Proposition 4.4.2 (Polarisation).label Let $E$ be a vector space over $\com$ then the map
\begin{align*}\hom{\com}{E}{\com}\to \hom{\R}{E}{\R}&&\phi\mto \Re(\phi)=:u\end{align*}
is a bijection whose inverse is
\begin{align*}\hom{\R}{E}{\R}\to \hom{\com}{E}{\com}&&u\mto (x\mto \inner{x}{u}_{E}-i\inner{ix}{u}_{E})=:\phi\end{align*}
If $E$ is normed then $\norm{\phi}{}=\norm{\Re(\phi)}{}$
Proof. Let $\phi\in \hom{\com}{E}{\com},x,y\in E,\lam\in \R$ then
\begin{align*}\Re(\inner{\lam x+y}{\phi}_{E})&=\Re(\lam \inner{x}{\phi}_{E}+\inner{y}{\phi}_{E})=\lam \Re(\inner{x}{\phi}_{E})+\Re(\inner{y}{\phi}_{E})\\ \Im(\inner{x}{\phi}_{E})&=\Re(-i\inner{x}{\phi}_{E})=-\Re(\inner{ix}{\phi}_{E})=-\inner{ix}{\Re(\phi)}_{E}\end{align*}
showing $\Re(\phi)\in \hom{\R}{E}{\R}$ and $\inner{x}{\phi}_{E}=\inner{x}{u}_{E}-i\inner{ix}{u}_{E}$.
Conversely let $u\in\hom{\R}{E}{\R},x,y\in E,\lam\in \com$ then
\begin{align*}\inner{\lam x+y}{\phi}_{E}&=\inner{\lam x+y}{u}_{E}-i\inner{i(\lam x+y)}{u}_{E}\\&=\lam(\inner{x}{u}_{E}-i\inner{ix}{u}_{E})+(\inner{y}{u}_{E}-i\inner{iy}{u}_{E})\\&=\lam\inner{x}{\phi}_{E}+\inner{y}{\phi}_{E}\end{align*}
In particular taking $y=0, \lam=i$ we have
\begin{align*}\inner{ix}{\phi}_{E}=\inner{ix}{u}_{E}-i\inner{i(ix)}{u}_{E}=\inner{ix}{u}_{E}-i\inner{-x}{u}_{E}=i\inner{x}{\phi}_{E}\end{align*}
Since $\abs{\Re(\inner{x}{\phi}_E)}\leq \abs{\inner{x}{\phi}_E}$ for any $x\in E$ we have $\norm{\phi}{}=\norm{u}{}$. On the other hand, if $\inner{x}{\phi}_{E}\neq0$ then let $\al=\cl{\text{sgn}\inner{x}{\phi}_E}$. Since
\begin{align*}\abs{\inner{x}{\phi}_E}=\al\inner{x}{\phi}_{E}=\inner{\al x}{\phi}_{E}=\inner{\al x}{u}_{E}\leq \norm{u}{}\norm{\al x}{}=\norm{u}{}\norm{x}{}\end{align*}
whence $\norm{\phi}{}\leq\norm{u}{}$.$\square$
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