大二下作业

有限群表示论

Sun, 04 May 2025 15:11:49 GMT

参见肖梁, Artin, Dummit.
A matrix representation of a group is a homomorphism

from to one of the complex general linear groups. The number is the dimension of the representation.
We use the notation instead of for the image of a group element . Each is an invertible matrix, and the statement that is a homomorphism reads

If a group is given by generators and relations, say , a matrix representation can be defined by assigning matrices that satisfy the relations. For example, the symmetric group can be presented as , so a representation of is defined by matrices and such that , and . Some relations in addition to these required ones may hold.

If is finite, then any finite dimensional representation is "completely reducible".

There is a natural question: can we find a "good" matrix representation for a given representation of ? Here "good" means there is a positive definite Hermitian form on so that every preserves this Hermitian form; and thus the image of in belongs to the unitary group .
Let be a Hermitian space -- a complex vector space together with a positive definite Hermitian form . A unitary operator on is a linear operator with the property

A representation on a Hermitian space is called unitary if is a unitary operator for every . We can write this condition as

Similarly, a matrix representation is unitary if . A unitary matrix representation is a homomorphism from to the unitary group:

Lemma. Let be a unitary representation of on a Hermitian space , and let be a -invariant subspace. The orthogonal complement is also -invariant, and is the direct sum of its restrictions to the Hermitian spaces and . These restrictions are also unitary representations.
Proof.
It is true that . Since is unitary, it preserves orthogonality: If is invariant and , then . This means that if , then .
□

In matrix terminology, two representation and are equivalent if there is a fixed invertible matrix such that

In -module terminology, two representation and are equivalent if there is a fixed -module isomorphism such that

The linear transformation or the matrix above is said to interwine the representation and .

Because they involve several matrices, each of which may have many entries, representations are notationally complicated. The secret to understanding them is to throw out most of the information that the matrices contain, keeping only one essential part, its trace, or character.
Our slogan is: characters determine the representation.

For , we have three representaions, , and .

Then the characters of these representations are displayed below in tabular form.

Several interesting phenomena can be observed in this table:

The rows form orthogonal vectors of length equal to six, which is also the order of . The columns are orthogonal too.
is the dimension (degree) of the representation, also called the dimension of the character.
Since a representation is a homomorphism, it sends the identity in the group to the identity matrix. So is the trace of the identity matrix.
The characters are constant on conjugacy classes.
The conjugacy classes in are the sets , and .
This is because conjugate matrices has the same trace. ()

FG-modules

Fri, 02 May 2025 08:58:54 GMT

See Representations and Characters of Groups (Gordon James, Martin Liebeck)
We will show that there is a connection between -modules and representations of over .
Let be a group and let be or .
Suppose that is a representation of . Write , as the vector space. For all and , the matrix product[1]

is a row vector in .
Now we list some basic properties

Motivatied by the above observations on the product , we now define an -modules.
Definition (-module). Let be a vector space over and let be a group. Then is an -module if a multiplication is defined, satisfying the following conditions for all and :
;
;
;
;
.

Definition (Matrix representation). Let be an -module, and let be a basis of . For each , let denote the matrix of the endomorphism of , relative to the basis .
If is a representation of over , and , then becomes an -module if we define the multiplication by

Moreover, there is a basis of such that

On the other hand, assume that is an -module and let a basis of . Then the function

is a representation of over .

Definition (faithful). An -module is faithful if the identity element of is the only element for which
For instance, the -module which appears in Example is faithful.
Our next aim is to construct faithful -modules by the basis for all subgroups of symmetric groups.
Definition (Permutation module). Let be a subgroup of . The -module with basis such that is called the permutation module for over . We call the natural basis of .

Exercise.
When , then . When , then

Now let , , . Then

Then

Omitted....
Exercise.
Let . Then

Check that is a -module. NTS: is a representation of , i.e.

which is routine.

Definition ( -submodule). Let be an -module. A subset of is said to be an -submodule of if is a subspace and for all and all .

Definition (Irreducible FG-module). An irreducible FG-module is said to be irreducible if it is non-zero and it has no -submodules apart from and .

Omitted....

↩︎

Group algebra

Fri, 02 May 2025 08:58:54 GMT

The group algebras are the source of all you need to know about representatio theory. In particular, the ultimate goal of representation theory -- that of understanding all the representations of finite groups -- would be achieved if group algebras could be fully analysed.
Definition. Let be a finite group and be or . The vector space , with the natural multiplication , is called the regular -module. The representation obtained by taking to be the natural basis of is called the regular representation of over .
Definition (group algebra). The vector space , with multiplication defined by is called the group algebra of over .
Recall that an -module is a vector space over , together with a multiplication for and . Now we can expend the definition of the multiplication so that we have an element of for all elements in the group algebra .
Suppose that is an -module, and that and ; say

Define by

Omitted...

group-representation

Sat, 10 May 2025 04:22:36 GMT

is a finite group.

Let be a 1-dimensional vector space over and make into an -module by letting for all and . This module affords the representation

to be trivial, i.e. taking to (the identity matrix). The trivial representation has degree 1 and if , it is not faithful.

Let be a -module, then affords a representation

where . has degree equal to , and it's always faithful. The matrix has a 1 in row and column if , and has 0's in all other positions.

Let , be an -dimensional vector space over with basis . Let act on by defining for each

This provides an injective homomorphism of into . Hence is an -module. has a 1 in row and column if .

If is any representation of and is any group homomorphism, then the composition is a representation of .

Any homomorphism of into the multiplication group is a degree 1 (matrix) representation.
For example, suppose is the cyclic group of order and is a fixed root of 1 in . Let , for all . This representation of is a faithful representation if and only if is a primitive root of 1.

For , it has the faithful representation

For , it has the faithful -linear representation

And a 4-dimensional faithful -linear representation is given by

Let , , .

Suppose and is an elementary abelian -group for some prime . Let be a vector space over , where scalar acts on the group element by . The conjugation action of on is -linear because

Then is an -module. The kernel of thsi representation is .

See Representations and Characters of Groups Gordon James.

Let . Define the matrices over by

Prove that each of the functions , given by

is a representation of . Which of these representations are faithful? Which are equivalent?
Proof.
It suffices to check that

which is routine. Since

are faithful.

Since their matrix representations have the same Jordan Canonical form.
□

Let be the dihedral group . Define the matrices and by

And check that

It follows (see Example 1.4) that the function which is given by

is a representation of over . The degree of is .
The matrices for in are given in the following table:

Let

Then

So we obtain another representation of for which

Let be a representation of the group . Suppose that and are elements of such that . Does it follow that

If is faithful then . Otherwise, may not commute.

Exercise (Yau-2024).
(1)
, . Let be a -vector space whose basis is . The dimension of is . More formally,

is a finite vector space over . Define the mapping (we will show that it's a representation),

where . We can extend the action linearly to any , by letting

To show that it's a homomorphism, we have

To show that the representation is faithful, consider the kernel

For ,

Then we have

Thus . The representation is faithful.
(2)
If is cyclic with order , and , then the mapping

is a faithful representation.
If has a faithful one-dimensional complex representation,

, then . Thus every lies in the set of roots of unity, which has size . Since is faithful, for . Then is cyclic. Hence is cyclic.
(3)
If can be generated by , then for any , we have

Assume that for every , then the mapping

is a faithful -dimensional complex representation.
If has a faithful -dimensional complex representation,

can be a diagonized matrix.
....
(4)
具有忠实的二维实表示的有限群的分类如下：

循环群（ n 阶循环群），对于任意。
二面体群（ 2 m 阶二面体群，即正 m 边形的对称群），对于任意。
Remark. Let be a finite dimensional -representation of a finite group . Consider the -invariant subspace
Show that is the same as the multiplicity of the trivial representation appearing in .
Show that .
Construct a surjective map , expressed in terms of a linear combination of linear operators for , such that (i.e. is a projection) and is a homomorphism.

Representation Theory and Character Theory

Sun, 04 May 2025 03:01:58 GMT

is a ring, is a group then the group ring is defined by

with the following properties:

is a commutative ring iff is an abelian group.
is a vector space over with the elements of as a basis. ()

Formally, a representation is faithful if the only element of that is mapped to the identity transformation is the identity element of , i.e.,

where is the identity element of .

An -module is finitely generated iff it is finite dimensional.

Tensor product

Sat, 03 May 2025 08:57:09 GMT

Let and be vector spaces over with bases and , respectively. For each with , we introduce a symbol . The tensor product space is defined to be the mn-dimensional vector space over with a basis given by

Thus consists of all expressions of the form

For and with and , we define by

For example,

Do not be misled by the notation into believing that every element of has the form , but linear combination of .
If and , then

If and , then

Next, we deine the tensor product of two -modules.
Let be a finite group and let and be -modules with bases and , respectively. We know that the elements

give a basis of . The multiplication of by an element of is defined in the following simple way, which is then extended linearly to a multiplication on the whole of .
Let . For all , , define

and, more generally, let

for arbitrary complex numbers .
Remark. You should be warned that for most elements in , e.g. when is a scalar multiple of .

See dummit&Foote Sec 10.4
Definition (-module). Let be a ring (associative with 1). An -module (or left -module) is an abelian group together with an operation , denoted by , such that for all and : A right -module is defined similarly, with the scalar multiplication on the right , denoted by , satisfying analogous axioms.
Our aim is to "extend" an -module to an -module.
To satisfy the relations necessary for an -module structure[1] and the compatibility relation with the action of on [2], we must take the quotient of this abelian group by the subgroup generated by all elements of the form

The resulting quotient group is denoted by (or jest if is clear from the context) and is called the tensor product of and over . By the definition of the quotient we have forced the relations

Definition (universal property for the tensor product).
Example.
Definition ( -balanced).
Definition ( -bilinear).
Example. We have , since for so that
and every simple tensor is reduced to 0. In particular .
Example. We have , since and generates .
Example. , where a simple tensor has the form for some rational numbers and . Then
Example. Similar to Example 9, for any divisible abelian group and torsion abelian group (an abelian group in which every element has finite order). For exmaple, .
Example (dimension). The structure of a tensor product can vary considerably depending on the ring over which the tensors are taken. and are isomorphic as left -modules (both are one dimensional vector spaces over ). Pick any simple tensor in , then
induces an isomorphism between and . Pick any simple tensor in , then
Thus . Hence,

Example. On the other hand, and are not isomorphic -modules (the former is a 1-dimensional vector space over and the latter is 2-dimensional over ). Every simple tensor in has the form , then
induces an isomorphism between and . Every simple tensor in has the form , then
induces an isomorphism between and . Hence

abstract algebra - Dimension of a tensor product as a vector space over complex number field

- Mathematics Stack Exchange
Exercise. Problem 2 (10 points). Consider the polynomial ring and its subring . Consider -modules What is the dimension as -vector space of each tensor product below? No need to prove or explain.
(a) ;
has every simple tensor of the form , then

induces an isomorphism between and (the construction is left as an exercise). Therefore

(b) ;
has every simple tensor of the form , then

induces a nature isomorphism between and . Therefore,

(c) .
has every simple tensor of the form , then

induces a nature isomorphism between and . Therefore,

We are done!

, .↩︎
↩︎

复变函数的积分

Fri, 02 May 2025 16:55:29 GMT

其实就是关于某个参数进行积分

柯西积分公式：若在一个包含圆周的区域内全纯，那么对于任意内的点，有

这个公式具有极其广泛的应用场景，比如计算导数，估计不等式....
Equation (1) 的证明思路就是类似于调和方程的平均值公式证明

对于 Equation (1) 两边同时求阶导数得到

这只需要根据导数的定义归纳验证导数的存在性即可.
柯西不等式 Equation (3) 是 Equation (2) 的直接推论

进一步我们有复变函数的幂级数表示

同时还有 Liouville 定理：若全纯且有界，那么是常值函数. (直接利用 Equation (2) )

然后可以证明代数基本定理.
根据幂级数表示 Equation (4) 可以证明复变函数的零点集是离散的.

参见《复变函数论》(第五版) 学习指导书 (钟玉泉编).
已知 Jordan 不等式

证明：

其中为圆周的上半圆周从到 .
选取参数化 . 于是

Remark. 不等式 Equation (5) 是一个重要不等式. 可以用来放缩很多围道积分（用来证明 Jordan 引理）
Exercise. 若，其中是从到沿的上半圆周，证明：

Proof.
时，直接利用 Equation (5) 放缩

时，利用留数定理显然，或者直接放缩估计 .
□

柯西积分定理对于解析函数的实部和虚部都不成立.
对于一般的题目，正常计算即可. 注意先将分母有理化，再判断奇点是否在积分曲线内部.
下面介绍一种思想：

复分析脉络

Fri, 02 May 2025 16:55:29 GMT

见 Rudin 《实分析与复分析》

A region is nonempty connected open subset of . Each open set in the plane is union of discs. For if exists then denote it . If exists , say . is the class of all holomorphic functions in , which is a ring. If is representable by power series in , then and is also representable. Check the convergence by evaluate

Fix , and choose s.t., where . For , if then

Then

An general result is for complex (finite) measure on a measurable space , is a plex measurable function on , is an open set in the plane which does not intersect and

Then is representable by power series in .
The proof is from the fact that for , every ,

Converges uniformly on . Then

Suppose a path, a continuous function defined on (the range of ), then

is unique under continuously differentaible one-to-one mapping.
Let be a closed path, the complement of , and define

Then is an integer-valued function on which is constant in each component of and which is 0 in the unbounded component of .
We call the index of w.r.t. , which is "the number of times winds around ".

The easy consequence of the above theorem is the representability of holomorphic functions by power series.
The Cauchy theorem has a useful convers:

Next we consider the power series representation of holomorphic functions. Suppose a region, and . Then either or has no limit point in . In the later case such that for , where and . Furthermore, is at most countable.
Analogous results hold of course ofor the set of -points of , i.e. the zero set of .

The proof is routine since

Here are some consequences:
Theorem (Liouvill's theorem). Every bounded entire function is constant.
Proof.
Suppose entire, for all , for all . Then

for all , which is possible only if .
□

利用柯西积分公式可以给出如下定理的证明.

Every nonconstant holomorphic function in a region is locally of the form (), except for an additive constant.

For , .

The first one concerns zeros of holomorphic functions, the second is the evaluation of a certain integral.

在处的极点阶数恰恰等于其在处的留数.

共形映射

Mon, 19 May 2025 13:49:11 GMT

See Dennis Zill. Appendex III for Table of Conformal Mappings

Some examples of conformal mapping:

Translations: , where is a complex number.
Rotations: , where is a real number.
Dilations (or scalings): , where is a real number.
Inversion: , .
Linear transformations: , where are complex numbers and .
Möbius transformations (or linear fractional transformations): , where are complex numbers and .

Definition (Simply Connected Domain). A domain is simply connected if it is connected and every closed curve in can be continuously deformed to a point in .
Then we introduce the Riemann Mapping Theorem:

Linear fractional transformations are also called Möbius transformations or bilinear transformations.

This matrix is called the adjoint matrix of .

See Section 7.3....

See Section 7.4....

See Section 7.5....

The expression has several important special properties in complex analysis, especially when is a point within the unit disk (i.e., ). This expression is commonly known as a Blaschke factor (or a component thereof) and is also a type of Möbius transformation.
Here are some of its key properties:

Maps the unit disk to itself: If , this transformation maps the open unit disk to itself. That is, if , then .
Maps the unit circle to itself: If , this transformation also maps the unit circle to itself. That is, if , then .
Zero: The transformation has a zero at . That is, when , .
Pole: The transformation has a pole at . If , then , so this pole lies outside the unit disk.
Conformal automorphism: When , this expression (or when multiplied by a constant with modulus 1) is a conformal automorphism of the unit disk onto itself (i.e., a biholomorphic map). This means it preserves angles and is a one-to-one and onto mapping.
Inverse Transformation: If , its inverse transformation is . This can be seen as a transformation of the same form, , where the parameter replaces and the sign in the denominator is adjusted (or more precisely, it is if we rewrite as ).
Derivative at : Let . Its derivative at is . This property is relevant in contexts such as Schwarz's Lemma.
In summary, this expression defines a very important complex transformation that plays a central role in the geometry and function theory of the unit disk. It allows for a standardized way to map any point inside the unit disk to the origin, while preserving the disk and its boundary, which is fundamental in many areas of complex analysis.

Exercise. 设 , 若使得 , 求全部 .
Proof.
若在有无穷个零点, 则若零点在内部有聚点, 则解析函数唯一性告诉我们 , 这不可能, 若这些零点的聚点只在边界, 那么在聚点处应该为 0, 这也不可能. 所以我们知道在只有有限个零点 , 这里允许有相同的项出现.
考虑无零点函数

显然且 . 故由最大模定理知 . 因此我们知道为常数, 所以

□

解析函数

Fri, 02 May 2025 16:55:29 GMT

判定可微
判定不可微
C-R 方程
在解析在可微
在解析在可微
存在可微但不解析的点
解析是一个局部的概念，指的是在附近都可微
可微是一个点态的概念
极坐标形式的 C-R 方程
若则 .
洛必达法则
Exercise (在可微但不解析).

Remark (对比).

考点：

. .

在不能断言 .
证明的一致连续性和不一致连续性
.
的零点是在的全部奇点
的零点，是在的全部奇点

Exercise. 证明

Proof.
只需证明

□
Exercise.
Proof.

□

这一节的主要内容是采用限制辐角或割破平面的方法，来分出根式函数和对数函数的单值解析分支。

Exercise.

由已给单值解析分支的初值，计算终值的公式为

其中为连接起点和终点且不穿过支割线的路线.
当把换成的动点时，得到的

也是此单值解析分支的解析表达式.
Exercise.

Exercise.

Exercise.

Exercise.

所有这些函数分成单值解析分支的方法，与我们前面用过的讨论方法是类似的，也要先讨论它们的支点，然后适当割破平面，只是较复杂些也较困难些。当然也可以像例 2.3.17 和例 2.3.19 一样，把他们视为复合函数来化简处理。
Exercise.

Exercise.
变量代换，化简，并将分成单值解析分支，然后就可以按照复合函数求导。

解析函数的幂级数表示

Wed, 14 May 2025 14:48:50 GMT

复级数的收敛性：

收敛半径：

解析函数的泰勒展开式：

Exercise. 求在的泰勒展开式.
因为

于是

Exercise. 求在的泰勒展式.
通过分析奇点可知，在内解析. 利用“大除法”：

Exercise. 求在的泰勒展式.

Exercise. 展开 .

Exercise. 若，且，则

Proof.
因为

所以

□

Exercise. 设函数列在区域上解析，且在中内闭一致收敛于函数 . 证明：
若不恒为零，是内可求长的简单闭曲线，其内部属于，且不经过的零点，则存在正整数 , 使得当时，在的内部和有相同个数的零点.
若在区域内部时单叶的，不为常数，则在内单叶解析.
Proof.
由 Weierstrass 定理，在内解析. 因为在上不为零，则

又在上一致收敛到，存在正整数，使得当时，在上有 . 即当时，在上有 . 由 Rouche 定理，在的内部，和有相同个数的零点.
对于第二问采用反证法，若在内不是单叶的，则存在使得，考虑函数，圆周，其中充分小使得其包含于内，而且在圆周内有唯一的零点. 于是存在，使得时，和在内有相同个数（1 个）零点，从而 . 这与的单叶性矛盾！
□
Exercise (cmc 高年级第 10 届). 设是复函数的阶极点. 证明：一定存在使得对于任意，函数在中必有个零点.
Proof.
定义

且在处补充定义（因为是可去奇点），于是在的一个小邻域内解析，故存在使得 . 再令

由最大模原理，对于有

由 Rouche 定理，函数在内零点个数与函数相等，即为个. 由于是在处的极点阶数，故，于是 . 故在存在个零点，故在中必有个零点.
□
Theorem (Picard's Big Theorem). Let be an analytic function on a punctured neighborhood of , say on . If has an essential singularity at , then takes on all possible complex values, with at most a single exception, infinitely often in any neighborhood of .
Theorem (Picard's Little Theorem). If is an entire function which is not constant, then the image of is either the whole complex plane or the complex plane minus a single point.

留数计算方法

Thu, 15 May 2025 02:20:55 GMT

参见《积分的方法与技巧》· 金玉明; A first course in complex analysis with applications by Dennis Zill
留数的表示

留数定理由柯西积分公式得到

无穷远点的留数定理积分曲线方向相反 (考虑黎曼球，这是显然的)

留数就是的 Laurent 展开式中系数，对于极点，有下面较为直接的方法，不需要 Laurant 展开.
Remark. Laurant 展开就是按照公式计算

在的邻域中有

于是

在点解析，且，但是，那么

Laurant 展开式为

那么

两边得到

于是

详细计算见《积分的方法与技巧》· 金玉明 p 328

Lemma (Jordan's Lemma). Let be a semicircular contour in the upper half-plane defined by for . Suppose is a function that is analytic in the upper half-plane for (for some constant ). If uniformly as for in the upper half-plane (i.e., as ), then for any real constant ,
A similar statement holds for a semicircular contour in the lower half-plane if .
Proof.
Let , denote

Estmate the value,

□

参见 A first course in complex analysis with applications by Dennis Zill, Section 6.6

Given the integrals of the form

To convert it to a complex integral on , we parametrize the contour by . We can then write

i.e.

Then Equation (1) becomes

where is the unit circle .

Cauchy principal value of an integral is defined by

If a integral diverges, it may still possess a Cauchy principal value.

Another powerful lemma is Lemma 2 (Jordan's Lemma).
Next we calculate

Indented contour is of great use.

In the next discussion we examine integration along a Branch cut (cf section 2.6 and 4.1 in Dennis Zill)

This technique also goes for .

Rouche's theorem is a consequence of Argument Principle.

洛朗级数

Fri, 02 May 2025 16:55:29 GMT

Theorem (Laurant 定理). 在圆环内解析的函数必可以展开成双边幂级数，即Laurant 级数
其中Laurant 系数
为圆周，并且展开式唯一.
展开洛朗级数直接暴力展开即可.

由于函数在总是无意义的，所以点总是的奇点.

对于函数

它是有理分式函数，分母的零点是这个函数的极点，下面考虑它们的阶.
是分母的一阶零点，且不是分子的零点，故是的一阶极点. 注意到 ,

于是也是的一阶极点.
接下来考虑，因为

其中在解析，且，因此是的一阶零点.

对于函数

它只有作为奇点，作为的三阶极点，在的主要部分 (即正幂) 为零，故是的可去奇点.
由于以为二阶极点，故以为二阶零点.

对于函数

解得到的零点

又因为，所以都是的一阶零点. 于是都是的一阶极点.
当时，. 故点是的非孤立奇点，即极点列的聚点.

Exercise. 设，当时 . 证明：必有一个零点.
Proof.
由于是整函数，故只有作为孤立奇点，于是可设

又由题设

可见为的可去奇点，从而为的一阶极点，从而

于是必然有且仅有一个零点.
□

Lemme de Schwarz — Wikipédia

Let be a holomorphic function on the open unit disk

Such that:

,
for all .
Then:

for all ,
,
Moreover, if equality holds at any point other than 0 (i.e., if for some ), or if , then for some real constant .

The proof is a straightforward application of the maximum modulus principle on the function

which is holomorphic on the whole of , including at the origin (because is differentiable at the origin and fixes zero). Now if denotes the closed disk of radius centered at the origin, then the maximum modulus principle implies that, for , given any , there exists on the boundary of such that

As we get .
Moreover, suppose that for some non-zero , or . Then, at some point of . So by the maximum modulus principle, is equal to a constant such that . Therefore, , as desired.

If , then let

Then . Apply Schwarz lemma,

i.e.

Exercise (cmc 第八届高年级决赛). 设函数在单位圆内解析，并且，为常数. 证明：

Proof.
令，做变换

则该变换将单位圆映射为 . 由 Schwarz 引理 . 由于

则

从而

□

LectureNotes
We have see that every rotation , for fixed , is an automorphism of the disk. For define

which is also an automorphism of .
When , we have

Observe that

Then

By the Schwarz lemma, , and . Thus is indentity map of . It follows that is bijection from to .
Theorem (Automorphism of ). A map is an automorphism of iff there are and s.t.

For , , , , evaluate the maximum of . Denote , then

Apply Schwarz lemma to , with then

At the same time, by Schwarz lemma, for any ,

Since is automorphism, replace by where , then

i.e.

The above two inequaltities is called Schwarz-pick lemma.

http://www.webpages.ttu.edu/jengwer/notes/SchwarzPickLemmas.pdf

最大模原理

Fri, 02 May 2025 16:55:29 GMT

参见 Rudin《实分析与复分析》.
This chaper contains further generealizations of the maximum modulus theorem, as well as some rather striking applications of it, and it concludes with a theorem which shows that the maximum property "almost" characterizes the class of holomorphic functions.

Suppose is an entire function and

for all . Then is a constant.
This follows immediately from the Cauchy estimates, since they show that for

hw1

Fri, 02 May 2025 16:55:29 GMT

p13 1(2)(3) 4 7 9(2) 11 14
Exercise.

Exercise.
, then

Hence

Exercise.
Let , then

Exercise.
Definition (棣莫弗公式).

Exercise.
Solve the equation:

Then

Exercise.

If , then

On the other hand

Therefore

If , we have then

And

We have

Consider the function

which is decreasing in . Since , we have , that is

Therefore,

hw2

Fri, 02 May 2025 16:55:29 GMT

p15 16 17(3)(4)(8)(9)(10) p38 3(3)(5) 4 5 6 8 10

在复平面拓扑下：
(1) 不含内点，故不是开集，没有聚点，故为闭集。由于同构于，故是紧集当且仅当是有界闭集，但并非有界，故不是紧集。
(2) 同理于 (1)，该集合不是开集，是闭集，是紧集。
(3) 令

所有点都是内点，所以是开集。而任意作为极限点，却不在中，故不是闭集，进而不是紧集。
(4) 显然是开集，不是闭集，不是紧集。是闭集，是开集，是紧集。是开集，是闭集，是紧集。
Remark. 的定义是黎曼球球极投影得到的，它是依照中的表面拓扑而定的。它是的紧致化。

(3)

单连通的闭圆盘
(4)

设，那么

平方得到

化简得到

平方得到

化简得到

是一个椭圆，显然单连通。
(8)

设，则

于是 . 平方得到

化简得到

也就是说

即

这是复平面挖掉一个圆盘，显然单连通。

(9)

这是单连通区域
(10)

设，则

于是，对于，，对于，则

若，则

也就是说

若，则

也就是说

总而言之，该区域为

是多连通的开区域

Definition (解析函数).
Theorem.
Theorem.
(3) so . If is differentiable at then implies that thus . Hence is the only point at which is differentiable. For any neighborhood of , is not differentiable in thus is not holomorphic over . Therefore is holomorphic at no point.
(5) so , and .

for , since then where , and where , we have , i.e. . (the differentiability is derived from and theorem 3.2) Therefore constant, and is constant in .

(1) for . means and , i.e. . If is constant in then , combined with the fact that and , we have . Therefore is constant in . There is a similar proof for in .
(2) If is a constant in then is holomorphic in . However, if both and satisfies the C-R equation, it is easy to get and thus is constant.

for . . By theorem 3.2 it suffices to show that and are differentiable in and for . By theorem 3.2 we have are differentiable in and , i.e. .

Therefore is holomorphic in the lower half plane.

. So we have

Then

And

The Cauchy-Riemann condition is , which means

where then

Therefore

Hence

which is Cauchy-Riemann equation.

hw3

Fri, 02 May 2025 16:55:29 GMT

p40 11 14 15 16
Exercise. 11．试求出的值．

Exercise. 14．设函数在解析，那么我们说在解析．下列函数中，哪些在无穷远点解析？
在无定义，故不解析，故在不解析。
在解析，故在解析。
，当且时，在解析，故在解析，否则在不解析。
是多值函数，在的每个分支内解析，故是多值解析函数。
Exercise. 15．在复平面上取上半虚轴作割线．试在所得区域内分别取定函数与在正实轴取正实值的一个解析分支，并求它们在上半虚轴左沿的点及右沿的点处的值．
对于，它有两个解析分支

令在正实轴取正实值，于是在的解析分支。在上半虚轴右沿，令，那么，.
在左沿，.
对于，它有无穷多个解析分支

令在正实轴取正实值，于是在的解析分支。在上半虚轴右沿，令，那么，.
在左沿，，.
Exercise. 16．在复平面上取正实轴作割线．试在所得的区域内：（1）取定函数在正实轴上沿取正实值的一个解析分支，并求这一分支在处的值；在正实轴下沿的值．（2）取定函数在正实轴上沿取实值的一个解析分支，并求这一分支在处的值；在正实轴下沿的值．
(1)

在正实轴上沿取正实值的解析分支为 . 令则

在正实轴下沿

(2)

在正实轴上沿取正实值的解析分支为，令则

在正实轴下沿

hw4

Fri, 02 May 2025 16:55:29 GMT

p40 17 18 19
Exercise.
函数的支点为

于是线段和可以将复平面分割为解析分支.

其中，取得到

于是

Exercise.
设函数
如果规定时，. 作两种适当的割线，求这函数的一个解析分支在的值.

令那么

故

第一种割法：

于是

第二种割法：

在割线左沿：

于是

在割线右沿：

于是

Remark. 计算的时候要注意，不是直接计算，而是注意到时，，然后做曲线，这条曲线不经过割线，然后算出这条曲线绕过的角度，就是的值.
Exercise.
但，这是因为

二者不是同一个多值函数.

hw5

Fri, 02 May 2025 16:55:29 GMT

p58 1 2 3 7 8 9 10
Exercise.
记

于是
(1)

(2)

(3)

Exercise.
记

(1)

(2)

Exercise.
Proof.
当时，

令得到

□
Exercise.

(1)
,

,

(2)
,

,

Exercise.
Let

Then

is holomorphic in , then for any

with . We have

where

Similarly,

Therefore

where .

Exercise.
(1)

(2)

(3)

Exercise.
Since is continuous in the neighborhood of 0, then for any , there exists , s.t.

Let , then

Since can be arbitrarily small, thus we have

hw6

Fri, 02 May 2025 16:55:29 GMT

第三章 11 12 14 15 17 18

(1)

(2)

(3)
Since is holomorphic in the unit disc and its boundary,

(4)

Apply Cauchy's integral formula to at , then

Therefore

Let

Then

Let then

Thus

On the other hand

Therefore

For , apply the Cauchy's integral formula

Clearly, reaches its maximum at . Let then

When ,
Let large enough such that is contained in , where is the circle centered at with radius , counterclockwise. Then

, s.t. , we have

Then for

Thus

Since is arbitrary,

When , let be the circle centered at with radius , clockwise. is small enough such that is contained in . Then

By Cauchy's integral formula

Hence

(1)
Fix a point and define

where is any path in connecting to and is a complex number so that . This definition is independent of the path since is simple connected. We find that is holomorphic with

and a simple calculation gives

so that is constant. Evaluating this expression at we find so that for all , and the proof is complete.
(2)
(1) implies (2). Put , then .

hw7

Fri, 02 May 2025 16:55:29 GMT

56789
Exercise.
(1) then .
(2) , where

Then , .
(3) then .
(4) then .
(5) . Then .
(6)

Lemma.

Thus .
Exercise.

Proof.
(1)

Then

(2)
If is entire, since is bounded, then for each , let ,

Therefore is constant.
(3)

□
Exercise.
Proof.

□
Exercise.
Proof.

Let , , then

. Thus

□
Exercise.
(1)

(2)

(3)

(4) 主值支：

(5)

hw8

Fri, 02 May 2025 16:55:29 GMT

10 11 12(3)(5)(6) 13
Exercise.
Proof.
由于是整函数，考虑它的幂级数表示形式

由 Cauchy 定理，

其中是以原点为中心，半径为的逆时针方向的圈. 对于任意，有

由的任意性，令，就有 . 于是是一个至多次的多项式或者常数.
□
Exercise.
Proof.
只需要证明

其中

当时，有

故得证！
□
Exercise.
(3)

(5)

(6) 在

在

Exercise.

(1)

有孤立奇点，其中为 1 阶极点，和为 2 阶极点. 为可去奇点.
(2)

为奇点当且仅当，即 . 也就是，即其中 . 于是，为的孤立奇点，都是 1 阶奇点.
(3)

于是孤立奇点为其中，若 , ，那么 , 是所有孤立奇点，都是 2 阶极点. 否则 , 是所有孤立奇点，都是 1 阶极点.
(4)

孤立奇点为和，其中为 1 阶极点，是本性奇点.
(5)

孤立奇点为，其中是本性奇点，是可去奇点.
(6)

, 即 . 也是奇点，但，故是可去奇点. 综上：, 是所有孤立奇点，其中是可去奇点，是 1 阶极点

hw9

Fri, 02 May 2025 16:55:29 GMT

第4章：14 17 18 19 第5章：1 2 3 7
Exercise.
Proof.

Since ,

is holomorphic at . Thus .

where

And .
□
Exercise.
Proof.
For , we have

where . Since for . Then for . Therefore

is constant in .
□
Exercise.
Proof.
(1)
不存在，由于非平凡解析函数零点孤立，而 , 在 0 的某个邻域内恒为 0. 当充分大的时候落在这个邻域内，但 .
(2)
可做函数

(3)
不存在，事实上点列，故在 0 的某个邻域内, ，这与题设矛盾.
□
Exercise.
Proof.
并不矛盾，因为 1 是函数的奇点，函数不在 1 处解析.
□
Exercise.
(1)

(2)

(3)

(4)

Exercise.

时，在处有可去奇点，在处有一阶极点.

时，在处有一阶极点，在处有一阶极点.

Exercise.
(1)

(2)

(3)

Exercise.
记奇点为，我们知道

由于在处解析，所以 . 从而

记

在扩充复平面上解析，因为洛朗展开没有正项，所以

因此

hw10

Tue, 06 May 2025 03:52:49 GMT

p110 8(1)(2)(4)
Exercise.
(1)
Let

Consider the toy contour

We have

On ,

Let in Equation (1), then

Hence

(2)

The singularities are and . Since , then

(4)

Let

Consider the toy contour, an indented semicircle in the upper plane.

On ,

where , thus .
Let in Equation (2) ,then

Thus

hw11

Tue, 13 May 2025 03:36:13 GMT

p110 8 (6)(7)(8)(10)(12)(13)(14) p111 11 12 14 15
Exercise.
(6)
(7)
(8)
Exercise.

Exercise.
(12) .
(13) .

Exercise.

Exercise.

Exercise.

Exercise.

Exercise.
假设有一个复变函数，它在一条简单闭合曲线的内部及边界上是解析的（除了有限个极点外），并且在上没有零点和极点。那么，当沿着曲线逆时针方向绕行一周时，的辐角变化量等于乘以函数在曲线内部零点的个数（计重数）减去极点的个数（计重数）。
用公式表示:
其中：

表示当沿曲线逆时针方向绕行一周时，的辐角变化量。
表示函数在曲线内部零点的个数（如果一个零点是重的，则计为个）。
表示函数在曲线内部极点的个数（如果一个极点是重的，则计为个）。
另一种等价的表述形式是利用对数导数:

这个积分实际上计算了的像在原点周围缠绕的圈数。

hw12

Tue, 20 May 2025 02:25:14 GMT

p134 1 3 4 5 6 7
Exercise.
Consider the parameterization of ,

Then the parameterization of is

Since is injective, the path is well-defined simple closed curve. Then

On the other hand,

Therefore

Denote ，where 。
The area of the region in the -plane can be obtained by variable substitution using the Jacobian determinant:

According to the variable substitution formula for double integrals, we have:

where is the Jacobian determinant:

Since is an analytic function, its real part and imaginary part satisfy the Cauchy-Riemann equations:

Substituting the Cauchy-Riemann equations into the Jacobian determinant:

We know that the complex derivative can be expressed as:

Therefore, the square of the modulus of the complex derivative is:

So, the value of the Jacobian determinant is equal to :

Now we substitute this result back into the area integral formula:

This proves the second conclusion.
Exercise.
(1)
For , if

Then

Thus . is injective.
(2)
For , if

Then

Thus . is injective.
(3)
For , if

Then

Thus . is injective.
Exercise.
Denote

Then

By Cauchy's formula,

where . Then

Let , then

Thus for any ,

By Rouche's theorem, and has the same numbers of zero's in , i.e. only zero. As , has no zero in .
Assume that is not injective in . There exists s.t.

