Seeking for a Simple Proof

nonsingular · #1 $Old$ July 3rd, 2009, 07:32 PM

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.

halbard · #2 $Old$ July 4th, 2009, 08:10 AM

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.