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 $A+B$ ?


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 $D(B)$.

| cite | improve this answer | |
  • $\begingroup$ 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 $B$? $\endgroup$ – Chr Apr 23 at 19:18
  • $\begingroup$ 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 $k$. $\endgroup$ – Giorgio Metafune Apr 23 at 22:16
  • $\begingroup$ 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 $D(B)$ $\endgroup$ – Chr Apr 24 at 7:50
  • $\begingroup$ Thank you for this quite elementary counter-example $\endgroup$ – Chr Apr 24 at 7:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.