# Periodic orbits and Sturm–Liouville theory

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, 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

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|.

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, 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

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

$M$ 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

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:

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

# Metric entropy 1

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:

1.

$\lim{n\to \infty}$ 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 $f$.

2.

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

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

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

# Interval map

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

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

\section{基本性质，例子}
\subsection{例子和基本性质}

$T:X\longrightarrow X$

\subsection{Transitivity}

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}

$\{x\in X|\{T^nx\}_{n\in Z}dense\}=\cap_{n\in N^*}\cap_{k\in N^*}\cup_{m\in Z} T^mB_{\frac{1}{k}}(x_n)$
\subsection{一个和矩阵有关的例子}

\subsection{minimality和Birkhoff回复定理}

minimality定义是动力系统$T:X\rightarrow X$所有的点都是transitivity point。\\

$T$不变集只有$X$和空集\\

1.zorn lemma，2.minimal性质的第二点\\

$\exists x\in X,\exists \{n_i\},\lim_{i\to \infty}T^{n_i}x=x$
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。\\

\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}

\newpage

\section{拓扑熵}

\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}