# 在新加坡的这五年—学术篇

### 基础书籍：

(1) Complex Analysis, 3rd Edition, Lars V. Ahlfors

(2) Complex Analysis, Elias M. Stein

(1) Lectures on Riemann Surfaces (GTM 81), Otto Forster

(2) Lectures on Quasiconformal Mappings, Lars V. Ahlfors

(1) Real Analysis and Complex Analysis, Rudin

(2) Real Analysis, Elias M. Stein

### 专业书籍：

(1) One Dimensional Dynamics, Welington de Melo & Sebastian VanStrien

(2) Mathematical Tools for One-Dimensional Dynamics (Cambridge Studies in

Advanced Mathematics), Edson de Faria / Welington de Melo

(3) Dynamics in One Complex Variable, John Milnor；Milnor 的教材总是写的清晰明确，容易上手，推荐初学者可以读这本书。

(4) Complex Dynamics, Lennart Carleson；Carleson 的教材偏向于分析学，读起来其实也有点难度，还是读 Milnor 的教材相对容易。

(5) Complex Dynamics and Renormalization, Curtis T. McMullen；McMullen 的书适合当做查阅，也不太适合从头到尾读下去。

(6) Renormalization and 3-Manifolds Which Fiber over the Circle, Curtis T. McMullen

(7) Iteration of rational functions (GTM 132), Alan F. Beardon

(8) An Introduction to Ergodic Theory (GTM 79), Walters Peter

### 论文

Question. 是否存在 $\ell\geq 4$ 的偶数和复数 $c\in\mathbb{C}$ 使得 $f(z)=z^{\ell}+c$ 的 Julia 集合 $J(f)$ 是正测度？

1. Combinatorics, geometry and attractors of quasi-quadratic maps，Pages 345-404 from Volume 140 (1994), Issue 2 by Mikhail Lyubich

2. Wild Cantor attractors exist，Pages 97-130 from Volume 143 (1996), Issue 1 by Hendrik Bruin, Gerhard Keller, Tomasz Nowicki, Sebastian van Strien

3. Quadratic Julia sets with positive area，Pages 673-746 from Volume 176 (2012), Issue 2 by Xavier Buff, Arnaud Chéritat

4. Polynomial maps with a Julia set of positive measure，Nowicki, Tomasz, and Sebastian van Strien，arXiv preprint math/9402215(1994).

### 参考资料：

1. 科研这条路
2. 维基百科：Julia 集合

# Hausdorff dimension of the graphs of the classical Weierstrass functions

In this paper, we obtain the explicit value of the Hausdorff dimension of the graphs of the classical Weierstrass functions, by proving absolute continuity of the SRB measures of the associated solenoidal attractors.

1. Introduction

In Real Analysis, the classical Weierstrass function is

$\displaystyle W_{\lambda,b}(x) = \sum\limits_{n=0}^{\infty} \lambda^n \cos(2\pi b^n x)$

with ${1/b < \lambda < 1}$.

Note that the Weierstrass functions have the form

$\displaystyle f^{\phi}_{\lambda,b}(x) = \sum\limits_{n=0}^{\infty} \lambda^n \phi(b^n x)$

where ${\phi}$ is a ${\mathbb{Z}}$-periodic ${C^2}$-function.

Weierstrass (1872) and Hardy (1916) were interested in ${W_{\lambda,b}}$ because they are concrete examples of continuous but nowhere differentiable functions.

Remark 1 The graph of ${f^{\phi}_{\lambda,b}}$ tends to be a “fractal object” because ${f^{\phi}_{\lambda,b}}$ is self-similar in the sense that

$\displaystyle f^{\phi}_{\lambda, b}(x) = \phi(x) + \lambda f^{\phi}_{\lambda,b}(bx)$

We will come back to this point later.

Remark 2 ${f^{\phi}_{\lambda,b}}$ is a ${C^{\alpha}}$-function for all ${0\leq \alpha < \frac{-\log\lambda}{\log b}}$. In fact, for all ${x,y\in[0,1]}$, we have

$\displaystyle \frac{f^{\phi}_{\lambda, b}(x) - f^{\phi}_{\lambda,b}(y)}{|x-y|^{\alpha}} = \sum\limits_{n=0}^{\infty} \lambda^n b^{n\alpha} \left(\frac{\phi(b^n x) - \phi(b^n y)}{|b^n x - b^n y|^{\alpha}}\right),$

so that

$\displaystyle \frac{f^{\phi}_{\lambda, b}(x) - f^{\phi}_{\lambda,b}(y)}{|x-y|^{\alpha}} \leq \|\phi\|_{C^{\alpha}} \sum\limits_{n=0}^{\infty}(\lambda b^{\alpha})^n:=C(\phi,\alpha,\lambda,b) < \infty$

whenever ${\lambda b^{\alpha} < 1}$, i.e., ${\alpha < -\log\lambda/\log b}$.

The study of the graphs of ${W_{\lambda,b}}$ as fractal sets started with the work of Besicovitch-Ursell in 1937.

Remark 3 The Hausdorff dimension of the graph of a ${C^{\alpha}}$-function ${f:[0,1]\rightarrow\mathbb{R}}$is

$\displaystyle \textrm{dim}(\textrm{graph}(f))\leq 2 - \alpha$

Indeed, for each ${n\in\mathbb{N}}$, the Hölder continuity condition

$\displaystyle |f(x)-f(y)|\leq C|x-y|^{\alpha}$

leads us to the “natural cover” of ${G=\textrm{graph}(f)}$ by the family ${(R_{j,n})_{j=1}^n}$ of rectangles given by

$\displaystyle R_{j,n}:=\left[\frac{j-1}{n}, \frac{j}{n}\right] \times \left[f(j/n)-\frac{C}{n^{\alpha}}, f(j/n)+\frac{C}{n^{\alpha}}\right]$

Nevertheless, a direct calculation with the family ${(R_{j,n})_{j=1}^n}$ does not give us an appropriate bound on ${\textrm{dim}(G)}$. In fact, since ${\textrm{diam}(R_{j,n})\leq 4C/n^{\alpha}}$ for each ${j=1,\dots, n}$, we have

$\displaystyle \sum\limits_{j=1}^n\textrm{diam}(R_{j,n})^d\leq n\left(\frac{4C}{n^{\alpha}}\right)^d = (4C)^{1/\alpha} < \infty$

for ${d=1/\alpha}$. Because ${n\in\mathbb{N}}$ is arbitrary, we deduce that ${\textrm{dim}(G)\leq 1/\alpha}$. Of course, this bound is certainly suboptimal for ${\alpha<1/2}$ (because we know that ${\textrm{dim}(G)\leq 2 < 1/\alpha}$ anyway).Fortunately, we can refine the covering ${(R_{j,n})}$ by taking into account that each rectangle ${R_{j,n}}$ tends to be more vertical than horizontal (i.e., its height ${2C/n^{\alpha}}$ is usually larger than its width ${1/n}$). More precisely, we can divide each rectangle ${R_{j,n}}$ into ${\lfloor n^{1-\alpha}\rfloor}$ squares, say

$\displaystyle R_{j,n} = \bigcup\limits_{k=1}^{\lfloor n^{1-\alpha}\rfloor}Q_{j,n,k},$

such that every square ${Q_{j,n,k}}$ has diameter ${\leq 2C/n}$. In this way, we obtain a covering ${(Q_{j,n,k})}$ of ${G}$ such that

$\displaystyle \sum\limits_{j=1}^n\sum\limits_{k=1}^{\lfloor n^{1-\alpha}\rfloor} \textrm{diam}(Q_{j,n,k})^d \leq n\cdot n^{1-\alpha}\cdot\left(\frac{2}{n}\right)^d\leq (2C)^{2-\alpha}<\infty$

for ${d=2-\alpha}$. Since ${n\in\mathbb{N}}$ is arbitrary, we conclude the desired bound

$\displaystyle \textrm{dim}(G)\leq 2-\alpha$

A long-standing conjecture about the fractal geometry of ${W_{\lambda,b}}$ is:

Conjecture (Mandelbrot 1977): The Hausdorff dimension of the graph of ${W_{\lambda,b}}$ is

$\displaystyle 1<\textrm{dim}(\textrm{graph}(W_{\lambda,b})) = 2 + \frac{\log\lambda}{\log b} < 2$

Remark 4 In view of remarks 2 and 3, the whole point of Mandelbrot’s conjecture is to establish the lower bound

$\displaystyle \textrm{dim}(\textrm{graph}(W_{\lambda,b})) \geq 2 + \frac{\log\lambda}{\log b}$

Remark 5 The analog of Mandelbrot conjecture for the box and packing dimensions is known to be true: see, e.g., these papers here and here).

In a recent paper (see here), Shen proved the following result:

Theorem 1 (Shen) For any ${b\geq 2}$ integer and for all ${1/b < \lambda < 1}$, the Mandelbrot conjecture is true, i.e.,

$\displaystyle \textrm{dim}(\textrm{graph}(W_{\lambda,b})) = 2 + \frac{\log\lambda}{\log b}$

Remark 6 The techniques employed by Shen also allow him to show that given ${\phi:\mathbb{R}\rightarrow\mathbb{R}}$ a ${\mathbb{Z}}$-periodic, non-constant, ${C^2}$ function, and given ${b\geq 2}$ integer, there exists ${K=K(\phi,b)>1}$ such that

$\displaystyle \textrm{dim}(\textrm{graph}(f^{\phi}_{\lambda,b})) = 2 + \frac{\log\lambda}{\log b}$

for all ${1/K < \lambda < 1}$.

Remark 7 A previous important result towards Mandelbrot’s conjecture was obtained by Barańsky-Barány-Romanowska (in 2014): they proved that for all ${b\geq 2}$ integer, there exists ${1/b < \lambda_b < 1}$ such that

$\displaystyle \textrm{dim}(\textrm{graph}(W_{\lambda,b})) = 2 + \frac{\log\lambda}{\log b}$

for all ${\lambda_b < \lambda < 1}$.

The remainder of this post is dedicated to give some ideas of Shen’s proof of Theorem1 by discussing the particular case when ${1/b<\lambda<2/b}$ and ${b\in\mathbb{N}}$ is large.

2. Ledrappier’s dynamical approach

If ${b\geq 2}$ is an integer, then the self-similar function ${f^{\phi}_{\lambda,b}}$ (cf. Remark 1) is also ${\mathbb{Z}}$-periodic, i.e., ${f^{\phi}_{\lambda,b}(x+1) = f^{\phi}_{\lambda,b}(x)}$ for all ${x\in\mathbb{R}}$. In particular, if ${b\geq 2}$ is an integer, then ${\textrm{graph}(f^{\phi}_{\lambda,b})}$ is an invariant repeller for the endomorphism ${\Phi:\mathbb{R}/\mathbb{Z}\times\mathbb{R}\rightarrow \mathbb{R}/\mathbb{Z}\times\mathbb{R}}$ given by

$\displaystyle \Phi(x,y) = \left(bx\textrm{ mod }1, \frac{y-\phi(x)}{\lambda}\right)$

This dynamical characterization of ${G = \textrm{graph}(f^{\phi}_{\lambda,b})}$ led Ledrappier to the following criterion for the validity of Mandelbrot’s conjecture when ${b\geq 2}$ is an integer.

Denote by ${\mathcal{A}}$ the alphabet ${\mathcal{A}=\{0,\dots,b-1\}}$. The unstable manifolds of ${\Phi}$through ${G}$ have slopes of the form

$\displaystyle (1,-\gamma \cdot s(x,u))$

where ${\frac{1}{b} < \gamma = \frac{1}{\lambda b} <1}$, ${x\in\mathbb{R}}$, ${u\in\mathcal{A}^{\mathbb{N}}}$, and

$\displaystyle s(x,u):=\sum\limits_{n=0}^{\infty} \gamma^n \phi'\left(\frac{x + u_1 + u_2 b + \dots + u_n b^{n-1}}{b^n}\right)$

In this context, the push-forwards ${m_x := (u\mapsto s(x,u))_*\mathbb{P}}$ of the Bernoulli measure ${\mathbb{P}}$ on ${\mathcal{A}^{\mathbb{N}}}$ (induced by the discrete measure assigning weight ${1/b}$ to each letter of the alphabet ${\mathcal{A}}$) play the role of conditional measures along vertical fibers of the unique Sinai-Ruelle-Bowen (SRB) measure ${\theta}$ of the expanding endomorphism ${T:\mathbb{R}/\mathbb{Z}\times\mathbb{R} \rightarrow \mathbb{R}/\mathbb{Z}\times\mathbb{R}}$,

$\displaystyle T(x,y) = (bx\textrm{ mod }1, \gamma y + \psi(x)),$

where ${\gamma=1/\lambda b}$ and ${\psi(x)=\phi'(x)}$. In plain terms, this means that

$\displaystyle \theta = \int_{\mathbb{R}/\mathbb{Z}} m_x \, d\textrm{Leb}(x) \ \ \ \ \ (1)$

where ${\theta}$ is the unique ${T}$-invariant probability measure which is absolutely continuous along unstable manifolds (see Tsujii’s paper).

As it was shown by Ledrappier in 1992, the fractal geometry of the conditional measures ${m_x}$ have important consequences for the fractal geometry of the graph ${G}$:

Theorem 2 (Ledrappier) Suppose that for Lebesgue almost every ${x\in\mathbb{R}}$ the conditional measures ${m_x}$ have dimension ${\textrm{dim}(m_x)=1}$, i.e.,

$\displaystyle \lim\limits_{r\rightarrow 0}\frac{\log m_x(B(z,r))}{\log r} = 1 \textrm{ for } m_x\textrm{-a.e. } z$

Then, the graph ${G=\textrm{graph}(f^{\phi}_{\lambda,b})}$ has Hausdorff dimension

$\displaystyle \textrm{dim}(G) = 2 + \frac{\log\lambda}{\log b}$

Remark 8 Very roughly speaking, the proof of Ledrappier theorem goes as follows. By Remark 4, it suffices to prove that ${\textrm{dim}(G)\geq 2 + \frac{\log\lambda}{\log b}}$. By Frostman lemma, we need to construct a Borel measure ${\nu}$ supported on ${G}$ such that

$\displaystyle \underline{\textrm{dim}}(\nu) := \textrm{ ess }\inf \underline{d}(\nu,x) \geq 2 + \frac{\log\lambda}{\log b}$

where ${\underline{d}(\nu,x):=\liminf\limits_{r\rightarrow 0}\log \nu(B(x,r))/\log r}$. Finally, the main point is that the assumptions in Ledrappier theorem allow to prove that the measure ${\mu^{\phi}_{\lambda, b}}$ given by the lift to ${G}$ of the Lebesgue measure on ${[0,1]}$ via the map ${x\mapsto (x,f^{\phi}_{\lambda,b}(x))}$satisfies

$\displaystyle \underline{\textrm{dim}}(\mu^{\phi}_{\lambda,b}) \geq 2 + \frac{\log\lambda}{\log b}$

An interesting consequence of Ledrappier theorem and the equation 1 is the following criterion for Mandelbrot’s conjecture:

Corollary 3 If ${\theta}$ is absolutely continuous with respect to the Lebesgue measure ${\textrm{Leb}_{\mathbb{R}^2}}$, then

$\displaystyle \textrm{dim}(G) = 2 + \frac{\log\lambda}{\log b}$

Proof: By (1), the absolute continuity of ${\theta}$ implies that ${m_x}$ is absolutely continuous with respect to ${\textrm{Leb}_{\mathbb{R}}}$ for Lebesgue almost every ${x\in\mathbb{R}}$.

Since ${m_x\ll \textrm{Leb}_{\mathbb{R}}}$ for almost every ${x}$ implies that ${\textrm{dim}(m_x)=1}$ for almost every ${x}$, the desired corollary now follows from Ledrappier’s theorem. $\Box$

3. Tsujii’s theorem

The relevance of Corollary 3 is explained by the fact that Tsujii found an explicittransversality condition implying the absolute continuity of ${\theta}$.

More precisely, Tsujii firstly introduced the following definition:

Definition 4

• Given ${\varepsilon>0}$, ${\delta>0}$ and ${x_0\in\mathbb{R}/\mathbb{Z}}$, we say that two infinite words ${u, v\in\mathcal{A}^{\mathbb{N}}}$ are ${(\varepsilon,\delta)}$-transverse at ${x_0}$ if either

$\displaystyle |s(x_0,u)-s(x_0,v)|>\varepsilon$

or

$\displaystyle |s'(x_0,u)-s'(x_0,v)|>\delta$

• Given ${q\in\mathbb{N}}$, ${\varepsilon>0}$, ${\delta>0}$ and ${x_0\in\mathbb{R}/\mathbb{Z}}$, we say that two finite words ${k,l\in\mathcal{A}^q}$ are ${(\varepsilon,\delta)}$-transverse at ${x_0}$ if ${ku}$, ${lv}$ are ${(\varepsilon,\delta)}$-transverse at ${x_0}$for all pairs of infinite words ${u,v\in\mathcal{A}^{\mathbb{N}}}$; otherwise, we say that ${k}$ and ${l}$ are${(\varepsilon,\delta)}$-tangent at ${x_0}$;
• ${E(q,x_0;\varepsilon,\delta):= \{(k,l)\in\mathcal{A}^q\times\mathcal{A}^q: (k,l) \textrm{ is } (\varepsilon,\delta)\textrm{-tangent at } x_0\}}$
• ${E(q,x_0):=\bigcap\limits_{\varepsilon>0}\bigcap\limits_{\delta>0} E(q,x_0;\varepsilon,\delta)}$;
• ${e(q,x_0):=\max\limits_{k\in\mathcal{A}^q}\#\{l\in\mathcal{A}^q: (k,l)\in E(q,x_0)\}}$
• ${e(q):=\max\limits_{x_0\in\mathbb{R}/\mathbb{Z}} e(q,x_0)}$.

Next, Tsujii proves the following result:

Theorem 5 (Tsujii) If there exists ${q\geq 1}$ integer such that ${e(q)<(\gamma b)^q}$, then

$\displaystyle \theta\ll\textrm{Leb}_{\mathbb{R}^2}$

Remark 9 Intuitively, Tsujii’s theorem says the following. The transversality condition ${e(q)<(\gamma b)^q}$ implies that the majority of strong unstable manifolds ${\ell^{uu}}$are mutually transverse, so that they almost fill a small neighborhood ${U}$ of some point ${x_0}$ (see the figure below extracted from this paper of Tsujii). Since the SRB measure ${\theta}$ is absolutely continuous along strong unstable manifolds, the fact that the ${\ell^{uu}}$‘s almost fill ${U}$ implies that ${\theta}$ becomes “comparable” to the restriction of the Lebesgue measure ${\textrm{Leb}_{\mathbb{R}^2}}$ to ${U}$.

Remark 10 In this setting, Barańsky-Barány-Romanowska obtained their main result by showing that, for adequate choices of the parameters ${\lambda}$ and ${b}$, one has ${e(1)=1}$. Indeed, once we know that ${e(1)=1}$, since ${1<\gamma b}$, they can apply Tsujii’s theorem and Ledrappier’s theorem (or rather Corollary 3) to derive the validity of Mandelbrot’s conjecture for certain parameters ${\lambda}$ and ${b}$.

For the sake of exposition, we will give just a flavor of the proof of Theorem 1 by sketching the derivation of the following result:

Proposition 6 Let ${\phi(x) = \cos(2\pi x)}$. If ${1/2<\gamma=1/\lambda b <1}$ and ${b\in\mathbb{N}}$ is sufficiently large, then

$\displaystyle e(1)<\gamma b$

In particular, by Corollary 3 and Tsujii’s theorem, if ${1/2<\gamma=1/\lambda b <1}$ and ${b\in\mathbb{N}}$ is sufficiently large, then Mandelbrot’s conjecture is valid, i.e.,

$\displaystyle \textrm{dim}(W_{\lambda,b}) = 2+\frac{\log\lambda}{\log b}$

Remark 11 The proof of Theorem 1 in full generality (i.e., for ${b\geq 2}$ integer and ${1/b<\lambda<1}$) requires the introduction of a modified version of Tsujii’s transversality condition: roughly speaking, Shen defines a function ${\sigma(q)\leq e(q)}$(inspired from Peter-Paul inequality) and he proves

• (a) a variant of Proposition 6: if ${b\geq 2}$ integer and ${1/b<\lambda<1}$, then ${\sigma(q)<(\gamma b)^q}$ for some integer ${q}$;
• (b) a variant of Tsujii’s theorem: if ${\sigma(q)<(\gamma b)^q}$ for some integer ${q}$, then ${\theta\ll\textrm{Leb}_{\mathbb{R}^2}}$.

See Sections 2, 3, 4 and 5 of Shen’s paper for more details.

We start the (sketch of) proof of Proposition 6 by recalling that the slopes of unstable manifolds are given by

$\displaystyle s(x,u):=-2\pi\sum\limits_{n=0}^{\infty} \gamma^n \sin\left(2\pi\frac{x + u_1 + u_2 b + \dots + u_n b^{n-1}}{b^n}\right)$

for ${x\in\mathbb{R}}$, ${u\in\mathcal{A}^{\mathbb{N}}}$, so that

$\displaystyle s'(x,u)=-4\pi^2\sum\limits_{n=0}^{\infty} \left(\frac{\gamma}{b}\right)^n \cos\left(2\pi\frac{x + u_1 + u_2 b + \dots + u_n b^{n-1}}{b^n}\right)$

Remark 12 Since ${\gamma/b < \gamma}$, the series defining ${s'(x,u)}$ converges faster than the series defining ${s(x,u)}$.

By studying the first term of the expansion of ${s(x,u)}$ and ${s'(x,u)}$ (while treating the remaining terms as a “small error term”), it is possible to show that if ${(k,l)\in E(1,x_0)}$, then

$\displaystyle \left|\sin\left(2\pi\frac{x_0+k}{b}\right) - \sin\left(2\pi\frac{x_0+l}{b}\right)\right| \leq\frac{2\gamma}{1-\gamma} \ \ \ \ \ (2)$

and

$\displaystyle \left|\cos\left(2\pi\frac{x_0+k}{b}\right) - \cos\left(2\pi\frac{x_0+l}{b}\right)\right| \leq \frac{2\gamma}{b-\gamma} \ \ \ \ \ (3)$

(cf. Lemma 3.2 in Shen’s paper).

Using these estimates, we can find an upper bound for ${e(1)}$ as follows. Take ${x_0\in\mathbb{R}/\mathbb{Z}}$ with ${e(1)=e(1,x_0)}$, and let ${k\in\mathcal{A}}$ be such that ${(k,l_1),\dots,(k,l_{e(1)})\in E(1,x_0)}$ distinct elements listed in such a way that

$\displaystyle \sin(2\pi x_i)\leq \sin(2\pi x_{i+1})$

for all ${i=1,\dots,e(1)-1}$, where ${x_i:=(x_0+l_i)/b}$.

From (3), we see that

$\displaystyle \left|\cos\left(2\pi x_i\right) - \cos\left(2\pi x_{i+1}\right)\right| \leq \frac{4\gamma}{b-\gamma}$

for all ${i=1,\dots,e(1)-1}$.

Since

$\displaystyle (\cos(2\pi x_i)-\cos(2\pi x_{i+1}))^2 + (\sin(2\pi x_i)-\sin(2\pi x_{i+1}))^2 = 4\sin^2(\pi(x_i-x_{i+1}))\geq 4\sin^2(\pi/b),$

it follows that

$\displaystyle |\sin(2\pi x_i)-\sin(2\pi x_{i+1})|\geq \sqrt{4\sin^2\left(\frac{\pi}{b}\right) - \left(\frac{4\gamma}{b-\gamma}\right)^2} \ \ \ \ \ (4)$

Now, we observe that

$\displaystyle \sqrt{4\sin^2\left(\frac{\pi}{b}\right) - \left(\frac{4\gamma}{b-\gamma}\right)^2} > \frac{4}{b} \ \ \ \ \ (5)$

for ${b}$ large enough. Indeed, this happens because

• ${\sqrt{z^2-w^2}>2(z-w)}$ if ${z+w>4(z-w)}$;
• ${z+w>4(z-w)}$ if ${z/w:=u < 5/3}$;
• ${\frac{2\sin(\frac{\pi}{b})}{\frac{4\gamma}{b-\gamma}}\rightarrow \frac{2\pi}{4\gamma} (< \frac{5}{3})}$ as ${b\rightarrow\infty}$, and ${2\sin(\frac{\pi}{b}) - \frac{4\gamma}{b-\gamma} \rightarrow (2\pi-4\gamma)\frac{1}{b} (>\frac{2}{b})}$ as ${b\rightarrow\infty}$ (here we used ${\gamma<1}$).

By combining (4) and (5), we deduce that

$\displaystyle |\sin(2\pi x_i)-\sin(2\pi x_{i+1})| > 4/b$

for all ${i=1,\dots, e(1)-1}$.

Since ${-1\leq\sin(2\pi x_1)\leq\sin(2\pi x_2)\leq\dots\leq\sin(2\pi x_{e(1)})\leq 1}$, the previous estimate implies that

$\displaystyle \frac{4}{b}(e(1)-1)<\sum\limits_{i=1}^{e(1)-1}(\sin(2\pi x_{i+1}) - \sin(2\pi x_i)) = \sin(2\pi x_{e(1)}) - \sin(2\pi x_1)\leq 2,$

i.e.,

$\displaystyle e(1)<1+\frac{b}{2}$

Thus, it follows from our assumptions (${\gamma>1/2}$, ${b}$ large) that

$\displaystyle e(1)<1+\frac{b}{2}<\gamma b$

This completes the (sketch of) proof of Proposition 6 (and our discussion of Shen’s talk).

# 低维动力系统

One Dimensional Real and Complex Dynamics需要学习的资料：

### 复分析基础：本科生课程

(1) Complex Analysis, 3rd Edition, Lars V. Ahlfors

(2) Complex Analysis, Elias M. Stein

### 进阶复分析：研究生课程

(1) Lectures on Riemann Surfaces (GTM 81), Otto Forster

(2) Lectures on Quasiconformal Mappings, Lars V. Ahlfors

### 实分析基础：本科生课程

(1) Real Analysis, Rudin

(2) Real Analysis, Elias M. Stein

### 实动力系统：

(1) One Dimensional Dynamics, Welington de Melo & Sebastian VanStrien

(2) Mathematical Tools for One-Dimensional Dynamics (Cambridge Studies in Advanced Mathematics), Edson de Faria / Welington de Melo

### 复动力系统：

(3) Dynamics in One Complex Variable, John Milnor

(4) Complex Dynamics, Lennart Carleson

(5) Complex Dynamics and Renormalization, Curtis T. McMullen

(6) Renormalization and 3-Manifolds Which Fiber over the Circle, Curtis T. McMullen

(7) Iteration of rational functions (GTM 132), Alan F. Beardon

### 遍历论：

(8) An Introduction to Ergodic Theory (GTM 79), Walters Peter

# Ergodic Properties

## One Dimensional Dynamics

— Welington De Melo, Sebastian van Strien

#### Chapter 5. Ergodic Properties and Invariant Measures.

##### 1. Ergodicity, Attractors and Bowen-Ruelle-Sinai Measures.

A distortion result for unimodal maps with recurrence

Given a unimodal map $f$, we say that an interval $U$ is symmetric if $\tau(U)=U$ where $\tau:[-1,1]\rightarrow [-1,1]$ is so that $f(\tau(x))=f(x)$ and $\tau(x)\neq x$ if $x\neq c$. Furthermore, for each symmetric interval $U$ let

$D_{U}=\{x: \text{ there exists } k>0 \text{ with } f^{k}(x)\in U\};$

for $x\in D_{U}$ let $k(x,U)$ be the minimal positive integer with $f^{k}(x)\in U$ and let

$R_{U}(x)=f^{k(x,U)}(x).$

We call $R_{U}: D_{U}\rightarrow U$ the Poincare map or transfer map to $U$ and $k(x,U)$ the transfer time of $x$ to $U$. The distortion result states that one can fined a sequence of symmetric neighbourhoods of the turning point such that the Poincare maps to these intervals have a distortion which is universally bounded:

Theorem 1.1.  Let $f:[-1,1]\rightarrow [-1,1]$ be a unimodal map with one non-flat critical point with negative Schwarzian derivative and without attracting periodic points. Then there exists $\rho>0$ and a sequence os symmetric intervals $U_{n}\subseteq V_{n}$ around the turning point which shrink to $c$ such that $V_{n}$ contains a $\rho-$scaled neighbourhood of $U_{n}$ and such that the following properties hold.

1. The transfer time on each component of $D_{U_{n}}$ is constant.

2. Let $I_{n}$ be a component of the domain $D_{U_{n}}$ of the transfer map to $U_{n}$ which does not intersect $U_{n}$. Then there exists an interval $T_{n}\supseteq I_{n}$ such that $f^{k}|T_{n}$ is monotone, $f^{k}(T_{n})\supseteq V_{n}$ and $f^{k}(I_{n})=U_{n}$. Here $k$ is the transfer time on $I_{n}$, i.e., $R_{U_{n}}|I_{n}=f^{k}$.

Corollary. There exists $K<\infty$ such that

1. for each component $I_{n}$ of $D_{U_{n}}$ not intersecting $U_{n}$, the transfer map $R_{U_{n}}$ to $U_{n}$ sends $I_{n}$ diffeomorphically onto $U_{n}$ and the distortion of $R_{U_{n}}$ on $I_{n}$ is bounded from above by $K$.

2. on each component $I_{n}$ of $D_{U_{n}}$ which is contained in $U_{n}$, the map $R_{U_{n}}:I_{n}\rightarrow U_{n}$ can be written as $(f^{k(n)-1}|f(I_{n}))\circ f|I_{n}$ where the distortion of $f^{k(n)}|f(I_{n})$ is universally bounded by $K$.

As before, we say that $f$ is ergodic with respect to the Lebesgue measure if each completely invariant set $X$ (Here $X$ is called completely invariant if $f^{-1}(X)=X$) has either zero or full Lebesgue measure. An alternative way to define this notation of ergodicity goes as follows: $f$ is ergodic if for each two forward invariant sets $X$ and $Y$ such that $X\cap Y$ has Lebesgue measure zero, at most one of these sets has positive Lebesgue measure. (Here $X$ is called forward invariant if $f(X)\subseteq X$.)

Theorem 1.2 (Blokh and Lyubich). Let $f:[-1,1]\rightarrow [-1,1]$ be a unimodal map with a non-flat critical point with negative Schwarzian derivative and without an attracting periodic points. Then $f$ is ergodic with respect to the Lebesgue measure.

Theorem 1.3.  Let $f:[-1,1]\rightarrow [-1,1]$ be a unimodal map with a non-flat critical point with negative Schwarzian derivative. Then $f$ has a unique attractor $A$, $\omega(x)=A$ for almost all $x$ and $A$ either consists of intervals or has Lebesgue measure zero. Furthermore, one has the following:

1. if $f$ has an attracting periodic orbit then $A$ is this periodic orbit;

2. if $f$ is infinitely often renormalizable then $A$ is the attracting Cantor set $\omega(c)$ (in which case it is called a solenoidal attractor);

3. $f$ is only finitely often renormalizable then either

(a) $A$ coincides with the union of the transitive intervals, or,

(b) $A$ is a Cantor set and equal to $\omega(c)$.

If $\omega(c)$ is not a minimal set then $f$ is as in case 3.a and each closed forward invariant set either contains intervals or has Lebesgue measure zero. Moreover, if $\omega(c)$ does not contain intervals, then $\omega(c)$ has Lebesgue measure zero.

Remark. Here a forward invariant set $X$ is said to be minimal if the closure of the forward orbit of a point in $X$ is always equal to $X$. The attractors in case 3.b is called a non-renormalizable attracting Cantor set, or absorbing Cantor attractor or wild Cantor attractor. Such an attractor really exists which is proven in [BKNS], and one has the following strange phenomenon: there exist many orbits which are dense in some finite union of intervals and yet almost all points tend to a minimal Cantor set of Lebesgue measure zero (this Cantor set is $\omega(c)$). The Fibonacci map is non-renormalizable and for which $\omega(c)$ is a Cantor set. It was shown by Lyubich and Milnor that the quadratic map with this dynamics has no absorbing Cantor attractors. More generally, Jakobson and Swiatek proved that maps with negative Schwarzian derivative and which are close to the map $f(x)=4x(1-x)$ do not have such Cantor attractors. Moreover, Lyubich has shown that these absorbing Cantor attractors can not exist if the critical point is quadratic. However, Bruin, Keller, Nowicki and Van Strien showed that the absorbing Cantor attractors exist for Fibonacci maps when the critical order $\ell$ is sufficiently large enough.

Theorem (Lyubich). If $f:[-1,1]\rightarrow [-1,1]$ is $C^{3}$ unimodal, has a quadratic critical point, has negative Schwarzian derivative and has no periodic attractors, then each closed forward invariant set $K$ which has positive Lebesgue measure contains an interval.

The next result, which is due to Martens (1990), shows that if these absorbing Cantor attractors do not exist then one has a lot of ‘expansion’. Let $x$ not be in the pre orbit of $c$ and define $T_{n}(x)$ to be the maximal interval on which $f^{n}|T_{n}(x)$ is monotone. Let $R_{n}(x)$ and $L_{n}(x)$ be the components of $T_{n}\setminus x$ and define $r_{n}(x)$ be the minimum of the length of $f^{n}(R_{n}(x))$ and $f^{n}(L_{n}(x))$.

Theorem 1.4 (Martens). Let $f$ be a $C^{3}$ unimodal map with negative Schwarian derivative whose critical point is non-flat. Then the following three properties are equivalent.

1. $f$ has no absorbing Cantor attractor;

2. $\limsup_{n\rightarrow \infty} r_{n}(x)>0$ for almost all $x$;

3. there exist neighbourhoods $U\subseteq V$ of $c$ with $cl(U)\subseteq int(V)$ such that for almost every $x$ there exists a positive integer $m$ and an interval neighbourhood $T$ of $x$ such that $f^{m}|T$ is monotone, $f^{m}(T)\supseteq V$ and $f^{m}(x)\in U$.

# Perron-Frobenius Operator

## Perron-Frobenius Operator

Consider a map $f$ which possibly has a finite (or countable) number of discontinuities or points where possibly the derivative does not exist. We assume that there are points

$\displaystyle q_{0} or $q_{0}

such that $f$ restricted to each open interval $A_{j}=(q_{j-1},q_{j})$ is $C^{2}$, with a bound on the first and the second derivatives. Assume that the interval $[q_{0},q_{k}]$ ( or $[q_{0},q_{\infty}]$ ) is positive invariant, so $f(x)\in [q_{0},q_{k}]$ for all $x\in [q_{0}, q_{k}]$ ( or $f(x)\in [q_{0},q_{\infty}]$  for all $x\in[q_{0},q_{\infty}]$ ).

For such a map, we want a construction of a sequence of density functions that converge to a density function of an invariant measure. Starting with $\rho_{0}(x)\equiv(q_{k}-q_{0})^{-1}$ ( or $\rho_{0}(x)\equiv(q_{\infty}-q_{0})^{-1}$ ),assume that we have defined densities up to $\rho_{n}(x)$, then define define $\rho_{n+1}(x)$ as follows

$\displaystyle \rho_{n+1}(x)=P(\rho_{n})(x)=\sum_{y\in f^{-1}(x)}\frac{\rho_{n}(y)}{|Df(y)|}.$

This operator $P$, which takes one density function to another function, is called the Perron-Frobenius operator. The limit of the first $n$ density functions converges to a density function $\rho^{*}(x)$,

$\displaystyle \rho^{*}(x)=\lim_{k\rightarrow \infty}\frac{1}{k}\sum_{n=0}^{k-1}\rho_{n}(x).$

The construction guarantees that $\rho^{*}(x)$ is the density function for an invariant measure $\mu_{\rho^{*}}$.

Example 1. Let

$\displaystyle f(x)= \begin{cases} x &\mbox{if } x\in(0,\frac{1}{2}), \\ 2x &\mbox{if } x\in(\frac{1}{2},1). \end{cases}$

We construct the first few density functions by applying the Perron-Frobenius operator, which indicates the form of the invariant density function.
Take $\rho_{0}(x)\equiv1$ on $[0,1]$. From the definition of $f(x)$, the slope on $(0,\frac{1}{2})$ and $(\frac{1}{2},1)$ are 1 and 2, respectively. If $x\in (\frac{1}{2},1)$, then it has only one pre-image on $(\frac{1}{2},1)$; else if $x\in(0,\frac{1}{2})$, then it has two pre-images, one is $x^{'}$ in $(0,\frac{1}{2})$, the other one is $x^{''}$ in $(\frac{1}{2},1)$. Therefore,

$\rho_{1}(x)= \begin{cases} \frac{1}{1}+\frac{1}{2} &\mbox{if } x\in(0,\frac{1}{2}), \\ \frac{1}{2} &\mbox{if } x\in(\frac{1}{2},1). \end{cases}$

By similar considerations,

$\displaystyle \rho_{2}(x)=\begin{cases}1+\frac{1}{2}+\frac{1}{2^{2}} &\mbox{if } x\in(0,\frac{1}{2}), \\ \frac{1}{2^{2}} &\mbox{if } x\in(\frac{1}{2},1).\end{cases}$

By induction, we get

$\displaystyle \rho_{n}(x)=\begin{cases}1+\frac{1}{2}+\cdot\cdot\cdot+\frac{1}{2^{n}} &\mbox{if } x\in(0,\frac{1}{2}), \\ \frac{1}{2^{n}} &\mbox{if } x\in(\frac{1}{2},1).\end{cases}$

Now, we begin to calculate the density function $\rho^{*}(x)$. If $x\in(0,\frac{1}{2})$, then
$\displaystyle \rho^{*}(x)=\lim_{k\rightarrow \infty}\frac{1}{k}\sum_{n=0}^{k-1}\rho_{n}(x) =\lim_{k\rightarrow \infty}\frac{1}{k}\sum_{n=0}^{k-1} \sum_{m=0}^{n}\frac{1}{2^{m}} =\lim_{k\rightarrow \infty}\frac{1}{k}\sum_{n=0}^{k-1}\left(2-\frac{1}{2^{n}}\right)=2.$
If $x\in(\frac{1}{2},1)$, then
$\displaystyle \rho^{*}(x)=\lim_{k\rightarrow \infty}\frac{1}{k}\sum_{n=0}^{k-1}\rho_{n}(x) =\lim_{k\rightarrow \infty}\frac{1}{k}\sum_{n=0}^{k-1}\frac{1}{2^{n}} =\lim_{k\rightarrow \infty}\frac{1}{k}\left(2-\frac{1}{2^{k}}\right)=0.$
i.e.

$\displaystyle \rho^{*}(x)= \begin{cases} 2 &\mbox{if } x\in(0,\frac{1}{2}), \\ 0 &\mbox{if } x\in(\frac{1}{2},1). \end{cases}$

Example 2. Let

$\displaystyle f(x)=\begin{cases} 2x &\mbox{if } x\in(0,\frac{1}{2}), \\ 2x-1 &\mbox{if } x\in(\frac{1}{2},1). \end{cases}$

Take $\rho_{0}(x)\equiv1$ on $(0,1)$. By induction, $\rho_{n}(x)\equiv1$ on $(0,1)$ for all $n\geq 0$. Therefore, $\rho^{*}(x)\equiv1$ on $(0,1)$.

Example 3. Let

$\displaystyle f(x)=\begin{cases} x &\mbox{if } x\in(0,\frac{1}{2}), \\ 2^{n+1}\cdot\left(x-\left(1-\frac{1}{2^{n}}\right)\right) &\mbox{if } x\in\left(1-\frac{1}{2^{n}},1-\frac{1}{2^{n+1}}\right) \text{ for all } n\geq 1.\end{cases}$

Take $\rho_{0}(x)\equiv1$ on $(0,1)$. Assume

$\displaystyle \rho_{n}(x)= \begin{cases} a_{n} &\mbox{if } x\in(0,\frac{1}{2}), \\ b_{n} &\mbox{if } x\in(\frac{1}{2},1). \end{cases}$

for all $n\geq 0$. It is obviously that $a_{0}=b_{0}=1$. By similar considerations,
$\displaystyle \rho_{n+1}(x)= \begin{cases} \frac{a_{n}}{1}+\frac{b_{n}}{4}+\frac{b_{n}}{8}+\frac{b_{n}}{16}+\cdot\cdot\cdot= a_{n}+\frac{b_{n}}{2} &\mbox{if } x\in(0,\frac{1}{2}), \\ \frac{b_{n}}{4}+\frac{b_{n}}{8}+\frac{b_{n}}{16}+\cdot\cdot\cdot = \frac{b_{n}}{2} &\mbox{if } x\in(\frac{1}{2},1). \end{cases}$
That means

$\displaystyle \left( \begin{array}{ccc} a_{n+1} \\ b_{n+1} \end{array} \right) =\left( \begin{array}{ccc} a_{n}+\frac{1}{2}b_{n} \\ \frac{1}{2}b_{n} \end{array} \right) = \left( \begin{array}{ccc} 1 & \frac{1}{2} \\ 0 & 1 \end{array} \right) \left( \begin{array}{ccc} a_{n} \\ b_{n} \end{array} \right)$

for all $n\geq 0$. From direct calculation, $\displaystyle a_{n}=2-\frac{1}{2^{n}}$ and $\displaystyle b_{n}=\frac{1}{2^{n}}$ for all $n\geq 0$. Therefore,

$\displaystyle \rho^{*}(x)=\lim_{k\rightarrow \infty}\frac{1}{k}\sum_{n=0}^{k-1}\rho_{n}(x)=\begin{cases} 2 &\mbox{if } x\in (0,\frac{1}{2}), \\ 0 &\mbox{if } x\in (\frac{1}{2},1). \end{cases}$

Example 4. Let

$\displaystyle f(x)=\begin{cases} 1.5 x &\mbox{if } x\in(0,\frac{1}{2}), \\ 2^{n+1}\cdot\left(x-\left(1-\frac{1}{2^{n}}\right)\right) &\mbox{if } x\in\left(1-\frac{1}{2^{n}},1-\frac{1}{2^{n+1}}\right) \text{ for all } n\geq 1.\end{cases}$

Take $\rho_{0}(x)\equiv1$ on $(0,1)$. Assume

$\displaystyle \rho_{n}(x)= \begin{cases} a_{n} &\mbox{if } x\in(0,\frac{3}{4}), \\ b_{n} &\mbox{if } x\in(\frac{3}{4},1). \end{cases}$

for all $n\geq 0$. It is obviously that $a_{0}=b_{0}=1$. By similar considerations,

$\displaystyle \left( \begin{array}{ccc} a_{n+1} \\ b_{n+1} \end{array} \right) =\left( \begin{array}{ccc} \frac{11}{12}a_{n}+\frac{1}{4}b_{n} \\ \frac{1}{4}a_{n}+\frac{1}{4}b_{n} \end{array} \right) = \left( \begin{array}{ccc} \frac{11}{12} & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{4} \end{array} \right) \left( \begin{array}{ccc} a_{n} \\ b_{n} \end{array} \right)$

for all $n\geq 0$. From matrix diagonalization , $\displaystyle a_{n}=\frac{6}{5}-\frac{1}{5}\cdot\frac{1}{6^{n}}$ and $\displaystyle b_{n}=\frac{2}{5}+\frac{3}{5}\cdot\frac{1}{6^{n}}$ for all $n\geq 0$.

Therefore,

$\displaystyle \rho^{*}(x)=\lim_{k\rightarrow \infty}\frac{1}{k}\sum_{n=0}^{k-1}\rho_{n}(x)=\begin{cases} \frac{6}{5} &\mbox{if } x\in (0,\frac{3}{4}), \\ \frac{2}{5} &\mbox{if } x\in (\frac{3}{4},1). \end{cases}$

## Perron-Frobenius Theory

Definition. Let $A=[a_{ij}]$ be a $k\times k$ matrix. We say $A$ is non-negative if $a_{ij}\geq 0$ for all $i,j$. Such a matrix is called irreducible if for any pair $i,j$ there exists some $n>0$ such that $a_{ij}^{(n)}>0$ where $a_{ij}^{(n)}$ is the $(i,j)-$th element of $A^{n}$. The matrix $A$ is irreducible and aperiodic if there exists $n>0$ such that $a_{ij}^{(n)}>0$ for all $i,j$.

Perron-Frobenius Theorem Let $A=[a_{ij}]$ be a non-negative $k\times k$ matrix.

(i) There is a non-negative eigenvalue $\lambda$ such that no eigenvalue of $A$ has absolute value greater than $\lambda$.

(ii) We have $\min_{i}(\sum_{j=1}^{k}a_{ij})\leq \lambda\leq \max_{i}(\sum_{j=1}^{k}a_{ij})$.

(iii) Corresponding to the eigenvalue $\lambda$ there is a non-negative left (row) eigenvector $u=(u_{1},\cdot\cdot\cdot, u_{k})$ and a non-negative right (column) eigenvector $v=(v_{1},\cdot\cdot\cdot, v_{k})^{T}$.

(iv) If $A$ is irreducible then $\lambda$ is a simple eigenvalue and the corresponding eigenvectors are strictly positive (i.e. $u_{i}>0$, $v_{i}>0$ all $i$).

(v) If $A$ is irreducible then $\lambda$ is the only eigenvalue of $A$ with a non-negative eigenvector.

Theorem.
Let $A$ be an irreducible and aperiodic non-negative matrix. Let $u=(u_{1},\cdot\cdot\cdot, u_{k})$ and $v=(v_{1},\cdot\cdot\cdot, v_{k})^{T}$ be the strictly positive eigenvectors corresponding to the largest eigenvalue $\lambda$ as in the previous theorem. Then for each pair $i,j$, $\lim_{n\rightarrow \infty} \lambda^{-n}a_{ij}^{(n)}=u_{j}v_{i}$.

Now, let us see previous examples, again. The matrix $A$ is irreducible and aperiodic non-negative matrix, and $\lambda=1$ has the largest absolute value in the set of all eigenvalues of $A$. From Perron-Frobenius Theorem, $u_{i}, v_{j}>0$ for all pairs $i,j$. Then for each pari $i,j$,
$\lim_{n\rightarrow \infty}a_{ij}^{(n)}=u_{j}v_{i}$. That means $\lim_{n\rightarrow \infty}A^{(n)}$ is a strictly positive $k\times k$ matrix.

## Markov Maps

Definition of Markov Maps. Let $N$ be a compact interval. A $C^{1}$ map $f:N\rightarrow N$ is called Markov if there exists a finite or countable family $I_{i}$ of disjoint open intervals in $N$ such that

(a) $N\setminus \cup_{i}I_{i}$ has Lebesgue measure zero and there exist $C>0$ and $\gamma>0$ such that for each $n\in \mathbb{N}$ and each interval $I$ such that $f^{j}(I)$ is contained in one of the intervals $I_{i}$ for each $j=0,1,...,n$ one has

$\displaystyle \left| \frac{Df^{n}(x)}{Df^{n}(y)}-1 \right| \leq C\cdot |f^{n}(x)-f^{n}(y)|^{\gamma} \text{ for all } x,y\in I;$

(b) if $f(I_{k})\cap I_{j}\neq \emptyset$, then $f(I_{k})\supseteq I_{j}$;

(c) there exists $r>0$ such that $|f(I_{i})|\geq r$ for each $i$.

As usual, let $\lambda$ be the Lebesgue measure on $N$. We may assume that $\lambda$ is a probability measure, i.e., $\lambda(N)=1$. Usually, we will denote the Lebesgue measure of a Borel set $A$ by $|A|$.

Theorem.  Let $f:N\rightarrow N$ be a Markov map and let $\cup_{i}I_{i}$ be corresponding partition. Then there exists a $f-$invariant probability measure $\mu$ on the Borel sets of $N$ which is absolutely continuous with respect to the Lebesgue measure $\lambda$. This measure satisfies the following properties:

(a) its density $\frac{d\mu}{d\lambda}$ is uniformly bounded and Holder continuous. Moreover, for each $i$ the density is either zero on $I_{i}$ or uniformly bounded away from zero.

If for every $i$ and $j$ one has $f^{n}(I_{j})\supseteq I_{i}$ for some $n\geq 1$ then

(b) the measure is unique and its density $\frac{d\mu}{d\lambda}$ is strictly positive;

(c) $f$ is exact with respect to $\mu$;

(d) $\lim_{n\rightarrow \infty} |f^{-n}(A)|=\mu(A)$ for every Borel set $A\subseteq N$.

If $f(I_{i})=N$ for each interval $I_{i}$, then

(e) the density of $\mu$ is also uniformly bounded from below.

# 2. 刚性定理

$\{ a\in[0,4]: f_{a} \text{ satisfies Axiom A} \}$ 是否在 [0,4] 中稠密？

$K(f_{4})=(R,L,L,L,...)=RLLL,$

$K(f_{1})=(L,L,L,L,...)=LLLL,$

$K(f_{2})=(c,c,c,c,...)=cccc,$

$K(f_{1.9})=(L,L,L,L,...)=LLLL,$

(1) $\varphi$ 是 ACL 的，也就是线段上绝对连续，absolutely continuous on lines.

(2) $| \frac{\partial \varphi}{\partial \overline{z}} | \leq \frac{K-1}{K+1} |\frac{\partial \varphi}{\partial z}|$ 几乎处处成立。

(i) $\varphi$ 几乎处处可微。对几乎所有的 $z_{0}\in \Omega$

$\varphi(z) = \varphi(z_{0}) + \frac{\partial \varphi}{\partial z}(z_{0})(z-z_{0}) + \frac{\partial \varphi}{\partial \overline{z}}(z_{0})\overline{(z-z_{0})}+ o(|z-z_{0}|).$

$| \frac{\partial \varphi}{\partial z}|>0$ 几乎处处成立。

(ii) Measurable Riemann Mapping Theorem ( Ahlfors-Bers )

Assume $f_{a}(x)=ax(1-x),$ $a_{0} \in (0,4]$

$Comb(a_{0})=\{ a\in(0,4]: K(f_{a})=K(f_{a_{0}}) \},$

$Top(a_{0})= \{ a\in (0,4]: f_{a} \text{ and } f_{a_{0}} \text{ are topological conjugate } \},$

$\Rightarrow Top(a_{0}) \subseteq Comb(a_{0}).$

$Qc(a_{0}) = \{ a\in (0,4]: f_{a} \text{ and } f_{a_{0}} \text{ are quasi-conformal conjugate} \},$

$Aff(a_{0}) = \{ a\in (0,4]: f_{a} \text{ and } f_{a_{0}} \text{ are linear conjugate} \},$

$\Rightarrow Aff(a_{0}) \subseteq Qc(a_{0}).$

# 1.一维动力系统中的双曲性

$\omega(x)=\{ y\in X: \exists n_{k} \rightarrow \infty, f^{n_{k}}(x)\rightarrow y\}$.

$|\lambda| \neq 1$称为$orb(x)=\{ f^{k}(x): k=0,1,2... \}$是双曲周期轨。

$|\lambda|=1$称为中性周期轨。

$|\lambda|<1$称为双曲吸引轨。

$|\lambda|>1$称为双曲斥性轨。

Axiom A： 假设 $f:[0,1]\rightarrow [0,1]$$C^{1}$ 映射，称 f 满足 Axiom A是指：

（1）f 有有限多个双曲吸引轨 $\theta_{1},...,\theta_{m}$,

（2）$B(\theta_{i})$ 是双曲吸引轨 $\theta_{i}$ 的吸引区域, $\Omega=[0,1]\setminus \cup_{i=1}^{m}B(\theta_{i})$ 是双曲集。

(1) f 的所有周期轨都是双曲的。

(2) Crit(f) 指的是 f 的临界点。$\forall c\in Crit(f)$, 则存在双曲吸引周期轨 $\theta_{c}$ 使得 $d(f^{n}(c),\theta_{c})\rightarrow 0, n\rightarrow \infty.$

$\Longleftrightarrow$ f 满足 Axiom A。

$U\subseteq Crit(f)\cup \text{ hyperbolic attracting orbits }\cup \text{ and neutral orbits }$,

$\Lambda_{U} = \{ x\in[0,1]: f^{n}(x)\notin U, \forall n\geq 0 \}$,

$\Rightarrow$ $\Lambda_{U}$ 是双曲集。

# The Cross Ratio Tool and the Koebe Principle

Let $j \subseteq t$ be intervals and let l, r be the components of $t \setminus j$. Then the Cross Ratio is defined as

$C(t,j) = (|t| \cdot |j|) / ( |l| \cdot |r|).$

Assume g is a $C^{3}$ monotone function on the interval t, and g(t)=T, g(j)=J, g(l)=L, g(r)=R. Then define

$B(g,t,j)=\frac{C(T,J)}{C(t,j)} = \frac{|T|\cdot |J|}{|L| \cdot |R|} \cdot \frac{|l|\cdot |r|}{|t|\cdot |j|}.$

Define the Schwarzian Derivative for $C^{3}$ function g,

$Sg(x)=\frac{D^{3}g(x)}{Dg(x)} -\frac{3}{2}(\frac{D^{2}g(x)}{Dg(x)})^{2}.$

Proposition 1. Assume f and g are $C^{3}$ functions, then

$S(f\circ g)(x)= Sf(g(x)) \cdot |Dg(x)|^{2}+ Sg(x).$

$S(f^{n})(x)= \sum_{i=0}^{n-1}(Sf(f^{i}(x)) \cdot |D(f^{i})(x)|^{2}.$

Proposition 2. If $f(x)=x^{\ell}+c$ for some $c\in \mathbb{R}$ and $\ell \geq 2$, then $Sf(x)<0$ for all $x \neq 0$.

Proposition 3. Minimum Principle.

Assume $I=[a,b]$, $f: I \rightarrow \mathbb{R}$ is a $C^{3}$ diffeomorphism with negative schwarzian derivative, then

$|Df(x)| > \min \{|Df(a), |Df(b)|\} \text{ for all } x \in (a,b).$

Theorem 1. Real Koebe Principle.

Let Sf<0. Then for any intervals $j \subseteq t$ and any n for which $f^{n}|t$ is a diffeomorphism one has the following. If $f^{n}(t)$ contains a $\tau-$scaled neighbourhood of $f^{n}(j)$, then

$(\frac{\tau}{1+\tau})^{2} \leq \frac{|Df^{n}(x)|}{|Df^{n}(y)|} \leq (\frac{1+\tau}{\tau})^{2} \text{ for all } x, y \in j.$

Moreover, there exists a universal function $K(\tau)>0$ which does not depend on f, n,  and t such that

$|l| / |j| \geq K(\tau),$

$|r| /|j| \geq K(\tau).$

Theorem 2. Complex Koebe Principle

Suppose that $D \subseteq \mathbb{C}$ contains a $\tau-$scaled neighbourhood of the disc $D_{1} \subseteq \mathbb{C}$. Then for any univalent function $f: D \rightarrow \mathbb{C}$ one has a universal function $K(\tau)>0$ which only depends on $\tau>0$ such that

$1/K(\tau) \leq \frac{|Df(x)|}{|Df(y)|} \leq K(\tau) \text{ for all } x, y \in D_{1}.$

Theorem 3. Schwarz Lemma (Original Form)

Assume $\mathbb{D}=\{ z: |z|<1\}$ is the unit disc on the complex plane $\mathbb{C}$, $f: \mathbb{D} \rightarrow \mathbb{D}$ is a holomorphic function with $f(0)=0$. Then $|f(z)|\leq |z|$ for all $z \in \mathbb{D}$ and $|f^{'}(0)| \leq 1.$ Moreover, if $|f(z_{0})|=|z_{0}|$ for some $z_{0}\neq 0$ or $|f^{'}(0)|=1,$ then $f(z)= e^{i\theta} z$ for some $\theta \in \mathbb{R}.$

Corollary 1.

Assume $\mathbb{D}$ is the unit disc on the complex plane $\mathbb{C}$, and $f: \mathbb{D} \rightarrow \mathbb{D}$ is a holomorphic function, then

$|\frac{f(z)-f(z_{0})}{1-\overline{f(z_{0})}f(z)}| \leq |\frac{z-z_{0}}{1-\overline{z}_{0}z}| \text{ for all } z, z_{0} \in \mathbb{D}.$

$\frac{|f^{'}(z)|}{1-|z|^{2}} \leq \frac{1}{1-|z|^{2}} \text{ for all } z \in \mathbb{D}.$

Corollary 2.

Assume $\mathbb{H}$ is the upper half plane of the complex plane $\mathbb{C}$, $f: \mathbb{H} \rightarrow \mathbb{H}$ is a holomorphic map. Then

$\frac{|f(z_{1})-f(z_{2})|}{|f(z_{1})-\overline{f(z_{2})}|} \leq \frac{|z_{1}-z_{2}|}{|z_{1}-\overline{z_{2}}|} \text{ for all } z_{1}, z_{2} \in \mathbb{H} .$

$\frac{|f^{'}(z)|}{\Im{f(z)}} \leq \frac{1}{\Im{z}} \text{ for all } z\in \mathbb{H} .$

Corollary 3. Pick Theorem

The hyperbolic metric on $\mathbb{D}$ is $\rho(z)= \frac{2}{1-|z|^{2}}dz$, assume $d(z_{1}, z_{2})$ denotes the hyperbolic distance between $z_{1}$ and $z_{2}$ on $\mathbb{D}$. Assume $f: \mathbb{D} \rightarrow \mathbb{D}$ is a holomorphic function, then

$d(f(z_{1}), f(z_{2}))\leq d(z_{1}, z_{2}) \text{ for all } z_{1}, z_{2} \in \mathbb{D}.$

Moreover, if $d(f(z_{1}), f(z_{2}))= d(z_{1}, z_{2})$ for some points $z_{1}, z_{2} \in \mathbb{D}$, then $f \in Aut(\mathbb{D})$, where

$Aut(\mathbb{D})=\{e^{i\theta}\frac{z-z_{0}}{1-\overline{z_{0}}z}: \theta \in \mathbb{R}, z_{0} \in \mathbb{D}\}.$

Background in hyperbolic geometry

Define

$\mathbb{C}_{J}=(\mathbb{C} \setminus \mathbb{R}) \cup J$

where $J \subseteq \mathbb{R}$ is an interval. It is easy to show that $\mathbb{C}_{J}$ is conformally equivalent to the upper half plane and define $D_{k}(J)$ as

$D_{k}(J)= \{ z: \text{the hyperbolic distance to J is at most k} \}.$

k is determined by the external angle $\alpha$ at which the discs intersect the real line. Moreover, $k=\ln \tan( \frac{\pi}{2}- \frac{\alpha}{4}) .$ Define

$D_{*}(J)=D(J,\frac{\pi}{2}) .$

Corollary 4. (NS) Schwarz Lemma

(1) Assume $G: \mathbb{C}_{I} \rightarrow \mathbb{C}_{J}$ is a holomorphic map, then $G((D_{*}{I})) \subseteq D_{*}(J).$

(2) Assume $F: \mathbb{C} \rightarrow \mathbb{C}$ is a real polynomial map, its critical points are on the real line. Assume $F: I \rightarrow J$ is a diffeomorphism, then there exists a set $D \subseteq D_{*}(I)$ such that $D\cap \mathbb{R} =I$ and

$F: D \rightarrow D_{*}(J)$ is a conformal map.

Corollary 5.

Assume $f: D \rightarrow \mathbb{C}$ is a univalent map and D contains $\tau-$scaled neighbourhood of $D_{1},$ and assume f maps the real line to the real line. For each $\alpha \in (\pi/2, \pi)$ there exists $\alpha^{'} \in (\alpha, \pi)$ such that if J is a real interval in $D_{1}$, then

$f(D(J,\alpha)) \supseteq D(f(J), \alpha^{'}).$

The Hyperbolic Metric On the Real Interval and Cross Ratio

As far as we know, the hyperbolic metric on the unit disc $\mathbb{D}=\{|z|<1\}$ is

$\rho_{D}(z)= \frac{2}{1-|z|^{2}}|dz| \text{ for all } z\in \mathbb{D}.$

Then the restriction to the real line is

$\rho_{I}(x)=\frac{2}{1-x^{2}} dx \text{ for all } x \in I=(-1,1).$

Moreover, from it, we can deduce the hyperbolic metric on the real interval $I=(a,b)$ is

$\rho_{I}(x)=\frac{b-a}{(x-a)(b-x)} dx \text{ for all } x \in I=(a,b).$

If $(c,d) \subseteq (a,b)$, then the hyperbolic length of the interval $(c,d)$ on the total interval $(a,b)$ is

$\ell_{(a,b)}((c,d))=\ell_{t}(j)=\ln(1+Cr(t,j)),$

where $l=(a,c), j=(c,d), r=(d,b), t=(a,b).$

Theorem 4. Assume $f: T \rightarrow f(T) \subseteq \mathbb{R}$ is a $C^{3}$ diffeomorphism with negative schwarzian derivative. Assume $J \subseteq T$, then

$\ell_{f(T)}(f(J)) \geq \ell_{T}(J).$

That means f expands the hyperbolic metric on the real interval.

Proof.  Since the schwarzian derivative of f is negative, $C(f(T),f(J)) \geq C(T,J).$

Therefore, $\ell_{f(T)}(f(J)) \geq \ell_{T}(J).$ That means f expands the hyperbolic metric on the real interval.

Remark. From Schwarz-Pick Theorem, for a holomorphic map $f: \mathbb{D} \rightarrow \mathbb{D}$, $f$ contracts the hyperbolic distance in the unit disc $\mathbb{D}$. Conversely, from above, for a $C^{3}$ diffeomorphism $f$ with negative schwarzian derivative, $f$ expands the hyperbolic distance in the real interval.

Exercise 1.  “Mathematical Tools for One Dimensional Dynamics” Exercise 6.5, Chapter 6

Let $f: I \rightarrow f(I) \subseteq \mathbb{R}$ be a $C^{3}$ diffeomorphism without fixed points ( $I$ being a closed interval on the real line). If $Sf(x)<0$ for all $x \in I$, then there exists a unique $x_{0} \in I$ such that $|f(x_{0})-x_{0}| \leq |f(x)-x|$ for all $x \in I$.

Proof.  If $f$ is a decreasing map, then the right boundary of the real interval I is the $x_{0}$. Therefore, assume that $f$ is an increasing map on the real interval I.

Since $f(x)$ has no fixed points on the real interval I, then $f(x)>x$ or $f(x) for all $x \in I$. Without lost of generality, assume $f(x)>x$ for all $x\in I$. Since $f(x)-x$ is a continuous function on the closed interval I, there exists $x_{0} \in I$ such that $|f(x_{0})-x_{0}| \leq |f(x)-x|$ for all $x\in I$.

By contradiction, there exist two distinct points $x_{0}, x_{1}$ $(x_{0} such that $|f(x_{0})-x_{0}| \leq |f(x)-x|$ and $|f(x_{1})-x_{1}| \leq |f(x)-x|$ for all $x\in I$. From here, we know that $|f(x_{0})-x_{0}|= |f(x_{1})-x_{1}|$.

From Langrange’s mean value theorem, there exists $\xi \in (x_{0}, x_{1})$ such that $(Df)(\xi)=1$. Since the schwarzian derivative of $f$ is negative, from the minimal principle, we get

$(Df)(\xi) > \min(Df(x_{0}), Df(x_{1})).$

i.e. $Df(x_{0})<1, Df(x_{1})<1$. However, from the definition of $x_{0}$ and $x_{1}$, we get

$Df(x_{0}) = \lim_{x\rightarrow x_{0}^{+}} \frac{f(x)-f(x_{0})}{x-x_{0}} \geq 1$

$Df(x_{1}) = \lim_{x\rightarrow x_{1}^{-}} \frac{f(x_{1})-f(x)}{x_{1}-x} \leq 1$

This is a contradiction. Therefore, the existence of $x_{0}$ is unique.

Assume $f: I \rightarrow f(I) \subseteq \mathbb{R}$ is a $C^{3}$ diffeomorphism, define the non-linearity of $f$ as

$f \mapsto Nf=\frac{D^{2}f}{Df} = D \ln Df \text{ whenever } Df \neq 0.$

Proposition 4.  $N(f \circ g)= (Nf \circ g) \cdot Dg+ Ng.$

Proposition 5. $Sf=\frac{D^{3}f}{Df} -\frac{3}{2}(\frac{D^{2}f}{Df})^{2}=D(Nf)-\frac{1}{2}(Nf)^{2}.$

Theorem 5. Koebe Non-linearity Principle.

Given $B, \tau>0$, there exists $K_{\tau,B}>0$ such that, if $f: [-\tau, 1+\tau] \rightarrow \mathbb{R}$ is a $C^{3}$ diffeomorphism into the reals and $Sf(t)\geq -B$ for all $t\in [-\tau,1+\tau],$ then we have

$|\frac{f^{''}(x)}{f^{'}(x)} | \leq K_{\tau,B}$

for all $0\leq x \leq 1.$ Show that $K_{\tau,B} \rightarrow 2/\tau$ as $B\rightarrow 0.$ (This recovers the classical Koebe non-linearity principle).