Norm of self-adjoint operator

It is clear that $|\langle Tx,x\rangle|\leq \|T\|$ for $\|x\|=1$. For the converse, it suffices to show that $|\langle Tx,y\rangle|\leq \alpha$ for all $\|x\|=\|y\|=1$, with $\alpha=\sup\bigl\{|\langle Tx,x\rangle|: \|x\|=1\bigr\}$. We can clearly assume $\langle Tx,y\rangle \in\mathbb R$. Then $$ \langle Tx,y\rangle = (\langle T(x+y),x+y\rangle - \langle T(x-y),x-y\rangle)/4. $$ But then $$ |\langle Tx,y\rangle|\leq\alpha(\|x+y\|^2+\|x-y\|^2)/4=\alpha, $$ by the parallelogram identity.

I have paraphrased this nice derivation from the book Essential Results of Functional Analysis, by Zimmer.