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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

    micromass

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    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

    micromass

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    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.
     
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