By the MVT, there exists , s.t.

By the convexity of ,

is a zero of , whihc is a contradiction. Hence, is injective in ; for any , it is an interior point of , thus has a neighborhood where is injective.
Exercise.

Then

Thus the inverse of is

Exercise.

Then

Exercise.

(1)

Then

(2)

i.e.

Then

primary-functions

Fri, 02 May 2025 16:55:29 GMT

See chapter 2 of A First Course in Complex Analysis with Applicaitons. Dennis

For , has infinite many different values. represents the principle value of .

is the complex number satisfying .

is the principle value of .
Note that for ,

We use instead of for the consideration of multiple values.

For , . . If then is -valued. If or , then is -valued.

阶支点：转圈回到最初的值。

无穷阶支点：转不回最初的值。

计算根式函数在某个给定的解析分支内的取值：

先计算出这个解析分支是啥，考虑极坐标
再代入给定的点到这个解析分支中

rudin-approximation-by-rational-functions

Thu, 22 May 2025 15:12:58 GMT

rudin-conformal-mapping

Thu, 22 May 2025 14:48:58 GMT

The direction of is .
Definition (preserves angles).
In less precise language, the requirement is that for any two rays and , starting at , the angle which their images and make at is the same as that made by and , in size as well as in orientation.
The property of preserving angles at each point of a region is characteristic of holomorphic functions whose derivative has no zero in that region.
Definition. Suppose , for some region . We call a normal family if every sequence of members of contains a subsequence which converges[1] uniformly on compact subsets of . The limit function is not required to belong to .
Theorem. Suppose and is uniformly bounded on each compact subset of the region . Then is a normal family.

The Riemann mapping theorem reduces the study of to , for any simply connected proper subregion of the plane.

Definition. is the class of all which are injective in and which satisfy Thus every has a power series expansion
The class is not closed under addition or multiplication, but has many other interesting properties. We shall develop only a few of these in this section. Theorem 14.15 will be used in the proof of Mergelyan's theorem, in Chap. 20.

Proof.
(a) is clear. To prove (b), write . Then , and has no zero in since has no zero in . Hence there exists , with , . Put

Then . NTS: . Clearly, , . For , , then, ; as is injective, . If , then ; it follows that , and since has no zero in , we have .
□
Theorem (area theorem (Theorem 14.13)). If , is one-to-one in , and then
Proof.
WLOG, assume , is real. Put , , . Then is a neighborhood of ; the sets , and are disjoint, as is injective. is in the interior of . Write

and

For , we then obtain

Then

There exists , if sufficiently small , we have

This says lies in the interior of ellipse , whose semiaxes are and , and which therefore bounds an area

Since is in the interior of , the area of is no larger than Equation (1). The Cauchy-Riemann equations show that the Jacobian of is , then

Thus

Let , then let . We are done!
□
Corollary. Under the same hypothesis in Theorem 5 (area theorem (Theorem 14.13)), .

The second assertion is that contains all with .
Proof.
(a) By theorem 14.12, there exists s.t. . Denote , then Theorem 5 (area theorem (Theorem 14.13)) applies to and this yields (a). Since

we have

and hence

By Corollary 6, .
To prove (b), suppose . Define

Then , is injective in , and

so that . Apply (a) to : we have and since , we finally obtain . So for every . This complete the proof.
□
Moreover, given any , one can find entire with , , that omit the value . For example,

Of course, by Theorem 14.14, that is injective in and cannot happen at the same time.

Proof.
Let , then , hence , so the image of under contains all with . Since is not in this image, we have .
□

may converges to ↩︎

rudin-harmonic-function

Thu, 22 May 2025 04:50:28 GMT

The Poisson kernel is

If ,

From Equation (1) ,we have , , , for . Then

Now denote by , by . Indentify the spaces and with the corresponding spaces of -period functions in .
Regard as , then

The Poisson integral is

If is real,

, then is harmonic.
Poisson integrals of continuous function behave well near the boundary of .

This theorem, using Poisson integral, solves the boundary value problem[1]: given , find harmonic in .
The harmonic is also unique.

Proof.
By theorem 10.7, . Let , is harmonic. NTS: on . Assume not, WLOG, , s.t. ; pick , , then . As , is subharmonic, by maximum principle, this is a contradiction.
□
For harmonic[2] ,

For , by Equation (1),

Let harmonic in open . For , satisfies Equation (2) [3], and for some , where is unique up to
The Poisson integral also yields information about sequences of harmonic functions.

Proof.
(a) As on , satisfies Equation (2), we have satisfies Equation (2), thus is harmonic.
(b) Assume that , , , . For any ,

Thus

Let , then Equation (3) also holds for . Thus are open. As is connected, or . We are done!
□

The theorem asserts can be extended to , where . , we extend to by letting , and .

Let , . Theorem 11.8 can be restated as: for , , then on as . i.e. .
We replace by :

Proof.

Then

Integrate ove , then we obtain (1).
As is dense in for [4], pick s.t. , , then

By theorem 11.8, , therefore as .
□

A considerably stronger uniqueness theorem will be obtained later.

the Dirichlet problem↩︎
in .↩︎
Equation (2) is the criterion of harmonic function, due to the above two theorems.↩︎
↩︎

rudin-Hp-spaces

Thu, 22 May 2025 16:02:48 GMT

This chapter studies certain subspaces of which are defined by certain growth conditions. These so-called -spaces have a large number of interesting properties concerning factorizations, boundary values, and Cauchy-type representations in terms of measures on the boundary of .
A convenient approach to the subject is via subharmonic functions, and we begin with a brief outline of their properties.

Definition. A function defined in an open set in the plane is said to be subharmonic if it has the following four properties.
for all .
is upper semicontinuous in .
Whenever , then
None of the integrals in (c) is .
Note that the integrals in (c) always exist and are not , since (a) and (b) imply that is bounded above on every compact . [1] Hence (d) says that the integrands in (c) belong to .
Theorem (Theorem 17.2). If is subharmonic in , and if is a monotonically increasing convex function on , then is subharmonic.
Theorem (Theorem 17.3). If is a region, , and is not identically 0, then is subharmonic in , and so are and .[2]
Theorem (Theorem 17.4). Suppose is a continuous subharmonic function in , is a compact subset of , is a continuous real function on which is harmonic in the interior of , and at all boundary points of . Then .
Clearly by maximum modulus principle. But we gives a proof without it.
Proof.
Put . Assume that for some . Since , attains its maximum on ; and since on the boundary of , the set is a nonempty compact subset of . Let be a boundary point of . Then for some , we have , but some subarc of the boundary of lies in the complement of . Hence

and this means is not subharmonic in , which is a contradiction.
□
Theorem (Theorem 17.5). Suppose is a continuous subharmonic function in , and If , then .
Proof.
Let be the continuous function on which conincides with on the boundary of and which is harmonic in . By Theorem 4 (Theorem 17.4), in . Hence

□

Let denote Lebesgue measure on , so normalized that .

and we also introduce

Definition. If and , we put
If , is defined to be the class of all for which . (Note that this coincides with our previously introduced terminology in the case .)
The class consists of all for which
It is clear that if

Proof: If is the set of all at which , then , so either for some , or , in which case for some .↩︎
if has no zero in , is harmonic.↩︎

rudin-maximum-modulus-principle

Thu, 22 May 2025 12:09:29 GMT

Theorem (Maximum Modulus Principle). If is the closure of a bounded region , if is continuous on and holomorphic in , then for every . If equality holds at one point , then is constant.
The equality , which is part of Theorem 11.32, implies that

This time boundedness on is enough; we do not need continuity on .
This chapter contains further generalizations of the maximum modulus theorem, as well as some rather striking applications of it, and it concludes with a theorem which shows that the maximum property "almost" characterizes the class of holomorphic functions.

Theorem (Schwarz lemma). Suppose , , and . Then if equality holds in (1) for one , or if equality holds in (2), then , where is a constant, .
Proof.
Apply Theorem 1 (Maximum Modulus Principle) to which has a removable singularity at 0.
□
For any , define

Then is bijection, , . The inverse is . We have

Suppose and are complex numbers, , and . How large can be if is subject to the conditions , and ?
To solve this, put . Clearly, , , . By Theorem 2 (Schwarz lemma), . The chain rule gives

Apply Equation (1), we have

The equality holds iff , in which case is a rotation, thus

for some constant with .
We imposed no smooth condition on the behavior of near .

Proof.
Let be the inverse of , . As is injective, , exists for any , thus . By the chain rule, . By Equation (2),

thus the equality holds. By Equation (3), with , for some .
□

For unbounded regions, for is no longer true.
Example. Let , the boundary is . Put , for real ,
, but since .
Theorem. Suppose is continuous on , , and suppose that for all and some fixed . If then we actually have i.e. is a convex function on .
Proof.
WLOG, assume that . Otherwise, replace by , which also satisfies the conditions, where .
We need to show for all . For each , define an auxiliary function

Since , in , thus . Also, , so that .
Let be the rectangle cut off from by , then on ; hence on by the maximum modulus theorem.
For fixed , for any sufficiently small , then ; let , then , . We are done!
□

The letter j will denote the identity function: . The function which assigns the number 1 to each will be denoted by 1.

Note that (c) is a rather weak form of the maximum modulus principle; (c) asserts only that the overall maximum of on is attained at some point of the boundary , but (c) does not a priori exclude the existence of local maxima of in .
Proof.
By (a) and (b), contains all polynomials. In conjunction with (c), this shows that satisfies the hypotheses of Theorem 5.25. Thus every is harmonic in . We shall use (b) to show that every actually satisfies the Cauchy-Riemann equation.
Let and be the differential operators introduced in Sec.11.1. The product rule for differentiation gives

Fix , and take . Then . Hence and are harmonic, so and . Also, and . The above identity therefore reduces to . Thus .
□

In particular, the theorem asserts that is at most countable, unless .

rudin-zeros-of-holomorphic-functions

Thu, 22 May 2025 15:19:56 GMT

See also Stein II, Chapter 5.

Lemma.
Proof.

□

This is known as Jensen's formula. The hypothesis causes no harm in applications, for if has a zero of order at , the formula can be applied to .
Proof.

□

the logic of Harmoic function

Fri, 02 May 2025 16:55:29 GMT

Suppose a complex function defined in a plane open set , regarding a transformation which maps into . Suppose differentiable at , WLOG , then

Since , then

which suggests the introduction of the differential operators

Now

If , then iff the Cauchy-Riemann equation

holds for every . In that case we have

If , and real. Then splits into .
The laplacian of is defined to be

If is continuous in and if then is said to be harmonic in .
Note that . If is holomorphic then , has continuous deivarives of all orders and , which means "Holomorphic functinos are harmonic."
Every real harmonic function is locally the real part of a holomorphic function, and it will yield information about the boundary behavior of certain classes of holomorphic functions in open discs.

常微分方程数值解

Thu, 15 May 2025 04:46:48 GMT

见 Fiddie 讲义 11,12,13

自治系统就是

其中与无关.

考虑以下形式的数值方法

其中和是实常数，是步长。这个公式被称为线性多步法。当时，该方法是显式的；否则，它是隐式的。
改写为

其中 .

对于线性步法 Equation (1), 考虑

Theorem (相容条件). 线性步法相容的充要条件是

Example. 判断求解常微分方程初值问题的线性多步法是否收敛，并说明理由.
首先判断零稳定性.
该方法的特征多项式为, 其根为, 均在单位圆上, 且绝对值为1的根是单根, 故该方法是零稳定的.
接下来判断相容性.
, , 故.
所以该方法收敛.

数值积分误差估计

Thu, 15 May 2025 03:42:02 GMT

好的，这是一份关于数值积分主要公式的备考资料，包括公式、误差分析、代数精度以及简要的推导思路。
数值积分备考资料

数值积分（Numerical Quadrature）用于逼近定积分的值，特别是当被积函数的原函数很难或无法用初等函数表达时，或者当是以离散数据点形式给出时。

牛顿-柯特斯公式的基本思想是用一个次数为的插值多项式来近似 , 然后计算插值多项式的积分。

节点在积分区间内等距分布。

使用1次插值多项式（连接和的直线）。

估计值:
误差分析 (截断误差):
代数精度: 1 (对于次数的多项式，结果精确)
简要推导思路:
用线性插值多项式近似。
计算得到。
误差项可以通过泰勒展开或插值多项式的误差公式推导。对于，积分后得到。

将分成个等长的子区间 , 每个子区间长度 ,。

估计值:
误差分析:
简要推导思路: 将各子区间上的基本梯形公式误差累加起来。。根据中值定理，存在使得，则误差为。

使用2次插值多项式。通常考虑区间和中点。

令（即每个小子区间的长度）。

估计值:
误差分析:
代数精度: 3 (对于次数的多项式，结果精确)。比预期的2次要高，这是因为选取节点的方式对称，使得三次项的误差积分为0。
简要推导思路:
用过点 , , 的二次插值多项式近似。
计算得到。
误差项可以通过更高阶的泰勒展开（Peano核方法）或插值多项式误差推导，但更标准的方法是使用Hermite插值或特定技巧显示其为。一个常用方法是先对在处泰勒展开到4阶。

将分成个等长的子区间（为偶数），每个子区间长度，。相当于应用次基本辛普森公式。

估计值:
误差分析:
简要推导思路:
累加个基本辛普森公式的误差。每个基本辛普森公式作用的区间长度是。所以单个误差是。总共个这样的应用。
总误差是

高斯求积公式的目标是选择节点和权重使得求积公式对于尽可能高次的多项式是精确的。
对于个节点，高斯求积公式可以达到的代数精度。
通用形式：

其中是权函数。节点是与权函数在区间上相关的次正交多项式的根。

高斯-勒让德求积 (Gauss-Legendre Quadrature) 这是最常见的高斯求积，其中权函数，积分区间通常标准化为。
节点 : 是次勒让德多项式的根。
权重 : 。
估计值 (例子):
: . (中点公式)
: .
: . .
误差分析: (对于一般的区间，需要做线性变换，然后误差项会包含因子)

代数精度:
简要推导思路:
目标：选择个节点和个权重 (共个自由参数)，使得公式对次数尽可能高的多项式精确。
令是一个次数的多项式。作带余除法，其中是次正交多项式（其根为），。
(由于与所有次数 < n 的多项式正交)。
(因为是的根)。
问题转化为选择使得对所有次数的精确。这可以通过构造一个次的插值多项式使得，然后令 (其中是拉格朗日基函数)。

高斯-切比雪夫求积 (Gauss-Chebyshev Quadrature)
这是高斯求积的一种特殊情况，权函数，积分区间。

节点 : 是次第一类切比雪夫多项式的根: , for .

权重 : 非常特殊，所有权重都相等 .
估计值:
误差分析:
代数精度:
简要推导思路: 同高斯求积一般思路，使用切比雪夫多项式作为正交多项式系。
下面给出误差的推导
设在上充分光滑，则可以将其展开为 Chebyshev 多项式级数：

其中是第阶 Chebyshev 多项式，是对应的系数。
将上述展开式代入积分：

利用 Chebyshev 多项式的正交性：

可以得到

因此，

另一方面，考虑数值积分公式：

利用 Chebyshev 多项式的离散正交性：

可以得到

因此，

所以误差为

当充分光滑时，有

其中为常数。因此，误差可以表示为

其中 .

中点公式 (Midpoint Rule): 的开型牛顿-柯特斯公式，或的高斯-勒让德公式（经过调整）。
. 误差: 。代数精度: 1.
龙贝格积分 (Romberg Integration): 对复合梯形公式使用理查森外推法，逐步提高精度。
自适应求积 (Adaptive Quadrature): 根据被积函数在不同子区间的行为自动调整步长或子区间划分，以在满足给定精度要求的情况下最小化计算量。

牛顿-柯特斯:
基于插值多项式的误差公式: 。
积分。如果，误差主项包含。
对于对称节点（如辛普森，中点），有时积分，使得误差阶比预期高一阶（即）。这需要使用更高阶的埃尔米特(Hermite)插值或Peano核方法。
Peano核方法: 误差对于所有次数的多项式，。。
高斯求积:
利用正交多项式的性质。关键在于次正交多项式与所有次数小于的多项式关于权函数正交。
误差基于次Hermite插值多项式，它在节点处与和都一致 (但高斯公式本身不使用的值)。实际推导更直接：构造一个在节点处取值为的次多项式。那么在处为0。误差。。但高斯公式误差是关于的，这是因为节点的特殊选择（正交多项式的根）。
一个更标准的误差推导是考虑一个在处与值相同的次Hermite插值多项式，它在处满足和（尽管公式不直接用）。积分得到高斯公式的权重。误差。
这份资料应该能帮助你复习数值积分的主要内容。祝你考试顺利！

正交多项式

Mon, 19 May 2025 14:50:53 GMT

Legendre多项式：
Laguerre多项式：
Hermite多项式：
Chebyshev多项式：
我们关心的是 Legendre 多项式和 Chebyshev 多项式.

Legendre多项式是以为定义域，以为权函数的正交多项式。满足递推关系

Chebyshev多项式是以为定义域，以为权函数的正交多项式。满足递推关系

交换代数

Wed, 21 May 2025 17:00:26 GMT

参见 Atiyah, Commutative_Algebra_Atiyah.pdf
Exercise. 设是环, 令证明:
是的可逆元是中可逆元且是幂零元。
幂零幂零。
是零因子存在着环的非 0 元使得。
如果 , 就叫本原多项式。证明: 如果 , 那么本原均为本原。
Proof.

: 设 , 不妨 , 否则考虑 , 则 , 记幂数为 , 从而。（实际上就是题 1）
: 设 , 且 , 令 , 可得 , 从而可逆。
从而有（线性代数的结论）, 特别的 , , 从而 , 如此往下, , 故幂零。对其余同理。（由于幂零，则幂零，故依然可逆，对重复上述论证可得幂零，依此类推）
□

Proposition. 极大理想都是素理想.
Exercise. .
这里使用而不是，是因为后者只在的情况下有定义，而前者总是良好定义的.
Exercise.

解释 3 的证明：
这个证明的目的是要说明：在一个局部环上，如果两个有限生成的 -模和的张量积，那么和中至少有一个是零模。
证明过程可以分解为以下几个步骤：

引入剩余域 (Residue Field)
设是一个局部环，其唯一的极大理想为。
考虑。因为是极大理想，所以是一个域，称为的剩余域。
张量积与剩余域
我们已知。
将这个等式两边都用在上作张量积，可以得到：

利用张量积的结合律和换环定理 (change of rings, ，其中是一个 -代数)，这里，我们有：

根据题目2的结论 ()，我们知道：

。我们记。
。我们记。
因此，上面的等式变为。注意和都是域上的向量空间。
向量空间的张量积
我们得到了，其中和是域上的向量空间。
对于域上的向量空间，如果，那么必然有或者。（如果均非零，它们分别有非空基底，则的基底由它们基底元素的张量积构成，因此也非零。）
证明中使用了Hom函子的性质来推导这一结论：

利用张量-Hom伴随 (tensor-hom adjunction)：
由于，那么左边的只包含零映射（从零空间到的唯一线性映射是零映射）。
因此，右边的也必须为零（即只包含零映射）。
对于域上的向量空间和，如果，那么或者。（如果且，我们可以构造一个非零的从到的线性映射）。
所以，从可以得出：
或者。
如果，这意味着从到的唯一线性映射是零映射。但如果，那么其上的恒等映射就不是零映射。所以，必然意味着。
综上，我们得出结论：或者。也就是或者。
应用中山正引理 (Nakayama's Lemma)
我们已知是一个局部环，是其唯一的极大理想。和是有限生成的 -模。
中山正引理的一个形式是：若是环上的有限生成模，是的一个理想且包含于的 Jacobson 根中。如果，则。
对于局部环，其 Jacobson 根就是其唯一的极大理想。
所以：

如果，这意味着。由于是有限生成的 -模，根据中山正引理，可得。
同理，如果，这意味着。由于是有限生成的 -模，根据中山正引理，可得。
结论
因为我们已经证明了或者，所以根据中山正引理，必然有或者。
这就完成了证明。这个证明巧妙地利用了局部环的性质、剩余域、张量积的性质以及关键的中山正引理。
Exercise.
证明中指出 ""。这里指的是由生成的 -子模，即。
作为 -模，是所有形如（其中）的多项式的集合。它可以看作是 -模，，，… 的直和：

这意味着是一个自由 -模，其基为。

环论

Fri, 02 May 2025 16:55:31 GMT

Theorem. 设为含的结合环，．若，且关于的方程在中有解．证明：．
Proof.
1）首先注意到

结果有：

因此有可逆且
2）现在考虑，则有，结合前面所证可逆，因此得．进而有

亦即．
□
Exercise (Kaplansky). 含幺环中某元若有多于一个右逆，则它必然有无限多个右逆.
Proof.
设有多于一个右逆，则它是左零因子，设理想

只需要证明：是无限集. 这是因为若是的一个右逆，则，也是的右逆.
考虑反证，假设是有限集，() 于是 . 考虑，它们显然也在内，但，两两不同，故是的一个置换.
取为的一个右逆，因为是零因子，故（若是的左右逆，则为 unit，不是零因子）从而，但是，故 . 于是 for some . 同时， for some . 于是

矛盾！
□
Exercise (注意环的乘法不包含可逆性). 设为整环，和为互素的正整数，. 如果，. 求证：.
Proof.
不妨设，由于，故存在整数使得，如果则故

由于是整环，，不是零因子，所以是零因子，故为 0. 也就是 .
若则考虑 .
□
Remark. 注意环关于乘法一般不可逆. 所以时，是没有意义的.

群论

Thu, 08 May 2025 07:15:30 GMT

见《近世代数三百题》
Proposition. 1．9．9．设为自然数，则
（1）当且仅当．
（2）如果两两互素，则．
Proposition. 1．9．10．试证阶群一定是循环群．
Proof.
首先利用 Sylow 定理证明 . 借助 Proposition 1 可知是循环群.
□
Definition. Let be a subgroup of . Define to be the intersection of all conjugates of .
Exercise.
Show that is a normal subgroup of .
Show that if is finite, then is finite. (Hint: first show that the intersection above defining is essentially a finite intersection.)
Proof.
(1) We check that for any ,

with in the notation. So is a normal subgroup of .
(2) We start with a lemma: if and are subgroups of of finite index. Then is a subgroup of of finite index. The easiest way to see this is to let act on the left cosets by left multiplication. Then the stabilizer group at is precisely . We know that the index of inside is precisely the number of elements in the orbit of the identity coset in under this action. In particular, . It then follows that .
Now, we come back to the proof of (2). As is assumed to be finite, we may choose a finite set of coset representatives of . Then for every element (writing ), we have

So is the intersection

which is the intersection of finitely many finite index subgroups. By the lemma above, is finite as well.
□

Broad classifications

Fri, 09 May 2025 02:43:00 GMT

Let be a group. The letter , , and are used for primes. Moreover, they are assumed to be distinct. is a positive integer.

is simple.

. has nontrivial center due to the class equation. Assume then

which is a contradiction. Then is a nontrivial normal subgroup of .

. is not simple since by Sylow III theorem.

Claim is not simple.

By Sylow III, we know . Then . The Sylow -subgroups is normal, thus is not simple.

By Sylow III, . If then we are done. As , , this cannot happen for . Now we consider the case .
Since each Sylow- subgroup has order , the intersection of two distinct Sylow- subgroups must be trivial. Thus there are elements of order . There left elements. Since by Sylow I, the remaining elements must form the unique Sylow- subgroup and we get . Thus is not simple.

is not simple.
Assume is simple, then . Since , and , then . Then there exists elements of order , of order and of order .

which is a contradiction.

.
Claim is not simple.
. If then we are done.
If then consider the action

Then . Since the action is transitive, . Since then , otherwise . Therefore is a proper nontrivial normal[1] subgroup of .

.
Claim is not simple.
. If then we are done.
If , consider the action

. as the action is transitive. as . Thus is proper nontrivial normal subgroup of .

The kernel is always normal.↩︎

Decomposition-of-Finite-Abelian-Groups

Fri, 02 May 2025 16:55:31 GMT

finite-abelian.pdf
Definition (indecomposable). Let be a nontrivial finite abelian group. Call indecomposable if we can't write for some nontrivial subgroups and . Call decomposable if we can write for two nontrivial subgroups and .

A group of prime order is abelian (it's cyclic) and is indecomposable.
A cyclic group of prime-power order is indecomposable because if it could be decomposed into for some nontrivial subgroups and , both and would contain a subgroup of order . This would imply that the original group has more than one subgroup of order , contradicting the property that a cyclic group has at most one subgroup of each size.
Theorem. Let be a cyclic group. Then, for every divisor of , there is a unique subgroup of of order .
Proof.
Let and . Suppose that , and let . Then , and so is a subgroup of of order .
Now suppose that is any subgroup of of order . Then for some integer . Also, , and so . Thus, for some integer , and so . Therefore, , and so .
□
Remark. The groups and are not isomorphic since is indecomposable. Or since has an element of order 4 and does not.
Theorem (Theorem 2.5). A nontrivial finite abelian group is a direct sum of indecomposable subgroups.
The proof is trivial.
Proof.
This argument will be the same as the standard proof of the existence of prime factorization in the positive integers. We argue by induction on the order of the group.
For the base case , abelian groups of order 2 are indecomposable since 2 is prime. Suppose and each nontrivial abelian group of order less than is a direct sum of indecomposable subgroups. Let be abelian of order .
Case 1: is indecomposable. We are done, since is a direct sum of itself (one term).
Case 2: is decomposable. We have for nontrivial subgroups and . Then with and being greater than 1 , so they are less than . By induction,

for indecomposable and . Then .
□

Lemma. If is an abelian group and where then for the subgroups and .
Proof.
The subsets and are subgroups because is abelian, e.g., if and then . Using multiplicative notation for a moment, we have when , but it might not be true if .
To show , write for since . For all ,

We have since , and similarly . Thus .
To show , if then and , so . Alternatively, the order of divides and , so the order divides and thus .
We have shown and , so .
□
Remark. Lemma 5 has a uniqueness aspect: , and these are the unique subgroups of with orders and . We will not need this.
Theorem. An indecomposable finite abelian group has prime-power order.
Proof.
Let be a nontrivial abelian group. We will prove the contrapositive of the theorem for : if is not a prime power then is decomposable. By Lemma 5, we are done.
□
Lemma (3.4). A nontrivial finite abelian -group with a unique subgroup of order is cyclic.
Proof.
Let be a finite abelian -group with a unique subgroup of order and let be the largest order of the elements of . Then and each element of has order where ( is a -group and is the maximal order), so all elements of have order dividing .
Let have order . Since has order is a subgroup of order , so it is the only one by assumption. To prove , we'll assume and get a contradiction.
The quotient group (this makes sense since is abelian) is nontrivial, abelian, and of -power order. By Cauchy's theorem, has an element of order , say . That means and . So we can write

for some . Since (all elements of have order dividing ) and ,

Since has order , so . Thus for some , so . Rewrite that as . The only subgroup of order is in , so . This contradicts , so .
□
Remark. Lemma 8 (3.4) is true without assuming is abelian when , but the quaternion group has a unique subgroup of order 2 and is not cyclic.
Theorem. Let be a finite abelian group and let have maximal order in . There is a subgroup such that .
See r.f. charthy.pdf section 5.

Elementary Field Theory

Thu, 15 May 2025 16:10:09 GMT

Exercise.
A perfect field can be characterized in several equivalent ways:

Separable Extensions: Every algebraic extension of is a separable extension. (An algebraic extension is separable if the minimal polynomial of every element in the extension over has distinct roots.)
Separable Polynomials: Every irreducible polynomial in (the ring of polynomials with coefficients in ) is separable (has distinct roots in its splitting field).
Characteristic Property:
If has characteristic 0 (like ), it is always perfect.
If has characteristic (where is a prime number), then is perfect if and only if the Frobenius endomorphism , defined by for all , is surjective. This means that every element in is a -th power of some element in (i.e., ). Since the Frobenius endomorphism is always injective for fields, this condition means it must be an automorphism.
Examples of perfect fields include:

All fields of characteristic 0 (e.g., the rational numbers , the real numbers , the complex numbers ).
All finite fields (e.g., for a prime , or ). In finite fields, any injective map from the field to itself must also be surjective.
All algebraically closed fields.
Now, an imperfect field is a field that does not satisfy these conditions. Based on the above:

An imperfect field must have characteristic for some prime .
In an imperfect field of characteristic , the Frobenius endomorphism is not surjective. This means there exists at least one element in that cannot be written as the -th power of any element in (i.e., ).
Imperfect fields have at least one irreducible polynomial that is inseparable (i.e., has multiple roots in an extension field).
There exist algebraic extensions of an imperfect field that are not separable.
Exercise (yau-2021-poblem-3).
NTS: the ideal is maximal, i.e. is irreducible. It suffices to show irreducible in , where is the fraction field of . Since is quadratic, if irreducible, it must have root in , denoted , where . Then

Consider the smallest orders of in the polynomials; the left is even, the right is odd; contradiction! Thus is an integral domain.

Elementary Ring Theory

Sun, 11 May 2025 05:39:13 GMT

See 肖梁
In this course, all rings are assumed to be unital and .

Field only has zero ideal.
If is an ideal such that then , is an ideal of the field thus is zero. Therefore, is maximal.
Exercise. Let be a commutative ring. If all submodules of finitely generated free modules over are free over , then is a PID.
Definition (Finitely generated free module over ). A module over a ring is called a finitely generated free module if it has a basis consisting of a finite number of elements. This means there exists a finite set of elements such that every element can be written uniquely in the form for some unique coefficients . Such a module is isomorphic to the direct sum of copies of , denoted .
The definition of as -module is similar to as -linear space.
Definition (free -module). An -module is a free -module if it has a basis, i.e., there exists a subset such that for every , there is a unique finite subset and unique scalars such that . Equivalently, is isomorphic to a direct sum of copies of , denoted for some index set .

Theorem. Let be a commutative ring. If all submodules of finitely generated free modules over are free over , then is a PID.
Proof.
First prove that is an integral domain. Suppose that with . Consider ; it is a submodule of free module; so is a free -module. However, we know that kills all of because ; this contradicts that is a free -module.
Let be an ideal; it is then a free -submodule of . We claim that is a free module of rank 1 over . Suppose not, then there exist forming an -basis of . But if , we know that , contradiction! This says that is free of rank one, i.e. is generated by one element so that is a principal ideal. Thus is a PID.
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Okay, let's break down this proof step by step.
The Statement to be Proven:
"Let be a commutative ring. If all submodules of finitely generated free modules over are free over , then is a PID."
Let's define the key terms:

Commutative Ring (): A set with addition and multiplication operations that satisfy certain properties (associativity, commutativity of both operations, distributivity, additive identity 0, additive inverses, multiplicative identity 1). "Commutative" means for all .
Module over (or -module): Similar to a vector space over a field, but scalars come from a ring .
Free -module: An -module that has a basis, i.e., a linearly independent spanning set. A finitely generated free module has a finite basis. itself is a free -module of rank 1 (with basis ). (tuples of elements from ) is a free -module of rank .
Submodule: A subset of a module that is itself a module under the same operations and with scalars from the same ring.
PID (Principal Ideal Domain): An integral domain in which every ideal is principal.
Integral Domain: A commutative ring with no zero divisors (if and , then either or ).
Ideal: A submodule of (when is considered as a module over itself). More formally, a subset such that is a subgroup of , and for any and , .
Principal Ideal: An ideal generated by a single element, i.e., an ideal of the form for some .
The Hypothesis: All submodules of finitely generated free modules over are free over .
The Goal: Prove that is a PID. This requires proving two things:

is an integral domain.
Every ideal in is principal.

Proof Breakdown:
Part 1: Prove that is an integral domain.

"Suppose that with ."
This is the standard way to start a proof that is an integral domain by contradiction. We assume is not an integral domain, which means there exist non-zero elements and whose product is zero. Our goal is to show this assumption leads to a contradiction, forcing us to conclude that such and cannot exist (meaning if , then or ).
"Consider ; it is a submodule of free module; so is a free -module."
itself can be considered as a (finitely generated) free -module of rank 1 (its basis is just the element ).
is the set . This is an ideal in , and every ideal is a submodule of (when is viewed as an -module).
Since is a finitely generated free module, and is a submodule of , the hypothesis of the theorem applies: " is a free -module."
"However, we know that kills all of because ; this contradicts that is a free -module."
"Kills" means for all . Let . Then for some .
So, . Since we assumed (and is commutative, so ), we have .
So, for all .
Now, why is this a contradiction to being a free -module?
If is a free -module, it has a basis.
Case 1: (the zero module). This means . But we started by assuming (i.e., ). So this case is ruled out by our initial assumption about .
Case 2: . Then is a non-zero free -module. Let be a basis for . Since , there is at least one basis element, say .
Since , we know from above that .
If an element annihilates a non-zero element in a free module, this means that the module cannot be free unless forces all coefficients to be zero in some sense.
More precisely: A free module over a ring has the property that if for some and , then either is a zero divisor for elements related to , or . If is free with basis , then any can be written as . If , then by linear independence, for all .
Here, is free. If , it has a non-empty basis, say . Take any basis element . We know . Since (it's a basis element of a non-zero module) and (by assumption), this implies that is a zero-divisor "acting on" .
The crucial property of a free module over a ring is that if annihilates every element of (i.e., for all ), and , then must be .
In our situation, and (since ). We have for all . This means is a non-zero element that annihilates the entire non-zero module . This can only happen if is not actually free (unless itself has issues). The only way a non-zero scalar can annihilate every element of a non-zero free module is if the module itself is the zero module (which we've ruled out because ) or if the ring itself isn't "nice."
The more direct contradiction: A non-zero free module cannot have a non-zero annihilator. The set of elements such that for all is called the annihilator of , denoted . If is a free -module and , then .
Here, . Since , (as ). We found that . But we assumed . This means , which contradicts the fact that is a non-zero free module.
This contradiction shows that our initial assumption ( with ) must be false. Therefore, must be an integral domain.
Part 2: Let be an ideal; it is then a free -submodule of . We claim that is a free module of rank 1 over .

"Let be an ideal; it is then a free -submodule of ."
An ideal of is, by definition, a submodule of (when is considered as an -module).
As established before, is a finitely generated free -module (rank 1, basis ).
By the hypothesis of the theorem, any submodule of a finitely generated free module is free. Thus, must be a free -module.
"We claim that is a free module of rank 1 over ."
This means should have a basis consisting of a single element. If for some , then would be its basis (assuming ).
"Suppose not, then there exist forming an -basis of . But if , we know that , contradiction!"
Let's assume the rank of (as a free -module) is not 1.
Could the rank be 0? If rank is 0, , which is the principal ideal . This fits the "rank 1 or 0" pattern, as is generated by one element. So the argument is primarily against rank .
So, assume the rank is greater than or equal to 2. This means there are at least two distinct basis elements in any basis for . Let and be two distinct elements from such a basis.
Consider the expression: .
Here, and are elements of the ideal , and also elements of the ring .
The coefficients are and , which are also elements of .
The expression is a linear combination of the basis elements and .
.
Since is a commutative ring (this was stated at the beginning), .
So, .
This is a linear combination of basis elements that equals zero:
, where and .
For a set of elements to be a basis, they must be linearly independent. This means if a linear combination of them is zero, all coefficients must be zero.
So, we must have and .
But if , then it cannot be a basis element of a free module (unless the module is the zero module, which would have an empty basis or rank 0, not ). Basis elements are non-zero.
The contradiction: We found a linear combination of distinct basis elements which is . For this to not violate linear independence, the coefficients must be zero. That is, and . But basis elements of a free module must be non-zero if the module itself is non-zero (and if has rank , it's non-zero).
This specific relation is always true in a commutative ring. If are part of a basis for an ideal , and these are considered as elements of the module , the coefficients in the linear combination are and .
If is non-zero, and its basis has at least two elements . Then .
For this to be a non-trivial relation, at least one of the coefficients or must be non-zero. Since are basis elements, they are non-zero (assuming ). So the coefficients and are non-zero (since is now an integral domain, if , then , which is not possible for a basis element).
So we have a linear combination of basis elements equal to zero, where the coefficients are non-zero elements from . This contradicts the definition of a basis for a free module over an integral domain (unless was the zero ring, which it isn't as it has a 1 and is an integral domain).
Therefore, the assumption that the rank of is must be false.
"This says that is free of rank one, i.e. is generated by one element so that is a principal ideal."
Since is a free -module and its rank cannot be , its rank must be 0 or 1.
If rank is 0, then , which is the principal ideal .
If rank is 1, then has a basis consisting of a single element, say , where . This means every element can be written uniquely as for some . This is precisely the definition of the principal ideal generated by , i.e., .
So, every ideal in is a principal ideal.
Conclusion:

We have shown that is an integral domain (Part 1).
We have shown that every ideal in is a principal ideal (Part 2).
By definition, an integral domain in which every ideal is principal is a Principal Ideal Domain (PID).
Thus, the proof is complete.

