Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers 2

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.

share|improve this answer
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.