Let $A,B$ be two positive bounded linear operators on a Hilbert space. If $A$ commutes with $B$, do we have $(AB)^p =A^p B^p$ for any $p>0$? Or more general, $f(AB) = f(A)f(B)$ for any Borel function $f$.


1 Answer 1


Yes for powers, but not for general $f$. For example, it's hardly ever true for $f(x) = 2x$. It's true for powers because $(xy)^p = x^p y^p$ for $x,y \in [0, \infty)$; apply the Gelfand isomorphism on $C(\Delta)$ where $\Delta$ is the maximal ideal space of the commutative C* algebra generated by $A$ and $B$.

  • $\begingroup$ Really great answer ! Thx $\endgroup$
    – user92646
    Nov 21, 2017 at 22:08

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