How to prove: $f \in C^1([0, \pi])$ with $f(0)= f(\pi) = 0$, then $\int_0^{\pi} |f|^2 dx \le \int_0^{\pi} |f'|^2 dx$. [duplicate]

Problem

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

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