Real Part of Linear Functional is Linear Functional

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Theorem

Let $X$ be a vector space over $\C$.

Let $f : X \to \C$ be a linear functional.

Define $g : X \to \R$ by:

$\map g x = \map \Re {\map f x}$

for each $x \in X$.


Then $f$ is $\R$-linear.


Proof

Let $x, y \in X$ and $\lambda, \mu \in \R$.

Then:

\(\ds \map g {\lambda x + \mu y}\) \(=\) \(\ds \map \Re {\map f {\lambda x + \mu y} }\)
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\map f {\lambda x + \mu y} + \overline {\map f {\lambda x + \mu y} } }\) Sum of Complex Number with Conjugate
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\lambda \map f x + \mu \map f y + \overline {\lambda \map f x} + \overline {\mu \map f y} }\) Sum of Complex Conjugates, linearity of $f$
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\lambda \map f x + \mu \map f y + \lambda \overline {\map f x} + \mu \overline {\map f y} }\) Product of Complex Conjugates
\(\ds \) \(=\) \(\ds \frac \lambda 2 \paren {\map f x + \overline {\map f x} } + \frac \mu 2 \paren {\map f y + \overline {\map f y} }\)
\(\ds \) \(=\) \(\ds \lambda \map g x + \mu \map g y\)

$\blacksquare$