# Sum of strongly commuting self-adjoint operators

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 $$A+B$$ ?

## 1 Answer

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 $$D(B)$$.

• Ok, so if I am correct, $D\cap D(A)$ is dense in $D(A)$ and $D\cap D(B)$ is dense in $D(B)$. Is it clear that $D\cap D(A)\cap D(B)$ is dense in $D(A)$ and in $D(B)$ ? Otherewise, how would you construct the common core to $A$ and $B$? – Chr Apr 23 at 19:18
• You are right, I did not check that, but it should be as follows. Take a sequence $(x_n) \subset D(A)$ with $x_n=0$ for $n > N$, let $s=x_1+\cdots +x_N$ and modify it by adding $-s/k$ at $k$ odd positions greater than $N$. The resulting sequence is in $D \cap D(B)$ and differs in the $D(A)$ norm from the old one $|s|/\sqrt {k}$ which can be made small taking a large $k$. – Giorgio Metafune Apr 23 at 22:16
• Ok, this proves that compactly supported sequences in $D(A)$ can be approached by sequences in $D\cap D(B)$. Then, since compaclty supported sequences of $D(A)$ are clearly dense in $D(A)$, this indeed proves that $D\cap D(A)\cap D(B)$ is dense in $D(A)$ and the same hold for $D(B)$ – Chr Apr 24 at 7:50
• Thank you for this quite elementary counter-example – Chr Apr 24 at 7:51