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Homework Help: Sequence of functions in L^2[-pi,pi]

  1. Apr 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider the functions [tex]f_n(x)=e^{inx},n\in\mathbb{Z},-\pi\leq x\leq\pi[/tex] viewed as points in [tex]\mathscr{L}^2[-\pi,\pi][/tex]. Prove that this set of functions is closed and bounded, but not compact.

    2. The attempt at a solution
    I'm first trying to prove that the set of functions is closed. Let g be a limit point of the set, so given any positive number epsilon, there exists fn such that [tex]\|f_n-g\|_2<\sqrt{\epsilon}[/tex]. Squaring both sides, I get on the left hand side

    2\pi -2\text{Re}\int_{-\pi}^{\pi}g(x)e^{-inx}dx+\int_{-\pi}^{\pi}|g(x)|^2 dx


    g\sim\sum\limits_{-\infty}^{\infty}c_n e^{inx}

    and by Parseval's Theorem,

    \sum\limits_{-\infty}^{\infty}|c_n|^2 = \frac{1}{2\pi}\int_{-\pi}^{\pi}|g|^2 dx

    I don't know where to go from here.
  2. jcsd
  3. Apr 16, 2009 #2
    What would [tex]\|f_n-f_m\|_2[/tex] be?

    Can you even have a Cauchy sequence?
  4. Apr 16, 2009 #3
    \|f_n-f_m\|_2 = 4\pi -\int_{-\pi}^{\pi}e^{i(m-n)x}dx-\int_{-\pi}^{\pi}e^{i(n-m)x}dx

    so when m and n are different, there exists a positive number less than the expression above. So I would say that no, you can't have a Cauchy sequence and so this shows the set of functions is not compact.

    But I'm still having trouble with proving the set is closed.
  5. Apr 16, 2009 #4
    Could show the set is complete as a metric space, hence closed.

    Note: I guess technically there are Cauchy sequences, but only the sequences that eventually become constant.
  6. Apr 16, 2009 #5
    Thanks for your help; I have to say though that I don't see how there are sequences that eventually become constant. I can't seem to find a sequence {fn(x)} that converges to a constant value given any x.
  7. Apr 16, 2009 #6
    I mean a sequence [tex](a_1, a_2, a_3,\dots)[/tex] in X with the property that there exists x in X and an integer N such that for all n>N, a_n=x.

    For example, the sequence [tex](f_2, f_8, f_8, f_8, f_8,\dots)[/tex] where all the rest of the "points" (i.e. functions) are f_8

    Such a sequence is trivially Cauchy.
  8. Apr 16, 2009 #7
    Okay, thanks again for your assistance.
  9. Apr 16, 2009 #8
    Sorry, just one last question: isn't any subset S of [tex]{\cal L}^2[-\pi,\pi][/tex] bounded by the zero function g(x)=0, since for any [tex]f\in S[/tex]

    \|f-g\|_2 = \left(\int_{-\pi}^{\pi}|f-g|^2 d\mu\right)^{1/2} = \left(\int_{-\pi}^{\pi}|f|^2 d\mu\right)^{1/2}

    which by definition is finite?
    Last edited: Apr 16, 2009
  10. Apr 16, 2009 #9
    No. Using g=0 is OK, but you would need one constant M such that for all f in S, [tex]\|f-g\|_2<M[/tex]

    Fortunately, in this case it is easy to find such an M, with your g.
  11. Apr 16, 2009 #10


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    Homework Helper

    All of your functions are orthogonal in L^2 and they are all of constant L^2 norm. Aren't you trying to prove the sequence is a closed set? It's a discrete set.
  12. Apr 17, 2009 #11
    iomtt6076, I'm just double checking. You do know what [tex]\int_{-\pi}^{\pi}e^{i(m-n)x}\,dx[/tex] is, right?
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