Borel Functional Calculus of Commuting Operators

Question

let $H$ be a Hilbert space, $T_1,T_2 \in L(H)$ two self-adjoint operators such that $T_1T_2=T_2T_1 \in L(H)$, i.e. they commute and their product is continuous. $E_1,E_2$ are their assigned spectral measures.

Let $f: \sigma(T_1)\rightarrow \mathbb{K}, g: \sigma(T_2)\rightarrow \mathbb{K}$ be borel-measurable functions. Show that the operators also commute: $\psi_1(f)\psi_2(g)=\psi_2(g)\psi_1(f), \ i=1,2$ and $ \psi_i$ being the borel functional calculus with respect to $T_i$.

An Idea would be:

Let $f=\chi_A,\ g=\chi_B$ be the indicatorfunctions for A,B each a borelset in the spectrum of $T_1,T_2$. $G:= \psi_1(f)\psi_2(g)$, with that I inspect $\left<Gx,x\right>=\left<x,G^*x\right>$. If I can show from there that $G=G^*$, one could approximate arbitrary measurable and bounded $f,g$ with a linear combination of simple functions. Would this be a good way?

Answer 1

Note that $\psi_1(\chi_A) = E_1(A)$ and $\psi_2(\chi_B)= E_2(B)$. So what you're really trying to prove is that the spectral measures commute, i.e. $E_1(A) E_2(B) = E_2(B)E_1(A)$. Proving that $G^* = G$ is of course equivalent to this. But it's easier to prove the following first:

If $T_1$ is bounded normal operator on a Hilbert space $H$ with associated spectral measure $E$ and $T_2 \in B(H)$. Then $T_1$ and $T_2$ commute if and only if $T_2$ commutes with $E(\Delta)$ for every Borel subset $\Delta$ of $\sigma(T_1)$.

To prove this, you first show that $T_2f(T_1) = f(T_1)T_2$ for every continuous function on $\sigma(T_1)$ using the Stone-Weierstrass theorem (you can approximate the continuous functions by a sequence of polynomials for which this trivially holds). Then you can approximate $\chi_\Delta$ by a sequence of continuous functions, proving that $T_2 \chi_\Delta(T_1) =\chi_\Delta(T_1) T_2$.

From this theorem it directly follows that $E_1(A)E_2(B)=E_2(B)E_1(A)$ for all Borel subsets $A,B$. You can then, as you suggested, approximate any two functions $f,g$ by a simple functions, from which your theorem then follows.