# Borel Functional Calculus of Commuting Operators

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?

• It appears that you are not assuming $T_1,T_2$ are bounded because of the assumption that $T_1T_2=T_2T_1$ are bounded. Is that correct? Nov 9, 2017 at 16:50
• yes. I forgot to include that.
– dba
Nov 9, 2017 at 21:41

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.

• In the theorem you write "Then $A$ and $B$ commute if and only if $B$ commutes with $E(\Delta)$". If $A$ and $B$ are sets, what do you mean with "commute"?
– dba
Nov 9, 2017 at 15:08
• I think the typo still exists in the following text: ....$Bf(A)=f(A)B$. And later BχΔ(A)=χΔ(A)B. Or is that correct?
– dba
Nov 9, 2017 at 15:20
• I think that in the paragraph after the theorem the use of $A$ and $B$ is still unclear. Are $A$ and $B$ sets or operators?
– dba
Nov 9, 2017 at 15:28
• If you write $\sigma(A)\,$ then $A$ must be an operator. And the expression $Bf(T_1)$ uses $B$ as an operator, right?
– dba
Nov 9, 2017 at 15:43
• What about "To prove this, you first show that $Bf(T_1)=f(T_1)B$"? Is $B$ a borel set here? I think $B$ must be replaced by some operator. And yes, the other change is good.
– dba
Nov 9, 2017 at 15:48