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I need some help with this question about Fourier Series.

1) If $f\in{L_{1}(T)}$ (that's $f$ periodic with period $2\pi$ and $|f|\in{L_{1}([-\pi,\pi]}$)) with Fourier series $\frac{a_0}{2}+\sum_{k=1}^{\infty}a_k cos(kt)+b_k sin(kt)$, then $f$ is even if and only if $b_k=0$ for all $k$, and $f$ is odd if and only if $a_k=0$ for all k.

I have no problem proving the "left to right" implications, just with simple integration. But i can't prove the "right to left" ones. ¿Can anyone help me at this?

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One should think of $f$ as an element of $L^1$, not as a pointwise-defined function (this avoids the messy things like "even almost everywhere"). Let $g(x)=f(-x)$. The Fourier series of $f-g$ is $2\sum b_k\sin kt$. If $f$ is not even, then $f-g$ is a nonzero element of $L^1$, therefore some $b_k$ is nonzero. (The fact that the only function with zero Fourier series is zero follows from the density of trigonometric polynomials in $L^1$).

Similarly, the Fourier series of $f+g$ is $a_0+2\sum a_k\cos kt$. If $f$ is not odd, then $f+g$ is a nonzero element of $L^1$, therefore some $a_k$ is nonzero.

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Just note that

$$ a_k = \int_{-\pi}^{\pi} f(x)\cos(kx)dx. $$

Now, if $f(x)$ is an odd function then the integrand is an odd function and this implies that the value of the above integral is zero. The same with the other case.

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Yes, I understand that implication. The doubt I have is how to prove the reverse ones. That is, if $b_k=0$, then $f$ is even; and if $a_k=0$ then $f$ is odd. – Mark_Hoffman Feb 19 '13 at 18:07

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