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For any $f \in L^2 (-\pi, \pi)$, prove that there exists unique orthonormal decomposition with even functions and odd functions : $ L^2 ( -\pi , \pi) = L^2 _{odd} (-\pi , \pi ) \oplus L^2_{even} (-\pi , \pi).$

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You can also think this way. Suppose the $f = g + h$, $g$ is even and $h$ is odd. Then you have

$f(x) = g(x) + h(x)$

and

$f(-x) = g(-x) + h(-x) = g(x) - h(x)),$ since $g$ is even and $h$ is odd. Add these two equations and you get $f(x) + f(-x) = 2g(x)$ and $ f(x) - f(-x) = 2h(x).$ Hence you have $g(x) = {f(x) + f(-x)\over 2} $ and $h(x) = {f(x) - f(-x)\over 2}.$ It is easy to check that $g$ is in fact even and that $f$ is in fact odd.

We see that, algebraically, $L^2$ decomposes as a direct sum into even and odd parts. Now if $g$ is any even function and $f$ is odd, $f\overline g$ is odd. So $\langle f, g\rangle_{L^2} = \int_{-\pi}^{\pi} f(x){\overline g}(x)\, dx = 0.$

The decomposition is orthogonal.
This is where Norbert's answer comes from.

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    Notice that uniqueness is built into this argument.2012-07-26
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Hint: For a given $f\in L^2((-\pi,\pi))$ consider functions $ \frac{f(x)+f(-x)}{2}\qquad\text{ and }\qquad \frac{f(x)-f(-x)}{2} $