Convergence of Fourier Series Coefficients for L2 Functions

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For any function f in L²[-π, π], the Fourier coefficients (f, e_n) converge to zero as |n| approaches infinity. The discussion emphasizes the use of Bessel's inequality to establish that the sum of squared coefficients is finite, leading to the conclusion that the coefficients must converge to zero. Concerns about the necessity of the completeness of the basis {e_n} are addressed, clarifying that it's not required for this proof. The conversation also highlights the distinction between L²-convergence and absolute convergence, prompting a re-indexing of the coefficients for clarity. Overall, the approach discussed effectively demonstrates the convergence of Fourier series coefficients for L² functions.
Kindayr
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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!
 
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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.
 

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