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Thread: Seeking for a Simple Proof

  1. #1
    Jun 2009

    Seeking for a Simple Proof

    Hi I found this in an Advanced Calculus book which does not go into the details of real analysis. Hence, I was wondering what would be an "elementary proof" of this result.

    Show that \displaystyle\sum_{k=-\infty}^{\infty}f(k)=\sum_{m=-\infty}^{\infty}\left[\int_{-\infty}^\infty e^{2\pi imx}f(x)dx\right]

    Yes, I understand that this is the Poisson Summation Formula.
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  2. #2
    Mar 2009
    Depends on what you mean by "elementary" I suppose...

    Assume suitable conditions on the function f, such as f\in \mathrm C_2(\mathbb R) with |f(x)|+|f'(x)|+|f''(x)|<A/(1+x^2) which ensure convergence of the relevant series.

    Let g(x)=\sum_{k=-\infty}^\infty f(x+2k\pi). This series converges uniformly to a function in \mathrm C_2[0,2\pi) and can be expanded in a uniformly convergent Fourier series:

    g(x)=\sum_{n=-\infty}^\infty \bar g(n)\mathrm e^{\mathrm i nx} where

    \bar g(n)=\frac1{2\pi}\int_0^{2\pi}g(x)\mathrm e^{-\mathrm inx}\mathrm dx=\frac1{2\pi}\sum_{k=-\infty}^\infty\int_0^{2\pi}f(x+2k\pi)\mathrm e^{-\mathrm inx}\mathrm dx {}=\frac1{2\pi}\sum_{k=-\infty}^\infty\int_{2k\pi}^{2(k+1)\pi}f(x)\mathrm e^{-\mathrm inx}\mathrm dx=\frac1{2\pi}\int_{-\infty}^\infty f(x)\mathrm e^{-\mathrm inx}\mathrm dx=\frac1{2\pi}\hat f(n).

    Here \hat f is the Fourier transform of f.

    Thus \sum_{k=-\infty}^\infty f(2k\pi)=g(0)=\sum_{n=-\infty}^\infty \bar g(n)=\frac1{2\pi}\sum_{n=-\infty}^\infty \hat f(n)=\frac1{2\pi}\sum_{n=-\infty}^\infty \int_{-\infty}^\infty f(x)\mathrm e^{-\mathrm inx}\mathrm dx.

    To obtain your result use h(x)=f(2\pi x) and change x to -2\pi x in the integral.
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