# Spectral theorem question

I am trying to understand how to develop the spectral measure of a bounded self-adjoint operator on a Hilbert space. For every continuous function on its spectrum, $f: C(\sigma(A)) \to \mathbb{C}$, and for every $w \in H$, by Riesz-Markov, we have:

$\langle w, f(A)w \rangle = \int_{\sigma(A)} f \mathrm{d}\mu_w$, for a finite regular Borel measure of total mass $||w||^2$.

Evidently, the left side makes sense for measurable $f$ (with respect to the Borel algebra). Since the values of $\langle w,Aw\rangle$ as $w$ spans the whole space uniquely determines the operator $A$, we define $\phi: M(\sigma(A)) \to B(H)$, $g \mapsto g(A)$ satisfying the above condition. Now, I want to check that $\phi$ is a unital *-homomorphism, but I can't seem to prove multiplicativity - that is,

$\langle w, f(A)g(A)w\rangle=\langle w, (fg)(A)w\rangle$

I tried using the polarization identity, writing the LHS as

$$\langle w, f(A)g(A)w\rangle = \frac{1}{4} \int_{\sigma(A)} f \mathrm{d}\mu_{w+g(A)w} - \frac{1}{4}\int_{\sigma(A)} f \mathrm{d}\mu_{w-g(A)w} + \frac{i}{4}\int_{\sigma(A)} f d\mu_{w+ig(A)w} -\frac{i}{4}\int_{\sigma(A)} f \mathrm{d}\mu_{w-ig(A)w}$$

But then I got completely lost, and I have no clue how to make the $g$ appear in the integrand. Any help would be greatly appreciated.

EDIT: Yes, sorry, I am assuming a self-adjoint operator. I added that above now.

I prefer the Mathematician's inner product which is linear in the first coordinate. For each $w \in H$, there is a unique regular Borel measure $\mu_{w}$ on $\sigma(A)$ such that $$(f(A)w,w) = \int_{\sigma} f\,d\mu_{w},\;\;\; f \in C(\sigma).$$ Because of this, $\|w\|^{2}=(Iw,w)=\mu_{w}(\sigma)$. Then \begin{align} \int_{\sigma}fd(\mu_{w+w'}+\mu_{w-w'}) & = (f(A)(w+w'),(w+w'))+(f(A)(w-w'),(w-w')) \\ & = 2(f(A)w,w)+2(f(A)w',w')\\ & = 2\int_{\sigma}fd(\mu_{w}+\mu_{w'}). \end{align} By uniqueness, the first identity below follows, and the second follows using a similar argument: $$\mu_{w+w'}+\mu_{w-w'}=2\mu_{w}+2\mu_{w'}\\ \mu_{\alpha w}=|\alpha|^{2}\mu_{w}.$$ For any bounded Borel function $f$ on $\mathbb{R}$, define $$q_{f}(w) = \int f d\mu_{w}.$$ Then $q_{f}$ is a bounded quadratic form because $$q_{f}(w+w')+q_{f}(w-w')= 2q_{f}(w)+2q_{f}(w'),\\ q_{f}(\alpha w)=|\alpha|^{2}q_{f}(w),\\ |q_{f}(w)| \le \|f\|_{\infty}\|w\|^{2}.$$ That's enough to guarantee the existence of a unique operator $T_{f} \in \mathcal{L}(H)$ such that $$q_{f}(w) = (T_{f}w,w).$$ By uniqueness, $T_{f}=f(A)$ if $f \in C(\sigma)$. So this extends the continuous functional calculus to bounded Borel functions. Of course--as you noted--there is no obvious reason that $T_{fg}=T_{f}T_{g}$ will continue to hold for bounded Borel functions $f$, $g$.
Consider the set of Borel subsets $S$ of $\mathbb{R}$ for which $$T_{f}T_{\chi_{S}} = T_{\chi_{S}}T_{f} = T_{\chi_{S}f}.$$ Start by showing that this holds for open intervals $S=(a,b)$. Define $g_{\epsilon} \in C(\sigma)$ such that $g_{\epsilon}$ is $0$ outside $[a,b]$, is $1$ on $(a-\epsilon,b+\epsilon)$ and is linear and continuous on the remaining intervals $(a,a+\epsilon)$ and $(b-\epsilon,b)$. As $\epsilon\downarrow 0$, the function $g_{\epsilon}\uparrow \chi_{(a,b)}$. Because of that, $g_{\epsilon}(A)$ converges in the strong operator topology up to an operator you can show to be $T_{\chi_{(a,b)}}$. After a little thought, you see that the above identity holds for $S=(a,b)$. The above also holds for the complement of $S$ if it holds for $S$. After a little work, you show that the above holds for a sigma algebra of sets containing the open intervals, which gives the identity for all $S$. Then another approximation argument will give you $T_{f}T_{g}=T_{g}T_{f}=T_{fg}$ for all bounded Borel functions $f$, $g$.