I'm having trouble understanding a technical argument of a proof similar to Theorem C.51,p 4 from this link, where the statement to prove is: $\|T\|=\sup\limits_{\|x\|\le 1}|<Tx,x>|$ for a bounded linear self-adjoint operator on a hilbert space $T\in L(H)$.
In a sketch it is defined that $M:=\sup\limits_{\|x\|\le 1}|<Tx,x>|$ and it shall be proved that $\|T\| \le M$. The line I don't understand is: $$4 Re(<Tx,x>)\le |<T(x+y),x+y>|+|<T(x-y),x-y>|\le M\|x+y\|^2+M\|x-y\|^2=M(\|x\|^2+\|y\|^2)$$
My question boils down to the second estimate where it should hold that $|<Tz,z>|\le \sup\limits_{\|u\|\le1 }|<Tu,u>|\|z\|^2 = M\|z\|^2$ for $z\in H$, $\|z\|\le 1$ ( or even for any $z\in H$?).
I tought about using the estimate $|<Tz,z>|\le \|Tz\|\|z\|\le \|T\|\|z\|^2$ but since $\|T\|\le M$ is what should be proved I can't use it. So I'm stuck here and don't see how to prove the estimate in question?