An equality in Fourier analysis

Question

I need at least a hint on proving this equality.

For $f\in C^1_c(\mathbb{R})$ we got $$\sum_{n=-\infty}^{\infty}\frac{1}{2\pi}\int_{-\infty}^{\infty} f(t)e^{int}dt=\sum_{m=-\infty}^{\infty}f(2m\pi).$$

I don't know how to start and i don't know how to use the fact that $f$ is differentiable. I'm studying Fourier analysis so I'm supose to solve that with Fourier instruments. Any hint i think can help me. Thanks!

Answer 1

First define $$F(x)=\sum_{k\in\Bbb Z}f(x+2\pi k),$$so $F$ has period $2\pi$.

You now want to show that $$F(0)=\sum_{n\in\Bbb Z}\hat F(n).$$

By standard arguments this follows if $\sum|\hat F(n)|<\infty$. That follows in turn by Cauchy-Schwarz, since $$|\hat F(n)|=\frac1{|n|}|\hat G(n)|\quad(n\ne0)$$if $G=F'\in L^2(\Bbb T).$

Answer 2

The identity can be rewritten in terms of the Dirac delta function

$$\sum_{n=-\infty}^\infty e^{int}=\sum_{m=-\infty}^\infty\delta\left(\frac{t}{2\pi}-m\right)\!.$$

The right-hand side is a $2\pi$-periodic delta train of the variable $t$. We can expand it into a Fourier series, which is the left-hand side. Then multiply by $f(t)$ and integrate over $\,t\in(-\infty,\infty)$ to get the equality to prove. A benefit of the delta function is that the result generalizes to $d$ dimensions (by raising both sides to power $d$).