If $H$ is a Hilbert space and $T:H\rightarrow H$ that is self-adjoint. How to prove that $T$ is bounded operator.

Does the following argument work?

For all $x,y\in \mathcal{H}$ $$\|Tx\|=\sup_{||y||=1}|\langle Tx,y\rangle|\leq \|Ty\|\cdot \|x\|$$ for any fixed $x$. So $\sup_y\|T\|<\infty$, UBP implies that there is a constant $c$. $||Tx||\leq c||x||$?

Any feedback? Thanks

  • $\begingroup$ This doesn't seem to make sense. You haven't defined $y$, $J_x$, or $c$, and the expression $\sup_x\|T\|$ is probably not what was meant since $\|T\|$ does not depend on $x$. $\endgroup$ – Brent Kerby Feb 27 '15 at 23:51
  • $\begingroup$ $J_x$ is a typo$ $\endgroup$ – amathnerd Feb 28 '15 at 0:00
  • $\begingroup$ How are you getting $\|Tx\|=|\langle Tx,y\rangle|$? $\endgroup$ – Brent Kerby Feb 28 '15 at 0:07
  • $\begingroup$ Look up Hellinger–Toeplitz theorem. $\endgroup$ – Robert Israel Feb 28 '15 at 0:10
  • 1
    $\begingroup$ As written, it still doesn't make sense; for example, $y$ appears free in $\|Ty\|\cdot\|x\|$ and yet it is bound in the sup on the other side of the inequality. I think you want to define a collection of linear functionals $T_y$ by $T_y(x)=\langle Tx, y\rangle$ and apply UBP to those. $\endgroup$ – Brent Kerby Feb 28 '15 at 0:25

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.