1
$\begingroup$

Say $A, B : H \supset D \to H$ are essentially self-adjoint operators on the dense common domain $D$. $H$ is some Hilbert space. Does it hold that $A + B$ is also essentially self-adjoint? If not, can you please give me a counterexample?

$\endgroup$
1
  • 2
    $\begingroup$ Why don't you just leaf through one of the hundreds of books on spectral theory and/or theory of operators (e.g., Reed and Simon's treatise)? a large part of the theory is devoted to answer precisely this question $\endgroup$ Jun 4, 2011 at 8:57

1 Answer 1

6
$\begingroup$

Let $D=H^4(0,1)\cap H^2_0(0,1)$, $Au=u''''$, $Bu=-u''''+u''$.

$\endgroup$
2
  • 2
    $\begingroup$ It may be helpful to note that the reason counterexamples exist is that the intersection of the two domains may be smaller than each of the domains; and if $A$ is ess. self-adjoint with domain $D$ and $D_0\subset D$ is dense, the closure of the restriction of $A$ to $D_0$ may not coincide with $A$. (If the closure of the restriction of $A$ to $D_0$ is $A$ one sometimes says that $D_0$ is a core for $A$). So your $A+B$ will be ess. self-adjoint if you assume that there is a common core for $A$ and $B$. $\endgroup$ Jun 4, 2011 at 7:43
  • 1
    $\begingroup$ In the example above D is a core for A and B. It is not a core for A+B. $\endgroup$ Jun 4, 2011 at 10:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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