Category Archives: Analysis

Area under holomorphic maps

Let $f$ be a map from $(x,y)\in \mathbb{R}^2$ to $(a,b)\in \mathbb{R}^2$. The area form $dA=dx\wedge dy$ gives the Jacobian $dA=da\wedge db= J(x,y)dx\wedge dy$, where $J(x,y)=a_xb_y- a_yb_x$.

Now consider the complex setting, where $\displaystyle dA=\frac{i}{2} dz\wedge d\bar z$. Let $f$ be a map from $z\in \mathbb{C}$ to $w\in \mathbb{C}$. Then $\displaystyle dA=\frac{i}{2} dw\wedge d\bar w= \frac{i}{2}f'(z)\overline{f'(z)} dz\wedge d\bar z$. So this time the Jacobian $J(z)$ becomes $f'(z)\overline{f'(z)}$.

Suppose $\displaystyle f(z)=\sum_{n\ge 0} a_n z^n$ is a holomorphic map on the unit disk $D$. Then
$\displaystyle J(z)=\sum_{n,m\ge 0}nm a_n \bar a_m z^{n-1} \bar z^{m-1}$, the area of $f(D)$ is $\displaystyle \int_D J_f(z) dA$.

Using polar coordinate, we have $dA= rdr\, d\theta$, $\displaystyle z^{n-1} \bar z^{m-1}=r^{n+m-2}e^{i\theta(n-m)}$,
and $\displaystyle \int_D r^{n+m-2}e^{i\theta(n-m)} rdr\, d\theta= 0$ if $n\neq m$, and $=\frac{\pi}{n}$ if $n=m$.

So $\displaystyle |f(D)|=\sum_{n\ge 0} n^2 |a_n|^2\cdot \frac{\pi}{n}=\pi \sum_{n\ge 0} n |a_n|^2$.

Regularity of center manifold

Let $X:\mathbb{R}^d\to \mathbb{R}^d$ be a $C^\infty$ vector field with $X(o)=0$. Then the origin $o$ is a fixed point of the generated flow on $\mathbb{R}^d$. Let $T_o\mathbb{R}^d=\mathbb{R}^s\oplus\mathbb{R}^c\oplus\mathbb{R}^u$ be the splitting into stable, center and unstable directions. Moreover, there are three invariant manifolds (at least locally) passing through $o$ and tangent to the corresponding subspaces at $o$.

Theorem (Pliss). For any $n\ge 1$, there exists a $C^n$ center manifold $C^n(o)=W^{c,n}(o)$.

Generally speaking, the size of the center manifold given above depends on the pre-fixed regularity requirement. Theoretically, there may not be a $C^\infty$ center manifold, since $C^n(o)$ could shrink to $o$ as $n\to\infty$. An explicit example was given by van Strien (here). He started with a family of vector fields $X_\mu(x,y)=(x^2-\mu^2, y+x^2-\mu^2)$. It is easy to see that $(\mu,0)$ is a fixed point, with $\lambda_1=2\mu<\lambda_2=1$. The center manifold can be represented (locally) as the graph of $y=f_\mu(x)$.

Lemma. For $n\ge 3$, $\mu=\frac{1}{2n}$, $f_\mu$ is at most $C^{n-1}$ at $(\frac{1}{2n},0)$.

Proof. Suppose $f_\mu$ is $C^{k}$ at $(\frac{1}{2n},0)$, and let $\displaystyle f_\mu(x)=\sum_{i=1}^{k}a_i(x-\mu)^i+o(|x-\mu|^{k})$ be the finite Taylor expansion. The vector field direction $(x^2-\mu^2, y+x^2-\mu^2)$ always coincides with the tangent direction $(1,f'_\mu(x))$ along the graph $(x,f_\mu(x))$, which leads to

$(x^2-\mu^2)f_\mu'(x)=y+x^2-\mu^2=f_\mu(x)+x^2-\mu^2$.

Note that $x^2-\mu^2=(x-\mu)^2+2\mu(x-\mu)$. Then up to an error term $o(|x-\mu|^{k})$, the right-hand side in terms of $(x-\mu)$: $(a_1+2\mu)(x-\mu)+(a_2+1)(x-\mu)^2+\sum_{i=3}^{k}a_i(x-\mu)^i$; while the left-hand side in terms of $(x-\mu)$:

$(x-\mu)^2f_\mu'(x)+2\mu(x-\mu)f_\mu'(x)=\sum_{i=1}^{k}ia_i(x-\mu)^{i+1}+\sum_{i=1}^{k}2\mu i a_i(x-\mu)^i$

$=\sum_{i=2}^{k}(i-1)a_{i-1}(x-\mu)^{i}+\sum_{i=1}^{k}2\mu i a_i(x-\mu)^i$.

So for $i=1$: $2\mu a_1=a_1+2\mu$, $a_1=\frac{-2\mu}{1-2\mu}\sim 0$;

$i=2$: $a_2+1=a_1+4\mu a_2$, $a_2=\frac{a_1-1}{1-4\mu}\sim -1$;

$i=3,\cdots,k$: $a_i=(i-1)a_{i-1}+2i\mu a_i$, $(1-2i\mu)a_i=(i-1)a_{i-1}$.

Note that if $k\ge n$, we evaluate the last equation at $i=n$ to conclude that $a_{n-1}=0$. This will force $a_i=0$ for all $i=n-2,\cdots,2$, which contradicts the second estimate that $a_2\sim -1$. Q.E.D.

Consider the 3D vector field $X(x,y,z)=(x^2-z^2, y+x^2-z^2,0)$. Note that the singular set $S$ are two lines $x=\pm z$, $y=0$ (in particular it contains the origin $O=(0,0,0)$). Note that $D_OX=E_{22}$. Hence a cener manifold $W^c(O)$ through $O$ is tangent to plane $y=0$, and can be represented as $y=f(x,z)$. We claim that $f(x,x)=0$ (at least locally).

Proof of the claim. Suppose on the contrary that $c_n=f(x_n,x_n)\neq0$ for some $x_n\to 0$. Note that $p_n=(x_n,c_n,x_n)\in W^c(O)$, and $W^c(O)$ is flow-invariant. However, there is exactly one flow line passing through $p_n$: the line $L_n=\{(x_n,c_nt,x_n):t>0\}$. Therefore $L_n\subset W^c(O)$, which contradicts the fact that $W^c(O)$ is tangent to plane $y=0$ at $O$. This completes the proof of the claim.

The planes $z=\mu$ are also invariant under the flow. Let’s take the intersection $W_\mu=\{z=\mu\}\cap W^c(O)=\{(x,f(x,\mu),\mu)\}$. Then we check that $\{(x,f(x,\mu))\}$ is a (in fact the) center manifold of the restricted vector field in the plane $z=\mu$. We already checked that $f(x,\mu)$ is not $C^\infty$, so is $W^c(O)$.

Doubling map on unit circle

1. Let $\tau:x\mapsto 2x$ be the doubling map on the unit torus. We also consider the uneven doubling $f_a(x)=x/a$ for $0\le x \le a$ and $f(x)=(x-a)/(1-a)$ for $a \le x \le 1$. It is easy to see that the Lebesgue measure $m$ is $f_a$-invariant, ergodic and the metric entropy $h(f_a,m)=\lambda(m)=\int \log f_a'(x) dm(x)=-a\log a-(1-a)\log (1-a)$. In particular, $h(f_a,m)\le h(f_{0.5},m)=\log 2 =h_{\text{top}}(f_a)$ and $h(f_a,m)\to 0$ when $a\to 0$.

2. Following is a theorem of Einsiedler–Fish here.

Proposition. Let $\tau:x\mapsto 2x$ be the doubling map on the unit torus, $\mu$ be an $\tau$-invariant measure with zero entropy. Then for any $\epsilon>0$, $\beta>0$, there exist $\delta_0>0$ and a subset $E\subset \mathbb{T}$ with $\mu(E) > 0$, such that for all $x \in E$, and all $\delta<\delta_0$: $\mu(B(x,\delta))\ge \delta^\beta$.

A trivial observation is $\text{HD}(\mu)=0$, which also follows from general entropy-dimension formula.

Proof. Let $\beta$ and $\epsilon$ be fixed. Consider the generating partition $\xi=\{I_0, I_1\}$, and its refinements $\xi_n=\{I_\omega: \omega\in\{0,1\}^n\}$ (separated by $k\cdot 2^{-n}$)….

Furstenberg introduced the following notation in 1967

Definition. A multiplicative semigroup $\Sigma\subset\mathbb{N}$ is lacunary, if $\Sigma\subset \{a^n: n\ge1\}$ for some integer $a$. Otherwise, $\Sigma$ is non-lacunary.

Example. Both $\{2^n: n\ge1\}$ and $\{3^n: n\ge1\}$ are lacunary semigroups. $\{2^m\cdot 3^n: m,n\ge1\}$ is a non-lacunary semigroup.

Theorem. Let $\Sigma\subset\mathbb{N}$ be a non-lacunary semigroup, and enumerated increasingly by $s_i > s_{i+1}\cdot$. Then $\frac{s_{i+1}}{s_i}\to 1$.

Example. $\Sigma=\{2^m\cdot 3^n: m,n\ge1\}$. It is equivalent to show $\{m\log 2+ n\log 3: m,n\ge1\}$ has smaller and smaller steps.

Theorem. Let $\Sigma\subset\mathbb{N}$ be a non-lacunary semigroup, and $A\subset \mathbb{T}$ be $\Sigma$-invariant. If $0$ is not isolated in $A$, then $A=\mathbb{T}$.

Furstenberg Theorem. Let $\Sigma\subset\mathbb{N}$ be a non-lacunary semigroup, and $\alpha\in \mathbb{T}\backslash \mathbb{Q}$. Then $\Sigma\alpha$ is dense in $\mathbb{T}$.

In the same paper, Furstenberg also made the following conjecture: a $\Sigma$-invariant ergodic measure is either supported on a finite orbit, or is the Lebesgue measure.

A countable group $G$ is said to be amenable, if it contains at least one Følner sequence. For example, any abelian countable group is amenable. Note that for amenable group action $G\ni g:X\to X$, there always exists invariant measures and the decomposition into ergodic measures. More importantly, the generic point can be defined by averaging along the Følner sequences, and almost every point is a generic point for an ergodic measure. In a preprint, the author had an interesting idea: to prove Furstenberg conjecture, it suffices to show that every irrational number is a generic point of the Lebesgue measure. Then any other non-atomic ergodic measures, if exist, will be starving to death since there is no generic point for them 🙂

Some simple dynamical systems

Dynamical formulation of Prisoner’s dilemma
Originally, consider the two players, each has a set of stratagies, say $\mathcal{A}=\{a_{i}\}$ and $\mathcal{B}=\{b_{j}\}$. The pay-off $P_k=P_k(a_{i},b_{j})$ for player $k$ depends on the choices of both players.

Now consider two dynamical systems $(M_i,f_i)$. The set of stratagies consists of the invariant probability measures, and the pay-off functions can be

$\phi_k(\mu_1,\mu_2)=\int \Phi_k(x,y)d\mu_1 d\mu_2$, where $\mu_i\in\mathcal{M}(f_i)$;

$\psi_k(\mu_1,\mu_2)=\int \Phi_k(x,y)d\mu_1 d\mu_2-h(f_i,\mu_i)$.

The frist one is related to Ergodic optimization. The second one does sound better, since one may want to avoid a complicated (measured by its entropy) stratagy that has the same $\phi$ pay-off.

Gambler’s Ruin Problem
A gambler starts with an initial fortune of $i, and then either wins$1 (with $p$) or loses \$1 (with $q=1-p$) on each successive gamble (independent of the past). Let $S_n$ denote the total fortune after the n-th gamble. Given $N>i$, the gambler stops either when $S_n=0$ (broke), or $S_n=N$ (win), whichever happens first.

Let $\tau$ be the stopping time and $P_i(N)=P(S_\tau=N)$ be the probability that the gambler wins. It is easy to see that $P_0(N)=0$ and $P_N(N)=1$. We need to figure out $P_i(N)$ for all $i=1,\cdots,N-1$.

Let $S_0=i$, and $S_n=S_{n-1}+X_n$. There are two cases according to $X_1$:

$X_1=1$ (prob $p$): win eventually with prob $P_{i+1}(N)$;

$X_1=-1$ (prob $q$): win eventually with prob $P_{i-1}(N)$.

So $P_i(N)=p\cdot P_{i+1}(N)+q\cdot P_{i-1}(N)$, or equivalently,
$p\cdot (P_{i+1}(N)-P_i(N))=q\cdot (P_i(N)-P_{i-1}(N))$ (since $p+q=1$), $i=1,\cdots,N-1$.

Recall that $P_0(N)=0$ and $P_N(N)=1$. Therefore $P_{i+1}(N)-P_i(N)=\frac{q^i}{p^i}(P_1(N)-P_{0}(N))=\frac{q^i}{p^i}P_1(N)$, $i=1,\cdots,N-1$. Summing over $i$, we get $1-P_1(N)=P_1(N)\cdot\sum_{1}^{N-1}\frac{q^i}{p^i}$, $P_1(N)=\frac{1}{\sum_{0}^{N-1}\frac{q^i}{p^i}}=\frac{1-q/p}{1-q^N/p^N}$ (if $p\neq .5$) and $P_1(N)=\frac{1}{N}$ (if $p= .5$). Generally $P_i(N)=P_1(N)\cdots\sum_{0}^{i-1}\frac{q^j}{p^j}=\frac{1-q^i/p^i}{1-q^N/p^N}$ (if $p\neq .5$) and $P_1(N)=\frac{i}{N}$ (if $p= .5$).

Observe that for fixed $i$, the limit $P_i(\infty)=1-q^i/p^i>0$ only when $p>.5$, and $P_i(\infty)=0$ whenever $p\le .5$.

Finite Blaschke products
Let $f$ be an analytic function on the unit disc $\mathbb{D}=\{z\in\mathbb{C}: |z|<1 \}$ with a continuous extension to $\overline{\mathbb{D}}$ with $f(S^1)\subset S^1$. Then $f$ is of the form

$\displaystyle f(z)=\zeta\cdot\prod_{i=1}^n\left({{z-a_i}\over {1-\bar{a_i}z}}\right)^{m_i}$,

where $\zeta\in S^1$, and $m_i$ is the multiplicity of the zero $a_i\in \mathbb{D}$ of $f$. Such $f$ is called a finite Blaschke product.

Proposition. Let $f$ be a finite Blaschke product. Then the restriction $f:S^1\to S^1$ is measure-preserving if and only if $f(0)=0$. That is, $a_i=0$ for some $i$.

Proof. Let $\phi$ be an analytic function on $\overline{\mathbb{D}}$. Then $\int_{S_1}\phi d(\theta)=\phi(0)$ and $\int_{S_1}\phi\circ f d(\theta)=\phi\circ f(0)$.

Significance: there are a lot of measure-preserving covering maps on $S^1$.

Kalikov’s Random Walk Random Scenery
Let $X=\{1,-1\}^{\mathbb{Z}}$, and $\sigma:X\to X$ to the shift $\sigma((x_n))=(x_{n+1})$. More generally, let $A$ be a finite alphabet and $p$ be probability vector on $A$, and $Y=A^{\mathbb{Z}}$, $\nu=p^{\times\mathbb{Z}}$. Consider the skew-product $T:X\times Y\to X\times Y$, $(x,y)\mapsto (\sigma x, \sigma^{x_0}y)$. It is clear that $T$ preserves any $\mu\times \nu$, where $\mu$ is $\sigma$-invariant.

