Proposition 11.2.7.label Let $E$ be a normed vector space over $\complex$, $*: E \to E$ be a complex conjugation map such that $\norm{x}_{E} = \normn{x^*}_{E}$ for all $x \in E$, and $\phi \in E^{*}$ be a Hermitian functional, then
\[\norm{\phi}_{E^*}= \sup\bracsn{\dpn{x, \phi}{E}|x \in E, x = x^*, \norm{x}_E = 1}\]
Proof. Since $\bracsn{x \in E|x = x^*}\subset E$, $\norm{\phi}_{E^*}\ge \sup\bracsn{\dpn{x, \phi}{E}|x \in E, x = x^*, \norm{x}_E = 1}$.
On the other hand, let $x \in E$ with $\norm{x}_{E} = 1$. Assume without loss of generality that $\dpn{x, \phi}{E}\in \real$, then
\begin{align*}\dpn{x, \phi}{E}&= \dpn{\text{Re}(x), \phi}{E}+ \underbrace{i\dpn{\text{Im}(x), \phi}{E}}_{\in \real}= \dpn{\text{Re}(x), \phi}{E}\\&\le \norm{\text{Re}(x)}_{E} \cdot \sup\bracsn{\dpn{y, \phi}{E}|y \in E, y = y^*, \norm{y}_E = 1}\end{align*}
where $\norm{\text{Re}(x)}_{E} = \norm{{(x + x^*)}/{2}}_{E} \le \norm{x}_{E}$. As the above holds for all $x \in E$,
\[\norm{\phi}_{E^*}\le \sup\bracsn{\dpn{x, \phi}{E}|x \in E, x = x^*, \norm{x}_E = 1}\]
$\square$
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