let $L>0$ and $f:[0, L] \rightarrow \mathbb{R}$ and odd function that satisfies $f(0)=0=f(L)$. also, $f^{\prime}$ is smooth and continously integrable in $[0, L]$. Also remember that: $$ b_n=\frac{2}{L} \int_0^L f(x) \sin \left(\frac{n \pi x}{L}\right) d x, n \geq 1 . $$
its the bn term of its fourier series
show that: $\sum_{n=1}^{\infty}\left|b_n\right|<+\infty$.
my attempt:
so, since f is odd we know that $a_n$ and $a_0$ are equal to zero, and so its fourier series is simply:
$f(x)=\sum_{n=1}^\infty b_{n}\sin\left(\dfrac{nx\pi}{L}\right)$
and by parsavel we have:
$\sum_{n=1}^{\infty} (b_n)^2 = \frac{2}{L} \int_0^L |f(x)|^2 \, dx$
so i thought, hey, maybe we can find that $(bn)^2$ with the fourier series (term by term), so we have:
$f(x)=\sum_{n=1}^\infty b_{n}\sin\left(\dfrac{nx\pi}{L}\right)$
and since f is smooth we can do its derivative as follows:
$f'(x) = \sum_{n=1}^{\infty} \frac{n\pi}{L} b_n \cos\left(\frac{n\pi x}{L}\right)$
and using the fact that ab is less or equal than (a^2 + b^2)/2:
$2f'(x) \leq \sum_{n=1}^{\infty} |b_n|^2 + \sum_{n=1}^{\infty} \left(\frac{n\pi}{L}\right)^2 \cos^2\left(\frac{n\pi x}{L}\right)$
ok, so up to here i am a bit stuck, I though that since |bn|^2 is still true I can bound it from either the left or right and that in conclusion will give me that $b_n$ is bounded and so finite, but then again I am a bit stuck. I am on the right track?