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Laplace transform/hilbert space

  1. Jul 25, 2006 #1
    I was just thinking back to my advanced linear algebra class and I remember the prof mumbling something about the fourier series being an orthonormal basis for the hilbert space of square-integrable real-valued functions, lebesgue something, etc., and demonstrating the properties of the integral inner product and projections into that space. It all made sense at the time, I think. :) Sorry if I'm kinda vague, I can provide more info if needed.

    I am curious, though, if something similar exists for laplace transform. It certainly looks like it fits the form, but I have been unable to find any detailed analysis of such. Anyone have insight?
     
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  3. Jul 25, 2006 #2

    quasar987

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  4. Jul 25, 2006 #3

    quasar987

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    Some of the following may be nonsense but...

    The way I interpret the article is that [itex]\{exp(inx)\}_{n\in \mathbb{Z}}[/itex] is a complete orthonormal basis for the space of real valued square integrable smooth function on [itex](-\pi,\pi)[/itex] of domain [itex](-\pi,\pi)[/itex] with inner product defined by

    [tex]\langle f,g \rangle =\int_{-\pi}^{\pi}f(x)g(x)dx[/tex].

    This allows us to write, for any function f of that space,

    [tex]f(x) = \sum_{n\in \mathbb{Z}}\langle f,\exp(inx) \rangle \exp(inx)[/tex]

    Using the language of vector spaces, <f,exp(inx)> is the projection of the vector f in the direction of the unit vector [itex]\exp(inx)[/itex], i.e. the component of f in the direction of [itex]\exp(inx)[/itex]

    More generally,

    [tex]\left\{ \exp \left( in\frac{2\pi}{P}x \right) \right\} _{n\in \mathbb{N}}[/tex]

    is a complete orthonomal basis for the space of real valued periodic functions of period P.

    For functions that are NOT periodic, but that have the property that the integral

    [tex]\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}f(x)\exp (-in\omega)dx = F(\omega)[/tex]

    converges, we can write them in a kind of "continuous" form of a Fourier series, i.e. as its Fourier transform:

    [tex]f(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}F(\omega)\exp (inx)d\omega[/tex]


    Now let's make the analogy with the Laplace transform. Suppose a function [itex]f:(0,+\infty)\subset D \rightarrow \mathbb{R}[/itex] can we written as the Laplace transform of some function a(t):

    [tex]f(s) = \int_0^{+\infty} a(t) \exp(-st)dt[/tex]

    Note that [itex]\exp(-st)=(e^{-s})^t[/itex]. Make the substitution [itex]e^{-s}=x[/itex] and the equation becomes a restriction of f to the positive real line:

    [tex]f|_{\mathbb{R}^+} = f(x) = \int_0^{+\infty} a(t) x^t dt[/tex]

    This is a kind of "continuous" form of a power series, right?

    So to answer your question...

    ...the analogue is the set [itex]\{x^n\}_{n\in \mathbb{N}}[/itex], which is a complete orthonormal basis for, say, the space of real valued function developable in a Taylor series of convergence radius R and of domain (-R,R), with the inner product defined as...

    ...as what exactly? Also, maybe [itex]\{x^n\}_{n\in \mathbb{N}}[/itex] is not orthoNORMAL, but just orthogonal. What inner product would yield

    [tex]\langle f,x^n \rangle = \frac{f^{(n)}(0)}{n!}[/tex]

    and

    [tex] \langle x^m,x^n \rangle = 0[/tex]

    except for m=n

    ??
     
    Last edited: Jul 25, 2006
  5. Jul 25, 2006 #4
    Ohh I see! I still have to process all that but just skimming it makes sense. Thanks!
     
  6. Jul 25, 2006 #5
    Hmm, I vaguely remember this. But isn't [itex]\{x^n\}_{n\in \mathbb{N}}[/itex] neither orthonormal nor orthogonal? You have to apply Gram-Schmidt for that to be true, right?

    As for which inner product...I have to think about it for a minute...

    hmm.the laplace transform? is that what I'm missing? I think I'm overthinking it...
     
    Last edited: Jul 25, 2006
  7. Jul 27, 2006 #6

    quasar987

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    At least it's a complete set, we know that much! :p

    (Also, the Fourier integrals in post #3 are with 'x' in the exponential and not 'n')
     
    Last edited: Jul 27, 2006
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