Exercise. Let be a finite field with elements.
How many monic irreducible polynomials are there in of each degree ?
What are the possible Galois groups of the splitting field of a (not necessarily irreducible) polynomial of degree 5 over ? Why?
Proof.
(1) Each irreducible polynomial of degree 2 has exactly two zeros in .
Denote one zero of the irreducible polynomial by , then , otherwise , thus , which is a contradiction. Claim that . Since , , then , ; but and the minimal polynomial of is unique (by definition); so . is the spliting field of over , with degree . , which means is a 2-dimensional vector space over ; , thus . Claim that . This refers to the construction of , more generally, , which is the splitting field of over . It suffices to verify that every field with elements is a[1] splitting field of over . Consider the multiplitcative group, ; it has elements. For any , by Lagrange's theorem from group theory, is divided by , the order of ; thus ; is the solution of . Including , then any element in is the solution of , which has distinct zeros. Since also has distinct elements, they correspond to different roots of . By the definition of , ; thus ; similarly, .
So there are irreducible polynomials of degree 2.
This is because, every element in the extension of has a unique minimal polynomial , which is the definition of minimal polynimial. We know that is the minimal polynomial of exactly two elements, in ; each irreducible polynomial of degree 2 corresponds to 2 different elements in . Thus there are irreducible polynomials of degree 2. We are done!
Similarly, there are irreducible polynomials of degree 3, and there are irreducible polynomials of degree 5.
Each irreducible polynomial of degree 4 corresponds to four elements of ; so there are irreducible polynomials of degree 4.
Each irreducible polynomial of degree 6 corresponds to six elements of ; there are such polynomials.
We know that
(2) The factorization of into irreducibles corresponds to partitions of 5:
If is irreducible, the splitting field is . The associated Galois group is .
If factors as the product of a degree 2 and a degree 3 polynomials, the splitting field is . The associated Galois group is .
In other cases, we can get similarly , and . The associated Galois groups are , respectively.
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Exercise. Suppose that is a UFD for which every nonzero prime ideal is maximal. Show that is a PID.
Proof.
We first prove that for two prime elements and , either they are associates, or there exists such that . Indeed, if and are not associates, the ideals and cannot have containment relations (otherwise, say , we must have ; which would immediately forces and to be associates). Now as nonzero prime ideals (such as and ) are maximal, the ideal must be the unit ideal, i.e. there exists such that .
Next, we show that if, in the factorization of two elements , no prime factors of are associates of prime factors of , then there exists such that . By induction, it suffices to prove that: if , then . Indeed, write and for , then

This implies that .
We finally prove that is a PID. Let be a nonzero ideal. Pick an element with minimal number of prime factors. We show that . If , then write and and with , and have distinct prime factors. By the discussion above, there exist such that . This implies that , contradicting with the minimality of prime factors of . Thus is a principal ideal.
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Proposition. PID is Bézout integral domain; UFD may not.
Definition (Bézout domain). A Bézout domain is an integral domain in which every finitely generated ideal is principal.
Equivalently, an integral domain is called Bézout domain provided that for any , exists, i.e. , i.e. s.t.

By the definition, PID is Bézout domain. But there exists non-Bézout UFD, e.g. . Consider in ; UFD admits gcd; , but no satisfy , as letting , then . (a contradiction!)

"a" means "the", since the splitting field is unique up to isomorphism.↩︎

Galois-theory

Fri, 02 May 2025 16:55:31 GMT

因为 Galois 理论本来是研究多项式的根的，所以我们知道 normal and separable 的条件是必然的. 我们称一个扩张是 normal，如果在内有一个根，那么的所有根都在内. 我们称一个扩张是 separable，如果任意的极小多项式都无重根. 这保证了我们在进行多项式的根的置换的时候是对于在中的个不同元素来进行的（若是有限扩张）.

Galois 理论的基本定理揭示了群和域之间存在的反序一一对应．确切地说，设是有限 Galois 扩张，是域同构．令的中间域的子群．考虑如下两个映射：
，其中，
，其中．
注意到也是有限 Galois 扩张，．从而有

Galois 理论的基本定理（主要）是说：
（1）和是互逆的反序的映射．于是有

（2）是的正规子群当且仅当是正规扩张；或等价地，是正规扩张当且仅当是的正规子群．在这种情况下，有

或等价地，

Lemma (Artin 引理). 设为域的一个有限自同构群，为它的不动域，则．

Theorem (Primitive element theorem). Theorem 16.3.1. A finite separable extension of fields is generated by one element.
In fact, we have a stronger statement: if with algebraic over and separable over . Then for some .

See 代数学引论·丁、聂 p.247.

下面说明域的 Galois 群同构于半直积

这个 Galois 群作用只需要考虑在生成元和上，只有如下可能：

恰好就是 .

Galois-theory-at-work

Sat, 17 May 2025 14:20:05 GMT

Exercise. 证明：
存在恰有个元素的域；
如果是一个元域，则；
设是有限域扩张，则映射是域自同构，且满足
域扩张的群是由生成的循环群；
关于中乘法成为一个循环群．特别地，存在使．
Proof.

Let be the splitting field of the polynomial . Since , has exactly roots in the splitting field , denoted as . Directly verifying that is a subfield, we conclude that .
If , then , which implies . Thus, is also an -th degree extension of , similar to . Since forms a group of order under multiplication, we have for all . This means that the field consists of the roots of in . Therefore, and are both splitting fields of the same polynomial, implying .
From the proof of conclusion (1), we know that elements in satisfy , i.e., . Because , we have . It's easy to prove that is a field isomorphism.
Let be an -th degree extension; then is a field with elements. From the proof of conclusion (2), we know that is a splitting field of the separable polynomial , so . On the other hand, for any is not an identity map (otherwise, the polynomial would have roots in , which contradicts ), so is a cyclic group generated by .
Let be an element of maximal order in the finite abelian group . We claim that Indeed, suppose the order of is . If , then . Let be the greatest common divisor of and . is the irreducible factorization, . Then , the order of is , and the order of is . It is easy to prove that the order of is , which contradicts the choice of . Therefore, all elements in are roots of the equation , so , which means is a cyclic group generated by .
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galoisapp.pdf
is a Galois extension of because:

, and , .
is a root of , and all roots of are in because .
is a root of , and all roots of are in .
is a splitting field of over , making it a normal extension.
Since has characteristic 0, is a separable extension of .
Therefore, is a Galois extension of .
Definition (normal extension). An algebraic field extension is a normal extension if every irreducible polynomial in that has a root in factors completely into linear factors in .
Definition (separable extension). An algebraic extension is a separable extension if every element of is separable over , i.e., is a root of a separable polynomial in .
Definition (separable polynomial). A polynomial is called separable if all its roots are distinct.
Task: why splitting field of over is separable.
Output:
The splitting field of over is separable because we are in characteristic 0. More generally, if is a field of characteristic 0 and , then the splitting field of over is separable. This is because every irreducible polynomial over a field of characteristic 0 is separable.
In this case, the splitting field is an extension of . Since has characteristic 0, also has characteristic 0. The polynomials and have distinct roots. The roots of are . The roots of are . Since these are all distinct, the extension is separable.
Theorem. Every irreducible polynomial over a field of characteristic 0 is separable.
Counterexample:
Let be the field of rational functions in one variable over . Then is irreducible by Eisenstein's criterion. However, , so . Thus, is not separable because a polynomial is separable if and only if it has no repeated roots in its splitting field, which is equivalent to the polynomial and its derivative being coprime (i.e., their greatest common divisor is 1). Since , has repeated roots and is therefore not separable.

The field extension has degree 8, not 16, because . The Galois group has at most 16 automorphisms, determined by and , but the choices of and are not independent. Since , is a common subfield of both and .
The effect of on partially determines it on , and conversely: , which gives the relation

This tells us that if then , so , while if then , so . For example, can't both fix and send to because Equation (1) would not hold.

Theorem. Let and be Galois over . There is an injective homomorphism given by . In particular, if and are finite abelian extensions then is a finite abelian extension of .
Proof.
We omit this proof
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Proposition. Consider a finite extension is normal if and only if any of the following equivalent conditions hold:
Every irreducible polynomial in that has at least one root in splits completely into linear factors over .
is the splitting field of some polynomial in .
Every -embedding of into an algebraic closure of has its image contained in . That is, if is a field homomorphism such that for all , then .
Task: The Galois group of over is . Its Galois group over is trivial. Why trivial?
Answer:
The Galois group of a polynomial over a field is the group of automorphisms of the splitting field of the polynomial that fix .
In this case, we are considering the polynomial over .
The roots of are , and the roots of are . Since and are both real numbers, all four roots are in . Thus, the splitting field of over is itself.
The Galois group consists of automorphisms of the splitting field that fix the base field. In this case, we seek automorphisms of that fix . The only such automorphism is the identity map. Therefore, the Galois group is trivial.

Theorem (Theorem 4.10). Let be a separable polynomial of degree . (a) If is irreducible in then its Galois group over has order divisible by .
(b) The polynomial is irreducible in if and only if its Galois group over is a transitive subgroup of .
Proof.
(a) For a root of in is a factor of the degree of the splitting field over , which is the size of the Galois group over .
(b) First suppose is irreducible. For two roots and of , we can write for some in the Galois group of over . Therefore the Galois group, as a subgroup of , sends to , so it is a transitive subgroup. Now suppose is reducible (so ). It is a product of distinct irreducibles since it is separable. Let and be roots of different irreducible factors of . These irreducible factors are the minimal polynomials of and over . For any in the Galois group of over has the same minimal polynomial over as , so we can't have . Therefore, as a subgroup of , the Galois group of does not send to , so it is not a transitive subgroup of .
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Definition (transitive subgroup). A subgroup of is called transitive if for every , there exists a such that .
Theorem. Let be an irreducible polynomial of prime degree with all but two roots in . The Galois group of over is isomorphic to .
Proof.
Let be the splitting field of over . The permutations of the 's by provide an embedding and is divisible by by Theorem 4.10, so contains an element of order by Cauchy's theorem. In , the only permutations of order are -cycles (why?). So the image of in contains a -cycle.
We may take to be a subfield of , since is algebraically closed. Complex conjugation restricted to is a member of . Since has only two non-real roots by hypothesis, complex conjugation transposes two of the roots of and fixes the others. Therefore contains a transposition of the roots of . (This is the reason for the hypothesis about all but two roots being real.)
We now show the only subgroup of containing a -cycle and a transposition is , so . By suitable labeling of the numbers from 1 to , we may let 1 be a number moved by the transposition, so our subgroup contains a transposition . Let be a -cycle in the subgroup. As a -cycle, acts on by a single orbit, so some with sends 1 to : . This is also a -cycle, because has order in and all elements of order in are -cycles, so writing as and suitably reordering the numbers (which replaces our subgroup by a conjugate subgroup), we may suppose our subgroup of contains the particular transposition (12) and the particular -cycle . For , it is a theorem in group theory that the particular transposition (12) and -cycle generate , so our subgroup is .
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Task: Why? In , the only permutations of order are -cycles
Output:
The order of a permutation written in cycle notation is the least common multiple of the lengths of its cycles. To get order , where is prime, we need the least common multiple of the cycle lengths to be . Since is prime, the cycle lengths must be or . Since the lengths must add up to , we must have one cycle of length and all the other cycles of length . Thus, we have a -cycle.
Task: why? in group theory that the particular transposition (12) and -cycle generate .
Output:
The symmetric group is the group of all permutations of elements. We want to show that the transposition (1 2) and the -cycle (1 2 ... ) generate .
Let and . We want to show that any permutation in can be written as a product of and .
First, we can generate any transposition of the form for . We have

For example,

So we can generate any transposition of the form .
Next, we can generate any transposition for . We have

Since we can generate any and , we can generate any .
Finally, since any permutation can be written as a product of transpositions, and we can generate any transposition using and , we can generate any permutation in using and .
Therefore, the transposition (1 2) and the -cycle (1 2 ... ) generate .

Example (4.13). The polynomial is irreducible in since it is irreducible mod 2 or since it is a cubic without any rational roots. It has one real root (approximately 1.3247), and one root of a cubic is all but two roots, so its Galois group over is isomorphic to .
Example (4.14). The polynomials and are both irreducible in since they are cubics without any rational roots. Each polynomial has three real roots (check!), so we can't use Theorem 4.11 to determine their Galois groups over .
Example (4.15). The quintic polynomial is irreducible in since it is irreducible mod 3. It has one real root, so Theorem 4.11 does not tell us the Galois group.
Example (4.16). The quintic polynomial is irreducible in since it is irreducible mod 3. It has three real roots, which is all but two roots, so its Galois group over is isomorphic to .

In general, if with and such that for any divisor of , , then the normal closure of is . The Galois group is always a subgroup of semidirect product .
In "majority" cases, the Galois group is isomorphic to .
is a normal subgroup of , which corresponds to that is a Galois extension of . On the other hand, is not a normal subgroup and thus is not normal over .
Definition (transitive). A subgroup of the symmetric group is said to be transitive if for every , there exists a such that . In other words, for any two elements in the set , there is a permutation in the subgroup that maps the first element to the second.

Definition (discriminant). For a nonconstant of degree that factors over a splitting field as the discriminant of is defined to be
The number disc is nonzero if is separable and is 0 if is not separable.
When is separable, is a symmetric polynomial in the 's, so it is fixed by and therefore by Galois theory.
Theorem. Let not have characteristic 2 and let be a separable cubic in with a root and discriminant . The splitting field of over is . Note we are not assuming is irreducible here.
Theorem (Theorem 4.23). Let be a separable polynomial of degree . If does not have characteristic 2, the embedding of the Galois group of over into as permutations of the roots of has image in if and only if disc is a square in .
Proof.
设，因此且。因此，是中的一个平方，当且仅当。
对于任何，设是其作为的排列的符号。根据排列符号的定义之一，

因此。由于且没有特征。当且仅当时，我们有，因此当且仅当时，。因此，在上的伽罗瓦群在中当且仅当被伽罗瓦群固定，这与相同。
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Theorem (Theorem 4.25). Let not have characteristic 2 and let be a separable irreducible cubic in . (a) If disc is a square in then the Galois group of over is isomorphic to . (b) If disc is not a square in then the Galois group of over is isomorphic to .
Remark (why require the characteristic ). If the characteristic of the field is 2, then , which means every permutation is even since .

In low-degree cases, explicit formulas for discriminants of some trinomials are

Theorem (Dedekind). Let be monic irreducible over of degree . For any prime not dividing disc , let the monic irreducible factorization of be and set , so . The Galois group of over , viewed as a subgroup of , contains a permutation of type .
The nicest proof of Theorem 21 (Dedekind) uses algebraic number theory and is beyond the scope of these notes.
Example. We compute the Galois group of over using Theorem 21 (Dedekind) .
This polynomial is irreducible , so it is irreducible over . Let its roots be . The extension has degree 4 , so the Galois group of over has order divisible by 4 . Since the Galois group embeds into , its size is either 4 , 8,12 , or 24 . The discriminant of is -283 , which is not a rational square, so the Galois group is not a subgroup of . This eliminates the possibility of the Galois group having order 12 , because the only subgroup of with order 12 is . (Quite generally, the only subgroup of index 2 in is for .) There are subgroups of with orders 4, 8 , and (of course) 24 outside of , so no other size but 12 is eliminated yet. Using Theorem 21 (Dedekind) with , This is an irreducible factorization, so Theorem 21 (Dedekind) says that the Galois group of over contains a permutation of the roots with cycle type , which means the Galois group has order divisible by 3 , and that proves the Galois group is .
Task: why implies irreducibility? We compute the Galois group of over using Theorem 21 (Dedekind). This polynomial is irreducible , so it is irreducible over .
A polynomial is irreducible over if it cannot be factored into non-constant polynomials with rational coefficients. Reducing a polynomial modulo a prime can be used to prove irreducibility over because if factors over , then it must also factor modulo for almost all primes . Thus, if is irreducible for some prime , then must be irreducible over .
In this specific case, since is irreducible modulo 2, it follows that is irreducible over . This is a consequence of the fact that if were reducible over , then it would also be reducible modulo 2, which contradicts the given information that it is irreducible modulo 2.

Let's determine the Galois group of over , which was left unresolved in Example 4.15. Its irreducible factorization mod 2 is

Because the polynomial is irreducible over divides the size of the Galois group. From the mod 2 factorization, the Galois group contains a permutation of the roots with cycle type , which has order 6 , so the Galois group has size divisible by . Since the Galois group is a subgroup of , its size is either 30,60 , or 120 .
It turns out that there is no subgroup of with order 30 and the only subgroup of order 60 is . The discriminant of is , which is not a rational square, so the Galois group is not in by Theorem 4.23. Therefore the Galois group is .

We want to understand why the size of the Galois group of the polynomial over the rational numbers must be divisible by .
This conclusion is derived from two key facts about the polynomial and fundamental theorems in Galois theory and group theory.

Divisibility by 5
Reason: The polynomial is irreducible over .
Theorem: For an irreducible polynomial of degree over a field , the order of the Galois group of its splitting field over is divisible by . Since is irreducible and has degree , the size of its Galois group must be divisible by 5. Equivalently, the Galois group acts transitively on the 5 roots, so is divisible by 5.
Divisibility by 6
Reason: The factorization of modulo 2 is given as: Both factors, and , are irreducible over the field .
Theorem (Dedekind's Theorem / Frobenius Elements): If a polynomial with integer coefficients factors modulo a prime into distinct irreducible factors of degrees , then its Galois group contains an element having the cycle structure . Thus, the Galois group , seen as a subgroup of , must contain a permutation with cycle structure .
The order of permutation is . Since contains an element of order 6, the order of the group must be divisible by 6.
Theorem (Lagrange's Theorem). In any finite group, the order of any element must divide the order of the group.
Proof.
By considering the partition of a finite group into disjoint left cosets of a subgroup , we can deduce that the order of divides the order of , and consequently, the order of any element in must also divide the order of .
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From Theorem 10 and Theorem 21 (Dedekind), we have
Corollary. Let be monic irreducible over of prime degree . If there is a prime number not dividing disc such that has all but two roots in , then the Galois group of over is isomorphic to .
Proof.
The proof of Theorem 10 can be used again except for the step explaining why the Galois group of over contains a transposition. In Theorem 10 this came from the use of complex conjugation to transpose two non-real roots, assuming there are only two non-real roots. We aren't assuming that anymore. By hypothesis the factorization of has all linear factors except for one quadratic irreducible factor. Therefore Theorem 4.29 says the Galois group contains a permutation of the roots with cycle type , which is a transposition in .
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If we seek an analogue of Theorem 10 for a Galois group to be isomorphic to , using 3 -cycles in place of transpositions, there is no analogue since an irreducible polynomial over can't have all but three roots in (the number of non-real roots is always even). But could have all but three roots in for some . This suggests the next result.
Corollary. Let be monic irreducible over of prime degree with disc a perfect square. If there is a prime number not dividing disc such that has all but three roots in , then the Galois group of over is isomorphic to .
Proof.
Let be the Galois group, so is a subgroup of since is a square. The Galois group has order divisible by , so it contains a -cycle. From the factorization of and Theorem 21 (Dedekind) , contains a 3 -cycle. It is a theorem of C. Jordan that for any prime , any -cycle and any 3 -cycle in generate , so .
□
Example. The polynomial has discriminant . It is irreducible , so it's irreducible over . Modulo 7, its irreducible factorization is This has all but three roots in , so the Galois group of over is isomorphic to .
Remark. It is a hard theorem of Chebotarev that the sufficient conditions for to have Galois group in Corollary 24 and in Corollary 25 are also necessary in a strong sense: if is monic irreducible of prime degree with Galois group over isomorphic to (resp., ) then there are infinitely many primes not dividing such that

To decompose , we want to find its roots in . We can test each element:

For , .
For , .
For , .
For , .
For , .
Since is a root, is a factor. Then we can perform polynomial division to find the other factor.
Note that , so we can write .
in .
Now we consider the quadratic . We want to see if it can be factored further. We can check its discriminant:
.
Since is not a quadratic residue modulo (i.e., there is no such that ), the quadratic has no roots in . The quadratic residues modulo 5 are , , , , .
Thus, is irreducible over .
Therefore, the decomposition of is .

Verifying whether a polynomial is irreducible in a given field is a fundamental problem in algebra. Here are several methods and theorems that can be used:

Theorem (Eisenstein's Criterion). Let be a polynomial with integer coefficients. If there exists a prime number such that:
for all ,
,
, then is irreducible over .
This criterion directly proves irreducibility over by checking divisibility conditions of the polynomial's coefficients by a prime number.

Choose a prime such that the degree of the polynomial with integer coefficients remains the same when reduced modulo .
If the reduced polynomial is irreducible over , then is irreducible over .
If the reduced polynomial is reducible, no conclusion can be made about the irreducibility of over .
This method involves reducing the polynomial modulo a prime and checking for irreducibility in the finite field .

For a polynomial with integer coefficients, any rational root must have the form , where divides and divides .
A polynomial of degree 2 or 3 with no rational roots is irreducible over .
This theorem helps find potential rational roots, and if none exist for polynomials of degree 2 or 3, it confirms irreducibility.

If is a polynomial over a field , and is a root of in some extension field, then is irreducible if and only if has degree equal to the degree of .
This method checks irreducibility by examining the degree of the field extension formed by adjoining a root of the polynomial.

Algorithms like Berlekamp's algorithm determine if a polynomial is irreducible over a finite field.
These algorithms provide computational methods to test for irreducibility, especially over finite fields.

Lemma (Gauss's Lemma). If a polynomial with integer coefficients can be factored into two polynomials with rational coefficients, then it can be factored into two polynomials with integer coefficients.

Suppose that is a polynomial with integer coefficients. Assume that it can be factored into two polynomials with rational coefficients, say , where and have rational coefficients.

Clear denominators: Multiply and by suitable integers and respectively, such that and have integer coefficients. Thus, we have , where and are polynomials with integer coefficients.
Remove common factors: Divide and by the greatest common divisor of their coefficients. Let and , where is the greatest common divisor of the coefficients of and is the greatest common divisor of the coefficients of . Then and are primitive polynomials (polynomials with integer coefficients whose coefficients have a greatest common divisor of 1). We then have .
Reduce to primitive polynomials: Let . Then . Since and have integer coefficients, their product also has integer coefficients. Furthermore, has integer coefficients. Hence must be a rational number such that has integer coefficients.
Show is an integer: Since and are primitive polynomials, their product is also a primitive polynomial. This is a key step and is often proven separately. Now, since and is a polynomial with integer coefficients, and is primitive, it follows that must be an integer.
Conclude the factorization: Since is an integer and , we have a factorization of into two polynomials with integer coefficients: , where and have integer coefficients after adjusting for the constant factor. We can then rewrite where .
Thus, can be factored into two polynomials with integer coefficients.
Remark. This lemma relates irreducibility over to irreducibility over , ensuring that if a polynomial is irreducible over the integers, it is also irreducible over the rationals.

Theorem (Hilbert's Irreducibility Theorem). If is an irreducible polynomial in , then there exist infinitely many tuples of rational numbers such that is irreducible in .
This theorem is used in advanced settings to prove the existence of irreducible polynomials by specializing variables.

Lemma (Capelli's Lemma). Let be a field and be an irreducible polynomial over . Let be a root of . If , then is irreducible over if and only if is irreducible over .
Given an irreducible polynomial over a field and a root of , this lemma relates the irreducibility of another polynomial over to the irreducibility of over .

Theorem (Irreducibility over Finite Fields). Let be a polynomial of degree over . Then is irreducible if and only if divides and for all with .
A polynomial of degree over a finite field is irreducible if and only if divides and for all with .

Galois-theory-at-work-1

Wed, 07 May 2025 17:01:14 GMT

Definition (trace, norm). If is a finite Galois extension with Galois group , the characteristic polynomial of is . In particular,

If and for a nonsquare in , the two elements of are determined by and , so

and

Also

For ,

The Galois group of is a cyclic group of order , generated by the Frobenius automorphism , where .

Theorem (Theorem 6.1). Let and be Galois over . a) The embedding given by is an isomorphism if and only if . In particular, if and only if . b) The image of the embedding in part a is the set of compatible pairs of automorphisms: on .
Proof.
We omit the proof.
□
Example. Let and . Both are Galois over , we know their Galois groups, and . Define and by the conditions These agree on since , so there is a unique that restricts to on and on .
Example (the Galois hypotheses cannot be removed). If , and , then but . See the field diagram below on the left.
If one of and is Galois, Theorem 2 (Theorem 6.1) still holds.
Theorem (Theorem 6.6). Let and be finite extensions with a Galois extension.
The extension is finite Galois and by restriction. In particular, so if and only if .
The sets of intermediate fields and are in bijection by , with inverse . In particular, every field between and has the form where , and if and correspond by the bijection then is Galois if and only if is Galois, in which case by restriction.
The extension is Galois, with Galois group . If we translate this extension by (which is not Galois over ), we get the Galois extension . What is its Galois group? By Theorem 5 (Theorem 6.6),

where . Obviously, , and since , , we have .

Definition. For a positive integer , where .
Define ; it is a finite field extension of , called the th cyclotomic extension of . A primitive th root of unity is a generator of ; it is equal to for some . Define it is called the th cyclotomic polynomial.
Example. We have .
Lemma. We have Each is a polynomial of degree with coefficients in .
Proof.
The first equality is easy:

We will prove that has coefficients in and its coefficients have gcd . Assume that this has been proved for smaller . Then Equation (1) and Gauss' lemma implies that has coefficients in and has coefficients' gcd .
□

Theorem (Theorem 7.1). If has characteristic 0 and is a finite extension such that and are both Galois and , then .
Proof.
Let . We will show this group is trivial.
Pick , so . Therefore

Since and are Galois over and , so and . This common difference is therefore in . Write , so

Applying repeatedly, for all integers . Choose such that is the identity (for instance, let ). Then , so . Since we are in characteristic 0 and is a positive integer, we must have , so and . Therefore is the identity on .
□
Theorem (Theorem 7.6). If has characteristic 0 and is a finite extension such that then .
Remark. The degree hypothesis in Theorem 10 (Theorem 7.6) is equivalent to when one of or is Galois over , so Theorem 9 (Theorem 7.1) is also true when only one of or is Galois over .
Example. Theorem 10 (Theorem 7.6) implies and .
Example. We know and this example does not fit Theorem 10 (Theorem 7.6) since while .
Example. Letting and be two roots of , we can't decide if equals from Theorem 10 (Theorem 7.6) since and . The two fields are not the same since has degree 6 over .
Example. Does ? Since (by Example 1.4) and , we can't answer this with Theorem 10 (Theorem 7.6). Since , you can check the Galois orbit of has size 8 , so in fact . (The minimal polynomial of over is .) Thus the degree hypothesis of Theorem 10 (Theorem 7.6) is sufficient to imply in characteristic 0 , but it is not necessary.

Theorem (Theorem 8.1). Every finite group is the Galois group of some finite Galois extension in any characteristic.
Theorem (Theorem 8.2). If every finite group can be realized as a Galois group over then every finite group can be realized as a Galois group over any finite extension of .

There are two important aspects of field extensions that are missing by a study of Galois theory of finite extensions, and we briefly address them:

Galois theory for infinite extensions
transcendental extensions
An example of an infinite Galois extension of is

the union of all -th power cyclotomic extensions of , where is a fixed prime.

Gauss's Lemma

Fri, 02 May 2025 08:58:54 GMT

hw1

Fri, 02 May 2025 16:55:31 GMT

习题1.2第11、12、16、22题习题1.3第1、3、7(1)（2）题
Exercise. 习题2．11．设是群的子群，且．证明或者 .
Assume that and , combined with the fact that , we have an element and an element . Since , then . If then for some , thus , which contradicts the fact . If then , thus , which contradicts the fact . Hence or .
Exercise. 习题2．12．在偶数阶群中，方程总有偶数个解．
is even. If then , . Thus is even. Therefore is even too.
Exercise. 习题2．16．设是群的两个子群．试证是的子群当且仅当 .

() If then for any , let , thus

Let then

Hence .
() If then for any , we have for some . Therefore for any , we have

Hence .
Exercise. 习题2．22．证明有理数加法群和乘法群不同构．
If then there is an isomorphism from to . Since is isomorphism, it's a bijective homomorphism. Then there exists an element in such that , then

Then , which is a contradiction. Hence .
Exercise. 习题 3．1．设试求和在中的阶．

Then .
Exercise. 习题 3．3．设是群的两个元素，的阶是 7 且．证明．

Exercise. 习题3．7．（1）的任意有限阶子群均为循环群．
（2）不是循环群，但它的任意有限生成子群都是循环群．
(1)
. For any subgroup with finite order, pick , then is finite. ( denotes the order of .) WLOG, let be the element with largest order , and let . We need to show that

If is an element in but . If , then while , which means . Thus , denoted by , where . for some . Then

and , thus , is not the element with largest order in .
Hence is cyclic.
(2)
If is cyclic group generated by , then is also a generator and . Hence is not cyclic.
For , where , , . Obviously,

is a subgroup of a cyclic group, thus a cyclic group.

hw1-wrong

Fri, 02 May 2025 16:55:31 GMT

习题1.2第11、12、16、22题习题1.3第1、3、7(1)（2）题
Exercise. 习题2．11．（1）验证均为的子群，且．
（2）验证均为的子群。
（3）验证是的子群。
Definition. （i）为对角线元全为 1 的阶上三角阵；
（ii）为阶可逆对角阵集合；
（iii）为阶可逆上三角阵集合。
(1)
It suffices to show that , . Since , is invertible, thus . Moreover, , thus .
For ,

Then

Therefore

For , we have

where are invertible in . Then

Therefore

where because is a field, and is invertible. Thus .
For , we have

where are invertible in . Then

Therefore

where because is a field, and is invertible. Hence .
Since , then

(2)

For , we have . Similarly, we have . Hence .
For , we have

Hence .
For , we have

Hence .
(3)

For , we have

Simliarly . Hence .
Exercise. 习题2．12．试证群的任意多个子群的交仍是的子群．
Let be an index set, . For any , we have for any . Thus , then . Hence .
Exercise. 习题2．16．（1）确定的所有子群．
（2）确定的所有子群，其中．
(1)

(2)

Exercise. 习题2．22．群的自同构称为没有不动点，是指对的任意元素，．如果有限群具有一个没有不动点的自同构且，证明一定是奇数阶阿贝尔群．
Consider the pair , where for and . Then is odd.
Consider , claim that . (othewise for some , then ) Thus for any , we have for some , then . , . Therefore is abel.
Exercise. 习题3．1．设试求和在中的阶．

Then .
Exercise. 习题3．3．设是群的两个元素，的阶是 7 且．证明．

Exercise. 习题3．7．设和是群的元素，阶数分别是和且．试证是的阶循环子群．

If for some where and , then

Since , we have , which means . Similarly , thus . Hence is a cyclic subgroup of of order .

hw2

Fri, 02 May 2025 16:55:31 GMT

习题1.2第17，19，24题习题1.3第8，9，13，16（1)(2)题
Exercise. 习题2．17．设和是有限群的两个非空子集．若，证明．特别地，如果是的一个子集，．证明对任意，存在使得．
Check: , s.t..
For any but fixed , let . If . Then

Contradiction! Thus , i.e. there exists , s.t., i.e. for some , i.e. , i.e. .
Let , then . Therefore for any , , s.t. .
Exercise. 习题2．19．证明：映射是的自同构当且仅当是阿贝尔群。
() If is automorphism in , then for any , we have

Therefore is Abel.
() If is Abel, then obviously is a bijection in . It suffices to show that is homomorphism. For any ,

Hence .
Exercise. 习题2．24．群的自同构称为没有不动点，是指对的任意元素，．如果有限群具有一个没有不动点的自同构且，证明一定是奇数阶阿贝尔群。
Consider the pair , where for and . Then is odd.
Consider , claim that . (othewise for some , then ) Thus for any , we have for some , then . , . Therefore is Abel.
Exercise. 习题3．8．设和是群的元素，阶数分别是和且．试证是的阶循环子群．

Denote , , .

Therefore , since , we have . Similarly, . . Hence .
Exercise. 习题3．9．设为奇素数，是阶整系数方阵．如果的阶有限，证明．

扩充数域到复数域，我们有 Jordan 分解

其中，显然这些若当块都是特征值为 0 的. 若不全是一阶若当块，必然有某个若当块使得

但这与

矛盾，故 .
Exercise. 习题3．13．（1）设是阿贝尔群，是中所有有限阶元素构成的集合．证明是的子群．
（2）＊举例说明上述结论对于一般群不正确．
(1)

For any , we have

Then .
(2)
反例：无穷二面体群
考虑无穷二面体群，它由以下生成元和关系定义：

其中：

的阶为 2 （即）。
具有无穷阶（即对任何）。
群的元素形如或，其中。
列出有限阶元素：

所有都是无穷阶的。
仅有和的阶为 2，因为

所以，所有的有限阶元素集合为：

检查封闭性：

，在中。
可能是无穷阶的，不在中！这说明不是封闭的，因此它不是一个子群。
这个例子说明了：对于一般群，有限阶元素的集合不一定构成子群。
Exercise. 习题3．16．回答下列问题：
（1）设是素数，方幂阶群是否一定含有阶元？
（2） 35 阶群是否一定同时含有 5 阶和 7 阶元素？
(1)
If , for any ,

Then . If , then . If then or , thus either or has order .
(2)
If , for any ,

Then or . If contains an element with order 35, then .
Assume there is no elements with order 35, if , then there are 34 elements with order 5. If then are distinct elements in with order 5, and . But .
If . Then , which is not true.
Hence contains element with order 5 and 7 at the same time.

hw3

Fri, 02 May 2025 16:55:31 GMT

习题1.3第20题习题1.4第3，4，6，9，11题
Exercise.
Proof.
若，则，. 故 . 同理 . 故 .
若，则

左乘，再右乘得到

于是，同理，因此 .
□
以后我们用表示，即的阶.
Exercise.
(1) 若，则，有，显然 .
(2) 不一定，比如

但是 .
Exercise.
Proof.
(1)

Check that ,

Fix , then for any ,

Therefore .
(2)
If with , then can be split to two orbit

Then the left and right cosets of are the same.
For any given , check that .
If , then . If , then . Therefore .
□
Exercise.
Proof.

Since is cyclic, then

for any nontrivial .
Fix , check that .
We know that must lie in for some . Then for some . Thus

Therefore is Abel.
□
Exercise.

Denote that

Then

Thus

Then

Since , we have . Then

Exercise.
Proof.
Since , for any , we have

Then for any ,

And

Thus

□

hw4

Fri, 09 May 2025 02:43:00 GMT

习题2.1第1，3，4，5，6，9，10，13题

If then

The number of transpositions in the factorization is . If then is even. If then is odd.
If then

The number of transposition in the factorization is . If then is odd. If then is even.
Therefore, if , is odd. If , is even.

Suppose that a permutation can be decomposed into disjoint cycles

Each has length , then . Therefore . By the definition of , we have

Therefore for each . Thus . Hence

先证明中型为的置换共有个. 我们进行如下操作：先将分到个盒子中：

然后无视这些盒子的次序，再考虑每个盒子内元素的对应关系，比如对于有个元素的盒子，要求

先从开始考虑，的选取有种（），的选取有种... 于是一共有种.
按照上述逻辑考虑置换的种类可以知道，中型为的置换共有

同时我们考虑的所有置换，按照分类计算可得

于是

Lemma. () 非单.
Proof.
Ansimple.pdf
□
Lemma. .
Proof.
Any permutation can decomposite to product of transformations. is defined to be the permutations in with even numbers of transformations in its decomposition. is the kernel of

Thus .
□
时，的全部正规子群为 .
When , evaluate all the normal subgroups of .

. There are 4 sylow 3-subgroups of :

Thus is not normal in . There is only one sylow 2-subgroup of :

Thus is normal in . C'est clair is simple.
的全部正规子群为 .

