Prove rayleigh quotient = operator norm without referring to eigenvalues Let $H$ be a Hilbert and $T \in \mathcal{L}(H,H)$ a symmetric operator. Prove 
$$
\|T\| 
=
\sup\{|(x,Tx)| : x \in H, \|x\| = 1\}
$$
without referring to the eigenvalues of $T$ (which is what all proofs I could find do). 

For convenience, let us set 
$$
M := \sup\{|(x,Tx)| : x \in H, \|x\| = 1\}
.
$$
$M \leq \|T\|$ follows straightforwardly from the Cauchy-Schwarz inequality and the definition of the operator norm, thus it remains to show $M \geq \|T\|$. 
Let $u,v \in H$ be arbitrary vectors. Then,
\begin{align}
4 \,\mathrm{Re} (u,Tv)
&=
(u+v,T(u+v)) - (u-v,T(u-v))
\\&\leq
M ( \|u+v\|^2 + \|u-v\|^2 )
\\ &=
2M(\|u\|^2 + \|v\|^2)
\end{align}
where the second line follows from Cauchy-Schwarz and the third from the parallelogram law. Now assume $\|Tu\| \neq 0$ and set $u = \frac{\|v\|}{\|Tv\|} Tv$. Inserting in the above yields
$$
4\|v\| \|Tv\| \leq 4 M \|v\|^2.
$$
This estimate is trivially satisfied if $\|Tv\| = 0$. We thus have $\|Tv\| \leq M$ for all $v \in H$ with $\|v\| = 1$, which implies $\|T\| \leq M$. 
