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}.

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