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Math Help - Test functions as a taylor series with integral remainder

  1. #1
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    Test functions as a taylor series with integral remainder

    Hi All,

    Given a test function \phi (C^infinity function with compact support), I have shown that we can write \phi = \sum_{j = 0}^{m - 1} \frac{x^j}{j!}\phi^{(j)}(0) + \frac{1}{(m - 1)!}\int_0^x (x - t)^{m - 1}\phi^{(m)}(t) \, \mathrm{d}t.

    Choose a test function \phi_0 on \R such that \phi_0(x) = 1 on a neighbourhood of x = 0. I have to show the following:
    On this neighbourhood, any test function \phi on \R can be written as \phi = \phi_0(x)\sum_{j = 0}^{m - 1} \frac{x^j}{j!}\phi^{(j)}(0) + \chi(x)

    where \chi(x) is a test function on \R such that \chi^{(j)}(0) = 0 for j = 1, \ldots m - 1.

    I cannot see where this is coming from.

    I have considered the function defined by \chi(x) = \phi(x) - \phi_0(x)\sum_{j = 0}^{m - 1} \frac{x^j}{j!}\phi^{(j)}(0), but I do not think this has the desired derivative properties.

    Can anyone help?
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  2. #2
    Super Member girdav's Avatar
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    You can show the derivative properties by the Leibniz's rule (notice that \phi_0^{(k)}(0)=0 if k\geq 1.
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  3. #3
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    Quote Originally Posted by girdav View Post
    You can show the derivative properties by the Leibniz's rule (notice that \phi_0^{(k)}(0)=0 if k\geq 1.
    Ah cheers, I tried this and thought I ended up with \chi'(x) = \phi'(x) - now I notice the term that cancels.
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  4. #4
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    I now have another problem.

    I have shown that if \phi is a test function on R, with the first m - 1 derivatives at the origin being zero, then \phi = x^m\psi for some test function \psi. (*)

    So it follows from my first post that \phi(x) = \phi_0(x)\sum_{j = 0}^{m - 1} \frac{x^j}{j!}\phi^{(j)}(0) + x^m\mu(\phi(x)) (**)
    for some mapping \mu from test functions to test functions.

    If v is a distribution, then apparently I can define u(\phi) = v(\mu(\phi)), and u is a distribution. Does anyone know how to prove this?

    The hint in the book is to use (*) and (**), but I just can't see it.
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