Odd Function Times Odd Function is Even
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Theorem
Let $S \subset \R$ be a symmetric set of real numbers:
- $\forall x \in S: -x \in X$
Let $f, g: X \to \R$ be two odd functions.
Let $f \cdot g$ denote the pointwise product of $f$ and $g$.
Then $\paren {f \cdot g}: S \to \R$ is an even function.
Proof
\(\ds \map {\paren {f \cdot g} } {-x}\) | \(=\) | \(\ds \map f {-x} \cdot \map g {-x}\) | Definition of Pointwise Multiplication of Real-Valued Functions | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-\map f x} \cdot \paren {-\map g x}\) | Definition of Odd Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \map f x \cdot \map g x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {f \cdot g} } x\) | Definition of Pointwise Multiplication of Real-Valued Functions |
Thus, by definition, $\paren {f \cdot g}$ is an even function.
$\blacksquare$