Why does this identity hold for Fejér Kernels?

I'm trying to read a proof for the existence of an $(\epsilon , \delta)$ approximation to the identity that is a trigonometric polynomial. For this, the Fejér Kernel is defined as $$F_N = \sum_{n = -N}^{N} (1 - \frac{|n|}{N})e_n$$

The author of the text also asserts that the following identity holds: $$F_N = \frac{1}{N} |\sum_{n = 0}^{N-1} e_n|^2$$

However, no proof is provided for this identity (since I suspect it should be easy to prove). I can't come up with my own proof for this nor could I find it by googling.

Is there any identity involving the characters $e_n$ that I should be using?

Thanks in advance

Solution 1:

I assume $e_n = e^{inx}$. Undoubtedly you can adapt the arguments here for other versions of $e_n$.

Here I present two ways to get the result. The direct method is essentially the fact that the convolution of rectangular functions is a triangle. After that, I will also give a method that passes through (IMO) more standard methods and explicit formulae that you might find in books on Fourier Analysis.

Direct Combinatorial Method

We start from the norm squared formula; note that $$ \left|\sum_{n=0}^{N-1} e_n\right|^2 = \sum_{n=0}^{N-1} e_n \sum_{m=0}^{N-1} e_{-m} = \sum_{n=0}^{N-1} e_n \sum_{m=0}^{N-1} e_{-(N-1)+m} = e_{-(N-1)} \left(\sum_{n=0}^{N-1} e_n\right)^2.$$

This has a well-known formula in the world of power series, given by the Cauchy Product (which is just a discrete convolution):

$$ \sum_{n=0}^\infty a_n x^n \sum_{n=0}^\infty b_n x^n = \sum_{n=0}^\infty c_n x^n \implies c_n = \sum_{k=0}^n a_k b_{n-k} = \sum_{k,l: k+l=n} a_k b_l.$$ In our situation, $a_n = b_n = \mathbb 1_{n\in [0,N-1]}$ takes only the values $0$ or $1$, so really this is:

$$c_n = \text{ # pairs $(k,l)\in[0,N-1]^2\cap \mathbb Z^2$ such that $k+l=n$}.$$ (Related Problem.) This is the length (as in number of terms) of the $n$th diagonal of an $N\times N$ matrix. (the first row is the $n=0$th row!)

There are $\text{#rows} + \text{# columns} - 1 = 2N-1$ diagonals. The shortest is the $n=0$th diagonal of length $1$, and the longest diagonal is the. $n=(N-1)$th diagonal of length $N$. The lengths are symmetric across $n=N-1$ and from $n=0$ to $n=N-1$, $c_n$ linearly changes in steps of 1. Thus $c_n = 0$ for $n\ge 2N-1$, and for $0\le n\le 2N-2$, this is $$ c_n= N - |n-(N-1)|.$$ Therefore, \begin{align} \frac1N\left|\sum_{n=0}^{N-1} e_n\right|^2 &= \frac{e_{-(N-1)}}N \sum_{n=0}^{2N-2} ( N - |n-(N-1)|) e_n \\ &= \frac{1}N \sum_{n=0}^{2N-2} ( N - |n-(N-1)|) e_{n-(N-1)} \\ &= \frac{1}N \sum_{n=N-1}^{N-1} (N-|n|) e_n \\ &= \frac{1}N \sum_{n=N}^{N} (N-|n|) e_n = \sum_{n=N}^{N} \left(1-\frac{|n|}N\right) e_n, \end{align} exactly as desired.

Second Method passing through more well-known results

Lets first put $F_N$ in a more usual form using the discrete Fubini theorem:

$$ F_N = \frac{1}{N} \sum_{n=-N}^N \underbrace{(N - |n|)}_{=\sum_{j=|n|}^{N-1} 1} e_n = \frac{1}{N}\sum_{\substack{(j,n)\in \mathbb N_0 \times \mathbb Z\\ 0\le|n|\le j\le N-1 }} e_n = \frac{1}{N} \sum_{j=0}^{N-1} \underbrace{\sum_{n=-j}^j e_n}_{=D_N},$$ so your $F_N$ is the average of the first $N$ Dirichlet kernels $D_N$, which is e.g. the Wikipedia form of the Fejér kernel. So you can now follow any usual textbook derivation (such as the one I just posted here) to find an explicit form of $F_N$: $$F_N(x) = \frac1{N} \left(\frac{\sin(Nx/2)}{\sin(x/2)}\right)^2$$ On the other hand, by the highschool geometric series formula, $$\sum_0^N e_n = \sum_0^N (e^{ix})^n=\frac{e^{iNx}-1}{e^{ix}-1}=\frac{e_N-1}{e_1-1},$$ and by elementary trigonometry, $$ |e^{ix} - 1|^2 = \cos^2x + \sin^2 x+1 - 2\cos x = 2(1-\cos x) = 4\sin^2(x/2).$$ This immediately implies the result- $$ \frac1N\left|\sum_0^N e_n\right|^2 = \frac{|e_N-1|^2}{N|e_1-1|^2} = \frac{1}{N} \frac{\sin^2(Nx/2)}{\sin^2(x/2)}.$$