Let $A,B$ be unbounded self-adjoint operators on a Hilbert space, that don't commute. Let $f,g : \mathbb{R} \to \mathbb{R}$ be functions such that $f(A)$ and $g(A)$ are well-defined by the spectral theorem, and similarly for $B$.
Suppose that $\| f(A) (f(B))^{-1} \| < \infty$ and $\| g(A) (g(B))^{-1} \| < \infty$.
I want to show that $b = \|f(A)g(A) (g(B))^{-1} (f(B))^{-1}\| < \infty$.
Of course one approach is to commute $f(A)$ with $(g(B))^{-1}$, but I wonder if
instead we can somehow cycle the operators - in the same way that one can do for the trace of matrices $tr(ABC) = tr(BCA)$ ... something like $\|ABC\| = \| BCA\|$ ?
So my question is if $b = \|g(A) (g(B))^{-1} (f(B))^{-1} f(A) \|$ ?