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## About a $2\pi$ periodical function

Real & Complex Analysis, Calculus & Multivariate Calculus, Functional Analysis,
Tsakanikas Nickos
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### About a $2\pi$ periodical function

Let $\displaystyle f$ be a $\displaystyle C^1 , 2\pi$-periodical function. If $\displaystyle \int_{0}^{2\pi}f(x)\mathrm{d}x = 0$show that

$\displaystyle \int_{0}^{2\pi} \left( f^{\prime}(x) \right)^{2} \mathrm{d}x \geq \int_{0}^{2\pi} \left( f(x) \right)^{2} \mathrm{d}x$

and the equality holds if and only if $\displaystyle f(x) = a\cos(x) +b\sin(x)$ for some constants $\displaystyle a,b \in \mathbb{R}$.
Tolaso J Kos Articles: 2
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### Re: About a $2\pi$ periodical function

Good morning Nickos,

We are basing the whole fact on Fourier series. Since all Dirichlet's conditions are met ,then $f$ can be expanded into a Fourier series. Therefore we can write it as:

$$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left [ a_n \sin nx+b_n \cos nx \right ]$$

However, since the integral of $f$ vanishes , so does $a_0$. Applying Pasheval's identity we get:

$\bullet \;\;\; \displaystyle \int_{0}^{2\pi}f^2(x)\,{\rm d}x=\sum_{n=1}^{\infty}\left ( a_n^2+b_n^2 \right )$ and

$\bullet \;\;\; \displaystyle \int_{0}^{2\pi}\left ( f'(x) \right )^2\,{\rm d}x=\sum_{n=1}^{\infty}n^2\left ( a_n^2+b_n^2 \right )$

Finally , since all summands are positive we get the desired inequality and the exercise comes to and end.
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