TITLE: Is every self-adjoint operator bounded?
QUESTION [9 upvotes]: 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?

REPLY [2 votes]: The operator you has mentioned 
$T : H \to H$ a linear operator such that $\forall x,y \in H \; \langle\,Tx,y\rangle=\langle\,x,Ty\rangle $
is an everywhere-defined symmetric operator on a Hilbert space so it is self-adjoint and by Hellinger–Toeplitz theorem  
it is a bounded operator, but this is not always the situation because there are many self-adjoint operators defined in a dense subspace of $H$ and they are not-bounded.
Note: The  self-adjoint operators defined in a dense subspace of $H$, we cannot extend them in an everywhere-defined symmetric operator, we can get only extension for them which is not symmetric
 the following example is a self-adjoint operator but not bounded.
Let $\Omega$ be any open subset of $ \mathbb{R}^n$ and $(T, D)$ such that $H= L^2(\Omega)$ and 
$$T= -\Delta \quad   and  \quad D=\{u\in L^2(\Omega) , \Delta u \in L^2(\Omega)\}$$