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Math Help - schwatz space and fourier transform: problem with induction

  1. #1
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    schwatz space and fourier transform: problem with induction

    I'll write everything down from the start.
    First, the notation:

    \alpha \in \mathbb{N}^n, \alpha=(\alpha_1, \ldots, \alpha_n)
    |\alpha|=\alpha_1 + \ldots \alpha_n
    x \in \mathbb{R}^n

    x^{\alpha}=x_1^{\alpha_1} \ldots x_n^{\alpha_n}

    \partial^{\alpha}=\partial_1^{\alpha_1} \ldots \partial_n^{\alpha_n}, where
    \partial_i^{\alpha_i}=\frac{\partial^{\alpha_i}}{\  partial x_i ^{\alpha_i}}


    Then the theorem:
    If x^{\alpha}f(x) is summable for every \alpha, |\alpha|\leq k, then
    \partial^{\alpha} \hat{f}={[(-2 \pi i x)^{\alpha})f]}^{\wedge}, where \hat{f} denotes Fourier transform of f.


    Now, the problem.

    We're supposed to prove this using induction on \alpha, and I just can't do it. The definition of \partial^{\alpha}f is somewhat confusing, and I would b really grateful if someone could help me out, just for the basis, and I will try to work from there.

    Thank you!
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  2. #2
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    Quote Originally Posted by marianne View Post
    If x^{\alpha}f(x) is summable for every \alpha, |\alpha|\leq k, then \partial^{\alpha} \hat{f}={[(-2 \pi i x)^{\alpha})f]}^{\wedge}, where \hat{f} denotes Fourier transform of f.
    [The equation in the statement of the theorem should read \partial^{\alpha} \hat{f}={[(-2 \pi i x)^{|\alpha|})f]}^{\wedge}, with |\alpha| rather than \alpha as the exponent on the right-hand side.]

    I'm assuming you know how to do this in the case n=1. (For a function f(x) of one variable, differentiating the Fourier transform of f corresponds to multiplying f by -2πix, provided that xf(x) is integrable. This is proved by integration by parts.)

    To prove the multi-variable case by induction, the inductive hypothesis will be that the result is true when |\alpha|=m-1 for some integer m>0. The inductive step goes like this. Let \alpha be a multi-index with |\alpha|=m. Then at least one component of \alpha, say \alpha_j, must have \alpha_j>0. Let \beta = (\alpha_1,\ldots,\alpha_{j-1},\alpha_j-1,\alpha_{j+1}\ldots,\alpha_n). Then |\beta|=m-1, so by the inductive hypothesis \partial^{\beta}\hat{f} = {[(-2 \pi i x)^{|\beta|})f]}^{\wedge}. To complete the inductive step, all you have to do now is to apply the single-variable result to the function {[(-2 \pi i x)^{|\beta|})f]}^{\wedge} considered as a function of its j'th variable.
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  3. #3
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    (there's a typo in the title: it should be Schwartz, from the name of Laurent Schwartz )
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  4. #4
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    Quote Originally Posted by Laurent View Post
    (there's a typo in the title: it should be Schwartz, from the name of Laurent Schwartz )
    It would have been more interesting if the poster mispelled it as Schwarz.
    Because that is a kinda well-known mathematician.
    Mispelling these two names is common.
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  5. #5
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    Opalg, thank you very much, I get it now!

    And sorry about the typo, latexing the question took me so much time I didin't pay much attention to the title.
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