Regarding Wirtinger's inequality

  • Context: Undergrad 
  • Thread starter Thread starter psie
  • Start date Start date
  • Tags Tags
    Fourier analysis
Click For Summary

Discussion Overview

The discussion revolves around Wirtinger's inequality, specifically examining the conditions under which the inequality holds for continuous real-valued functions defined on the interval (0, π) with specific boundary conditions. Participants explore the implications of the function's derivative being in L² space and the requirements for the Fourier series representation.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes a proof of Wirtinger's inequality by extending the function to an odd function and using its Fourier series representation.
  • Another participant questions whether the condition of the derivative being in L² space guarantees integrability of the derivative.
  • There is a suggestion that square integrability of the derivative is necessary to phrase the statement of the inequality.
  • A later reply indicates a need for the Fourier coefficients to be well-defined and questions if the L² condition ensures this.
  • One participant references a version of the inequality that includes an additional condition and mentions the Dirichlet conditions related to Fourier series convergence.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the implications of the derivative being in L² space and whether it guarantees integrability. There is no consensus on the necessity of additional conditions for the Fourier series representation.

Contextual Notes

Some participants note the importance of the Cauchy-Schwarz inequality in the context of their arguments, but the discussion remains focused on the implications of the L² condition without resolving the mathematical steps or assumptions involved.

psie
Messages
315
Reaction score
40
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?
 
Physics news on Phys.org
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.
 
  • Like
Likes   Reactions: psie
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
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 16 ·
Replies
16
Views
2K