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Strange convolution equation

  1. Jun 5, 2012 #1
    I've arrived at the following equation involving the convolution of two functions:

    [itex]
    f(x) = \int_{-\infty}^{\infty} f(t) g(t-x) dt = f(x) \ast g(x)
    [/itex]

    Where:

    [itex]
    g(z) = e^{-z^2/2}
    [/itex]

    In other words, a function convoluted with a Gaussian pdf results in the same function.

    I've tried taking fourier transforms, realizing that the FT of a gaussian results in another Gaussian:

    [itex]
    F[f(x)] = F[f(x) \ast g(x)] = F[f(x)] \cdot F[g(x)]
    [/itex]

    But this results in the [itex] F[f(x)] [/itex] cancelling out, leaving me with just:

    [itex]
    1 = F[g(x)] = e^{-w^2/2}
    [/itex]

    Any suggestions?
     
  2. jcsd
  3. Jun 5, 2012 #2

    micromass

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    What if f=0?
     
  4. Jun 5, 2012 #3
    Yes that is the trivial solution.

    Perhaps this can be casted as an eigenvalue problem - as it seems to imply that the convolution operator (wrt to the gaussian) may have certain eigenvalues and corresponding eigenfunctions f(x) being one of them
     
  5. Jun 6, 2012 #4
    edit - doh - this obviously implies that f(x) must be equal to 0 (no other solution satisfies:

    f(x)=f(x)g(x) unless g(x) = 1, which in this case, it isn't)
     
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