Regarding Wirtinger's inequality

  • #1
psie
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TL;DR Summary
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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  • #2
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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  • #3
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?
 
  • #4
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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