Convergence of Fourier Series Coefficients for L2 Functions

Click For Summary

Homework Help Overview

The discussion revolves around the convergence of Fourier series coefficients for functions in the space L² over the interval [-π, π]. The original poster seeks to demonstrate that the Fourier coefficients of a function f in L² converge to zero as the index n approaches infinity.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the use of dominant convergence and Bessel's inequality in relation to the completeness of the basis functions. There is a discussion on whether the completeness of the basis is necessary for the convergence of the coefficients to zero.

Discussion Status

Participants are actively engaging with each other's ideas, questioning assumptions about the completeness of the basis and the implications of L² convergence. Some guidance has been offered regarding the use of Bessel's inequality, and the conversation reflects a mix of attempts to clarify concepts and explore different approaches.

Contextual Notes

There is a noted constraint regarding the prerequisite knowledge of measure theory, which may limit some approaches to the problem. Additionally, participants express confusion about the relationship between L² convergence and absolute convergence.

Kindayr
Messages
159
Reaction score
0

Homework Statement


Let e_{n}(t)= \frac{1}{ \sqrt{2\pi}}\cdot e^{int} for n\in\mathbb{Z} and -\pi\le t\le\pi.

Show that for any f\in L^{2}[-\pi,\pi] we have that (f,e_{n})=\int_{-\pi}^{\pi}f(t)\cdot e^{-int}dt\to0 as |n|\to \infty.

The Attempt at a Solution


I want to use dominant convergence, but unfortunately measure theory isn't a prerequisite for this course. Any help will be awesome!
 
Physics news on Phys.org
Wait, if {e_n} is complete, then \lim_{n\to\infty}\sum_{m=-n}^{n}(f,e_m)e_m converges to f absolutely, so the coefficients necessarily converge to zero. Would this work?
 
Have you seen and proved Bessel's inequality??
 
Yep we have. Could I say that:

Since \{e_n\} is complete it follows that f=\lim_{n\to\infty} \sum_{m=-n}^{n}(f,e_m)e_m. Thus, the latter sum converges and hence \lim_{n\to\infty}\sum_{m=-n}^{n}|(f,e_m)|^{2}<\infty. Thus |(f,e_m)|^2 \to 0 so |(f,e_m)|\to 0 and whence (f,e_m)\to 0, as required.

Does this work?
 
Kindayr said:
Yep we have. Could I say that:

Since \{e_n\} is complete it follows that f=\lim_{n\to\infty} \sum_{m=-n}^{n}(f,e_m)e_m. Thus, the latter sum converges and hence \lim_{n\to\infty}\sum_{m=-n}^{n}|(f,e_m)|^{2}<\infty. Thus |(f,e_m)|^2 \to 0 so |(f,e_m)|\to 0 and whence (f,e_m)\to 0, as required.

Does this work?

You don't really need \{e_n\} to be complete for that. Bessel's inequality works fine.
Also, the limit you mention is L^2-convergence. You might want to be careful with that.
 
micromass said:
You don't really need \{e_n\} to be complete for that. Bessel's inequality works fine.
Also, the limit you mention is L^2-convergence. You might want to be careful with that.

What should I be careful with? Does L^2-convergence not imply absolute convergence? Sorry, just a little confused by your statement.
 
What if I re-index so that n\in \mathbb{N} so that d_0=e_0, d_1=e_1, d_2=e_{-1},\dots. Then by Bessel's inequality we have \sum_{n=0}^{\infty}|(f,d_n)|^2\le \parallel f\parallel ^{2}<\infty. Hence \sum_{n=0}^{\infty}|(f,d_n)|^2 converges absolutely and whence (f,d_n)\to 0 and thus (f,e_n)\to\ 0 as |n|\to\infty.

Would this be better?

I'm far from an analyst as you can probably tell hahahah.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K