Let $A,B$ be two positive unbounded, self-adjoint operators on some Hilbert space that strongly commute. Let $D(A)$ and $D(B)$ denote their respective domain. Then, using for instance the spectral theorem, A+B is self-adjoint on $D(A)∩D(B)$.


If we furthermore assume that $A$
and B are essentially self-adjoint on some common core D, is it also the case of $A+B$ ? 



Sum of strongly commuting self-adjoint operators
The answer is no. Take $X=l^2$, $Ax=(a\_n x\_n)$, $Bx=(b\_n x\_n)$, where $a\_n=n$, $b\_n=1$ if $n$ is even, $a\_n=1$, $b\_n=n$ if $n$ is odd. Then $A+B$ is the multiplication by $(n+1)$ on $$D(A+B)=\{(x\_n): \left((n+1)x\_n\right) \in l^2\}.$$ The non-zero functional $F(x)=\sum\_n x\_n$ is continuous on $D(A+B)$ but discontinuous both on $D(A)$ and $D(B)$, hence $D=Ker F$ is closed in $D(A+B)$ but dense both in $D(A)$ and $D(B)$.


