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 1


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 $D(B)$.

  • $\begingroup$ 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 $B$? $\endgroup$
    – Chris
    Apr 23, 2020 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 $k$. $\endgroup$ Apr 23, 2020 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 $D(B)$ $\endgroup$
    – Chris
    Apr 24, 2020 at 7:50
  • $\begingroup$ Thank you for this quite elementary counter-example $\endgroup$
    – Chris
    Apr 24, 2020 at 7:51

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