TITLE: $\|S\|=\sup\{|\langle Sx\;,\;y\rangle |;\;\|x\| \leq 1,\, \|y\| \leq 1\}\,?$
QUESTION [9 upvotes]: Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$. Let $S\in \mathcal{B}(F)$. The norm of $S$ is defined us
$$\|S\|:=\sup_{\substack{x\in F\\ x\not=0}}\frac{\|Sx\|}{\|x\|}$$
Why
$$\|S\|=\sup_{\substack{\|x\| \leq 1,\\ \|y\| \leq  1}}|\langle Sx\;,\;y\rangle |?$$
Thanks.

REPLY [6 votes]: For any $x, y \in X$ with $\|x\|, \|y\| \le 1$ we have
$$|\langle Sx, y\rangle| \le \|Sx\|\|y\| \le \|S\|\|x\|\|y\| \le \|S\|$$
Now, for any $x \in X$  with $Sx \ne 0$ we have:
$$\left|\left\langle Sx, \frac{Sx}{\|Sx\|}\right\rangle\right| = \frac{\|Sx\|^2}{\|Sx\|} = \|Sx\|$$
So taking the supremum over $\|x\| \le 1$ gives:
$$\sup_{\|x\|, \|y\| \le 1}|\langle Sx, y\rangle| \ge \sup_{\|x\| \le 1} \left|\left\langle Sx, \frac{Sx}{\|Sx\|}\right\rangle\right| = \sup_{\|x\| \le 1} \|Sx\| = \|S\|$$
We conclude $\displaystyle \sup_{\|x\|, \|y\| \le 1}|\langle Sx, y\rangle| = \|S\|$.