Since ,

When , the simplicity and noncommutativity of Lemma 1 implies thus . When , maybe or . Since

Then is trivial.

由于指数为 2 的子群必然是正规子群，而且时的非平凡正规子群只有交错群，时，的非平凡正规子群只有和 . 故得证！

hw5

Fri, 09 May 2025 16:52:23 GMT

习题2.3 第3、7、10题习题2.4第6、8、11、14题

Proof.
There are 3 cases. or .
If then . Consider the homomorphism

acts on by left multiplication,

, otherwise then is not simple. Thus , .
If then . Consider the homomorphism

acts on by left multiplication,

, otherwise then is not simple. Thus . Since is not simple, .
If then . Consider the homomorphism

acts on by left multiplication,

, otherwise then is not simple. Thus , . Also, implies that

is divisible by . Thus or . If then or , not simple. If then or or , not simple. If , it has a Sylow 3-subgroup, then is not simple.
□

Proof.
Assume that , then . Consider

acts on by left multiplication,

Then , thus . For any , we have . For any in , for some , thus , . But which is a contradiction.
□

Proof.

□

Proof.
(1) then then . By Sylow II theorem, the Sylow 37-subgroup is normal in . Thus is not simple.
(2) , then then . By Sylow II theorem, the Sylow 5-subgroup is normal in . Thus is not simple.
(3) then . If , denote the Sylow 2-subgroups by . Then consider the homomorphism

Assume that is simple, then , , thus . But , which is a contradiction. Therefore is not simple.
□

Proof.
, and are relatively prime, then for some . By Sylow I theorem, contains a Sylow -subgroup with order . is also a Sylow -subgroup of . By Sylow II theorem, all Sylow -subgroups are conjugate to , i.e. have the form for some . NTS: . Since , is equivalent to , which is trivial. Hence contains all the Sylow -subgroups of .
□

Proof.
. Since , we have . Thus the Sylow -subgroup is normal in . When , the Sylow -subgroup si proper normal subgroup of . When , if then has no proper normal subgroup. If then is proper normal subgroup of .
□

. .
If , then by Sylow II theorem, the Sylow 3-subgroup is normal in .
Unsolved...
下面证明仅供参考而已，不是这样证明，但我太困了，下次再补. 下面的证明完全错误！

hw6

Fri, 02 May 2025 16:55:31 GMT

习题2.5第4题习题2.6第1，3，7，11，13，15题
Exercise.
Definition (product of two groups). The (direct) product of two groups and is the group whose underlying set is the Cartesian product of and , and whose group operation is defined componentwise: for all and .
Proof.
, because has nontrivial center

But is trivial.
□
Exercise.

(1) For , by Sylow III theorem, . Thus has unique Sylow 3-subgroup and Sylow 11-subgroup , which are normal in . For any , we have , since . is a group of order 33, since for any . Since , , we have . Therefore is an abelian group of 33 order. (Moreover, is cyclic since , are cyclic.) By fundamental theorem of finitely generated abelian groups,

Remark. The direct product and direct sum construct new groups from a collection of groups. For a finite number of groups , the direct product consists of tuples where , with component-wise operation. The direct sum is the same as the direct product in this case. For an infinite collection of groups , the direct product consists of all tuples where . The direct sum is a subgroup of the direct product, containing only tuples where finitely many are not the identity element in their respective groups . For example, with , the direct product can have elements like , while the direct sum requires elements with finitely many 1s, such as .
(2) For , by Sylow III theorem, , .
Let be the Sylow 3-subgroup of , then it's normal in (by Sylow ii theorem), with order 9. Pick a Sylow 2-subgroup of . Since , , is a subgroup of of order , so and the recognition theorem for semidirect products tells us is isomorphic to a semidirect product . The group is isomorphic to or . The group is isomorphic to . Then we are going to classify all semidirect products and .
To classify , let be a homomorphism. It is determined by , which is a solution of . By listing all the conditions, . When , the homomorphism is trivial and . When , we get a semidirect product

by

It is isomorphic to , with an isomorphism

being given by . It's clearly a homomorphism. Check the relationships:

Thus is an isomorphism.
To classify , let be a homomorphism. It is determined by , which is a pair of solution of . The solutions are . When , the homomorphism is trivial, then . When , we get a semidirect product

by

It is isomorphic to , with an isomorphism

being given by . Similarly, when , is isomorphic to .
When , we get a semidirect product

by

There are 5 different groups with order 18:

, by .
Exercise.
Definition (自由阿贝尔群的一组基). 设是一个自由阿贝尔群。的一组基是一个子集, 满足
是的生成集，即。
是线性无关的，即如果对于和, 使得那么。
(1) 记全体素数集合为，那么，这是因为任意都存在素因子分解

所以，同时由于，所以，因此 .

Exercise.

Exercise.

不变因子为，初等因子为 .
Exercise.
Proof.

The only -subgroup of is itself. Since is cyclic, the unique -subgroup of is . List all the subgroups.

Thus has 5 -subgroups.
□
Exercise.
Proof.
Since are finite abel groups,

There is an isomorphism between and . Restrict to the second component by left composing , then is an isomorphism between and . Thus

□

hw7

Fri, 09 May 2025 02:43:00 GMT

习题3.1第3，10，11，12题习题3.2第1，2，5，7，9题
Exercise.
Definition (Integral domain). A commutative ring containing no zero-divisor is called an integral domain.
Definition (unit). is called a unit in if there exists such that
Proof.
显然是交换环. 假如满足 . 那么

在 Equation (1) 中，考虑，则 . 考虑，则 . 从而是整环.
对于，它是单位元当且仅当存在使得

也就是

这蕴含着

但是，这要求

当非零时有

这意味着 .
若，则 . ( 否则为零因子，非单位元.)
若，则，由取等条件可知 , .
综上：当时，单位元群为 . 当时，单位元群为 .
□
Exercise.
Proof.
(1)

(2)

is a ring and thus is bool.
□
Exercise.
Proof.
An integral domain is a commutative ring with no zero-divisor. To verify that any nontrivial finite integral domain is field, it suffices to find the inverse for every . Consider the ring homomorphism of additive groups

The kernel , due to the property of integral domain. Thus is injective, and hence is isomorphism by counting the number of elements. In particular, is a multiplicative inverse of . Then we are done.
□
Exercise.
Proof.
(1) Since are nilpotent, there exists , s.t.

Then

Either or happens. Then . Hence is also nilpotent.
(2) The conclusion in (1) may fail when is not commutative. Consider the nilpotent elements in the ring ,

But

is invertible, not nilpotent.
□
Exercise.
A subset is called an ideal provided that

, we have .
, we have .
Proof.
Assume not, i.e. there exists an ideal . For , pick , then , which is a contradiction.
□
Exercise.
Proof.
If is a ring automorphism, we have

The equation has solution in . If , then for any , , which means is not automorphism. Thus .
Then consider , we have

Then .
For any , is determined by and , since for any ,

Then

where .
(2) For , it is determined by , since for any ,

Since is automorphism, we have . Otherwise, is not unit, then is zero-divisor, i.e. for some , which means .
We construct an isomorphism between and ,

The kernel is trivial. Then is injective, and hence is isomorphism by counting the number of elements. Therefore,

□
Exercise.
Proof.
Since

we have

Therefore,

Since and , the homomorphism is surjective. Then

Consider the dimension . And

Hence,

Any proper ideal of containing is generated by 2 elements, otherwise it's trivial.
□
Exercise.
Proof.
(1)
For any , we have for some . Since is commutative, is a two-side ideal. Then

Therefore, .
For any , we have for some . Then

Hence is an ideal.
(2)
If , then it's trivial that . If , then implies that for some . Thus .
(3)
Obviously, . Conversely, for any , for some . Then for some . Hence . We are done.
(4)

For , for some . Thus . For any , for some , i.e. , where for some . Thus

Therefore . Hence .

If then for some . Then . If then for some . Then , thus . Hence

□
Exercise.
Proof.
For any , consider the sequence of principle ideals

Since the number of ideals is finite, we must have for some . Then there exists s.t.

Since is integral domain, is not zero-divisor, then . Since is commutative, is the inverse of . Due to the arbitrarity of , each nonzero element in admits an inverse. Thus is a field.
□

hw8

Sun, 11 May 2025 07:04:33 GMT

习题3.3第1，2，9题习题3.4第1，2，3，4，7，13题
Exercise.
Proof.
, for some ; then ; thus . , for some ; then ; thus . is an ideal in .
Consider the quotient ring ; if is nilpotent, for some ; then ; thus . in the quotient ring.
□
Exercise.
is the ring of matrices with entries in the ring .
Proof.
Clearly, is an ideal of the ring . Construct the ring homomorphisms

Then , . Thus

□
Exercise.
Proof.
(1)
The center of a ring , denoted , is defined as

Since for any , then ; .
(2)
Check that is an ideal in .
For any and ,

For any ,

Thus is an ideal in . Similarly, is an ideal in .
Construct the mappings

We have

Thus and are homomorphisms. Also , . Then

□
Exercise.
Proof.
We prove by induction on that

The base case is clear, and suppose that we have prove Equation (1) for , and now we prove this for . By induction hypothesis, for each , we may find $x_i\in $ such that

If for some , , we have already verified Equation (1) using this element . Now it suffices to treat the case when for every . Then consider the element

Clearly . We show that for every . Indeed, all elements in the sum except the th term belongs to . But as is a prime ideal, this product does not belong to . So . This completes the inductive proof.
□
Exercise.
Proof.
Let be a finite commutative ring with prime ideal . Then is an integral domain. As finite integral domain is field, is maximal in .
□

Exercise.
Proof.
If for some , pick , then

which is a contradiction.
□
Exercise.
Proof.
for any . Claim for some . Assume not, pick for each , then for each ; . But for each , as is prime, , which is a contradiction.
□
Exercise.
Proof.

□

Exercise. is PID; is its quotient domain; is subring of . Show that is PID.
Proof.
Let be an ideal in , . Then for any , and , thus . For any , and , then . is an ideal in . As is PID, for some in .
Claim that . Clearly, . We need to show . For any , , . Since is fraction field over PID, WLOG, assume that ; i.e. for some . Then

Since and , we have thus for some . Then

Therefore . Hence is PID.
□

hw9

Wed, 21 May 2025 16:35:09 GMT

习题4.1第3，6，8题习题4.3第6，8，9，13题
Exercise. 设为无平方因子的整数，.
证明和在上均为不可约元
证明和在上不能同时为素元
Example.
The mapping

is clearly a norm. , if , are not unit, then , satisfies , which cannot happen. , if , if , then , ; thus , , but cannot happen. , if , if , then , , then , thus is unit. Then , if is reducible, then , which cannot happen.
Definition (素元). 设是一个整环，是一个非零非单位元，如果对任意的，若，则或，那么称是的一个素元.
Example. 证明和在上不能同时为素元。
我们将分两种情况讨论：是偶数和是奇数。
情况 1：是偶数。
由于是无平方因子的偶数，且，所以可以写成，其中是奇数。我们有（因为）。所以。如果是素元，那么或者（即）。

是否可能？如果，那么。比较系数，我们得到且。这意味着整除。所以或。但我们已知。因此不整除。（另一种方法是使用范数：如果，则，即。因为是偶数、无平方因子且，所以不可能是的因子，能满足这些条件的不存在，也被排除。）
是否可能？如果，那么，即。这意味着，所以。但是无平方因子的，所以不能被整除。因此不整除。
因为整除乘积，但它既不整除也不整除，所以当是偶数（且，无平方因子）时，不是素元。
情况 2：是奇数。
我们有等式。因为是奇数，所以是偶数。设，其中是一个整数。于是。
假设是素元。
因为，所以或者。

是否可能？如果，则对某些整数成立。展开得到。比较系数，有（所以）且。代入到第二个方程：。因为是奇数且，所以。因此不是一个整数（因为）。所以不整除。
因此，如果是素元，它必须整除。这意味着对某些整数成立。将此代入原方程：。因为，所以。由于是一个整环（integral domain），我们可以消去：。这意味着。所以。比较系数：且。这对于整数是不可能的。
这个矛盾表明，如果是奇数（且，无平方因子），那么不是素元。
Exercise. 设是整环. 证明多项式环是 PID 当且仅当是域.
若为域，那么存在 , 故是 ED，故为 PID. 若为 PID，那么对任意理想，考虑理想，由于是 PID，故存在使得 . 由于，故，，由于，故为单位元. ，由于，故，是单位，故 . 的任意理想平凡，故是域.
Exercise.
在 PID 中，素理想都是极大理想，故为域.
是素元当且仅当是素理想，当且仅当是整环.
Exercise.
反证而设 , 记 . 若 , 则

于是 , 由于为奇数，为偶数，故，于是是的非平凡因子，这与不可约矛盾. 故 , 的前系数为 0，，于是，这与不可约矛盾.
Exercise.
是 PID, 的不可约元为一次多项式、无实根的二次多项式，或 . 于是的所有素理想（即极大理想）为 , ，, 其中 .
的素理想总结：

(零理想)。
, 其中是中的素数。
, 其中是中的非零本原不可约多项式。
, 其中是中的素数，且其系数模后得到的是中的不可约多项式。
的极大理想总结：
中的极大理想只有上述第 4 类理想：
, 其中是中的素数，且其系数模后得到的是中的不可约多项式。好的，我们来确定和的所有素理想和极大理想。
Exercise.

Exercise.
若 , , ，则 , 其中 , . 反之，若可约，将前面的推导应用到上，就有可约. 故在中可约当且仅当在中可约. 故在中不可约当且仅当在中不可约.

integral-ring

Thu, 15 May 2025 16:51:31 GMT

See 肖梁
Today, all rings are commutative.

We point out that, since the polynomial ring over a general ring is no longer a PID, we do not have the notion of “minimal polynomial” here.

Theorem (The Determinant Trick). Let be a commutative ring.
Let be an -module.
Let be an matrix with entries .
Let be a column vector where each . Suppose that , i.e.,
Then, for all .
In other words, the determinant of the matrix annihilates each element of the vector .
Proof.

Recall the Adjugate Matrix: For any matrix with entries in a commutative ring , its adjugate (or classical adjoint), denoted , satisfies the property:
where is the identity matrix. The entries of are also in .
Multiply the System by the Adjugate:
We are given the system of equations .
Multiply both sides of this equation on the left by :

Simplify:
The right side is simply the zero vector:
Using the associative property of matrix multiplication and the property of the adjugate:

Conclusion:
The product is the vector:
Since this vector is equal to the zero vector , each of its components must be zero:

□
This trick is a cornerstone in various proofs in module theory, including versions of Nakayama's Lemma and the Cayley-Hamilton theorem for modules over commutative rings. It elegantly connects the "linear algebra" of matrices over a ring with the structure of modules over that ring.
Definition (Adjoint of ). Let be a matrix with entries in a ring . The adjoint of , denoted , is the transpose of the matrix of cofactors of .
In ring theory, the definition of the cofactor of a matrix is analogous to its definition in linear algebra over fields. Let be an matrix with entries in a commutative ring .
The -cofactor of , denoted as , is defined as times the determinant of the matrix formed by deleting the -th row and -th column of .
Specifically:

Let be the matrix obtained by removing the -th row and the -th column from .
The -cofactor is .
The matrix of cofactors is the matrix whose -entry is . The adjugate (or classical adjoint) of , denoted as , is the transpose of the cofactor matrix. That is, the -entry of is .

Irreducibility-of-polnomial

Fri, 02 May 2025 16:55:31 GMT

irredselmerpoly.pdf
For a general field there is no simple way to determine if an arbitrary polynomial in is irreducible. Here we will focus on the case and describe two useful irreducibility tests in for monic polynomials in . Let

The two tests are

Reduction : for a prime , reducing coefficients of modulo leads to If is irreducible in for some , then is irreducible in .
Eisenstein criterion: call Eisenstein at if for all and . If is Eisenstein for some , then is irreducible in .
Remark. But the converse is not true, considering .

Theorem (Theorem 1.1 (Selmer)). For all , the polynomial is irreducible in .
The proof is very classic.
Proof.
Definition. For nonzero of degree , let be its reciprocal polynomial:
We call the reciprocal polynomial because its roots are the reciprocals of the roots of . More precisely, if has leading coefficient and then

The following properties of this construction will be used below without comment:

if then and ,
if then ,
for every nonzero constant ,
if has degree then the -coefficient of is .
Check all of these properties yourself. The last one is the most interesting.
The proof of the theorem will be presented in three steps.
Step 1: For and its reciprocal polynomial have no common root in characteristic 0 .
The reciprocal polynomial is . If this shares a root with in characteristic 0 , say , then

so . Thus . Substituting this into either equation in (2.1) gives us , which implies , so every power of is either , or . If then the first equation in (2.1) becomes , which is false since . If then , which is absurd. If then the two equations in (2.1) become and , so , but . Thus a common root in characteristic 0 doesn't exist.
Step 2: For , assume and and have no common roots in characteristic 0 . If for some nonconstant and in , then there is a in with such that and or .
Since , both and are not 0 , so and . Define

Then and . If and are equal up to sign then and are equal up to sign, so and are equal up to sign, but then every root of (it has roots since is nonconstant) would be a common root of and , which is a contradiction. The proof that and are not equal up to sign is similar with in place of .
Step 3: The polynomial , for , is irreducible in .
We argue by contradiction, and can assume since the case can be checked directly. If is reducible in then it factors into a product of two nonconstant polynomials in . Set . By Steps 1 and for some of degree where is not or . Write , so . Then in , so and . Replacing with doesn't change or , so with a sign change we can make monic: . Then .
Comparing the coefficients of in both and ,

Since and , we get in , so exactly one of is and the rest are with and .
Let's look at the terms of and in degrees above :

and

where means terms of degree or less. Thus

The terms on the left have distinct degrees since (the last inequality uses ), so there are three terms on the left with degree greater than . Therefore on the right side (otherwise the right side would have only one term with degree above ). If then , so and , so . If then , so and , so . Recalling that is not or , we have reached a contradiction.
□

The irreducibility argument we gave for can be applied to nearly all trinomials of the form , in the sense that it tells us exactly when they are irreducible.
Theorem (Theorem A.1). For with , and and equal to , the polynomial is irreducible in if and only if it has no root in common with its reciprocal polynomial.
Example (Example A.2.). Let's apply Theorem 4 (Theorem A.1) to . Computer data suggest that is reducible if and only if with , and that in this case is a factor of . For example,

is irreducible unless with (e.g., ), when is a factor,
is irreducible unless , when is a factor.

Corollary (Corollary A.3). For all and , the polynomial is irreducible in .
Proof.
The proof of Theorem 2 (Theorem 1.1 (Selmer))
still works when it is applied to both and (here the degree is ). Details are left to the reader.
□
Theorem (Theorem A.4). For is irreducible over if and only if is a power of 3 and is irreducible over if and only if for some nonnegative integers and .

p-adic

Thu, 15 May 2025 15:30:43 GMT

See Zuniga.pdf
Definition (-adic norm). Let be a fixed prime number, and let be a nonzero rational number. Then, , with , and . The -adic absolute value (or -adic norm) of is defined as
Definition (non-Archimedean metric). Let be a metric space. The metric is called non-Archimedean if
Exercise. Let be a valued field, where is a non-Archimedean absolute value. Assume that is complete with respect to . Then, the series converges if and only if .
Theorem (Ostrowski). Any non trivial absolute value on is equivalent to or to the standard absolute value .

Lemma. Consider the set
endowed with the p-adic norm . Then, the following assertions hold.
is a complete metric space;
is dense in ;
is a field of characteristic zero;
the completion of is .
(i)
I don't understand.
(ii)
We set
Then any can be written as , with and . Given , , we have to show the existence of such that . We take and satisfying .
(iii) omitted
(iv) combine (i), (ii), (iii).
The field of -adic numbers is defined as the completion of w.r.t. the distance induced by .
The unit ball

is a PID. Any ideal of has the form

Indeed, let be an ideal. Set , and let such that . Then .
From a geometric point of view, the ideals of the form , , constitute a fundamental system of neighborhoods around the origin in .
The residue field of is (the finite field with elements). The group of units of is

Exercise. is a unit iff . Moreover if , then , , .

Define

as the ball with center and radius , and

as the sphere with center and radius .
The radius are always integer powers of .
Remark. Notice that and .
We declare that the , , , are open subsets. These sets form a basis for the topology of .
Proposition. , are open and closed sets in the topology of .
Proof.
We first show that is open. Notice that

Consequently, is open.

□
Lemma. If then , i.e. any point of the ball is its center.
Proof.
Let , then

i.e. .
Since (i.e. ), we can repeat the previous argument to show that .
□
Exercise. Any two balls in are either disjoint or one is contained in another.
Exercise. The boundary of any ball is the empty set.
Theorem. A set is compact iff closed and bounded.

We extend the -adic norm to by taking

We define , then .

Note that but .
As a topological space is totally disconnected, i.e. the only connected subsets of are the empty set and the points.

Exercise (yau-2020-problem-5).

Permutation-Groups

Fri, 02 May 2025 16:55:31 GMT

See 丘维声, 肖梁, dummit&foote.
Theorem. Any permutation can be decomposited to disjoint transpositions. (Unique without considering the order)

And

Generally,

Since

Then

When ,

When ,

Remark. An element of odd order in a symmetric group is an even permutation, i.e. lies in . That's because the odd permutation can be decomposed to product of even transformations.

Semidirect-product

Fri, 02 May 2025 09:47:45 GMT

See semidirect-product.pdf

Theorem (recognition theorem). Let be a group with subgroups and where
; that is, every element of has the form for some and ,
in ,
for all and . Then the map by is an isomorphism.

In , let

Since

we have , and clearly is trivial, but matrices in and in often do not commute with each other. You can find your own such matrices (nearly any random choice will work), but also observe that if elements of and of always commute with one another then by Theorem 1 (recognition theorem) , but since is abelian ( and are abelian) while is nonabelian.

In , let be a 2-Sylow subgroup and be a 3-Sylow subgroup (so and is cyclic of order 3). In the Sylow theorems for and , so and are not normal in . The set can be written as -cosets and as -cosets , so is divisible by and by , so . Therefore . The subgroups and intersect trivially. We have since has a nontrivial center ( and both have nontrivial center) while has a trivial center.
The center of depends on whether is even or odd. Recall that .

If is odd, then .
Proof.
Suppose that . Write or . Assume that . Then . Hence . But , so . Thus , which implies that . Thus , so or . If , then . If , then is even. So if is odd, then . Now assume that . Then . Thus , so . Hence , so . Thus , so . But . Thus , so . This implies that , so or . But is odd and , a contradiction. Therefore, .
□
If is even, then .
Proof.
Suppose that . Write or . Assume that . Then . Hence . But , so . Thus , which implies that . Thus , so or . If , then . If , then . Now assume that . Then . Thus , so . Hence , so . Thus , so . But . Thus , so . This implies that , so or . If , then , so , a contradiction since . If , then . Thus .
□
In summary:

For subgroups of , might not be a subgroup. However, if or is normal in , then is a subgroup. Take , e.g.

Consider induced by , then we have the natural definition:
Definition (semidirect product). For two groups and and an action of on by automorphisms, the corresponding semidirect product is defined as follows: as a set it is . The group law on is
Note that , might not be subgroups of the same group, e.g. , thus fails to be defined, but has definition.
Remark. The notation means is normal "", and has a "twisted" action on .
Example. In , so if and only if and .
Example (Example 3.4). Take , and where by . Note is an automorphism of as an additive group and since for all . The group has the operation This resembles the multiplication in , where . In the affine matrices we multiply in the upper left, while in , components multiply in the second coordinate. That suggests turning into by . In the group you have to be careful about how you decompose a matrix: The nice decomposition puts the matrix associated to first, before that associated to .
Example (Example 3.5). In the previous example, replace with and with . We have since automorphisms of the additive group are the mappings for . Let by making for each be multiplication by . The semidirect product has operation and is isomorphic to by .
Example (Example 3.6). Since acts as additive automorphisms on , we have a semidirect product where . The homomorphism given by leads to a semidirect product by .
Theorem (Theorem 3.7). Inside , we have and . The copy of in is a normal subgroup with conjugation by being described with : In particular, commutes with each if and only if , and every and commute if and only if is trivial on for all .
^6742ba
Proof.
The proof is routine.
□
Example (Example 3.8.). Let , and by . We saw in Example 3.4 that by .
^6742ba says the effect of on looks like conjugation in , and this is related to the affine group conjugation formula:

Example (The semiproduct isomorphic to .). Example 3.10. Let be an abelian group written additively, so negation neg: is an automorphism of order 2. Let by and . Then is a homomorphism (tip: when a group contains an element of order , we always get a homomorphism by , and here we're using the special case and is inversion on . The group has operation (3.6) . Then Thus and commute in if and only if . If all nonzero elements of have order 2 (so negation on is the identity) then . If some nonzero element of does not have order 2 then is nonabelian. Consider the case where . The group has order and the group law (3.6) in this special case is This may look like a weird group of order , but in fact it is isomorphic to . If we identify with and with in then by and . For example, (3.7) says , which matches the familiar dihedral relation . (The general multiplication rule in is , where the exponents on and look like (3.7).)

Similar to Theorem 1 (recognition theorem), we have a recognition theorem for semidirect products.
Theorem (Theorem 4.1). Let be a group with subgroups and such that
,
,
. Let be conjugation: . Then is a homomorphism and the map where is an isomorphism.
Proof.
That makes sense at all is due to (3). That it is a homomorphism means , and this is left to the reader to check. The function

where is surjective by (1), and is injective by (2) using the same argument for injectivity as in the proof of Theorem 2.1. To show is a homomorphism, calculate

Hence is an isomorphism.
□
Example. We will show for odd that the direct product is isomorphic to a nontrivial semidirect product .
We have met several examples of groups that are isomorphic to a semidirect product of two groups:

;
;
;
In these respective groups,

acts on by multiplication maps for ;
is identified with for any transposition in ;
is identified with and is identified with in ;
is identified with the group of matrices (not with the matrices );

The case when and are finite with is boring, since has a trivial image and thus the only semidirect product is the direct product .
Example. A semidirect product has order 15 and it must be a direct product: has order 4 and that is relatively prime to 3. In fact, all groups of order 15 are cyclic; groups of order will be studied below.
Example (Construct a semiproduct). Example 5.2. Is there a nontrivial ? Since has size 20 and , there might be nontrivial examples. We seek a nontrivial homomorphism It is necessary that goes to a such that . If , then , so (by testing ) . Then Choose . Check (in fact, if then , so ). Let by , so (e.g., ). We get a nontrivial semidirect product with the group law This is a nonabelian group of order built from two cyclic groups of order 15 and 25.

We'll use the Sylow theorems and semidirect products to find all groups of order up to isomorphism.
Theorem (Theorem 6.1). If primes satisfy then all groups of order are cyclic.
A useful lemma:
Lemma (Lemma 6.2). A semidirect product is unchanged up to isomorphism if the action is composed with an automorphism of : for automorphisms .
Moreover, if , then

e.g. if , then

Theorem (Theorem 6.3). If primes satisfy then there are two groups of order up to isomorphism: one is cyclic and one is nonabelian.
Explicitly, for , a nonabelian matrix group of order is

If a group contains subgroups and such that and , then and are called complementary subgroups. For a normal subgroup , is there always a complementary subgroup ? If so, we'd then have .
The answer is no!
Example (Example 9.1.). Suppose is a cyclic -group with . Every subgroup of is normal. Pick a subgroup with . Suppose , so . Then and have subgroups of order by Cauchy's theorem (or check this directly in cyclic -groups). The cyclic group has at most one subgroup per size, so the subgroups of order in and are the same, and thus . This contradiction shows is not a semidirect product of two proper subgroups.
Theorem (Schur-Zassenhaus). If and , then has a complementary subgroup in and all complementary subgroups to in are conjugate.

Theorem. Let and be isomorphisms. For each homomorphism there is a corresponding homomorphism such that .

Simplicity-of-An

Fri, 09 May 2025 13:38:13 GMT

Ansimple.pdf
Some examples of groups:

A finte group of prime order is simple.
A finite abelian gorup not of prime order is not simple.
Let prime , by Cauchy's theorem, contains a subgroup of order , which is abelian[1] thus normal in .
When , is not simple since with index 2.
A meaningful result is as follows.
Theorem. For , is simple.
Proof.
See Ansimple.pdf, where 5 proofs are listed.
□
Next we show 3 useful lemmas and a theorem.
Lemma. For is generated by 3-cycles. For is generated by permutations of type .
The 3-cyclics in are all conjugate in , since permutations of cycle type in are conjugate. Are 3-cycles conjugate in ?
Lemma. For , all 3-cycles in are conjugate in .
Lemma. For , the only nontrivial proper normal subgroup of is . In particular, the only subgroup of with index 2 is .
Corollary. By Lemma 4 , for each homomorphic image of not isomorphic to has order 1 or 2. So there is no surjective homomorphism for .

Theorem. For , no subgroup of has index strictly between 2 and . Moreover, each subgroup of index in is isomorphic to .
Proof.
Let be a proper subgroup of and let , so . If then by Lemma 2.4. If then we will show . The left multiplication action of on gives a group homomorphism

By hypothesis , so is not injective. Let be the kernel of , so and is nontrivial. Since , Lemma 2.4 says or . Since , we get or , which implies . Therefore we can't have .
Now let be a subgroup of with index . Consider the left multiplication action of on . This is a homomorphism . Since has order , is isomorphic to . The kernel of is a normal subgroup of that lies in (why?). Therefore the kernel has index at least in . Since the only normal subgroups of are , and , the kernel of is trivial, so is an isomorphism. What is the image in ? Since if and only if is the group of permutations of that fixes the "point" in . The subgroup fixing a point in a symmetric group isomorphic to is isomorphic to . Therefore .
□
Corollary. Let be a field. If and , the number of different polynomials we get from by permuting its variables is either 1, 2, or at least .
Proof.
Letting act on by permutations of the variables, the polynomials we get by permuting the variables of is the -orbit of . The size of this orbit is , where . By Theorem 2.7, this index is either 1, 2, or at least .
□

Corollary. For , each proper subgroup of has index at least .
The proof is similar to Theorem 6
Proof.
Let be a proper subgroup of , with index . Consider the left multiplication action of on . This gives a group homomorphism

Let be the kernel of , so (why?) and . By simplicity of is trivial. Therefore injects into , so !, which implies .
□

As is abelian↩︎

sylow-more

Fri, 02 May 2025 16:55:31 GMT

sylowmore.pdf
It is natural to ask how the SYlow theorems can be extended to -subgroups that are not -Sylow subgroups.
Theorem (Theorem 4.1). If then there is a subgroup of with size .
First we directly prove Theorem 1 (Theorem 4.1) .
Proof.
We induct on the size of . The case when or prime is trivial. Now suppose and the theorem is proved for all groups of smaller size. That is, we assume each group with has a subgroup of size equal to an arbitrary prime power dividing .
Choose a prime power dividing , with . We seek a subgroup of with size . If has a proper subgroup such that , then we're done: has a subgroup of size by induction (since ) and this subgroup is in too.
Now we suppose every proper subgroup has size not divisible by . Since is divisible by , we see every proper subgroup of has index divisible by . Consider the class equation

where represent the conjugacy classes of size greater than 1. We have and for each since each subgroup of is proper (if then would be in a conjugacy class of size 1, which isn't true). Therefore . By Cauchy's theorem, has an element of order , say . As .
We now consider the quotient group , which is a group with size less than that of . Since , by induction has a subgroup with size . Its inverse image under is a subgroup of with size .
□
Theorem (Cauchy's Theorem). Let be a finite group and let be a prime. If divides , then contains an element of order .
Here is a proof: cauchypf.pdf.
Theorem. Let be a finite group. If and then each subgroup of with size has index in a subgroup of .
From Theorem 3 we can build a nested chain

where , so . Theorem 1 (Theorem 4.1) is a special case of Theorem 3, so it suffices to prove Theorem 3.
Proof.
The case says there is a subgroup of size in . This is Cauchy's theorem.
Now take . Let be a subgroup of with size . We want to find a subgroup in which has index . Consider the left multiplication action of (not ) on . Since is a non-trivial -group,

The left side of the congruence is , which is divisible by . Which cosets in are fixed points? They are

Therefore the set of fixed points of acting on is , which has the structure of a group since . By (4.2), , so Cauchy tells us there is a subgroup of order . Its inverse image under is a subgroup of with size , and it contains with index .
□
Theorem (Orbit-Stabilizer Theorem). Let be a group acting on a set . For any , we have where is the orbit of and is the stabilizer of . In other words,
Explain the reason why Equation (1) holds.
Let's consider the orbits of the action of on . For any , the orbit of under the action of is given by . The size of this orbit is given by the Orbit-Stabilizer Theorem: ,
If is not a fixed point, then there exists at least one such that , which means is a proper subgroup of . Since is a -group of order , the size of any proper subgroup of must be a power of strictly less than . Therefore, the index must be a multiple of .
We can partition the set into disjoint orbits under the action of . Let be the set of fixed points. Then,

We know that , and for every orbit not in , its size is a multiple of . Therefore, when we take this equation modulo , the sum on the right-hand side becomes 0 , and we are left with:

Corollary. Let be a -subgroup of the finite group . Then In particular, if then .
Proof.
The congruence here is (4.2). When , so .
□
Theorem (Frobenius). If , the number of subgroups of with size is .

Sylow-theorems

Sat, 17 May 2025 14:55:41 GMT

Sylow exercises | Aryaman Maithani
sylowapp.pdf
Expository papers by K. Conrad
Dummit&Foote 4.5
It should be pointed out that the definition of cosets does not require normality, but the definition of quotient group requires. What's the difference? The former (as set) admits disjoint orbits, but need not be . The latter (as group) admits both.

The affine group

It's defined by the transformation

The Heisenberg group over , denoted by , is the set of upper triangular matrices of the form:

where:

, meaning they are elements of the finite field (the integers modulo ).
The group operation is matrix multiplication.

sylowapp.pdf

Sylow exercises | Aryaman Maithani
Exercise. Prove that if and is a subgroup of containing , then . Give an example to show that, in general, a Sylow -subgroup of a subgroup of need not be a Sylow -subgroup of .
Proof.
We are given with and such that . As , we get that for some .
Moreover, and thus, . As , we get that .
Thus, is a Sylow -subgroup of .
For the second part, consider and . Then is a Sylow 2subgroup of but not one of .
□
Exercise. Prove that if is a subgroup of and then for all .
Proof.
We first note that is indeed a subgroup of .
The assertion follows from the fact that and .
□
Exercise. Use Sylow's theorem to prove Cauchy's theorem.
Proof.
Recall that Cauchy's theorem says that if is a finite group such that for some prime , then has an element of order .
To prove this, let and be as above and be a Sylow -subgroup of . (Existence of is given by the Sylow theorems.)
Let with usual meanings. Then, .
Consider such that . Then, the order of is for some . Consider . Then, order of is . (How? It is clear that . Thus, the only other possibility for the order is 1 but that is not possible since as , the order of .)
□
Exercise. Exhibit all Sylow 2-subgroups and Sylow 3-subgroups of and .
Proof.
:
Note that . Thus, the Sylow 2-subgroup(s) will have order 4 and Sylow 3-subgroup(s) will have order 3 .
Sylow 3:
Note that a Sylow 3-subgroup must necessarily be isomorphic to . Since \underline{all elements not of the form have order 2}, we just find powers of which have order 3. These turn out to be and . As , we have a unique Sylow 3-subgroup:

Sylow 2:
Note that no element of has order 4. (Any such element would have to be a power of but 4 does not divide 6, the order of .)
Thus, any Sylow 2-subgroup must be isomorphic to .
Every element of this group must have order 2. is the only power of which has order 2. Thus, the other two elements must be of the form . Noting that the multiplication of two such elements is:

we get that . This gives us three such groups which are all the Sylow 2subgroups:

(It can be checked that all of these are distinct subgroups.)

