Proposition 10.4.1 (Polarisation).label Let $E$ be a vector space over $\complex$, $\phi \in \hom(E; \complex)$, and $u = \re{\phi}$, then
- (1)
$u \in \hom(E; \real)$ when $E$ is viewed as a vector space over $\real$.
- (2)
For any $x \in E$, $\dpb{x, \phi}{E}= \dpb{x, u}{E}- i \dpb{ix, u}{E}$.
Conversely, if $u \in \hom(E; \real)$ and $\phi \in \hom(E; \complex)$ is defined by $\dpb{x, \phi}{E}= \dpb{x, u}{E}- i \dpb{ix, u}{E}$ for all $x \in E$, then $f \in \hom(E; \complex)$.
Proof [Proposition 5.5, Fol99]. (1): Given that $\phi \in \hom(E; \real)$, $u \in \hom(E; \real)$. For any $x \in E$,
\[\im{\dpb{x, \phi}{E}}= \re{-i \dpb{x, \phi}{E}}= -\dpb{ix, u}{E}\]
so (2) holds.
(2): Since $u \in \hom(E; \real)$, so is $\phi$. On the other hand, for any $x \in E$,
\[\dpb{ix, \phi}{E}= \dpb{ix, u}{E}- i\dpb{-x, u}{E}= i(\dpb{x, u}{E}- i \dpb{ix, u}{E}) = i \dpb{x, \phi}{E}\]
$\square$