Proposition. Let $\mu=(.5,.5)^{\times\mathbb{Z}}$. Then $h(T,\mu\times \nu)=h(\sigma,\mu)=\log 2$ for all $(A,p)$.

Proof. Note that $T^n(x,y)\mapsto (\sigma^n x, \sigma^{x_0+\cdot+x_{n-1}}y)$. CLT tells that $\mu(x:|x_0+\cdot+x_{n-1}|\ge\kappa\cdot \sqrt{n})< \delta(\kappa)$ as $n\to\infty$, where $\delta(\kappa)\to 0$ as $\kappa\to\infty$. There are only $2^{n+\kappa \sqrt{n}}$ different $n$-strings (up to an error).

Significance: this gives a natural family of examples that are K, but not isomorphic to Bernoulli.

Creation of one sink. 1D case. Consider the family $f_t:x\mapsto x^2+2-t$, where $0\le t\le 2$. Let $t_\ast$ the first parameter such that the graph is tangent to the diagonal at $x_\ast=f_{t_\ast}(x_\ast)$. Note that $x_\ast$ is parabolic. Then for $t\in(t_\ast,t_\ast+\epsilon)$, $f_t(x)=x$ has two solutions $x_1(t), where $x_1(t)$ is a sink, and $x_2(t)$ is a source.

2D case. Let $B=[-1,1]\times[-\epsilon,\epsilon]$ be a rectangle, $f$ be a diffeomorphism such that $f(B)$ is a horseshoe lying  above $B$ of shape ‘V’. Moreover we assume $|\det Df|<1$. Let $f_t(x,y)=f(x,y)-(0,t)$ such that $f_1(B)$ is the regular horseshoe intersection:  V . Clearly there exists a fixed point $p_1$ of $f_1$ in $B$. We assume $\lambda_1(1)<-1<\lambda_2(1)<0$. Then Robinson proved that $f_t$ admits a fixed point in $B$ which is a sink.

First note that for any $t$, and any fixed point of $f_t$ (if exists), it is not on the boundary of $B$. Since $p_1$ is a nondegenerate fixed point of $f_1$, the fixed point continues to exist for some open interval $(t_1,1)$ (assume it is maximal, and denote the fixed point by $p_t$). Clearly $t_1>0$. Note that $p_{t_1}$ is also fixed by $f_{t_1}$, since it is a closed property. If there is some moment with $\lambda_1(t)=\lambda_2(t)$ for the fixed point $p_t$ of $f_t$, then it is already a sink, since $\det Df=\lambda_1\cdot\lambda_2<1$. So in the following we consider the case $\lambda_1\neq\lambda_2$ for all $p_t$, $t\in[t_1,1]$. Then the continuous dependence of parameters implies that both are continuous functions of $t$. The fixed point $p_{t_1}$ must be degenerate, since the fixed point ceases to exist beyond $t_1$, which means: $\lambda_i(t_1)=1$ for some $i\in\{1,2\}$.

Case 1. $\lambda_1(t_1)=1$. Note that $\lambda_1(1)<-1$. So $\text{Re}\lambda_1(t_\ast)=0$ for some $t_\ast\in(t_1,1)$, which implies that $\lambda_1(t_\ast)=ai$ for some $a\neq 0$. In particular, $\lambda_2(t_\ast)=-ai$, and $a^2=|\det Df|<1$. So $p_{t_\ast}$ is a (complex) sink.

Case 2. $\lambda_2(t_1)=1$. Note that $\lambda_2(1)<0$. Similarly $\text{Re}\lambda_2(t_\ast)=0$ for some $t_\ast\in(t_1,1)$.

