed unit ball of $X$ is $\sigma(X^{**},X^*)$ dense in the closed unit ball of $X^{**}$.

For every bounded measurable function $g\in B(\sigma(A))\subset C(\sigma(A))^{**}$, there is a net $g_i\in C(\sigma(A))$ such that $$g_i\xrightarrow{\sigma(C(\sigma(A))^{**},C(\sigma(A))^*)}g$$
by theorem 2.
That is precisely $$\int g_i d \mu\to\int g d \mu$$
for every $\mu \in M(\sigma(A))$ by theorem 1. Note that $fd\mu\in M(\sigma(A))$ for $f\in B(\sigma(A))$ and $\mu\in M(\sigma(A))$.
