Q:

Calculation of operator norm

$H$ is a Hilbert space, $T: H \to H$ linear bounded operator, 
$||T||$ is the norm of $T$ given by
$$||T||=\sup\{||T(x)||;||x||\le 1 \}. $$
Is it true that
$$||T||=\sup\{|\langle Tx,y\rangle|;||x||\le 1 , ||y||\le 1\} ? $$

A:

Have you tried using Cauchy-Schwarz?
$$|\langle Tx, y \rangle | \ \leq \ ||Tx|| \, || y || \ \leq \ ||Tx||$$
Is there a $y$ for which this inequality is an equality?

