$\|T\|= \sup_{\| f\|=1} |\langle Tf,f \rangle|$ for complex Hilbert spaces 
Let $H$ be a hilbert space over the complex numbers. Let $T:H\to H$ be a linear bounded operator prove that $$\| T\|= \sup_{\| f\|=1} |\langle Tf,f \rangle|.$$

Obviously $\sup_{\| f\|=1} |\langle Tf,f\rangle| \le \| T\|$ by C-S inequality. But how about the other inequality?

As user8268 points out, this is wrong even in complex hilbert spaces.
But there is at least the following substitute:
Define $B(u,v) := \langle Tu, v\rangle$. Then $B$ is sesquilinear (i.e. linear in the first argument and antiliear in the second). Define $Q(u) := B(u,u)$. The sesquilinearity of $B$ implies the polarization identity
$$B(u,v) = \frac{1}{4} \sum_{j=0}^3 i^j Q(u + i^j v),$$
where $i = \sqrt{-1}$.
We thus derive
$$|\langle Tu, v\rangle| = |B(u,v)| \leq \frac{1}{4} \sum_{j=0}^3 |Q(u + i^j v)|.$$
The right hand side can be estimated by $4 \cdot \sup_{\Vert f \Vert = 1}|\langle Tf, f\rangle|$, because $\Vert u + i^j v\Vert \leq 2$ if $\Vert u \Vert, \Vert v \Vert \leq 1$.
Then use $\Vert Tu \Vert = \sup_{\Vert v \Vert \leq 1} |\langle Tu, v\rangle|$.
