Wirtinger’s inequality

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 \bigintsss_0^{2\pi} f(x) \, {\rm d}x =0 . Prove that:

\displaystyle \int_{0}^{2\pi}\left ( f' (x)\right )^2\,{\rm d}x\geq \int_{0}^{2\pi}f^2(x)\,{\rm d}x

Solution

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:

\displaystyle 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 have 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 )

as well as

\displaystyle \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. Equality holds if and only if  a_n = b_n =0 .

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