## Correlation functions and power spectrum

Let $(X, f, \mu)$ be a mixing system, $\phi\in L^2(\mu)$ with $\mu(\phi)=0$.
The auto-correlation function is defined by $\rho(\phi,f,k)=\mu(\phi\circ f^k\cdot f)$.
In the following we assume $\sigma^2:=\sum_{\mathbb{Z}}\rho(\phi, f,k)$ converges. Under some extra condition, we have the central limit theorem $\frac{S_n \phi}{\sqrt{n}}$ converges to a normal distribution.

The power spectrum of $(f,\mu,\phi)$ is defined by (when the limit exist)
$\displaystyle S:\omega\in [0,1]\mapsto \lim_{n\to\infty} \frac{1}{n}\int |\sum_{k=0}^{n-1} e^{2\pi \i k\omega} \phi\circ f^k|^2 d\mu$.
Note that $S(0)=S(1)=\sigma^2$ whenever $\sum_{\mathbb{Z}}\rho(\phi, f,k)$ converges.
Proof. Let $T_k=\sum_{n=-k}^k \rho(n)$. Then $T_k\to \sigma^2$. So
$\mu|\sum_{k=0}^{n-1}\phi\circ f^k|^2 =\sum_{k,l=0}^{n-1} \mu(\phi\cdot \phi\circ f^{k-l})=\sum_{k,l=0}^{n-1}\rho(k-l)$
$=n\cdot \rho(0)+(n-1)\cdot (\rho(1)+\rho(-1))+\cdots +2\cdot (\rho(n-1)+\rho(1-n))$
$=T_0+ T_1+\cdots + T_{n-1}$. Then we have
$S(0)=S(1)=\lim \frac{1}{n}(T_0+ T_1+\cdots + T_{n-1})=\sigma^2$ since $T_k\to \sigma^2$.

More generally, we have $\mu|\sum_{k=0}^{n-1} e^{2\pi \i k\omega} \phi\circ f^k|^2 =\sum_{k,l=0}^{n-1} e^{2\pi \i (k-l)\omega} \mu(\phi\cdot \phi\circ f^{k-l})=T_0^\omega + T_1^\omega +\cdots + T_{n-1}^\omega,$
where $T_k^\omega=\sum_{n=-k}^k e^{2\pi \i n\omega}\rho(n)$. So $S(\omega)$ exists whenever $\sum_{\mathbb{Z}} e^{2\pi \i n\omega}\rho(n)$ converges.

This is the power spectrum of $(X,f,\mu,\phi)$. Some observations:

Proposition. Assume $\sum |\rho(n)|<\infty$.
Then $S(\omega)$ is well-defined, continuous function on $0\le \omega\le 1$. Moreover,
$S(\cdot)$ is $C^{r-2}$ if $|\rho(k)|\le C k^{-r}$ for all $k$;
$S(\cdot)$ is $C^{\infty}$ if $\rho(k)$ decay rapidly;
$S(\cdot)$ is $C^{\omega}$ if $\rho(k)$ decay exponentially.

Some preparations.

Hardy: Let $a_n$ be a sequence of real numbers, $A_n=a_1+\cdots +a_n$ such that
$|n\cdot a_n|\le M$ for all $n\ge 1$;
$\sigma_n=\frac{A_1+\cdots +A_n}{n}\to A$.
Then $\sum_{n\ge 1}a_n$ also converges to $A$.

Proof. Let $\epsilon>0$ be given, $N$ large such that $|\sigma_n-A|\le \epsilon$ for all $n\ge N$.
Then for any $p\ge 1$, we have
$(n+p)\sigma_{n+p}-n\sigma_n=A_{n+1}+\cdots + A_{n+p}$;
$R_{n,p}=(n+p)(\sigma_{n+p}-A)-n(\sigma_n-A)-p(A_n-A)=A_{n+1}+\cdots + A_{n+p}-pA$
$=a_{n+1}+(a_{n+1}+a_{n+2})+\cdots +(a_{n+1}+\cdots+ a_{n+p})$.
Note that $|R_{n,p}|\le \frac{M}{n+1}+(\frac{M}{n+1}+\frac{M}{n+2})+\cdots +(\frac{M}{n+1}+\cdots+ \frac{M}{n+p}) \le \frac{p(p+1)}{2}\cdot \frac{M}{n}.$
Then $|A_n-A|\le \frac{n+p}{p}|\sigma_{n+p}-A|+\frac{n}{p}|\sigma_{n}-A|+\frac{|R_{n,p}|}{p} \le \epsilon+2n\epsilon/p+\frac{p+1}{2n}M.$ So we can pick $p\sim n\sqrt{\epsilon}$, which leads to
$|A_n-A|\le \epsilon+3\sqrt{\epsilon}+M\sqrt{\epsilon}$ for all $n\ge N$.

Let $f\in C(\mathbb{T})$ be differentiable, and $f'\in L^1(\mathbb{T})$. Let $f(t)\sim \sum_n f_n e^{2\pi\text{i} nt}$ and $f'(t)\sim \sum_n d_n e^{2\pi\text{i} nt}$ be the Fourier series. Then $d_n=\text{i} n f_n$.
Proof: integrate by parts.

Let $f\in C(\mathbb{T})$ be differentiable, and $f'\in L^2(\mathbb{T})$. Then $\sum_n |f_n|$ converges.
In particular, $\sum_n |f_n|$ converge for all $f\in C^1(\mathbb{T})$.

Proof. Note that $\sum_{n\neq 0} |f_n|=\sum_{n\neq 0} |\frac{1}{n}\cdot d_n| \le (\sum_{n\ge 1} \frac{2}{n^2})^{1/2}\cdot(\sum_n d_n^2)^{1/2}=\frac{\pi}{\sqrt{3}}\cdot\|f'\|_{L^2}.$