Linear Functional on Complex Vector Space is Uniquely Determined by Real Part
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Theorem
Let $X$ be a vector space over $\C$.
Let $f : X \to \C$ be a linear functional.
Define a function $g : X \to \R$:
- $\map g x = \map \Re {\map f x}$
for each $x \in X$.
Then:
- $\map f x = \map g x - i \map g {i x}$
for each $x \in X$.
Proof
For brevity, define a function $h : X \to \R$ by:
- $\map h x = \map \Im {\map f x}$
for each $x \in X$.
Note that:
- $\map f x = \map \Re {\map f x} + i \map \Im {\map f x} = \map g x + i \map h x$
so that:
- $\map f {i x} = \map g {i x} + i \map h {i x}$
for each $x \in X$.
On the other hand, by the linearity of $f$, we have:
\(\ds \map f {i x}\) | \(=\) | \(\ds i \map f x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds i \paren {\map g x + i \map h x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds i \map g x - \map h x\) |
for each $x \in X$.
Comparing real parts, we have:
- $-\map h x = \map g {i x}$
so $\map h x = -\map g {i x}$.
So we have:
- $\map f x = \map g x - i \map g {i x}$
for each $x \in X$.
$\blacksquare$