Given an operator $T:D_1(T)\subset L^2 \rightarrow L^2$ and the same operator $T:D_2(T) \subset L^2 \rightarrow L^2$, such that the operator is both times self-adjoint and closed, with $D_1(T) \subset D_2(T)$ both being dense in $L^2$. Does this mean that for any $A \subset L^2$, the operator $T:A \rightarrow L^2$ is self-adjoint and closed, too?-if we have that $D_1(T) \subset A \subset D_2(T)$.


Let's denote the operator defined on $D_i(T)$ by $T_i$. Then since $T_1 \subset T_2$ we have $T_2^\ast \subset T_1^\ast$.

Now use that $T_i$ is self-adjoint to conclude that there is not much choice for $A$.

  • $\begingroup$ so $D_1(T)=D_2(T)$? $\endgroup$ – user66906 Jun 30 '14 at 17:25
  • 2
    $\begingroup$ Yes, that's the thing. If you have two self-adjoint operators, one extending the other, they are necessarily the same operator. $\endgroup$ – Daniel Fischer Jun 30 '14 at 17:32
  • $\begingroup$ so there is only just one appropriate domain for an operator where this operator is self-adjoint? In that case: Is it a non-trivial question to figure out the domain on which an operator is self-adjoint? Like, if you are looking at Sturm-Liouville problems with periodic boundary conditions. Do you coincidentially know the appropriate domain for such an operator? $\endgroup$ – user66906 Jun 30 '14 at 17:44
  • 1
    $\begingroup$ Yes, being self-adjoint is a strong condition with strong consequences. I don't know how (non-)trivial it is in general to find the domain of self-adjointness. In practice, most examples I've met were simple, the straightforward definably operator was essentially self-adjoint, so one just needed to find the adjoint (= closure, then) of the operator. I guess that abstractly, it is a hard problem. No idea about Sturm-Liouville, sorry. $\endgroup$ – Daniel Fischer Jun 30 '14 at 17:54

Your Answer

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