1
$\begingroup$

I think I have a proof that if $A: H\rightarrow H$ is a self adjoint operator on a Hilbert space $H$, then $A$ is bounded:

We can use the closed graph theorem. Let $x_n \rightarrow x$ and $Ax_n \rightarrow y$. For every $z \in H$, $$\langle z,Ax -y\rangle=\langle z, Ax\rangle -\langle z,y \rangle=\langle Az,x\rangle-\langle z,y\rangle$$ $$=\lim_n \langle Az,x_n\rangle -\lim_n \langle z,Ax_n \rangle=0,$$ by self adjointness. Thus, $Ax=y$.

First of all, is this correct? Second, if it is, then why is there a GTM text titled "Unbounded Self Adjoint Operators on Hilbert Space"? I think it might have to do with the fact that I am assuming here that $A$ is defined on all of $H$, not just on some subspace.

Thanks for the help.

$\endgroup$
  • $\begingroup$ Unbounded self adjoint operators are partially defined only. $\endgroup$ – user251257 Apr 19 '16 at 2:10
  • $\begingroup$ Meaning not defined on all of $H$? $\endgroup$ – Rick Sanchez Apr 19 '16 at 2:12
  • $\begingroup$ Yes. ${}{}{}{}$ $\endgroup$ – user251257 Apr 19 '16 at 2:13
  • $\begingroup$ An added note, if an operator $A$ is densely defined, we get that $A^*$ is closed and since we have that the operator is self adjoint, we have $A^*=A$ everywhere $A$ is defined. Usually this is written $A\subseteq A^*$. We call $A^*$ the closure of $A$ in this case. $\endgroup$ – user244643 Apr 19 '16 at 2:22
  • $\begingroup$ Yes, you are correct. If $A$ is symmetric and defined everywhere, then it is closed and, hence, bounded. So an unbounded $A$ cannot be defined everywhere; the best you can hope for is to be densely-defined (meaning that the domain is dense in $H$,) which is the typical assumption. $\endgroup$ – DisintegratingByParts Apr 19 '16 at 5:46

Your Answer

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

Browse other questions tagged or ask your own question.