0
$\begingroup$

Let $A$ be a densely defined symmetric operator on some Hilbert space $\mathcal{H}$. Let's say that we know that there exists $(A + 1)^{-1}$ and that this is a self-adjoint operator.

How can I prove that $A$ must be self-adjoint?

Any help will be very much appreciated!

$\endgroup$
5
  • $\begingroup$ How do you know it is a "simple question"? $\endgroup$ Sep 15, 2018 at 14:41
  • $\begingroup$ I suspect it should not be too hard, but I haven't done any functional analysis in a very long time, so it's just a guess :) $\endgroup$
    – Onil90
    Sep 15, 2018 at 14:44
  • $\begingroup$ You suspect it should not be too hard, yet you have no clue. So "hard" is quite subjective. $\endgroup$ Sep 15, 2018 at 14:45
  • $\begingroup$ Of course it is. Ok, let me edit the title. $\endgroup$
    – Onil90
    Sep 15, 2018 at 14:59
  • $\begingroup$ Are you assuming that $-1$ is in the resolvent set, meaning that $(A+I)^{-1}$ is defined on the full space, and is bounded? If so, then $A$ is selfadjoint. $\endgroup$ Sep 16, 2018 at 18:02

1 Answer 1

0
$\begingroup$

Let $y \in \mathcal{D}(A^*)$ so that $$ \langle Ax,y\rangle = \langle x,A^*y\rangle,\;\; \forall x\in\mathcal{D}(A),\\ \langle (A+I)x,y\rangle = \langle x,(A^*+I)y\rangle,\;\;\forall x\in\mathcal{D}(A). $$ All you need to assume is that $A$ is densely-defined, $A$ is symmetric, and $A+I$ is surjective. If $A+I$ is surjective, then $(A^*+I)y=(A+I)z$ for some $z\in\mathcal{D}(A)$. This gives $$ \langle (A+I)x,y\rangle = \langle x,(A+I)z\rangle = \langle (A+I)x,z\rangle,\;\;\forall x\in\mathcal{D}(A). $$ Because $A+I$ is surjective, then $y=z$, which implies that $y\in\mathcal{D}(A)$. Hence $\mathcal{D}(A^*)\subseteq\mathcal{D}(A)$. So $\mathcal{D}(A)=\mathcal{D}(A^*)$ and $A=A^*$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .