Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a $2\pi$ periodic function which is continuous and has a continuous derivative throughout $\mathbb{R}$ such that $\displaystyle \int_{0}^{2\pi}f(x)\,{\rm d}x=0$ . Prove that:
$$\int_{0}^{2\pi}\left ( f' (x)\right )^2\,{\rm d}x\geq \int_{0}^{2\pi}f^2(x)\,{\rm d}x$$
(This version of the Wirtinger inequality is the one-dimensional Poincare Inequality, with optimal constant)
Proof
We are basing the whole fact on Fourier series. Since Dirichlet's conditions are met then $f$ can be expanded into a Fourier series. Therefore we can write:
$$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left [ a_n \sin nx+b_n \cos nx \right ]$$
However, since the integral of $f$ vanishes we get that $a_0=0$. Using Parseval's identity we get that $\displaystyle \int_{0}^{2\pi}f^2(x)\,{\rm d}x=\sum_{n=1}^{\infty}\left ( a_n^2+b_n^2 \right ) $ and $$\int_{0}^{2\pi}\left ( f'(x) \right )^2\,{\rm d}x=\sum_{n=1}^{\infty}n^2\left ( a_n^2+b_n^2 \right )$$
Since all summands are positive we get the desired inequality. The equality holds if $a_n=b_n=0, \; n \geq 2$.
$$\int_{0}^{2\pi}\left ( f' (x)\right )^2\,{\rm d}x\geq \int_{0}^{2\pi}f^2(x)\,{\rm d}x$$
(This version of the Wirtinger inequality is the one-dimensional Poincare Inequality, with optimal constant)
Proof
We are basing the whole fact on Fourier series. Since Dirichlet's conditions are met then $f$ can be expanded into a Fourier series. Therefore we can write:
$$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left [ a_n \sin nx+b_n \cos nx \right ]$$
However, since the integral of $f$ vanishes we get that $a_0=0$. Using Parseval's identity we get that $\displaystyle \int_{0}^{2\pi}f^2(x)\,{\rm d}x=\sum_{n=1}^{\infty}\left ( a_n^2+b_n^2 \right ) $ and $$\int_{0}^{2\pi}\left ( f'(x) \right )^2\,{\rm d}x=\sum_{n=1}^{\infty}n^2\left ( a_n^2+b_n^2 \right )$$
Since all summands are positive we get the desired inequality. The equality holds if $a_n=b_n=0, \; n \geq 2$.
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