I'm given that the function $f(x)$ is smooth and $2\pi$ periodic on $-\pi \le x \le \pi$.
I want to use Parseval's theorem to find: $$\frac{1}{2\pi} \int_{-\pi}^{\pi}{f'(x)^2} dx$$
In terms of $$||f|| = \int_{-\pi}^{\pi}{f(x)^2}dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty}(a_n^2+b_n^2)$$
My attempt:
Let $f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}{a_n cos(nx)+ b_n sin(nx)}$
Then differentiating:
$f'(x) = \sum_{n=1}^{\infty}{cos(nx)(nb_n)+sin(nx)(-na_n)}$
Applying Parseval's theorem for the first integral yields:
$\sum_{n=1}^{\infty}{(nb_n)^2}+(na_n)^2$
How do I express this in terms of $||f||$?