why $\|Sx\|=\sup_{\|y\| \le 1} |\langle Sx, y \rangle|,\;\forall x\in E?$

Question

Let $E$ be complex Hilbert space.

If $S\in \mathcal{L}(E)$, why $$\|Sx\|=\sup_{\|y\| \le 1} |\langle Sx, y \rangle|,\;\forall x\in E?$$

Answer 1

This is because of the Cauchy-Schwartz inequality: For $\|y\|\leq 1$ we have $$ |\langle Sx, y \rangle| \leq \sqrt{|\langle Sx, Sx \rangle|}\sqrt{|\langle y, y \rangle|} = \|Sx\| \|y\| \leq \|Sx\|. $$ On the other hand, if we choose $y=\frac{1}{\|Sx\|}Sx$, then clearly $\|y\|=1$ and $$ |\langle Sx, y \rangle| = \frac{1}{\|Sx\|}|\langle Sx, Sx \rangle| = \frac{1}{\|Sx\|}\|Sx\|^2 = \|Sx\|. $$