Some functional calculus of commuting positive operators Let $A,B$ be two positive operators on a complex Hilbert space. We know that we can define $A^a$ for any $a\geq0$. If $A$ commutes with $B$, then do we have $(AB)^a=A^aB^a$? I believe this is correct but I am not too sure whether my proof is correct (it uses the spectral theorem)? 
I would appreciate any hint.
Thanks in advance.
 A: This answer uses several times the fact that positive square roots of positive operators are unique. 
You have 
$$\tag{1}
(AB)^n=A^nB^n,\ \ n\in\mathbb N. 
$$
It follows, by $(1)$, that 
$$\tag{2}
(A^{1/n}B^{1/n})^n=AB,
$$
so 
$$\tag{3}
(AB)^{1/n}=A^{1/n}B^{1/n}.
$$
Now, combining $(1)$ and $(3)$, 
$$\tag{4}
(AB)^q=A^qB^q,\ \ q\in\mathbb Q. 
$$
For $r>0$ arbitrary, taking limits in $(4)$ we get 
$$
(A^rB^r)^{1/r}=AB,
$$
and so 
$$
A^rB^r=(AB)^r.
$$
A: Note that
\begin{align*}
e^{A+B} &= \sum_{n=0}^\infty \frac{(A+B)^n}{n!} = \sum_{n=0}^\infty \sum_{j=0}^n\binom{n}{j}\frac{A^jB^{n-j}}{n!} + \sum_{n=0}^\infty \sum_{j=0}^n\frac{n!}{j!(n-j)!}\frac{A^jB^{n-j}}{n!}\\
& = \sum_{n=0}^\infty \sum_{j=0}^n \frac{A^jB^{n-j}}{j!(n-j)!} = \sum_{j=0}^\infty \sum_{n=j}^\infty \frac{A^jB^{n-j}}{j!(n-j)!} = \sum_{j=0}^\infty \sum_{k=0}^\infty \frac{A^jB^{k}}{j!k!} = e^A e^B.
\end{align*}
The second equality follows from the binomial theorem, which we may employ since $A$ and $B$ commute. Since the functional calculus ensures $e^{\ln(A)}=A$, we conclude that
$$
\ln(AB)=\ln(e^{\ln(A)}e^{\ln(B)}) =\ln(e^{\ln(A)+\ln(B)})=\ln(A) + \ln(B),
$$
where we used that $\ln(A)$ and $\ln(B)$ commute as well. Furthermore, it also follows from functional calculus that $\ln(A^a)=a\ln(A)$. But then
$$
\ln((AB)^a)=a\ln(AB)=a(\ln(A)+\ln(B))=a\ln(A)+a\ln(B)=\ln(A^a)+\ln(B^a)=\ln(A^aB^a),
$$
and therefore $(AB)^a=A^aB^a$.