So in the orientation-preserving case there always exists a complex sink. In the orientation-reversing case ($\lambda_2(1)\in(0,1)$), we need modify the argument for case 2:

Case 2′. $\lambda_2(t_1)=1$. Note that $\lambda_2(1)\in(0,1)$. So $|\lambda_1(t_\ast)|<1$ for some $t_\ast\in(t_1,1)$. We pick $t_\ast$ close to $t_1$ in the sense that $|\lambda_1(t_\ast)|>|\det Df|$, which implies $|\lambda_1(t_\ast)|\ast<1$, too. So $p_{t_\ast}$ is also a sink.

Playing pool with pi

This is a short note based on the paper

Playing pool with π (the number π from a billiard point of view) by G. Galperin in 2003.

Let’s start with two hard balls,  denoted by $B_1$ and $B_2$, of masses $0 on the positive real axis with position $0, and a rigid wall at the origin. Without loss of generality we assume $m=1$. Then push the ball $B_2$ towards $B_1$, and count the total number $N(M)$ of collisions (ball-ball and ball-wall) till the $B_2$ escapes to $\infty$ faster than $B_1$.

Case. $M=1$: first collision at $y(t)=x$, then $B_2$ rests, and $B_1$ move towards the wall; second collision at $x(t)=0$, then $B_1$ gains the opposite velocity and moves back to $B_2$; third collision at $x(t)=x$, then $B_1$ rests, and $B_2$ move towards $\infty$.

Total counts $N(1)=3$, which happens to be first integral part of $\pi$. Well, this must be coincidence, one might wonder.

However, Galperin proved that, if we set $M=10^{2k}$, then $N(M)$ gives the integral part of $10^k\pi$. For example, $N(10^2)=31$; and  $N(10^4)=314$.

Notes-09-14

4. Borel–Cantelli Lemma(s). Let $(X,\mathcal{X},\mu)$ be a probability space. Then

If $\sum_n \mu(A_n)<\infty$, then $\mu(x\in A_n \text{ infinitely often})=0$.

If $A_n$ are independent and $\sum_n \mu(A_n)=\infty$, then for $\mu$-a.e. $x$, $\frac{1}{\mu(A_1)+\cdots+\mu(A_n)}\cdot|\{1\le k\le n:x\in A_k\}|\to 1$.

The dynamical version often involves the orbits of points, instead of the static points. In particular, let $T$ be a measure-preserving map on $(X,\mathcal{X},\mu)$. Then

$\{A_n\}$ is said to be a Borel–Cantelli sequence with respect to $(T,\mu)$ if $\mu(T^n x\in A_n \text{ infinitely often})=1$;

$\{A_n\}$ is said to be a strong Borel–Cantelli sequence if $\frac{1}{\mu(A_1)+\cdots+\mu(A_n)}\cdot|\{1\le k\le n:T^k x\in A_k\}|\to 1$ for $\mu$-a.e. $x$.

3. Let $H(q,p,t)$ be a Hamiltonian function, $S(q,t)$ be the generating function in the sense that $\frac{\partial S}{\partial q_i}=p_i$. Then the Hamilton–Jacobi equation is a first-order, non-linear partial differential equation

$H + \frac{\partial S}{\partial t}=0$.

Note that the total derivative $\frac{dS}{dt}=\sum_i\frac{\partial S}{\partial q_i}\dot q_i+\frac{\partial S}{\partial t}=\sum_i p_i\dot q_i-H=L$. Therefore, $S=\int L$ is the classical action function (up to an undetermined constant).

2. Let $\gamma_s(t)$ be a family of geodesic on a Riemannian manifold $M$. Then $J(t)=\frac{\partial }{\partial s}|_{s=0} \gamma_s(t)$ defines a vector field along $\gamma(t)=\gamma_0(t)$, which is called a Jacobi field. $J(t)$ describes the behavior of the geodesics in an infinitesimal neighborhood of a given geodesic $\gamma$.