The order of the group is . Thus, the Sylow 2-subgroup(s) will have order 4 and Sylow 3-subgroup(s) will have order 9.
Sylow 2:
Note that there are three subgroups of order 2 of . Let these be . (Note that these are indeed distinct.)
Then, the nine products:

are subgroups of have order 4. This gives us nine Sylow 2-subgroups. However, by the Sylow theorems, and thus, there can't be any more.
Sylow 3:
Let be a Sylow 3-subgroup. Note that no element of has order 9 . Thus, every non-identity element of must have order 3 . Now, note that the order of an element is the lcm of the orders of and . This gives us that there are 8 elements of order 3 . These elements together with the identity do form a subgroup and moreover, there can't be any other. Thus, there is a unique Sylow 3-subgroup which is:

□
Exercise. Show that a Sylow -subgroup of is cyclic and normal for every odd prime .
Proof.
Let as usual. As is odd and is even, we must have that for some . Thus, with .
Note that has order and hence, has order . Hence, is Sylow p-subgroup of
We now show that is normal. To do this, it is enough to work with the generators of . Clearly, . Also, . Thus, is normal.
As all Sylow p-subgroups are conjugates, we get that the above Sylow p-subgroup is the unique Sylow p-subgroup. As it is cyclic and normal, we are done.
□
Exercise. Exhibit all Sylow 3-subgroups of and all Sylow 3-subgroups of .
Proof.
: Clearly, . We exhibit 4 such Sylow 3-subgroups now:

(Note that these are indeed distinct subgroups.)
These were easy to find as a Sylow 3-subgroup must have order 3 and thus, must be isomorphic to . It was then a matter of finding elements of order 3 .
: Same as earlier. Note that there are 8 elements of order 3. If we consider all the 8 subgroups generated by them, we see that we get \underline{only four distinct ones}, listed above.
□
Exercise. Exhibit all Sylow 2-subgroups of and find elements of which conjugate one of these into each of the others.
Proof.
A little bit of experimenting with elements of order 2 and 4 gives the following subgroups:

It can be verified that these are distinct. (For example, the 2 cycles listed in the generators don't appear in any of the other subgroups.) It can also be verified that all of these have order 8. Moreover, there can't be any more as .
As for the conjugation question, (34) conjugates the first to the second and (24) the second to the third.\footnote{Note that the conjugate of works by transforming the places of the elements.}
□
Exercise. Exhibit two distinct Sylow 2-subgroups of and an element of that conjugates one into the other.
Proof.
Note that and thus, any two of the Sylow 2-subgroups listed earlier will work again. (With the understanding that the elements are now elements of )
□
Lemma. The order of is .
Proof.
Consider the choice of the first column. There are choices for the first column. The second column must be a vector \underline{independent} of the first column. There are choices for the second column.
□
Exercise. Exhibit all Sylow 3-subgroups of .
Proof.
Note that the order of is . It can be easily shown that the order of is half of that.
Thus, or 4 . Also, note that any Sylow 3-subgroup will have order 3 and hence, is isomorphic to .
It is easy to find two distinct subgroups of order 3 by observing the following matrices to have order 3:

Also, these matrices generate distinct subgroups. Thus, which forces . \textbf{We may now conjugate the above matrices to get the other two.} In any case, we are left with the following:

□
Exercise. Prove that the subgroup of generated by and is the unique Sylow 2-subgroup of .
Proof.
Hmm.
□
Exercise. Prove that a group of order 351 has a normal Sylow -subgroup for some prime dividing its order.
Proof.
Note that . By the Sylow theorems, and 13. This forces . If , then we are done. Let us assume that . Thus, .
Thus, there are 27 Sylow 13 -subgroups of . As 13 is a prime, this forces the intersection of two distinct Sylow 13 -subgroups to be trivial. Thus, the number of elements having order 13 equals .
This leaves us with remaining elements which are not part of any Sylow 13-subgroup. Now, by Sylow Theorems, we know that . However, no non-identity element can be part of a Sylow 3 as well as a Sylow 13 -subgroup. Thus, the remaining elements form the unique Sylow 3 -subgroup which gives us that and thus, we are done.
□
Exercise. Let , where , and are primes with . Prove that G has a normal Sylow subgroup for either or .
Proof.
Let , and be distinct prime numbers. WLOG, we may assume that . Let be a group with order .
We shall be appealing to the Sylow theorems without mentioning it explicitly. If any of , or is equal to 1 , then we know that is not simple. For sake of contradiction assume that each of the above is strictly greater than 1. As and , we get that . (Since we have assumed that .) Thus, there are Sylow- subgroups of .
Now, note that each such Sylow- subgroup has order , a prime and thus, the intersection of two distinct Sylow- subgroups must be trivial, id est, (1). Thus, the number of elements having order equals .
Now, and . Thus, . However, and thus, . This gives us that . Thus, the number of elements having order equals .
Lastly, similar argument as earlier gives us that . Note that , and are counting distinct non-identity elements and thus,

Thus, we have a contradiction as and we are done!
□
Exercise. Prove that if , then has a normal Sylow 5-subgroup and a normal Sylow 7 subgroup.
Proof.
Assume not. Then, and .
Note that . The Sylow theorems force and .
Note that the intersection of any two of these Sylow subgroups must be trivial (by Lagrange's Theorem). Thus, their union contains exactly

This is clearly a contradiction. (As .)
□
Exercise. Prove that if then is not simple.
Proof.
Note that . For the sake of contradiction, let us assume that is not simple. Thus, , and are all strictly greater than 1 . Now, looking at the possibilities for each by considering the factors of the "complimentary part", we get that

Note that the intersection of any two of these Sylow subgroups must be trivial (by Lagrange's Theorem). Thus, their union contains exactly

This is clearly a contradiction. (As .)
□
Exercise. Prove that if then is not simple.
Proof.
Note that . For the sake of contradiction, let us assume that is not simple. Thus, , and are all strictly greater than 1
.
Now, looking at the possibilities for each by considering the factors of the "complimentary part", we get that

Note that the intersection of any two of these Sylow subgroups must be trivial (by Lagrange's Theorem). Thus, their union contains at least

This is clearly a contradiction.
□
Exercise. Prove that if then is not simple.
Proof.
Let be a group with order 132 . We show that is not simple.
Note that . Let us assume that is simple and arrive at a contradiction. By simplicity, we know that , and .
By Sylow Theorem (3), it is forced that . Also, and .
Note the following:

Intersection of any two Sylow- 11 subgroups is trivial.
Intersection of any two Sylow-3 subgroups is trivial.
Intersection of any Sylow- subgroup with any Sylow- subgroup is trivial.
The above facts follow by considering the fact that the intersection would be a subgroup of the two bigger subgroups and would have to divide their orders.
Now, if we consider the union of all the Sylow-11 and Sylow- 3 subgroups, it contains at least elements. Thus, the remaining elements are at most 3 . However, we do need at least 3 more elements to form a Sylow- 2 subgroup.
These 3 elements, along with the identity must form the unique Sylow- 2 subgroup of order 4 . However, we have reached a contradiction as we get that .
Thus, we are done!
□
Exercise. Prove that if then is not simple.
Proof.
Note that . Considering the factors of 42 , we see that is forced to be 1 .
□
Exercise. Prove that if is a group of order 231 then contains a Sylow 11-subgroup of and a Sylow 7-subgroup is normal in .
Proof.
Note that . The restrictions from Sylow Theorems force that . In particular, this shows that has a (unique) Sylow 7 -subgroup which is normal in it.
Now, we show that the Sylow 11-subgroup is contained in . Let us denote this subgroup by . Note that is normal. Consider the action of on given by conjugation, that is, .
This induces a homomorphism . Note that is cyclic and thus, . (The last equality may be manually verified by noting that is a generator for which has 10 elements.)
Thus, is isomorphic to a subgroup of (First Isomorphism Theorem). In particular, . Looking at the divisors of , we see that this forces or .
This means that given any , the map is the identity map. In other words, for all and . The above is equivalent to or for all and . Thus, , by the definition of the center.
Remark. We didn't really require the fact that . Only would've been sufficient (which is more straightforward as well).
□
Exercise. Let be a group of order 105. Prove that if a Sylow 3-subgroup of is normal then is abelian.
Proof.
Let be the Sylow 3-subgroup.
Claim 1: .
Note that must divide both and . This forces or .
(We have used the fact that is normal in the above by considering the defined similarly in the previous exercise.)
Claim 2: is cyclic.
Note that has order . Noting that proves the claim. (See this.)
Claim 3: is cyclic.
By the third isomorphism theorem, we have that

(Note that Claim 1 (and the fact that ) tells us that makes sense.) By the previous claim, is cyclic. The claim follows as quotients of a cyclic group are cyclic.
Claim 4: is abelian.
This is a consequence of the above claim.
□
Lemma. is isomorphic to .
Proof.
Consider as a 2-dimension vector in . Then is a nondegenerate transform above .
□
Exercise. Let be a group of order 315 which has a normal Sylow 3-subgroup. Prove that contains a Sylow 3-subgroup of and deduce that is abelian.
Proof.
Let be the normal Sylow 3-subgroup. (It must necessarily be unique.) Consider the action of on given by conjugation. (This is an action as is normal.) Consider the induced homomorphism .
Now, note that is a group of order 9 . Thus, it is isomorphic to either or . Accordingly, has order either 6 or 48.
As by the first isomorphism theorem, we get that divides 48. (6 divides 48 anyway.)
Also, . Thus, is either 1 or 3.
If , then . However, note that and hence which is a contradiction as . Hence, we get that or .
The above shows that the image of is just the identity map. In other words, for all and . This is precisely what it means for .
The rest of the solution is now the same as the earlier one. We note that has order 35 and hence, is cyclic.
We also have

and thus, is cyclic. This proves that is abelian.
□
Exercise. Let be a group of order 1575 . Prove that if a Sylow 3 -subgroup of is normal then a Sylow 5-subgroup and a Sylow 7 -subgroup are normal. In this situation prove that is abelian.
Proof.
By the same arguments as earlier, we get that is abelian. Now, by the Sylow Theorems, there do exist Sylow 5 and 7-subgroups. As is abelian, they must be normal.
□
Exercise. How many elements of order 7 must there be in a simple group of order
Proof.
Note that . The Sylow Theorems force to be either 1 or 8 . If , then there is a unique Sylow 7 -subgroup which must be normal. This contradicts the assumption that the group is simple.
Thus, . This means that every Sylow 7 -subgroup is isomorphic to . In particular, every non-identity has order 7.
Moreover, two distinct Sylow 7-subgroups can only intersect trivially, by Lagrange's theorem. Thus, the total number of order 7 elements contained in these Sylow subgroups is

Now, we show that there are no more. That is, any order 7 element is contained in some Sylow 7-subgroup.[1]
Let be any order 7 element. Then, is a 7-group and thus, must be contained in some Sylow 7 -subgroup, as desired. (In fact, is itself a Sylow 7 -subgroup.)
□
Exercise. For , and 5 , find and .
Proof.
:
Note that Sylow Theorems force and .
As is simple, cannot be 1 for any and hence, and .
Now, we show that . Note that is a Sylow 2-subgroup. (The way I found this was by taking a Sylow 2-subgroup from . And intersecting it with .) Note that its elements are

That is, identity along with all possible elements of the cycle type 1-2-2 formed using the numbers .
Recalling that all Sylow 2-subgroups are conjugates and also recalling how conjugates in look, we see that any Sylow 2-subgroup is of this form:

(Where are distinct elements of .)
This shows us that picking any four numbers from uniquely defines Sylow 2 subgroup. (In the sense, that there's no over-counting or under-counting.) Thus, the number of such subgroups is .
To conclude, we have, for :

:

For , we have the possibilities 1,10 , and 40 . Note that it can't be 1 as any Sylow 3 subgroup of is again a Sylow 3-subgroup of and we already had 10 there.
Now, if were 40 , then there would be a total of 80 elements of order 3 . However, it can be verified that only elements of the cycle type ( ) have order 3 and there are only 20 of those. Thus, .
For , the discussion is even simpler as there are no new possibilities and we have as before.
As before, we have . We show that by demonstrating fifteen such subgroups. Let be distinct. Assume that is the smallest amongst the four.
Consider the subgroup:

It can be verified that the above does have 8 elements. Moreover, to determine such a subgroup, we first need to pick 4 elements from and then from those 4 , pick an element to be paired with the smallest.
It can be verified that different such pickings will give different subgroups. (Simply by writing explicitly the elements of the above subgroup.)
This gives us different subgroups and hence, we are done.
To conclude, we have, for :

□

Order 90 | Aryaman Maithani
Order 144 | Aryaman Maithani

Order 180 | Aryaman Maithani
Exercise. Let be a group with order 180. Show that is not simple.
Proof.
Assume that is simple, note that ; by Sylow III theorem, , .
Claim that . Suppose not, , then there is a homomorphism

recall that . By Sylow II theorem, is transitive, thus . As is simple, is trivial, otherwise the nontrivial ; thus is not simple. So , then

which is a contradiction.
Claim that . Suppose not, , then there is a homomorphism

Similarly, , then

We have , strictly between and , which is impossible by the following lemma.
Lemma. For , no subgroup of has index strictly between 2 and . Moreover, each subgroup of index in is isomorphic to .
Applying the idea in the proof of Lemma 27, we consider the left multiplication action of on ,

Then

Clearly, , then is trivial or . Since (by the definition of ), we have , thus must be . Then

which is a contradiction. Hence .
Note that the intersection of any two Sylow-5 subgroups is trivial, since is subgroup of , then ; .
Remark. When , the intersection of Sylow -subgroups may be nontrivial. A counterexample is ; consider , , which are Sylow 2-subgroups of ; but .
Thus we now arrive the conclusion that and . Then there are two cases,

Intersection of any two Sylow-3 subgroups is trivial.
Intersection of some two Sylow-3 subgroups is non-trivial.
In the first case,

The number of elements of order 9 is .
The number of elements of order 5 is .
; this is a contradiction.
In the second case, let and have nontrivial intersection. Note that and thus must contain exactly 3 elements. As groups of order are abelian, we have . By the definition of group, the set is contained in , as , . This gives

Thus the coset

Let act on by left multiplication,

Since is simple, is trivial; since the action is transitive, ; thus .

Therefore . But by Lemma 27, , as is a subgroup of , .
□

This topic involves extensive discussion and various techniques in group theory.
Let . Then or 4 . We will show that if then . Assume and . Since has no subgroup of size 18 (it would have index 2 and therefore be normal, so a 3-Sylow subgroup of it would be normal in by Lemma 5.8, which contradicts ). Since is nonabelian. Our goal is to get a contradiction. We will try to count elements of different orders in and find the total comes out to more than 36 elements. That will be our contradiction.
Let be a 3-Sylow in , so . Left multiplication of on gives a homomorphism . Since , the kernel is nontrivial. Since , either or . Since does not equal , so .
Since , we can make act on by conjugations. This is a homomorphism . If this homomorphism is onto (that is, some element of conjugates on in a nontrivial way) then the kernel is a subgroup of with size 18 , which does not have. So the conjugation action of on is trivial, which means every element of commutes with the elements of , so . Then , so the size of is one of the numbers in . Since is nonabelian and a group is abelian when the quotient by its center is cyclic, can't be 12,18 , or 36 . Since there is no normal subgroup of size 9 , so . If then the product set is a subgroup of size 18 , a contradiction. So we must have , which means .
Now we start counting elements with various orders. The center is a 3 -subgroup of , so by the conjugacy of 3 -Sylow subgroups every 3 -Sylow subgroup contains . Each pair ofdifferent 3-Sylow subgroups have as their intersection, so we can count the total number of elements of 3 -power order: .
Let have order 2. Then is a subgroup of order 6 by Lemma 4.1, and it is is abelian and in fact cyclic since . The cyclic group has a unique element of order 2, which must be . Therefore when and are different elements of order 2 in , the groups and have has their intersection. So each element of order 2 in provides us with 2 new elements of order 6 . Let be the number of elements of order 2 in , so there are at least elements of order 6 , giving at least elements in total with order 2 or 6 . Since we already found 27 elements with 3-power order (including the identity), , or . We can get an inequality on in the other direction: . Indeed, no element of order 2 lies in , so some conjugate of an element of order 2 is a second element of order 2 . Thus .
Since is even (by McKay's proof of Cauchy's theorem) and this number is is odd, so . Therefore has 3 elements of order 2 , so at least elements of order 2 or 6 . Adding this to 27 from before gives , so each element of has 3-power order or order 2 or 6 . In particular, the 2-Sylow subgroup of is isomorphic to (no elements of order 4 in . Then different 2-Sylow subgroups meet at most in a group of order 2 , which gives us 5 elements of order 2 from both subgroups. We saw before that there are only 3 elements of order 2 . This is a contradiction.

This is important.↩︎

ture-or-false

Fri, 09 May 2025 06:55:17 GMT

Exercise. 如果是群的正规子群，则是和的半直积．
If is a normal subgroup of ，then is a semi-direct product of with ．
False．It is not true in general that one can embed back to ．Semi－direct product requires that can be realized as a subgroup of ．（This is a hard question．）
Simple Conclusion:
The statement is false because having a normal subgroup in a group does not guarantee that contains a subgroup that is both isomorphic to the quotient group and acts as a necessary "complement" to for the semidirect product structure.
Exercise. For , let be a normal subgroup of satisfying and , then .
False. If one wants , one needs the isomorphism to induce the corresponding isomorphism . A typical example is to take and and . Clearly and are abstractly isomorphic, but and .
Exercise. In a UFD, every nonzero element can be uniquely written as products of prime elements.
False. In a UFD, every nonzero nonunit element can be written as products of prime elements up to associates. Here "up to associates" means that the prime element factorization can be modified by units; and thus not literally unique.
Exercise. Every cyclic extension over of degree is of the form for some ．
False．On needs to contain th roots of unity for this to be true．
Definition (Cyclic Extension). A field extension is called a cyclic extension if it is a finite Galois extension and its Galois group is a cyclic group.
Exercise. Let be a field extension of a field with intermediate fields and . Suppose that and are Galois over . Then .
False. It is a theorem that . So the statement holds if and only if .
Exercise. For any finite field and any positive integer , there exists a unique (up to isomorphism) cyclic extension of of degree .
True. If for a power of a prime, then the cyclic extension of of degree is .

常微分方程周期性

Thu, 15 May 2025 08:07:01 GMT

Exercise (yau-2020).
Proof.
There exists constant , s.t.

Then is bounded since

Then for some .
If , we have , then is monotonic.

Thus there exists , s.t.

Then

Therefore ; ; thus .
If , there exists , s.t. . There is a sequence , s.t. . Then

which has only finite solutions; thus there exists (WLOG) s.t.

Then and has the same initial condition at . The solution is unique since

Let , then

is Lipschitz on , which guarantees the uniqueness. Hence is periodic with period .
□

场论初步

Sun, 04 May 2025 16:37:03 GMT

Definition (等值面). 设是定义在空间区域上的一个数量场．对其值域中的任意一个实数，称点集为的一个等值面．

Proof.
证明考虑使用一大堆正交变换.
□
Note 根据这个引理，如下定义是合理的

数量场的梯度有以下性质

梯度运算的基本公式如下：

为常数；
为常数）；
；
；
；
；
更一般地，有其中表示函数关于第个变元的偏导数．

对于封闭区面，本章总规定其单位法向量指向其外侧，即外侧为正侧，内侧为负侧.

以流体的速度场为例子，当流体速度场穿过的通量大于零时，意味着流体从所包围区域之内穿过流向该区域之外的流量大于从其外部流向区域之内的流量，表明所包围区域内必然有供给流体的源泉. 反之亦然.
因此，向量场穿过封闭曲面的通量反映了所包围区域中产生向量场的物质的正源与负源的某种代数和.
但我们不能了解这种源物质在局部的分布与强弱，因此我们通过取极限来达到这样的目的. 给出散度的定义.

当向量场在的散度时，在必有产生向量场的物质的正源，并且越大这种正源的强度越大，反之亦然.
散度不依赖于坐标系的选取.

散度运算具有以下基本性质：

（为常数）；
；
．

由公式可以看出，向量场沿封闭曲线的环量反映了沿的旋转情况.
对于流体而言, 速度场沿封闭曲线的环量是否为零反映了曲线是否包围了漩涡的涡管; 当环量非零时, 其正负号反映了曲线的方向是否与它所包围漩涡的旋转方向一致, 其大小则反映了该漩涡的旋转强度.
与散度概念类似，为了了解向量场涡旋的局部分布情况，我们给出方向旋量的概念

方向旋量也不依赖于坐标系的选取.

定义

则有

Definition (旋度). 由 Equation (1) 定义的向量为向量场在点的旋度.
旋度也记作，.

旋度运算具有以下基本性质：

（为常数）；
；
．
引力场和静电场都是无旋场.

定义

下面推导一个重要定理 ——亥姆霍兹分解定理. 粗略地说, 这个定理告诉我们, 每个向量场都可分解为一个无源场和一个无旋场的和.

对于给定的一组标准正交基，对于标量场，定义梯度

对于，定义散度

定义旋度

对于，定义叉乘

分布

Sat, 03 May 2025 15:37:38 GMT

参见于品数学分析讲义

我们用表示在上定义并且有紧支集的光滑函数所组成的集合，我们把它称作是为试验函数空间。按照定义，对于任意的，存在紧集（这是中的紧集），使得，即对任意的。

以下是一些常见的闭包及其对应的函数空间：

在空间中（使用范数）：
试验函数空间在空间中是稠密的，对于。这意味着在范数下的闭包是整个空间。
简单来说，任何一个函数都可以通过一列试验函数在该范数下逼近。

在连续函数空间或中（使用一致范数）：
试验函数是连续的，并且具有紧支集。在一致范数下，试验函数空间的闭包是具有紧支集的连续函数空间。这是因为任何一个紧支集连续函数都可以通过光滑的，紧支集函数（例如通过卷积光滑化）来一致逼近。

在 Sobolev 空间中（使用 Sobolev 范数）：
试验函数空间在 Sobolev 空间中是稠密的，对于和非负整数。这意味着在范数下的闭包是整个 Sobolev 空间。Sobolev 空间是研究偏微分方程非常重要的函数空间，它包含了直到阶的广义导数（或弱导数）存在且属于空间的函数。

Definition (分布). 所谓上的一个分布(也称作广义函数)指的是上的一个线性泛函(线性映射): 满足如下两个条件
对任意的和，我们有
对任意的紧集，存在非负整数和正常数( 和依赖于)，使得对任意的，都有如果上述的的选取不依赖于紧集的选取，那么，我们就把最小的这样的非负整数称作是分布的阶。
Example (Dirac 函数). 对任意的，我们可以定义分布。其中，对于任意的，我们定义我们来验证实际上是分布：
对任意的紧集，如果，那么，对任意的，我们都有如果，那么，使得对任意的，我们有所以，我们在分布的定义中取即可。特别地，我们还知道的阶为 0。
Example (局部可积的函数). 给定开集（总是装配了 Borel 代数和 Lebesgue 测度），局部可积的函数指的是在每个紧的局部上都可积的函数，即可测函数（所对应的几乎处处相等的函数的等价类），对于任意紧集，函数。我们用表示上局部可积的函数。对于任意的，我们定义上的线性泛函：由于在它的支集上有界，所以，上面的积分是良好定义的。我们证明是上的阶为 0 的分布：对任意的紧集，对任意的，我们有所以，我们在分布的定义中取即可.
Remark. 为了方便起见，我们通常把直接写成。
Proposition. 任意选定（我们通常偏爱之前所构造的那个），我们假定对任意的，我们定义那么，在分布的意义下，当时，我们有。
Example (Radon 测度). 假设是上的测度，其中，是开集，是 Borel 代数（包含所有开集的最小 -代数）。如果每个紧集，我们就把这种测度称作是一个 Radon 测度。比如说，对任意的正函数（几乎处处），对任意的，我们可以定义这就是一个 Radon 测度。
任意给定一个 Radon 测度，我们可以定义一个分布：对于，我们要求我们证明是上阶为 0 的分布：
对任意的紧集，对任意的，我们有所以，我们在分布的定义中取即可。
特别地，我们可以把中的元素看作是某个 Radon 测度的密度函数，从而，定义出了同样的分布。
Remark. 利用所谓的 Riesz 表示定理，我们可以证明，上所有的 0 阶分布都是（由如上方式给出的）Radon 测度。
Proposition. 给定开集，我们已经定义如下的线性映射（把局部可积函数视为分布）这是单射。
Remark (广义函数的命名由来). 根据这个命题，局部可积的函数可以看做是分布的子集合。在分析中，我们把的元素称作是上的“函数”（这个类已经足够大了），由于某些分布不是“函数”，所以我们也经常把分布称作是“广义函数”。
Lemma. 假设并且，那么，也是光滑函数。
所以是光滑函数（自然是局部可积的）。从而，当时，上述积分的极限存在：

我们现在定义

为了证明这是分布，我们利用中值定理：对任意的紧集，对任意的支集在上的光滑函数，我们有

所以，

所以是一个阶不超过 1 的分布。在作业中，我们将证明的阶恰好是 1 。
另外，vp 是法语 valeur principale 的缩略，英文文献经常用，因为他们把主值写为 principal value。

在微积分的学习中，我们可以对一个函数做特定的操作，比如可以把一个函数限制到比较小的定义域上、可以对一个函数求导数、两个函数可以相乘等等。我们现在讨论如何对分布做一些特定的操作。
我们通常用表示试验函数.

假设是开子集，那么，我们可以定义限制映射

其中，对于每个，它自然可以看作是中的元素，从而，我们可以要求

通过分部积分，我们有

这个计算启发我们对于，我们可以用下面的等式来定义它的导数：

Definition. 假设是有界开集，给定，对于任意的多重指标，我们定义
显然根据分布的定义，Definition 1 (分布)，第二条，可知分布的导数依然是分布。
Example. 我们计算上的 Dirac 函数的导数，其中。任给，我们有

对光滑函数和分布，我们可以定义它们的乘积：

下一步验证，进而分布具有 -模结构.

根据 Leibniz 法则，我们有

所以，

这表明。

对于，我们有如下的平移变换：

对于局部可积的函数，我们可以定义（这是一种特殊的变量替换）：

对一般的分布 , ，我们定义

容易验证 . （承自上一节中分布的 -模结构）
给定的两个开集和，我们假定

是微分同胚。对任意一个上的局部可积的函数和上的试验函数，根据换元积分公式，我们有

根据这个计算，我们定义：

其中，对于和，对于我们定义如下：

简单验证可知：.
Remark. 注记（记号）．因为在光滑函数情况下，就是函数的复合，我们还把上面的拉回映射写成
给定微分同胚，它把上的 Dirac 函数拉回，得到上的 Dirac 函数，这给出了 Jacobi 行列式的一个精确的解释：这是一点处体积的变化。

给定微分同胚，如果是的函数，我们可以对求导数运算运用链式法则：

对于一般的分布，我们实际上（后来）可以先用光滑函数逼近这个分布，然后上面的链式法则在极限的情况下仍然成立。
我们现在给出一个直接的证明：设是的逆映射。按照分布与一个微分同胚的复合的定义，我们有

另外，对任意，我们有

根据到嵌入的单射性，上面的等式等价于说

从而

这就证明如下关于分布的链式法则：

Theorem. (Stokes 公式) 假设是一个有界带边光滑区域，为的单位外法向量，为上的曲面测度。对任意的，我们有

给定上面 Stokes 公式中所述的，它的边界的曲面测度在如下的意义下定义了上一个分布：

这是一个 0 阶的分布，我们把证明的细节留给不放心的同学来验证。类似地，对每一个，如下的公式也定义了一个分布：

我们可以把 Stokes 公式改写成如下的形式：

所以，用分布的语言来写，我们有
Theorem (Stokes 公式). 假设是一个有界带边光滑区域，为的单位外法向量，为上的曲面测度，作为分布，我们有等式如果用向量值分布的语言（可以望文生义地定义）来写，我们有
我们现在回到 1 维的情形，此时的 Stokes 公式就是 Newton-Leibniz 公式。
Lemma. 对于，我们定义其原函数为那么，是连续函数。在分布的意义下，我们有
证明只需照章办事.（利用 Fubini 定理）

这里跳过第 60 节

Theorem (单位分解). 任意给定中的紧集，假设被有限个开集所覆盖。那么，对每个，存在光滑函数，满足
对任意，有；
存在包含的开集，对任意，我们有
单位分解的证明并没有任何启发性的意义，我们只要能够运用该结论即可。
我们现在证明一个分布在（所有）小的开集上的限制决定了，这表明分布是可以局部定义的：
Theorem. 给定开集。任意给定一族开集，其中对任意的。我们假定对每个，我们在上指定一个分布。如果这一族分布满足如下的相容关系：那么，存在唯一的，使得对任意的，我们都有进一步，当且仅当对每个 ; 当且仅当对每个，，其中。
这个定理表明，在的开集上所定义的分布可以构成上的一个层。

在代数几何中，仿射空间上的层通常指的是在某个拓扑空间（比如赋予通常的欧几里得拓扑）上的一个结构，它将每个开集关联到一个代数对象（比如一个环、一个模），并且这些关联满足某些相容性条件。更具体地说：
一个层由以下部分构成：

对于每个开集，有一个代数对象，例如一个交换环。我们称为在上的截面。
对于每一对开集，有一个限制映射，它是一个代数同态。这个映射描述了如何将上的截面限制到上。
这些数据需要满足以下公理：

恒等性：对于每个开集，是上的恒等映射。
传递性：如果是中的开集，那么。换句话说，从限制到，可以直接进行，也可以先限制到再限制到，结果是一样的。
粘合性：设是开集的一个开覆盖。如果是两个截面，并且对于每个，都有，那么。也就是说，如果两个截面在每个开覆盖的子集上都相等，那么它们在整个集合上相等。
整体截面的存在性：设是开集的一个开覆盖。假设我们有一族截面，使得对于每一对，都有。那么存在唯一的截面，使得对于每个，都有。也就是说，如果一族截面在开覆盖的交集上相容，那么它们可以粘合成一个整体截面。
最常见的例子是连续函数层，其中是上所有连续实值函数的集合，限制映射就是通常的函数限制。另一个重要的例子是全纯函数层，其中是上所有全纯函数的集合，限制映射也是通常的函数限制。

给定开集上的分布，我们来定义它的支集。
假设是开子集，如果，我们就说在上为零，也就是说，对于每个，我们有

我们现在来说明，存在中使得在其上为零的最大开集。为此，我们定义

令

按照定义，为开集。我们要证明，为此，只要证明对任意的，我们都有

即可。实际上，令，那么存在限个覆盖。我们取与这个覆盖相应的单位分解。从而，

最后一步，我们用到了。很明显，是这种开集中最大的。
Definition (分布的 (紧) 支集). 我们把称作是的支集并仍然用符号表示。如果是紧集，我们就说是有紧支集的分布。我们用来表示上有紧支集的分布的全体。

傅里叶变换的一致连续性

Tue, 13 May 2025 07:03:44 GMT

Definition. 设 , 即 . 其傅里叶变换定义为: 其中表示和在中的标准内积（点积）。
我们要证明在上是一致连续的。也就是说，对于任意的 , 存在一个 , 使得对于所有 , 若 , 则 .
Proof.
考虑差值 :

由于 , 我们可以得到:

令 . 我们的目标是证明当时,

我们使用勒贝格控制收敛定理 (Lebesgue's Dominated Convergence Theorem)。定义函数序列（或族） .

逐点收敛 (Pointwise Convergence): 对于任意固定的 , 当时, . 因此, . 所以, . 从而, 对于几乎所有的（在定义的意义上）。
控制函数 (Dominating Function): 对于任意和任意 , 我们有:
. 因此, . 由于 , 函数在上是可积的, 即 .
根据勒贝格控制收敛定理，由于逐点收敛到并且被一个可积函数控制，我们有:

这意味着，对于任意 , 存在一个 , 使得只要 , 就有

将代回，如果 , 则

这正是函数在上一致连续的定义。
□
因此, 上的函数的傅里叶变换是一致连续函数。这个性质也被称为 Riemann-Lebesgue 引理的一个推论或相关性质（尽管 Riemann-Lebesgue 引理本身通常指当）。

高维常微分方程

Tue, 13 May 2025 10:11:31 GMT

参见 Arnold，他使用了一种很高的观点.

Consider , defined by the vector field . We define the Picard mapping

Geometrically, passing from to means constructing w.r.t. a curve a new curve whose tangent for each is parallel to a given direction field, only not on the curve itself -- for then would be a solution -- but at the corresponding point of the curve .

To prove convergence of the successive approximations we shall construct a complete metric space in which the Picard mapping is a contraction.
The mapping satisfies a Lipschitz condition with constant provided that

Let be a - mapping (). Naturally, the derivative of at

is a linear operator. If we choose a basis for each tangent space, has a matrix representation.
provides the best linear approximation of at the neighborhood of , i.e. .

Remark. By hypothesis , is continuous; then attains a maximum value on any compact set .
See arnold for the proof of theorem of existence and uniqueness.
Next we consider the theorem on differentiablility. The motivation is as follows.
Associated with a differentiable mapping is a linear operator on the tangent space

At each point , .
In exactly the same way with a differential equation

there is associated a system of differential equations

called the system of equations of variations for Equation (1), and linear w.r.t. the tangent vector .

We can correspond the unknown vector to a unknown linear operator .

Suppose of Equation (1) is in some neighborhood of . Then the solution of Equation (1) with initial condition depends on the condition in a continuously differentiable manner as and vary in some neighborhood of :

This theorem explains the meaning of the equations of variations: they describe the action of the transformation over the time from to on the vectors tangent to the phase space (Fig. 227).

More generally,

We prove by induction

In fact, we have

Lemma. The solution of a linear equation whose right-hand side depends continuously on , exists; is unique; is determined uniquely by the initial conditions ; is a continuous function of and ; is a linear function of and a continuously differentiable function of , thus in and .
Lemma. The solution of a linear equation where is continuous, is a continuous function of , , and .
We now apply Lemma 3 to the equation of variations, Equation (2) .
The system of equations of variations

has a solution uniquely determined by its initial data and depends continuously on them provided .

缓增分布的Fourier变换

Fri, 02 May 2025 10:03:11 GMT

参见于品数学分析讲义

Diffusion

Sun, 04 May 2025 15:32:10 GMT

See Partial Differential Equations in Action, Complements and Exercises.
The diffusion equation (heat equation) is

defined on a cylindrical domain , where is a domain[1] of , and the Laplacian is taken w.r.t. the spatial variables only.

为了保证一个函数的傅里叶级数收敛，该函数需要满足一定的条件。以下是一些常见的条件：

狄利克雷条件（Dirichlet Conditions）：
一个周期为的函数，如果满足以下条件，则其傅里叶级数收敛：
在任何有限区间内，只有有限个不连续点。
在任何有限区间内，只有有限个极大值和极小值。
在任何有限区间内是绝对可积的，即。如果在处连续，则傅里叶级数收敛到；如果在处有跳跃不连续点，则傅里叶级数收敛到，其中和分别表示在处的右极限和左极限。
更强的条件：
如果是连续的，并且存在且连续，那么的傅里叶级数一致收敛到。
如果是平方可积的，即，那么的傅里叶级数在均方意义下收敛到。
总的来说，狄利克雷条件是最经典和常用的判断傅里叶级数收敛的条件。

Definition (Bessel's equality). 令是一个希尔伯特空间，并设是的一个正交基。那么，对于中的任何，Bessel 不等式成立
Definition (Bessel's equality). Let be a function on such that . Then where
Theorem (Poincaré's inequality). 设是一个有界开集。存在一个常数，使得对于所有，有
Proof.
设是一个有界开集。那么存在一个，使得。设。我们可以将延拓到上，使得。那么在上为 0。因此，我们可以将延拓到上，使得。
对于，有

因此，

利用 Cauchy-Schwarz 不等式，我们有

由于是有界的，所以。因此，

在上积分，我们有

因此，

取平方根，我们有

□
Theorem (Poincaré不等式（n=1）). 设，其中。那么
Proof.
通过延拓为上的奇函数，其中，我们有。因此，的傅里叶级数为

其中

Parseval恒等式给出

此外，形式上的导数为

因此

由于，我们有

或者

由于在外为零，且，我们得到

□

If on , then reaches its maximum and minimum on the boundary .

