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 \( \displaystyle a,b \in \mathbb{R} \).
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About a $2\pi$ periodical function

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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.
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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