Reading here page 8 and 9 the author says the following :
For a sequence $x=x_0,x_1,...,x_{n-1}$ where $x_i \in \{+1, -1\}$ compute the $cos$ related Discrete Fourier coefficient as :
$$c_j = \sum_{k=0}^{n-1} x_k \cdot \cos\frac{2k\pi j}{n}$$
The first paragraph of the page 9 says that $c_j$ converge to $\mathcal{N}(0, \frac{n}{2})$ under the assumption that $x$ is random. Let's try to prove this.
Attempt proof:
Assume that $C_j$ is the random variable that generates $c_j$ and $X_0,\dots X_{n-1} $ generates $x_0,\dots,x_1$ (w.r.t to uniform distribution over $1$ and $-1$).
We use the characteristic function as follows: \begin{align} \psi_{C_j}(t) &= \psi_{\sum_{k=0}^{n-1} X_k \cdot \cos\frac{2k\pi j}{n}} \ (t) \\ &=\prod_{k=0}^{n-1}\psi_{X_k}\left(t \cdot \cos\frac{2k\pi j}{n}\right) \\ &=\prod_{k=0}^{n-1}\frac{1}{2} + \frac{1}{2} e^{i t \cos\frac{2k\pi j}{n}} \end{align} The last equality sign follows from the fact that $x_k$ are supposed to be random (hence have a Bernoulli distribution whose characteristic function is $1 - p + pe^{it}$ with $p = \frac{1}{2}$). See the table for this value. Also the linearity property used in the second equality sign is also described there.
Letting $n \rightarrow \infty$ , we have to prove that $\psi_{C_j}(t) \rightarrow e^{-\frac{1}{2}\frac{n^2}{4}t^2}$. See the table for
$\mathcal{N}(0, \frac{n}{2})$.
Now take log of the above expression : \begin{align} \log\psi_{c_j}(t) = \sum_{k=0}^{n-1} \log\left(\frac{1}{2} + \frac{1}{2} e^{i t \cos\frac{2k\pi j}{n}}\right) \end{align}
Taking the $log$ would help me to use a Riemann sum and use the integral, proving that this value goes to $-\frac{1}{2}\frac{n^2}{4}t^2$ as $n \rightarrow \infty$.
I don't see exactly how to do it. Any help from here ?
First, saying that a process $X_n$ is asymptotically $\mathcal{N}(0,n/2)$ is a loose way of saying that $\sqrt{2/n}X_n$ is asymptotically standard normal; note in some of your statements you have $n$ sent to infinity one side of an equation but $n$ showing up on the other side. Also note you are using the ch. function for a bernoulli 0/1 variable rather than -1/1 ("rademacher") variable. You will then want to show that $\Pi_j\cos(t\sqrt{2/n}\cos(2\pi k j/n))$ tends to the standard normal ch. function, in $t$.
Alternatively you can apply the Lindeberg-Feller CLT. In Wikipedia's notation, \begin{align} s_n^2=(2/n)\sum_{k=0}^n\cos^2(2\pi kj/n)\to2\int_0^1\cos^2(2\pi j x)dx=1\\ \end{align}
($j\in\mathbb{N}$, right?). The Lindeberg condition is
\begin{align} \lim_{n\to\infty}(1/\sigma_n^2)\sum_j \mathbb{E}((2/n)\cos^2(2\pi k j)/n)\{\sqrt{2/n}|\cos((2\pi k j)/n)|>\sigma_n\epsilon\})=0 \end{align} and the terms all vanish once $n$ is so large that $\sqrt{n/2}>1/(\sigma_n\epsilon)$.