I want to compute the following product
$$\frac{1}{N}\sum_{t=1}^{N}\left(\sum_{s=-\infty}^{\infty}a_{s}\exp(2\pi is\frac{t}{N}\right)\left(\sum_{z=-\infty}^{\infty}a_{z}\exp(-2\pi iz\frac{t}{N})\right)$$
If $N\rightarrow\infty$ we can approximate it with integral:
$$\int_{0}^{1}\left(\sum_{s=-\infty}^{\infty}a_{s}\exp(2\pi isx\right)\left(\sum_{z=-\infty}^{\infty}a_{z}\exp(2\pi izx)\right)dx=\sum_{s=-\infty}^{\infty}a_{s}\sum_{z=-\infty}^{\infty}a_{z}\int_{0}^{1}\exp(2\pi i(s-z))dx$$
Hence $s=z$ and we have
$$\sum_{s=-\infty}^{\infty}a_{s}^{2}$$
On the other hand this product is equal to:
$$\frac{1}{N}\left(\sum_{s=-\infty}^{\infty}a_{s}\right)\left(\sum_{z=-\infty}^{\infty}a_{z}\right)\left(\sum_{t=1}^{N}\exp(2\pi i(s-z)\frac{t}{N})\right)=\sum_{w=-\infty}^{\infty}\sum_{s=-\infty}^{\infty}a_{s}a_{s+Nw}$$
We have
$$\sum_{s=-\infty}^{\infty}a_{s}^{2}+\sum_{w=-\infty,w\neq0}^{\infty}\sum_{s=-\infty}^{\infty}a_{s}a_{s+Nw}$$
Now I don't see why the second term will disappear.
Which is correct?
Best Answer
This is a consequence of the Dominated Convergence Theorem, see This question.
In details: $$ \sum_{w\neq0}\sum_{s} a_s a_{s+Nw} $$ is better written $$ \sum_{s}\sum_{z} b_{s,z}^N $$ where $$ b_{s,z}^N = a_{s}a_{z} \mbox{ when } z=s \mbox{ mod } N,\, N\neq0 \mbox{ and } b_{s,z}^N=0 \mbox{ otherwise.} $$ Now, $|b_{s,z}^N|\leq |a_{s}||a_{z}|$ for all $N$, and $$\sum\sum|a_{s}||a_{z}|=(\sum |a_{s}|)^2<\infty.$$ On the other hand, for any fixed $z, s$ $$ \lim_{N\to \infty} b_{s,z}^N = 0 \,(\mbox{ since } a_{s+Nw} \to 0 \mbox{ with }N) $$ Now apply the DCT to conclude.