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.

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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)<x_2(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<m\le M on the positive real axis with position 0<x< y, 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.

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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.

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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.

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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<r. So the angle \theta(x) satisfies \cos\theta=\frac{a}{r}.

difference

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.

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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}}.

ellipse

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.

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