Let $A$ and $B$ be two unbounded self-adjoint operators. From this mathoverflow post, for instance, we know that $A + B$ is self-adjoint on $\mathcal{D}(A) \cap \mathcal{D}(B)$ if $A$ and $B$ are commuting and positive operators (Putnam's book is cited as a reference there). My question is: assume $A$ is positive and $A$ and $B$ commute. $B$ is not positive though, but $B$ is a relatively bounded perturbation of $A$. Could we still say $A + B$ is self-adjoint?
[Math] Sum of two unbounded self-adjoint operators
fa.functional-analysisoperator-theoryreference-requestunbounded-operators
Best Answer
This will work if you take the assumption that $A,B$ commute in a sufficiently strong sense (commuting resolvents would be enough). Then no extra assumption is needed.
There is a version of the spectral theorem that says that there is a projection valued measure that represents both $A$ and $B$: $$ A = \int s\, dE(s,t) , \quad B = \int t\, dE(s,t) $$ See here. Clearly this makes $\int (s+t)\, dE(s,t)$ self-adjoint on a suitable domain. Note, however, that this domain is not necessarily $D(A)\cap D(B)$, as simple examples show (multiplication by $|x|$ and $x$).