Poisson summation formula clarification regarding Fejer kernel

Define $$\mathbf{F}_R(t) = \begin{cases} R \left(\dfrac{\sin(\pi R t)}{\pi R t}\right)^2 & t \neq 0\\[10pt] R & t = 0 \end{cases}$$

A problem in Stein's Fourier Analysis asks us to prove that the periodization of $\mathbf{F}_R(t)$ is equal to the Fejer kernel on the circle

i.e.

$$\sum_{n=-\infty}^{\infty}\mathbf{F}_N(x+n) = \sum_{n=-N}^{N}\left(1-\frac{|n|}{N}\right)e^{2 \pi i n x} = \frac{1}{N} \frac{\sin^2(N \pi x)}{\sin^2(\pi x)}$$

for $N \geq 1$ an integer

This strongly suggests an application of Poisson summation is needed, which would mean that we need to calculate $$\sum_{n=-\infty}^{\infty}\hat{\mathbf{F}}_N(n)e^{2 \pi i n x}$$

correct?

However, as I don't see how we can go from an infinite series to a finite series using Poisson, I assume we have to show that the above series converges to the closed form expression above? I'm still a bit unclear as to how apply Poisson in this case: if this is correct, any hints as to how to tackle the integral $$\int_{-\infty}^{\infty} \mathbf{F}_N(x)e^{-2 \pi i n x}dx$$ would be appreciated.

• I was just going through the same problem, but I think it may be possible to compute the integral but seems too convoluted and the authors intended a solution of the form given in the below answer. After all, the identity we need to use is already given in Exercise 9 of Chapter 3. – takecare Oct 8 '16 at 11:34

How about trying the following alternative approach. First, note that for integer $N$, the numerator of $F_N(t)$ factors out since: $$\sin^2(\pi N (t + n)) = \sin^2(\pi N t + \pi N n) = \sin^2(\pi N t)$$ The remaining part is given by the identity: $$\sum_{n= -\infty}^{\infty} \frac{1}{\pi^2 (t+n)^2} = \frac{1}{\sin^2(\pi t)}$$ I am not entirely sure how to prove this identity...but I hope this will help anyway.
• That's actually the problem immediately following this one, and I was able to prove it (apply Poisson to the function $f(x) = 1-|x|$ for $|x|<1$ and $0$ elsewhere). Thanks for the help, it definitely works. I'll probably try looking a bit more for a solution independent of this identity given the order of the exercises before going down this way. – Lost Apr 3 '14 at 5:46