This is a problem our teacher gave us and I have a feeling he forgot to mention some additional data.
Let $H$ be a Hilbert space and $T : H \to H$ a everywhere-defined linear operator such that $T$ is self-adjoint, i.e. $\forall x,y \in H : \langle Tx,y\rangle=\langle x,Ty\rangle$.
Show that $T$ is bounded!
Using Cauchy-Schwarz inequality it is clear that if $T$ is idempotent then it's bounded, but with just the information initially provided I don't feel like it's possible to prove the boundedness.
Can somebody shed some light on this? Perhaps with a counter-example?
This follows from the Uniform Boundedness Principle. $\newcommand{\ip}[2]{\langle #1,#2\rangle}$ Say $B$ is the closed unit ball of $H$. For $x\in H$ define $\Lambda_x:H\to\Bbb C$ by $$\Lambda_x y=\ip{Tx}y.$$Note that each $\Lambda_x$ is bounded, in fact with norm $\|Tx\|$. For every $y\in H$ we have $$\sup_{x\in B}|\Lambda_xy|=\sup_{x\in B}|\ip x{Ty}|=\|Ty\|<\infty.$$So Uniform Boundedness implies that $$\sup_{x\in B}\|Tx\|=\sup_{x\in B}||\Lambda_x||<\infty,$$hence $T$ is bounded.