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不确定性原理：局部化、Bernstein 估计与 Gaussian
旧博客原文

原题：Uncertainty principle

The pdf version is Uncertainty principle. The nice note of terrence tao seems given a nice answer for the problem below.

1. Introduction

Is there a Brunn-Minkowski inequality approach to the phenomenon charged by uncertainty principle? More precisely, is it possible to say some thing about the Gaussian distribution

$\displaystyle G(x)=e^{-|x|^2} \ \ \ \ \ (1)$

to be the best choice that ${\|\hat G-G\|_2}$ arrive minimum?

Remark 1 Or some other suitable distance space on reasonable function (may be some gromov hausdorff distance? Any way, to say the guassian distribution is the best function to defect the influence of uncertain principle.

I do not know the answer of the problem 1, but this is a phenomenon of a universal phylosphy, aid, uncertainty principle, heuristic:

It is not possible for both function ${f}$ and its Foriour transform ${\hat f}$ to be localized on small set.

Now let me give some approach by intuition to explain why the phenomenon of “uncertainty principle” could happen.

The approach is based on:
1. level set decomposition.
2. area formula (or coarea formula), anyway, some kind of change variable formula.
3. integral by part.
4. Basic understanding on exponential sum.
Let our function ${f\in S}$ the Shwarz space, we begin with a intuition (not very rigorous) calculate:

$\displaystyle \begin{array}{rcl} \hat f(\xi) & = & \int e^{2\pi i<\xi, x>}f(x)dx\\ & \overset{integral \ by \ part}= &\int \frac{1}{-2\pi i\xi}e^{-2\pi i<x,\xi>}\cdot \nabla f(x)\\ & \overset{Fubini}= & \int_{inf |f|}^{max |f|}\int_{level set A(t)} \frac{-e^{2\pi i<\xi,x>}}{-2\pi i\xi}\nabla f(x)dH^{n-1}(A)dt \end{array}$

Now we try to understanding the result of the calculate, it is,

$\displaystyle \hat f(\xi) =\int_{inf |f|}^{max |f|}\int_{level set A(t)} \frac{-e^{2\pi i<\xi,x>}}{-2\pi i\xi}\nabla f(x)dH^{n-1}(A)dt \ \ \ \ \ (2)$

$\displaystyle \pounds(A(t),\xi)=\int_{level set A(t)} \frac{-e^{2\pi i<\xi,x>}}{-2\pi i\xi}\nabla f(x)dH^{n-1}(A) \ \ \ \ \ (3)$

The calculate is wrong, but not very far from the thing that is true, the key point is now the exponential sum involve. We could use the pole coordinate in the frequence space and get some very rough intuition of why the the uncertainty principle could occur.

Remark 2 Why we consider the level set decomposition, due to the integral is a combination of linear sum of the integral on every level set, so shape of level set is the key point.

The part of ${\frac{-e^{2\pi i<\xi,x>}}{-2\pi i\xi}}$ in 2 is a rotation on the level set, a wave correlation of it and the christization function ${\chi_{A_t}}$ of level set ${A_t}$ in the whole space, this is of course a exponential sum.

Now we can begin the final intuition explain of the phenomenon of uncertainty principle. If the density of function ${f}$ is very focus on some small part of the physics space, then it is the case for level sets of ${f}$ , but we could say some thing for the exponential sum ${\pounds(A(t),\xi)}$ 3 related to the level set, just by very simply argument with hardy litterwood circle method or Persaval identity? Any way, something similar to this argument will make sense, due to if the diameter of level set focus ois small, then we can not get a decay estimate for ${\pounds(A(t),\xi)}$ when ${\xi\rightarrow \infty}$ along one direction in frequency space, in fact we could say the inverse, i.e. it could not decay very fast.

2. Bernstein’s bound and Heisenberg uncertainty principle

2.1. Motivation and Bernstein’s bound

There is two different Bernstein’s bound, we discuss the first with the motivation, and proof the second rigorously. \paragraph{Form 1} ${A}$ is a invertible affine map, then for a ball ${B}$ , ${A(B)=\epsilon}$ is a ellipsoid.

$\displaystyle \epsilon=\{x\in {\mathbb R}^d|\sum_{j=1}^{d}r_j^{-2}(x_j-y_j)^2\leq 1\} \ \ \ \ \ (4)$

By a orthogonal transform we could make ${A}$ to be a diagonal matrix, i.e. ${A=diag(r_1,...,r_d)}$ . It is said, for ${\forall f\in S}$ or ${f}$ is a smooth bump function, ${f_A=f\circ A^{-1}}$ , so we have,

$\displaystyle \hat f_A(\xi)=\int e^{2\pi i<x,\xi>}\cdot f\circ A^{-1}(x)dx \ \ \ \ \ (5)$

We define dual of ${\epsilon}$ , ${\epsilon^*:=\{\xi\in {\mathbb R}^d| \sum_{j=1}^d\xi_j^2r_j^2\leq 1\}}$ .

Remark 3 Why there we use the metric ${\xi_j^2r_j^2\leq 1}$ but not the standard inner product ${<\xi,x>}$ ? How to understand the choice?
Proposition 1 We have the following property:
1. ${f_A\in L^{\infty}\Longrightarrow \|\hat f_A\|_{1}\leq +\infty}$ .
2. ${|\hat f_A(\xi)|\leq c_N|\xi|(1+|\xi|^2_{\epsilon^*})^{-N}}$
Remark 4

$\displaystyle |\xi|^2_{\epsilon^*}=\sum_{i=1}^d\xi_j^2r_j^2$

This is a norm of ${{\mathbb R}^d}$ related to ${\epsilon^*}$ .

Proof: Suffice to proof 2.

$\displaystyle \begin{array}{rcl} |\hat f_A(\xi)| & = & |\int_{{\mathbb R}^d}e^{2\pi i<x,\xi>}f_A(x)dx|\\ & \overset{integral \ by \ part}\sim & \frac{1}{(1+|\xi|)^N}\int |e^{2\pi i<x,\xi>}\partial^N f_A(x)dx|\\ & = & c_N |\xi|(1+|\xi|_{\epsilon^*}^2)^{-N} \end{array}$

$\Box$

More quantitative we have rigorous one: \paragraph{Form 2} If ${f\in L^2({\mathbb R}^d)}$ , ${supp f\in B(r,0)}$ , then it is not possible for ${\hat f}$ to be concentrate on a scale much less than ${R^{-1}}$ .

Proposition 2 (Bernstein’s bound) Suppose ${f\in L^2({\mathbb R}^d)}$ , ${supp f\subset B_R(0)}$ . Then,

$\displaystyle \|\partial^{\alpha}\hat f\|_2\leq (2\pi r)^{|\alpha|}\|f\|_2, \forall \alpha. \ \ \ \ \ (6)$

Proof: ${\alpha=0}$ case is trivial by Paserval identity, which said on ${L^2({\mathbb R}^d)}$ , fourier transform is a isometry, ${\|f\|_2=\|\hat f\|_2}$ . For general case, integral by part, and use trivial estimate,

$\displaystyle \begin{array}{rcl} \|\partial^{\alpha}\hat f\|_2 & \overset{integral \ by \ part}= & \|x^{\alpha} f\|_2\\ & \leq & (2\pi r)^{\alpha}\|f\|_2 \end{array}$

$\Box$

2.2. Heisenberg inequality

Theorem 3 (Heisenberg uncertain principle) ${f\in L^2({\mathbb R}^d)}$ , so ${\hat f\in L^2({\mathbb R}^d)}$ , ${\|f\|_2=\|\hat f\|_2}$ . then for any ${x_0,\xi_0\in {\mathbb R}^d}$ , every direction, we have

$\displaystyle \|f\|_2^2=\|f\|_2\|\hat f\|_2\leq \|(x-x_0)f\|_2\|(\xi-\xi_0)\hat f\|_2 \ \ \ \ \ (7)$

Remark 5 We could understand the inequality by the following way. suffice to prove it with ${f\in S}$ and then by approximation argument. ${f\otimes \hat f\in S({\mathbb R}^d\times {\mathbb R}^d)}$ , define ${\|f\otimes \hat f\|_2:= \|f\|_{L^2({\mathbb R}^d)}\cdot \|\hat f\|_{L^2({\mathbb R}^d)}}$ . then we have the following:

$\displaystyle \|f\otimes \hat f\|_{L^2}\leq 4\pi \|xf\otimes \hat{xf}\|_{L^2} \ \ \ \ \ (8)$

Remark 6 The inequality is shape, the extremizers being precisely given by the modulated Gaussians: arbitrary

$\displaystyle f(x)= c e^{2\pi i\xi_0x}e^{-\pi \delta(x-x_0)^2} \ \ \ \ \ (9)$

There are two proof strategies I have tried, I try them for several hour but not work out with a satisfied answer, the method more involve, I explain what happen in section 1, I have not tried, I will try it later. Both this two strategies i face some difficulties, I explain why I can not work out them with a proof: \paragraph{Strategy 1} The first one is, we could work with ${f\in S}$ of course, by approximation, then we find, by Paserval, ${\|f\|_2=\|\hat f\|_2, \|\partial_x f\|_2=\|\xi \hat f\|2}$ and are both true. then we use our favourite way to use Cauchy-Schwarz, the difficulty is we can not use a integral by part argument directly, even after restrict ourselves with monotonic radical symmetry inequality and by a rearrangement inequality argument, it seems reasonable due to rearrangement decreasing the kinetic energy as said in Lieb’s book. But even work with monotonic one, then one involve with some complicated form, try to use Fubini theorem to rechange the order of integral try to say something, it is possible to work out by this way but I do not know how to do. There is some calculate under this way,

$\displaystyle \begin{array}{rcl} \|xf\|_2\|\xi \hat f\|_2 & \overset{Cauchy-Schwarz}\geq & \int xf\cdot \partial_x f\\ & \sim & \int f^2 \end{array}$

but you know, at a point we have ${\partial_x(Xf)=f+x\partial_x f\neq f}$ , the reasonable calculate is following,

$\displaystyle \partial_x(xf)=f+x\partial_x f \ \ \ \ \ (10)$

We want ${\partial_x P(x,f) =f}$ , Then

$\displaystyle \begin{array}{rcl} \partial_x P(x,f) & = & f\\ & = & \partial_x(xf)-x\partial_x f\\ & = & \partial_x(xf)-\partial_x(\frac{1}{2}x^2\partial_x f)+\frac{1}{2}x^2\partial_{x^2}f\\ & = & \partial_x(\frac{1}{6}x^3\partial_{x^2}f)-\frac{1}{6}x^3\partial_{x^3}f\\ ...\\ & = &\partial_x(\sum_{i=1}^{\infty}(-1)^{i+1}x^i\frac{1}{i!}\partial_{x^i}f)+(-1)^{i+1}x^i\frac{1}{i!}\partial_{x^{i+1}}f \end{array}$

Seems to be ${f=\partial_x(ln(f))}$ … I do not know.

\paragraph{Strategy 2} The second strategy is, in the quantity ${\|xf\|_2\|\xi \hat f\|_x}$ we lose two cone very near ${x_0,\xi}$ , we need use the extra thing to make up them. May be effective argument come from some geometric inequality.

3. The Amerein-Berthier theorem

Next we investigate following problem, the problem is following: if ${E,F\subset {\mathbb R}^d}$ are of finite measure, can there be a nonzero ${f\in L^2({\mathbb R}^d)}$ with ${supp (f)\subset E}$ and ${supp(\hat f)\subset F}$ ? Some argument is folowing: Observe that:

$\displaystyle \chi_{F}\hat f=\hat f \Longrightarrow \chi_{E}(\chi_F \hat f)^{\vee}=f. \ \ \ \ \ (11)$

Assume that: ${Tf:=\chi_{E}(\chi_F \hat f)^{\vee}}$ then ${Tf=f}$ . So we have, at least ${\|T\|_{2-2}\geq 1}$ . Some dirty calculate show:

$\displaystyle \begin{array}{rcl} (Tf)(x) & = & \int e^{2\pi i\xi x}\chi_F\hat f(\xi)\chi_E(x)d\xi\\ & = & \int \int e^{2\pi i\xi(x-y)}f(y)\chi_F(\xi)\chi_E(x)dy d\xi\\ & \overset{Fubini}= &\int_{{\mathbb R}^d}\chi_E(x)\chi_F^{\vee}(x-y)f(y)fy \end{array}$

So we can define kernel of ${T}$ ,

$\displaystyle K(x,y)=\chi_E(x)\chi_F(x-y)^{\vee} \ \ \ \ \ (12)$

By Fubini, we calculate the Hilbert-Schmidt norm:

$\displaystyle \int_{{\mathbb R}^{2d}}|K(x,y)|^2dxdy=|E||F|=\sigma^2<+\infty \ \ \ \ \ (13)$

So ${T}$ is a compact operator and its ${L^2}$ operator norm satisfied ${\|T\|=\min(\sigma,1)}$ . So if ${\sigma<1}$ then we can canculate we can not have ${f\neq 0}$ in the original question.

The story is in fact more interesting, the answer of the question is no even for ${\sigma\geq 1}$ , so in all case. We have the following quatitative theorem:

Theorem 4 ${E,F}$ finite measure in ${{\mathbb R}^d}$ , then

$\displaystyle \|f\|_{L^2({\mathbb R}^d)}\leq c(\|f\|_{L^2(E^c)}+\|\hat f\|_{L^2(F^c)}) \ \ \ \ \ (14)$

for some constant ${c=c(E,F,d)}$ .

Remark 7 There is a naive approach for this theorem: Area formula trick, the shape of level set. Obvioudly we have:

$\displaystyle \|f\|_{L^2({\mathbb R}^d)}\leq \|f\|_{L^2(E)}+\|f\|_{L^2(E^c)} \ \ \ \ \ (15)$

Key point is proof:

$\displaystyle \|f\|_{L^2(E)}\leq c(E,F,d)\|\hat f\|_{L^2(F^c)} \ \ \ \ \ (16)$

Let us do some useless further calculate:

$\displaystyle \begin{array}{rcl} |f\|_{L^2(E)} & = & \|\chi_E \cdot f\|_{L^2({\mathbb R}^d)}\\ & = & \|\widehat {\chi_E\cdot f}\|_{L^2({\mathbb R}^d)}\\ & = & \|\hat \chi_E \cdot \hat{f^{\vee}}\|_{L^2({\mathbb R}^d)} \\ & = & \|\chi_E^{\vee} * f^{\vee}\|_{L^2({\mathbb R}^d)} \end{array}$

So suffice to have:

$\displaystyle \|\chi_E^{\vee} * f^{\vee}\|_{L^2({\mathbb R}^d)}\leq c(E,F,d)\|\hat f\|_{L^2(F^c)} \ \ \ \ \ (17)$

But there is connter example given by modified scaling Gaussian distribution… The point is form 15 to 16 is too loose.

Following I given a right approach, following by my sprite on level set and area formula argument and discritization.

Proof: The story is the same for a discretization one. We need point out, change the space ${{\mathbb R}^d}$ to ${{\mathbb Z}^d}$ , then every thing become a discretization one, and the change could been argue as a approximation way. What happen then, we have a naive picture in mind which is:

$\displaystyle \delta \rightarrow wave , \ wave \rightarrow \delta$

What is the case with ${L^2}$ norm, it become the standard nner product on ${{\mathbb Z}^d}$ , and the scale involve, i.e. we have the following basic estimate:

$\displaystyle \|\chi_E f\|_1^2\leq \|\chi_E f\|_2 \cdot|E| \ \ \ \ \ (18)$

Now image if the density of ${f}$ concentrate in a very small area, then by a cut off argument we consider the supp of ${f}$ , ${supp f=E}$ is very small, then use the argument 18, we could conclute the density of ${\hat f}$ could not very concentrate in the fraquence space. The constant ${c(E,F,d)}$ could be given presicely by this way, but I do not care about it. $\Box$

4. Logvinenko-Sereda theorem

Next we formulate some result that provide further evidence of the non-concentration property of functions with Fourier support on ${B_1}$ .

4.1. A toy model

Theorem 5 Let ${\alpha>1}$ an suppose that ${S\subset {\mathbb R}^d}$ satisfies,

$\displaystyle |S\cap B|<\alpha |B|, \ for \ all \ balls \ B \ of \ radius\ 1. \ \ \ \ \ (19)$

If ${f\in L^2({\mathbb R}^d)}$ satisfies ${supp(\hat f)\subset B(0,1)}$ then

$\displaystyle \|f\|_{L^2(S)}\leq \delta(\alpha)\|f\|_2 \ \ \ \ \ (20)$

Where ${\delta(\alpha)\rightarrow 0}$ as ${\alpha \rightarrow 0}$ .

Proof: This is a easy corollary of the argument I give in the proof of Amerein-Berthier theorem 4. $\Box$

4.2. A refine version

Theorem 6 Suppose that a measurable set ${E\subset {\mathbb R}^d}$ satisfies the following “thinkness” condition: there exists ${\gamma\in (0,1)}$ such that

$\displaystyle |E\cap B|>\gamma |B| \ for \ all \ balls \ B \ of \ radius\ R^{-1}. \ \ \ \ \ (21)$

where ${R>0}$ is arbitrary but fixed. Assume that ${supp(\hat f)\subset B(0,R)}$ . Then

$\displaystyle \|f\|_{L^2({\mathbb R}^d)}\leq C\|f\|_{L^2(E)}. \ \ \ \ \ (22)$

where the constant ${C}$ depends only on ${d}$ and ${\gamma}$ .

Remark 8 This proof need some very good estimate come from several complex variables.

5. The Malgrange-Ehrenpreis theorem

Theorem 7 Let ${\Omega}$ be a bounded domain in ${{\mathbb R}^d}$ and let ${p\neq 0}$ be a polynomial, Then, for all ${g\in L^3(\Omega)}$ , there exists ${f\in L^2(\Omega)}$ such that ${p(D)f=g}$ in a distribution sence.

补充说明

以下是新整理的中文说明；上方旧博客原文保持不变。

不确定性原理的核心直觉是：一个函数和它的 Fourier transform 不可能同时高度集中。这里的“集中”可以有许多精确版本：支撑大小、方差、指数衰减、带限函数的空间尺度，或者通过 level sets 来衡量的几何集中。

不确定性原理的基本图像：物理空间中的强集中会迫使频率空间展开，反之亦然。

1. 基本直觉

Fourier transform 把函数分解成平面波：

$$\widehat f(\xi)=\int_{\mathbb R^n}f(x)e^{-2\pi i x\cdot \xi}\,dx.$$

如果 $f$ 集中在很小的物理区域里，那么积分里的相位 $x\cdot \xi$ 在许多方向上变化不够快，$\widehat f$ 就不可能在频率空间里迅速衰减。反过来，如果 $\widehat f$ 被限制在很小的频率球内，$f$ 就必须在物理空间里足够平滑、足够展开。

从 level set 的角度看，$f$ 的每一层都贡献一个振荡积分。level set 的几何形状越小、越刚性，相位平均带来的抵消就越弱。

2. Bernstein 估计

一个非常实用的形式是 Bernstein inequality。若 $\widehat f$ 支撑在半径 $R$ 的球内，那么

$$\|\nabla f\|_{L^p}\lesssim R\|f\|_{L^p}.$$

这说明带限函数不能在尺度远小于 $R^{-1}$ 的区域内剧烈变化。频率支撑越小，物理空间越平滑；频率支撑越大，函数才有能力产生更细尺度的振荡。

更一般地，如果频率支撑在一个椭球中，那么物理空间的自然尺度由对偶椭球控制。这个“对偶尺度”正是不确定性原理的几何版本。

3. Heisenberg 不等式

最经典的方差形式是

$$\left(\int_{\mathbb R^n}|x|^2|f(x)|^2\,dx\right)
\left(\int_{\mathbb R^n}|\xi|^2|\widehat f(\xi)|^2\,d\xi\right)
\ge C_n\|f\|_2^4.$$

一维情形可以通过分部积分和 Cauchy-Schwarz 证明。核心计算是把 $\|f\|_2^2$ 写成 $-\int x\partial_x(|f|^2)$ 的形式，再把 $x f$ 和 $\partial_x f$ 配对。Plancherel 把 $\partial_x f$ 转成频率侧的 $\xi\widehat f$。

4. Gaussian 为什么是极值

Gaussian 同时在物理空间和频率空间保持 Gaussian 形状：

$$f(x)=e^{-\pi |x|^2},\qquad \widehat f(\xi)=e^{-\pi |\xi|^2}.$$

它正好平衡了两个方向的集中，因此在 Heisenberg 型不等式中给出等号情形。这个现象提示一个更几何的问题：是否可以用类似 Brunn-Minkowski 或 entropy convexity 的语言解释 Gaussian 的最优性？

从现代观点看，答案常常与热流、log-Sobolev 不等式和最优输运有关。热流让任意函数逐渐 Gaussian 化；entropy 的凸性则记录了这个过程中的不确定性增加。

5. 一个可继续追问的问题

如果把“集中程度”换成某种几何距离，例如 level sets 的体积增长、分布之间的 transport distance，或者函数图像诱导的度量结构，那么 Gaussian 是否仍然是最自然的最优对象？这个问题不一定有唯一答案，但它把 Fourier 分析里的不确定性和凸几何、概率、几何测度论连在了一起。
2017年12月28日
Brunn-Minkowski 不等式：集合加法、离散化与几何 PDE 的影子
旧博客原文

原题：Brunn-Minkowski inequality

In this short note, I posed a conjecture on Brunn-Minkwoski inequality and explain why we could be interested in this inequality, what is it meaning for further developing of some fully nonlinear elliptic equation come from geometry. The main part of the note devoted to discuss several different proof of classical Brunn-Minkowski inequality.

Brunn-Minkowski inequality

1. Introduction

I believe, every type of Brunn-Minkowski inequality, type of Brunn-Minkowski inequality is in some special sense and will be explained later, will be crucial with a corresponding regularity result of a fully nonlinear elliptic equation which could be realizable by geometric way which will also explained in further note.

So the key point is that Brunn-Minkowski inequality is crucial and have potential application, I posed a problem there and then consider the classical Brunn-Minkowski inequality, we give several proof of the classical Brunn-Minkowski inequality, everyone could help us to have a more refine understanding of the original difficulty with different angle.

Theorem 1 (conjecture) We have a map

$\displaystyle f:{\mathbb R}^d\times {\mathbb R}^d \longrightarrow {\mathbb R}^d \ \ \ \ \ (1)$

$\displaystyle (x,y) \longrightarrow f(x,y)$

We are willing to called the function ${f}$ as the hamiltonian function. then we could consider the hamiltonian flow of the function ${f}$ , but this could only true for a even dimension manifold to make there exists ${f}$ that ${df}$ is a non-degenerate closed ${2-}$ form.

Anyway we consider the level set of ${f}$ , we get a foliation i.e ${{\mathbb R}^d=\{\amalg_{p\in {\mathbb R}}f^{-1}(p)\}}$ . we consider the gradient flow with ${f}$ , called the gradient flow begin with ${q\in {\mathbb R}}$ as ${\{\phi_q(t)\}_{t\in {\mathbb R}}}$ . And we wish the gradient flow have a addition structure on itself then we could consider what is the Brunn-Minkowski inequality in this setting, the condition is a group structure on the space of level set ${L_f=\{\amalg_{p\in {\mathbb R}}f^{-1}(p)\}}$ , i.e.

$\displaystyle \forall t_1,t_2\in {\mathbb R}, \forall q\in R^d, \phi_{q}(t_1+t_2)=\phi_{q}(t_1)\circ \phi_{q}(t_2) \ \ \ \ \ (2)$

Remark 1 take ${f(x,y)=x+y}$ in 1, this conjecture reduce to the toy model, i.e. classical Brunn-Minkowski inequality.

Remark 2 We could generate the problem to the problem which is charged by several energy function ${f_1,...,f_k}$ , if the induced gradient flow is amenable, then this is somewhat similar with the one dimension case, I wish if we could do something for the single function ${f_1}$ , then we can say something for the several functions involved case.

Remark 3 This could also generate to amenable group action case and quantization of it.

Meaning, the cohomology induce by a hamiltonian system on some special foliation on fiber of geometric bundle. This type of result could help to establish the vanish of the cohomology, the get the existence theory and regularity result for corresponding elliptic nonlinear differential equation. And solve the original problem I consider.

Now we given the statement of Brunn-Minkowski inequality.

Theorem 2 (brunn minkowski inequality) For ${A,B}$ measurale set in ${{\mathbb R}^d}$ . we have following,

$\displaystyle \mu^{\frac{1}{d}}(A+B)\geq \mu^{\frac{1}{d}}\mu(A)+\mu^{\frac{1}{d}}\mu(B) \ \ \ \ \ (3)$

“>

The general spproach of Brunn-Minkwoski inequality is following,
1. divide the measurable set ${A,B}$ into small cubes.
2. Shinking trick, transform the set into convex one.
for the first one, we have the following lemma,

Lemma 3 ${\forall 0<\lambda<1}$ , ${\exists \epsilon>0}$ ${A_{\epsilon}}$ measurable set, ${A_{\epsilon}\subset A}$ , ${\mu(A-A_{\epsilon})<\epsilon}$ , and ${A=(A-A_{\epsilon})\amalg \cup_{i\in I}(c_i\cap A_{\epsilon} ) }$ , and

$\displaystyle \frac{\mu(c_i\cap A_{\epsilon})}{\mu(c_i)}>\lambda, \forall i\in I \ \ \ \ \ (4)$

Proof: The proof of the lemma is a easy corollary of the construction of Lesbegue(or Borel) measurable ${\sigma}$ algebra. $\Box$

Remark 4 The existence of the property given in the lemma is not the key point, the key point is ${d(A_{\epsilon},A)\rightarrow 0, as\ \ \epsilon\rightarrow 0}$ .

Has this two simplify in hand, we could give several approach to proof the inequality and these proof carry information more than just a proof, they carry some information with the structure of space ${(\{ 0,1 \}_{{\mathbb R}^d},\mu^{\frac{1}{d}},+)}$ . \newpage

2. A proof with discretization

There is a lots of ways to attack the Brunn-Minkowski inequality, the most natural one is discretization. But unfortunately there is some technique obstacle for proof or even state the discretization version of “Brunn-Minkowski” inequality.

The “boundary” and “area” should not compatible.

$\displaystyle \mu_{d-1}(\partial E)<<\mu_{d}(\mu(E)) \ \ \ \ \ (5)$

And we need use the fact,

$\displaystyle A_{\epsilon}\overset{G-H \ sence}\longrightarrow A \ \ \ \ \ (6)$

Now we just state what we expect it should transform in, because we have a fully understanding with the discretization model, there is a result named Cauchy-Daveport inequality.
Theorem 4 (cauchy-daveport inequality) There are two case, one in ${{\mathbb Z}}$ , one in finite field ${{\mathbb Z}_p}$ .
1. ${{\mathbb Z}}$ case, ${\forall A,B\subset {\mathbb Z}}$ are finite set,we have,
  $\displaystyle |A+B|\geq |A|+|B|-1. \ \ \ \ \ (7)$
2. ${{\mathbb Z}_p}$ case, ${\forall A,B\subset {\mathbb Z}}$ are finite set,we have,
  $\displaystyle |A+B|\geq \max\{|A|+|B|-1.p\}. \ \ \ \ \ (8)$
Proof: for the ${{\mathbb Z}}$ case, the story is more or less trivial, just do to a observation, if ${A=\{a_i,a_1<a_2<...<a_n\},B=\{b_i,b_1<b_2<...<b_m\}}$ , then

$\displaystyle a_1+b_1<\min\{a_1+b_2+a_2+b_1\}<....<a_n+a_m$

There exists a strictly increasing chain of length at least ${n+m-1}$ .

For the ${{\mathbb Z}_p}$ case, following is a graph to explain what happen, basically we define a operation on tuples, i.e. ${T:(A,B)\rightarrow (T(A),T(B))}$ , and make the additive energy ${E_{A,B}:=|A+B|}$ decreasing. after induction with this transform and the transform from a tuple to the minimum additive energy by translation, the additive energy decreasing and decreasing then arrive the global minimum. But it is easy to conclude in this case one of ${A,B}$ become null set and then the inequality 8 follows. $\Box$

But when we discrete the Brunn-Minkowski inequality, we expect a high dimension generation of the inequality 4. Naively we wish,

Theorem 5 (naive generation of cauchy-daveport inequality) For ${d\in {\mathbb N}}$ , and ${\forall A,B\subset {\mathbb Z}_d}$ are finite sets,

$\displaystyle |A+B|^{\frac{1}{d}}\geq |A|^{\frac{1}{d}}+|B|^{\frac{1}{d}} \ \ \ \ \ (9)$

But this is not the case, there is a counterexample for 5. We could construct some ${A,B}$ such that ${|A+B|\sim |A|+|B|}$ , consider they be very thin line.

So why we are in this worse situation? because we lose the information of ${A_{\epsilon}\overset{G-H}\longrightarrow A }$ , ${B_{\epsilon}\overset{G-H}\longrightarrow B }$ . So they have the trend tending to make the “boundary” campatible with “area”. Two thin line in the same direction is exactly the worst case, which is just a equal condition of 1-dimension case.

One natural way to except the situation is to bounded the “isperimetric constant”, to assume ${A,B}$ varies in a subset of measurable set, with addition condition that ${\frac{\mu_{d-1}(\partial A)}{\mu_d(A)}}$ is bounded by some constant. But this is also not the suitable set for our inequality, I explain how to capture the information of the G-H coverage.

Now assume ${A,B}$ are convex bounded set, and we take a global orthogonal basis in ${{\mathbb R}^d}$ . named ${(e_1,...,e_d)}$ . We give the definition of ${\epsilon-}$ discretization of ${A}$ , named ${A_{\epsilon}}$ .
Definition 6 ( ${\epsilon-}$ discretization) The construction of ${A_{\epsilon}}$ from ${A}$ is following:
1. divide ${A}$ into ${\amalg_{i\in I}c_i\cap A}$ , ${c_i}$ is the ${\epsilon}$ cubes.
2. use ${c_i}$ or ${\emptyset}$ instead of ${c_i\cap A}$ depending on iff ${\frac{\mu(A\cap c_i)}{\mu(c_i)}>\lambda}$ , where ${\lambda<1}$ is a given number only rely on ${\epsilon}$ . i.e.
  $\displaystyle A\cap c_i\rightarrow D_{\epsilon} (A\cap c_i) \ \ \ \ \ (10)$
3. glue them, define ${A_{\epsilon}:=\amalg_{i\in I}D_{\epsilon}(c_i\cap A)}$ .
Now we describe the condition of ${A_{\epsilon}\overset{G-H}\rightarrow A}$ rigorous meaning campatible with ${\epsilon-}$ discretization.

Under the basis, there is a coordinate we could know iff ${c_i=c(\lambda_1,...,\lambda_d)}$ is the cube center at ${(\epsilon\lambda_1,...,\epsilon\lambda_d )}$ if it is in ${A_{\epsilon}}$ . Due to ${A}$ is convex, ${\partial A}$ is lipchitz. So you will have some locolization property, said, at every fix discretization scale ${\epsilon}$ , the position of ${(\lambda_1\epsilon ,...,\lambda_d\epsilon)}$ is morally known so the number of cubes in ${A_{\epsilon}}$ in the one dimensional affine space ${\Omega^{\epsilon}_{a_1,...,a_{k-1},a_{k+1},...,a_n}}$ which is the subspace of ${{\mathbb Z}^d}$ the number ${\Omega^{\epsilon}_{a_1,...,a_{k-1},a_{k+1},...,a_n}}$ is asymptopic to the ${1-}$ dimensional hausdorff measure of ${A\cap \Omega_{a_1,...,a_{k-1},a_{k+1},...,a_n}}$ . So at least,

$\displaystyle |\Omega^{\epsilon}_{a_1,...,a_{k-1},a_{k+1},...,a_n}|=O((\frac{1}{\epsilon})) \ \ \ \ \ (11)$

Property 11 is crucial, which mean ${A_{\epsilon}}$ is really a n-dimensional space and automatically we have the bounded on isoperimetric constant ${\frac{\mu_{d-1}(\partial A)}{\mu_d(A)}}$ .

Now we can look at every ${A_{\epsilon}}$ and take limit ${\epsilon\rightarrow \infty}$ . In fact we a in the situation with Accumulation of wood to make the product have smallest volume. Not to optimized the tuples ${(A,B)}$ but fix one of it, said ${B}$ , optimized the other one, said ${A}$ . This is the key point of proof, a little bit different from the argument of ${1-}$ dimensional 4 where we optimized the tuple.

Key point:
1. we can ignore “small core”.
2. This inequality is said, due to ${A+B=(\frac{A+B}{2})+\frac{A+B}{2}}$ , the convex of the functional ${\mu^{\frac{1}{d}}}$ on convex set.
The way of discretization could not handle the problem but definitely said that the difficulty occur with the shape of boundaries ${\partial A,\partial B}$ .

3. A proof with “central of mass” and Minkowski functional

Definition 7 (Central of mass) The central of mass ${p}$ of measurable set ${A}$ , if exist, satisfied, ${\forall e\in S^d}$ , there is a subspace ${L_e}$ with codimension 1 divide ${A}$ into two connected part ${A_{1,L_e},A_{2,L_e}}$ such that

$\displaystyle \mu(A_{1,L_e})=\mu(A_{2,L_e}) \ \ \ \ \ (12)$

then ${p\in l_e}$ .

Remark 5 For a measurable set ${A}$ , if central of mass ${p}$ exists, then there exist only one. This is a easy observation do to the definition of ${p}$ , i.e. the intersection of suitable affine subspace in every direction.

Theorem 8 (existence of central of mass)

Proof: It is easy to attain ${p}$ by take ${n+1}$ different directions in ${S^d}$ , then easy to proof every line ${L_e}$ across it be definition of ${L_e}$ . $\Box$

Definition 9 (Minkowski functional) for a measurable set ${X\subset {\mathbb R}^d}$ and a point ${p}$ , define ${M_{X,p}}$ on ${S^{d-1}}$ , such that

$\displaystyle M_X(e)=\sup_{\lambda>0,\lambda e+p\in X}\lambda \ \ \ \ \ (13)$

.

Remark 6 If ${A}$ is convex, then ${M_A}$ is a convex function on ${S}$ , so it is lipchitz.

we have following formula for the measure of ${A}$ .

Theorem 10

$\displaystyle \mu(A)=\frac{1}{\mu(S^{d-1})}\int_{S^{d-1}}M_A(e)de \ \ \ \ \ (14)$

Proof: trivial. $\Box$

Now the task reduce fixing ${B}$ and ${\mu(A)}$ to optimized ${A}$ make ${\mu(A+B)}$ small. It is the same as make ${\frac{1}{\mu(S^{d-1})}\int_{S^{d-1}}M_{A+B}(e)de}$ small when fix ${\frac{1}{\mu(S^{d-1})}\int_{S^{d-1}}M_A(e)de}$ and ${M_B}$ . Due to

$\displaystyle M_{A+B}(x)=\sup_{x_1,x_2\in S^{d-1}}M_A(x_1)+M_B(x_2) \ \ \ \ \ (15)$

This lead to the whole story, given a proof of 3.

4. A proof with multi-scale analysis

This approach is a nonstandard one, due to I believe the renormlization or continue fractional or multilinear estimate is everywhere. We first play with a toy model, the rectangle.

Theorem 11 Brunn-Minkowski inequality is right for ${A,B}$ are rectangles.

Proof:

$\displaystyle (\Pi(a_1+b_i) )^{\frac{1}{d}}\geq (\Pi(a_1) )^{\frac{1}{d}}+(\Pi(a_1+b_i) )^{\frac{1}{d}}$

$\displaystyle \Leftrightarrow 1\geq [\frac{\Pi a_i}{\Pi(a_i+b_i)}]^{\frac{1}{d}}+[\frac{\Pi a_i}{\Pi(a_i+b_i)}]^{\frac{1}{d}} \ \ \ \ \ (16)$

In fact,

$latex \displaystyle \begin{array}{rcl} 16 RHS & \overset{A-G}\leq &\frac{1}{d}\sum_i\frac{a_i}{a_i+b_i}+ \frac{1}{d}\sum_i\frac{b_i}{a_i+b_i}\\ & = & 1. \end{array} &fg=000000$

$\Box$

The story is following,

5. connection of Brunn-Minkowski inequality and Sobolev inequality, the firth proof

We begin with a calculate based on intuition and it is not rigorous.

$\displaystyle \begin{array}{rcl} \mu(A+B)^{\frac{1}{d}} & = & [\int_{{\mathbb R}}(\int_{{\mathbb R}^{d-1}}\chi_{A+B}(\xi_1,...,\xi_{n-1})d\xi_1...d\xi_{n-1} )d\xi_n]^{\frac{1}{d}}\\ & \sim & \frac{1}{\mu(A_n)}\int_{{\mathbb R}}(\int_{{\mathbb R}^{d-1}}\chi_{A+B}(\xi_1,...,\xi_{n-1})d\xi_1...d\xi_{n-1})^{\frac{1}{d-1}}d\xi_n\\ &\overset{induction \ on \ d} \geq & \frac{1}{\mu(A_n)}(\mu^{\frac{1}{d-1}}(A(\xi_n))+\mu^{\frac{1}{d-1}}(B(\xi_n)))d\xi \\ &\overset{A-G \ inequality} \geq & \mu^{\frac{1}{d}}(A)+\mu^{\frac{1}{d}}(B). \end{array}$

The second line is due to I believe there ${\exists \ A,B}$ such that it is a equality, by the equal condition of Minkowski inequality, in fact this is morally inverse of Minkowski inequality. The second reason in general case why the second inequality is true is due to a rescaling argument, change ${(\xi_1,...,\xi_n)\rightarrow (\lambda\xi_1,...,\lambda\xi_n ), \forall \lambda\in {\mathbb R}}$ , by the rescaling argument we conclude if there is a such inequality, the index of it must be the case.

补充说明

以下是新整理的中文说明；上方旧博客原文保持不变。

Brunn-Minkowski 不等式看起来是凸几何中的体积不等式，但它背后的思想更宽：一种“加法结构”如果和体积或能量兼容，就会产生凹性。这篇笔记从经典不等式出发，说明离散化、Cauchy-Davenport 型现象以及几何 PDE 之间为什么会出现在同一条线上。

Brunn-Minkowski 不等式说明 Minkowski 加法下体积的 $1/n$ 次方具有凹性。

1. 经典陈述

给定 $\mathbb R^n$ 中的可测集合 $A,B$，定义 Minkowski 和

$$A+B=\{a+b:a\in A,\ b\in B\}.$$

Brunn-Minkowski 不等式说

$$|A+B|^{1/n}\ge |A|^{1/n}+|B|^{1/n}.$$

也可以写成插值形式：对 $0
$$|(1-t)A+tB|^{1/n}\ge (1-t)|A|^{1/n}+t|B|^{1/n}.$$

所以体积的 $1/n$ 次方在集合加法下是凹的。

2. 为什么这个不等式重要

它直接推出等周不等式，也连接到 Prékopa-Leindler 不等式、凸体理论和最优输运。对几何分析来说，更重要的是它提供了一种模板：几何对象的“混合”如果存在，那么与之对应的体积、容量或能量往往应该满足某种凹性。

这正是完全非线性椭圆方程中常见的结构。很多正则性问题并不只是局部估计问题，也和底层几何量的凹性有关。

3. 离散化视角

一种直观证明路线是把集合近似成小立方体的并。连续体积问题被转成格点集合的加法问题。最简单的一维模型是 Cauchy-Davenport 型不等式：

$$|A+B|\ge |A|+|B|-1.$$

这个不等式说明，集合加法至少应该增加自由度。高维情形不能天真地照搬，因为很细的集合可能集中在低维方向上，边界和体积也不再同步。正是这些失败之处提示我们：Brunn-Minkowski 真正使用的是凸化、投影和切片结构。

4. 凸化与测度证明

另一路线是先归约到凸体，再用切片或 Prékopa-Leindler 思想证明。Prékopa-Leindler 可以看成函数版本的 Brunn-Minkowski：如果

$$h((1-t)x+ty)\ge f(x)^{1-t}g(y)^t,$$

那么

$$\int h\ge \left(\int f\right)^{1-t}\left(\int g\right)^t.$$

这说明 Brunn-Minkowski 不只是集合命题，也是积分、概率和凸性之间的共同结构。

5. Hamiltonian 或 level-set 方向的问题

一个自然的问题是：如果我们有一个能量函数 $H$，它的 level sets 形成某种 foliation，并且梯度流或 Hamiltonian flow 在这些 level sets 上诱导出合理的“加法”，是否存在对应的 Brunn-Minkowski 型不等式？

在欧氏空间中，level set 的平移和集合加法非常明确；在一般几何背景中，加法结构可能来自群作用、Hamiltonian flow 或某种可交换的 transport。若这样的不等式成立，它很可能对应某个完全非线性椭圆方程的存在性或正则性。

所以经典 Brunn-Minkowski 在这里扮演的是 toy model：先弄清楚体积凹性怎样从线性加法中出现，再问类似结构能否在更复杂的几何流中重现。
2017年12月27日
伪微分算子与奇异积分：从 Fourier multiplier 到 symbol calculus
旧博客原文

原题：Pesudo differential opertor and singular integral

I already understand this material 3days ago but it is a little difficult for me to type the latex…

1. Introduction

There is two space to understand a function’s behaviour, the physics space and the frequency space (Why thing going like this? Why there is such a duality?). Namely, we have:

$\displaystyle \hat f(\xi)=\int_{{\mathbb R}^d}e^{2\pi i\xi x}f(x)dx \ \ \ \ \ (1)$

The key point is, waves is a parameter group of scaling of definition of a constant fraquence wave, so it connected the multiplication and addition. Basically due to it can be look as the correlation of a function and the scaling of wave with carry all the information about ${f}$ . A generation of this obeservation is the wavelet theory.

So as we well know, the key ingredient of Fourier transform is to image function as a sum of series waves. A famous theorem of Mikhlion said that a translation-invariant operator ${T}$ on ${R^n}$ could be represented by a multiplication operator on the Fourier transform side. translation is the meaning, ${h\circ T=T\circ h, \forall h}$ is a translation.

In a formal level, consider it as distribution (compact distribution or temperature distribution is both OK). We have:

$\displaystyle T(e^{2\pi ix\xi})=a(\xi)e^{2\pi ix\xi}, \forall \xi \in {\mathbb R}^n \ \ \ \ \ (2)$

the meaning is if we consider ${T}$ is a operator on distribution space, ${T:S'\rightarrow S'}$ , then ${\forall f\in S}$ ,

$\displaystyle \int T(e^{2\pi ix\xi})f=\int a(\xi)e^{2\pi ix\xi}f$

due to the linear combination of ${e^{2\pi ix\xi}}$ will consititue a dense set in ${S}$ . So this could extend to the whole distribution space by dual and give the definition of ${T}$ , i.e.

$\displaystyle (Tf)(x)=\int_{{\mathbb R}^n}a(x,\xi)e^{2\pi ix\xi}\hat f(\xi)d\xi \ \ \ \ \ (3)$

Remark 1 ${T}$ is bounded on ${L^2({\mathbb R}^n)}$ when ${a}$ is a bounded function, thanks to Parevel theorem. When ${a}$ is a bounded function, the composition of two such operator could be defined, and the symbol of composition operator corresponding to the composite of their symbol, i.e.

$\displaystyle T_a\circ T_b(e^{2\pi ix\xi})=b(\xi)a(\xi)e^{2\pi ix\xi} \ \ \ \ \ (4)$

Remark 2 For parenval theorem, i.e. ${\|\hat f\|_2=\|f\|_2}$ , there is two approach, heat kernel approximation approach and discretization.

We wish to investigate the operator given by multiplier, i.e.

$\displaystyle (Tf)(x)=\int_{{\mathbb R}^n}a(x,\xi)e^{2\pi ix\xi}\hat f(\xi)d\xi \ \ \ \ \ (5)$

When it is satisfied ${\|T\|_{p-p}<\infty}$ ?

Intuition, the following calculate is only morally true, not rigorous.

$\displaystyle \begin{array}{rcl} \|Tf\|^p_p & = & \int_{{\mathbb R}^d }|\int_{{\mathbb R}^d}a(x,\xi) e^{2\pi ix\xi} \hat f(\xi)d\xi|^pdx \\ & \overset{\exists f}\sim & \int_{{\mathbb R}^d }\int_{{\mathbb R}^d}|a(x,\xi) e^{2\pi ix\xi} \hat f(\xi)|^p d\xi dx \\ & \overset{Fubini}\sim & \int_{{\mathbb R}^d}\int_{{\mathbb R}^d}|\widehat{a(x,\xi)}\hat f(\xi) |^pdxd\xi \\ & \sim & \int_{{\mathbb R}^n}\int_{{\mathbb R}^n}|{a(x,\xi)}^{\vee}*f(\xi)|^pdxd\xi \end{array}$

So we need some restriction on ${ {a(x,\xi)}^{\vee}}$ , namely ${\widehat{a(x,\xi)}}$ , so we need some decay condition on ${|\partial_x^{\alpha}\partial_{\xi}^{\beta}a(x,\xi)|}$ , why this, just consider integral by part for ${a(x,\xi)\in S}$ . The rigorozaton of this intuition inspirit us to the definition of symbol calss.

Definition 1 we say ${a(x,\xi)}$ is in symbol class ${S_m}$ iff,

$\displaystyle |\partial_x^{\beta}\partial_{\xi}^{\alpha}a(x,\xi)|\leq A_{\alpha,\beta}(1+|\xi|)^{m-|\alpha|} \ \ \ \ \ (6)$

for all ${\alpha,\beta}$ is multi-indece.
Remark 3
1. we note that all partial differential operator, whose coefficient, together with all their derivatives are bounded belong to this class, In this particular circumstance, the symbol is a polynomial in ${\xi}$ , essentially the “characteristic polynomial” of the operator.
2. The general operator of this class have a parallel description in terms of their kernels. That is, in a suitable sense,
  $\displaystyle (Tf)(x)=\int_{{\mathbb R}_x}K(x,y)f(y)dy \ \ \ \ \ (7)$
  
  besides enjoying a cancellation property, ${K}$ is here characterized by differential inequalities “dual” to those for ${a(x,\xi)}$ . In the key case where the order ${m=0}$ , this kernel representation makes ${T}$ a singular integral operator.
3. The crucial ${L^2}$ estimate, when ${m=0}$ , is atelatively simple consequences of Plancherel’s theorem for the Fourier transform. With this, the ${L^p}$ theory introduce in previous note is therefore applicable.
4. The product identity that holds in the translation-invariant case generalized to the situation treated here as a symbolic calculus for the composition of operators. That is, there is an asymptotic formula for the composition of two such operators, whose main term is the point-wise product of their symbols.
5. The succeeding terms of the formula are of decreasing orders. These orders measure not only the size of the symbols, but determine also the increasing smoothing properties of the corresponding operators. The smoothing properties are most neatly expressed in terms of the Sobolev space ${W_k^p}$ and the Lipschitz space ${\Lambda_{\alpha}}$ .
2. Pseudo-differential operator

“Freezing principle”: from variable coefficient differential equation to constant coefficient differential equation by approximation. divide into 2 steps:
1. divide space into small cubes.
2. take average of the coefficient of differential equation in every cubes.
Suppose we are interested in study the solution of the classical elliptic second order equation.

$\displaystyle (Lu)(x)=\sum a_{ij}(x)\frac{\partial^2 u(x)}{\partial x_i\partial x_j}=f(x) \ \ \ \ \ (8)$

Where the coefficient matrix ${\{a_{ij}(x)\}}$ is assume to be real, symmetric, positive definite and smooth in ${X}$ . Understanding ${P}$ , such that,

$\displaystyle PL=I \ \ \ \ \ (9)$

Looking for a ${P}$ . Such that ${PL=I+E}$ . ${E}$ is a error term which have good control. To do this, fix an arbituary point ${x_0}$ , freeze the operator ${L}$ at ${x_0}$ :

$\displaystyle L_{x_0}=\sum a_{ij}(x_0)\frac{\partial^2}{\partial x_i\partial x_j} \ \ \ \ \ (10)$

In Fourier sense ( ${L^2}$ sence).

$\displaystyle \begin{array}{rcl} L_{x_0}f(x) & = & \int e^{2\pi ix\xi}(\widehat{ \sum a_{ij}(x_0)\frac{\partial^2 f(\xi)}{\partial x_i\partial x_j}}) d\xi\\ & = & \int e^{2\pi ix\xi}\int e^{-2\pi i \xi y}\sum a_{ij}(x_0)\frac{\partial^2}{\partial x_i\partial x_j}f(y)dyd\xi\\ & = & \int e^{2\pi ix\xi}(-4 \pi^2)\sum_{i,j}a_{ij}(x_0)\xi_i\xi_j \end{array}$
Remark 4 The remark is, morally speaking, for application of fourier transform in PDE. morally we could only solve the problem with linear differential equation (although we could consider the hyperbolic type). The main obstacle for Fourier transform application into PDE:
1. it only make sense with Schwarz class or its dual, this is not main obstacle, in principle could be solved by rescaling.
2. the main obstacle is it only compatible with linear differential equation.
Cut-off function: ${\eta}$ vanish near the origin,

$\displaystyle (P_{x_0}f)=\int_{{\mathbb R}^n}(-4\pi^2\sum_{i,j}a_{ij}(x_0)\xi_i\xi_j)^{-1}\eta(\xi)\hat {f(\xi)}e^{2\pi ix\xi}d\xi. \ \ \ \ \ (11)$

then:

$\displaystyle L_{x_0}P_{x_0}=I+E_{x_0}.$

${E_{x_0}}$ is actually a smoothing operator, because it is given by convolution with a fixed test function. It should be seasonable when ${x}$ near ${x_0}$ , ${(Pf)(x)}$ is well approximated by ${(P_{x_0}f)(x)}$ , it is actually the case, define ${((Pf)(x):=(P_xf)(x)}$ , i.e.

$\displaystyle (Pf)(x)=\int_{{\mathbb R}^n}(-4\pi^2\sum_{i,j}a_{ij}(x)\xi_i\xi_j)^{-1}\eta(\xi)\hat f(\xi)e^{2\pi ix\xi}d\xi \ \ \ \ \ (12)$

The operator ${P}$ so given is a propotype of a pesudo-differential operator. Moreover, one has ${LP=I+E}$ , where the error operator ${E}$ is “smoothing of order 1”. That this is indeed the case is the main part of the symbolic calculus described.

Definition 2 (symbol class) A function ${a(x,\xi)}$ belong to ${S^m}$ and is said to be of order ${m}$ of ${a(x,\xi)}$ is a ${C^{\infty}}$ function of ${(x,\xi)\in {\mathbb R}^n\times {\mathbb R}^n}$ and satisfies the differential inequality:

$\displaystyle |\partial_x^{\beta}\partial_{\xi}^{\alpha}a(x,\xi)|\leq A_{\alpha,\beta}(1+|\xi|)^{m-|\alpha|} \ \ \ \ \ (13)$

For all ${\alpha,\beta}$ are multi-indece.

Now we trun to the exact meaning of pesudo-differential operator, i.e. how them action on functions. Under some suffice given regularity condition, for ${a\in S^m}$ , ${T_a:S\rightarrow S}$ .

$\displaystyle (Tf)(x)=\int_{{\mathbb R}^n}a(x,\xi)\hat f(\xi)e^{2\pi ix\xi}d\xi \ \ \ \ \ (14)$

Remark 5 ${T_a:S\rightarrow S}$ is continuous and for ${a_k\rightarrow a}$ pointwise, ${a_k\in S,\forall k\in {\mathbb N}^*}$ , ${T_{a_k}(f)\rightarrow T_a(f)}$ in ${S}$ .

then expense it, we get:

$\displaystyle (T_af)(x)=\int\int a(x,\xi)e^{2\pi i\xi(x-y)}f(y)dyd\xi \ \ \ \ \ (15)$

This could be diverge, even when ${f\in S}$ . The key point is we do not have control with the second integral, morally speaking, this phenomenon is the weakness of Lesbegue integral which would not happen in Riemann integral, so sometime we need the idea from Riemann integral, this phnomenon is settle by multi a cut off function ${\eta_{\epsilon}}$ and take ${\epsilon\rightarrow \infty}$ , the same deal also occur as the introduced of P.V. integral in Hilbert transform. The precise method to deal with the obstacle is following: ${a_{\epsilon}(x,\xi)=a(x,\xi)\gamma(\epsilon x,\epsilon \xi)}$ , if ${a\in S^m}$ , ${a_{\epsilon}\in S^m}$ . ${T_{a_{\epsilon}}\rightarrow T_a}$ in the sense:

${\forall f\in S}$ , ${T_{a_{\epsilon}}(f)\rightarrow T_a(f)}$ ,

$\displaystyle (T_af)(x)=\lim_{\epsilon\rightarrow 0}\int\int a_{\epsilon}(x,\xi)e^{2\pi i\xi(x-y)}f(y)dyd\xi \ \ \ \ \ (16)$

We also have:

$\displaystyle <T_af,g>=<f,T_a^*g>, \forall f,g\in S. \ \ \ \ \ (17)$

Then we have:

$\displaystyle (T^*_ag)(y)=\lim_{\epsilon\rightarrow 0}\int\int \bar a_{\epsilon}(x,\xi)e^{2\pi i\xi(y-x)}g(x)dxd\xi \ \ \ \ \ (18)$

and ${<f,g>}$ denotes ${\int_{{\mathbb R}^n}f(x)\bar g(x)dx}$ . Thus the pesudo-differential operator ${T_a}$ initially defined as a mapping from ${S}$ to ${S}$ , extend via the identity 17 to a mapping from the space of temperatured distribution ${S'}$ to itself ${S'}$ . Notice also that ${T_a}$ is automatically continuous in this space. \newpage

3. ${L^p}$ bounded theorem

We first introduce a powerful tools, called dyadic decomposition,

Lemma 3 (dyadic decomposition) In eculid space ${{\mathbb R}^n}$ there exists a function ${\phi\in C^{\infty}({\mathbb R}^n)}$ such that,

$\displaystyle \sum_{i\in {\mathbb Z}}\phi(2^{-i}x)=1 \ \ \ \ \ (19)$

and ${\forall x\in {\mathbb R}^n}$ , there is only two of ${i\in {\mathbb Z}}$ such that ${\phi(2^{-i}x)\neq 0}$ , and we can choose ${\phi}$ to be radical and ${\phi(x) \geq 0,\forall x\in {\mathbb R}^n}$ .

Remark 6

So for a given mutiplier ${a(x,\xi)}$ , we will have ${a(x,\xi)=\sum_{i\in {\mathbb Z}}a_i(x,\xi)=\sum_{i\in Z}\phi(2^{-ix})a(x)}$ .

Proof: The proof is easy, after rescaling we just need observed there is a bump function satisfied whole condition. $\Box$

Theorem 4 Suppose ${a}$ is a symbol of order 0, i.e. that ${a\in S^0}$ Then the operator ${T_a}$ , initially defined on ${S}$ , extends to a bounded operator from ${L^2({\mathbb R}^n)}$ to itself.

Remark 7 Suffice to show ${\|T_a(f)\|_{L^2}\leq A\|f\|_{L^2}, \forall f\in S}$ and by dual.

In fact we can directly proof a more general theorem:

Theorem 5 Let ${m:{\mathbb R}^d-\{0\}\rightarrow {\mathbb C}}$ satisfy, for any multi-index ${\gamma}$ of length ${|\gamma|\leq d+2}$ ,

$\displaystyle |\partial^{\gamma}m(\xi)|\leq B|\xi|^{-|\gamma|}$

For all ${\xi\neq 0}$ . Then, for any ${0<p<\infty}$ , there is a constant ${C=C(d,p)}$ such that,

$\displaystyle \| (m\hat f)^{\vee}\|_p\leq C(p,d)\|f\|_p \ \ \ \ \ (20)$

for all ${f\in S}$ .

Proof: ${a\in S^0}$ , so we have:

$\displaystyle |\partial_x^{\beta}\partial_{\xi}^{\alpha}a(x,\xi)\leq a_{\alpha,\beta}(1+|\beta|)^{-|\alpha|} \ \ \ \ \ (21)$

${\forall \alpha,\beta}$ are multi indeces. Then we consider dyadic decomposition, the is a function ${\phi}$ satisfied the condition in 19, define ${a_i(x,\xi)=\phi(2^{-i}x)a(x)}$ . then ${supp a_i(x,\xi)}$ cpt, ${\|a_i\|<\infty}$ . So ${a_i\in L^p({\mathbb R}^n)}$ , we have,

$\displaystyle \begin{array}{rcl} \|T_{a_i}f\|_p^p & = & \int_{{\mathbb R}^d}|K_i*f(x)|^pdx \\ & = & \int_{{\mathbb R}^d}|\int_{{\mathbb R}^d}K_i(x-y)f(y)dy|^pdydx\\ & \leq &\int_{{\mathbb R}^d}\int_{{\mathbb R}^d}|K_i(x-y)f(y)|^pdydx\\ & = & \|K_i\|_p^p\|f\|_p^p \end{array}$

${\|K_i\|_p}$ have good decay estimate, thanks to ${u_i=\phi(2^{-i}x)a(x)\in S^0}$ , this estimate is deduce morally along the same ingredient of “station phase”, it is come from a argument combine “counting point” argument and a rescaling argument. So,

$\displaystyle \begin{array}{rcl} \|Tf\|_p & = & \|\sum T_if\|_p\\ & \leq & (\sum \|K_i\|_p^p)\|f\|_p \end{array}$

But we have ${\sum\|K_i\|_p^p\leq \infty}$ , ending the proof. $\Box$

Remark 8 this method also make sense of restrict the condition to be:

$\displaystyle |\partial_x^{\beta}\partial_{\xi}^{\alpha}a(x,\xi)|\leq A_{\alpha,\beta}(1+|\xi|)^{-|\alpha|}, \forall |\alpha|\leq d+2. \ \ \ \ \ (22)$

Where ${d}$ is the dimension of the space, and we could change ${2}$ to ${1+\epsilon}$ .

Remark 9

$\displaystyle \widehat{\frac{\partial^2 u}{\partial x_i\partial x_j}}|\xi|=\frac{\xi_i\xi_j}{|\xi|^2}\widehat \Delta u(\xi), m(\xi)=\frac{\xi_i\xi_j}{|\xi|^2} \ \ \ \ \ (23)$

is a counter example for ${p=1,\infty}$ .

Remark 10 The key point is the estimate

$\displaystyle \int|\int e^{2\pi i\xi x}\phi(2^{-j}\xi)m(\xi)d\xi |^pdx \ \ \ \ \ (24)$

Correlation of taylor expension and wavelet expension. This is also crutial for the theory of station phase.

4. Calculus of symbols

This calculus of symbols would imply there is some structure on this set.

Theorem 6 Suppose ${a,b}$ are symbols belonging to ${S^{m_1}}$ and ${S^{m_2}}$ respectively. Then there is a symbol ${c}$ in ${S^{m_1+m_2}}$ so that:

$\displaystyle Tc=T_a\circ T_b$

Moreover,

$\displaystyle c\sim \sum_{\alpha}\frac{(2\pi i)^{-|\alpha|}}{\alpha}(\partial_{\xi}^{\alpha}a)(\partial_x^{\alpha}b). \ \ \ \ \ (25)$

in the sense that,

$\displaystyle c-\sum_{|\alpha|<N}\frac{(2\pi i)^{-|\alpha|}}{\alpha !}\partial_{\xi}^{\alpha}\partial_x^{\alpha}b\in S^{m_1+m_2-N} \ \ \ \ \ (26)$

For all ${N>0}$ .

The following “proof” is not rigorous, we just calculate it formally, we could believe it is true rigorously, by some approximation process. Proof: We assume ${a,b}$ have compact support so that our manipulations are justified. We use the alternate formula 15 to write,

$\displaystyle (T_af)(y)=\int b(y,\xi)e^{2\pi i\xi(y-z)}f(z)dzd\xi \ \ \ \ \ (27)$

Then we apply ${T_a}$ , again in the form 15, but here with the variable ${\eta}$ replacing in the integration. The result is,

$\displaystyle T_a(T_bf)(x)=\int a(x,\eta)b(y,\xi)e^{2\pi i\eta(x-y)}e^{2\pi i\xi(y-z)}f(z)dzd\xi dyd\eta. \ \ \ \ \ (28)$

This calculate is easy to derive, but the following is more tricky. Now ${e^{2\pi i\eta(x-y)}\cdot e^{2\pi i\xi(y-z)}=e^{2\pi i(x-y)(\eta-\xi)}\cdot e^{2\pi i(x-z)\xi}}$ , so

$\displaystyle T_a(T_bf)(x)=\int c(x,\xi)e^{2\pi i(x-z)\xi}f(z)dz d\xi \ \ \ \ \ (29)$

with

$\displaystyle c(x,\xi)=\int a(x,\eta)b(y,\xi)e^{2\pi i(x-y)(\eta-\xi)}dyd\eta \ \ \ \ \ (30)$

we can also carry out the integration in the y-variable. This leads to the corresponding Fourier transform of ${b}$ in that variable, and allows us to rewrite 30 as,

$\displaystyle c(x,\xi)=\int a(x,\xi+\eta)\hat b(\eta,\xi)e^{2\pi i x\eta}d\eta. \ \ \ \ \ (31)$

With this form in hand, use taylor expense to the symbol ${a(x,\xi+\eta)}$ , i.e.

$\displaystyle a(x,\xi+\eta)=\sum_{|\alpha|<N}\partial_{\xi}^{\alpha}a(x,\xi)\eta^{\alpha}+R_N(x,\xi,\eta) \ \ \ \ \ (32)$

with a suitable error term ${R_N}$ , due to

$\displaystyle \frac{1}{\alpha !}\int \partial_{\xi}^{\alpha}a(x,\xi)\hat\eta(\eta,\xi)e^{2\pi ix\eta}d\eta=\frac{(2\pi i)^{|\alpha|}}{\alpha !}(\partial_{\xi}^{\alpha}a(x,\xi))(\partial_x^{\alpha}b(x,\xi)). \ \ \ \ \ (33)$

we only need to proof ${R_N\in S^{m_1+m_2-N}}$ and it is definitely the case, we get the theorem. $\Box$

Remark 11 We need replace ${a,b}$ with ${a_{\epsilon},b_{\epsilon}}$ , where

$\displaystyle a_{\epsilon}(x,\xi)=a(x,\xi)\cdot \gamma(\epsilon ,\epsilon \xi), b_{\epsilon}(x,\xi)=b(x,\xi)\cdot \gamma(\epsilon ,\epsilon \xi). \ \ \ \ \ (34)$

we note that ${a_{\epsilon},b_{\epsilon}}$ satisfy the same differential inequalities that ${a}$ and ${b}$ do, uniformly in ${\epsilon, 0<\epsilon\leq 1}$ .passage to the limit as ${\epsilon\rightarrow 0}$ will then give us our desired result.

5. Estimate in ${L^p}$ , Sobolev, and Lipchitz space

We now take up the regularity properties of our pesudo-differential operator as expressed in terms of the standard function spaces, we begin with the ${L^p}$ boundedness of an operator of order ${0}$ .

5.1. ${L^p}$ estimate

Suppose ${a}$ belongs to the symbol class ${S^0}$ . Then, we can express ${T=T_a}$ as

$\displaystyle (Tf)(x)=\int K(x,y)f(y)dy=\int K(x,x-y)f(y)dy \ \ \ \ \ (35)$

due to ${a\in S^0}$ , we know, with some approximation argument and first do it with a cutoff symbol of ${a}$ , i.e. ${a_{\epsilon}}$ , that,

$\displaystyle |K(x,y)|\leq A|x-y|^{-n} \ \ \ \ \ (36)$

So that the integral coverage whenever ${f\in S}$ and ${x}$ is away from the support of ${f}$ . Since we know that ${T}$ is bounded on ${L^2({\mathbb R})}$ , this representation extends to all ${f\in L^2({\mathbb R})}$ for almost every ${x\notin supp f}$ . More generally, we have,

$\displaystyle |\partial_{x}^{\alpha}\partial_{y}^{\beta}K(x,y)|\leq A_{\alpha,\beta}|x-y|^{-n-|\alpha|-|\beta|} \ \ \ \ \ (37)$

hence ${K}$ satisfies,

$\displaystyle \int_{|x-y|\geq 2\delta}|K(x,y)-K(x,\bar y)|dx\leq A, \ if\ |y-\bar y|\leq \delta, all \ \ \delta>0. \ \ \ \ \ (38)$

Use the general singular integral theory we get the following ${L^p}$ estimate.

Theorem 7 Suppose ${T_a}$ is the pseudo-differential operator corresponding to a symbol ${a}$ in ${S^0}$ , then ${T_a}$ extends to a bounded operator on ${L^p({\mathbb R}^n)}$ to itself, for ${1<p<\infty}$ .

5.2. Sobolev spaces

We first recall the definition of the Sobolev spaces ${W_k^p}$ , where ${k}$ is a positive integer. A function ${f}$ belongs to ${W_k^p({\mathbb R}^n)}$ if ${f\in L^p({\mathbb R}^n)}$ and the partial derivatives ${\partial_x^{\alpha}f}$ , taken in the sense of distribution, belong to ${L^p({\mathbb R}^n)}$ , whenever ${0\leq |\alpha|\leq k}$ . The norm in ${W_k^p}$ is given by,

$\displaystyle \|f\|_{W_k^p}=\sum_{|\alpha|\leq k}\|\partial_x^{\alpha}f\|_{L^p} \ \ \ \ \ (39)$

the following result is the directly corollary of 7.

Theorem 8 Suppose ${T_a}$ is a pseudo-differential operator whose symbol ${a}$ belongs to ${S^m}$ . If ${m}$ is an integer and ${k\geq m}$ , then ${T_a}$ is a bounded mapping from ${W_k^p}$ to ${W_{k-m}^p}$ , whenever ${1<p<\infty}$ .

Remark 12 This theorem remain valid for arbitrary real ${k,m}$ .

5.3. Lipschitz spaces

Theorem 9 Suppose ${a}$ is a symbol in ${S^m}$ . Then the operator ${T_a}$ is a bounded mapping from ${\Lambda_{\gamma}}$ to ${\Lambda_{\gamma-m}}$ , whenever ${\gamma>m}$ .

Lemma 10 Suppose the symbol ${a}$ belongs to ${S^m}$ , and define ${T_{a_j}=T_a\Delta_j}$ . Then, as operator from ${\L^{\infty}({\mathbb R}^n)}$ to itself, the ${T_{a_j}}$ have norms that satisfy

$\displaystyle \|T_{a_j}\|\leq A2^{jm} \ \ \ \ \ (40)$

We shall now point out a very simple but useful alternative characterization of ${\Lambda_{\gamma}}$ . This is in terms of approximation by smooth functions; it is also closely connected with the definition of ${\Lambda_{\alpha}}$ space as intermediate spaces, using the “real” method of interpolation.

Corollary 11 A function ${f}$ belongs to ${\Lambda_{\gamma}}$ if and only if there is a decomposition,

$\displaystyle f=\sum_{j=0}^{\infty}f_j \ \ \ \ \ (41)$

with ${\|\partial_x^{\alpha}f_j\|_{L^infty}\leq A2^{-j\gamma}\cdot 2^{j|\alpha|}}$ , for all ${0\leq |\alpha|\leq l}$ , where ${l}$ is the smallest integer ${>\gamma}$ .

When ${f\in \Lambda_{/\gamma}}$ , the argument prove 10, with ${T_a=I}$ , ${f_j=F_j=\Delta_j(f)}$ , gives the required estimate for the ${f_j}$ .

A second consequence of 9 is the following:

Corollary 12 The operator ${(I-\Delta)^{\frac{m}{2}}}$ gives an isomorphism from ${\Lambda_{\gamma}}$ to ${\Lambda_{\gamma-m}}$ , whenever ${\gamma>m}$ .

This is clear because ${(I-\Delta)^{\frac{m}{2}}}$ is continuous from ${\Lambda_{\gamma}}$ to ${\Lambda_{\gamma-m}}$ , and its inverse, ${(I-\Delta)^{\frac{-m}{2}}}$ , is continuous from ${\Lambda_{\gamma-m}}$ to ${\Lambda_{\gamma}}$ .

补充说明

以下是新整理的中文说明；上方旧博客原文保持不变。

理解伪微分算子的一个自然入口，是先理解 Fourier multiplier。平移不变算子在频率侧通常只是乘以一个函数；一旦系数依赖空间位置，乘子就必须升级为同时依赖 $x$ 和 $\xi$ 的 symbol。这篇笔记沿着这个想法，从 multiplier 走到 kernel，再走到 Calderon-Zygmund 型奇异积分。

伪微分算子把位置变量 $x$ 和频率变量 $\xi$ 同时放进 symbol，kernel 的奇异性由 symbol 的阶数控制。

1. Fourier multiplier

若 $T$ 与平移可交换，那么在 Fourier 侧它应当满足

$$\widehat{Tf}(\xi)=m(\xi)\widehat f(\xi).$$

这里 $m$ 是 multiplier。Plancherel 定理直接给出一个基本结论：如果 $m\in L^\infty$，那么 $T$ 在 $L^2$ 上有界，并且

$$\|Tf\|_2\le \|m\|_\infty\|f\|_2.$$

这解释了为什么频率空间是处理常系数线性方程的自然语言：微分在频率侧变成乘法，算子代数变成函数代数。

2. 从 multiplier 到 symbol

变系数方程的困难在于，算子不再平移不变。伪微分算子的基本形式是

$$Af(x)=\int_{\mathbb R^n}e^{ix\cdot\xi}a(x,\xi)\widehat f(\xi)\,d\xi.$$

函数 $a(x,\xi)$ 称为 symbol。它描述算子在点 $x$ 附近、频率 $\xi$ 方向上的局部行为。常系数情形是 $a$ 不依赖 $x$ 的特例。

标准 symbol class $S^m_{1,0}$ 要求对所有多重指标 $\alpha,\beta$，有

$$|\partial_x^\alpha\partial_\xi^\beta a(x,\xi)|\le C_{\alpha\beta}(1+|\xi|)^{m-|\beta|}.$$

对 $\xi$ 求导会降低阶数，这正对应物理空间 kernel 的衰减和光滑性。

3. Freezing principle

变系数椭圆方程可以先在一个点附近“冻结系数”。例如二阶椭圆算子

$$Lu=-\partial_i(a^{ij}(x)\partial_j u)$$

在点 $x_0$ 附近可以用常系数算子

$$L_{x_0}u=-a^{ij}(x_0)\partial_i\partial_j u$$

作为第一近似。Fourier transform 可以求解冻结后的模型；误差则由系数的变化控制。伪微分算子把这个“每个点冻结一次”的过程系统化了。

4. Kernel 与奇异积分

伪微分算子也可以写成 kernel 形式

$$Af(x)=\int K(x,y)f(y)\,dy.$$

当 symbol 的阶数 $m=0$ 时，kernel 在 $x=y$ 附近有奇异性，但又满足 cancellation 和光滑估计。这就把问题接到 Calderon-Zygmund 理论：$L^2$ 有界性来自 Plancherel，弱 $(1,1)$ 或 $L^p$ 有界性则依赖实变量分解。

因此，伪微分算子不是孤立的一套语言，而是 Fourier 分析和实变量奇异积分之间的桥。

5. Symbolic calculus

平移不变情形下，两个 multiplier 的复合对应乘子相乘。伪微分算子也有类似结论，只是要加入低阶修正。形式上，如果 $A$ 的 symbol 是 $a$，$B$ 的 symbol 是 $b$，那么 $AB$ 的 symbol 有渐近展开

$$a\# b(x,\xi)\sim \sum_\alpha \frac1{\alpha!}\partial_\xi^\alpha a(x,\xi)D_x^\alpha b(x,\xi).$$

主项是 $ab$，后面的项阶数越来越低，表示越来越强的 smoothing。这个公式是椭圆正则性、参数构造和微局部分析的基本计算规则。
2017年12月26日
辛几何笔记：Hamiltonian flow、Darboux 坐标与 Poisson bracket
旧博客原文

原题：Symplectic geometry

1. Introduction

This is the first note of a series of notes concert on semiclassical analysis. Given the basic material on symplectic geometry. Including the following material,
1. The case at a point, or we can look it as the case in ${{\mathbb R}^{2n}}$ .
2. The standard material in symplectic geometry, i.e. Hamiltonian mechanics, two approach, global one concentrating on lie derivative, and a locally one concentrating on the power of Darboux theorem, i.e. the existence of a canonical coordinate.
3. The basic facts on Poission bracket.
4. The basic facts on Lagrange sub-manifold, and the involve of Liouville measure.
2. Case of a point, or ${{\mathbb R}^{2n}}$

Let ${V: {\mathbb R}^n \rightarrow {\mathbb R}^n}$ be a vector field, at once we have a vector field, we could consider the associated flow of it,

$\displaystyle \left\{ \begin{aligned} \overset{\cdot}\omega & = & V(\omega)\\ \omega(0) & =&z \end{aligned} \right. \ \ \ \ \ (1)$

express the trajectory start from ${z}$ along the vector field.

Remark 1 There ${\overset{\cdot}\omega:=\frac{\partial \omega}{\partial t}}$ . One the other hand, due to the locally existence theorem of ODE, if the regularity of ${V}$ is enough, then the solution exist and is uniqueness.

Definition 1 ${\psi_t z=\omega(t,z)}$ or more convenient ${\psi_t:= exp(tV)}$ . We call ${\{\psi_t\}_{t\in {\mathbb R}}}$ the flow map or the exponential map generated by ${V}$ .
Lemma 2 For flow map, we have following:
1. ${\psi_0z=z}$ for all ${z\in {\mathbb R}^N}$ .
2. ${\psi_{t+s}=\phi_t\psi_s}$ for all ${s,t\in {\mathbb R}}$ .
3. for each time ${t\in {\mathbb R}}$ , the mapping ${\psi_t: {\mathbb R}^n\rightarrow {\mathbb R}^n}$ is a diffeomorphism with ${(\psi_{t})^{-1}=\psi_{-t}}$
So it is a group action on ${{\mathbb R}^n}$ , with units as diffeomorphism of ${{\mathbb R}^n}$ . Proof:

This lemma is the direct corollary of the theory of ODE. $\Box$

Now let us special to the case ${{\mathbb R}^{2N}={\mathbb R}^n\times {\mathbb R}^n}$ . In local coordinate we have ${z=(x,\xi)}$ , ${x\in {\mathbb R}^n}$ express position of particle, ${\xi \in {\mathbb R}^n}$ express momentum of particle.

Definition 3 ${z=(x,\xi),w=(y,\eta )}$ in ${{\mathbb R}^{2n}}$ define their symplectic product,

$\displaystyle \sigma(z,w):=<\xi,y>-<x,\eta> \ \ \ \ \ (2)$

In a matrix form, ${\sigma}$ coincide with a ${2n\times 2n}$ matrix

$\displaystyle J=\begin{pmatrix} 0 & I \\ -I & 0\end{pmatrix} \ \ \ \ \ (3)$

Following lemma given the basic property of ${\sigma, J}$ .
Lemma 4 The following basic property are true.
1. ${\sigma(z,\omega)=<Jz,w>}$ , ${\forall z,w \in {\mathbb R}^{2n}.}$
2. the bilinear form ${\sigma}$ is antisymmetric, ${\sigma(z,w)=-\sigma(w,z)}$ and degenerate, i.e. if ${\sigma(z,w)=0}$ for all ${w\in {\mathbb R}^{2n}}$ , then ${z=0}$ .
3. ${J^2=-I}$ , ${J^{T}=-J=J^{-1}}$ .
Proof:
1. trivial calculate get this.
2. trivial.
3. ${JJ^{-1}=-I}$ , by basic linear algebra everything follows.
$\Box$

3. Hamiltonian mechanics

Definition 5 Symplectic form: non-degenerate closed 2 form in a standard coordinate(Darboux coordinate, coordinate like ${{\mathbb R}^{2n}}$ ) looks like,

$\displaystyle \begin{pmatrix} 0 & -I \\ I & 0 \end{pmatrix} \ \ \ \ \ (4)$

${\forall x\in X}$ , map

$\displaystyle T_xX\rightarrow T_x^*X$

$\displaystyle v\rightarrow w(\cdot, v)$

is an isomorphism. ${w}$ is called the symplectic form.

There is locally coordinate for ${TM,T^*M}$ , i.e.,

$\displaystyle TM:\frac{\partial}{\partial x_1},...,\frac{\partial}{\partial x_n}$

$\displaystyle T^*M:dx_1,...,dx_n$

So ${w=f^{ij}dx_i\wedge dx_j}$ , roughly we have ${w(\frac{\partial}{\partial x_k},\frac{\partial}{\partial x_i})=f^{ik}-f^{ki}}$ , this is of course not true, but morally true. Now let us give the definition of symplectic manifold and the relationship of Hamiltonian mechanics.
Definition 6 We have the following definition,
1. A symplectic manifold is a pair ${(X,\omega)}$ where ${X}$ is a smooth manifold and ${w}$ is a closed two-form on ${X}$ such that ${\forall x\in X}$ the map,
  $\displaystyle T_xX\rightarrow T_x^*X$
  
  $\displaystyle v\rightarrow w(\cdot,v)$
  
  is an isomorphism, ${w}$ is called the symplectic form.
2. If ${(X,w)}$ is symplectic, and ${f:X\rightarrow {\mathbb R}}$ is differentiable, the hamiltonian vector field of ${f}$ is the field ${\Xi_f}$ on ${X}$ whose image under the previous map is ${df}$ . In other word, ${\Xi_f}$ is characticed by the property,
  $\displaystyle w(\cdot,\Xi_f)=df(\cdot)$
3. The flow of ${\Xi_f}$ will be referenced to as the hamiltonian flow of ${f}$ .
Lemma 7 If ${X=R^{2n}}$ coordinate ${(x_1,...,x_n,p_1,...,p_n)=(x,p)}$ and the symplectic form

$\displaystyle w=\sum_{j=1}^ndp_j\wedge dx_j$

then,
1. If ${f:{\mathbb R}^{2n}\rightarrow {\mathbb R}}$ is differentiable, then the integral curves of ${\Xi_f}$ are the solutions to the system of ODEs,
  $\displaystyle \left\{ \begin{aligned} \overset{\cdot}x_j & = & \frac{\partial f}{\partial p_j}\ \ \ j=1,2,...,n \\ \overset{\cdot}p_j & =& -\frac{\partial f}{\partial x_j} \ \ \ j=1,2,...,n \\ \end{aligned} \right. \ \ \ \ \ (5)$
2. Moreover, if
  $\displaystyle f(x,p)=\frac{1}{2m}\|p\|^2+V(x). \ \ \ \ \ (6)$
  
  Where ${V}$ is a smooth solution (“potential”), and ${(x(t),p(t))}$ is a trajectory of the Hamiltonian flow of ${f}$ , then
  
  $\displaystyle m\overset{\cdot\cdot}x=-\nabla \ \ \ \ \ (7)$
  
  This is the Newton’s second law for the force ${-\nabla V}$ .
Proof: ${w=\sum_{j=1}^n dp_j\wedge dx_j}$ , due to we have ${w(\cdot, \Xi_f=)=df}$ . So ${\forall j}$ we have:

$\displaystyle w(\frac{\partial}{\partial p_j},\Xi_f) = df(\frac{\partial}{\partial p_j})=\frac{\partial f}{\partial p_j} \ \ \ \ \ (8)$

Assume ${\Xi_f=\lambda^i \frac{\partial}{\partial x_i}+\gamma^i\frac{\partial}{\partial p_i}}$ . Then we have,

$\displaystyle \begin{array}{rcl} \frac{\partial f}{\partial p_j} & = & w(\frac{\partial}{\partial p_j},\lambda^i\frac{\partial}{\partial x_i}+ \gamma^i\frac{\partial}{\partial p^i})\\ & = & \lambda^j\cdot (-1)^{\sigma(...)} \end{array}$

and also,

$\displaystyle \begin{array}{rcl} \frac{\partial f}{\partial x_j} & = & w(\frac{\partial}{\partial p_j},\lambda^i\frac{\partial}{\partial x_i}+ \gamma^i\frac{\partial}{\partial p^i})\\ & = & \gamma^j\cdot (-1)^{\sigma(...)} \end{array}$

Combine with the definition of integral curve we derive the integral curves of ${\Xi_f}$ i.e. ${\gamma(t)}$ such that ${\overset{\cdot}\gamma(t)=\Xi_f}$ , ${\gamma(t)=(x_1(t),...,p_n(t))}$ is given by 5.

Now we begin to proof the Newton second law for the force ${-\nabla V}$ . We consider the 2-dimensional case at first. We have,

$\displaystyle \begin{array}{rcl} m\overset{\cdot\cdot}x & = & m(\overset{\cdot} {\frac{\partial f}{\partial p}})\\ & = & m\frac{\partial}{\partial t}(\frac{\partial f}{\partial p})\\ & = & m\frac{\partial}{\partial p}[\frac{\partial}{\partial t}(\frac{\|p\|^2}{m}+V(x))]\\ & = & m\frac{\partial }{\partial p}(\frac{2p\overset{\cdot}p}{m}+\frac{\partial V(x)}{\partial x}\cdot \frac{\partial f}{\partial p}) , due\ to \ -\frac{\partial f}{\partial x}=-\frac{\partial V(x)}{\partial x}.\\ & = & m\frac{\partial }{\partial p}(\frac{2p(-\frac{f}{\partial x})}{m}+\frac{\partial V(x)}{\partial x}\cdot \frac{\partial f}{\partial p})\\ & = &-\frac{\partial V(x)}{\partial x} \end{array}$

The high dimension case is similar, thanks to the linearity of ${V}$ and ${\overset{\cdot\cdot}x}$ . $\Box$

Remark 2 Two make Newton’s second law to be true, the form 6 play a crucial role. Is there some generalization of this type of result to more general case, roughly speaking, it is reasonable to expect this could still be true if the hamiltonian function could be divide into potential energy part and kinetic energy part. And the describe of potential energy part is that it is given by a quadratic form.
Lemma 8 In general, for any Hamilton field ${\Xi_f}$ one has:
1. ${\pounds_{\Xi_f}f=0}$ , conservation of energy. In orther word, ${\Xi_f}$ is everywhere tangent to the level sets of ${f}$ .
2. ${\pounds_{\Xi_f}\omega=0}$ , so the Hamiltonian flow of ${f}$ consists of automorphism of ${M,\omega}$ .
General speaking, to proof a theorem on manifold, there always have two choice, coordinate free proof and proof in a careful choose coordinate. If we choose to believe the Darboux theorem 9 is true, the meaning of it is that locally the symplectic manifold are the same.

Proof: If we believe the Darboux theorem 9 is true. then consider in a standard coordinate ${(x_1,...,x_n,p_1,...,p_n)}$ , we have,

$\displaystyle \Xi_f=\sum_{i}\frac{\partial f}{\partial x_i}\frac{\partial}{\partial p_i}-\sum_{i}\frac{\partial f}{\partial p_i}\frac{\partial}{\partial x_i} \ \ \ \ \ (9)$

So of course ${\pounds_{\Xi_f}f=\Xi f=0}$ . In general case, i.e. coordinate free proof, ${\omega(\cdot,\Xi_f)=df}$ . Use identity if lie derivative. The second thing is also easy to proof by look in a local canonical coordinate, involve the indentity of lie derivative. $\Box$

Remark 3 I need more understanding on the lie derivative, see wiki.

Theorem 9 (Darboux theorem) Near any point there exist coordinate: ${(x_1,...,x_n,p_1,...,p_n)}$ usually called Darboux coordinates, such that the sympletic form ${w}$ has the form,

$\displaystyle w=\sum_{j=1}^ndp_j \wedge dx_j. \ \ \ \ \ (10)$

Remark 4 This theorem means there do not exist local invariant in symplectic manifold.

Proof: $\Box$

\newpage

Theorem 10 If ${M}$ is any smooth manifold, then its cotangent bundle ${X=T^*M}$ has a natural symplectic structure.

Proof: we have local coordinate on ${T^M}$ derive from ${(x_1,...,x_n,dx_1,...,dx_n)}$ , it is ${(x_1,...,x_n,p_1,...,p_n)}$ . Remember we have Riemann metric: ${g^{ij}dx_idx_j}$ on ${T^*M\otimes T^*M}$ , the existence of Riemann metric involve a unit decomposition argument and bump function, I just recall it there. Now we move on, consider the relationship between ${M,TM,T^*M}$ .

$\displaystyle M{\longleftrightarrow} T^*M\overset{pairing, (X,f)=X(f)} {\longleftrightarrow} TM \ \ \ \ \ (11)$

$\Box$

Remark 5 It need not be the case that ${\alpha}$ non-degenerate ${\Longrightarrow}$ ${d\alpha}$ non-degenerate. This case in the lemma is a example to show that could be the case: ${\alpha}$ degenerate ${\Longrightarrow}$ ${d\alpha}$ non-degenerate. We glue something together on the space pf differential operator to understand the topology of it but not deifferential structure or more refinement structure. Quntalization could be look as a way to glue, this could be down if there is a differential equation with some special condition (come from a flow take charge of it suffice).

Lemma 11 (The proof of ${d\alpha}$ is non-degenerate) Let ${(x_1,...,x_n)}$ be local coordinates on ${U\subset M}$ . Define a coordinate system ${(x_1,...,x_n,p_1,...,p_n)}$ on ${T^*U}$ by the condition:

$\displaystyle \forall \xi\in T^*_{x}U, p_j(\xi)=\xi(\frac{\partial}{\partial x_j}) \ \ \ \ \ (12)$

Prove that in Darboux coordinate, ${\alpha=\sum_{j=1}^n p_jdx_j}$ and therefore ${\omega=\sum_{j=1}^n dp_j\wedge dx_j}$ .

Proof: $\Box$

Theorem 12 Let ${(M,g)}$ be a smooth Riemann manifold and let ${f:T^*M\rightarrow {\mathbb R}}$ be one half of the square of the Riemann norm, so that in local coordinate,

$\displaystyle f(x,p)=\frac{1}{2}\sum_{i,j}g^{ij}p_ip_j \ \ \ \ \ (13)$

Then the trajectorics of the hamiltonian flow of ${f}$ , projected down to ${M}$ , are geodesic aries in this fashion.

Newton’s second law+ energy vanish.

Proof:

$\displaystyle m\overset{\cdot\cdot}x=\nabla V=0$

So second variation formula describe of geodesic give us the fact that the trajective is geodesic. $\Box$

Remark 6 We could directly calculate in local coordinate.

4. Poisson brackets

${f}$ is the Hamiltonian generating the dynamic ${g}$ is any smooth function on phase space (the symplectic manifold), then the rate of change of ${g}$ along the trajectraries of d is the function

$\displaystyle \overset{\cdot}g=\pounds_{\Xi_f}g=d_g(\Xi_f)=\omega(\Xi_f,\Xi_g) \ \ \ \ \ (14)$

Definition 13 If ${(X,\omega)}$ is symplectic and ${f,g\in C^{\infty}(X)}$ , the poisson bracket of ${f}$ and ${g}$ is defined to be the function on ${X}$ .

$\displaystyle \{f,g\}:=\omega(\Xi_f,\Xi_g) \ \ \ \ \ (15)$

Lemma 14 In canonical (Darboux) coordinate where ${\omega=\sum_j dp_j\wedge dx_j}$ , one has,

$\displaystyle \{f,g\}=\sum_{j=1}^n\frac{\partial f}{\partial p_j}\frac{\partial g}{\partial x_j}-\frac{\partial f}{\partial x_j}\frac{\partial g}{\partial p_j} \ \ \ \ \ (16)$

In particular, ${\{p_j,x_j\}=\delta_{ij}}$ .

Proof:

$\displaystyle \begin{array}{rcl} \{f,g\} & = & \omega(\Xi_f,\Xi_g)\\ & = & \sum_{j=1}^n dp_j\wedge dx_j(-\frac{\partial f}{\partial p_j}\frac{\partial}{\partial x_j}+\frac{\partial f}{\partial x_j}\frac{\partial}{\partial p_j}, -\frac{\partial g}{\partial p_j}\frac{\partial}{\partial x_j}+\frac{\partial g}{\partial x_j}\frac{\partial}{\partial p_j} )\\ & = & \sum_{j=1}^n\frac{\partial f}{\partial p_j}\frac{\partial g}{\partial x_j}-\frac{\partial f}{\partial x_j}\frac{\partial g}{\partial p_j}. \end{array}$

$\Box$

Theorem 15 If ${(X,\omega)}$ is a symplectic manifold then ${\{C^{\infty}(X),\{\cdot\}\}}$ is a Lie algebra.

Proof: Bilinearty, skew-symmetric come form,

$\displaystyle \{f,g\}=\omega(\Xi_f,\Xi_g) \ \ \ \ \ (17)$

Jacobi identity:

$\displaystyle \{\{f,g\},h\}+\{\{g,h\},f\}+\{\{h,f\},g\}=0 \ \ \ \ \ (18)$

could be proved by calculate under a local coordinate. $\Box$

补充说明

以下是新整理的中文说明；上方旧博客原文保持不变。

辛几何是半经典分析和 Hamiltonian mechanics 的共同语言。它研究的不是长度和角度，而是相空间中的面积形式、Hamiltonian flow 以及由此产生的守恒结构。这篇笔记从向量场的 flow 开始，逐步进入 symplectic form、Darboux theorem 和 Poisson bracket。

Hamiltonian flow 在相空间中沿能量面运动，并保持 symplectic form。

1. 从向量场到 flow

给定光滑向量场 $V$，它生成常微分方程

$$\frac{d}{dt}\phi_t(x)=V(\phi_t(x)),\qquad \phi_0(x)=x.$$

若解存在且唯一，那么 $\phi_t$ 满足群性质

$$\phi_{t+s}=\phi_t\circ \phi_s,\qquad \phi_0=\operatorname{id}.$$

因此一个向量场不仅给出每个点的速度，也给出整个空间上的一参数 diffeomorphism group。

2. 标准相空间

在 Hamiltonian mechanics 中，标准相空间是 $\mathbb R^{2n}$，坐标写成

$$(q_1,\ldots,q_n,p_1,\ldots,p_n),$$

其中 $q$ 是位置，$p$ 是动量。标准 symplectic form 是

$$\omega=\sum_{j=1}^n dq_j\wedge dp_j.$$

它是闭的、非退化的二形式。非退化性意味着每个一形式都可以唯一对应一个向量场。

3. Hamiltonian vector field

给定能量函数 $H$，Hamiltonian vector field $X_H$ 由

$$\iota_{X_H}\omega=dH$$

定义。在标准坐标下，这给出 Hamilton 方程

$$\dot q_j=\frac{\partial H}{\partial p_j},\qquad
\dot p_j=-\frac{\partial H}{\partial q_j}.$$

若 $H(q,p)=\frac12|p|^2+V(q)$，那么 Hamilton 方程化成 Newton 方程

$$\ddot q=-\nabla V(q).$$

所以经典力学可以看成辛几何中的一条 flow。

4. 守恒与 Darboux theorem

Hamiltonian flow 保持能量：

$$\frac{d}{dt}H(\phi_t(x))=dH(X_H)=\omega(X_H,X_H)=0.$$

它也保持 symplectic form，即 $\phi_t^\ast\omega=\omega$。这比保持体积更强，是 Hamiltonian 系统的基本刚性。

Darboux theorem 说，任意 symplectic manifold 在局部都可以选坐标，使得

$$\omega=\sum dq_j\wedge dp_j.$$

这说明辛流形没有类似曲率那样的局部不变量；它的困难主要来自全局拓扑和 Lagrangian 子流形的相互位置。

5. Poisson bracket

两个函数 $f,g$ 的 Poisson bracket 定义为

$$\{f,g\}=\omega(X_f,X_g).$$

在标准坐标下，

$$\{f,g\}=\sum_j\left(\frac{\partial f}{\partial q_j}\frac{\partial g}{\partial p_j}
-\frac{\partial f}{\partial p_j}\frac{\partial g}{\partial q_j}\right).$$

沿 Hamiltonian flow 的演化满足

$$\frac{d}{dt}f(\phi_t(x))=\{f,H\}(\phi_t(x)).$$

这就是从几何到动力系统、再到半经典分析的接口：量子力学里的 commutator 在半经典极限中对应 Poisson bracket。
2017年12月22日
若干有趣问题：线排列染色、单纯形结构与度量畸变
旧博客原文

原题：Some interesting problems

There are some interesting problem, I post them at there in case I forget them. Excuse me if they are trivial, I have not took enough time to consider them about I think they are valuable to be consider.
Problem 1:

This problem is stated by graph coloring. there are two prat of it, in fact the first part I heard from someone else and I try to generate it to high dimension.
1. there are finite lines $\{l_i\}_{i\in I}, l_i\subset \mathbb R^2$ , crossing each other and the is a set $J$ of crossing point. for technique reason, assume the position of lines are generic, i.e. no three of them intersect at one point. Then we could use 3 different colors to color $J$ make Neighbor points have different color. And to proof 3 is smallest.
2. generate it to high dimension, to prove $\mathbb R^n$ case, $n+1$ is the number.
This seems to be a graph problem, but the underlying structure is linear structure and some topological obstacle. I am not very sure. But it seems we can use an energy decrement argument with the obesevation:

The existence of a reasonable definition of “energy of correlation”.

the simplex arrive with the maximum of “correlation energy” in a very symmetric way, and this situation is easy to handle (coloring).

If make sense, this argument could also generate to high dimension.

Problem 2:

Let us consider some example of map between two metric space, a toy model is a line and two parallel lines, I called two parallel lines by $X_1\cup X_2$ , the single line by $X_3$ . The problem is try to find a tuple $(d,f)$ , where $d$ is a metric define on $X_1\cup X_2$ and $f: X_1\cup X_2\to X_3$ . such that the distortion of $f^* d$ and the standard metric on $X_3$ arrive at a infimum, this of course could not be the case, such like the situation of Yamabe problem on manifold with conners. So, let us ask a more general problem, could we describe the behavior of $f$ in some sense? what could we say with this kind of $f$ ?

补充说明

以下是新整理的中文说明；上方旧博客原文保持不变。

这里记录两个表面上很初等、但背后可能带有线性结构和拓扑障碍的问题。第一个问题从直线排列的交点染色开始，第二个问题从度量空间之间的最佳畸变开始。它们的共同点是：单纯的图论语言可能太粗，真正控制问题的是隐藏的几何结构。

直线排列染色看似是图论问题，但一般位置和线性排列结构给了它额外的几何约束。

1. 直线排列上的三染色

考虑平面中有限条直线，并假设一般位置：没有三条直线共点。交点构成一个图，若两个交点在同一条直线上相邻，就连一条边。问题是：能否总用三种颜色给交点染色，使相邻交点颜色不同？

三色的必要性很容易从三角形构型看出来；困难在于证明三色总是足够。这个图不是任意平面图，它来自直线的全局排列，因此有额外的线性约束。

2. 高维推广

若把直线换成高维中的超平面，交点换成更高余维的相交胞腔，问题自然变成：需要多少种颜色？猜想的数目应当与单纯形的顶点数有关。

一种可能思路是定义某种“相关能量”。当能量接近最大时，构型应当逼近对称单纯形，而这种对称情形反而容易染色。若能建立能量递降，就可能把一般情形归约到对称模型。

3. 度量畸变问题

第二个问题是：给定一条直线 $L$ 和两条平行线 $L_1\sqcup L_2$，是否可以在两边选择合适度量，使某个自然映射的畸变达到最小？

这有点像带角点流形上的 Yamabe 型问题：最优对象可能不存在，但极小化序列会呈现某种退化形态。于是更合理的问题不是“是否达到最小”，而是“接近最优时几何如何坍缩或分裂”。

4. 可能的共同结构

染色问题和畸变问题都可以看作约束优化：前者优化颜色冲突，后者优化距离拉伸。若存在合适的能量或紧性定理，就能把直觉转化成证明。真正值得追问的是：这些模型中的对称构型是不是唯一的极值障碍。
2017年12月21日
奇异积分一般理论一瞥：近似恒等、Fourier multiplier 与 square function
旧博客原文

原题：A glimpse to the general theory

1. Introduction

We have talked about a very basic result in singular integral, i.e. if we have an additional condition, i.e. ${q-q}$ bounded condition, then by interpolation theorem we only need to establish the weak ${1-1}$ bound then we establish the ${p-p}$ bound of ${T}$ , ${\forall 1< p< q }$ .

The category of of singular integral is very general, in fact the singular integral we interested in always equipped more special structure. We discuss following 3 types result which world be the central role in this further series note.
1. Approximation of the identity.
2. Singular integral with ${L^2}$ bounded translation invariant operator.
3. Maximal function, singular integral, and square functions.
The underlying object we consider in both the three case is some special singular integral, in the first case, it looks like a ${T=sup_{t>0} \Phi_{t}*f}$ , this, among the other thing, has a close relationship with the maximal operator ${Mf}$ . This is discussed in 2. For the singular integral with ${L^2}$ bound, the Fourier transform or its discretization version, Fourier series is natural involved. And there is a “representation theorem” similar to the sprite of Reisz representation theorem, said, roughly speaking, if we consider the ${L^2}$ bound operator adding the condition of transform invariant, then it is really coinside with the case of our image, the operator must behaviour as a Fourier multiple. This is the contant of famous Mikhlin multiplier theorem, and we discuss some technique difficulty in the process of establishing such a theorem, this is the contact of 3. At last we discuss some deep relationship between three basic underlying intution and objects in harmonica analysis, the Maximal function, singular integral, and square functions. They could all be understanding as tools to understanding the variant complicated emerging in singular integral. But there is definitely some common points. This is the theme of 4. Of course there are some further topic which are also interesting, but I do not want to discuss them here, maybe somewhere else.

2. Approximation of the identity

First topic, we discuss the approximation of the identity, this play a central role in understanding solution of PDE, why, I think a key point is this tools carry a lots of information about the scaling of the space, as it well known, analysis could roughly divide into two parts, “hard analysis” and “soft analysis”, approximation of the identity supply a way to transform a result form “hard analysis” side to “soft analysis” side and reverse. And when it shows its whole power always along with the involving of following Dominate convergence theorem:

Theorem 1 (DCT)

Let ${\{f_n\}_{n=1}^{\infty}}$ be a series of function on measure space ${(X,\Sigma,\mu)}$ , and ${f_n \rightarrow f, a.e. x\in X}$ , and ${\{f_n\}_{n=1}^{\infty}}$ satisfied a controlling condition, i.e. we can find a integrable function ${g\in L^{1}(X)}$ , such that ${|f_n(x)|\leq |g(x)|, a.e. x\in X, \forall n\in {\mathbb N}^*}$ , then we know,

$\displaystyle \lim_{n\rightarrow \infty} \int_{X}f_n(x)d\mu\rightarrow \int f(x)d\mu \ \ \ \ \ (1)$

In fact we have even stronger,

$\displaystyle \lim_{n\rightarrow \infty} \int_{X}|f_n(x)-f(x)|d\mu=0 \ \ \ \ \ (2)$

This is a standard theorem in real analysis, we give the proof.

Proof: ${f}$ is the point-wise limit of ${f_n}$ so we know f is measurable and also dominate by ${g}$ , so by triangle inequality we have:

$\displaystyle |f-f_n|\leq 2|g|$

Then the 1 is trivially true, due to a diagonal taking subsequences trick. For more subtle result 2, we need use reverse Fatou theorem to show it is true, roughly speaking we have,

$\displaystyle \limsup_{n\rightarrow \infty}\int_{X}|f_n-f|\leq \int_{X}\limsup_{n\rightarrow \infty}|f_n-f|=0$

The key point is the first inequality above used the reverse Fatou theorem. $\Box$

Now we discuss of the main result of the approximation identity. So first we need to define what is a approximation identity. a key ingredient is scaling. i.e. we given a function ${\Phi}$ and consider ${\Phi_t=t^{-n}\Phi(\frac{x}{t})}$ , and we wish,

$\displaystyle \lim_{t\rightarrow 0}(f*\Phi_t(x))=f(x), for a.e. x\ \in {\mathbb R}^n \ \ \ \ \ (3)$

Whenever ${f\in L^p, 1\leq p\leq \infty}$ , but there need some technique assume to make this intuition to be tight, this lead the following definition.

Definition 2 (Approximation of the identity) Suppose ${\Phi}$ is a fixed function on ${{\mathbb R}^n}$ that is appropriated small at infinity (have good enough decay rate), for example, take,

$\displaystyle |\Phi(x)|\leq A(1+|x|)^{-n-\epsilon} \ \ \ \ \ (4)$

Then we define ${\{\Phi_t:\Phi_t(x)=t^{-n}\Phi(\frac{x}{t})\}}$ to be an approximation of the identity.

The key theorem is the following, related the approximation of the indentity with the maximal operator.

Theorem 3

$\displaystyle \sup_{t>0}|(\Phi_t*f)(x)|\leq c_{\Phi}Mf(x) \ \ \ \ \ (5)$

For heat kernel, the thing is more subtle.

Theorem 4 [Heat kernel estimate]

$\displaystyle \|f-e^{t\Delta}f\|_2\leq \|\nabla f\|_2\sqrt{t} \ \ \ \ \ (6)$

Remark 1 I know this theorem from Lieb’s book. The power of 4 combine with Plancherel theorem could use to establish the Sobolev inequality, at least for the index ${p=2}$ .

There are 3 ingredients which cold be useful.
1. the power of Rearrangement inequality involve in the Approximation of indentity operator. we could consider the relationship between ${f*\Phi_t}$ and ${f*\overline \Phi_t}$ , where ${\overline \Phi}$ is constructed by take the average of ${\Phi}$ on the level set but the foliation of scaling. Intuition seems some monotonic property natural emerge.
2. There is a discretization model, i.e. the toy model on gragh, or we think it as correlation between particles, the key point is the rescaling deformation could be instead by semi group or renormalization property.
3. We consider the more general case, now there is not only one ${\Phi}$ but a group of them, i.e. ${\Phi_k, k\in A}$ , this will involve some amenable theory I think.
We give two of the original and most important examples, First, if

$\displaystyle \Phi(x)=c_n(1+|x|^2)^{\frac{-(n+1)}{2}}$

where

$\displaystyle c_n=\frac{\Gamma(\frac{n+1}{2})}{\pi^{\frac{n+1}{2}}}$

then ${\Phi_t(x)}$ is the possion kernel, and,

$\displaystyle u(x,t)=(f*\Phi_t)(x)$

Gives the solution of the Dirichlet problem for the upper half space,

$\displaystyle {\mathbb R}^{n+1}_{+}=\{(x,t):x\in {\mathbb R}^n,t>0\}$

Namely

$\displaystyle \Delta u=(\frac{\partial^2}{\partial t^2}+ \sum_{j=1}^n\frac{\partial^2}{\partial x_j^2})u(x,t)=0,\ u(x,0)\equiv f(x) \ \ \ \ \ (7)$

The second example is the Gaussian kernel,

$\displaystyle \Phi(x)=(4\pi)^{-\frac{n}{2}}e^{-\frac{|x|^2}{4}}.$

This time, if ${u(x,t)=(f*\Phi_{t^{\frac{1}{2}}})(x)}$ , then ${u}$ is a solution of the heat equation,

$\displaystyle (\frac{\partial}{\partial t}- \sum_{j=1}^n\frac{\partial^2}{\partial x_j^2})u(x,t)=0,\ u(x,0)\equiv f(x) \ \ \ \ \ (8)$

3. Singular integral with ${L^2}$ bounded translation invariant operator

The main result proved in last note about singular integral is a conditional one, guaranteeing the boundedness on ${L^p}$ for a range ${1<p\leq q}$ , on the assupution that the boundedness on ${L^q}$ is already known; the most important instance of this occurs when ${q=2}$ . In keeping with this, we consider bounded linear transformation ${T}$ from ${L^2({\mathbb R}^n)}$ to itself that commute with translation. As is well known, such operator are characterized by the existence of a bounded function ${m}$ on ${{\mathbb R}^n}$ (the “multiper”), so that ${T}$ can be realized as,

$\displaystyle \widehat{Tf(\xi)}=m(\xi)\widehat f(\xi) \ \ \ \ \ (9)$

Where ${\widehat{}}$ denotes the Fourier transform. Alternatively, at least on test function ${f\in S}$ , ${T}$ can be realized in terms of convolution with a kernel ${K}$ ,

$\displaystyle Tf=f*K \ \ \ \ \ (10)$

Where ${K}$ is the distribution given by ${\hat K=m}$ . We shall now examine how the theorem with condition on singular integral weill lead to some result of this type of operator. Roughly speaking, it is due to now we know the boundedness on ${L^2}$ , for technique condition, we need to assume the distribution ${K}$ agree away from the origin with a function that is locally integrable away from the origin with a function that is locally integrable away from the origin; in this case we define the function by ${K(x)}$ . Then 10 implies that,

$\displaystyle Tf(x)=\int K(x-y)f(y)dy,\ for \ a.e. x\notin supp f. \ \ \ \ \ (11)$

Whenever ${f}$ is in ${L^2}$ and ${f}$ has campact support. Tis is the representation of singular integral in the present context. Next, the crucial hormander condition is then equivalent with,

$\displaystyle \int_{|x|\geq c|y|}|K(x-y)-K(x)|dx\leq A \ \ \ \ \ (12)$

for all ${y\neq 0}$ , where ${c>1}$ . In this case, the condition 12 have a further understanding, in fact,

Lemma 5

$\displaystyle |(\frac{\partial}{\partial x}^{\alpha}K(x))|\leq A_{\alpha}|x|^{-n-|\alpha|},\ for\ all \ \ \alpha \ \ \ \ \ (13)$

or its weaker form, (here ${\gamma>0}$ is fixed )

$\displaystyle |K(x-y)-K()|\leq A\frac{|y|^{\gamma}}{|x|^{n+\gamma}}, whenever \ |x|\geq c|y|. \ \ \ \ \ (14)$

imply the hormander condition 12

Proof: Integral by part. $\Box$

So, now the key point is how do ${K}$ , satisfied such conditions, come about? It turns out that, toughly speaking, such condition on ${K}$ have equivalent versions when sated in terms of the Fourier transform of ${K}$ , namely the multiper ${m}$ . This is transform the difficulties from physics space to fractional space In the future note, we will find a proof of the following Theorem:

Theorem 6 For ${m=\hat K}$ .

If we assume that,

$\displaystyle |(\frac{\partial}{\partial \xi}^{\alpha}m(\xi))|\leq A'_{\alpha}|\xi|^{-n-|\alpha|},\ for\ all \ \ \alpha \ \ \ \ \ (15)$

holds for all ${\alpha}$ , then ${K}$ satisfied 5 for all ${\alpha}$ .

If we assume that ${m}$ satisfied the above inequality for all ${0\leq |\alpha| \leq l}$ , where ${l}$ is the smallest integer ${>\frac{n}{2}}$ , then ${K}$ satisfied 12

Remark 2 The multiplier ${m}$ satisfied the second part condition of 6, are called Marcinkiewicz mulltiplier.

4. Maximal function, singular integral, and square functions.

补充说明

以下是新整理的中文说明；上方旧博客原文保持不变。

奇异积分的基本理论不只是 Calderon-Zygmund 分解。真正进入一般理论以后，会反复出现三类对象：approximation of the identity、translation invariant singular integrals，也就是 Fourier multipliers，以及 maximal function 和 square function。

奇异积分一般理论中，近似恒等、Fourier multiplier 和 square function 分别控制尺度极限、平移不变结构和多尺度正交性。

1. 从弱型估计到 $L^p$ 有界性

如果一个奇异积分算子 $T$ 已知在 $L^2$ 上有界，并且满足弱 $(1,1)$ 估计

$$|\{x:|Tf(x)|>\lambda\}|\lesssim \frac{\|f\|_1}{\lambda},$$

那么 Marcinkiewicz interpolation 给出 $1
2. Approximation of the identity

取一族核 $\phi_t(x)=t^{-n}\phi(x/t)$，若 $\int\phi=1$，则

$$\phi_t*f\to f\quad(t\to0).$$

这类算子看起来温和，但它们和 maximal operator 紧密相连。控制

$$\sup_{t>0}|\phi_t*f(x)|$$

本质上就是控制函数在不同尺度上的平均行为。

3. Translation invariant operators

若 $T$ 与平移可交换，那么 Fourier transform 会把它对角化：

$$\widehat{Tf}(\xi)=m(\xi)\widehat f(\xi).$$

Mikhlin multiplier theorem 给出一套可检验条件：若

$$|\partial^\alpha m(\xi)|\lesssim |\xi|^{-|\alpha|}$$

到足够阶数成立，则 $T$ 在 $L^p$ 上有界。

4. Square functions

square function 把函数分解到不同频率或尺度：

$$Sf(x)=\left(\sum_j |P_jf(x)|^2\right)^{1/2}.$$

它不是只估计每一块，而是用正交性追踪所有尺度的总能量。这是 Littlewood-Paley 理论的核心。

5. 一条总线

近似恒等处理尺度极限，multiplier theory 处理平移不变结构，square function 处理多尺度正交性。奇异积分的一般理论，就是在这三种结构之间来回切换。
2017年12月21日
Sturm-Liouville theory 与周期轨道：谱分解、边界条件和 monodromy

旧博客原文

原题：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 $A=\{k_1<k_2<...<k_l\}$ , 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.

补充说明

以下是新整理的中文说明；上方旧博客原文保持不变。

Sturm-Liouville theory 把二阶线性微分方程变成谱问题。若再加入周期边界条件，就会自然出现 monodromy matrix、周期解和一维动力系统的交界。

Sturm-Liouville 周期谱可以通过一阶系统的 monodromy matrix 来刻画。

1. Sturm-Liouville 系统

标准形式为

$$-(p(x)y’)’+q(x)y=\lambda w(x)y.$$

在合适边界条件下，这是自伴特征值问题，特征函数构成正交基。

2. 周期边界条件

若区间为 $[0,T]$，周期解满足

$$y(0)=y(T),\qquad y'(0)=y'(T).$$

这等价于一阶系统的 monodromy matrix 有特征值 $1$。

3. 从二阶到一阶系统

令 $Y=(y,py’)$，二阶方程可写为

$$Y’=A_\lambda(x)Y.$$

基本解矩阵 $M_\lambda(T)$ 描述一个周期后的变化。周期谱由

$$\det(M_\lambda(T)-I)=0$$

刻画。

4. 反问题

一个自然问题是：能否指定某个集合 $S$，使得周期特征值恰好落在 $S$ 中？这类问题接近 inverse spectral theory，需要理解势函数 $q(x)$ 如何控制 monodromy。

5. 拓扑与分析

周期轨道是动力系统对象，Sturm-Liouville 是谱分析对象。monodromy matrix 把二者联系起来：谱参数变化时，monodromy 在矩阵群中运动，周期解对应它穿过特定子集。

2017年12月21日
Calderon-Zygmund 分解：奇异积分的实变量入口
旧博客原文

原题：Calderon-Zygmund theory of singular integrals.

1. Calderon-Zygmund decomposition

The Calderon-Zygmund decomposition is a key step in the real variable analysis of singular integrals. The idea behind this decomposition is that it is often useful to split an arbitrary integrable function into its “small” and “large” parts, and then use different technique to analyze each part.

The scheme is roughly as follows. Given a unction ${ f}$ and an altitude ${ \alpha}$ , we write ${ f=g+b}$ , where ${ |g|}$ is point wise bounded by a constant multiple ${\alpha}$ . While ${ b}$ is large, it does enjoy two redeeming features: it is supported in a set of reasonable small measure, and its mean value is zero on each of the ball that constitute its support. To obtain the decomposition ${ f=g+b}$ , one might be tempted to “cut” ${ f}$ at the height ${ \alpha}$ ; however, this is not what works. Instead, one bases the composition on the set where the maximal function of ${ f}$ has height ${ \alpha}$ .
Theorem 1 (Calderon-Zygmund decomposition)

Suppose we are given a function ${ f\in L^1}$ and a positive number ${ \alpha}$ , with ${\alpha>\frac{1}{\mu(R^n)}\int_{R^n}|f|d\mu}$ . Then there exists a decomposition of ${ f}$ , ${f=g+b}$ , with ${ b=\sum_{k}b_k}$ , and a sequences of balls ${\{B_k^*\}}$ , so that,
1. ${ |g(x)|\leq c\alpha}$ , for a.e. ${ x}$ .
2. Each ${latex b_k}$ is supported in ${B_k^*}$ , ${ \int|b_k(x)|d\mu(x)\leq c\alpha\mu(B_k^*)}$ , and ${ \int b_k(x)d\mu(x)=0}$ .
3. ${ \sum_k\mu(B_k^*)\leq \frac{c}{\alpha}\int|f(x)|d\mu(x)}$ .
Before proof this theorem, I explain the geometric intuition why this theorem could be true first. Merely speaking, this is just base on cut off the function into two part, the part with high altitude and the part with low altitude and extension the part with high altitude to make the extension one satisfied the condition 2 and 3.

Proof: In fact this decomposition have a good geometric explain, we just divide the part ${\{x: |f(x)|>\alpha\}}$ and extension it carefully to make they behaviour like several balls, to satisfied the special condition on this part. $\Box$

Remark 1 Remark 1: A Calderon-Zygmund decomposition for ${L^p}$ function was done in Charlie Fefferman’s thesis; see Section II of ams.org/mathscinet-getitem?mr=257819 One can also find this in Loukas Grafakos’s Classical Fourier Analysis Classical Fourier Analysis page 303 exercise 4.3.8. The question is broken up into parts that should be easy to handle.

Several people have considered with this question. An excellent paper that comes to mind is Anthony Carbery’s Variants of the Calderon–Zygmund theory for ${ L^p}$ -spaces which appeared in Revista Matematica Iberoamericana, Volume 2, Number 4 in 1986. There are also several useful references that appear in Carbery’s paper.

Remark 2 We could also consider a variant of Calderon-Zygmund decomposition, such as equipped with a nontrivial weight function ${ w}$ or find some different way to decomposition for some special purpose.

Remark 3 Consider suitable decomposition of the physics space or even both the physics space and fractional space try to gain some reasonable estimate is a fundamental philosophy in harmonic analysis, beside the Calderon-Zegmund decomposition,

Whitney decomposition. Which is important trick in the proof of fefferman-stein restriction theorem and differential topology.

Wave packet decomposition. The wave packet decomposition. This decomposition underlies the proof of Carleson’s theorem (this is more explicit in Fefferman’s proof than Carelson’s original proof), Lacey and Thiele’s proof of the boundedness of the bilinear Hilbert transform, as well as a host of follow-up work in multilinear harmonic analysis. The idea of the wave packet decomposition is to decompose a function/operator in terms of an overdetermined basis. This allows one to preserve symmetries (such as modulation symmetries) that aren’t preserved by a classical Calderon-Zygmund decomposition (which endows the frequency with a distinguished role). One might consider using a wave packet decomposition if is working with an operator that has a modulation symmetry. This is discussed in more detailed in Tao’s blog post on the trilinear Hilbert transform.

Polynomial decomposition. The application of polynomial decomposition to harmonic analysis is more recent, and its full potential still seems unclear. Applications include Dvir’s proof of the finite field Kakeya conjecture, Guth’s proof of the endpoint multilinear Kakeya conjecture (and, indirectly, the Bourgain-Guth restriction theorems), Katz and Guth’s proof of the joints problem and Erdos distance problem, among many other results. Generally, the idea behind the polynomial decomposition is to partition a subset of a vector space over a field into a finite number of cells each of which contains roughly the same fraction of the original set. One further wishes that no low degree algebraic variety can intersect too many of the cells. In Euclidean space, the polynomial ham sandwich decomposition does exactly this. This allows one to, for instance, control linear (or, more generally, `low algebraic degree’) interactions between points in distinct cells. This has so far proven the most useful in incidence-type problems, but many problems in harmonic analysis, thanks to the translation symmetry of the Fourier transform, are inextricably linked with such incidence-type problems. See (again) Tao’s survey of this topic for a more detailed account.

2. Singular integrals

Have the Calderon-Zegmund decomposition in hand, now we proof a conditional one bounded result for singular integrals.

The singular integral one is interested in are operator ${ T}$ , expressible in the form

$\displaystyle (Tf)(x)=\int_{R^n}K(x,y)f(y)d\mu(y) \ \ \ \ \ (1)$

Where the kernel ${ K}$ is singular near ${ x=y}$ , and so the expression is meaningful only if ${ K}$ is treated as a distribution or in some limiting sense. Now the particular regularization of ${ (Tf)(x)}$ may be appropriate depends much on the context, and a complete treatment of the issues thereby raised take us quite far afield.

Let us limit ourselves to two closely related ways of dealing with the questions concerning the definability of the operator. One is to prove estimates for the (dense) subspace where the operator is initially defined. The other is to regularize the given operators by replacing it with a suitable family, and to prove the uniformly estimates for this family. This idea is similar occurring in spectral geometry when we wish to investigate the spectrum of some operator we try to consider some deformation, so deduce to control the spectrum of a seres of paramatrix, for example, consider the wave kernel or heat kernel rather than the passion kernel itself. Common to both methods is a priori approach: We assume some additional properties of the kernel, but then prove estimates that are independent of these “regularity” properties.

We now carry out the first approach in detail. There will be two kinds of assumptions made about the operator. The first is quantitative: we assume that we are given a bound ${ A}$ , so that the operator ${ T}$ is defined and bounded on ${ L^q}$ with norm ${ A}$ ; that is,

$\displaystyle \|T(f)\|_q\leq A\|f\|_q, \forall f, f\in L^q \ \ \ \ \ (2)$

Moreover, we assume that there is associated to ${ T}$ a measurable function ${ K}$ (that plays the role of its kernel), so that for the same constant ${ A}$ and some constant ${ c>1}$ ,

$\displaystyle \int_{R^n-B(y,c\delta)}|K(x,y)-K(x,\bar y)|d\mu(x)\leq A, \forall \bar y\in B(y,\delta) \ \ \ \ \ (3)$

for all ${ y\in R^n, \delta>0}$ .

The further regularity assumption on the kernel ${ K}$ is that for each ${ f}$ in ${L^q}$ that has compact surppot, the integral coverages absolutely for almost all ${ x }$ in the complement of the support of ${ f}$ , and that equality holds for these ${ x}$ .

Theorem 2 (Bounded of singular integral with condition)

Under the condition 1 and 3 made above on ${ K}$ , the operator ${ T}$ is bounded in ${ L^p}$ norm on ${ L^p\cap L^q}$ , when ${ 1<p<q}$ . More precisely,

$\displaystyle \|T(f)\|_p\leq A_p\|f\|_p$

For ${ f\in L^p\cap L^q}$ with ${ 1<p<q}$ , where the bound ${ A_p}$ depends only on the constant ${ A}$ appearing in 1 and 3 and on ${ p}$ , but not on the assumed regularity of ${ K}$ , or on ${ f}$ .

Proof:

Now let us begin to prove the conditional theorem. The key point is to use the potential of ${T}$ has been a bounded operator from ${L^q\rightarrow L^q}$ . Said, it already assumed ${\exists A>0}$ such that ${\forall f\in L^q}$ we have ${\|T(f)\|_q\leq A\|f\|_q}$ . Now let us look at the singular integral expression:

$\displaystyle (Tf)(x)=\int_{R^n}K(x,y)f(y)d\mu(y). \ \ \ \ \ (4)$

The key point is to proof the mapping ${f\rightarrow T(f)}$ is a weak-type ${1-1}$ ; that is,

$\displaystyle \mu\{x:|Tf(x)|>\alpha\}\leq \frac{A'}{\alpha}\int |f|d\mu. \ \ \ \ \ (5)$

At once we establish 5, then the theorem followed by interpolation. Now we use theorem 1 on ${f}$ get ${f=g+b}$ , thanks to the triangle inequality and something similar we have ${g,b \in L^q}$ , in fact ${R^n= A\amalg B, B\cup_{k}B_k}$ , ${g=\chi_A g+\chi_{B}g, b=\chi_A b+\chi_{B} b}$ , by triangle inequality and ${f=g+b}$ , to proof ${g,b \in L^q}$ , we only need to proof ${\chi_A g, \chi_B g, \chi_A b, \chi_B b\in L^q}$ , but this is easy to proof.

Now we know the ${L^q}$ bounded of ${g,b}$ , we divide the difficult of establish the weak 1-1 bound of ${f}$ into the difficult of establish the weak 1-1 bound for ${g}$ and ${b}$ . i.e.

$\displaystyle \mu\{x:|Tf(x)|>\alpha\}\leq \mu \{x:|Tg(x)|>\alpha\} +\mu\{x:|Tb(x)|>\alpha\} \ \ \ \ \ (6)$

For ${g}$ , if this weak 1-1 bound is not true, we have,

$\displaystyle \mu \{x:|Tg(x)|>\alpha\}\geq \frac{A'}{\alpha}\int |g|d\mu \ \ \ \ \ (7)$

thanks to the trivial estimate ${\|g\|_q \leq c\alpha^{q-1}\|g\|_1 }$ . combine this two estimate we have:

$\displaystyle c\alpha^{q-1}\|g\|_1\geq \|g\|^q_q\geq c\|Tg\|^q_q \geq A'\alpha^{q-1} \|g\|_1 \ \ \ \ \ (8)$

The first estimate is true on ${A}$ due to ${|g|\leq \alpha, a.e. x\in R^n}$ . But compare the left and the right of 8 lead a contradiction, so 7 follows. For ${b}$ , the thing is more complicated and in fact really involve the structure of the convolution type of the singular integral. The key point is controlling near the diagonal of ${K(x,y)}$ . we warm up with a more refine decomposition ${b=\sum b_k}$ , ${\forall k, b_k=b\cdot \chi_{B_k}}$ . For a large constant ${c>>1}$ choose later define ${B^*_k=c B_k}$ . We know ${b\in L^q}$ , but the really difficult thing occur in the how to combine the following 5 condition to lead a contradiction:
1. ${\|Tb\|_q\leq \|b\|_q}$ .
2. property come from the Calderon-Zegmund decomposition, ${\int_{B_k}\|b\|\leq c\alpha \mu(B_k),\forall k}$ and ${\int_{B_k}b=0}$ .
3. Hormander condition 3 , ${\int_{R^n-B(y,c\delta)}|K(x,y)-K(x,\bar y)|d\mu(x)\leq A, \forall \bar y\in B(y,\delta)}$
4. the reverse of weak 1-1 of ${b}$ , ${\mu\{x:b(x)>\alpha\}> \frac{A'}{\alpha}\|b\|_1}$ .
5. the structure ${Tb(x)=\int_{R^n} K(x,y)b(y)dy}$
The first step is to break ${b}$ into ${b_k}$ , and reduce the case of several balls to the case of only one ball, this could be done by triangle inequality or more may be we could do it derectly, but any way it is not difficult.

Then the thing become intersting, we focus on ${b_1}$ , divide ${Tb_1=T\chi_{B_1} b_1+ T\chi_{{\mathbb R}^n-B_1}b}$ . thanks to the hormander condition 3 we have good control on ${T\chi_{{\mathbb R}^n-B_1^*}}$ , in fact we can proof a weak 1-1 bound on it,

$\displaystyle \mu\{x:|T_{{\mathbb R}^n-B_1^*}b_1|>\alpha\}< \frac{A'}{\alpha}\|b_1\|_1 \ \ \ \ \ (9)$

$\displaystyle \begin{array}{rcl} T_{{\mathbb R}^n-B_1^*}b_1(x) & = & \int_{{\mathbb R}^n-B_1^*}K(x,y)b_1(y)dy\\ & = & \int_{{\mathbb R}^n-B_1^*}[K(x,y)-K(x,\bar y)]b_1(y)dy+\int_{{\mathbb R}^n-B_1^*}K(x,\bar y)b_1(y)dy\\ & \leq & \int_{{\mathbb R}^n-B_1^*}Ab_1(y)dy+\int_{{\mathbb R}^n-B_1^*}K(x,\bar y)b_1(y)dy \end{array}$

So we conclude,

$\displaystyle \begin{array}{rcl} \|T_{{\mathbb R}^n-B_1^*}b_1\|_1 & = & \int_{{\mathbb R}^n}|\int_{{\mathbb R}^n-B_1^*}K(x,y)b_1(y)dy|dx\\ & = & \int_{{\mathbb R}^n}\int_{{\mathbb R}^n-B_1^*}|[K(x,y)-K(x,\bar y)]b_1(y)dy|dx+\int_{{\mathbb R}^n}|\int_{{\mathbb R}^n-B_1^*}K(x,\bar y)b_1(y)dy|dx\\ & \leq & A\int_{{\mathbb R}^n}b_1(y)dy+\int_{{\mathbb R}^n-B_1^*}K(x,\bar y)b_1(y)dy=A\int_{{\mathbb R}^n}b_1(y)dy \end{array}$

The last equality used the condition ${\int b_1=0}$ .

$\Box$

补充说明

以下是新整理的中文说明；上方旧博客原文保持不变。

Calderon-Zygmund 分解是实变量调和分析里最基本的动作之一。它不是简单地把函数按高度截断，而是用 Hardy-Littlewood maximal function 找到真正危险的区域，再把函数分成一个有界的好部分和一族有 cancellation 的坏部分。

Calderon-Zygmund 分解用 maximal function 找到坏区域，把函数拆成有界部分和带 cancellation 的局部坏块。

1. 为什么不能直接截断

给定 $f\in L^1(\mathbb R^n)$ 和高度 $\lambda>0$，我们希望写成

$$f=g+\sum_j b_j.$$

其中 $g$ 应该满足 $|g|\lesssim \lambda$，而 $b_j$ 虽然可能很大，但支撑在小集合上，并且有零平均。直接令 $g=f\mathbf 1_{\{|f|\le \lambda\}}$ 并不够，因为奇异积分不是逐点算子，它会感受到坏集合附近的空间结构。

正确做法是看 maximal function 的超水平集：

$$\Omega=\{x:Mf(x)>\lambda\}.$$

然后对 $\Omega$ 做 Whitney decomposition，得到一族彼此控制重叠的 cubes 或 balls。

2. 分解定理

Calderon-Zygmund 分解给出

$$f=g+\sum_j b_j,$$

并且满足：

$$|g(x)|\lesssim \lambda \quad\text{a.e.},$$

$$\operatorname{supp} b_j\subset Q_j,\qquad \int b_j=0,$$

以及

$$\sum_j |Q_j|\lesssim \frac{\|f\|_1}{\lambda},\qquad \sum_j\|b_j\|_1\lesssim \|f\|_1.$$

这里的零平均是关键。奇异积分核在远离 $Q_j$ 的地方可以用光滑性做差分估计，从而把 $b_j$ 的大振幅抵消掉。

3. 用它证明弱 $(1,1)$

设 $T$ 是 Calderon-Zygmund singular integral，并已知 $T$ 在 $L^2$ 上有界。为了证明

$$|\{x:|Tf(x)|>\lambda\}|\lesssim \frac{\|f\|_1}{\lambda},$$

把 $f=g+b$。好部分用 $L^2$ 有界性处理：

$$|\{|Tg|>\lambda/2\}|\lesssim \lambda^{-2}\|g\|_2^2\lesssim \frac{\|f\|_1}{\lambda}.$$

坏部分先丢掉扩大后的坏 cubes，它们总体积可控；在外面利用 $\int b_j=0$ 写

$$Tb_j(x)=\int_{Q_j}\bigl(K(x,y)-K(x,c_j)\bigr)b_j(y)\,dy,$$

再用 kernel 的 Holder 光滑性求和。

4. 更广的分解哲学

Calderon-Zygmund 分解的精神是：在物理空间中识别坏区域，把坏区域局部化，并为每个坏块制造 cancellation。类似思想在 weighted theory、Hardy space、Whitney decomposition、wave packet decomposition 和 polynomial partitioning 中都会出现。

不同分解保留不同对称性。Calderon-Zygmund 分解突出空间局部性；wave packet 分解同时追踪空间和频率；polynomial partitioning 则把几何 incidence 信息放进分析估计里。调和分析很多证明，本质上都是在寻找适合当前算子的分解方式。
2017年12月19日
Large sieve 与 Bombieri-Vinogradov theorem：几乎正交性的数论形态

旧博客原文

原题：The large sieve and the Bombieri-Vinogradov theorem

-1.Motivation-

Large sieve a philosophy reflect as a large group of inequalities which is very effective on controlling some linear sum or square sum of some correlation of arithmetic function, some idea of which could have originated in harmonic analysis, merely rely on almost orthogonality.

One fundamental example is the estimate of the quality,

$\sum_{n\leq x}|\Lambda(n)\overline{\chi(n)}|$

One naive idea of control this quality is using Cauchy-schwarz inequality. But stupid use this we gain something even worse than trivial estimate. In fact by triangle inequality and trivial estimate we gain trivial bound: $\sum_{n\leq x}|\Lambda(n)\overline{\chi(n)}|\leq x$ . But by stupid use Cauchy we get following,

$\sum_{n\leq x}|\Lambda(n)\overline{\chi(n)}|\leq ((\sum_{n\leq x}|\Lambda(n)|^2)(\sum_{n\leq x}|\chi(n)|^2))^{\frac{1}{2}}\leq xlog^{\frac{1}{2}}x$

But this does not mean Cauchy-Schwarz is useless on charge this quality, we careful look at the inequality and try to understand why the bound will be even worse. Every time we successful use Cauchy-Schwarz there are two main phenomenon, first, we lower down the complexity of the quantity we wish to bound, second we almost do not loss any thing at all. So we just reformulate the quantity and find it lower down the complexity and the change is compatible with the equivalent condition of Cauchy-Schwarz. For example we have following identity,

$\sum_{n\leq x}|\Lambda(n)\overline{\chi(n)}|=\sqrt{ \sum_{n\leq x}|\Lambda(n)\overline{\chi(n)}| \sum_{m\leq x}|\Lambda(m)\overline{\chi(m)}|}=\sqrt{ \sum_{k_1,k_2\in \mathbb F_p^{\times}}\sum_{n',m'\leq \frac{x}{p}}|\Lambda(n')\Lambda(m')\overline{\chi(k_1)\chi(k_2)}| }$

So we could understand this quality as the Variation of primes in arithmetic profession constructed by $\{pn+b| b\in\{1,2,...,p-1\}\}$ . But this is still difficult to estimate, merely because of we need to control the variation of convolution of $\Lambda$ with itself on $\mathbb F_p^{\times}\simeq \{pn+b| b\in\{1,2,...,p-1\}\}$ .

Now we change our perspective, recall a variant of Cauchy-Schwarz inequality, which called Bessel inequality, as following,

Bessel inequality

Let ${g_1,\dots,g_J: {\bf N} \rightarrow {\bf C}}$ be finitely supported functions obeying the orthonormality relationship,

$\displaystyle \sum_n g_j(n) \overline{g_{j'}(n)} = 1_{j=j'}$

for all ${1 \leq j,j' \leq J}$ . Then for any function ${f: {\bf N} \rightarrow {\bf C}}$ , we have,

$\displaystyle (\sum_{j=1}^J |\sum_{n} f(n) \overline{g_j(n)}|^2)^{1/2} \leq (\sum_n |f(n)|^2)^{1/2}$ .

Pf: The proof is not very difficult, we just need to keep an orthogonal picture in our mind, consider $\{g_{j}(n)\}, 1\leq j\leq J$ to be a orthogonal basis on $l^2(\mathbb N)$ , then this inequality is a natural corollary.

Have this inequality in mind, by the standard argument given by transform from version of orthogonal to almost orthogonal which was merely explained in the previous note. We could image the following corresponding almost orthogonal variate of “Bessel inequality” is true:

Generalised Bessel inequality

Let ${g_1,\dots,g_J: {\bf N} \rightarrow {\bf C}}$ be finitely supported functions, and let ${\nu: {\bf N} \rightarrow {\bf R}^+}$ be a non-negative function. Let ${f: {\bf N} \rightarrow {\bf C}}$ be such that ${f}$ vanishes whenever ${\nu}$ vanishes, we have

$\displaystyle (\sum_{j=1}^J |\sum_{n} f(n) \overline{g_j(n)}|^2)^{1/2} \leq (\sum_n |f(n)|^2 / \nu(n))^{1/2} \times ( \sum_{j=1}^J \sum_{j'=1}^J c_j \overline{c_{j'}} \sum_n \nu(n) g_j(n) \overline{g_{j'}(n)} )^{1/2}$

for some sequence ${c_1,\dots,c_J}$ of complex numbers with ${\sum_{j=1}^J |c_j|^2 = 1}$ , with the convention that ${|f(n)|^2/\nu(n)}$ vanishes whenever ${f(n), \nu(n)}$ both vanish.

补充说明

以下是新整理的中文说明；上方旧博客原文保持不变。

Large sieve 是一族控制算术序列在线性相位或剩余类中分布的强大不等式。它的核心精神和调和分析中的 almost orthogonality 非常接近。

large sieve 用几乎正交性控制许多模数和 residue phases 上的平方和。

1. 基本形状

典型 large sieve inequality 形如

$$\sum_{q\le Q}\sum_{\substack{a\bmod q\\(a,q)=1}}
\left|\sum_{n\le N}a_n e(an/q)\right|^2
\le (N+Q^2)\sum_{n\le N}|a_n|^2.$$

它说明不同模数和不同 residue phases 之间几乎正交。

2. 为什么 Cauchy-Schwarz 要小心用

粗暴使用 Cauchy-Schwarz 可能比平凡估计还差。成功的关键是先把要求的量重写成具有正交结构的平方和，再用 Cauchy-Schwarz 或 Parseval。

3. 算术级数中的素数

Bombieri-Vinogradov theorem 控制素数在平均模数意义下的分布：

$$\sum_{q\le Q}\max_{(a,q)=1}\left|\psi(x;q,a)-\frac{x}{\varphi(q)}\right|$$

在 $Q\le x^{1/2}$ 附近仍有强估计。它可以看作广义 Riemann hypothesis 在平均意义下的替代。

4. Large sieve 的作用

large sieve 给出对字符和指数和的均方控制。结合 Vaughan identity、零点密度估计或双线性分解，可以处理素数在很多模数下的平均误差。

5. 哲学

Large sieve 的力量在于：我们不逐个模数证明最优分布，而是在整个模数族上利用几乎正交性。这个思想和调和分析中 square function、wave packet 的平均控制是一脉相承的。

2017年12月19日
自由群 $F_2$ 上的线性度量：Cayley graph、范数空间与 pullback

旧博客原文

原题：Linear metric on F2, free group with two generator.

I may have made a stupid mistake, but if not, we could construct a metric by pullback a metric on a suitable linear normalized space $H$ which we carefully constructed. Let we define the generators of free group $F_2$ by $a,b$ .

Step 1.

Constructed the linear normalized space $H$ . the space $H$ was spanned by basis $\Lambda=\Lambda_a \coprod \Lambda_b$ , $\Lambda_a, \Lambda_b$ are defined by look at the Cayley graph of $F_2$ , there is a lot of vertical vector and horizontal vector in the Cayley graph, for every level set of vertical vector we put a basis in $\Lambda_a$ , because there is only countable many vertical vectors (for example, $a,a^2,a^{-5}$ are in the same vertical level, $bab^{-1}, ba^{10}b^{-1}$ are in the same vertical level, $bab^{-1},a$ are not in the same vertical level), we put a basis in $\Lambda_a$ for every vertical level and claim we accomplished the construct of $\Lambda_a$ , we do the same operation for $\Lambda_b$ but only change the vertical level with horizontal level. Now we accomplished the construction of $\Lambda$ , We spanned this with coefficient $\mathbb Z$ and we get a linear space $V$ . by Zorn’s lemma there exists a norm on the space, take one norm $\|\cdot\|$ we accomplished the construction of $H=(V,\|\cdot\|)$ .

Step 2:

Pullback the norm $\|\cdot\|$ on $H$ to the free group $F_2$ . In fact there is a natural bijection $T: F_2\to H$ , which is given by following: On the Cayley graph (imaged it is embedding in $\mathbb R^2$ ), identity $1$ in the group $F_2$ corresponding to the original, and more general every element in $F_2$ exactly identify with a point in the Cayley graph, thanks to there is no relation between $a,b$ . And then there is of course infinity many of path from original to the point, but there is only one shortest path , thanks to there is no loop in the Cayley graph. We identify the elements in $F_2$ with the point in Cayley graph with the shortest path. Now we could explain why the path lies $H$ . This path only across to finite vertical level and horizontal level and on every level it only pass finite step, this already given a representation $\sum_{e_i\in \Lambda}c_i\cdot e_i, c_i\in \mathbb Z$ , the key point is there is only finite $c_i\neq 0$ . So we have defined the bijection $T:F_2\to H$ , and we could use the bijection to pullback the norm on $H$ to a norm on $F_2$ .

Step 3:

Now we begin to proof the norm we get by pullback satisfied the condition we need. We need only to proof the condition of linear growth and triangle inequality. The conjugation invariance is automatically by linear growth by the comments of Tobias Fritz. The triangle inequality is automatically, due to the bijection $T$ stay the structure in fact, the multiplier of elements $x_1,x_2 \in_2$ could be view as put the two path together but this is not true… merely because of the addition operation is not commutative.

The space we should consider is the path space equipped with the composition operation. I image there exists a “big space” such that the natural metric on the “big space” restrict on the embedding image of $\mathbb F_2$ is a linear growth metric.

补充说明

以下是新整理的中文说明；上方旧博客原文保持不变。

自由群 $F_2=\langle a,b\rangle$ 的 Cayley graph 是一棵正则树。一个自然想法是：能否把这棵树嵌入某个线性赋范空间，然后把范数距离 pull back 回群上，得到一种由线性构造出来的 metric？

自由群的 Cayley graph 是树，把它映到赋范空间后可以 pull back 得到由线性构造出来的 metric。

1. Cayley graph 图像

$F_2$ 的每个元素都是一个约化字。Cayley graph 的顶点是群元素，边对应左乘或右乘生成元。因为自由群没有非平凡关系，这个图没有环，是一棵树。

2. 从边方向构造线性空间

可以尝试给 Cayley tree 中不同“水平”的水平边、垂直边分配基向量。设这些基张成一个向量空间 $V$，再在 $V$ 上选择一个范数 $\|\cdot\|$。

每个群元素对应从单位元到该点的唯一 geodesic path，于是可把路径上的边向量相加，得到映射

$$\Phi:F_2\to V.$$

3. Pullback metric

定义

$$d(g,h)=\|\Phi(g)-\Phi(h)\|.$$

如果 $\Phi$ 是单射，这确实给出 metric。若范数选得合适，它可能与 word metric 有可比较关系；若选得太退化，则会丢失树的几何。

4. 需要注意的问题

真正困难在于，这种构造是否自然、是否左不变、是否 quasi-isometric 于 word metric。普通 word metric 满足

$$d_S(g,h)=|g^{-1}h|_S,$$

具有明显的左不变性；pullback metric 未必自动保留这个性质。

5. 几何意义

这个问题可以看作自由群嵌入 Banach space 的 toy model。它连接 Cayley graph、tree metric、coarse embedding 和 geometric group theory 中的线性化思想。

2017年12月19日