# Seeking for a Simple Proof

• Jul 3rd 2009, 05:32 PM
nonsingular
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 $\displaystyle \displaystyle\sum_{k=-\infty}^{\infty}f(k)=\sum_{m=-\infty}^{\infty}\left[\int_{-\infty}^\infty e^{2\pi imx}f(x)dx\right]$

Yes, I understand that this is the Poisson Summation Formula.
• Jul 4th 2009, 06:10 AM
halbard
Depends on what you mean by "elementary" I suppose...

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

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

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

$\displaystyle \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$$\displaystyle {}=\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 $\displaystyle \hat f$ is the Fourier transform of $\displaystyle f$.

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

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