Composition of spectral measures

Question

Let $f: \mathbb{R}\rightarrow \mathbb{C}$ be a measurable function, $H$ some Hilbert space and

$$ f_E := \int_{\mathbb{R}} f dE$$ for some spectral measure $E$.

Now, my question is: When do we have $f_E\circ g_E = (fg)_E$ for two measurable functions $f$ and $g$?

My idea was that $g$ should be bounded.

The domain of $f_E$ is defined as $D(f_E):=\{x \in H: f \in L^2( \mu_x)\}$, where $\mu_x$ is the Borel measure $\mu_x(\Delta):= \langle E(\Delta)x,x\rangle.$

Then we should have $D(f_E \circ g_E)= D(g_E) \cap D((fg)_E).$ Now let's assume that $g$ is bounded, then $D(g_E)= H$ by the finiteness of the measure. Thus, we have equality of domains.

Is this correct or is also $f$ bounded a sufficient condition?- I don't think so, but is there a better way to say when we get equality?

Answer 1

The best reference I know on these questions is Fell & Doran, Representations of ${}^*$-algebras, locally compact groups, and Banach ${}^*$-algebraic bundles. Vol. 1 (1988).

First they show your equality for $f$ and $g$ "$E$-measurable and $E$-essentially bounded" (11.8.III), and then I quote from (11.21):

Let $f:X\to\mathbf C$ and $g:X\to\mathbf C$ be $E$-measurable. Then: (...)

(III) $\left(\int f\,dE\right)\left(\int g\,dE\right)\subset\int fg\, dE$

(...) Equality holds in (III) provided that there exist positive numbers $k_1$, $k_2$ satisfying $$ \left|g(x)\right|\geqslant k_1\Rightarrow\left|f(x)\right|\geqslant k_2 $$ for $E$-almost all $x$ in $X$. In particular, equality holds in (III) if $g$ is $E$-essentially bounded. (...)

(V) If $\xi$ belongs to the domains of both $\int g\,dE$ and $\int fg\,dE$, then $(\int g\,dE)(\xi)$ belongs to the domain of $\int f\,dE$, that is, $\xi$ belongs to the domain of the left side of (III).