Regarding Wirtinger's inequality

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SUMMARY

The discussion centers on Wirtinger's inequality, specifically proving that for a continuous function \( f \) defined on \( (0, \pi) \) with boundary conditions \( f(0) = f(\pi) = 0 \) and \( f' \in L^2([0, \pi]) \), the inequality \( \int_0^{\pi}(f(x))^2 dx \leq \int_0^{\pi} (f'(x))^2 dx \) holds. The equality condition is established for functions of the form \( f(x) = C \sin x \). The participants clarify that the square integrability of \( f' \) is essential for the validity of the inequality and for the coefficients in the Fourier series representation to be well-defined.

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psie
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TL;DR
I have a doubt about about an assumption in Wirtinger's inequality.
Consider the following problem:
Let ##f## be a continuous real-valued function on ##0<x<\pi## such that ##f(0)=f(\pi)=0## and ##f' \in L^2([0,\pi])##.
a) Prove that $$\int_0^{\pi}(f(x))^2 dx \leq \int_0^{\pi} (f'(x))^2 dx.$$
b) For what functions does equality hold?
What I struggle with in this exercise is why ##f'## is in ##L^2([0,\pi])##. Does this make sense? The way I prove this inequality is that I extend ##f## to an odd function on ##[-\pi,\pi]##. I find its Fourier series, namely ##f(x)\sim\sum_{n=1}^\infty b_n\sin nx##. Then, since ##f'## will be even, its Fourier coefficients are $$a_n=\frac{2}{\pi}\int_0^\pi f'(x)\cos(nx)dx=\frac{2n}{\pi }\int_0^\pi f(x)\sin(nx)=nb_n,$$ since ##f(0)=f(\pi)=0## by partial integration. Then I use Parseval's formula $$\int_0^\pi f^2= \frac{\pi}{2}\sum_{n=1}^\infty b_n^2 \quad \text{and} \quad \int_0^\pi (f')^2= \frac{\pi}{2}\sum_{n=1}^\infty n^2b_n^2.$$ And the inequality follows from this. Equality holds for ##a_n=0## for ##n\neq 1##, i.e. when ##f(x)=C\sin x##.

In order for this solution to work, we need ##f'## to be integrable, right? Does ##f'\in L^2([0,\pi])## guarantee this?
 
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psie said:
In order for this solution to work, we need ##f'## to be integrable, right? Does ##f'\in L^2([0,\pi])## guarantee this?
We need square integrability of ##f'## to even phrase the statement and ##f'\in L^2## is just that.
 
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fresh_42 said:
We need square integrability of ##f'## to even phrase the statement and ##f'\in L^2## is just that.
I oversaw something there :smile: In my solution, however, I also need ##a_n=\frac2{\pi}\int_0^\pi f’(x)\cos(nt)dt## to be well-defined. Does ##f’\in L^2## ensure this?
 
psie said:
I oversaw something there :smile: In my solution, however, I also need ##a_n=\frac2{\pi}\int_0^\pi f’(x)\cos(nt)dt## to be well-defined. Does ##f’\in L^2## ensure this?
Cauchy-Schwarz
 

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