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Prove that if $f \in C^1([0, \pi])$ with $f(0)=f(\pi) = 0$, then

$$ \int_0^\pi |f|^2 dx \le \int_0^\pi |f'|^2 dx.$$

I want to prove it using Fourier sine series of $f$. Let $f \sim \sum_{n=1}^\infty A_n \sin nx$ be Fourier sine series of $f$. (obtained over $(-\pi, \pi).$) By Parseval's theorem, $ \int_0^\pi |f|^2 dx = \frac{1}{\pi} \sum_{n=1}^\infty |A_n|^2$. If $\sum_{n=1}^\infty |A_n| < \infty$, series converges uniformly so $$f' = \sum_{n=1}^\infty n A_n \cos nx.$$ Thus, $\int_0^\pi |f'|^2 dx = \frac{1}{\pi}\sum_{n=1}^\infty |A_n|^2 n^2$ so the inequality holds. However, there is no assumption that $\sum_{n=1}^\infty |A_n| < \infty$. My question is that whether the series of $f$ converges uniformly. Is the equation $f' = \sum_{n=1}^\infty n A_n \cos nx$ valid? At least, I want to know that whether applying Parseval's theorem to conclude $\int_0^\pi |f'|^2 dx = \frac{1}{\pi} \sum_{n=1}^\infty n^2|A_n|^2$ is possible.


marked as duplicate by HAMIDINE SOUMARE, Community Apr 28 at 16:39

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    $\begingroup$ $f'\in L^2$ means that $\sum n^2|A_n|^2 < \infty $, so $\sum |A_n| = \sum n|A_n| \frac 1n \le \sqrt{\sum n^2 |A_n|^2 \sum \frac1{n^2}} < \infty $ $\endgroup$ – Calvin Khor Apr 28 at 16:32