Alternatively, A vector field $J(t)$ along a geodesic $\gamma$ is said to be a Jacobi field, if it satisfies the Jacobi equation:

$\frac{D^2}{dt^2}J(t)+R(J(t),\dot\gamma(t))\dot\gamma(t)=0,$

where $D$ denotes the covariant derivative with respect to the Levi-Civita connection, and $R$ the Riemann curvature tensor on $M$.

Exponential map on the complex plane

Let $f(z)=e^z=e^x(\cos y+\i\sin y)$ (for $z=x+\i y$) be the exponential map. Note that $f(x)>0$ for all real numbers and $f^{n+1}(x):=f(f^nx)$ goes to $\infty$ really fast: the dynamics of $f$ on $\mathbb{R}$ is trivial. But the dynamics of $f$ on $\mathbb{C}$ is completely different. First note that $e^{2k\pi\i}=1$: the map is not a diffeomorphism, but a covering map branching at the origin. The following theorem was conjectured by Fatou (1926) and proved by Misiurewicz (1981).

Theorem (Orbits of the complex exponential map).
Let $\mathcal{O}_e(z)$ be the orbit of a point $z\in\mathbb{C}$ under the iterates of $f(z)=e^z$. Then each of the following sets is dense in the complex plane:
1. the basin of $\infty$, $B_e(\infty)=\{z\in\mathbb{C}: f^n(z)\to\infty\}$;
2. the set of transitive points, $\text{Tran}(e)=\{z\in\mathbb{C}: \mathcal{O}_e(z)\text{ is dense}\}$;
3. the set of periodic points, $\text{Per}(e)=\{z\in\mathbb{C}: \mathcal{O}_e(z)\text{ is finite}\}$.

So the exponential map is chaotic on the complex plane.

Reference:

The exponential map is chaotic: An invitation to transcendental dynamics,
Zhaiming Shen and Lasse Rempe-Gillen arXiv

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

Some notations

7. Let $f$ be an Anosov diffeomorphism and $g\in\mathcal{U}(f)$ be close enough, which leads to a Holder continuous conjugate $h_g:M\to M$ with $g\circ h_g=h_g\circ f$. Ruelle found an explicit formula of $h_g$.

Let $f,g:M\to M$ be two homeomorphisms, $d(f,g)=\sup_M d(fx,gx)$, and $\mathcal{U}(f,\epsilon)=\{g \text{ homeo and }d(f,g)<\epsilon\}$. Let $g\in \mathcal{U}(f,\epsilon)$. Then the map $X_g:x\in M \mapsto \exp^{-1}_{fx}(gx)\in T_{fx}M$ gives a shifted-vector field on $M$, which induces a diffeomorhism $\mathcal{U}(f,\epsilon)\to \mathcal{X}(0_f,\epsilon), g\mapsto X_g$.
Let $f$ be a $C^r$ diffeomprhism. Then $\mathcal{X}^r(0_f,\epsilon)\to \mathcal{U}^r(f,\epsilon), g\mapsto X_g$ induces the local Banach structure and turns $\mathrm{Diff}^r(M)$ into a Banach manifold.

Let $X_g\circ f^{-1}=X_g^s+X_g^u$ be the decomposition of the correction $X_g\circ f^{-1}$ with respect to the hyperbolic splitting $TM= E_g^s\oplus E_g^u$. Then the derivative of $g\mapsto h_g$ in the direction of $X_g$ is given by the vector field $\displaystyle \sum_{n\ge 0}Dg^n X^s_g-\sum_{n\ge1}Dg^{-n}X^u_g$.

6. Let $M$ be a compact orientable surface of genus $g\ge1$, $s\ge1$ and let $\Sigma=\{p_1,\cdots,p_s\}$ be a subset of $M$. Let $\kappa= (\kappa_1,\cdots,\kappa_s)$ be a $s$-tuple of positive integers with $\sum (\kappa_i-1) =2g-2$.

