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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac \lambda 2 \paren {\map f x + \overline {\map f x} } + \frac \mu 2 \paren {\map f y + \overline {\map f y} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lambda \map g x + \mu \map g y\) |
$\blacksquare$