Suppose $S,T$ are two commuting normal operators in $B(H)$,how to use the continuous functional calculus to show that $f(S)g(T)=g(T)f(S)$,where $f\in C(\sigma(S)),g\in C(\sigma(T))$?
There is an isometric $*$ isomorphism from $C(\sigma(S))$ to $C^*(S)$ which maps $f$ to $f(S)$ and there is also another isometric $*$ isomorphism from $C(\sigma(T))$ to $C^*(T)$ which maps $g$ to $g(T)$.$f(z)$ is the norm limit of polynomials in $z$ and $\bar{z}$,then $f(S)$ is the norm limit of polynomials in $S$ and $S^*$.Similarly,$g(T)$ is the norm limit of polynomials in $T$ and $T^*$.Polynomials in $S,S^*,T,T^*$ commute,passing to the norm limit,we have $f(S)g(T)=g(T)f(S)$.
Is my thought correct?
Best Answer
Yes, but I'm not sure if you are correct to the last detail, because you are not specific enough.
To go from $ST=TS$ to $ST^*=T^*S$ you need Fuglede's Theorem.
Once you know that, you get that all polynomials in $X,X^*,T,T^*$ commute.
To go to continuous $f,g$, you need to do it in two steps (to do it in one step you would need to guarantee uniform boundedness of the polynomials approaching $f$ and $g$; it can be done this way, but it requires its own set of deails): fist you take limits to show that $f(S)T=Tf(S)$. Now you repeat all the above with $T$ and $f(S)$ (you also need to justify the easy fact that $f(S)$ is normal) to get $f(S)g(T)=g(T)f(S)$.