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Fourier Basis functions question

  1. Jun 3, 2009 #1
    the continuous Fourier Transform is often defined on a finite interval, usually [tex][-\pi,\pi][/tex]:

    [tex]\hat{f_k} = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-ikx}dx[/tex]

    If I understood correctly, this allows the basis functions to be defined so that they have norm=1, and they form an orthonormal basis for [tex]L^2([-\pi,\pi])[/tex].
    Now, I get confused when one tries to compute the FT of a function f in the whole [tex]\mathcal{R}[/tex] because:

    1) The 2-norm of the basis functions goes to [tex]+\infty[/tex]
    2) They are not square integrable
    3) Should I conclude that the basis-functions are orthogonal but NOT orthonormal?
    4) If (1,2,3) are correct, then what is the space spanned by the basis-functions?
    5) Is it possible to define an orthonormal basis for [tex]L^2(\mathcal{R})[/tex] with Fourier basis ?
    Last edited: Jun 3, 2009
  2. jcsd
  3. Jun 3, 2009 #2
    I'm not familiar with a 2-norm but the basis is normalised "in my book" like so:

    [tex]\int_{-\pi}^{\pi}\phi_{k}^{*}\phi_{k}dx = \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{+ikx}e^{-ikx}dx = \frac{1}{2\pi}\int_{-\pi}^{\pi}dx = 1 [/tex]

    Last edited by a moderator: Aug 6, 2009
  4. Jun 3, 2009 #3
    Maybe I used an incorrect term, but indeed, you proved that the squared norm of the basis-functions are always 1.
    The important points of my question (1,2,3,4,5) arise when the bounds of the integral are not anymore [tex][-\pi,\pi][/tex] but they become [tex](-\infty,+\infty)[/tex]

    Basically, I wanted you to consider:
    and then, answer those questions.
  5. Jun 4, 2009 #4
    I see what you mean now. In this case, I think you cannot define the basis.
    But maybe there is a way to expand functions in an unnormalised basis.

    Last edited by a moderator: Aug 6, 2009
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