# Self adjoint map are bounded

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

• 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 $x$. – Brent Kerby Feb 27 '15 at 23:51
• $J_x$ is a typo$– amathnerd Feb 28 '15 at 0:00 • How are you getting$\|Tx\|=|\langle Tx,y\rangle|$? – Brent Kerby Feb 28 '15 at 0:07 • Look up Hellinger–Toeplitz theorem. – Robert Israel Feb 28 '15 at 0:10 • As written, it still doesn't make sense; for example,$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. – Brent Kerby Feb 28 '15 at 0:25