We separate the variables twice.

connected, open subset↩︎

first-order-equation

Sun, 04 May 2025 15:32:10 GMT

Fourier 分析初步

Sat, 03 May 2025 16:31:13 GMT

参见于品数学分析讲义 65,66节
Definition ( 函数的 Fourier 变换). 对于函数，我们定义它的 Fourier 变换（或者 ) 为如下的上的函数：其中。我们通常把的定义域称作是频率空间，它的变量通常用来表示。
我们考虑的线性子空间

我们知道是 Banach 空间. (取一列内于，验证即可)
Lemma (引理 335). 给定赋范线性空间和它们之间的线性映射。那么，如下三个论断是等价的：
在处连续；
是连续的；
是有界，即存在，对每个，我们都有
Theorem (Theorem 4.48). 对任意的 , . 特别地, 线性映射是连续线性映射, 即存在常数 , 使得对任意的 , 我们有
显然

由于在中稠密，故只需要证明连续，由 Lebesgue 控制收敛定理可知：假设中的点列，有

控制函数为 .
对于 , 我们有如下两个重要观点：

物理空间的求导等价于频率空间的乘法：
物理空间的乘法等价于频率空间的求导：
实际上，对任意的，对任意的正整数，利用第一个原则，我们有

其中是 Laplace 算子.
很明显，，所以，我们有

这表明是衰减的并且我们对于求的导数越多，那么衰减速度就越快. 特别地，我们还证明了 .
通常为了说明在无穷远处趋于 0，我们任取，再选取使得

从而对于任意的，我们有

当很大的时候，有 .
然⽽，这些过程实际上都被包装在开始的命题上，所以我们通常只对光滑有紧支集的函数（在中稠密）来证明即可.

Theorem (卷积对应于频率空间的乘积). 对任意的，我们有
Proposition. 假设 . 那么，

Proposition (命题 451 （仿射坐标变换与 Fourier 变换之间的关系）). 任给和可逆线性变换我们有
1）物理空间的平移对应频率空间乘相应的频率，即 2）频率空间应该视作是余切丛，即
证明：第一部分是显然的；为了证明第二部分，我们直接利用换元公式：

命题成立。

Theorem (Gauss 函数的 Fourier 变换). 对任意的正数，我们有
利⽤上⾯ Guass 函数的 Fourier 变换，我们可以证明关于 Fourier 逆变换的定理。
Definition. 对任意的，对任意的我们定义
Theorem (Fourier 逆变换). 给定，如果，那么，我们有其中，上面等号成立是在的意义下的。

注意到, 是稠密的子空间。我们任意选取 , 很明显, (因为光滑性意味着衰减很快, 所以可积)。另外, 我们有

所以，我们有

从而,

这表明定义在是稠密的子空间上的 Fourier 变换

是连续的。

根据连续线性映射扩张的定理，我们就证明了
Theorem (定理 4.55 ( 上的 Fourier 变换与 Planchrel 公式)). 我们可以定义 Fourier 变换 : 使得是等距同构。特别地，对于，我们有通过极化，我们有对任意的，我们有
Proof.
上述一切叙述对是成立的。对一般的，用中函数逼近即可。
□
Remark. 上面的定理定义了的Fourier变换，为了行文清楚，我们暂且把意义下的Fourier变换记作 . 另外，对于 , 它的Fourier变换是可以用Fourier积分表示的，我们把它记做 , 也就是说那么，对于, 我们有
Example. 考虑上的－函数其中，。很明显，我们知道，所以我们不能直接用－函数的 Fourier 积分来写它的 Fourier 变换。然而，我们知道是的，我们可以先显式写下的 Fourier 变换。由于序列在中逼近，所谓，我们有

对任意的给定的函数，对任意的多重指标，我们采用如下的符号：

Definition (Schwartz 空间，Schwarz 函数（速降函数）). 函数是上的光滑函数。如果满足如下的条件：对任意的多重指标，我们都有那么，我们就称是一个 Schwartz 函数或者是一个速降的函数。我们把上所有的 Schwartz 函数所构成的线性空间称作是 Schwartz 空间，并记作。对于每个非负整数，我们定义如下的（一族）范数：在上，我们规定如下的收敛性（拓扑）：给定 Schwartz 函数的序列，它收敛到 Schwartz 函数，指的是对任意的非负整数，我们都有我们把这个极限简写成
Example. 我们已经见过很多 Schwartz 函数
；
；
对于 Schwartz 函数，对它求若干次导数或者乘以一个多项式仍然是一个 Schwartz函数，即对任意的多重指标，我们有上面例子的验证我们留作作业。
给定一个 Schwarz 函数，我们对它有如下的估计：对于任何的多重指标和，其中，我们有

其中，是空间的维数。
从而，对任意的，我们有

上式右边的函数是可积的，所以，

特别地，我们可以对用傅里叶积分来定义其傅里叶变换。作为推论，我们还知道

另外，以上的估计是常用的技巧，在后面的不少场合都会用到。
Theorem. 在中是稠密的，即对任意的，存在函数序列，使得
意味着对于任意非负整数，有 .
光滑函数乘以一个函数，结果仍为光滑函数，当且仅当另一个函数也是光滑函数.
Proof.
我们选取有紧支集的光滑函数，使得

对于，我们令

我们只要证明，对任意的非负整数，我们有

即可。对于满足的多重指标，我们有

上式的第二个求和部分有这样的衰减因子，所以极限为 0。对于第一项，由于在半径为 1 的球内部为 1，所以然而，我们有

这就完成了证明。
□

Theorem ( 上的 Fourier 变换). 如果是 Schwartz 函数，那么，。在 Schwartz 函数空间上的 Fourier 变换：满足如下的性质：对任意的，存在常数，使得对每个，我们都有特别的，是连续的线性同构，即对任意的在中收敛的函数序列我们有另外，对任意的，我们还有公式
Proof.
我们首先证明叙述中的最后两个恒等式。对任意的 Schwartz 函数，对任意的，利用分部积分，我们有

第二个等式要用 Lebesgue 控制收敛的推论（积分与求导数可交换），我们有

现在证明定理中的不等式（从而证明了 Fourier 变换的像也落在中）。固定两个多重指标和，其中。利用已经证明的公式，我们就有

Fourier 变换的连续性可以通过这个不等式得到：对任意给定，我们有

按照定义，我们就有

最后，我们来说明是同构。实际上，我们可以定义直接考虑 Fourier 变换的逆，因为，所以之前定义的在此也是良好定义的。此时，我们已经证明了与互为逆映射，所以命题得证（也是连续的）。
□

Definition (缓增的分布). 假设是一个分布。如果存在非负整数和常数，使得对每个，我们都有我们就说是一个缓增的分布。我们用来表示所有缓增分布的集合，很明显是线性子空间。

Functional-Analysis

Wed, 14 May 2025 04:34:06 GMT

Let be a domain. The Sobolev spaces we will use most are: the Hilbert space

(all derivatives are meant in distributional sense) with the inner product

Theorem. 证明 .[1]
Proof.
设。我们需要证明在上连续。由于 , 则且 (弱导数) 。
根据Sobolev嵌入定理，如果是一个有界开集，且足够光滑，那么当时，。
在我们的例子中，，所以。我们有，因此且。因为 , Sobolev嵌入定理告诉我们。
这意味着存在一个连续函数使得几乎处处成立。由于我们通常将中的元素视为等价类（即，如果两个函数在Lebesgue测度意义下相等，则认为它们是等价的），我们可以选取这个连续的代表作为。
因此，在上有一个连续的代表，这意味着。
□
Theorem (finit-dimensional space). Let be a normed linear space (over or ). Then is finite-dimensional if and only if its closed unit ball is compact.

Imagine you have:

A special kind of "space" where you can measure distances and angles (this is the Hilbert space, ).
A rule that takes any two items from this space and gives you a number. This rule needs to be "nice" in two ways:
It doesn't blow up: The number you get isn't excessively large compared to the "size" of the two items (this is boundedness).
It's "positive" in a strong way: When you use the rule with the same item twice, the number you get is significantly positive and related to the "size" of that item squared (this is coercivity). This rule is your bilinear form, .
Another rule that takes any one item from your space and gives you a number, and this rule is also "nice" (it doesn't blow up) (this is your bounded linear functional, ).
The Lax-Milgram Theorem then says:
If your "combining rule" () is "nice" (bounded and coercive), then for any "measuring rule" () you pick, you can always find one and only one special item () in your space such that:
When you combine this special item () with any other item () using your rule , you get the exact same number as when you measure that other item () using your rule .
In short:
Under certain "good behavior" conditions for how you combine and measure things in a special space, a specific type of equation () is guaranteed to have exactly one solution () in that space.
Why it matters (the even simpler version):
It's a powerful math tool that tells us that certain types of problems (often coming from physics and engineering, like how heat spreads or structures bend) definitely have a unique answer, as long as the problem is "well-behaved" in a specific mathematical sense.

The Big Idea: Does My Problem Even Have an Answer?
Imagine you're an engineer or a scientist. You've written down a mathematical equation (often a "partial differential equation" or PDE) that you believe describes a real-world situation. Before you spend a lot of time (and computer power) trying to find the answer, you'd want to know:

Does an answer even exist?
If it exists, is there only one unique answer? (If there are many, which one is the "right" one for your physical situation?)
This is where the Lax-Milgram theorem comes in. It's a powerful mathematical tool that can answer "YES!" to both questions for a specific class of problems.
How it Works (Simplified):

Hard Problem: The original PDE is often tricky to solve directly.
Easier Version (Weak Formulation): Mathematicians have a way to rephrase the hard problem into an "easier" (though still abstract) form. This easier form is an equation that looks like .
Lax-Milgram's Job: If the parts of this "easier" equation ( and ) satisfy certain "good behavior" conditions (boundedness and coercivity), the Lax-Milgram theorem guarantees that there's exactly one solution to this easier version.
Good News: This "weak solution" to the easier version is often good enough for practical purposes and, in many cases, can be shown to be the actual solution to the original hard problem.
Why is this "guarantee" important?

For Scientists/Engineers: It tells them their mathematical model of the physical world is well-posed (meaning it has a unique solution). This gives them confidence that their model makes sense.
For Computer Simulations: Many complex engineering and science problems are solved using computer simulations (e.g., Finite Element Method - FEM). These methods essentially try to find an approximate solution to the "easier" weak formulation. The Lax-Milgram theorem provides the theoretical backbone, assuring us that the problem these numerical methods are trying to solve actually has a unique solution to begin with. Without this, the computer might be chasing a ghost!
So, while you might not use the Lax-Milgram theorem directly in everyday calculations, it's a crucial piece of foundational mathematics that makes a lot of advanced engineering analysis and scientific computation possible and reliable. It gives a green light for many solution-seeking processes.

Exercise (非线性算子的连续性；紧性).
一个关键的定理是：对于线性算子，以下说法是等价的：

在上是连续的。
在中的某一点处是连续的。
在原点处是连续的。
是有界的 (Bounded)。这是最常用和最重要的等价条件。
有界线性算子的定义：
线性算子称为有界的，如果存在一个常数 , 使得对于所有的 ,都有：
这个常数称为算子的一个界。最小的这样的称为算子的范数，记为。

我们可以证明：对于任意的 ,

故良定义. 接下来直接验证连续性：对于 ,

若 , 那么 . 故连续.
接下来验证紧性

Exercise.
If any subsequence of a sequence converges to , then the sequence converges to . (Proved by limsup and liminf)

Exercise (yau-22-problem 4). Problem 4. Let be the space of all continuous -valued functions equipped with -norm. Let be a closed linear subspace. Assume that the elements of are polynomials. Prove that .
Main Idea:
The core strategy is to demonstrate that the closed unit ball in the subspace is compact. A fundamental theorem in functional analysis states that a normed linear space is finite-dimensional if and only if its closed unit ball is compact. By proving the compactness of the unit ball of , the proof concludes that must be finite-dimensional.
Sketch of the Proof (Logical Flow):

Problem Setup: We have a closed linear subspace of (continuous functions on with -norm), and all elements of are polynomials. We want to prove .
Introduce Slope Operators:
Define a family of linear operators (where is the complex numbers) for distinct :

This is essentially the slope of the polynomial between and .
Show Pointwise Boundedness of these Operators:
For any fixed polynomial , its derivative is continuous on and thus bounded (i.e., ).
By the Mean Value Theorem, for some between and .
This means for each , the set is bounded.
Apply Banach-Steinhaus Theorem (Uniform Boundedness Principle):
is a closed subspace of a Banach space (), so itself is a Banach space.
Since the family of continuous linear operators is pointwise bounded on the Banach space , the Banach-Steinhaus Theorem implies that their operator norms are uniformly bounded.
So, there exists a constant such that for all .
Establish Equicontinuity of the Unit Ball of P:
Consider the unit ball .
For any :

This implies for all .
This is the definition of uniform Lipschitz continuity for the functions in , which in turn implies that the family is equicontinuous.
Apply Arzelà-Ascoli Theorem:
The unit ball is defined on the compact set .
is equicontinuous (from step 5).
is uniformly bounded (since for , , meaning for all ).
The Arzelà-Ascoli Theorem states that a family of functions that is equicontinuous and pointwise (or uniformly) bounded on a compact set is relatively compact. Since is also closed (it's a closed unit ball), it is compact.
Conclude Finite Dimensionality:
A normed linear space is finite-dimensional if and only if its closed unit ball is compact.
Since the closed unit ball of has been shown to be compact, must be finite-dimensional.
In essence, the proof cleverly uses the properties of polynomials (differentiability) to construct a family of operators. The Banach-Steinhaus theorem then provides a crucial uniform bound. This bound leads to the equicontinuity of the unit ball, which, combined with its boundedness, allows the Arzelà-Ascoli theorem to establish its compactness, ultimately proving finite dimensionality.

This is false in dimension .↩︎

fundamental-solution

Wed, 14 May 2025 07:35:34 GMT

Okay, let's deduce the fundamental solution of the Laplacian more intuitively. The fundamental solution is defined as the solution to the equation:

where is the Laplacian operator and is the Dirac delta function, representing a point source at the origin. The previous proof concerning the harmonic function in used the representation:

This implies that the fundamental solution satisfying is in . Let's see how we can arrive at this.

This approach is very physical and geometric, especially for an operator like the Laplacian.
a. Radial Symmetry:
The Dirac delta function is a point source at the origin. It's spherically symmetric (its value depends only on whether ). We can expect the potential generated by this point source to also be spherically symmetric. Thus, should depend only on the distance from the origin:

b. Laplacian in Spherical Coordinates:
For a radially symmetric function in , the Laplacian is:

In (which is the context of the original problem), , so:

c. Solution Away from the Origin ():
Away from the origin (, so ), the Dirac delta is zero, . So, we must have .

This implies that must be a constant, let's call it :

So,

Integrating with respect to gives:

where is another constant. For potentials that vanish at infinity (which is a common physical condition, and matches in the problem), we typically set . So, .
d. Determining the Constant A (using the source):
To find , we use the defining equation . Integrate both sides over a small ball of radius centered at the origin:

The right side, by definition of the Dirac delta, is .
For the left side, we use the Divergence Theorem (Gauss's Law):

where is the sphere of radius , and is the outward unit normal vector.
Since , its gradient is , where is the radial unit vector. The outward normal on the sphere is also .
So, . At , this is .
The surface area of the sphere in is .
So the surface integral becomes:

Equating this to the integral of the Dirac delta:

Substituting this back into :

Thus, the fundamental solution in is:

This is precisely the kernel that appeared in the integral representation for in the problem you referenced.
Intuition from Flux Spreading:
Imagine a "flux" emanating from the point source at the origin.

The term (which is in ) can be thought of as proportional to the total flux of through a sphere of radius .
The condition for means this flux is constant for any .
The strength of the field (gradient) must therefore decrease as to keep the flux constant, because the surface area of the sphere is proportional to .
In , surface area , so field . Potential .
In , surface area (circumference) , so field . Potential .
The constant value of this flux is determined by the strength of the source , which is when integrated. This directly gives the constant .

This method is more algebraic and general for constant-coefficient linear PDEs.
Let be the Fourier transform.
The equation is .
Taking the Fourier transform of both sides:

FT of Dirac Delta: .
FT of Laplacian: The Fourier transform of is .
So, .
Putting these together, the equation in Fourier space becomes:

So, the Fourier transform of the fundamental solution is:

Now, we need to find the inverse Fourier transform :

This integral needs to be computed.
For (as in the problem):

To evaluate this, we can use spherical coordinates for . Let . By rotating coordinates, we can align with the -axis, so , and , where is the angle between and . The volume element is (where is the azimuthal angle for ).

The inner integral is

So,

Let , then . The integral becomes

This is the Dirichlet integral, which evaluates to .

This matches the result from the radial symmetry method.
Intuition from Fourier Transform:

The Laplacian acts as a multiplication by in Fourier space. It heavily attenuates high frequencies.
The Dirac delta has a flat spectrum (all frequencies equally present: ).
means the fundamental solution has a spectrum that decays fairly slowly for small (it's singular at ) and faster for large . This behavior in Fourier space is characteristic of type potentials in real space for 3D. The singularity at in hints at the long-range nature of the potential .
Both methods yield the same result. The radial symmetry/Gauss's law approach is often more physically intuitive for the Laplacian, while the Fourier transform method is a very powerful and general algebraic tool applicable to a wider range of constant-coefficient linear differential operators.

The fundamental solution for the Laplace equation satisfies

where is the Laplacian operator.

Fourier Transform:
Applying the Fourier transform with respect to to both sides:

We know that . For the left side, the derivative property of Fourier transforms gives . Thus, .
So, .
The transformed equation is:

Solving for :

Inverse Fourier Transform:
Now we need to compute the inverse Fourier transform:

Case :
We use the Schwinger parametrization . Here :

Substituting this into the integral for :

Assuming we can interchange the order of integration:

The inner integral is an -dimensional Gaussian integral. The exponent is .
Completing the square for each component : .

Since , the -dimensional integral is .
Substituting this back:

Let . Then , so .
The integral becomes (for , so for convergence of ):

So, for :

For example, if , .

Case :
The above derivation for the -integral requires , so it doesn't directly apply to . The direct Fourier inversion of in 2D is subtle. The known result is:

This can be verified by showing in the distributional sense.

Case :
Similarly, for . Direct inversion is subtle. The known result is:

This satisfies .

The fundamental solution (or heat kernel) satisfies the homogeneous heat equation with a Dirac delta initial condition:

where is the thermal diffusivity.

Spatial Fourier Transform:
Let . Applying the Fourier transform to the PDE:

The initial condition transforms to .

Solve the ODE in Time:
This is a first-order ODE for with respect to :

Using the initial condition , we find . So, for :

Inverse Spatial Fourier Transform: The exponent is .
We complete the square for each component : The integral becomes: Each 1D Gaussian integral is .
Therefore:

The fundamental solution for the wave equation satisfies:

where is the wave speed. We seek the retarded Green's function, satisfying for .

Space-Time Fourier Transform:
Let .

The transformed equation is:

Inverse Fourier Transform with respect to (Time): Let . The integral is .
The poles are at . For the retarded Green's function, this integral evaluates to: where is the Heaviside step function. This ensures for .
Definition (Heaviside step function). The Heaviside step function is defined by

Inverse Fourier Transform with respect to (Space): The evaluation of this integral depends heavily on the dimension .

For : The integral if , and if . Here . This means for and , and otherwise.
For : The integral evaluates to .
For : The integral can be evaluated using spherical coordinates. Let . This can also be written as .
These derivations show how Fourier transforms can be systematically used to find fundamental solutions. The main challenge often lies in evaluating the resulting inverse Fourier transform integrals, which can be quite involved depending on the dimension and the complexity of the function in Fourier space.

heat-and-Schrodinger-equation

Fri, 02 May 2025 08:58:52 GMT

The irreversibility of its semiflow and the need to impose a growth condition as in order to pick out a unique solution.

The finite-dimensional phase sapce of the ODE is replaced by the infinite-dimensional function space ; then we think of a solution of the heat equation as a parametrized curve in the vector space .

Note Formally, the Schrodinger equation is obtained by the transformation of the heat equation to 'imaginary time.' The analytical properties of the heat and Schr¨odinger equations are, however, completely different and it is interesting to compare them.

Hyperbolic-equations

Fri, 02 May 2025 08:58:52 GMT

The prototypical example of a hyperbolic PDE is the wave equation

To begin with, consider the one-dimensional wave equation on

The general solution is the d'Alembert solution

where , are arbitrary functions. This solution describes a superposition of two traveling waves with arbitrary profiles, one propagating with speed one to the right, the other with speed one to the left.
Let us compare this solution with the general solution of the one-dimensional heat equation

which is given for by

Some of the qualitative properties of the wave equation that differ from those of the heat equaion, which are evident from these solutions.

To obtain the basic energy estimatie for the wave equation, we multiple Equation (1) ny and write

to get

This is the differential form of conservation of energy. The quantity is the energy density and is the energy flux.

Laplace

Sat, 03 May 2025 09:39:07 GMT

见 Evans 偏微分方程笔记
Remark. 这里很多证明都非常符合正常人的思考过程，其中还用到了经典的分析学思想，中既开又闭的集合平凡。非常不错。
解题参见 Partial Differential Equations in Action, Complements and Exercises

Parabolic-equations

Fri, 02 May 2025 08:58:52 GMT

Some results about the integration and differentiation of Banach-space valued functions of a single variable.
Vector-valued integrals of integrable functions have similar properties, often with similar proofs, to scalar-valued -integrals.
The existence of different topologies (such as the weak and strong topologies) in the range space of integrals that take values in an infinite-dimensional Banach space introduces significant new issues that do not arise in the scalar-valued case.
Suppose a real Banach space with and dual space .

Definition.
Definition (strongly measurable).
Measurability is preserved under natural operations on functions.

Definition (weakly measurable). A function is said to be weakly measurable if the real-valued function is measurable for every .
This amounts to a 'coordinatewise' defintion of measurablility, in which we represent a vector-valued function by its ral-valued coordinate functions.
For finite-dimensional, or separable, Banach spaces these definitions coincide, but for non-separable sapces a weakly measurable function need not be strongly measurable.
The relationship between weak and strong measurability is given by the following Pettis theorem.
Definition (almost separably valued).
Theorem (Pettis theorem). A function is strongly measurable iff it is weakly measurable and almost separably valued.
Definition (weak continuous).
Since a continuous function is measurable, every almost separably valued, weakly continuous function is strongly measurable.
Example (weak but not strong measurable).
Example ( separable but not).

The definition of the Lebesgue integral as a supremum of integrals of simple functions does not extend directly to vector-valued integrals because it uses the ordering properties of in and essential way.
One can use duality to define -valued integrals in terms of the corresponding real-valued integrals where , but we will not consider such weak definitions of an integral here.
Instead, we define the integral of vector-valued functions by completing the space of simple functinos with respect to the -norm. The resulting integral is called the Bochner integral, and its properties are similar to those of the Lebesgue integral of integrable real-valued functions.
Definition (simple function).
Definition (Bochner integral).
The value of the Bochner integral of is independent of the sequence of approximating simple functions, and

Moreover, if is a bounded linear operator between Banach spaces and is integrable, then is integrable and

More generally, this equality holds whenever is a closed linear operator and , in which case .
Definition (closed linear operater).
Example ( as an operator).
Theorem.
The DCT for Bochner integrals with proof same as the scalar-valued case omitted.
Theorem.
The definition and properties of -spaces of -valued functions are analogous to the case of real-valued functions.
Definition ().
As usual, functions equal pointwise a.e. are regarded equivalent.
Theorem. If is a Banach space and then is a Banach space.
Simple functions of the form are dense in . By mollifying these functions with respect to , we get the following density result.
Theorem.
Theorem.
The proof is more complicated than in the scalar case and some condition on is required. Reflexivity is sufficient.

Definition (strong continuous and strong differentiable).

Poincare Inequality

Wed, 14 May 2025 15:19:36 GMT

庞加莱不等式（Poincaré Inequality）通常叙述于有界区域上。对于整个空间，标准形式的庞加莱不等式（即）通常不成立，除非对函数施加特定条件，例如具有紧支撑（compact support）或者属于特定的加权 Sobolev 空间。
一个在上下文中最直接理解的庞加莱不等式版本是针对具有紧支撑的函数。我们将证明这个版本。
Theorem ( 上的庞加莱不等式 - 紧支撑函数版本). 设，即是上具有紧支撑的连续可微函数（也可以推广到 Sobolev 空间）。令表示的支撑集。则存在一个常数，使得
其中常数依赖于的支撑集的大小（例如，其直径或包含该支撑集的某个立方体的边长）。
为了简便，我们考虑的情况。
因为具有紧支撑，所以存在一个足够大的开立方体，使得。这意味着在的边界上为零。立方体的边长为。
对于任意，由于（因为点在的边界上，而在边界外为零），我们可以沿方向使用微积分基本定理：
取绝对值：
由于积分区间包含于，并且被积函数非负，我们可以扩大积分区间：
现在，我们对在立方体上积分。首先固定，对积分：
令。这是一个关于的常数。
所以，
现在应用 Jensen 不等式（或 Hölder 不等式：，这里，）：
对于，我们有
由于 , 。
所以，
代回到关于的积分不等式：
现在，我们在的其余维度上积分 (即对积分)：
这给出：
由于，我们可以得到：
因为，所以在之外为零，也在之外（几乎处处）为零。因此，积分可以扩展到整个：
取次根：
这里的常数是包含支撑集的立方体的边长。更一般地，常数可以取为支撑集在某个方向上的最大长度，或者其直径。
重要说明：

常数的依赖性：这个版本的庞加莱不等式的常数依赖于函数支撑集的大小。对于支撑集越来越大的函数序列，这个常数会趋于无穷，这与有界区域上的庞加莱不等式常数仅依赖于区域的几何性质不同。
一般情况：如果不对的行为（如紧支撑或在无穷远处衰减）做任何假设，标准形式的庞加莱不等式在上是不成立的。例如，一个非零常数函数的梯度为零，但其范数为无穷大。
相关不等式：在上，与函数及其梯度相关且不依赖于支撑集大小（或具有不同类型依赖）的更典型不等式是 Sobolev 不等式（例如 Gagliardo-Nirenberg-Sobolev 不等式），它通常将空间嵌入到某个空间中，其中可能不等于。
上面证明的是在上针对紧支撑函数最直接和常见的“庞加莱不等式”形式。

Second-Order-Elliptic-Equations

Sat, 03 May 2025 17:05:45 GMT

We will exploit two essentially distinct techniques:

Energy methods within Sobolev spaces ()
Maximum principle methods ()

We will in this chapter mostly study the boundary-value problem

where is an open, bounded subset of , and is the unknown, .

for given coefficient functions ()
Definition (elliptic).
Ellipticity thus means that for each point , the symmetric matrix is positive definite, with smallest eigenvalue greater than or equal to .
An obvious example is , , , in which case the operateor is .

Sobolev-Spaces

Fri, 02 May 2025 08:58:52 GMT

To apply ideas of functional analysis to glean information concerning PDF, theories of Sobolov spaces is developed, which is sometimes subtle and unmotivated but ultimately will prove extremely useful.

We defined the Hölder spaces and the supremum norm and the -Hölder seminorm of to be

And the -Hölder norm to be

The Hölder space consists of all functions for which the norm

is finite. The space of functions is a Banach space. (left as an exercise)

If denotes a real linear space, then a mapping is called a norm provided

for all .
for all .
if and only if .
A norm provides us with a notion of convergence: we say a sequence converges to , written , if .
A Banach space is then a normed linear space which is complete, i.e. within which each Cauchy sequence converges.

Hölder spaces is not often suitable. We want other spaces containing less smooth functions. In practice we must strike a balance.

denotes the space of infinitely differentiable functions with compact support in . is sometimes called a test function. We have

There are no boundary terms, since has compact support in and thus vanishes near . Then we define the weak derivative of .
Let be a multiindex. Say that is the -weak partial derivative of , written , provided

The weak derivative is well-defined. (it's unique.)

is not strongly differentiable at but weakly differentiable.

This is an example showing that the weak derivative does not always exists.
Let , and

We assert does not exists in the weak sense. To check this, we must show that there does not exist any function satisfying

for all . Suppose, to the contrary, Equation (1) were valid for some and all . Then

Choose a sequence of smooth functions satisfying

Replace by in Equation (2) and send , we discover

a contradiction.

Fix and let be a nonnegative integer. The Sobolev space consists of all locally summable function such that for each multiindex with , exists in the weak sense and belongs to .
Remark. If , we usually write . is a Hilbert space. Note that .
If , we define its norm to be

We write

to mean

for each . 意指为紧集即有界闭集，且 .

The following example shows what condition the parameters should satisfy such that .

The following example illustrates a fundamental fact of life, that although a function belonging to a Sobolev space possesses certain smoothness properties, it can be rather badly behaved in other ways.

We verify some certain properties of weak derivatives, which is obvious for smooth functions but relies proof solely upon the defintion of weak derivatives.

We prove the Leibniz's formula by induction on . Suppose first . Choose any . Then

Check in by checking that in for all .

In order to study the deeper properties of Sobolev spaces, we therefore need to develop some systematic procedures fo rapproximating a function in a Sobolev space by smooth functinos.
Fix positive integer and . .

Claim that if then

Next we show that we can find smooth functions which approximate in and not just in . Notice that the smoothness of is not neccessary.

使用了单位分解的思想。

见崔尚斌

Theorem (Trace Theorem).
Definition (trace). We call the trace of on .

Sobolev 空间理论的核心部分是三个嵌入定理:

Sobolev 嵌入定理
Morrey 嵌入定理
Kondrachov-Rellich 嵌入定理
Theorem.
只需证明对于任意都成立上述不等式，因为由此通过取极限便可以得到这个不等式对任意也都成立，并进而得到包含关系 .
当时，把零延拓到上便得到，这样对于任意满足的，有

关于在单位球面上积分，注意到，就得到

记 , . 由所设条件知道 , 进而 . 注意到

以及 , 应用推广的 Holder 不等式得

其中 . 因此

这里用到 . 选取使得，则对任意有

把代入，得

注意到，便从（1．8．2）得到了（1．8．1）．证毕．
把 Theorem 4 应用到的特殊情况，就有如下著名不等式
Theorem (Poincare 不等式). 对任意和有界开集成立不等式

把 Theorem 4 应用到的特殊情况，则有如下不等式
Corollary. 设，则对任意有界开集有，且成立不等式
其中 .
Theorem.
Remark. 我还是不大理解这个证明

 Theorem 4 和 Theorem 7 说明，在关于和的一定条件下，中的函数可以有比大的某些幂次的可积性，而且它们关于这些幂次的范数可以用其一阶弱导数的范数界定。后一性质是所特有的；当时，中的函数一般不具有这种性质，即中函数的范数一般不能被其一阶弱导数的范数界定. 例如当是有界开集时，非零的常值函数都属于；而对于这些函数，不等式和显然都不可能成立. 至于前一性质，则不是中的函数所特有的，事实上中的函数也具有类似的这种性质. 这一事实由以下定理所保证：
Theorem.
反复使用 Theorem 9 便可以得到
Theorem (Sobolev 嵌入定理).

本节介绍 Morrey 嵌入定理，它解释了 Sobolev 空间到 Holder 空间的嵌入关系。
Theorem.
Proof.
由于是凸开集，所以其边界是 Lipschitz 连续的，进而是可延拓开集. 因此根据定理 1.7.10 可知，只需证明上述不等式对任意成立即可. 这时，由的凸性可知对任意有

关于在上积分并除以，得

令表示把的各个分量都作零延拓所得到的上的向量函数，并记，则有

据此应用与定理 1.8.1 的证明类似的方法即可得到（1.9.1）．证毕．
□

Theorem (Holder 不等式).

Theorem (内插不等式).

Sobolev空间基本性质

Sat, 03 May 2025 16:53:38 GMT

参见于品数学分析讲义

在后面的课程中，我们会经常用所谓的 Planchrel 公式：对任意的，我们有

它的另外一个版本是说对任意的，我们有

这个公式我们之前已经证明过。
我们现在引入上的 Sobolev 空间的定义。
Definition (指标为的 Sobolev 空间). 给定，我们将把这个数称为是 Sobolev 空间的指标。我们考虑满足如下性质的缓增分布：
是局部可积的函数；
是平方可积的函数。对于这样的函数，我们定义其 Sobolev 范数为：我们把所有满足上述条件的缓增分布的集合称作是一个指标为的 Sobolev 空间，这显然是一个复线性空间，我们用来表示。
在上所赋予的范数与下面的内积是相容的：对任意的，令所以，是内积空间。
我们注意到，当时，我们实际上就是，这由 Plancherel 公式立即就可以得到：

所以，

类似的，如果我们在频率空间上考虑测度

那么，当且仅当。利用这个观察，我们现在证明：
Theorem. 对任意的是 Hilbert 空间（即完备的内积空间）。
Proof.
假设是 Cauchy 列，那么，根据定义，是 Cauchy 列。利用－空间的完备性，存在作为上述序列的极限。我们用表示它的 Fourier 逆变换，即

那么，

这就证明了完备性。
□
根据Sobolev空间的定义，我们知道构成了一个下降的链，即对任意的

我们观察到，Schwartz 函数生活在所有的Sobolev空间中：

Proposition. 假设为正整数，那么，有如下的等价刻画：
这个证明是初等的.

受到的启发，我们定义算子，简记为 . 形式上，对一个分布的作用在频率空间上看来就是乘以 .
Definition (Fourier 乘子). 给定频率空间上的函数，我们假设它是多项式增长的。对于任意的缓增分布，我们定义由于是多项式增长的，所以，仍然是缓增分布，所以，如下的算子是良好定义的：
Example. 我们先看几个简单的例子：
当时，其中，我们有
当时，我们有
给定线性微分算子它可以被视作是一个 Fourier 乘子，其中
算子表示的是函数所对应的 Fourier 乘子。
Example. 下面的Sobolev空间都定义在上。
对任意的，我们有实际上，我们只要说明下面的积分有限即可：这在时是成立的。同样的推理表明，当时，。
常数值函数1不在任何的中。特别地，这表明

Proposition (Sobolev 空间的映射性质). 给定多项式增长的乘子函数，其中，我们假设存在常数和，使得对任意的，我们都有那么，对任意的，对任意的。这就定义出有界（连续）线性映射：特别地，对任意的－阶的微分算子，对任意的，我们有连续线性映射另外，对任意的，我们还有连续的线性同构：其中，上述映射的逆映射是。
Proof.
对任意的，我们首先证明，其中具有命题中所要求的多项式增长。根据 Planchrel 公式，我们有

所以，存在常数，使得

这表明是从到的连续线性映射。
微分算子的情形是一个特例。为了说明有逆，我们用

作为乘子即可，这是因为

命题得证。
□

Proposition. 对每个指标，光滑有紧支集的函数在中是稠密的。
Proof.
我们首先证明是稠密的，其中：这个论断对是正确的，因为在中是稠密的。由于

是连续可逆的线性映射（是同胚）（因为 Proposition 7 (Sobolev 空间的映射性质) ），所以在这个算子下的像也是稠密的，然而，

所以，是稠密的。
在上面的论证中，不一定把有紧支集的函数映射为有紧支集的函数，所以，我们的推理是对进行的。
为了证明命题，我们只要说明在的意义下，中的任意一个函数都可以被的函数逼近。我们上次证明了存在常数，使得对任意的，我们有不等式

所以，对任意的，我们先选取，使得

再利用的稠密性，选取，使得

此时，我们有

这就证明了命题。
□
我们下⾯证明著名的 Sobolev 嵌⼊定理（的⼀种形式）：
Theorem (Sobolev 嵌入定理). 假设指标，那么，每个都落在中。进一步，我们有连续的线性嵌入即存在，使得对任意，我们都有进一步，是连续函数（可以在它的代表类中选到一个连续函数）并且在无穷远处的极限为零，即。
Remark. 证明的想法比较简单：我们只要说明是一个函数即可，因为 Fourier 逆变换就把它还原成一个在处衰减的连续函数，从而是的函数。
Proof.
根据，我们知道

我们现在说明被所控制。根据，我们可以写成两个平方可积的函数的乘积：

前一部分根据所以是的；后一部分根据所以是的。利用 Cauchy-Schwarz 不等式，我们有

所以，

连续性的部分是明显的，因为

证明完毕。
□
Corollary. 假设，其中为非负整数，那么，对任意的，我们都有。
Proof.
对任意的多重指标，如果，那么是连续函数，从而， (用归纳法来证明会更严格一点)。
□
Remark. 这个版本的 Sobolev 嵌入定理说的是，如果指标足够大，那么，函数就会非常光滑。
Remark. 我们在作业中将构造函数局部可积的，使得。换句话说，如下的嵌入并不成立：这表明 Sobolev 嵌入的指标至少是，其中可以任意小。

下一个定理说的是如果指标足够大，那么，两个的函数的乘积也是的。这个定理在证明非线性偏微分方程的解的局部存在性时很有用。
Theorem. 如果，那么，是一个代数，即对任意的，我们有实际上，存在常数，使得对任意的，我们有
Sobolev 不等式的⼀个重要的观点就是⽤函数以及它的导数的积分来控制函数的最⼤值.
Proof.
我们来计算的－范数。由于在 Fourier 变换下，乘积变化为卷积，所以按照定义，我们有

我们要把因子进行拆分。首先，对任意的，我们显然有

所以，对任意的，我们有如下的不等式

所以，我们就得到了

上面的表达式中本质上是两项，它们的结构是类似的，我们只要处理一项就好。我们现在利用第一项中的卷积结构来控制它。根据的定义，我们有

根据，我们在 Sobolev 不等式的证明中已经证明了。特别地，存在常数和，使得

我们观察到，上面就是控制的范数的大小。我们回忆上学期（5月 9 日的课程，利用 Fubini 定理）已经证明了

其中对任意的，我们有

我们对和运用这个不等式，就得到

□
实际上，我们还可以证明更强的结论：对任意的是一个代数。我们之后将利用频率空间的二进分解进行证明。

Yau contest

Sun, 04 May 2025 15:32:10 GMT

Exercise.
算子的可逆性和它的范数有密切的关系。
一般来说，如果一个算子的范数“足够小”，那么“恒等算子+该算子”是可逆的。更具体地：
Theorem. 设是巴拿赫空间上的有界线性算子。如果，则算子可逆，且

Theorem (Hölder's Inequality). 设且 , 则对于任意函数和 , 有, 且
只需要证明：

Unsolved...

