Periodic orbits and Sturm–Liouville theory

21 12月 201721 12月 2017 hxypqr留下评论

I thinks there is some problem related to the solution of a 2 order differential equation given by Sturm-Liouville system which is nontrivial.

It is well-know that the power of Sturm-Liouville theory see wiki, is due to it is some kind of “spectral decomposition” in the solution space.

Two kind of problem is interesting, one is the eigenvalue estimate, both upper bound and lower bound, this already investigated in ESTIMATING THE EIGENVALUES OF STURM-LIOUVILLE. PROBLEMS BY APPROXIMATING THE DIFFERENTIAL EQUATION.

I post two problem here, this is a product due to a random walk along the boundary of topology and the analysis,

Problem 1.

Fix a set $A=\{k_1<k_2<...<k_l\}$ , is there a 2 order ordinary differential equation given by Sturm–Liouville theory such that the eigenfunction $f_{k}$ is periodic if and only if $k\in A$ ?

There is also some weak version of this and a infinity version of this.

Of course we have the following map, from the high order ordinary differential equation to the 1 order differential equation in high dimension. But the key point is that it is not a bijection! The Frobenius condition play a crucial role.

Problem 2.

There is a homotopy in the moduli space of differential equation, and we could define a direct product operator in this space, and we consider the topology defamation of the eigenfunction, could there be some equality, one side of it explain the topology information, the other side explain the spectral (or analysis) information?

There is another interesting problem.

Rotation number

4 12月 20175 12月 2017 hxypqr留下评论

Consider compact 1 dimension dynamic system.

We focus on $S_1$ , it does not mean $S_1$ is the only compact 1 dimensional system , but it is a typical example.

$T: S_1\to S_1$ .

If $T$ is a homomorphism then $T$ stay the order of $S_1$ (by continuous and the zero point theorem). That is just mean:

(may be do a reflexion $e^{2\pi i\theta}\to e^{-2\pi i\theta}$ ).

In the homomorphism case. We try to define the rotation number to describe the expending rate of the dynamic system.

$T: \mathbb S_1\to \mathbb S_1$ i.e. $T:\mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ . Lifting to,

$\hat T:\mathbb R\to \mathbb R$ .

How to realize the lifting?

Step1: Periodic extend $T:\mathbb S_1\to \mathbb S_1$ to $T': \mathbb R\to \mathbb S_1$ . (Regard $\mathbb S_1$ as $[0,2\pi]$ ).

Step2: Consider the “flow” of $T':\mathbb R\to \mathbb S_1$ . we get $\hat T:\mathbb R\to \mathbb R$ .

The rotation number is defined as:

$\rho (T)=\limsup_{n\to \infty}\frac{\hat T^n(x)}{n}$ .

Following we will shall it is independent of the choice of $x$ and $\rho (T)=\lim_{n\to \infty}\frac{\hat T^n(x)}{n}$ in fact.

It is not difficult to proved the following property:

Property:

1.If $T'$ is conjugate (in fact semi-conjugate is enough ). Then 1.If $T'$ is conjugate (in fact semi-conjugate is enough ). Thenrotation number of $T'$ equal to rotation number of $T$ .

2.If $T'$ is conjugate to $T$ . Then rotation number of $T'$ equal to
rotation number of $T$ .

$T'(y)=\Psi\circ T\circ \Psi^{-1}(y)=\Psi(\Psi^{-1}(y)+t(\Psi^{-1}(y)))$ )

Example: $T: x\to x+\alpha$ , $\alpha\in \mathbb R$ . It is not difficult to prove the rotation number of $T$ is $\alpha$ .

Propersion:

1)For $n\geq 1$ we have that $\rho(T^n)=\rho(T) (mod 1)$ .

2).If $T$ has a periodic point, i.e. $x\in S_1,\exists n\in \mathbb N^*, T^{n}(x)=x$ . Then $latex\rho (T)$ is rational.

3) $T:\mathbb R/\mathbb Z \to \mathbb R/\mathbb Z$ has no periodic point then $\rho(T)$ is irrational.

4) The limit actually exists and we have: $\rho(T)=\lim_{n\to \infty}\frac{\hat T^n(x)}{n} (mod 1)$ .

pf of 1):

$\rho(T^n)=\lim_{k\to \infty}\frac{(\hat T^n)^{k}(x)}{k}$

$=\lim_{k\to \infty}n\frac{(\hat T)^{nk}(x)}{nk}$

$=n\rho (T)$ .

Used the property $|x-y|<k \leftrightarrow |T^{\omega}(x)-T^{\omega}(y)|<k+1, \forall \omega\in N^*, \forall k\in \mathbb Z^{+}$ .

pf of 2):

It is not difficult to prove $\rho(T)$ is independent with the choice of $x$ . So choose $x$ to be the periodic point.

Remark: but the inverse of 2) is not true. For example:

$x\to x+\frac{1}{2}+\frac{1}{100}sin(4\pi x)$ .

This dynamic system has both periodic points( $\{0,\frac{1}{2}\},\{\frac{1}{4},\frac{3}{4}\}$ ) and non-periodic pint (maybe orbits generated by $\{\frac{1}{\sqrt{2}}\})$ .

pf of 3):

If not. Assume $\rho(T)$ is rational number $\frac{q}{p}$ . Take any point $x\in \mathbb S_1$ , then:

$\lim_{n\to \infty}\frac{\hat T^n(x)}{n}=\frac{q}{p}$ .

$\Longrightarrow \lim_{n\to \infty}\frac{(\hat T^p)(x)}{n}=q$ .

$\Longrightarrow \lim_{n\to \infty}\frac{(\hat T^p-q)^n(x)}{n}=0$ .

Now assume $\hat T^p-q=\widetilde T$ .

Then $\widetilde x>x$ . $\forall x\in \mathbb S_1$ (if $\widetilde x<x, \forall x\in \mathbb S_1$ , take reflection $x\to -x$ ).

And there do not exists $n\in \mathbb N^*$ such that $\widetilde T^nx>x+1$ . If not, we could prove rotation number is large than $\frac{1}{n}$ lead a contradiction.

So $\{\widetilde T^nx\}_{n=1}^{\infty}$ is a bounded monotonically increasing sequences in $\mathbb S_1$ , it limits point $z\in \mathbb S_1$ must satisfied $\widetilde T^n (z)=z$ .

pf of 4):

Using the point wise approximation inequality induced from the monotonically and stay ordering property of $\mathbb S_1$ by $T$ .

Corollary:

Assume $\rho(T)$ is irrational.

1. Let $n_1,n_2,m_1,m_2\in \mathbb Z$ , and $x,y\in \mathbb R$ . If $\hat T^{n_1}(x)+m_1<\hat T^{n_2}(x)+m_2$ , then $hat T^{n_1}(y)+m_1<\hat T^{n_2}(y)+m_2$ .

2. The bijection $n\rho (T)+m\to \hat T^n(0)+m$ between the set $\Omega=\{n\rho(T)+m| n,m\in \mathbb Z\}$ and $\Gamma=\{\hat T^{n}(0)+m,n,m\in \mathbb Z\}$ precise the natural ordering on $\mathbb R$ .

This corollary is not difficult to prove use the established property.

Denjoy’s theorem

Proposition:

If $T: \mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ is a minimal orientation presenving homomorphism with irrational rotation number $\rho$ then $T$ is topologically conjugate to the standard rotation $R_{\rho}: \mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ .

leave as a ex.

For $T: \mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ , $T': \mathbb R/\mathbb Z\to \mathbb R$ . We define the variation of $log|T'|: \mathbb R/\mathbb Z\to \mathbb R$ by:

$Var(log(|T'|))=$

$sup\{\sum_{i=0}^{n-1}|log|T'|(x_{i+1})-log|T'|(x_i)|: 0=x_0<x_1<...<x_n=1\}$

We say that the logarithm of $|T'|$ has bounded variation if this value $Var(log|T'|)$ is finite.

Denjoy’s theorem:

If $T: \mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ is a $C^1$ orientation preserving homomorphism of the circle with derivative of standard variation and irrational rotation number $\rho=\rho(T)$ then $T:\mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ is topologically conjugate to the standard rotation :

$R_{\rho}:\mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ .

Due to the upper proposition we only need show $T:\mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ is minimal. Proof pf minimal is splitting to following two sub lemmas.

Sublemma1:

If $T$ has irrational rotation number and there are a constant $C>0$ and a sequences of integers $q_n\to \infty$ such that the map: $T: \mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ Satisfy : $|(T^{q_n})'(x)||(T^{-q_n})'(x)|\geq C$ Then $T: \mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ is minimal.

Sublemma2:

Fix $x\in \mathbb R/\mathbb Z$ and write $x_n=T^n(x)$ , for $x\in \mathbb Z$ There exists an increasing sequences $q_n\to \infty$ of natural number such that the intervals $(x_0,x_{q_n}),(x_1,x_{q_n+1}),...,(x_i,x_{q_n+i}),...,(x_{q_n},x_{2q_n})$ are all disjoint.

Paradox and problem

Graph:

$\rho(T)>0$ because of existence of fix point.

$T_{\alpha}=T+\alpha$ for $\alpha< sup_x|T_x-x|$ .

Is $\rho(T_{\alpha})=0$ always true for $\alpha \in R$ ?

If it is right, then there is a contradiction with argument $(*)$ , but for what type of dynamic system $T$ ?

$T_{\alpha}=T+\alpha$ satisfied $\rho(T_{\alpha})=\rho(T)+\alpha$ . for all $\alpha\in \mathbb R$ ?

Problem:

If $T$ is not homomorphism but $T:x\to x+g(x)$ induced $g(x)=x-f(x)$ , $f(x)$ is striating increasing, Is the limit of $\lim_{x\to \infty}\frac{\hat T(x)}{n}$ always exists? it could not be increase with $x$ .

Baragar-Bourgain-Gamburd-Sarnak conjecture

22 11月 201722 11月 2017 hxypqr留下评论

$M$ is the markov triple $(x,y,z)$ :

$x^2+y^2+z^2=xyz$ and $(x,y,x)\in \mathbb Z^3 \ \ \ \ (*)$ .

It is easy to see:

$R_1: (x,y,z)\to (3yz-x,y,z)$ .

map markov triple to markov triple.

This is also true for $R_2,R_3$ . and the transform $R_1,R_2,R_3$ and permutation a classical result of markov claim that all solution of (*) could be generated from $(1,1,1)$ . I get a similar result for a similar algebraic equation 1 half years ago when consider a $Q$ version of problem about 1-form given by Xu Bin.

Now we know the graph with root $(1,1,1)$ and with node generate by transform $R_1\cup R_2 \cup R_3 \cup S_3$ is connected.

The B-B-G-S conjecture is is the connected property still true for prime $p$ surfficed large?

Metric entropy 2

21 11月 201722 11月 2017 hxypqr留下评论

I am reading the article “ENTROPY THEORY OF GEODESIC FLOWS”.

Now we focus on the upper semi-continuouty of the metric entropy map. The object we investigate is $(X,T,\mu)$ , where $\mu$ is a $T-$ invariant measure.

The insight to make us interested to this kind of problem is a part of variational problem, something about the existence of certain object which combine a certain moduli space to make some quantity attain critical value(maximum or minimum). The most simple example maybe Isoperimetric inequality and Dirichlet principle of Laplace. Any way, to establish such a existence result a classical approach is to proof the upper semi-continuouty and bounded for associate energy of the problem. In our case the semi-continuouty will be some thin about the regularity of the entropy map:

$E:M(X,T)\to h_{\mu}.$

We define the entropy at infinity:

$sup_{(\mu_n)}limsup_{\mu_n\to 0}h_{\mu_n}(T)$

Where $(u_n)_{n=1}^{\infty}$ varies in all sequences of measure coverage to $0$ in the sense for all $A\subset M$ , $A$ measurable then $\lim_{n\to \infty} \mu_{n}(A)=0$ .

Compact case

we say some thing about the compact case, In this case we have finite partition with smaller and smaller cubes, this could be understand as a sequences of smaller and smaller scales. A example to explain the differences is $\mathbb N^{\mathbb N},\sigma$ , shift map on countable alphabet.

Because of this thing, there is a good sympolotic model, i.e. h-expension, and it generalization asymptotically h-expension equipped on a compact metric space $X$ have been proved to be that the corresponding entropy map is upper semi-continous.

In particular $C^{\infty}$ diffeomorphisms on compact manifold is asymptotically h-expensive.

Natural problem but I do not understand very well:

Why it is natural to assume the measure to be probability measure in the non-compact space?

Non-compact case

$(X,d)$ metric space

$T:X\longrightarrow X$ is a continuous map.

$d_n(x,y)=\sup_{0\leq k\leq n-1}d(T^kx,T^ky)$ , then $d_{n}$ is still a metric.

Easy to see $\frac{1}{n}h_{\mu}(T^n)=h_{\mu}(T)$ . This identity could be proved by the cretition of entropy by $\delta$ -seperate set and $\delta$ -cover set.

Kapok theorem:

$X$ compact, for every ergodic measure $\mu$ the following formula hold:

$h_{\mu}(T)=\lim_{\epsilon \to 0}limsup_{n\to \infty}\frac{1}{n}logN_{\mu}(n,\epsilon,\delta)$ .

Where $h_{\mu}(T)$ is the measure theoretic entropy of $\mu$ .

Riquelme proved the same formula hold for Lipchitz maps on topological manifold.

Let $M_e(X,T)$ defined the moduli space of $T$ -invariant portability measure.

Let $M_(X,T)$ defined the moduli space of ergodic $T$ -invariant probability measure.

Simplified entropy formula:

$(X,d,T)$ satisfied simplified entropy formula if $\forall \epsilon >0$ surfaced small and $\forall \delta\in (0,1)$ , $\mu\in M _e(X,T)$ .

$h_{\mu}(T)=\limsup_{n\to \infty}\frac{1}{n}log(N_{\mu}(n,\epsilon,\delta))$ .

Simplified entropy inequality:

If $\epsilon>0$ suffciently small, $\mu \in M_{e}(X,T)$ , $\delta\in (0,1)$ .

$h_{\mu}(T)\leq \limsup_{n\to \infty}\frac{1}{n}log(N_{\mu}(n,\epsilon,\delta))$ .

Weak entropy dense:

$M_e(X,T)$ is weak entropy dense in $M(X,T)$ . $\forall \lambda>0$ , $\forall \mu\in M(X,T)$ , $\exists \mu_n\in M_e(X,T)$ , satisfied:

$\mu_n\to \mu$ weakly.
$h_{\mu_n}(T)>h_{\mu}(T)-\lambda$ , $\forall \lambda>0$ .

Metric entropy 1

21 11月 201721 11月 2017 hxypqr留下评论

Some basic thing, include the definition of metric entropy is introduced in my early blog.

Among the other thing, there is something we need to focus on:

1.Definition of metric entropy, and more general, topological entropy.

2.Spanning set and separating set describe of entropy.

3.amernov theorem:

$h_{\mu}(T)=\frac{1}{n}h_{\mu}(T^n)$ .

Now we state the result of Margulis and Ruelle:

Let $M$ be a compact riemannian manifold, $f:M\to M$ is a diffeomorphism and $\mu$ is a $f$ -invariant measure.

Entropy is always bounded above by the sum of positive exponents;i.e.,

$h_{m}(f)\leq \int_{i}\lambda_i^{+}(x)dimE_i(x)dm(x).$

Where $dimE_i(x)$ is the multiplicity of $\lambda_i(x)$ and $a^{+}=max(a,0)$ .

Pesin show the inequality is in fact an equality if $f\in C^2$ and $m$ is equivalent to the Riemannian measure on $M$ . So this is also sometime known as Pesin’s formula.

F.Ledrappier and L.S.Young generate the result of Pesin.

One of their main result is:

$f:M\to M$ is a $C^2$ diffemoephism, where $M$ is a compact riemanian manifold, f is compatible with the Lesbegue measure on $M$ , and

$h_m({f,\mu})=\int_{M}\lambda_idim(V_i)dm$

If and only if on the canonical defined quation manifold $M/W_{\mu}$, i.e. the manifold mod unstable manifold $W_{\mu}$, the induced conditional measure $m_{\xi}$ is absolute continuous.

Remark: according to my understanding, the equality just mean in some sense we have the inverse estimate:

$h_{m}(f,\mu)\geq \int_{M}\lambda_idim(V_i)dm.$

This result maybe just mean near the fix point of $f$ ,i.e. the place charge the topology of the foliation, we have the inverse estimate. Such a inverse estimate will lead a control of the singularity of the push forward measure $m_{\xi}$ on the quation manifold. So $m_{\xi}$ have good regularity. But this idea is not complete to solve the problem.

Now we begin to get a geometric explain and which will lead a rigorous proof of the inequality:

$h_{m}(f)\leq \int_{i}\lambda_i^{+}(x)dimE_i(x)dm(x).$

At first we could observe that the long time average $\lim_{n\to \infty}\frac{1}{n}log||Df^n||$ of $Df$ could be diagonal. Assume after diagonal the eigenvalue is

$\lambda_1\leq \lambda_2\leq \lambda_3\leq...\leq \lambda_{n-1}\leq \lambda_n$ .

This eigenvalue could divide into 3 parts: <0,=0,>0.

This will lead to a direct sum decomposition of the tangent bundle $TM$ :

$TM\simeq E_{u}\otimes E_s\otimes E_c.$

Where $E_u$ is the part corresponding to the eigenvalue>0, For this part we consider the more refinement decomposition:

$E_u=\otimes_{k=1}^rV_k$ , $V_k$ is the eigenvector space of $\lambda_k$ . The dimension of $V_k$ is $dim V_k$.

On the other hand, we have a equality of metric entropy:

$h_{m}(f)=\frac{1}{n}h_{m}(f^n)=\sup_{\alpha\in partition \ set}\frac{1}{n}h_m(f^n,\alpha)$ .

For the later one, $\alpha$ is a measurable partition of $M$ , then $\alpha$ could always be refine to a smaller partition $\beta$ , and we have:

$h_{m}(f,\alpha)\leq h_{m}(f,\beta)$ .

Now we arrive the central place of the proof:

every partition could be refine by a partition with boundary of almost all cubes is parallel to the foliation. So we focus ourselves on the portion $\beta$ and all boundary of cubes in $\beta$ is parallel to the eigenvector.

Under this situation, we need only estimate the numbers of $\vee_{i=1}^nT^i\beta$ . Estimate it is not very difficult. we need only observe the following two thing:

$\lim{n\to \infty}$ exists a.e. in $M$ . So this lead to the definition of foliation almost everywhere, and except a measurable zero set. In fact this set is the set of fix point of $M$ under $f$ .

After a rescaling, every point which is not a fix point of $f$ could be understand as it is far away from fix points. Then the foliation could be understand as a product space locally. The flow with the direction which the eigenvalue is less than 1 cold not change $\vee_{i=1}^nT^i\beta$ . The direction with eigenvalue equal to 1 is just transition and just change the number of $\vee_{i=1}^nT^i\beta$ with polynomial growth. But the central thing is the direction with eigenvalue large than one and will make $\vee_{i=1}^nT^i\beta$ change with viscosity $e^{\lambda_i}$ . and we product it and get :

$h_{m}(f,\mu)\leq \int_{M}\lambda_i dim(V_i)dm$ .

In fact the proof only need $f$ to be $C^1$

Covering a non-closed interval by disjoint closed intervals

20 10月 2017 hxypqr留下评论

this note will talk about the Ostrowski representation and approximation by continue fraction.

As well-known,by the Weyl criterion, $\{n\alpha\}$ is uniformly distribution in $[0,1]$ iff $\alpha\in R-Q$ .

i.e. we have: $\forall 0\leq a\leq b\leq 1$ ,we have:

$\lim_{N\to \infty}|\{1\leq n\leq N|\{n\alpha\}\in [a,b]\}|=(b-a)N+o(N)$ .

but this will not give the effective version.i.e. we do not the the more information about the decay of $o(N)$ .

we will give a approach of effective version of $\alpha$ with smooth condition by give another proof of the uniformly distribution (in fact to to decomposition the interval $[a,b]$ in to a finite sums of special intervals).and get the result:

$D_N=\int_{M}D_N(\theta)d\mu=\int_Msup_{0<a<b<1}|\sum_{n=1}^{N}\chi_{(a,b)}(\{\theta n \})-N(b-a)|d\mu\sim O(log N)$

if the term in the continuous fraction of $\alpha$ have a up bound.this is so called $\alpha$ is smooth.

Interval map

27 9月 201724 10月 2017 hxypqr留下评论

1.period 3 induce chaos

theorem:if a interval map $T:I\to I$ have a period 3 point $x$ ,then $\forall n\in N^*$ ,there is a period n point for $T$ .

proof:

n=1 case. trivial

$n>1,n\neq 3$ case:

the key point is to consider the structure of monotone interval contain previous one with fix length.

this will easy to lead a proof.

2.a work of J.Milnor and W.Thurston.

$N(T^n)$ defined as the number of monotone interval of the map $T^n$ .

theorem: $h(T)=lim_{n\to infty}\frac{1}{n}log N(T^n)$ .

3.monotone Markov map

this structure have two property:

1.piesewise monotone and $C^1$,the derive has control!

there is a relative dynamic system with this map.is a shift map with a relative $n\times n$ matrix $A$ .

this two dynamic system have a lot of relation,the key one is:

the topological entropy of monotone Markov map is just the unique maximum eigenvalue of $A$ .

and some byproduct…

Flat surface 1

26 9月 201724 10月 2017 hxypqr留下评论

Topological point of view:

in topological point of view a flat surface is a topological space $M$ with a (ramified in nontrivial case) map $\pi:M\longrightarrow T^2$ .

and the map satisfied: $\pi$ is not ramified on $\pi^{-1}(T^2-\{0\})$ is not ramified and defined a covering map.

Geometric-analytic point of view:

we begin with compact connected oriented surface $M$ ,and a nonempty finite subset $\Sigma=\{A_1,...,A_n\}$ of M.and

translation surface of type $k$ :

translation structure:

complex structure:

动力系统笔记

25 9月 201724 10月 2017 hxypqr留下评论

\section{基本性质，例子}
\subsection{例子和基本性质}
在这一章的第一节引入了我们的研究对象，一般是一个紧的度量空间$X$装备上了一个同胚
$T:X\longrightarrow X$
介绍了三个简单例子，包括 $S_1$ 上的加倍映射，旋转映射以及 $X_k=\Pi_{n\in Z}\{1,2,...,k\}$ 上的平移映射。
加倍映射会出现在微分流形中一些函数 $f$ 的singular point，也就是 $hess f=0$ 的地方附近的环绕数计算，还有一些scalling变换或者是一些多尺度的问题里。\\
旋转映射会和旋转数是有理数还是无理数有关,相关的wely准则告诉我们如果是无理数的话会是每个点的轨道均匀分布的，稠密性在动力系统里面说就是这个动力系统是minimal的。相关的问题有sarnack猜想在Torus上的特殊情形，目前半解析的$T^2$情形已经解决，这是最近的工作，后续很多工作在进行，本质困难来自解析数论。\\
平移映射我不是很懂，第二章中讲的Van der warden定理的证明是一个好例子，动力系统中的回复定理主要是用来刻画这些动力系统内蕴的算术性质的，basic ideal是如下事实：\\
将一个大的集合分类，同一类有序的出现的存在性。

\subsection{Transitivity}
这个性质是指一个动力系统中存在轨道在动力系统中稠密。\\
动力系统往往具有transitivity的性质，加倍映射的例子用二进制分解构造，平移映射构造transitivity point的方法与之雷同，旋转映射情形这是初等的。\\
transitivity会有很多等价的刻画，包括四种：

\begin{thm}(transitivity的等价定义)\\
1.transitivity\\
2. $U$ open, $TU=U \Longrightarrow U=\emptyset$ 或者 $U$ 是一个稠密集\\
3. $U,V$ 开集, $\exists N\in N^*$ , $T^n U \cap V\neq \emptyset$ \\
4. $\{x\in X|\{T^nx\}_{n\in Z}dense\}$ 是一个 $G_{\delta}$ 集合
\end{thm}
证明都是标准的，提两个关键点，第一点是注意到 $\cup_{n\in Z}T^n U$ 这个集合是$T$不变的，第二点是注意到transitive point可以通过选取一组开集集进行描述从而有集合等式：
$\{x\in X|\{T^nx\}_{n\in Z}dense\}=\cap_{n\in N^*}\cap_{k\in N^*}\cup_{m\in Z} T^mB_{\frac{1}{k}}(x_n)$
\subsection{一个和矩阵有关的例子}
定义了一个和矩阵有关的动力系统，并且说明了这个动力系统是transitive的当且仅当底层的矩阵是不可约的，对于矩阵不可约这个概念不熟所以这个例子没有仔细看。
\subsection{minimality和Birkhoff回复定理}
我认为这部分内容是Pollicott书第一章最有趣的部分。\\
minimality定义是动力系统 $T:X\rightarrow X$ 所有的点都是transitivity point。\\
也有三个等价定义，其他两个是：\\
$T$ 不变集只有 $X$ 和空集\\
任何开集通过T作用生成的集合是全空间\\

可以看出来这三个定义都是transitivity情形对应定义的加强版。这些证明也是标准的，接下来一个定理表明 $X$ 这个空间可以在 $T$ 不变的意义下分解成很多小的空间，每个都是不能再分解的，这个定理的证明的两个关键点是：
1.zorn lemma，2.minimal性质的第二点\\

那么马上我们就可以得到minimality的定理系统满足birkhoff回复定理：\\
$\exists x\in X,\exists \{n_i\},\lim_{i\to \infty}T^{n_i}x=x$
Birkhoff回复定理在高维情形也会很有趣，我们这时就需要多个可交换的动力系统（为什么一定要可交换？一种解释是可交换大幅降低复杂度）一旦这些动力系统被证明是minimal的，我们用类似的路线建立起以上定理是没有本质困难的。\\
步骤一：建立起transitive的相关定理\\
步骤二：建立起minimal的相关定理\\
步骤三：说明 $T^i$ 不变的集合满足zorn lemma，所以有最小元\\
整个过程在乘积空间中进行
\newpage

\section{Birkhoff回复定理蕴含Van Der Warden定理}
\subsection{Van Der Warden定理与它的动力系统解释}
这是一个组合定理，原始证明是很trick的，单遵老先生有一个证明，很trick，高中的时候尝试过证明，自己证了一个星期证明不出来就看掉了，现在回想起来应该跟当时的工具太原始了有关系。我想强调的是并不是数学思想的飞跃，而是数学工具的升级使得这个问题变简单了。\\

原始问题是将 $N^*$ 分成若干个类，一定存在一个类存在任意长等差数列。\\

怎么转化成一个组合问题呢，其实用第一章中的 $X_k$ 装备上平移这个同胚构成的动力系统就够了，等差数列的存在性等价于若干个可以交换的映射，其实就是平移的步长不一样下的都回到原始值附近，这是birkhoff回复定理能够告诉我们的。\\

关于细节的建立
第一步是简单的，问题出在第二步，也就是证明整个动力系统是minimal的这一步上，我们知道这个动力系统是初始状态通过平移生成再取闭包得到的，所以天然是transitivity的，如果是minimality的，那么就可以用birkhoff定理得到结果了。这其实不难，因为这个距离空间是non-archimeadian的，用初始状态的平移去逼近就好了。\\

上面这一段划去，实际上要想真正建立一个动力系统本身是minimal的性质，本质上需要比连续性更强的某种正则性，比如一个lipchitz连续性的动力系统就是minimal的。但是仅仅找到一个紧集是minimal的时简单是事情，用minimal等价定义第二条加zorn引理就可以做到。好了现在我们有了一个minimal的动力系统，我们只需要建立多重birkhoff回复定理就完成了证明。\\

在思考多重birkhoff回复定理的过程中我发现了几种方式来构建整个框架，pollicott上标准的证明是利用乘积空间的对角线作为低空间加上归纳法，我尝试过将对角线作为低空间证明但是失败了，主要原因是对角线在乘积空间中是低维子集我不知道怎么将合适映射限制到这个空间上，事实也证明做归纳法的话我们可以转而对映射而不是空间做文章而规避这个困难。\\

但是在这个过程中我发现了另外一个有意思的现象，就是我们可以归纳的构造出一个集合，至少有限步的构造在逻辑上式对的，利用算子 $T^i$ 之间的交换性得到一个很好地X的子空间， $T^i$ 在上面的作用也有很好的性质，但是还不够好。具体的说，是一种纤维结构的空间， $T^1$ 在底空间上作用是transitive的， $T^2$ 在第一层纤维上的作用是transitive的，依次类推。由于交换性可以导致在每一个section上 $T^i$ 的作用都是trsnsitive的。但是不好的地方在于每个 $T^i$ 想要在全空间中transitive都必须借助别的 $T^i$ ，换而言之每个 $T^i$ 都只管一层。所以这并不是我们想要的空间。\\

这样构造出来的空间在这里可能没有用，但是这个空间本身具备很好的性质，而就算我们知道多重回复定理这样的空间也是构造不出来的，注意我们并不是因为构造了一个 $T^i$ 在上面”一致的”transitivity的空间而把回复定理证明出来了，而是用了一些更弱的argument达到目的。这个空间可能在计算全空间上某些可交换的映射的特征时有用，尤其在可以证明这个空间和全空间只差一个零测集的情况下。\\

猜想：存在一个 $T^i:X\longrightarrow X$ ,$T^iT^j=T^jT^i$，并不存在满足某种”一致”minimal的子空间。但是我们知道多重回复定理是对的。\\

总之用归纳的方法加上一些拓扑的标准的方法我们可以得到多重回复定理从而完成证明。
\newpage

\section{拓扑熵}
拓扑熵的定义可复杂了，拓扑熵是一个描述拓扑动力系统复杂程度的量。顺序是先引入标准定义和基本性质，然后给出一个计算方法，再然后引入spanning set和separeting set，利用这两种集合引入等价的定义方法，再证明amernov定理： $h(T^m)=mh(T)$ ,最后证明动力系统之间的半共轭会导致熵之间的不等式。

\newpage
$f(x),g(x)\in C_{c}^{\infty}(R^n)$
\[f*g(x)=\int_{R^n}f(\xi)g(x-\xi)d\xi \]
\[x=(x_1,x_2,…,x_n)\in R^n\]
\[g(x)=\frac{1}{x_1^2+x_2^2…+x_n^2+1}\]
\[f*g(x)=\int_{R^n}f(\xi)g(x-\xi)d\xi\sim\sum_{k_1=-\infty}^{\infty}…\sum_{k_n=-\infty}^{\infty}\frac{f(x_1-k_1,x_2-k_2,…,x_n-k_n)}{k_1^2+k_2^2+…+k_n^2+1} \]
\[\sum_{k_1=-\infty}^{\infty}…\sum_{k_n=-\infty}^{\infty}\frac{f(x_1-k_1,x_2-k_2,…,x_n-k_n)}{k_1^2+k_2^2+…+k_n^2+1}=\sum_{\xi\in Z^n}f(x-\xi)g(\xi)=\int_{\xi\in R^n}f(x-\xi)\delta g(\xi) d\xi\]\\
(Young inequality)
$f\in L_1(R^n)$,$g\in L^p(R^n)$:
\[ ||f*g||_{p}\leq ||f||_1||g||_{p} \]

(hardy-litterwood-soblev inequality)
$p,r>1$,$0<\lambda<n$,$\frac{1}{p}+\frac{\lambda}{n}+\frac{1}{r}=2$,$f\in L^p(R^n),h\in L^r(R^n)$.exists a constant C,$C\sim n,\lambda,p$.
\[|\int_{R^n}\int_{R^n} f(x)|x-y|^{\lambda}g(y)dxdy|\leq C(n,\lambda,p)||f||_p||h||_r\]

\section{热核正则性}
$t \to 0^+$ 情况的技巧
gap太多了，主要集中在两条，第一条是需要研究billiard上热核的正则性，这需要建立大量的耗散性先验估计。连续是显然的，我目前连C1都证明不出来，因为其中需要处理一个级数和。如果这一条对了，那么我们集中看t趋于0正时的热核。

\section{Caldron Zygmund算子的谱}
我们需要刻画Caldron Zygmund算子作用在某个区域上之后产生的谱会携带多少区域的形状的信息。通过在热核中令 $t\longrightarrow 0$ 这个会化简为简单的情况，再加上凸性。
第二条是热核对t求任意次导以后是Caldero ́ n Zygmund算子，对这个算子卷积上一个具备 $C^1$ Boudary正则性的区域上特征函数的的谱我们有没有好的刻画，这其中能不能蕴含这个区域的几何信息。

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分类： Dynamic system