Proposition 12.1.5.label Let $E$ be a normed vector space, then for any $x \in E$,
\[\norm{x}_{E} = \sup_{\substack{\phi \in E^* \\ \norm{\phi}_{E^*} = 1}}\dpn{x, \phi}{E}\]
Proof. For any $\phi \in E^{*}$ with $\norm{\phi}_{E^*}= 1$, $\dpn{x, \phi}{E}\le \norm{x}_{E} \cdot \norm{\phi}_{E^*}= \norm{x}_{E}$. On the other hand, by the Hahn-Banach Theorem, there exists $x \in E^{*}$ such that $\dpn{x, \phi}{E}= \norm{x}_{E}$ and $\norm{\phi}_{E^*}\le 1$.$\square$