A translation structure on $(M,\Sigma)$ of type $\kappa$ is an atlas on $M\backslash\Sigma$
for which the coordinate changes are translations, and such that each singularity $p_i$
has a neighborhood which is isomorphic to the $\kappa_i$-fold covering of a neighborhood
of $0$ in $\mathbb{R}^2\backslash\{0\}$.

The Teichmüller space $Q_{g,\kappa}= Q(M,\Sigma,\kappa)$ is the set of such structures modulo isotopy relative to $\Sigma$. It has a canonical structure of manifold.

Area of the symmetric difference of two disks

This post goes back to high school: the area $\delta_d$ of the symmetric difference of two $d$-dimensional disks when one center is shifted a little bit. Let’s start with $d=1$. So we have two intervals $[-r,r]$ and $[x-r,x+r]$. It is easy to see the symmetric difference is of length $\delta_1(x)=2x$.

Then we move to $d=2$: two disks $L$ and $R$ of radius $r$ and center distance $x=2a. So the angle $\theta(x)$ satisfies $\cos\theta=\frac{a}{r}$.

The symmetric difference is the union of $R\backslash L$ and $L\backslash R$, which have the same area: $\displaystyle (\pi-\theta)r^2+2x\sqrt{r^2-x^2}-\theta r^2=2(\frac{\pi}{2}-\arccos\frac{x}{r})r^2+2x\sqrt{r^2-x^2}$. Note that the limit
$\displaystyle \lim_{x\to0}\frac{\text{area}(\triangle)}{2a}=\lim_{a\to0}2\left(\frac{r^2}{\sqrt{1-\frac{a^2}{r^2}}}\cdot\frac{1}{r}+\sqrt{r^2-a^2}\right)=4r$.
So $\delta_2(x)\sim 4rx$.

I didn’t try for $d\ge3$. Looks like it will start with a linear term $2d r^{d-1}x$.

—————–

Now let ${\bf r}(t)=(a\cos t,\sin t)$ be an ellipse with $a>1$, and ${\bf r}'(t)=(-a\sin t,\cos t)$ be the tangent vector at ${\bf r}(t)$. Let $\omega$ be the angle from ${\bf j}=(0,1)$ to ${\bf r}'(t)$.
Let $s(t)=\int_0^t |{\bf r}'(u)|du$ be the arc-length parameter and $K(s)=|{\bf l}''(s)|$ be the curvature at ${\bf l}(s)={\bf r}(t(s))$. Alternatively we have $\displaystyle K(t)=\frac{a}{|{\bf r}'(t)|^{3}}$.

The following explains the geometric meaning of curvature:

$\displaystyle K(s)=\frac{d\omega}{ds}$, or equivalently, $K(s)\cdot ds=d\omega$. $(\star)$.

Proof. Viewed as functions of $t$, it is easy to see that $(\star)$ is equivalent to $K(t)\cdot \frac{ds}{dt}=\frac{d\omega}{dt}$.

Note that $\displaystyle \cos\omega=\frac{{\bf r}'(t)\cdot {\bf j}}{|{\bf r}'(t)|}=\frac{\cos t}{|{\bf r}'(t)|}$. Taking derivatives with respect to $t$, we get
$\displaystyle -\sin\omega\cdot\frac{d\omega}{dt}=-\frac{a^2\sin t}{|{\bf r}'(t)|^3}$. Then $(\star)$ is equivalent to

$\displaystyle \frac{a}{|{\bf r}'(t)|^{3}}\cdot |{\bf r}'(t)|=\frac{a^2\sin t}{\sin\omega\cdot |{\bf r}'(t)|^3}$, or
$\displaystyle \sin\omega\cdot |{\bf r}'(t)|=a\sin t$. Note that $\displaystyle \sin^2\omega=1-\cos^2\omega=1-\frac{\cos^2 t}{|{\bf r}'(t)|^2}$. Therefore $\displaystyle \sin^2\omega\cdot |{\bf r}'(t)|^2= |{\bf r}'(t)|^2-\cos^2 t=a^2\sin^2 t$, which completes the verification.