函数空间

Fri, 02 May 2025 16:55:29 GMT

参见《实变函数习题精选》徐森林
Remark. 当时， in 与互不蕴含. 当时，前者蕴含后者.
Theorem ( 空间包含关系). 设，且，则，且

Theorem ( 空间两边夹住). 设，则
由此

Exercise (逐点收敛加上什么条件能推出收敛). 设在上， . 证明： in .
Proof.
由于 . 于是逐点考虑

又因为 , , 故

注意到，由 Lebesgue 控制收敛定理可知

也就是

□
Exercise (有关 ). 设且．证明：
Proof.
因为，故．于是

另一方面，对．

从而可得

又因，故，当时，有

综上，当时，有

因此

□
Exercise ( 空间的包含关系). 设，若证明对于有

注意这里的包含关系和 Theorem 2 ( 空间包含关系) 是相反的.
Proof.
先证明 . 反证而设，取中无交可测子集列，s.t. . 作函数

易见

因此 . 这与矛盾.
由于这个包含关系是反的，就会有一些神奇的性质，对于，

于是可以取使得，再用 Theorem 3 ( 空间两边夹住) 得到 . 因此 .
事实上，利用 Theorem 2 ( 空间包含关系) 可知这里 .
□
Exercise (利用 Holder 不等式). 设，令
证明：当时，有 .
Proof.
对于，记，则，且

□

. 于是可以定义内积

定义的夹角 (进而有垂直的概念)

Theorem ( 收敛蕴含弱收敛). 若则 .
Remark. 弱收敛不蕴含收敛，也不蕴含几乎处处收敛，也不蕴含依测度收敛.
设是中的规范正交系[1][2] 对于定义关于的广义 Fourier 系数为

广义 Fourier 级数为

此处省略习题....（太多了😭）

可以证明该集合至多可数↩︎
正交意味着任意两个元垂直，规范意味着模为 1.↩︎

积分理论

Thu, 15 May 2025 06:44:13 GMT

参见《实变函数习题精选》徐森林

Theorem (用连续函数刻画 Lebesgue 可测函数). 设为 Lebesgue 可测集，为上的 Lebesgue 可测函数，则对于任意，存在的闭子集，使得，且为上的连续函数.
将延拓到上，我们有定理 Theorem 1 (用连续函数刻画 Lebesgue 可测函数) 的另一种表述形式：
Theorem. 设为 Lebesgue 可测集，为上的 Lebesgue 可测函数，则对于任意，必有上的连续函数，使得
如果 (或 ) 则上述可以同样 (或 ).
Remark. 可以被选为具有紧 [1]的支撑的函数.
一个推论是:
Corollary. 设为上几乎处处有限的 Lebesgue 可测函数，则存在上的连续函数列使得在上

连续函数复合上 Lebesgue 可测函数得到的依然可测，这由 Lebesgue 可测性的“任意开集逆像为可测集”定义可得. 但不一定可测.
Exercise (逼近·几乎处处有限的可测函数). 设为测度空间，为上的几乎处处有限的可测函数，. 证明：对于任意，存在上的有界可测函数，使得

Proof.
因为为上的几乎处处有限的可测函数，所以

都是可测集，且，关于单调递减，. 由于，所以有测度的下连续性

因此对于任意，存在，使得当时，, 令

即可.
□
Remark. 也可以反证.
Proof.
假设存在，使得对于任意上的有界可测函数有

记，取

简单验证可知有界可测. 于是

这与几乎处处有界矛盾！
□
Definition (Baire 函数 (Borel 可测函数)).
Exercise (Baire 函数很接近有限的 Lebesgue 可测函数). 设是上有限的 Lebesgue 可测函数，则一定存在全直线上的 Borel 可测函数使得

Proof.

□

Theorem (Levi 定理). 设为测度空间，为上的非负广义可测递增函数列，且 . 则

Exercise. 为 Lebesgue 可测集上的几乎处处大于零的 Lebesgue 可测函数，且满足
证明：.
Proof.
考虑反证，我们知道

由题意，若则存在使得，那么

矛盾！
□
Exercise. 设为上非负 Lebesgue 可积函数
证明：.
Proof.
记 , 则，于是

□
Exercise. 设为上的 Lebesgue 可测函数，．证明：
为上的 Lebesgue 可积函数．如果，举例说明充分性不成立．
Proof.
令，则

由此推得
在上 Lebesgue 可积．
举出反例：. 于是但是 .
□
Exercise. 设
证明：.
Proof.
设 . 则当时，

于是

于是 .
□

Theorem (勒贝格微分定理). 设 , 则对于几乎处处的 , 我们有
Definition (Lebesgue 点). 设 . 对中的 , 如果那么称为的 Lebesgue 点.
若，那么几乎处处为的 Lebesgue 点.

在中，由海涅定理，等价于有界闭集.↩︎

可测函数列的收敛

Fri, 02 May 2025 16:55:29 GMT

See Royden.
Theorem (Egoroff's Theorem). Assume has finite measure. Let be a sequence of measurable functions on that converges pointwise on to the real-valued function . Then for each , there is a closed set contained in for which
Proof.
Lemma (Lemma 10). Under the assumptions of Egoroff's Theorem, for each and , there is a measurable subset of and an index for which
Proof.
For each , the function is properly defined, since is real-valued, and it is measurable, so that the set is measurable. Therefore

is ascending. And , since converges pointwise to on . We infer from the continuty of measure that

Since , we may choose an index for which . Define , then we are done.
□
In the preceding lemma, let and , i.e.

Define , then

For any , pick such that , then

There exists closed s.t. . Therefore,

Thus converges to uniformly on .
□
Theorem (Lusin's Theorem). Let be a real-valued measurable function on . Then for each , there is a continuous function on and a closed set contained in for which

细节

Fri, 02 May 2025 16:55:29 GMT

Theorem (Theorem in Rudin). If and if is a Cauchy sequence in with limit , then has a subsequence which converges pointwise almost everywhere to .
This occurs in many proofs, e.g.
Theorem (the inversion theorem (Rudin )). If and and if
then and a.e.
, then

The integrands on the right side are bounded by , and since as , the right side converges to for every , by the dominated convergence theorem.
If we combine theorem and , we see that there is a sequence such that and

Hence a.e. That follows from theorem .

周民强

Fri, 02 May 2025 08:58:52 GMT

参见周民强

Cardinality

Fri, 02 May 2025 16:55:29 GMT

Theorem (theorem 2.12 in baby rudin). Let be a sequence of countable sets, and put
Then is countable.

Theorem 1 (theorem 2.12 in baby rudin) can be written as

Theorem (theorem 2.13 in baby rudin). Let be a countable set, and let be the set of all -tuples , where () and the elements need not be distinct. Then is countable.
Theorem 2 (theorem 2.13 in baby rudin) is proved by induction from Theorem 1 (theorem 2.12 in baby rudin), which means

But

In particular, we have
Theorem (theorem 2.14 in baby rudin). Let be the set of all sequences whose elements are the digits and . This set is uncountable.
The elements of are sequences like
The proof is very classic to show that is uncountable.
Let be a countable subset of , and let consist of the sequences . We construct a sequence as follows. If the th digit in is , we let the th digit of be , and vice versa. Then the sequence differs from every member of in at least one place; hence . But clearly , so that is a proper subset of .
We have shown that every countable subset of is a proper subset of . It follows that is uncountable (for otherwise would be a proper subset of , which is absurd).

It should be noted that the mapping

is surjective but not injective. We denote the sets by

It's easy to check that are countable. Then

is injective, thus bijective. Therefore

Construction-of-measure

Fri, 02 May 2025 16:55:29 GMT

is a countable collection of nonempty open, bounded intervals that covers . Define the outer measure of to be

The proof of countably subadditivity is very classic, i.e. check that

By the definition of outer measure, for any given , there is a collection for each , such that

Then is a open cover of , thus by the definition of ,

A set is said to be measurable provided for any set ,

Clearly, , it suffices to check that

Clearly any set of outer measure 0 is measurable, since , .
The proof of measurablity preserving under countable union is classic. Let be a countable union of measurable sets . WLOG, assume that are disjoint, then the proof is routine.
The collection of measurable sets is a sigma-alegbra containing all the Borel sets, i.e. each and sets.

If is a measurable set of finite outer measure contained in , then

If is a measurable set then its Lebesgue measure is defined by

i.e.

It's natural to be interested about the sets that is not measurable.
Theorem (Vitali). Any set of real numbers with positive outer measure contains a subset that fails to be measurable.

For example, is -finite under the Lebesgue measure on .

The Lebesgue measure restricted to Borel sets on is not complete, since a Borel sets with measure zero contains a subsets that is not Borel, see royden page 52.

See evans 偏微分方程笔记 Chap 3 Appendix Measures.

We now define the general concept of an outer measure and of measurability of a set with respect to an outer measure, and show that the Carathéodory strategy for the construction of Lebesgue measure on the real line is feasible in general.

The union of a countable collection of measurable sets is measurable.

The definition of outer measure before in this section is by properties, while the following is by construction. We can construct by a set function (not measure) .

We show that a semiring has the property that every premeasure on has a unique extension to a premeasure on a collection of sets that is closed with respect to the formation of relative complements.

hw1

Fri, 02 May 2025 16:55:29 GMT

(1)

(2)

(3) take completion of both side of (1)

(4) take completion of both side of (2)

(5) for any , means that for any , there exists a such that . In the set language,

(6) for any , means that for any , there exists a such that . In the set language,

(7) combining with the result in (2) and (5) yields

(8) similar to (7)

(9)

(10) similar to (9)

(11)

(12) Since : , combining with the results of yields the desired conclusion.

Find an error of the question above.
Errata:
If for , , show that for each fixed , exists and

Fix then exists iff and exists and equals.

Since exists, thus . Hence exists and equals to . Obviously, .

Then

If then for . Therefore .
For

Then

Claim that

For

Then

Claim that

Denote that

For , we have . Since is dense in , there exists for some , thus . Therefore .
For , we have for some then , i.e. . Therefore . Hence .

For any , denote that

Since on pointwisely and ,we know that for any fixed , we have i.e. there exists such that for all .

Therefore exists and equals to .

Since for any , . Therefore .
If , then there exists such that . By the definition of , we have for some thus . Therefore . Hence .
For , there exists such that . For any , then . Therefore , i.e. .

(4) 对任意有 .
(1) means is one-one. Since is defined to be onto, then (1) means if for then . In other words, if then .
(1)=>(2): for any , we have and then and thus . Therefore .
It suffices to show that . We argue by contradiction. If there is an element not containd in , then there exists and such that , . Since is one-one, we have . Therefore , , which is a contradiction. Hence .
(2)=>(3): Trivial.
(3)=>(4): For fixed and , , subtract from . Since , then we have . But , so i.e. .
(4)=>(1): Pick , , and let , then means , which implies i.e. .

原题有误，应改为 .
does not always hold. A conterexample is as follow.
Let be groups, is a subgroup of and is a subgroup of . The relation on groups is defined to be group isomorphism. Then we have and . Since is set minus, or is not a group! (without the identity) Therefore .

hw2

Fri, 02 May 2025 16:55:29 GMT

Exercise. 若是无穷集，是可数集，则．
只需要建立一个一一映射，用来说明和的元素个数一样多。
Theorem.
不妨设 . 由于可数，不妨设为，由于是无穷集，根据 Theorem 2 我们知道，其中，是一个可数集。那么我们构造映射如下

显然，. 因此是一一映射，故 .
Exercise. 证明：可列集的所有有限子集构成的集合是可列集．
不妨设，构造一个嵌入映射

其中有限，显然是一个单射，因此是一个一一映射，由于，所以至多可数，故至多可数。由于，所以不可能有限，因此可数。
Exercise. 所谓代数数，是指满足某个整系数多项式方程的复数．显然所有的有理数都是代数数，但代数数也可能是无理数，甚至虚数，例如和都是代数数。证明：全体代数数构成的集合是可列集．
设全体代数数构成的集合为，全体次整系数多项式构成的集合为，全体满足次整系数多项式方程的代数数构成的集合为，我们构造如下的映射

显然，由于可数集的可数并也是可数集，故只需要证明都是至多可数集即可。由代数数的定义可知，是一个满射，由于可数，所以至多可数。得证！
Exercise. 设．证明：
（1）若存在常数，使得中任意有限个不同的数之和都不超过，则是可数集；
（2）若是无穷集，且中任意选取可列个不同的数所组成的正项级数都收敛，则是可列集．
(1) 显然，不妨设，否则用代替。由于中任意有限个不同的数之和不超过，那么，又因为

是一列至多可数集的可数并，故至多可数。
(2) 考虑

由 (1) 可知：存在常数使得中任意有限个不同的数之和都不超过，于是是至多可数集。由于至多可数集的可数并可数。所以可数。
Exercise. 证明：上的凸函数的不可导的点只有可数个．
根据凸函数的定义，对于任意，对于任意，我们有

任意给定，考虑函数

显然在和上分别单调递减，显然在上单调递减且有下界，于是存在，由于在上单调递减且有上界，于是存在。定义集合

由于的凸性，我们知道。因此，对于，考虑区间，由于在中稠密，，任取的一个代表元，我们得到如下的嵌入映射

显然这是一个单射，那么，于是至多可数。
Exercise. 证明：上的函数的局部极小值只有可数个．
考虑集合

由局部极小值的定义可知，是的全部局部极小值构成的集合，显然 . 只需证明集合至多可数即可。
对于任意的，将中离最近的极小值点记作，将中离最近的极小值点记作 . 由于是一个局部极小值点，所以存在的一个充分小的邻域，使得。让取遍中的所有元素，这样得到的所有两两无交。对于任意给定的，由于在中稠密，可以任取一个作为代表元，构造如下嵌入映射

这显然是一个单射，于是，于是至多可数。
Exercise. 是否存在集合，使得是可列集？
表示的幂集，这是一个代数，若有限，那么显然有限。若无限，假设是的一个可数子集，不是一般性，我们不妨假设中集合两两无交，否则令

由于表示的幂集，必然有。断言不可数，我们现在考虑的任意一列子集，构造一个新的子集：若，则，否则。于是，但 . 这意味着的任意一个可数子集都是真子集，若可数，那么便是它自身的真子集，这与真子集的定义矛盾！故不可数，因此包含一个不可数集，故也不可数。
综上，不存在集合使得可列。
Exercise. 证明：上的函数的第一类间断点只有可数个．
记

固定，记是左侧最近的第一类间断点，是右侧最近的第一类间断点。，根据有理数的稠密性，任取，构造如下映射

这显然是一个单射，于是至多可数。
Exercise. 证明：上的函数的不可导但单侧可导的点只有可数个．
记

固定，记是左侧最近的不可导但单侧可导的点，是右侧最近的不可导但单侧可导的点。，根据有理数的稠密性，任取，构造如下映射

这显然是一个单射，于是至多可数。
Exercise. 教材第一章习题第 19 题．
19．设且为可数集，试构造的一个分解，使得平行于轴的直线与的交点为可数个，平行于轴的直线与的交点为可数个．
这是显然的！

hw3

Fri, 02 May 2025 16:55:29 GMT

Exercise. 1．判断下列集合的基数是，还是，并说明理由：
（1）所有维开矩体＊构成的集合；
（2）所有以有理点（即坐标皆为有理数的点）为顶点的维开矩体构成的集合；
(1)

(2)

Exercise. 2．是否存在集合族，使得对任意集合，都存在，使得？
I don't understand.
Exercise. 3．设且．证明：中存在经过和两点且不含有理点的圆周．
Proof.
假设任意经过两点的圆周都包含有理点，考虑的中垂线，这是一条直线，上面全部点的势为 . 取任意为圆心，作过的圆，可以找到一个不同于的点，我们有如下自然映射

这显然是个单射，因为任意两个不完全重合的圆至多有两个交点，而刚才取的圆都交于，故 . 因此

矛盾！
□
Exercise. 4．设是可数集．证明：存在，使得．
Proof.
Assume, for contradiction, that for any , there exists such that

That is

Construct the following map

which is 1-1, sisnce . Thus

which is a contradiction.
□
Exercise. 5．教材第一章习题第 42 题（1）（2）．

Proof.
(1)
If then which means that is a limit point of . By definition,

Errata: .

(2)
For any , check that

Denote that

Then s.t.

Pick , then

Thus

Since is arbitrary, we have

WLOG, assume that is bounded (otherwise we replace by )
Pick in each for . Since is closed and bounded, thus compact, thus sequentially compact. Then for some . Obviously, . We have

That is

Hence

□
Exercise. 6．教材第一章习题第 43 题．

Since open set remains open under any union. That is, for any set , is open.
Now we consider .
If is open, pick any , then for some . There exists for some , then . Thus is an interior point of . Hence is open.
If is closed. Check that is open.
Pick , then . Pick in each , then for some , thus . Let , then . is an interior point of . Hence is open.
If then is neither open nor closed.
Exercise. 7．若都是的闭集，是否一定存在，使得？予以证明或举出反例．
不妨设为有界闭集，则为紧集，故列紧. 由的定义可知，存在使得，其中 .
Exercise. 8．试构造下列各函数（列）．
（1）设是中互不相交的非空闭集．试作上的连续函数，使得
（i）；
（ii）；
（iii）．
（2）设，且满足．试作上的连续函数，使得
（i）；
（ii）．
（3）设是的闭集．试作上的连续函数序列，使得，．
（4）设是的开集．试作上的连续函数序列，使得， .
(1)

(2)

(3)

(4)

Exercise. 9．教材第一章习题第 44 题．

Note that all the sets must lies in a metric space. And the property satisfy is called Normal separation property.
Let

Then .
Exercise. 10．教材第一章习题第 45 题．

证不出来
Exercise. 11．教材第一章习题第 46 题．

必要性显然，考虑充分性，假设不是闭集，则存在一个包含在中的极限点，于是存在使得

于是，矛盾！
Exercise. 12．教材第一章习题第 48 题．

下面证明是开集. 任意给定，记

对于任意给定的，存在，使得

那么对于任意的，有

因此

故 . 得证！
再用表示的连续点集，显然 . 另一方面，对于任意，

也就是说，对于任意，存在，使得

这就是在处连续的定义.

hw4

Fri, 02 May 2025 16:55:29 GMT

(1) yes, yes.
(2) , s.t.. Since is dense in , then . Construct

is clearly one-one, then

Thus is at most countable.

(1) .
(2) .
(3) pick then , , .
(4) , .

Claim that .
First to show that . Let

Denote

. Therefore . Pick in . It must lie in for some . Then pick the sequence , which lies in , with limit point which is the nearest point in . So is for the sequence . Therefore .
If , then there is a point in with a sequence in convergent to , but . Since each is midpoint of then can be any closed to the points in , as . Hence which is a contradiction.

Clearly: and .
, for . Suppose that and . Then

By induction, we have

Counterexample:

Pick , then is constant on each , thus continuous. But is not continuous on .

Denote , . Construct a sequence

Since , it suffices to chech that for each . Clearly, . When ,

Since , then covering such that

. WLOG are positive or negative. Let

Then , must lie in for some . Then lies in , which means that covers , thus

Thus

Denote that

Then is continuous thus uniformly continuous on . For any given , let

Then is an open cover of . Since is closed and bounded, thus compact. Then there exists such that

Then

Then

Since is arbitrary, . Therefore

It suffices to check that . By the definition of outer measure, since , then covering , s.t.

Pick covering s.t.

Put , then is a cover of , then

Since is arbitrary, thus

Therefore

Since , then

Since , then

Since then

Thus

NTS: . , check that .
NTS: .
Fix , NTS: .

Therefore .

Hence

hw5

Fri, 02 May 2025 16:55:29 GMT

(1) false. Let , where denotes the cantor set.
(2) false. Let .
(3) false. Let .
(4) true. By the definition of outer measure,

Then for any , there exists a collection of nonempty open, bounded intervals covering , with

In the general definition of outer measure defined on a sigma-algebra and by a set function, can be chosen to be closed. But here we construct a closed set from each . Denote by , let

Then

Thus is a collection of open bounded nonempty intervals that covers . Therefore

Since is arbitrary, we have

Since , ,

(5) false. Let , then is countable thus has measure zero, but has measure .

Pick , then such that . Therefore

Since

Then for any set ,

Therefore

Thus .

Errata: .
For any set ,

Therefore

Hence .

If , then the equality is trivial. If , since , then

Hence

Errata: .
Since , then

Then

For any set ,

When exists, we have , then

Thus , and

(1)
Claim that:

.
.
Firstly, is trivial. , since is bijection, s.t. , s.t.. But the preimage of is unique, thus , thus . Therefore .
Secondly, , for some , then , then , . , . Since is bijection, s.t.. Clearly , thus , . Therefore .
Then for any set , its preimage is , i.e.. Then

Thus is measurable.
(2) Since the outer measure is invariant under translation, the measure is also invariant under translation.

hw6

Fri, 02 May 2025 16:55:29 GMT

Exercise. 设为可测集，若，试证；若，试证．
Proof.
反证而设, ，则存在，并非的聚点，故存在区间使得 . 故，矛盾！
若，则对于，存在的邻域（某个区间），使得，于是矛盾！
□
Exercise.
Proof.
(1)

又因为

故

(2)

(3)

要使得只需要 .
要使得只需要 .
□
Exercise.
开集和闭集的边界可能有正测度！
Proof.

若可测，由于，所以 .
若可测，不会证明！
□
Exercise.
Proof.
不一定. 当，考虑，显然，故可测，，对于任意包含的开集，对于任意，的某个充分小小邻域都在内，由于在内稠密，故对于充分小的，，从而

于是

故

□
Exercise.
Proof.
(1) 只需要验证对于任意集合，满足

我们知道 . 于是

又因为

故

由任意性，. 因此可测.
(2) 只需要验证对于任意集合，满足

因为

所以

由任意性，

故可测.
(3) 充分性由 (2) 显然，下面证明必要性.
假设可测，则对于任意给定的，可作开集且 , 从而

此外

取充分大使得

于是

得证！
□
Exercise.
Proof.
Lemma.
Proof.
若，根据可测的定义，存在一列开区间 , 使得 (开集的任意并还是开集)，且

于是

于是

若，定义 (有限测度)，则对于任意给定的，存在开集使得

那么定义 (仍然是开集)，则

故得证！
接下来对于集合，考虑，对于任意给定的，存在开集，使得 . 现在令，由于，所以 . 故得证！
□
(1)
若有限，只需要证明：若可测，那么对于任意给定的，存在有界闭集，使得

也就是

对于任意给定的，存在，使得 . 由 Lemma 7 可知，存在闭集，使得

由于

于是

若，那么定义，于是

由于，那么根据 Lemma 7 ，存在，使得，于是

(2) 由 Lemma 7 可知，对于任意给定的存在包含的开集，使得 .
与此同时，，从而存在有界闭集 ,使得

从而，且 . 由 Exercise 5 (1) 可知可测.
□
Exercise.
Proof.
由 Lemma 7 可知，对于任意，存在集合使得

记，那么是集，且

令，那么，且 . 是的等测包，存在！
□
Exercise.
Proof.
Lemma (From Royden).
(1)
确实可能严格大于零，因为不一定可测，而等价于可测. 该等价性的证明如下：

其中可测，由于零外测集一定可测，所以，由于，所以 . 反之，若有可测，则根据 Lemma 7 可知，存在，，使得 .
接下来考虑的任意可测子集，有，由于，那么 . 由于，所以 . 由于，所以，由于，于是

(2) 作的等测包，于是，故，于是 . 故是的等测包.
(3) 对于的等测包，考虑的任一可测子集，那么

故，由 Exercise 9 (2) 可知，.
□
Exercise.
Proof.
(1) 对每个作等测包满足

从而

(2) 由于 , 所以 . 从而

因此存在. 由 (1) 可知：

同时，因为，所以

因此

故

□
Exercise.
Proof.
设分别是的等测包，那么

由于

从而

由于，所以

因此

从而

□
Exercise.

Exercise.

hw7

Fri, 02 May 2025 16:55:29 GMT

Exercise.
Proof.
for some , let

For , we have

Thus . We have . Applying the intermediate value theorem to , for any , there exists s.t. , i.e. . Let then we are done.
□
Exercise.
Proof.
Since and are measurable functions over , we have for any . Then

Similarly,

Then

□
Exercise.
Proof.
(1) Since is open in , we have disjoint s.t.

(2) For open or closed in , we have

Then

Therefore for any Borel set ,

□
Exercise.
Proof.
.

If is countable, then for any , is countable, thus measurable. Hence is measurable.
□
Exercise.
Proof.

Since is measurable, then . For , . Since , then

Then

Therefore for any , we have

Hence is measurable.
□
Exercise.
Proof.
There is a set contained in for which . is measurable on (it's trivial). For any ,

Thus ,

□
Exercise.
Proof.
Suppose that is measureable, then for any . Conversely, for any then for any , there is a rational sequence with . Then

Therefore is measurable on .
□
Exercise.
Proof.
当指标集为不可数集，的上确界函数不一定可测，反例可取为不可数集，对，作函数

则并不可测.
如果加上了连续的条件，那么对于任意的 ,

由于，故是开集，开集的任意并也是开集，故是开集，于是是集，故可测. 于是可测.
□
Exercise.
Proof.
(1)
Remark (Royden Chap 3 Ex 8).
3.8_哔哩哔哩_bilibili
Lemma. If is Lebesgue measurable, and is a Borel set, then .
Proof.
For any open interval , we have . Given , open, then . In open interval

Let . Every open set in is contained in . In particular, . is closed under completion and coutable union. (left as an exercise) Hence the conclusion follows.
□
To prove that is Lebesgue measurable function, it suffices to check that for any open set . Note that

Since is Borel function and is Borel set, . By Lemma 11, , which yields the conclusion.
(2) For any open set ,

Hence the conclusion follows.
□
Exercise.
Proof.
Let

Then

Since , is measurable on . Thus is measurable.
For any ,

Lemma. 设，若存在中可测集 , 使得，则 , 且有 .
Proof.
Trivial.
□
Put and , then

By Lemma 13, we have . Hence the conclusion follows.
□
Exercise.

Proof.
(1)

That is, for any , there exists s.t.

Let and , then , and .
(2) No. Consider and

If , then . Thus , .
□

hw8

Fri, 02 May 2025 16:55:29 GMT

Exercise.
Proof.
Denote , . Then for any open set

□
Exercise.

(1)
Theorem (Egoroff's Theorem). Assume has finite measure. Let be a sequence of measurable functions on that converges pointwise on to the real-valued function . Then for each , there is a closed set contained in for which
By Egoroff's theorem, for each , there exists closed s.t.

Thus

Let then

(2) The reasoning is wrong.

Exercise.
Let , then . Apply Egoroff's theorem to , for any , there exists s.t.

There exists s.t.

Since is finite a.e. on , there exists s.t.

And

There exists s.t.

Therefore

where . Define

Then

Then we are done.
Exercise.
意味着几乎处处收敛，但依测度收敛并不蕴含几乎处处收敛.

Exercise.
由 Riesz 定理，存在的子列，使得在上几乎处处收敛于 . 也就是说

也就是对于任意给定的，存在 , 使得对于任意，有

由于 , a.e. 于，我们有

从而对于任意给定的，

从而

于是对于任意，我们有

因此

Exercise.
Proof.
By Exercise 8, iff , a.e., that is, a.e.
Suppose that , then a.e., i.e. for any , there exists s.t.

Then for any , we have

Thus a.e.
Suppose that a.e. For any given , we have to find its subsequence a.e. There exists with measure 0, s.t. on . There exists s.t. on , for any . Thus

Thus as on , i.e. a.e. Then we are done!
□
Exercise.
Proof.
若，则，由 Riesz 定理，存在 a.e.
若 , a.e. 假设，也就是存在，使得

于是存在子列 , 其中 . 又由于 a.e. 那么，但是

矛盾！
□
Exercise.

(1)
If , then for any , by Riesz's theorem, a.e. That is, with measure 0, s.t. on . Since is continuous, we have on . Thus a.e. By Exercise 8 we are done.
If is uniformly continuous on , then , , s.t.

which means implies and implies . Since , we also have , as . Then

If is uniformly bounded on , i.e. , s.t. , . We restrict to , then , and is uniformly continuous on the closed interval . Apply (2), we have . For any ,

(2)
(3)
Since then , as . Due to the inequality,

We have the inclusion of sets.

Thus

Let , then . Hence .
By Riesz's theorem, there exists a subsequence a.e. Since a.e. on , we have

Then

Therefore

Hence a.e. on .
Exercise.

并不一定.
Exercise.
(1) 若有一者为零，则显然. 若非零，则

(2)
利用 Riesz 定理，我们有对于任意，存在 a.e.，存在 a.e. 从而 a.e. 再利用 Exercise 8 可知 .
(3)
时存在反例.

时存在反例，但时结论成立，证明如下

(4)
利用 Riesz 定理，我们有对于任意，存在 a.e.，存在 a.e. 从而 a.e. 再利用 Exercise 8 可知 .
(5)

断言：，这是因为

再结合 (1) 可知，, .

hw9

Tue, 06 May 2025 07:17:17 GMT

Exercise.
Proof.
If , then given , for any , there exists , s.t.

Then

Thus

Conversely, suppose that

i.e. for any , there exists , s.t.

Then there exists s.t.

Therefore,

Given , for any , we have

Therefore

□
Exercise.
Proof.
Obviously, is finte a.e. Since , then for given , for any , there exists s.t.

For any , we have

Then for any ,

Thus is Cauchy in measure.
□
Exercise.
结论：这两种说法既不等价，也不存在蕴含关系。
Proof.

证明
（即：几乎处处连续不能推出几乎处处等于某个连续函数）
反例：考虑函数定义在上：
分析（1）：函数在上是连续的，因为它是由连续函数复合而成的。它仅在处不连续，因为当时，在 -1 和 1 之间振荡，极限不存在。因此，间断点集合。点集的勒贝格测度为零，所以。故函数满足说法（1）。
分析（2）：假设存在一个在上连续的函数，使得几乎处处成立。这意味着集合的测度为零，。因为和都在上几乎处处相等，并且它们都在上连续 ( 在整个连续，因此在连续；在连续），那么它们必须在整个上都相等。（如果存在使得，由连续性，必然存在的一个邻域使得，这个区间的测度为，与几乎处处矛盾）。因此，必然有对所有成立。但是，函数在时极限不存在，所以不可能将它在处定义某个值使得在处连续。这与是上的连续函数的前提相矛盾。所以，不存在这样的连续函数。故函数不满足说法（2）。
结论：（1）不能推出（2）。
证明
（即：几乎处处等于某个连续函数不能推出几乎处处连续）
反例：考虑狄利克雷函数（Dirichlet function）定义在上：
分析（2）：令对所有。显然是上的连续函数。集合。有理数集合是可数集，其勒贝格测度为零，即。因此，与连续函数几乎处处相等。故函数满足说法（2）。
分析（1）：狄利克雷函数在上的每一点都不连续。对于任意，其任意小的邻域内都既包含有理数（值为 1 ）也包含无理数（值为 0 ）。因此不存在。所以，间断点集合。该集合的勒贝格测度。故函数不满足说法（1）。
结论：（2）不能推出（1）。
□
Exercise.
Proof.
There exists with measure 0, s.t. is continuous thus measurable on . is clearly measurable. For any open set ,

Since and , then

Thus we are done!
□
Exercise.
Proof.
Let

List these functions

Then on , because for any ,

Thus

But disconverges everywhere on , since for given , then for any , lies in for some large enough (), there exists , s.t.

Thus does not converges at , by Cauchy's criterion.
We are done!
□
Exercise.
Proof.
For any given , there exists closed with , s.t. is continuous on . Then for any open set , we have

Note that

For any , we have

Note that

Then

Since is arbitrary, we have

for any open , thus we are done!
□
Exercise.

错误在于并不在上连续，反例考虑

令，显然在上连续，但

，并不连续.
接下来，我们讨论 Lusin 定理的结论是否能改为：存在的可测子集，使得且在上连续。并给出反例。
Lusin 定理原文保证存在闭集使得且连续。修改后的版本要求：

是可测集 (比闭集更弱的要求)。
(比更强的要求)。
在上连续。
这个修改后的命题不成立。我们可以举出反例。
Example. 令。构造一个 "Fat Cantor set" (胖康托集) 。这种集合具有以下性质：
是闭集 (因此是可测集)。
是无处稠密的 (不包含任何区间)。
的勒贝格测度。 (例如，在康托三分集的构造中，每次去掉中间长度的开区间，改为去掉长度为的开区间，其中足够小，使得，则剩余集合的测度为）。现在定义函数，即由于是可测集，是上的可测函数。且。
证明不满足修改后的命题：
假设存在一个可测集使得且在上连续。

由于，这意味着。
因为且，所以必须与相交，且交集的测度。(因为，所以。又因为。故）。
同样地，因为且 (假设 )，所以必须与相交，且。
取一个点。因为在上连续，所以在点连续。这意味着：由于，我们有。所以必须有。
然而，是无处稠密的。这意味着在点的任意小邻域内都存在不属于的点。也就是说，存在序列使得。
由于，集合不能包含一个测度为的集合的 "大部分"。更确切地说，对于，由于无处稠密，是的极限点。因为，所以不可能是空集，并且它必须在附近积累。因此，我们可以找到一个序列，其中。
对于这样的序列：

，所以。

根据在处连续的要求，我们必须有。但是，而。这导致，产生矛盾。
这个矛盾说明，我们最初的假设（存在满足条件的可测集）是错误的。
结论：对于（其中是胖康托集），不存在可测子集使得且在上连续。这表明 Lusin 定理的结论不能简单地将“闭集 , ”替换为“可测集 , ”。关键在于 Lusin 定理保证的集合必须是闭集，这提供了更强的拓扑性质，使得限制在其上的函数得以连续。
Exercise.
Definition (近一致收敛). 任给，存在的可测子集，使在上一致收敛于，而 .
(1)
只需要是可测函数列.
反证而设 a.e. 则存在集合，于，且 . 由于近一致收敛于，故存在集合，使得

且 . 断言，这是因为若，那么，故

矛盾. 故存在，这意味着

矛盾！故近一致收敛蕴含几乎处处收敛.
Remark. 注意：当时，几乎处处收敛并不蕴含依测度收敛.
反证而设 , 则存在，使得，也就是存在，使得

由于近一致收敛于，故存在集合，使得

且 . 也就是存在，使得

我们任取，固定，由于 , ，那么 , 于是存在，故

矛盾！故近一致收敛蕴含依测度收敛.
Exercise.
判断：不一定存在这样的连续函数。
Proof.
(举反例) 考虑的情况。令可测集。
令为一个“胖”康托集（generalized Cantor set）。这样的集合具有以下性质：

是闭集。
是无处稠密集（nowhere dense），即的闭包（就是本身）不包含任何开区间。因此其内部为空。
的勒贝格测度。
定义上的可测函数为，即的特征函数。

这个函数在上显然是可测的（因为是闭集，故为可测集），并且几乎处处有限（实际上处处有限，其值仅为0或1）。
现在，我们假设存在一个在上连续的函数，使得与在上几乎处处相等。这意味着集合的勒贝格测度为0。
即。
令。由于是闭集且，所以是中的一个开集（它是一些互不相交的开区间的并集）。
对于，我们有。
由于在上几乎处处相等，所以对于几乎所有成立。
设。则，所以。
因此，在上成立，其中。
由于是连续函数，且是开集，若在的一个具有全测度的子集上为0，则必须在整个上为0。
（详细说明：假设存在一点使得。由于连续，存在一个以为中心的小开区间使得对所有成立。但这意味着，这与对几乎所有矛盾。因此，对所有成立。）
因为是中的无处稠密集，所以在中是稠密的。这意味着 ( 在中的闭包) 包含。更准确地说， (因为不包含任何区间，且 )。
对于任何一点，由于是无处稠密集，是的一个聚点。因此，存在一个序列，其中，使得。
由于是连续函数，且对于所有，我们有：

这表明对于所有成立。
结合对所有和对所有，我们得到对所有成立。
由于在上是连续的，所以对所有成立。
也就是说，对所有成立。
现在我们比较和在上的差异：

对于所有。
那么，当且仅当，这意味着当且仅当。
所以，集合。
为了使得和在上几乎处处相等，这个集合的测度必须为0，即。
但这与我们选择为一个具有正测度（）的胖康托集相矛盾。
因此，我们的初始假设（即存在这样的连续函数）是错误的。
所以，不一定存在上的连续函数与在上几乎处处相等。
这个反例可以推广到 ()。例如，取。令为如上定义的胖康托集。令。则是中的闭集、无处稠密集，且其维勒贝格测度。定义。通过与上述类似的论证，可以证明不存在满足题设条件的连续函数。核心步骤是证明若存在，则必须在上为0，进而利用的连续性推广到在上为0，从而导致，产生矛盾。
□
Exercise.
Proof.
由于可测，则由 Lusin 引理，存在闭集和上的连续函数，使得

于是

这样构造出一列，从而

由的任意性可知

□

hw10

Mon, 12 May 2025 10:52:52 GMT

Exercise.
Proof.
Assume that . Since every point in belongs to at least of , we have

Then

By the linearity of integral,

which is a contradiction.
□
Exercise.
Proof.
(1)
, denote , then

Since ,

(2)

□
Exercise.
Proof.
Let , then

Since , we have ; and

It suffices to show

For any , , as is monotonically increasing, . Similarly, for any , . Thus

We are done!
□
Exercise.
It's similar to the proof of Egorov theorem.
Proof.
WLOG, assume that on , and on , for any ; thus on . Denote , then

is measurable, as are measurable. Construct sets

Note that given , we have for any , thus is descending for each ; then

Recall the continuity of measure for descending chain, we just need to show that[1]

If , then , which is Egorov theorem; if , but is controlled by , we can also prove .

