Odd Function Times Odd Function is Even

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Theorem

Let $X \subset \R$ be a symmetric set of real numbers:

$\forall x \in X: -x \in X$

Let $f, g: X \to \R$ be two odd functions.

Let $f \cdot g$ denote the pointwise product of $f$ and $g$.


Then $\left({f \cdot g}\right): X \to \R$ is an even function.


Proof

\(\displaystyle \left({f \cdot g}\right) \left({-x}\right)\) \(=\) \(\displaystyle f \left({-x}\right) \cdot g \left({-x}\right)\) Definition of Pointwise Multiplication
\(\displaystyle \) \(=\) \(\displaystyle \left({-f \left({x}\right)}\right) \cdot \left({-g \left({x}\right)}\right)\) Definition of Odd Function
\(\displaystyle \) \(=\) \(\displaystyle f \left({x}\right) \cdot g \left({x}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({f \cdot g}\right) \left({x}\right)\) Definition of Pointwise Multiplication

Thus, by definition, $\left({f \cdot g}\right)$ is an even function.

$\blacksquare$


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