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PostPosted: Thu Jul 14, 2016 6:56 pm 
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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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PostPosted: Thu Jul 14, 2016 6:57 pm 
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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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This inequality is known as Wirtinger's Inequality and more specifically as Poincare's Little Inequality. There are also many generalizations of this in \( \mathbb{R}^n \) and not only. The generalization in \( \mathbb{R}^n \) is known as Sobolev's inequality.

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