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About a $2\pi$ periodical function

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

#1

Post by Tsakanikas Nickos » Thu Jul 14, 2016 6:56 pm

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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Tolaso J Kos
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Re: About a $2\pi$ periodical function

#2

Post by Tolaso J Kos » Thu Jul 14, 2016 6:57 pm

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.
Hidden Message
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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