1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Functional Analysis HMWK

  1. Nov 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex]e_{n}(t)= \frac{1}{ \sqrt{2\pi}}\cdot e^{int}[/itex] for [itex]n\in\mathbb{Z}[/itex] and [itex]-\pi\le t\le\pi[/itex].

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

    3. 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!
  2. jcsd
  3. Nov 12, 2012 #2
    Wait, if [itex]{e_n}[/itex] is complete, then [itex]\lim_{n\to\infty}\sum_{m=-n}^{n}(f,e_m)e_m[/itex] converges to [itex]f[/itex] absolutely, so the coefficients necessarily converge to zero. Would this work?
  4. Nov 12, 2012 #3
    Have you seen and proved Bessel's inequality??
  5. Nov 12, 2012 #4
    Yep we have. Could I say that:

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

    Does this work?
  6. Nov 12, 2012 #5
    You don't really need [itex]\{e_n\}[/itex] to be complete for that. Bessel's inequality works fine.
    Also, the limit you mention is [itex]L^2[/itex]-convergence. You might want to be careful with that.
  7. Nov 12, 2012 #6
    What should I be careful with? Does [itex]L^2[/itex]-convergence not imply absolute convergence? Sorry, just a little confused by your statement.
  8. Nov 12, 2012 #7
    What if I re-index so that [itex]n\in \mathbb{N}[/itex] so that [itex]d_0=e_0, d_1=e_1, d_2=e_{-1},\dots[/itex]. Then by Bessel's inequality we have [itex]\sum_{n=0}^{\infty}|(f,d_n)|^2\le \parallel f\parallel ^{2}<\infty[/itex]. Hence [itex]\sum_{n=0}^{\infty}|(f,d_n)|^2[/itex] converges absolutely and whence [itex](f,d_n)\to 0[/itex] and thus [itex](f,e_n)\to\ 0[/itex] as [itex]|n|\to\infty[/itex].

    Would this be better?

    I'm far from an analyst as you can probably tell hahahah.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook