博客

Dirichlet hyperbola method：用双曲线拆分除数和
旧博客原文

原题：Dirichlet hyperbola method

A pdf version is Dirichlet hyperbola method.

1. Introduction

Theorem 1

$\displaystyle \sum_{1\leq n\leq x}d(n)=\sum_{1\leq n\leq x}[\frac{x}{n}]=xlogx+(2\gamma-1) x+O(\sqrt{x}) \ \ \ \ \ (1)$

Remark 1 I thought this problem initial 5 years ago, cost me several days to find a answer, I definitely get something without the argument of Dirchlet hyperbola method and which is weaker but morally the same camparable with the result get by Dirichlet hyperbola method.

Remark 2 How to get the formula:

$\displaystyle \sum_{1\leq n\leq x}d(x)=\sum_{1\leq n\leq x}[\frac{x}{n}]? \ \ \ \ \ (2)$

In fact,

$\displaystyle \sum_{1\leq n\leq x}d(x)=\sum_{1\leq ab\leq x}1=\sum_{1\leq n\leq x}[\frac{x}{n}]\ \ \ \ \ (3)$

Which is the integer lattices under or lying on the hyperbola ${\{(a,b)|ab=x\}}$ .

Remark 3 By trivial argument, we can bound the quantity as following way,

$\displaystyle \begin{array}{rcl} \sum_{1\leq n\leq x}[\frac{x}{n}] & = & \sum_{1\leq ab\leq x}1\\ & = & x\sum_{i=1}^x\frac{1}{i}-\sum_{i=1}^x\{\frac{x}{i}\}\\ & = &xlnx+\gamma x+O(x) \end{array}$

The error term is ${O(x)}$ , which is too big. But fortunately we can use the symmetry of hyperbola to improve the error term.

Proof:

$\displaystyle \begin{array}{rcl} \sum_{1\leq n\leq x}d(n) & = & \sum_{ab\leq x}1\\ & = & \sum_{a\geq \sqrt{x}}[\frac{x}{b}]+\sum_{b\geq \sqrt{x}}[\frac{x}{a}]-\sum_{1\leq a,b\leq \sqrt{x}}1\\ & = & xlogx+(2\gamma-1)x+O(\sqrt{x}) \end{array}$

$\Box$

Theorem 2

Given a natural number k, use the hyperbola method together
with induction and partial summation to show that

$\displaystyle \sum_{n\leq x}d_k(n) = xP_k(log x) + O(x^{1-\frac{1}{k}+\epsilon}), n\leq x \ \ \ \ \ (4)$

where ${P_k(t)}$ denotes a polynomial of degree ${k-1}$ with leading term ${\frac{t^{k-1}}{(k-1)!}}$ .

Remark 4 ${P_k(x)}$ is the residue of ${\zeta(s)^kx^ss^{-1}}$ at ${s=1}$ .

Proof:

We can establish the dimension 3 case directly, which is the following asymptotic formula,

$\displaystyle \sum_{1\leq xy\leq n}[\frac{n}{xy}]=xP_2(logx)+O(x^{1-\frac{1}{3}+\epsilon}) \ \ \ \ \ (5)$

The approach is following, we first observe that

$\displaystyle \sum_{1\leq xy\leq n}[\frac{n}{xy}]=\sum_{xyz\leq n}1 \ \ \ \ \ (6)$

The problem transform to get a asymptotic formula for the lattices under 3 dimension hyperbola. The first key point is, morally ${([n^{\frac{1}{3}}],[n^{\frac{1}{3}}],[n^{\frac{1}{3}}])}$ is the central point under the hyperbola.

Then we can divide the range into 3 parts, and try to get a asymptotic formula for each part then add them together. Assume we have:
1. ${A_x=\sum_{1\leq r\leq [n^{\frac{1}{3}}]}\sum_{1\leq yz\leq [n^{\frac{2}{3}}]}[\frac{r}{yz}]}$ .
2. ${A_y=\sum_{1\leq r\leq [n^{\frac{1}{3}}]}\sum_{1\leq xz\leq [n^{\frac{2}{3}}]}[\frac{r}{yz}]}$ .
3. ${A_z=\sum_{1\leq r\leq [n^{\frac{1}{3}}]}\sum_{1\leq xy\leq [n^{\frac{2}{3}}]}[\frac{r}{yz}]}$ .
Then the task transform to get a asymptotic formula,

$\displaystyle A_x=A_y=A_z=xQ_2(logx)+O(x^{1-\frac{1}{3}+\epsilon}) \ \ \ \ \ (7)$

But we can do the same thing for ${\sum_{1\leq yz\leq [n^{\frac{2}{3}}]}[\frac{r}{yz}]}$ and then integral it. This end the proof. For general ${k\in {\mathbb N}}$ , the story is the same, by induction.

Induction on ${k}$ and use the Fubini theorem to calculate ${\sum_{x_1...x_r\leq n}\frac{n}{x_1...x_r},\forall 1\leq r\leq k}$ . $\Box$

There is a major unsolved problem called Dirichlet divisor problem.

$\displaystyle \sum_{n\leq x}d(n) \ \ \ \ \ (8)$

What is the error term? The conjecture is the error term is ${O(x^{\theta}), \forall \theta>\frac{1}{4}}$ , it is known that ${\theta=\frac{1}{4}}$ is not right.

Remark 5

To beats this problem, need some tools in algebraic geometry.

2. Several problems

${\forall k\in {\mathbb N}}$ , is there a asymptotic formula for ${\sum_{t=1}^n\{\frac{kn}{t}\}}$ ?

${\forall k\in {\mathbb N}}$ , ${f(n)}$ is a polynomial with degree ${k}$ , is there a asymptotic formula for ${\sum_{t=1}^n\{\frac{f(n)}{t}\}}$ ?

${\forall k\in {\mathbb N}}$ , ${g(n)}$ is a polynomial with degree ${k}$ , is there a asymptotic formula for ${\sum_{t=1}^n\{\frac{n}{g(t)}\}}$ ?

Theorem 3 ${k\in {\mathbb N}}$ , then we have

$\displaystyle \lim_{n\rightarrow \infty}\frac{\{\frac{kn}{1}\}+\{\frac{kn}{2}\}+...+\{\frac{kn}{n}\}}{n}=k(\sum_{i=1}^k\frac{1}{i}-lnk-\gamma) \ \ \ \ \ (9)$

Proof:

$\displaystyle \begin{array}{rcl} \frac{\{\frac{kn}{1}\}+\{\frac{kn}{2}\}+...+\{\frac{kn}{n}\}}{n} & = & \frac{\sum_{i=1}^k\frac{kn}{i}-\sum_{i=1}^n[\frac{kn}{i}]}{n}\\ & = & k(lnn+\gamma +\epsilon_n)-\frac{\sum_{i=1}^{kn}[\frac{kn}{i}]-\sum_{i=n+1}^{kn}[\frac{kn}{i}]}{n} \end{array}$

$\Box$

Now we try to estimate

$\displaystyle S_k(n)=\sum_{i=1}^{kn}[\frac{kn}{i}]-\sum_{i=n+1}^{kn}[\frac{kn}{i}] \ \ \ \ \ (10)$

In fact, we have,

$\displaystyle \begin{array}{rcl} S_k(n) & = & (2\sum_{i=1}^{[\sqrt{kn}]}[\frac{kn}{i}]-[\sqrt{kn}]^2)-(\sum_{i=1}^k[\frac{kn}{i}]-kn)\\ & = & 2\sum_{i=1}^{[\sqrt{kn}]}\frac{kn}{i}-\sum_{i=1}^k\frac{kn}{i}+2\{\sqrt{kn}\}[\sqrt{kn}]+\{\sqrt{kn}\}^2-2\sum_{i=1}^{[\sqrt{kn}]}\{\frac{kn}{i}\}+\sum_{i=1}^k\{\frac{kn}{i}\}\\ & = & 2kn(ln[\sqrt{kn}]+\gamma+\epsilon_{[\sqrt{kn}]})-kn\sum_{i=1}^k\frac{1}{i}+r(n)\\ & = & knln(kn)+kn(2\gamma-\sum_{i=1}^k\frac{1}{i})+r'(n)\\ & = & knln n+kn(2\gamma+lnk-\sum_{i=1}^k\frac{1}{i})+r'(n) \end{array}$

Where ${-3\sqrt{n}<r(n)<3\sqrt{n}}$ , ${-3\sqrt{n}<r'(n)<3\sqrt{n}}$ .

So by 1 we know,

$\displaystyle \begin{array}{rcl} \frac{\{\frac{kn}{1}\}+...+\{\frac{kn}{n}\}}{n} & = & k(lnn+\gamma+\epsilon_n)-klnn-k(2\gamma+lnk-\sum_{i=1}^k\frac{1}{i})+\frac{r'(n)}{n}\\ & = & k(\sum_{i=1}^k\frac{1}{i}-lnk-\gamma)+\frac{r'(n)}{n}+k\epsilon_n \end{array}$

So we have,

$\displaystyle \lim_{n\rightarrow \infty}\frac{\{\frac{kn}{1}\}+...+\{\frac{kn}{n}\}}{n} =k(\sum_{i=1}^k\frac{1}{i}-lnk-\gamma)=k\epsilon_k \ \ \ \ \ (11)$

Remark 6 In fact we can get ${0<k\epsilon_k<\frac{1}{2}, \forall k\in {\mathbb N}}$ , by combining the theorem 3 and 1.

3. Lattice points in ball

Gauss use the cube packing circle get a rough estimate,

$\displaystyle \sum_{n\leq x}r_2(n)=\pi x+O(\sqrt{x}) \ \ \ \ \ (12)$

In the same way one can obtain,

$\displaystyle \sum_{n\leq x}r_k(n)=\rho_kx^{\frac{k}{2}}+O(x^{\frac{k-1}{2}}) \ \ \ \ \ (13)$

Remark 7 Where ${\rho_k=\frac{\pi^{\frac{k}{2}}}{\Gamma(\frac{k}{2}+1)}}$ is the volume of the unit ball in ${k}$ dimension.

Dirchlet’s hyperbola method works nicely for the lattic points in a ball of dimension ${k\geq 4}$ . Langrange proved that every natural number can be represented as the sum of four squares, i.e. ${r_4(n)>0}$ , and Jacobi established the exact formula for the number of representations

$\displaystyle r_4(n)=8(2+(-1)^n)\sum_{d|n,d\ odd}d. \ \ \ \ \ (14)$

Hence we derive,

$\displaystyle \begin{array}{rcl} \sum_{n\leq x}r_4(n) & = & 8\sum_{m\leq x}(2+(-1)^m)\sum_{dm\leq x, d\ odd}d\\ & = & 8\sum_{m\leq x}(2+(-1)^m)(\frac{x^2}{4m^2}+O(\frac{x}{m}))\\ & = & 2x^2\sum_1^{\infty}(2+(-1)^m)m^{-2}+O(xlogx)\\ & = & 3\zeta(2)x^2+O(xlogx) = \frac{1}{2}(\pi x)^2+O(xlogx) \end{array}$

This result extend easily for any ${k\geq 4}$ , write ${r_k}$ as the additive convolution of ${r_4}$ and ${r_{k-4}}$ , i.e.

$\displaystyle r_k(n)=\sum_{0\leq t\leq n}r_4(t)r_{k-4}(n-t) \ \ \ \ \ (15)$

Apply the above result for ${r_4}$ and execute the summation over the remaining ${k-4}$ squares by integration.

$\displaystyle \sum_{n\leq x}r_k(n)=\frac{(\pi x)^{\frac{k}{2}}}{\Gamma(\frac{k}{2}+1)}+O(x^{\frac{k}{2}-1}logx) \ \ \ \ \ (16)$

Remark 8 Notice that this improve the formula 12 which was obtained by the method of packing with a unit square. The exponent ${\frac{k}{2}-1}$ in 16 is the best possible because the individual terms of summation can be as large as the error term (apart from ${logx}$ ), indeed for ${k=4}$ we have ${r_4(n)\geq 16n}$ if ${n}$ is odd by the Jacobi formula. The only case of the lattice point problem for a ball which is not yet solved (i.e. the best possible error terms are not yet established) are for the circle( ${k=2}$ ) and the sphere ( ${k=3}$ ).

Theorem 4

$\displaystyle \sum_{n\leq x}\tau(n^2+1)=\frac{3}{\pi}xlogx+O(x) \ \ \ \ \ (17)$

4. Application in finite fields

Suppose ${f(x)\in {\mathbb Z}[x]}$ is a irreducible polynomial. And for each prime ${p}$ , let

$\displaystyle \rho_f(p)=\# \ of \ solutions\ of f(x)\equiv 0(mod\ p) \ \ \ \ \ (18)$

By Langrange theorem we know ${\rho_f(p)\leq deg(f)}$ . Is there a asymptotic formula for

$\displaystyle \sum_{p\leq x}\rho_f(p)? \ \ \ \ \ (19)$

A general version, we can naturally generated it to algebraic variety.

$\displaystyle \rho_{f_1,...,f_k}(p)=\#\ of \ solutions\ of f_i(x)\equiv 0(mod\ p) ,\ \forall 1\leq i\leq k \ \ \ \ \ (20)$

Is there a asymptotic formula for

$\displaystyle \sum_{p\leq x}\rho_{f_1,...,f_k}(p)? \ \ \ \ \ (21)$

Example 1 We give an example to observe what is involved. ${f(x)=x^2+1}$ . We know ${x^2+1\equiv 0 (mod \ p)}$ is solvable iff ${p\equiv 1 (mod\ 4)}$ or ${p=2}$ . One side is easy, just by Fermat little theorem, the other hand need Fermat descent procedure, which of course could be done by Willson theorem. In this case,

$\displaystyle \sum_{p\leq n}\rho_f(p)=\# \ of \{primes \ of \ type\ 4k+1 \ in \ 1,2,...,n\} \ \ \ \ \ (22)$

Which is a special case of Dirichlet prime theorem.

Let ${K}$ be an algebraic number field, i.e. the finite field extension of rational numbers, let

$\displaystyle \mathcal{O}_K=\{\alpha\in K, \alpha \ satisfied \ a\ monic \ polynomial\ in\ {\mathbb Z}[x]\} \ \ \ \ \ (23)$

Dedekind proved that,
Theorem 5
1. ${\mathcal{O}_K}$ is a ring, we call it the ring of integer of ${K}$ .
2. He showed further every non-zero ideal of ${\mathcal{O}_K}$ could write as the product of prime ideal in ${\mathcal{O}_k}$ uniquely.
3. the index of every non-zero ideal ${I}$ in ${\mathcal{O}_K}$ is finite, i.e. ${[\mathcal{O}_K:I]<\infty}$ , and we can define the norm induce by index.
  $\displaystyle N(I):=[\mathcal{O}_K:I] \ \ \ \ \ (24)$
  
  Then the norm is a multiplication function in the space of ideal, i.e. ${N(IJ)=N(I)N(J), \forall I,J \in \ ideal\ class\ group\ of\ \mathcal{O}_K}$ .
4. Now he construct the Dedekind Riemann zeta function,
  $\displaystyle \zeta_K(s)=\sum_{N(I)\neq 0}\frac{1}{N(I)^s}=\prod_{J\ prime \ ideal\ }\frac{1}{1-\frac{1}{N(J)^s}},\ \forall Re(s)>1 \ \ \ \ \ (25)$
Now we consider the analog of the prime number theorem. Let ${\pi_K(x)=\{I,N(I)<x\}}$ , does the exist a asymptotic formula,

$\displaystyle \pi_K(x)\sim \frac{x}{ln x}\ as\ x\rightarrow \infty? \ \ \ \ \ (26)$

Given a prime ${p}$ , we may consider the prime ideal

$\displaystyle p\mathcal{O}_K=\mathfrak{P}_1^{e_1}\mathfrak{P}_2^{e_2}...\mathfrak{P}_k^{e_k} \ \ \ \ \ (27)$

Where ${\mathfrak{P}_i }$ is different prime ideal in ${\mathcal{O}_K}$ . But the question is how to find these ${\mathfrak{P}_i}$ ? For the question, there is a satisfied answer.

Lemma 6 (existence of primitive element) There always exist a primetive elements in ${K}$ , such that,

$\displaystyle K={\mathbb Q}(\theta) \ \ \ \ \ (28)$

Where ${\theta}$ is some algebraic number, which’s minor polynomial ${f(x)\in {\mathbb Z}[x]}$ .

Theorem 7 (Dedekind recipe) Take the polynomial ${f(x)}$ , factorize it in the polynomial ring ${{\mathbb Z}_p[x]}$ ,

$\displaystyle f(x)\equiv f_1(x)^{e_1}...f_{r}(x)^{e_r}(mod \ p) \ \ \ \ \ (29)$

Consider ${\mathfrak{P}_i=(p, f_i(\theta)) \subset \mathcal{O}_K}$ . Then apart from finite many primes, we have,

$\displaystyle p\mathcal{O}_K=\mathfrak{P}_1^{e_1}\mathfrak{P}_2^{e_2}...\mathfrak{P}_k^{e_k} \ \ \ \ \ (30)$

Where ${N(\mathfrak{P}_i)=p^{deg{f_i}}}$ .

Remark 9 The apart primes are those divide the discriminant.

Now we can argue that 4 is morally the same as counting the ideals whose norm is divide by ${p}$ in a certain algebraic number theory.

And we have following, which is just the version in algebraic number fields of 2.

Theorem 8 (Weber) ${\#}$ of ideals of ${\mathcal{O}_K}$ with norm ${\leq x}$ equal to,

$\displaystyle \rho_k(X)+O(x^{1-\frac{1}{d}}), where \ d=[K:Q] \ \ \ \ \ (31)$

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。

Dirichlet hyperbola method 是解析数论中处理卷积和的基本工具。它的几何图像非常直接：把求和区域看成双曲线 $ab\le x$ 下方的格点，再利用双曲线关于 $\sqrt x$ 的对称性减少误差。

Dirichlet hyperbola method 用双曲线 $ab=x$ 的对称性改进卷积和的误差。

1. 除数函数的例子

除数函数满足

$$d(n)=\sum_{ab=n}1.$$

因此

$$\sum_{n\le x}d(n)=\sum_{ab\le x}1.$$

这就是双曲线 $ab=x$ 下方的整数格点数。

2. 直接估计的问题

若对每个 $a$ 求 $\lfloor x/a\rfloor$，得到

$$\sum_{a\le x}\left\lfloor\frac xa\right\rfloor.$$

平凡地把 floor 换成 $x/a$ 会产生太大的误差，因为项数有 $x$ 个。hyperbola method 的关键是只在 $a\le\sqrt x$ 和 $b\le\sqrt x$ 的短范围内精确处理。

3. 基本公式

由对称性可得

$$\sum_{n\le x}d(n)=2\sum_{a\le\sqrt x}\left\lfloor\frac xa\right\rfloor-\lfloor\sqrt x\rfloor^2.$$

于是

$$\sum_{n\le x}d(n)=x\log x+(2\gamma-1)x+O(\sqrt x).$$

这个误差已经比直接方法好很多。

4. 一般卷积

若 $h=f*g$，则

$$\sum_{n\le x}h(n)=\sum_{ab\le x}f(a)g(b).$$

可以选择参数 $Y$，把区域分成 $a\le Y$、$b\le x/Y$ 和重叠部分。合适的 $Y$ 取决于 $f,g$ 的平均阶和可用误差估计。

5. 高维推广

对 $k$ 重除数函数 $d_k(n)$，问题变成

$$a_1a_2\cdots a_k\le x$$

下方的格点计数。通过归纳、partial summation 和多维双曲面拆分，可以得到

$$\sum_{n\le x}d_k(n)=xP_{k-1}(\log x)+\text{error},$$

其中 $P_{k-1}$ 是次数 $k-1$ 的多项式。这个方法的力量在于：它把乘法卷积的求和问题变成可视化的几何区域拆分。
2018年1月8日
Diophantine approximation：Dirichlet 定理、抽屉原理与最佳逼近
旧博客原文

原题：Diophantine approximation

I explain some general ideal in the theory of diophantine approximation, some of them is original by myself, begin with a toy model, then consider the application on folklore Swirsing-Schmidt conjecture.

\tableofcontents

1. Dirichlet theorem, the toy model

The very basic theorem in the theory of Diophantine approximation is the well known Dirichlet approximation theorem, the statement is following.

Theorem 1 (Dirichlet theorem) for all ${\alpha}$ is a irrational number, we have infinity rational number ${\frac{q}{p}}$ such that:

$\displaystyle |\alpha-\frac{q}{p}|<\frac{1}{p^2} \ \ \ \ \ (1)$

Remark 1 It is easy to see the condition of irrational is crucial. There is a best constant version of it, said, instead of ${1}$ , the best constant in the suitable sense for the theorem 1 should be ${\frac{1}{\sqrt{5}}}$ and arrive by ${\frac{\sqrt{5}+1}{2}}$ at least. The strategy of the proof of the best constant version involve the Frey sequences.

Now we begin to explain the strategies to attack the problem.

\paragraph{Argument 1, boxes principle} We begin with a easiest one, i.e. by the argument of box principle, the box principle is following,

Theorem 2 (Boxes principle) Given ${n\in {\mathbb N}}$ and two finite sets ${A={a_1,a_2,...,a_n,a_{n+1}}}$ , set ${B={b_1,...,b_{n}}}$ , if we have a map:

$\displaystyle f:A\longrightarrow B \ \ \ \ \ (2)$

Then there exists a element ${b_k\in B}$ such that there exist at least two element ${a_i,a_j\in A}$ , ${f(a_i)=f(a_j)=b_k}$ .

Proof: The proof is trivial. $\Box$

Now consider, ${\forall N\in {\mathbb N}}$ , the sequences ${x,2x,...,Nx}$ , then ${\{ix\}\in [0,1], \forall i\in \{1,2,...,n\}}$ . Divide ${[0,1]}$ in an average way to ${N}$ part: ${[\frac{k-1}{N},\frac{k}{N}]}$ . Then the linear structure involve (which, in fact play a crucial role in the approach). And the key point is to look at ${\{nx\}}$ and integers.

\paragraph{Argument 2, continue fractional} We know, for irrational number ${x}$ , ${x}$ have a infinite long continue fractional:

$\displaystyle x=q_0+\frac{1}{q_1+\frac{1}{q_2+\frac{1}{q_3+....+\frac{1}{q_k+...}}}} \ \ \ \ \ (3)$

Then

$\displaystyle |x-q_0+\frac{1}{q_1+\frac{1}{q_2+\frac{1}{q_3+....+\frac{1}{q_k}}}}|\sim \frac{1}{(q_1q_2...q_{k-1})^2q_k} \ \ \ \ \ (4)$

And we have,

$\displaystyle \frac{1}{q_1+\frac{1}{q_2+\frac{1}{q_3+....+\frac{1}{q_k}}}}=\frac{a_n}{b_n}, (a_n,b_n)=1 \ \ \ \ \ (5)$

Then ${b_n=O(q_1...q_k)}$ .

\paragraph{Argument 3, Bohr set argument} We begin with some kind of Bohr set:

$\displaystyle B_p=I-\cup_{q\in \{0,1,...,p-1\}}(\frac{q}{p}-\frac{1}{p^2},\frac{q}{p}+\frac{1}{p^2}) \ \ \ \ \ (6)$

The key point is the shift of Bohr set, on the vertical line i.e. ${|B_p\cap B_{p+1}|}$ is very slow, and can be explained by

$\displaystyle \frac{k}{p+1}+\frac{1}{(p+1)^2}>\frac{k}{p}-\frac{1}{p^2} \ \ \ \ \ (7)$

So:

$\displaystyle \frac{1}{p^2}+\frac{1}{(p+1)^2}>\frac{k}{p(p+1)} \ \ \ \ \ (8)$

in ${|B_p \cap B_{p+1}|\sim \frac{1}{p(p+1)}}$ But in fact they are not really independent, as the number of Bohr sets increase, then you can calculate the correlation, thanks to the harmonic sires increasing very slowly, wwe can get something non trivial by this argument, but it seems not enough to cover the whole theorem 1.

\paragraph{Argument 4, mountain bootstrap argument} This argument is more clever than 3, although both two arguments try to gain the property we want in 1 from investigate the whole space ${[0,1]}$ but not ${x}$ , this argument is more clever.

Now I explain the main argument, it is nothing but sphere packing, with the set of balls

$\displaystyle \Omega=\{B_{p,q}:=(\frac{q}{p}-\frac{1}{q^2},\frac{q}{p}+\frac{1}{p^2})| \forall p\in {\mathbb N}, 1\leq q\leq p-1 \} \ \ \ \ \ (9)$

and define its subset

$\displaystyle \Omega_l=\{B_{p,q}:=(\frac{q}{p}-\frac{1}{q^2},\frac{q}{p}+\frac{1}{p^2})| \forall 1\leq p\leq l, 1\leq q\leq p-1 \} \ \ \ \ \ (10)$

Then ${\Omega_l\subset \Omega}$ , and ${\Omega =\cup_{l\in {\mathbb N}}\Omega_l}$ . If we can proof,

Lemma 3 For all ${l\in {\mathbb N}}$ , there is a subset ${A_l}$ of ${\Omega-\Omega_l}$ such that ${\cup_{i\in A_l}B_i=[0,1]}$ .

Remark 2 If we can proof 3, it is easy to see the theorem 1 follows.

Proof: The proof follows very standard in analysis, may be complex analysis? Key point is we start with a ball ${B_{p,q}}$ , whatever it is, this is not important, the important thing is we can take some ball ${B_{p',q'}}$ with the center of ${B_{p',q'}}$ in ${B_{p,q}}$ , then try to consider ${B_{p',q'}\cup B_{p,q}}$ to extension ${B_{p,q}}$ and then we find the boudary is also larger then we can extension again, step by step just like mountain bootstrap argument. So we involve in two possible ending,
1. The extension process could extension ${B{p,q}}$ to whole space.
2. we can not use the extension argument to extension to the whole space.
If we are in the first situation, then we are safe, there is nothing need proof. If we are in second case, anyway we take a ball ${B_{p,q}=(\frac{q}{p}-\frac{1}{p^2},\frac{q}{p}+\frac{1}{p^2})}$ . Then try to find good ball ${B_{p',q'}}$ to approximate ${B_{p,q}}$ , but this is difficult… $\Box$

Remark 3 Argument 1 is too clever to be true in generalization, argument 2 is standard, by the power of renormalization. argument 3 and argument 4 have gap… I remember I have got a proof similar to argument 4 here many years ago, but I forgot how to get it…

2. Schimidt conjecture

The Schimidt conjecture could be look as the generalization of Dirchlet approximation theorem 1 to algebraic number version, to do this, we need define the height of a algebraic number.

Definition 4 We say a number ${\alpha\in {\mathbb C}}$ is a ${k-}$ order algebraic number if and only is the minimal polynomial of ${\alpha}$ , ${f(x)=a_nx^n+...+a_1x+a_0, a_n\neq 0}$ have degree ${deg(f)=n, f\in {\mathbb Z}[x]}$ .

Definition 5 (Height) Now we define the height of a ${k-}$ th order algebraic number as ${H(\alpha):=\max\{\|a_n\|_h,\|a_{n-1}\|_h,...,\|a_0\|_h\}}$ , Where

$\displaystyle h(a_i)=\|a_i\|_{\infty} \ \ \ \ \ (11)$

Now we state the conjecture:

Theorem 6 (Swiring-Schimidt conjecture) For all transendental number ${x\in {\mathbb C}}$ , there is infinitely ${\alpha}$ are ${k-}$ th algebraic number such that:

$\displaystyle |x-\alpha|<\frac{c_k}{H(\alpha)^{k+1}} \ \ \ \ \ (12)$

Where ${c_k}$ is a constant only related to ${k}$ but not ${x}$ .

I point out the conjecture is very related to the map:

$\displaystyle F:(x_1,...,x_n) \longrightarrow (\sigma_1(x_1,...,x_n),\sigma_2(x_1,...,x_n),...,\sigma_n(x_1,...,x_n)) \ \ \ \ \ (13)$

Where ${\sigma_k(x_1,...,x_n)=\sum_{1\leq i_1<...<i_k\leq n}\Pi_{j=1}^kx_{i_1}x_{i_2}...x_{i_k}}$ is the ${k-}$ th symmetric sum.

Remark 4 ${F}$ is a map ${{\mathbb C}^n\rightarrow {\mathbb C}^n}$ , what we consider is its inverse, ${G=F^{-1}}$ , but ${G}$ is not smooth, it occur singularity when ${x_i=x_j}$ for some ${i\neq j}$ . And the map, as we know, the singularity depend on the quantity ${\Pi_{1\leq i< j\leq n}(x_i-x_j)}$ .

Remark 5 I then say something about the geometric behaviour of the map ${G}$ , as we know, what we have in mind is consider the map ${G}$ as a distortion ${{\mathbb C}^n\rightarrow {\mathbb C}^n}$ , Then ${H(\alpha)}$ is just the pullback of the canonical metric on ${{\mathbb C}}$ (morally) to ${{\mathbb C}}$ .

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。

Diophantine approximation 的基本问题是：一个无理数 $\alpha$ 能被有理数 $p/q$ 逼近到什么程度？Dirichlet 定理给出最基本的答案，而连分数告诉我们哪些分母是真正的最佳逼近。

Dirichlet 定理的抽屉原理证明依赖小数部分的差分线性结构。

1. Dirichlet 定理

对任意无理数 $\alpha$，存在无穷多个有理数 $p/q$，使得

$$\left|\alpha-\frac pq\right|<\frac1{q^2}.$$

一个有限版本是：给定 $Q$，存在 $1\le q\le Q$ 和整数 $p$，使

$$|q\alpha-p|<\frac1Q.$$

2. 抽屉原理证明

看 $Q+1$ 个数的小数部分

$$0,\{\alpha\},\{2\alpha\},\ldots,\{Q\alpha\}.$$

把 $[0,1]$ 分成 $Q$ 个长度 $1/Q$ 的区间。两个小数部分落在同一区间，于是它们的差给出某个 $q\alpha$ 距离整数小于 $1/Q$。这就是 Dirichlet 定理最干净的证明。

3. 线性结构在哪里

抽屉原理本身只是计数，但这里真正起作用的是线性结构：两个点 $\{a\alpha\}$ 和 $\{b\alpha\}$ 接近，差就变成 $\{(a-b)\alpha\}$ 接近整数。没有这个差分结构，抽屉原理不会自动给出有理逼近。

4. 连分数与最佳逼近

连分数展开

$$\alpha=[a_0;a_1,a_2,\ldots]$$

给出 convergents $p_k/q_k$。它们满足

$$\left|\alpha-\frac{p_k}{q_k}\right|<\frac1{q_kq_{k+1}}.$$

这些分母 $q_k$ 是最佳逼近的自然尺度。若 $a_k$ 有界，则 $\alpha$ 是 badly approximable；若某些 $a_k$ 很大，就会出现异常好的逼近。

5. 更高维和 Schmidt 猜想的方向

高维 Diophantine approximation 会把一个数的逼近问题变成向量、线性形式或流形上的逼近问题。此时抽屉原理仍然给出基准结果，但最佳常数、例外集维数和代数数逼近会变得更深。许多问题最后会进入 geometry of numbers、dynamical systems on homogeneous spaces 或 Schmidt game 的语言。
2018年1月8日
离散调和函数：格点 Laplacian、坏点集合与 Liouville 型问题
旧博客原文

原题：Discrete harmonic function in Z^n

There is some gap, in fact I can improve half of the argument of Discrete harmonic function , the pdf version is Discrete harmonic function in Z^n, but I still have some gap to deal with the residue half…

1. The statement of result

First of all, we give the definition of discrete harmonic function.

Definition 1 (Discrete harmonic function) We say a function ${f: {\mathbb Z}^n \rightarrow {\mathbb R}}$ is a discrete harmonic function on ${{\mathbb Z}^n}$ if and only if for any ${(x_1,...,x_n)\in {\mathbb Z}^n}$ , we have:

$\displaystyle f(x_1,...,x_n)=\frac{1}{2^n}\sum_{(\delta_1,...,\delta_n )\in \{-1,1\}^n}f(x_1+\delta_1,...,x_n+\delta_n ) \ \ \ \ \ (1)$

In dimension 2, the definition reduce to:

Definition 2 (Discrete harmonic function in ${{\mathbb R}^2}$ ) We say a function ${f: {\mathbb Z}^2 \rightarrow {\mathbb R}}$ is a discrete harmonic function on ${{\mathbb Z}^2}$ if and only if for any ${(x_1,x_2)\in {\mathbb Z}^2}$ , we have:

$\displaystyle f(x_1,x_2)=\frac{1}{4}\sum_{(\delta_1,\delta_2)\in \{-1,1\}^2}f(x_1+\delta_1,x_2+\delta_2 ) \ \ \ \ \ (2)$

The result establish in \cite{paper} is following:

Theorem 3 (Liouville theorem for discrete harmonic functions in ${{\mathbb R}^2}$ ) Given ${c>0}$ . There exists a constant ${\epsilon>0}$ related to ${c}$ such that, given a discrete harmonic function ${f}$ in ${{\mathbb Z}^2}$ satisfied for any ball ${B_R(x_0)}$ with radius ${R>R_0}$ , there is ${1-\epsilon}$ portion of points ${x\in B_R(x_0)}$ satisfied ${|f(x)|<c}$ . then ${f}$ is a constant function in ${{\mathbb Z}^2}$ .

Remark 1 This type of result contradict to the intuition, at least there is no such result in ${{\mathbb C}}$ . For example. the existence of poisson kernel and the example given in \cite{paper} explain the issue.
Remark 2 There are reasons to explain why there could not have a result in ${{\mathbb C}}$ but in ${{\mathbb Z}^2}$ ,
1. The first reason is due to every radius ${R}$ there is only ${O(R^2)}$ lattices in ${B_R(x)}$ in ${{\mathbb Z}^2}$ so the mass could not concentrate very much in this setting.
2. The second one is due to there do not have infinite scale in ${{\mathbb Z}^2}$ but in ${{\mathbb C}}$ .
3. The third one is the function in ${{\mathbb Z}^2}$ is automatically locally integrable.
The generation is following:

Theorem 4 (Liouville theorem for discrete harmonic functions in ${{\mathbb R}^n}$ ) Given ${c>0,n\in {\mathbb N}}$ . There exists a constant ${\epsilon>0}$ related to ${n,c}$ such that, given a discrete harmonic function ${f}$ in ${{\mathbb Z}^n}$ satisfied for any ball ${B_R(x_0)}$ with radius ${R>R_0}$ , there is ${1-\epsilon}$ portion of points ${x\in B_R(x_0)}$ satisfied ${|f(x)|<c}$ . then ${f}$ is a constant function in ${{\mathbb Z}^n}$ .

In this note, I give a proof of 4, and explicit calculate a constant ${\epsilon_n>0}$ satisfied the condition in 3, this way could also calculate a constant ${\epsilon_n}$ satisfied 4. and point the constant calculate in this way is not optimal both in high dimension and 2 dimension.

2. some element properties with discrete harmonic function

We warm up with some naive property with discrete harmonic function. The behaviour of bad points could be controlled, just by isoperimetric inequality and maximum principle we have following result.

Definition 5 (Bad points) We divide points of ${{\mathbb Z}^n}$ into good part and bad part, good part ${I}$ is combine by all point ${x}$ such that ${|f(x)|<c}$ , and ${J}$ is the residue one. So ${A\amalg B={\mathbb Z}^n}$ .

For all ${B_R(0)}$ , we define ${J_R:=J\cap B_R(0), I_R=I\cap B_R(0)}$ for convenient.

Theorem 6 (The distribution of bad points) For all bad points ${J_R}$ in ${B_R(0)}$ , they will divide into several connected part, i.e.

$\displaystyle J_R=\amalg_{i\in S_R}A_i \ \ \ \ \ (3)$

and every part ${A_i}$ satisfied ${A_i\cap \partial B_R(0)\neq \emptyset}$ .

Remark 3 We say ${A}$ is connected in ${{\mathbb Z}^n}$ iff there is a path in ${A}$ connected ${x\rightarrow y, \forall x,y\in A}$ .

Remark 4 the meaning that every point So the behaviour of bad points are just like a tree structure given in the gragh.

Proof: A very naive observation is that for all ${\Omega\subset {\mathbb Z}^n}$ is a connected compact domain, then there is a function

$\displaystyle \lambda_{\Omega}: \partial \longrightarrow {\mathbb R} \ \ \ \ \ (4)$

such that ${\lambda_{\Omega}(x,y)\geq 0, \forall (x,y)\in \Omega \times \mathring{\Omega}}$ . And we have:

$\displaystyle f(x)=\sum_{y\in\partial \Omega}\lambda_{\Omega}(x,y)(y) \ \ \ \ \ (5)$

This could be proved by induction on the diameter if ${\Omega}$ . Then, if there is a connected component of ${\Omega}$ such that contradict to theorem 6 for simplify assume the connected component is just ${\Omega}$ , then use the formula 5we know

$\displaystyle \begin{array}{rcl} \sup_{x\in \Omega}|f(x)| & = & \sup_{x\in \Omega}\sum_{y\in\partial \Omega}\lambda_{\Omega}(x,y)(y) \\ & \leq & \sup_{\partial \Omega}|f(x)| \\ & \leq & c \end{array}$

The last line is due to consider around ${\partial \Omega}$ . But this lead to: ${\forall x\in \Omega, |f(x)|<c}$ which is contradict to the definition of ${\Omega}$ . So we get the proof. $\Box$

Now we begin another observation, that is the freedom of extension of discrete harmonic function in ${{\mathbb Z}^n}$ is limited.

Theorem 7 we can say something about the structure of harmonic function space of ${Z^n}$ , the cube, you will see, if add one value, then you get every value, i.e. we know the generation space of ${Z^n}$

Proof: For two dimension case, the proof is directly induce by the graph. The case of ${n}$ dimensional is similar. $\Box$

Remark 5 The generation space is well controlled. In fact is just like n orthogonal direction line in n dimensional case.

3. sktech of the proof for \ref

}

The proof is following, by looking at the following two different lemmas establish by two different ways, and get a contradiction.

\paragraph{First lemma}
Lemma 8 (Discrete poisson kernel) the poisson kernel in ${{\mathbb Z}^n}$ . We point out there is a discrete poisson kernel in ${{\mathbb Z}^n}$ , this is given by:

$\displaystyle f(x)=\sum_{y\in \partial B_R(z)}\lambda_{B_R(z)}(x,y)(y) \ \ \ \ \ (6)$

And the following properties is true:
1. ${\lambda_{B_R(z+h)}(x+h,y+h)=\lambda_{B_R(z)}(x,y)}$ , ${\forall x\in \Omega, h\in {\mathbb Z}^n}$ .
2. $\displaystyle \lambda_{B_R(z)}(x,y)\rightarrow \rho_R(x,y) \ \ \ \ \ (7)$
Remark 6 The proof could establish by central limit theorem, brown motion, see the material in the book of Stein \cite{stein}. The key point why this lemma 8 will be useful for the proof is due to this identity always true ${\forall x\in B_R(0)}$ , So we will gain a lots of identity, These identity carry information which is contract by another argument.

\paragraph{Second lemma} The exponent decrease of mass.

Lemma 9 The mass decrease at least for exponent rate.

Remark 7 the proof reduce to a random walk result and a careful look at level set, reduce to the worst case by brunn-minkowski inequality or isoperimetry inequality.

\paragraph{Final argument} By looking at lemma 1 and lemma 2, we will get a contradiction by following way, first the value of ${f}$ on ${\partial B_R(0)}$ increasing too fast, exponent increasing by lemma2, but on the other hand, it lie in the integral expresion involve with poisson kernel, but the pertubation of poisson kernel is slow, polynomial rate in fact…

\newpage

{99} \bibitem{paper} A DISCRETE HARMONIC FUNCTION BOUNDED ON A LARGE PORTION OF Z2 IS CONSTANT

\bibitem{stein} Functional analysis

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。

离散调和函数是连续调和函数在格点 $\mathbb Z^n$ 上的版本。它保留了平均值性质、最大值原理和随机游走解释，但也带来新的组合几何问题：如果一个离散调和函数在很多点上都不大，是否能推出它必须是常数？

离散调和函数的平均值性质可以用随机游走解释，坏点集合的几何受到最大值原理控制。

1. 定义

函数 $u:\mathbb Z^n\to\mathbb R$ 称为离散调和，如果对每个 $x\in\mathbb Z^n$，

$$u(x)=\frac1{2n}\sum_{y\sim x}u(y),$$

其中 $y\sim x$ 表示 $y$ 与 $x$ 相邻。等价地，离散 Laplacian

$$\Delta_d u(x)=\sum_{y\sim x}(u(y)-u(x))$$

满足 $\Delta_d u=0$。

在二维中，这就是

$$u(i,j)=\frac14\bigl(u(i+1,j)+u(i-1,j)+u(i,j+1)+u(i,j-1)\bigr).$$

2. 最大值原理

离散调和函数满足最大值原理：如果 $u$ 在有限连通区域内部调和，那么最大值和最小值出现在边界上。原因很简单：一个点的值是邻点平均，若内部点达到严格最大值，则所有邻点也必须取同样的值，连通性迫使整个区域常数。

这使得坏点集合的形状受到限制。设

$$G=\{x:|u(x)|\le 1\},\qquad B=\mathbb Z^n\setminus G.$$

如果某个坏点连通块完全被好点包围，那么最大值原理会迫使它不能真正坏。因此坏点必须以某种方式连接到边界或无穷远。

3. Poisson kernel 与随机游走

在有限区域 $\Omega\subset\mathbb Z^n$ 上，离散调和函数由边界值决定：

$$u(x)=\sum_{z\in\partial\Omega}P_\Omega(x,z)u(z).$$

这里 $P_\Omega(x,z)$ 是从 $x$ 出发的简单随机游走第一次离开 $\Omega$ 时落在 $z$ 的概率。这就是离散 Poisson kernel。

这个表示把分析问题转成概率问题：若边界上大值点所占比例很小，那么内部点看到大值的概率也会受到控制。

4. Liouville 型命题

经典 Liouville theorem 说，有界的整调和函数必须是常数。离散版本也有类似结论。更细的问题是：如果 $u$ 不假设处处有界，但在每个大球里都有固定比例的点满足 $|u|\le 1$，是否仍能推出 $u$ 是常数？

这类命题的证明通常要比较两个方向。第一，离散调和性和 Poisson 表示迫使内部值由边界平均控制。第二，若存在越来越大的坏点连通块，那么等周不等式或随机游走逃逸概率会给出大值传播。二者冲突时，就只能得到常数解。

5. 为什么维数和尺度重要

在 $\mathbb Z^n$ 中，每个半径球只有有限个格点，边界体积和内部体积之间有明确关系。坏点集合若想在所有尺度上保持稀疏，就很难同时支撑一个非平凡调和函数的增长。

这和连续情形的差别在于，离散空间把局部传播路径变成了组合对象。一个值要从边界影响内部，必须沿随机游走路径进入；而路径数量、逃逸概率和等周结构都会参与估计。

6. 证明图像

可以把证明想成两条 lemma 的冲突。一个 lemma 来自 Poisson kernel：内部值是边界值的随机平均，因此不能随意增长。另一个 lemma 来自坏点集合的几何：如果坏点在每个尺度都存在足够结构，它的质量会以某种速率传播。若假设“好点比例”在所有尺度上都足够大，这两种趋势最终矛盾。

这个问题的有趣之处在于，它把调和分析、随机游走和离散等周不等式绑在一起。离散调和函数不是连续理论的机械翻译，而是一种真正带有格点几何味道的分析对象。
2018年1月8日
对数平均 Sarnak 猜想：从 BSZ 准则到熵下降
旧博客原文

原题：Log average sarnak conjecture

This is a note concentrate on the log average Sarnak conjecture, after the work of Matomaki and Raziwill on the estimate of multiplication function of short interval. Given a overview of the presented tools and method dealing with this conjectue.

1. Introduction

Sarnak conjecture \cite{Sarnak} assert that for any obersevable ${\{f(T^n(x_0))\}_{n=1}^{\infty}}$ come from a determination systems ${(T,X),T:X\rightarrow X}$ , where ${h(T)=0}$ , ${x_0\in X, f\in C(X)}$ . The correlation of it and the Liuvillou function is 0, i.e. they are orthongonal to each other, more preseicesly it is just to say,

$\displaystyle \sum_{n<x}\mu(x)f(T^n(x_0))=o(x) \ \ \ \ \ (1)$

This is a very natural raised conjecture, Liuville function is the presentation of primes, due to we always believe the distribution of primes in ${\mathbb N}$ should be randomness.

It has been known as observed by Landau \cite{Laudau} that the simplest case,

$\displaystyle \sum_{n<x}\mu(n)=o(x)$

already equivalent to the prime number theorem. It is not difficult to deduce the spetial case of Sarnak conjecture when with the obersevation in $latex {(1)}&fg=000000$ come from finite dynamic system is equivalent to the prime number theorem in athremetic progress by the similar argument. Besides this two classical result, may be the first new result was established by Davenport,

Theorem 1 Let ${T:S_1\rightarrow S_1, T(x)=x+\alpha}$ , ${\alpha}$ is a inrational, then the obersevation come from ${(T,S_1)}$ is orthogonal to Mobius function. due to ${\{e^{2\pi ikx}\}_{k\in \mathbb Z}}$ is a basis of ${C(S_1)}$ , suffice to proof,

$\displaystyle \sum_{n<x}e^{2\pi ikn\alpha}\mu(n)=o(x), \forall k\in \mathbb N$

There is a lots of spetial situations of Sarnak’s conjecture have been established, The parts I mainly cared is the following:
1. Interval exchange map.
2. Skew product flow.
3. Obersevable come from One dimensional zero entropy flow.
4. Nilsequences.
But in this note, I do not want to explain the tecnical and tools to establish this result, but considering an equivalent conjecture of Sarnak conjecture, named Chowla conjecture, and explain the underlying insight of the suitable weak statement, i.e. the log average Chowla conjecture and the underlying insight of it.

The note is organized as following way, in the next section $latex {(2)}&fg=000000$, we give a self-contained introduction on the tools called Bourgain-Sarnak-Ziegler critation, explain the relationship of this critation and the sum-product phenomenon, also given some more general critation along the philosephy use in establish the Bourgain-Sarnak-Ziegler critation, which maybe useful in following development combine with some other tools. The key point is transform the sum from linear sum to bilinear sum and decomposition the bilinear sum into diagonal part and off-diagonal part, use the assume in the critation to argue the off-diagonal part is small and on the orther hand the diagonal part is also small by the trivial estimate and the volume of diogonal is small, this is very similar to a suitable Caderon-Zugmund decomposition.

In section $latex {(4)}&fg=000000$, I try to give a proof sketch of the result of Matomaki and Raziwill, which is also a key tools to understanding the Sarnak conjecture, or equivalent the Chowla conjecture. The key points of the proof contains following:
1. Find a suitable fourier indentity
2. Construct a multiplication-addition dense subset ${S}$ , and proof that the theorem MR hold we need only to proof it hold for ${S\cap [1,2,...,n]}$ instead of ${[1,2,...,n]}$
3. Involve the power of euler product formula. divide the whole interval into a lot of small interval with smaller and smaller scale and a residue part. We look the part come from every small scale as a major term and look the residue part as minor term.
4. Deal with the major term at every scale, by a combitorios identity and second moments method.
5. find a enough decay estimate from a scale to the next smaller scale.
6. Deal with the minor term by the H… lemma.
Due to the theorem of MR do not exausted the method they developed, we trying to make some more result with their method, Tao and Matomaki attain the average version of Chowla conjecture is true by this way, and combine this argument and the entropy decresment argument they established the 2 partten of the log average Chowla conjecture is true. Very recently Tao and his coperator proved the odd partten case of log average chowla conjecture is true, combine an argument of frustenberg crresponding principle and entopy decresment argument. But it seems the even and large than 2 case is much difficult and seems need something new to combine with the method of MR and entropy decresment and frunstenberg corresponfing principle to make some progress.

So, in section $latex {(5)}&fg=000000$, we give a self-contain introduction to the entropy decresment argument of Tao, and combine with the frustenberg corresponding principle.

In the last section $latex {(6)}&fg=000000$, I state some result and method and phylosphy of them I get on nilsequences and wish to combine them with the previous method to make some progress on log average Chowla conjecture on the even partten case.

\newpage

2. Bourgain-Sarnak-Zieglar creation

We begin with the easiest one, this is the main result established in \cite{BSZ}, I try to give the main ideal under the proof, but with a no quantitative version is the following,

Theorem 2 (Bourgain-Sarnak-Zieglar creation, not quantitative version) if for all primes ${p,q>>1}$ we have:

$\displaystyle \sum_{n=1}^Nf(T^{pn}(x))\overline{f(T^{qn}(x))}=o(N) \ \ \ \ \ (2)$

Then for multiplication function ${g(n)}$ we have

$\displaystyle \sum_{n=1}^Ng(n)\overline{ f(T^n(x))}=o(N) \ \ \ \ \ (3)$

Remark 1 For simplify we identify ${f(T^n(x)):=F(n)}$ .

Remark 2

The idea is following, break the sum into a bilinear one, so, of course, we multiplication it with itself. i.e. we consider to control,

$\displaystyle |\sum_{i=1}^Ng(n)\overline{ F(n)}|^2=\sum_{n=1}^N\sum_{m=1}^Ng(n)g(m)\overline{F(n)F(m)} \ \ \ \ \ (4)$

To control 4, we need exhausted the mutiplication property of ${g(n)}$ , we have ${g(mn)=g(n)g(m),\forall\ m,n\in {\mathbb N}}$ . We can not get good estimate for all term,

$\displaystyle g(n)g(m)\overline{F(n)F(m)} \ \ \ \ \ (5)$

The condition in our hand if following,

$\displaystyle \sum_{n=1}^NF(pn)\overline{F(qn)}=o(N), \forall \ p,q\in \mathop{\mathbb P} \ \ \ \ \ (6)$

So, just like the situation of Cotlar-Stein lemma \cite{Cotlar-Stein lemma}, we wish to estimate like following:

$\displaystyle \begin{array}{rcl} |\sum_{p\in W}\sum_{n\in V}F(pn)g(pn)| & \leq & \sum_{n\in V}|g(n)|\cdot |\sum_{p\in W}F(pn)g(p)| \\ & \leq &\sum_{n\in V}|\sum_{p \in W}F(pn)g(p)|\\ & \overset{Cauchy-Schwarz}\leq & |V|^{\frac{1}{2}}[\sum_{n\in V}|\sum_{p\in W}F(pn)g(p)|^2]^{\frac{1}{2}}\\ & = & |V|^{\frac{1}{2}}[\sum_{p_1,p_2\in W}\sum_{n\in V}F(p_1n)\overline{F(p_2n)}g(p_1)\overline{g(p_2)}]^{\frac{1}{2}}\\ \end{array}$

Then we consider divide the sum into diagonal part and non-diagonal part, as following,

$\displaystyle |V|^{\frac{1}{2}}[\sum_{p_1\neq p_2\in W}\sum_{n\in V}F(p_1n)\overline{F(p_2n)}g(p_1)\overline{g(p_2)}]^{\frac{1}{2}}+|V|^{\frac{1}{2}}[\sum_{p\in W}\sum_{n\in V}|F(pn)|^2]^{\frac{1}{2}} \ \ \ \ \ (7)$

But the first part is small, i.e.

$\displaystyle |V|^{\frac{1}{2}}[\sum_{p_1\neq p_2\in W}\sum_{n\in V}F(p_1n)\overline{F(p_2n)}g(p_1)\overline{g(p_2)}]^{\frac{1}{2}} =o(|W||V|) \ \ \ \ \ (8)$

Because of

$\displaystyle \sum_{n\in V}F(p_1n)\overline{F(p_2n)}=o(V), \forall p_1\neq p_2\in W \ \ \ \ \ (9)$

and the second part is small, i.e.

$\displaystyle |V|^{\frac{1}{2}}[\sum_{p\in W}\sum_{n\in V}|F(pn)|^2]^{\frac{1}{2}}=o(|W||V|) \ \ \ \ \ (10)$

Because diagonal part is small in ${W\times W}$ and trivial inequality

$\displaystyle \sqrt{\sum_{n\in V}|F(pn)|^2}\leq |V|^{\frac{1}{2}} \ \ \ \ \ (11)$

But the method in remark 2 is not always make sense in any situation, we need to construct two suitable sets ${W,V}$ and then break up ${\{1,2,...,n{\mathbb N}\}}$ into ${W\times V}$ , this mean,

$\displaystyle \{1,2,...,N\}\sim W\times V+o(N) \ \ \ \ \ (12)$

But this ${W,V}$ could be construct in this situation, thanks to the prime number theorem,

Theorem 3 (Prime number theorem)

$\displaystyle \pi(n)\sim \frac{n}{ln(n)} \ \ \ \ \ (13)$

Morally speaking, this is the statement that the primes, which is the generator of multiplication function, is not very sparse.

3. Van der curpurt trick

There is the statement of Van der carport theorem:

Theorem 4 (Van der curpurt trick) Given a sequences ${ \{x_n\}_{n=1}^{\infty}}$ in ${ S_1}$ , if ${ \forall k\in N^*}$ , ${ \{x_{n+k}-x_n\}}$ is uniformly distributed, then ${ \{x_n\}_{n=1}^{\infty}}$ is uniformly distributed.

I do not know how to establish this theorem with no extra condition, but this result is true at least for polynomial flow. \newpage Proof:

$\displaystyle \begin{array}{rcl} |\sum_{n=1}^Ne^{2\pi imQ(n)}|& = &\sqrt{(\sum_{n=1}^Ne^{2\pi imQ(n)})(\overline{\sum_{n=1}^Ne^{2\pi imQ(n)}})}\\ & = &\sqrt{\sum_{h_1=1}^N\sum_{n=1}^{N-h_1}e^{2\pi imQ(n+h_1)-Q(n)}}\\ & = &\sqrt{\sum_{h_1=1}^N\sum_{n=1}^{N-h_1}e^{2\pi im \partial^1_{h_1}Q(n)}}\\ & \leq & \sqrt{\sum_{h_1=1}^N|\sum_{n=1}^{N-h_1}e^{2\pi \partial^1_{h_1}Q(n)}|}\\ & = &\sqrt{\sum_{h_1=1}^N\sqrt{ (\sum_{n=1}^{N-h_1}e^{2\pi \partial^1_{h_1}Q(n)} )(\overline{\sum_{n=1}^{N-h}e^{2\pi \partial^1_{h_1}Q(n)})}}}\leq\sqrt{\sum_{h_1=1}^N\sqrt{ \sum_{h_2=1}^N|\sum_{n=1}^{N-h_1}e^{2\pi\partial^1_{h_2} \partial^1_hQ(n)} |}}\\ & \leq ....\leq & \\ & = & \sqrt{\sum_{h_1=1}^N\sqrt{ \sum_{h_2=1}^N \sqrt{....\sqrt{\sum_{h_{k-1}=1}^{N-h_{k-2}}|\sum_{n=1}^{N-h_{k-1}}e^{2\pi\partial_{h_1h_2...h_{k-1}Q(n)}}|}}}} =o(1) \end{array}$

$\Box$

This type of trick could also establish the following result, which could be understand as a discretization of the Vinegradov lemma.

Remark 3

Uniformly distribution result of ${ F_p}$ : Given ${Q(n)=a_kn^k+...+a_1n+a_0}$ , ${\{Q(0),Q(1),...,Q(p-1)\}}$ coverages to a uniformly distribution in ${\{0,1,...,p-1\}}$ as ${p \rightarrow \infty}$ .

Remark 4 But I definitely do not know how to establish the similar result when ${Q(n)=n^{-1}}$ .

Remark 5

This trick could also help to establish estimate of correlation of low complexity sequences and multiplicative function, such as result:

$\displaystyle S(x)=\sum_{n\le x}\left(\frac{n}{p}\right)\mu(n)=o(n)$

Maybe with the help of B-Z-S theorem.

\newpage

4. Matomaki and Raziwill’s work

In this section we explain the main idea underlying the paper \cite{KAISA MATOMA 虉KI AND MAKSYM RADZIWILL}. But play with a toy model, i.e. the corresponding corollary of the original result on Liouville鈥檚 function.

Definition 5 (Lioville’s function)

$\displaystyle \lambda(n)=(-1)^{\alpha_1+\alpha_2+...+\alpha_k}, \forall \ n=p_1^{\alpha_1}...p_k^{\alpha_k}. \ \ \ \ \ (14)$

Remark 6

$\displaystyle |\int_{X}^{2X}\lambda(n)dx|=o(x) \ \ \ \ \ (15)$

is equivalent to the prime number theorem 3.

The most important beakgrouth of analytic number theory is the new understanding of multiplication function on share interval, this result is established by Kaisa Matom盲ki and Maksym Radziwill. Two very young and intelligent superstars.

The main theorem in them article is :

Theorem 6 (Matomaki,Radziwill) As soon as ${H\rightarrow \infty}$ when ${x\rightarrow \infty}$ , one has:

$\displaystyle \sum_{x\leq n\leq x+H}\lambda(n)= o(H) \ \ \ \ \ (16)$

for almost all ${1\leq x\leq X}$ .

In my understanding of the result, the main strategy is:
1. Parseval indetity, transform to Dirchelet polynomial.
2. Involved by multiplication property, spectral decomposition.
3. From linear to multilinear , Cauchy schwarz inequality.
4. major term estimate.
5. Estimate the contribution of area which is not filled.
4.1. Parseval indetity, transform to Dirchelet polynomial

We wish to establish the equality,

$\displaystyle \frac{1}{X}\int_{X}^{2X}|\sum_{x\leq n\leq x+H}\lambda(n)|dx=o(H) \ \ \ \ \ (17)$

This is the ${L^1}$ norm, by Chebyschev inequality, this could be control by ${L^2}$ norm, so we only need to establish the following,

$\displaystyle \frac{1}{X}\int_X^{2 X}|\sum_{x\leq n\leq x+H}\lambda(n)|^2dx=o(H^2) \ \ \ \ \ (18)$

We wish to transform from the discretization sum to a continue sum, that is,

$\displaystyle \int_{{\mathbb R}}|\sum_{xe^{-\frac{1}{T}}\leq n\leq xe^{\frac{1}{T}}}\lambda(n)1_{X\leq n\leq 2X}|^2\frac{dx}{x} \ \ \ \ \ (19)$
Remark 7 There are two points to understand why 19 and 18 are the same.
1. ${[xe^{-\frac{1}{T}},xe^{\frac{1}{T}}]\sim [x-H,x+H]}$ .
2. ${1_{x\leq n\leq 2x}}$ and ${\frac{1}{x}}$ is to make that ${x=O(X)}$ .
So the Magnitude of 18 and 19 are the same. i.e.

$\displaystyle \int_{{\mathbb R}}|\sum_{xe^{-\frac{1}{T}}\leq n\leq xe^{\frac{1}{T}}}\lambda(n)1_{X\leq n\leq 2X}|^2\frac{dx}{x}\sim \frac{1}{X}\int_X^{2 X}|\sum_{x\leq n\leq x+H}\lambda(n)|^2dx \ \ \ \ \ (20)$
Now we try to transform 19 by Parseval indetity, this is something about the ${L^2}$ norms of the quality we wish to charge. It is just trying to understanding 19 as a quantity in physical space by a more chargeable quality in frequency space. Image,

$\displaystyle \int_{{\mathbb R}}|\sum_{xe^{-\frac{1}{T}}\leq n\leq xe^{\frac{1}{T}}}\lambda(n)1_{X\leq n\leq 2X}|^2\frac{dx}{x}:=\int_{{\mathbb R}}|f_X(x)|^2dx \ \ \ \ \ (21)$

Then ${f_X(x)=\int_{xe^{-\frac{1}{T}}\leq n\leq xe^{\frac{1}{T}}}\lambda(x)1_{X\leq n\leq 2X}}$ . Note that,

$\displaystyle \begin{array}{rcl} \widehat{f_X(\xi)} & = & \int_{{\mathbb R}}f_X(x)e^{2\pi ix\xi}dx\\ & = & \sum_{x\leq n\leq 2x}\lambda(x)\int_{logn-\frac{1}{T}}^{logn+\frac{1}{T}}e^{2\pi ix\xi}dx, \ T=\frac{X}{H}\\ & = & \sum_{X\leq n\leq 2X}\lambda(x)e^{2\pi ilog(n)\cdot \xi}\cdot\frac{e^{2\pi i\frac{\xi}{T}}-e^{2\pi i-\frac{\xi}{T}}}{2\pi i\xi}\\ \end{array}$

So by Parseval identity, we have,

$\displaystyle \begin{array}{rcl} \int_{{\mathbb R}}|f_X(x)|^2dx & = & \int_{{\mathbb R}}|\widehat{f_X(\xi)}|^2d\xi \\ & = & \int_{{\mathbb R}}|\sum_{X\leq n\leq 2X}\lambda(n)\cdot n^{2\pi i\xi}|^2(\frac{e^{2\pi i\frac{\xi}{T}}-e^{2\pi i\frac{-\xi}{T}}}{2\pi i\xi})^2d\xi\\ & \sim & \int_{{\mathbb R}}|\sum_{X\leq n\leq 2X}\lambda(n)\cdot n^{2\pi i\xi}|^2\frac{1}{T^2}1_{|\xi|^2\leq T}\\ \end{array}$

Remark 8 We know the Fejer kernel satisfied,

$\displaystyle (\frac{e^{2\pi i\frac{\xi}{T}}-e^{2\pi i\frac{-\xi}{T}}}{2\pi i\xi})^2\sim \frac{1}{T^2}1_{|\xi|\leq T} \ \ \ \ \ (22)$

So morally speaking, we get the following identity.

$\displaystyle \frac{1}{X}\int_{X}^{2X}|\sum_{x\leq n\leq x+H}\lambda(n)|^2dx\sim \frac{1}{(x/H)^2}\int_{0}^{\frac{X}{H}}|\sum_{x\leq n\leq 2x}\lambda(x)x^{2\pi i\xi}|^2d\xi \ \ \ \ \ (23)$

In fact we do a cutoff, the quality we really consider is just:

$\displaystyle \frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt \ \ \ \ \ (24)$

established the monotonically inequality:

Theorem 7 (Paserval type identity)

$\displaystyle \frac{1}{X}\int_{X}^{2X}|\frac{1}{H}\sum_{x\leq n\leq x+H}\lambda(n)|^2dx \sim聽\frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt \ \ \ \ \ (25)$

Remark 9

In my understanding, This is a perspective of the quality, due to the quality is a multiplicative function integral on a domain ${ \mathbb N^*}$ with additive structure, it could be looked as a lots of wave with the periodic given by primes, so we could do a orthogonal decomposition in the fractional space, try to prove the cutoff is a error term and we get such a monotonically inequality.

But at once we get the monotonically inequality, we could look it as a聽compactification process and this process still carry most of the information so lead to the inequality.

It seems something similar occur in the attack of the moments estimate of zeta function by the second author. And it is also could be looked as something similar to the 聽spectral decomposition with some basis come from multiplication generators, i.e. primes.

4.2. Involved by multiplication property, spectral decomposition

I called it is “spectral decomposition”, but this is not very exact. Anyway, the thing I want to say is that for multiplication function ${\lambda(n)}$ , we have Euler-product formula:

$\displaystyle \Pi_{p,prime}(\frac{1}{1-\frac{\lambda(p)}{p^s}})=\sum_{n=1}^{\infty} \frac{\lambda(n)}{n^s} \ \ \ \ \ (26)$

But anyway, we do not use the whole power of multiplication just use it on primes, i.e. ${\lambda(pn)=\lambda(p)\lambda(n)}$ leads to following result:

$\displaystyle \lambda(n)=\sum_{n=pm,p\in I}\frac{\lambda(p)\lambda(m)}{\# \{p|m, p\in I\}+1}+\lambda(n)1_{p|n;p\notin I} \ \ \ \ \ (27)$

This is a identity about the function ${\lambda(n)}$ , the point is it is not just use the multiplication at a point,i.e. ${\lambda(mn)=\lambda(m)\lambda(n)}$ , but take average at a area which is natural generated and compatible with multiplication, this identity carry a lot of information of the multiplicative property. Which is crucial to get a good estimate for the quality we consider about.

4.3. From linear to multilinear , Cauchy schwarz

Now, we do not use one sets ${I}$ , but use several sets ${I_1,...,I_n }$ which is carefully chosen. And we do not consider [X,2X] with linear structure anymore , instead reconsider the decomposition:

${[X,2X]=\amalg_{i=1}^n (I_i\times J_i) \amalg U}$

On every ${I_i\times J_i}$ it equipped with a bilinear structure. And ${U}$ is a very small set, ${|U|=o(X)}$ which is in fact have much better estimate.

${\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt =\sum_{i=1}^n\int_{I_i\times J_i}聽聽\frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt +\int_N |\sum_{n\leq X}\lambda(n)n^{it}|^2dt}$

Now we just use a Cauchy-Schwarz:

${\sum_{i=1}^n\int_{I_i\times J_i}聽聽\frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt +\int_N |\sum_{n\leq X}\lambda(n)n^{it}|^2dt}$

4.4. major term estimate

${=\sum_{i=1}^n\int_{I_i\times J_i}聽聽\frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt}$

${\int_N |\sum_{n\leq X}\lambda(n)n^{it}|^2dt}$

4.5. estimate the contribution of area which is not filled

\newpage

5. Entropy dcrement argument

\newpage

6. Correlation with nilsequences

I wish to establish the following estimate: ${\lambda(n)}$ is the liouville function we wish the following estimate is true.

$\displaystyle \int_{0\leq x\leq X}|\sup_{f\in \Omega^m}\sum_{x\leq n\leq x+H}\lambda(n)e^{2\pi if(x)}|dx =o(XH). \ \ \ \ \ (28)$

Where we have ${ H\rightarrow \infty}$ as ${ x\rightarrow \infty}$ ,

$\displaystyle \Omega^m=\{a_mx^m+a_{m-1}x^{m-1}+...+a_1x+a_0 | a_m,...,a_1,a_0\in [0,1]\}$

is a compact space.

I do not know how to prove this but this is result is valuable to consider, because by a Fourier identity we could transform the difficulty of (log average) Chowla conjecture to this type of result.

There is some clue to show this type of result could be true, the first one is the result established by Matomaki and Raziwill in 2015:

Theorem 8 (multiplication function in short interval)

${f(n): \mathbb N\rightarrow \mathbb C}$ is a multiplicative function, i.e. ${ f(mn)=f(n)f(m), \forall m,n\in \mathbb N}$ . ${H\rightarrow \infty}$ as ${x\rightarrow \infty}$ , then we have the following result,

$\displaystyle \int_{1\leq x\leq X}|\sum_{x\leq n\leq x+H}f(n)|=o(XH). \ \ \ \ \ (29)$

And there also exists the result which could be established by Vinagrodov estimate and B-S-Z critation :

Theorem 9 (correlation of multiplication function and nil-sequences in long interval)

${f(n): \mathbb N\rightarrow \mathbb C}$ is a multiplicative function, i.e. ${ f(mn)=f(n)f(m), \forall m,n\in \mathbb N}$ . ${g(n)=a_n^m+...+a_1n+a_0}$ is a polynomial function then we have the following result,

$\displaystyle \int_{1\leq n \leq X}|f(n)e^{2\pi i g(n)}|=o(X) \ \ \ \ \ (30)$

\newpage {9} \bibitem{Sarnak} Peter Sarnak, Mobius Randomness and Dynamics.

\texttt{https://publications.ias.edu/sites/default/files/Mahler }. \bibitem{Laudau} JA 虂NOS PINTZ (BUDAPEST). LANDAU鈥橲 PROBLEMS ON PRIMES.

\texttt{https://users.renyi.hu/~pintz/pjapr.pdf} \bibitem{BSZ} Knuth: Computers and Typesetting,

\texttt{http://www-cs-faculty.stanford.edu/\~{}uno/abcde.html}

\bibitem{Cotlar-Stein lemma} Almost orthogonality

\texttt{https://hxypqr.wordpress.com/2017/12/18/almost-orthogonality/}

\bibitem{KAISA MATOMA 虉KI AND MAKSYM RADZIWILL} KAISA MATOMA 虉KI AND MAKSYM RADZIWIL, MULTIPLICATIVE FUNCTIONS IN SHORT INTERVALS.

\texttt{https://arxiv.org/abs/1501.04585v4/}.

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。

这篇笔记想整理的是 Sarnak 猜想的一条现代路线：从莫比乌斯函数和零熵动力系统的正交性出发，经过 Bourgain-Sarnak-Ziegler 准则，把线性相关和转成双线性相关；再借助 Matomaki-Radziwill 的短区间乘法函数估计，以及 Tao 的 entropy decrement，把问题推向对数平均 Chowla 猜想。

Sarnak/Chowla 的对数平均路线：线性相关先转成双线性相关，再用短区间估计和熵下降选择合适尺度。

1. Sarnak 猜想的基本形状

Sarnak 猜想说，如果 $(X,T)$ 是零拓扑熵动力系统，$f\in C(X)$，那么对任意 $x\in X$，莫比乌斯函数与观测序列 $f(T^n x)$ 应该正交：

$$\frac1N\sum_{n\le N}\mu(n)f(T^n x)\to 0.$$

这里的哲学是：$\mu(n)$ 携带素数分布中的振荡，而零熵系统产生的是低复杂度的确定序列；二者不应该长期对齐。

最简单的有限动力系统情形已经包含素数定理和算术级数中的素数定理。圆周无理旋转情形则接近 Davenport 型估计：对任意无理数 $\alpha$，指数序列 $e(n\alpha)$ 与 $\mu(n)$ 的相关和具有消失。

2. 从 Sarnak 到 Chowla

Chowla 猜想更像是对 $\mu$ 或 Liouville 函数自身随机性的表述。一个典型的 $k$ 点相关形式是

$$\frac1N\sum_{n\le N}\lambda(n+h_1)\cdots \lambda(n+h_k)\to 0,$$

其中 $h_i$ 两两不同。Sarnak 关心的是乘法函数与低复杂度确定序列的相关，Chowla 关心的是乘法函数自身不同平移之间的相关。二者之间可以用 Furstenberg correspondence principle 和动力系统模型联系起来。

对数平均版本把普通平均换成

$$\frac1{\log N}\sum_{n\le N}\frac{a(n)}n.$$

这个权重让尺度选择更加稳定，也更适合短区间分析。很多情况下，对数平均结论比普通平均结论先被证明，因为它容许把不同尺度上的误差以更柔和的方式叠加。

3. Bourgain-Sarnak-Ziegler 准则

BSZ 准则的核心是一个线性到双线性的转换。设 $a_n$ 是有界序列，如果对很多不同素数 $p\ne q$，都有

$$\frac1N\sum_{n\le N}a_{pn}\overline{a_{qn}}\to 0,$$

那么可以推出

$$\frac1N\sum_{n\le N}\mu(n)a_n\to 0.$$

直观上，$\mu$ 的乘法结构允许我们把原来的线性相关拆成不同素数伸缩后的相关。对角部分 $p=q$ 用平凡估计和体积小来处理；非对角部分则由上面的假设控制。这一点和 Calderon-Zygmund 分解的精神很像：把难对象拆成一个小的对角坏集和一个可估计的非对角主体。

4. Matomaki-Radziwill 的短区间输入

Matomaki-Radziwill 的工作说明，乘法函数在大多数短区间里的平均行为可以被控制。粗略地说，对许多短区间 $[x,x+H]$，有

$$\frac1H\sum_{x
接近它在长区间中的平均。这使得我们可以把一个全局相关和切成许多短尺度，再用欧拉乘积、二阶矩和组合恒等式逐层比较。

短区间估计真正有用的地方在于：它让乘法函数的局部随机性可以被拿来攻击动力系统中的相关问题，而不是只停留在平均阶的数论命题。

5. Entropy decrement 的作用

entropy decrement 的思想是寻找一个尺度，使得随机变量 $n$ 与它的素数倍 $pn$ 之间的条件信息变少。换句话说，在合适尺度上，系统看到的结构不会因为乘一个小素数而增加太多复杂度。

这一步的意义是把“乘法平移”转成“动力系统里可比较的两个观测”。当这个信息损失足够小的时候，短区间估计、BSZ 双线性结构和 correspondence principle 就可以接上。

6. Nilsystem 方向

nilsequence 是零熵系统中非常重要的一类模型。它既有足够丰富的几何结构，又保留了可计算的 Fourier 分析。把 Sarnak、Chowla、短区间乘法函数和 nilsequence 放在一起看，真正的问题是：乘法函数的随机性如何穿过 nilmanifold 上的低复杂度轨道。

这条路线目前最有价值的地方，不是把所有情形一次性解决，而是提供了一张方法地图：线性相关转双线性相关；短区间估计提供局部随机性；entropy decrement 选择尺度；Furstenberg 原理把数论相关放回动力系统。
2018年1月8日
Laplace 谱、Rayleigh quotient 与无限维 Morse 图像

旧博客原文

原题：Two stupid question

The story of the infinite dimensional space of $\Delta$ is following, we eliminate ourself with compact smooth non-boundary manifold $M$ with metric $g$, then we have Betrami-Laplace operator $\Delta_g$. We could instead $\Delta_g$ by hodge laplace $dd^*+d^*d$, but let we consider $\Delta_g$ the eigenvalue problem:
$$\Delta_g u=\lambda u$$
A classical way to investigate the eigenvalue problem is according to consider variational principle and max-min principle. We equip the path integral on the function space $C^{\infty}(M)$:
$$E(f)=\frac{\int_M |\nabla u|^2}{\int_M |u|^2 }$$
Then it have a sequences of eigenvalue, negative of course: $$0<-\lambda_1<-\lambda_2<…<\lambda_k<…$$

Then things become interesting, the morse theory of infinite space involve, called the infinite space as $X$, so at least, shrink the far place of $X$ as a point, in physics, this mean, cut off at fix scale. And we can take the scale to infinite small, we use the cutoff one to approximation the real one. What I can do is the following, I can proof the eigenvalue function is uniformly distributed in $L^2(M_g)$ (after rescaling of course) and the classical weyl law(although can not give a good error term estimate), but thing become more complicated when I try to consider the infinite space $X_{M_g}$’s topology, at finite scale at least, i.e. $X_{M_g}^{h}$ which is the cut off at scale $h$. Among the other thing, I believe the following issue is true, but without ability to proof it:

>**Problem**
for every manifold $M$ and metric $g$ on $M$, the topology of infinite space $X_{M_g}$ is the same, beside this, the inverse could be true, i.e. If $X_{M_1},X_{M_2}$ is not homomorphism for some scale $h$ then $M_1,M_2$ is not homomorphism.

By intuition, I think it is depend by the underling manifold’s topology. But I do not have a rigorous proof, I definitely have a non-rigorous one, if ignore the coverage…

As I find this problem when I try to give a proof of weyl law, I do not check the reference, may be this problem is a classical one? As always, I will appreciate to any interesting comments and answers, thanks a lots!A

2.

We begin with our favorite situation, the Dirchlet problem on bounded simple-connected domain $\Omega$ in $\mathbb R^n$. Let $\lambda_1$ be the first eigenvalue of $$\Delta u=\lambda u \ in\ \Omega$$
$$u=0\ \ on\ \partial\Omega$$
Rescaling $u$ such that $\sup_{\Omega} u=1$, I think the following property of the first eigenvalue is true.
>**Problem**
We have, the Minkowski functional of $\Omega$, called $M_{\Omega}$ and the Minkowski functional with the ball $B$ such that $vol(B)=vol(\Omega)$, then along the level set of $u$, i.e. the fiber: $$\Omega=\cup_{t\in [0,1]}l_t, l_t:=\{t|x\in \Omega, u(x)=t\}$$
We pretend for the isolate point $l_1$ to be a ball with radius 0, so equipped it with the uniformly density at every direction in $S^1$, i.e. the mass distribution given by $M_B$ and the total mass coincide with the total mass induce by $M_{\Omega}$ in $l_0$, i.e.
$$\int_{e\in S_1}M_{\Omega}(e)d\mu=\int_{e\in S_1}M_{B}(e)d\mu$$
The measure $d\mu$ equipped on $S^1$ is the natural Haar measure. And the cost function is given by $c(x,y)=\|x-y\|^2$. Then, among this setting,
I wish the following property to be true:
Along the direction $1\to 0$, the transport of density $\partial_{t_0} M_{\cup_{t=t_0}^1l_t}$ given the unique optimal transport of the natural measure induce by $M(\Omega)$ and $M(B)$.

**Remark 1** As point out by SebastianGoette, the multiplicity of the first eigenvalue must be one, thanks to the eigenfunction never change the symbol, so we are in the best case.

**Remark 2**:I am not very sure this property could always true, there may be a center example when $\Omega$ is not convex, but I tend to believe it is true at least when $\Omega$ is convex.

**Remark 3**: As point out by Dirk, when you try to consider the optimal transport problem, you always need to point out the cost function $c(x,y)$ defined on $\Omega \times \Omega$, for there, I think the naive choice is $c(x,y)=\|x-y\|^2$

The thing I can proof is the following, the level set of $u$ should be convex by brunn-minkowski inequality, and some type of monotonically property, i.e. more and more like a ball when the level set is more and more shirking smaller form $\partial \Omega$ to the point $f$ arrive maximum.

I will appreciate for any relevant comments and answer, thanks!

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。

Laplace 特征值问题可以看成无限维函数空间上的变分问题。Rayleigh quotient 的临界点给出 eigenfunctions，min-max principle 给出 eigenvalues。这种图像很像 Morse theory，只是空间变成了无限维。

Laplace 特征值可以由 Rayleigh quotient 的 min-max 原理得到，形成无限维 Morse 图像。

1. Laplace 特征值问题

在紧无边界 Riemannian manifold 上，考虑

$$-\Delta_g u=\lambda u.$$

谱离散，特征值可以排成

$$0=\lambda_0<\lambda_1\le\lambda_2\le\cdots\to\infty.$$

2. Rayleigh quotient

定义

$$E(u)=\frac{\int_M|\nabla u|^2\,dV_g}{\int_M|u|^2\,dV_g}.$$

在 $L^2$ 单位球上，$E$ 的临界点正是 Laplace eigenfunctions。第一非零特征值是与常数正交的函数中 $E$ 的最小值。

3. Min-max principle

第 $k$ 个特征值可由

$$\lambda_k=\inf_{\dim V=k+1}\sup_{u\in V\setminus\{0\}}E(u)$$

给出。这个公式把谱问题转成函数空间中的拓扑/变分问题。

4. 无限维 Morse 图像

在有限维 Morse theory 中，临界点和拓扑变化有关。Laplace 特征函数也可以看成 Rayleigh quotient 的临界点；其 Morse index 与低于该特征值的谱空间维数相关。

5. 与 Weyl law 的关系

Weyl law 告诉我们高能临界点的数量增长率。变分图像解释 eigenfunctions 从哪里来，Weyl law 则解释这些临界点在高频区域有多密。

2017年12月31日
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