# Self adjoint operators on Hilbert spaces are bounded

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.

• Unbounded self adjoint operators are partially defined only. – user251257 Apr 19 '16 at 2:10
• Meaning not defined on all of $H$? – Rick Sanchez Apr 19 '16 at 2:12
• Yes. ${}{}{}{}$ – user251257 Apr 19 '16 at 2:13
• 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. – user244643 Apr 19 '16 at 2:22
• 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. – DisintegratingByParts Apr 19 '16 at 5:46