# Functions of unbounded operators: do they commute or not?

Given two unbounded commuting self-adjoint operators $A$ and $B$. Then all bounded Borel functions of $A$ and $B$ commute (in the sense that all the projections in their associated projection-valued measures commute). But what is when we have an unbounded function? More precise, for example, do $g(A)$ and $f(A)$ commute, where $f$ is a bounded Borel function and $g$ a unbounded one (e.g. $g(x)=x$).

• Have a look at Rudin's functional analysis text. I believe the answer is true even for unbounded functions (at least continuous ones). Rudin has a good treatment for the unbounded functional calculus. – voldemort Jan 16 '14 at 18:01

You should specify if the operators commute in strong or weak sense. By definition, s.a. operators commute strongly iff all spectral projections $E_A(\cdot)$ and $E_B(\cdot)$ commute.
But then every spectral projection of $f(A)$ $E_{f(A)}(M)=\chi_M(f(A))=\int\chi_M(f(\cdot))dE_A(\cdot)=\int\chi_{f^{-1}(M)}dE_A=E_A(f^{-1}(M))$ commutes with $E_{g(B)}(N)=E_B(g^{-1}(N))$ for every Borel $M,N\subseteq\mathbb C.$