A homework problem:
Let $H$ be a Hilbert space. Let $T:H\rightarrow H$ be a symmetric linear map ($\langle Tx,y\rangle=\langle x,Ty\rangle$).
Show that $S$ is bounded.
My attempt: I'd like to use the closed graph theorem. I take $(x_n)\subset H$ and assume $x_n \rightarrow x$ and $Tx_n\rightarrow y$. I'd like to show $Tx=y$. So I calculate: $\|Tx_n-Tx\|^2=\|T(x_n-x)\|^2=|\langle T(x_n-x), T(x_n-x)\rangle|=$ $|\langle x_n-x, T(T(x_n-x))\rangle|\leq \|x_n-x\|\cdot \|T(T(x_n-x))\|$.
So, it's enough to show that $\|T(T(x_n-x))\|$ is bounded. The fact that $T(x_n)$ converges tells me that $\|T(x_n-x)\|$ is bounded, but I don't know what about $\|T(T(x_n-x))\|$.