The following steps are routine.
For any , there exists for each , s.t.

Let , then

Let , then , we verify that on .
For any given , pick s.t. . For any , for any ; thus ; , i.e.

i.e. on . We are done!
□
Exercise.
Proof.

Then

□
Exercise.

Proof.
(1) false, consider , , . This is true when

(2) fasle, consider , , . This is true when

(3) true, if , then

(4) true, from (3) we know that a.e. and a.e. thus a.e.
□
Exercise.
Proof.

Then

By Beppo Levi's theorem,

i.e.

Therefore,

□
Exercise.
Proof.

□

we haven't use till now↩︎

hw11

Tue, 20 May 2025 15:43:28 GMT

Exercise.
Proof.
When , ,

□
Exercise.
Proof.
(1)

Let , then

(2)
Claim that

Case 1:
If , then by the definition of the indicator function, .
By the definition of the limit inferior of sets, this means there exists a natural number such that for all
If for all , then for all
Now consider :

For , we have Since for all , the infimum of these values is 1.
So,
Then, must be at least 1. Since values are at most 1, their infima and suprema are also at most 1.
Therefore,
Thus, if , then
Case 2:
If , then
By the definition of the limit inferior of sets, this means that for every natural number , there exists some such that .(ie, is not in all from any index onwards; for any we can find a later that does not contain
If for any , there exists a such that , then for any , there exists a such that
This implies that for any (Because if the infimum were 1, all for would have to be 1, which contradicts that there's one equal to 0)
Now consider

Since for all , the supremum of a sequence of zeros is 0.
So,
Thus, if , then

By Fatou's lemma,

Similarly,

Hence,

(3)
As exists, by Equation (1),

□
Exercise.
Proof.
(1)
By Equation (1), if ,

which is a contradiction.
(2)
If ,

Then

which is a contradiction.
Exercise.
(1)

(2)

Let , then

Thus

□
Exercise.
Proof.
By the definition of , there exists s.t.

, then , thus there exists s.t. a.e. by Fatou's lemma,

□
Exercise.
Proof.
By Exercise 5 ,

As a.e. on ,

Then

On the other hand, as a.e., we have

Thus

exists.

□
Exercise.
Proof.
(1)

Let then

On the other hand, since ,

Let , then

Hence,

(2)

As is integrable, is finite a.e.

Therefore

(3)

□
Exercise.
Proof.
For , we have , thus . We need to show that

By Levi's theorem,

If , then or . If , then

Thus

If , then

Thus

If , then for , for any , is integrable on . By Lebesgue dominant theorem,

Then

□
Exercise.
Proof.

Hence .
□
Exercise.

Proof.
(1)
The integration of is defined as

It makes sense iff the integration of and make sense. WLOG, we suppose that is nonnegative. For , on iff on . Therefore, by the linearity of the integral of bounded functions of finite support, .
If , then . For , , then . By the definition of integral of negative function, .
(2)(3)

□

hw12

Wed, 21 May 2025 10:00:16 GMT

Exercise.

(1)
If on , , then

Contradiction!
(2)
Apply (1). a.e. on , a.e. on ; thus a.e. on .
(3)
Put in (2), then a.e. on .
Exercise.
iff . As ,

, then .

By DCT,

Exercise.

(1) 对任意上可测的 , , 由于，则，故，故和不同时为，故有定义. 由于，故 , 有定义. 由于于 , 任意，故，对右边关于取下极限，得到 .
(2) 考虑 , , . 于是 , 有定义，其中 , . 因此不同时为 , 不同时为 , 故有意义. 由 (1) 可知 . 故
.
(3) 对于数列，显然有，根据 , ，结合 (1)(2) 可知

Exercise.
类似 Exercise 3 中的讨论，可知在上积分有意义，对于非负递增可测函数列，我们有 Levi 定理可知 . 由积分的线性性可知

Exercise.
由比值判别法，可积. 记几乎处处收敛于，则 , 故 . 令，则 , 由比值判别法，. 考虑非负可测函数列，它几乎处处收敛于 . 利用 Fatou 引理，, 于是

于是 . 再考虑非负可测函数列得到 . 因此 .
若，对于任意子列，存在子列 a.e. 于 . 从而 , . 对于数列，它的任意子列都含有子列，不妨考虑的上下极限序列，它们都含有子列收敛于，故 .
若 , 同样考虑的任意子列，再选取子列使得 a.e. 于 , 重复上述论述即可得证.
Exercise.
令，于是，利用 Exercise 5 立得 , 也就是 .
Exercise.
若，则 . 若 , 注意到，由 DCT

Exercise.
考虑反证，假设 , 则存在使得

记 . 于是 , 从而存在使得，故

令，就有 , 存在使得，故

矛盾！
Exercise.
, then

Then . Let , then

Thus a.e. on . As , by BCT,

Since is nonnegative measurable, is integrable.

Invariance of measure

Fri, 02 May 2025 08:58:52 GMT

See royden Chapter 22.

Lebesgue测度-可测集-测度-Borel集

Wed, 14 May 2025 14:39:29 GMT

参见《实变函数解题指南》·周民强.

Definition (Lebesgue outer measure). Let . If is a countable collection of open rectangles in , and then is called an -covering of (obviously there are many such coverings, and each -covering determines a non-negative generalized real value (can be , represents the volume of ), we call the Lebesgue outer measure of the point set .
Exercise. （5）设，则存在的子集，使得．
Proof.
Let . Then . Consider and . Without loss of generality, assume that . Then, by

we have , that is,

Similarly, for , we can also prove similar inequalities. In summary, we have

This indicates that . According to the intermediate value theorem for continuous functions, for , there must exist such that . Thus, taking yields the desired result.
□
Definition (可数覆盖空间). 设是一个拓扑空间. 如果存在一个可数集 , 使得的任何开覆盖都有一个子覆盖, 其元素个数不超过的元素个数, 那么被称为可数覆盖空间.
一个拓扑空间被称为可数覆盖空间, 如果存在一个可数子集 , 使得的任何开覆盖都有一个子覆盖, 其元素个数不超过的元素个数. 换句话说, 存在一个可数集合 , 使得对于的任何开覆盖, 都可以从中选择一个可数子覆盖.
由于具有可数基, 因此是第二可数空间. 根据定义, 第二可数空间是 Lindelöf 空间, 这意味着的每个开覆盖都有一个可数子覆盖. 因此, 是一个可数覆盖空间.
Exercise. （5）设．若对任意的，存在开球，使得，则．
Proof.
(5) By assumption, there exists a countable cover of by balls such that , and . It follows that

□
Proposition. 设，且，则
Definition (Limit superior and Limit inferior of sets). Let be a sequence of sets. The limit superior is defined by The limit inferior is defined by
Exercise. （2）设．若，则．（3）设定义在上的函数列满足则存在且，使得．
Proof.
（2）注意，且依题设知，对任给，存在，使得．从而对任意，有

由此知．证毕．
（3）令，则由题设知．因此当时，必存在，使得．从而有．
□

Definition (Lebesgue 可测集，Carathéodory 条件). 设．若对任意的点集，有则称为 Lebesgue 可测集（或－可测集），简称为可测集，其中称为试验集（这一定义可测集的等式也称为 Carathéodory 条件）；可测集的全体称为可测集类，简记为．
Proposition. 设是可测集列，则．
Exercise. 设 , 的充要条件是：对任给 , 存在可测集 : , 使得 .
Exercise. （3）设是中的可测集列，且满足，试证明对任意的，必存在，使得．
Proof.
（3）由题设知，对任意的，存在，使得

由此知．从而得到

故有．
□
Exercise. （5）设有中可测集列，且当时，．若存在，试证明：．
Proof.

□
Exercise. 构造零测的第二纲集.
Proof.
（2）令，以及

则点集在中无处稠密. 我们有

是第二纲集.
□
Exercise. （3）在中作点集
在十进位小数表示式中的所有都不出现 10 个数字中的某一个，
试证明是不可数集，且．
Proof.
Let be a set defined as follows:

where .
The set is constructed by iteratively removing intervals from . In the first step, the interval is divided into 10 equal subintervals, and the -th subinterval is removed. This ensures that the remaining points have a decimal expansion where the first digit is not equal to .
In the second step, each of the remaining 9 subintervals is divided into 10 equal subintervals, and the -th subinterval is removed from each. This process continues indefinitely.
The total length of the removed intervals can be calculated as the sum of a geometric series:

Since the total length of the intervals removed is 1, the measure of the remaining set is:

□
Exercise. （5）将中的点用十进位小数展开，令试问：（i）是闭集？（ii）是开集？（iii）是可数集？（iv）？（v）是可测集？？
Proof.
(5) (i) Suppose , and let . If , then or 7. Hence , i.e., is a closed set.
(ii) Noticing (i) and , we know that is not an open set.
(iii) .
(iv) Since is a closed set and , we know that is not dense in .
(v) is a measurable set (see ). .
□
Exercise. （1）设，且存在，使得对任一区间，都有开区间列：则．
Proof.
证明（1）因为，所以只需指出对任意的，有．由题设知，存在，使得．再对每个作覆盖，其总长度小于．依此程序继续作下去，可得（对任意）

由此易知．
□
Exercise (yau-21-analysis-problem-3).
For , , show that their convolution is a well-defined continuous function.
Let be a Lebesgue measurable set with Lebesgue measure . Prove that contains an open neighborhood of .
Proof.
(a) This is standard: In fact, we have . Therefore, by the continuity argument, it suffices to prove the theorem for . In this case, we have

Now let , the integrand converges to 0 uniformly. This yields (a).
(b) It suffices to consider the case where . We take , , thus is a continuous function. In particular, . Therefore, there exists an open set such that and for some . For , by definition,

Therefore, there must be some , such that . This implies . Hence .
□

Positive definite function

Tue, 13 May 2025 06:55:53 GMT

验证：对于正定函数

.
非负函数的 Fourier 变换函数是正定函数.
好的，我们来分别验证这两个属性。
Definition (正定函数 (Positive Definite Function)). 一个函数被称为正定函数（或半正定函数，因为通常定义为），如果对于任意有限个点和任意复数，都有：
其中是的复共轭。

验证正定函数的复共轭性：
设是一个正定函数。
(a) 首先证明是实数且。
在正定函数的定义中，取。则对于任意和任意：

取，得到。由于和式的结果必须是实数（因为它），并且是实数，所以必须是实数。
(b) 证明。
在正定函数的定义中，取。则对于任意和任意：

令，则。

因为是实数，且和都是实数（因为是实数），所以也必须是实数。即。

设，则。上式变为：

现在我们选择特定的 (从而选择特定的 )：

情况 1：取。则。令和。所以。这意味着虚部是奇函数。
情况 2：取。则。。除以：比较实部：这意味着实部是偶函数。
结合和，我们得到：

证毕。

验证函数的傅里叶变换是正定函数 (在的条件下)
设。其傅里叶变换定义为：

我们要验证是否为正定函数。即，对于任意有限个点和任意复数，验证：

将的定义代入：

由于且求和是有限的，我们可以交换求和与积分的顺序（Fubini定理）：

现在处理括号内的双重求和：

注意到。
所以括号内的和可以写作：

令。则上式变为。
所以，

这个平方项显然是非负的。
代回到积分表达式中：

为了使对于任意和成立，我们需要被积函数几乎处处成立。
由于总是成立的，因此，如果对于几乎所有的都成立，那么。
结论：
如果并且 (几乎处处)，那么它的傅里叶变换是一个正定函数。
重要注记：
如果不满足非负条件，那么不一定是正定函数。例如，如果 (几乎处处)，那么将是负定函数（即）。Bochner 定理更精确地指出，一个连续函数是正定函数当且仅当它是一个有限非负 Borel 测度的傅里叶变换。对于且的情况，就是这样一个测度。
所以，原命题“函数的傅里叶变换是正定函数”需要加上条件“该函数几乎处处非负”。

Topological preliminaries

Fri, 02 May 2025 16:55:29 GMT

Open sets remains to be open under arbitrary unions, while closed sets remains to be closed under arbitrary intersections. Imagine the special case of intervals to remember it.
The definition of compact is not technique to prove a set compact but a property.

Another definition of closure:

Definition (relatively open). Suppose , where is a metric space. Motivated by the idea that can also be a metric space, we say is open relative to if to each there is associated an such that whenever and .
A set may be open relative to without being an open subset of , e.g. .
Theorem. Suppose . A subset of is open relative to iff for some open subset of .
Theorem 2 can be another definition of relatively open.

Every neighborhood is an open set.
If is a limit point of a set , then every neighborhood of contains infinitely many points of .
A finite point set has no limit points.
Closed subsets of compact sets are compact.
Perfect set in is uncountable.
If is a collection of compact subsets of a metric space such that the intersection of every finite subcollection of is nonempty, then is nonempty.
If is a decreasing sequence of nonempty compact sets, then is not empty.

Theorem (Heine-Borel theorem). The compact subsets of a enclidean space are precisely those that are closed and bounded.
Moreover, the theorem is true for any locally compact Hausdorff space. Note that metric spaces are locally compact Hausdorff space.
Theorem (Closed subsets of compact sets are compact.). Suppoes is compact and is closed, in a topological space . If then is compact.
Proof.
If is an open cover of and then covers ; hence there is a finite collection such that

Then .
□
Corollary. If and if has compact closure, so does .
Theorem. hausdorff, , compact, and . Then there are open sets and such that , , and .
Proof.
If , the Hausdorff separation axiom implies the existence of disjoint open sets and such that and . Since is compact, there are points such that

Our requirements are then satisfied by the sets

□
Corollary. Compact subsets of Hausdorff spaces are closed. (the inverse is not true.)
Corollary. If closed, compact in a Hausdorff space, then is compact.

Use the definition of compact set.

Use the definition of compact set.

See Royden Chapter 11.

The Tychonoff Separation Property:
For each two points and in , there is a neighborhood of that does not contain and a neighborhood of that does not contain .
The Hausdorff Separation Property:
Each two points in can be separated by disjoint neighborhoods.
The Regular Separation Property:
The Tychonoff separation property holds and, moreover, each closed set and point not in the set can be separated by disjoint neighborhoods.
The Normal Separation Property:
The Tychonoff separation property holds and, moreover, each two disjoint closed sets can be separated by disjoint neighborhoods.

Proposition. A topological space is a Tychonoff space iff every set consisting of a single point is closed.
Proposition. Every metric space is normal.
Proposition. Let be Tychonoff. Then is normal iff whenever is a neighborhood of a closed subset of , there is another neighborhood of whose closure is contained in , that is, there is an open set for which

In a topological space that is not first countable, it is possible for a point to be a point of closure of a set and yet no sequence in the set converges to the point.

In view of the definition of the subspace topology, a subset of is compact provided every covering of by a collection of open subsets of has a finite subcover.
Definition (finite intersection property). A collection of sets is said to be finite intersection property provided every finite subcollection has nonempty intersection.

Additionally, homeomorphism requires continuous inverse, so hypothesis upon space is neccessary.

Regard a compact set as a compact topological space...

Definition (countably compact). A topological space is said to be countably compact provided every countable open cover has a finite subcover.

Definition (separate). Two nonempty open subsets of a topological spaces are said to separate if they are disjoint and their union is .
Definition (connected). A topological space which cannot be separated by such a pair is said to be connected.
Connectness is preserved under continuous mapping.

For a set of real number, the following are equivalent:

is an interval.
is convex.
is connected.

二叉树

Wed, 14 May 2025 07:47:19 GMT

好的，我们来解释一下完全二叉树（Complete Binary Tree）。
在了解完全二叉树之前，先回顾一下满二叉树（Full Binary Tree）和完美二叉树（Perfect Binary Tree）：

满二叉树：除了叶子节点外，所有节点都有两个子节点。
完美二叉树：是一种特殊的满二叉树，所有叶子节点都在同一层，并且所有内部节点都有两个子节点。完美二叉树的形状是一个完全对称的三角形。
现在来说完全二叉树：
完全二叉树是指这样一种二叉树：

除最后一层外，所有层都是完全填满的。
最后一层上的节点从左到右依次排列，不能有空缺。也就是说，如果最后一层不满，那么节点必须连续地靠左排列。
关键特点：

它的节点分布非常紧凑，除了最后一层，每一层都达到最大节点数。
最后一层的节点尽可能靠左。
举例说明：
这是一个完全二叉树：
A / \ B C / \ / D E F
解释：第0层 (A) 是满的，第1层 (B, C) 是满的，第2层 (D, E, F) 是最后一层，虽然不满，但节点从左到右是连续排列的 (D、E 在左子树，F 在右子树且靠左)。
这不是一个完全二叉树（因为最后一层有空缺，且不是最右边的空缺）：
A / \ B C / \ \ D E G
解释：节点 F 缺失了，导致最后一层的节点 E 和 G 之间有空缺。
这不是一个完全二叉树（因为倒数第二层不满）：
A / B / \ C D
解释：第1层只 B 一个节点，没有右子节点 C 的位置（如果是完全二叉树，C 应该在 B 的右边或者 B 的左子树被填满后才到 C），直接跳到了第2层。
为什么完全二叉树很重要？
完全二叉树的一个重要特性是它们可以非常高效地用数组来表示，而不需要使用指针。

如果根节点存储在数组索引 0 (或 1) 的位置，那么它的左子节点将存储在索引 2 * i + 1 (或 2 * i) 的位置，右子节点存储在索引 2 * i + 2 (或 2 * i + 1) 的位置（取决于数组从 0 还是 1 开始索引）。
反过来，对于索引为 i 的非根节点，其父节点存储在索引 (i - 1) / 2 (或 i / 2) 的位置。
这种基于数组的表示使得完全二叉树成为实现堆（Heap）数据结构的最佳选择，因为堆需要快速访问父子节点，并且需要紧凑地存储数据。
简而言之，完全二叉树是一种结构紧凑、节点靠左排列的二叉树，它非常适合用数组存储，常用于实现堆。完美二叉树和满二叉树都是完全二叉树的特例。

对于二叉树：
A / \ B C / \ D E

前序遍历：根左右
对于上述二叉树，前序遍历的结果是：A, B, D, E, C
中序遍历：左根右
对于上述二叉树，中序遍历的结果是：D, B, E, A, C
后序遍历：左右根
对于上述二叉树，后序遍历的结果是：D, E, B, C, A
前序遍历加上中序遍历序列，可以唯一确定一颗二叉树.

好的，将一颗完全二叉树存储到数组（或 C++中的 std::vector）中，利用了完全二叉树的紧凑性，使得节点位置和数组索引之间存在固定的数学关系。
最常用的映射方式是：

如果根节点存储在索引 i (通常是 0 或 1) 的位置。
那么它的左子节点存储在索引 2 * i + 1 的位置（如果从 0 开始索引）。
它的右子节点存储在索引 2 * i + 2 的位置（如果从 0 开始索引）。
反过来，对于非根节点存储在索引 i 的位置，它的父节点存储在索引 (i - 1) / 2 的位置（如果从 0 开始索引，且 i > 0）。
由于完全二叉树除了最后一层外都是满的，且最后一层节点靠左排列，这意味着如果一棵完全二叉树有 N 个节点，那么它们可以恰好填满数组的前 N 个位置，而不会有空缺（前提是按照层序从左到右排列）。
因此，将完全二叉树存入数组最直接的方法就是进行层序遍历（Level-Order Traversal），并依次将遍历到的节点值存入数组。层序遍历通常使用队列（Queue）来实现。
下面是一个使用 C++ 语言，将一个给定根节点的完全二叉树存储到 std::vector 中的程序示例：
#include #include #include #include // for calculating parent/child index if needed, but direct level-order filling is easier // 定义二叉树的节点结构 struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; // 函数：将完全二叉树存储到 vector 中 std::vector storeCompleteBinaryTree(TreeNode* root) { std::vector result; // 用于存储节点值的vector if (root == NULL) { return result; // 如果树为空，返回空vector } // 使用队列进行层序遍历 std::queue q; q.push(root); while (!q.empty()) { TreeNode* current = q.front(); q.pop(); // 将当前节点的值存入 vector result.push_back(current->val); // 将当前节点的左孩子入队 // 对于完全二叉树的数组表示，即使某个位置理论上是左孩子， // 但如果原始树中该节点不存在，我们也不会将其加入队列或vector。 // 这里我们假设输入的TreeNode结构已经代表了一棵完全二叉树的结构。 if (current->left != NULL) { q.push(current->left); } // 将当前节点的右孩子入队 if (current->right != NULL) { q.push(current->right); } } return result; } // 辅助函数：创建一个示例完全二叉树 // 结构: // 1 // / \ // 2 3 // / \ / // 4 5 6 TreeNode* createExampleCompleteBinaryTree() { TreeNode* root = new TreeNode(1); root->left = new TreeNode(2); root->right = new TreeNode(3); root->left->left = new TreeNode(4); root->left->right = new TreeNode(5); root->right->left = new TreeNode(6); // 注意：这里特意不创建右子树的右孩子(7)，使其成为非完美的完全二叉树 return root; } // 辅助函数：释放树的内存 void deleteTree(TreeNode* root) { if (root == NULL) return; deleteTree(root->left); deleteTree(root->right); delete root; } int main() { // 创建一个示例完全二叉树 TreeNode* root = createExampleCompleteBinaryTree(); // 将完全二叉树存储到 vector std::vector treeArray = storeCompleteBinaryTree(root); // 打印存储在 vector 中的节点值 std::cout << "完全二叉树存储在数组中的顺序 (层序遍历): "; for (int i = 0; i < treeArray.size(); ++i) { std::cout << treeArray[i] << (i == treeArray.size() - 1 ? "" : " "); } std::cout << std::endl; // 验证数组索引和节点位置的关系 (例如，索引0的左孩子在索引1，右孩子在索引2) if (!treeArray.empty()) { std::cout << "--- 验证索引关系 (0-based) ---" << std::endl; std::cout << "节点 " << treeArray[0] << " (索引 0):" << std::endl; if (2*0 + 1 < treeArray.size()) { std::cout << " 左孩子 (索引 1): " << treeArray[2*0 + 1] << std::endl; } if (2*0 + 2 < treeArray.size()) { std::cout << " 右孩子 (索引 2): " << treeArray[2*0 + 2] << std::endl; } if (treeArray.size() > 1) { std::cout << "节点 " << treeArray[1] << " (索引 1):" << std::endl; std::cout << " 父节点 (索引 0): " << treeArray[(1 - 1) / 2] << std::endl; if (2*1 + 1 < treeArray.size()) { std::cout << " 左孩子 (索引 3): " << treeArray[2*1 + 1] << std::endl; } if (2*1 + 2 < treeArray.size()) { std::cout << " 右孩子 (索引 4): " << treeArray[2*1 + 2] << std::endl; } } if (treeArray.size() > 2) { std::cout << "节点 " << treeArray[2] << " (索引 2):" << std::endl; std::cout << " 父节点 (索引 0): " << treeArray[(2 - 1) / 2] << std::endl; if (2*2 + 1 < treeArray.size()) { std::cout << " 左孩子 (索引 5): " << treeArray[2*2 + 1] << std::endl; } if (2*2 + 2 < treeArray.size()) { std::cout << " 右孩子 (索引 6): " << treeArray[2*2 + 2] << std::endl; // 这里会输出 6 } } if (treeArray.size() > 3) { std::cout << "节点 " << treeArray[3] << " (索引 3):" << std::endl; std::cout << " 父节点 (索引 1): " << treeArray[(3 - 1) / 2] << std::endl; // Left and Right children checks... } } // 释放内存 deleteTree(root); return 0; }
代码解释：

TreeNode 结构体：定义了二叉树节点的基本结构，包含一个值 (val) 和指向左右子节点的指针 (left, right)。
storeCompleteBinaryTree 函数：
接收 TreeNode* root 作为输入（树的根节点）。
创建一个 std::vector 叫做 result 用来存放遍历结果。
如果输入的 root 是 NULL (空树)，直接返回空的 result。
创建一个 std::queue 队列 q，用于实现层序遍历。
将根节点 root 入队。
进入 while 循环，当队列不为空时继续：
取出队列头部的节点 (current) 并出队。
将 current 节点的值 (current->val) 添加到 result 向量的末尾。
如果 current 的左子节点 current->left 不为空，将其入队。
如果 current 的右子节点 current->right 不为空，将其入队。
循环结束后，result 向量中就按层序从左到右存储了完全二叉树所有节点的值。
返回 result 向量。
createExampleCompleteBinaryTree 函数：这是一个辅助函数，用于手动构建一个符合完全二叉树定义的示例树，方便测试。
deleteTree 函数：辅助函数，用于递归地删除树节点，释放内存，防止内存泄漏。
main 函数：
调用 createExampleCompleteBinaryTree 创建示例树。
调用 storeCompleteBinaryTree 函数将树存储到 treeArray 向量中。
打印 treeArray 的内容，展示存储结果。
演示了如何根据数组索引计算父子节点的索引，并打印出对应的值，验证了完全二叉树的数组索引特性。
调用 deleteTree 释放内存。
运行此程序，你将看到完全二叉树的节点值按照层序从左到右的顺序被打印出来，并验证了父子节点在数组中的索引关系。

存储两个孩子
找双亲的时间复杂度为，找孩子的时间复杂度为 .

存储两个孩子和一个双亲

例如创建下面的二叉树：
A / \ B C \ D
使用前序遍历创建的过程：

输入前序序列：A B # D # # C # #，每次输入一个字符
输入 A：创建根节点 A
递归创建 A 的左子树:
输入 B：创建节点 B 作为 A 的左子节点
递归创建 B 的左子树:
输入#：B 的左子节点为 NULL
递归创建 B 的右子树:
输入 D：创建节点 D 作为 B 的右子节点
递归创建 D 的左子树:
输入#：D 的左子节点为 NULL
递归创建 D 的右子树:
输入#：D 的右子节点为 NULL
递归创建 A 的右子树:
输入 C：创建节点 C 作为 A 的右子节点
递归创建 C 的左子树:
输入#：C 的左子节点为 NULL
递归创建 C 的右子树:
输入#：C 的右子节点为 NULL
这样，通过序列 A B # D # # C # # 就完成了整棵二叉树的创建。

举例说明 root 的作用：
假设我们创建一棵如下的二叉树：
A / \ B C

创建过程：
构造函数BiTree()调用Creat(root)
用户输入A B # # C # #
Creat函数递归构建树并返回指向根节点A的指针
这个指针被赋值给root
访问和遍历：
执行T.PreOrder()时，内部会调用PreOrder(root)
root作为起点，从A开始访问整棵树
如果没有root，就无法知道从哪里开始遍历
树的判空： if (root == NULL) return; //判断是否为空树
内存释放：
析构函数调用Release(root)
从root开始递归释放所有节点内存
root 就像是树的"把手"，有了它才能对整棵树进行操作。如果 root 丢失，即使树的节点还存在内存中，也无法访问到它们，会导致内存泄漏
。

是的，通过后序遍历和中序遍历序列可以唯一地确定一棵二叉树。
核心思想：

后序遍历 (Left-Right-Root)：序列的最后一个元素一定是当前（子）树的根节点。
中序遍历 (Left-Root-Right)：一旦找到了根节点，那么在在中序遍历序列中，根节点左边的所有元素都属于左子树，右边的所有元素都属于右子树。
步骤：
假设我们有后序遍历序列 postOrder 和中序遍历序列 inOrder。

确定根节点：
postOrder 的最后一个元素就是整棵树的根节点。设这个根节点为 rootVal。
划分左右子树：
在 inOrder 序列中找到 rootVal 的位置。
inOrder 中 rootVal 左边的所有元素构成了左子树的中序遍历序列 (leftInOrder)。
inOrder 中 rootVal 右边的所有元素构成了右子树的中序遍历序列 (rightInOrder)。
确定左右子树的后序遍历序列：
左子树的节点数量等于 leftInOrder 的长度。设为 leftTreeSize。
右子树的节点数量等于 rightInOrder 的长度。设为 rightTreeSize。
在 postOrder 序列中（除去最后一个根节点 rootVal）：
前 leftTreeSize 个元素构成了左子树的后序遍历序列 (leftPostOrder)。
接下来的 rightTreeSize 个元素构成了右子树的后序遍历序列 (rightPostOrder)。
递归构建：
现在我们有了左子树的 leftPostOrder 和 leftInOrder，以及右子树的 rightPostOrder 和 rightInOrder。
递归地对左子树和右子树执行上述步骤，直到处理的序列为空（表示子树为空）。
将递归构建出的左子树连接到根节点的左孩子，右子树连接到根节点的右孩子。
举例说明：
假设：

后序遍历 (postOrder): [D, E, B, F, G, C, A]
中序遍历 (inOrder): [D, B, E, A, F, C, G]
构建过程：

第一次递归 (构建整棵树):
根节点: postOrder 的最后一个元素是 A。所以 A 是根。
划分中序: 在 inOrder [D, B, E, A, F, C, G] 中，A 的位置：
leftInOrder = [D, B, E] (左子树的中序遍历)
rightInOrder = [F, C, G] (右子树的中序遍历)
划分后序:
左子树节点数 leftTreeSize = 3 (来自 leftInOrder 的长度)。
右子树节点数 rightTreeSize = 3 (来自 rightInOrder 的长度)。
postOrder (除去 A) 是 [D, E, B, F, G, C]。
leftPostOrder = [D, E, B] (前 3 个元素)
rightPostOrder = [F, G, C] (后 3 个元素)
现在，根节点 A 的左孩子通过 leftPostOrder: [D, E, B] 和 leftInOrder: [D, B, E] 构建。
根节点 A 的右孩子通过 rightPostOrder: [F, G, C] 和 rightInOrder: [F, C, G] 构建。
第二次递归 (构建 A 的左子树):
输入: postOrder_L = [D, E, B], inOrder_L = [D, B, E]
根节点: postOrder_L 的最后一个元素是 B。所以 B 是该子树的根。
划分中序: 在 inOrder_L [D, B, E] 中，B 的位置：
leftInOrder_L_sub = [D]
rightInOrder_L_sub = [E]
划分后序:
左子树节点数 = 1。右子树节点数 = 1。
postOrder_L (除去 B) 是 [D, E]。
leftPostOrder_L_sub = [D]
rightPostOrder_L_sub = [E]
B 的左孩子通过 post: [D], in: [D] 构建。
B 的右孩子通过 post: [E], in: [E] 构建。
第三次递归 (构建 B 的左子树):
输入: postOrder = [D], inOrder = [D]
根节点: D。
划分中序: leftInOrder 为空，rightInOrder 为空。
D 是叶子节点。返回 D。
第三次递归 (构建 B 的右子树):
输入: postOrder = [E], inOrder = [E]
根节点: E。
划分中序: leftInOrder 为空，rightInOrder 为空。
E 是叶子节点。返回 E。
此时，B 的左右孩子都已确定：B 的左孩子是 D，右孩子是 E。
第二次递归 (构建 A 的右子树):
输入: postOrder_R = [F, G, C], inOrder_R = [F, C, G]
根节点: postOrder_R 的最后一个元素是 C。所以 C 是该子树的根。
划分中序: 在 inOrder_R [F, C, G] 中，C 的位置：
leftInOrder_R_sub = [F]
rightInOrder_R_sub = [G]
划分后序:
左子树节点数 = 1。右子树节点数 = 1。
postOrder_R (除去 C) 是 [F, G]。
leftPostOrder_R_sub = [F]
rightPostOrder_R_sub = [G]
C 的左孩子通过 post: [F], in: [F] 构建 (得到叶子节点 F)。
C 的右孩子通过 post: [G], in: [G] 构建 (得到叶子节点 G)。
此时，C 的左右孩子都已确定：C 的左孩子是 F，右孩子是 G。
最终结果：
将所有构建好的子树连接起来，得到的二叉树结构如下：
A / \ B C / \ / \ D E F G
这个过程通常用递归函数来实现，基本情况是当传入的遍历序列为空时，返回 null (或表示空子树的标记)。

是的，通过前序遍历和中序遍历序列同样可以唯一地确定一棵二叉树。
核心思想：

前序遍历 (Root-Left-Right)：序列的第一个元素一定是当前（子）树的根节点。
中序遍历 (Left-Root-Right)：一旦找到了根节点，那么在中序遍历序列中，根节点左边的所有元素都属于左子树，右边的所有元素都属于右子树。
步骤：
假设我们有前序遍历序列 preOrder 和中序遍历序列 inOrder。

确定根节点：
preOrder 的第一个元素就是整棵树的根节点。设这个根节点为 rootVal。
划分左右子树（在中序遍历中）：
在 inOrder 序列中找到 rootVal 的位置。
inOrder 中 rootVal 左边的所有元素构成了左子树的中序遍历序列 (leftInOrder)。
inOrder 中 rootVal 右边的所有元素构成了右子树的中序遍历序列 (rightInOrder)。
确定左右子树的前序遍历序列：
左子树的节点数量等于 leftInOrder 的长度。设为 leftTreeSize。
右子树的节点数量等于 rightInOrder 的长度。
在 preOrder 序列中（除去第一个根节点 rootVal）：
接下来的 leftTreeSize 个元素构成了左子树的前序遍历序列 (leftPreOrder)。
再往后的元素构成了右子树的前序遍历序列 (rightPreOrder)。
递归构建：
现在我们有了左子树的 leftPreOrder 和 leftInOrder，以及右子树的 rightPreOrder 和 rightInOrder。
递归地对左子树和右子树执行上述步骤，直到处理的序列为空（表示子树为空）。
将递归构建出的左子树连接到根节点的左孩子，右子树连接到根节点的右孩子。
举例说明：
假设：

前序遍历 (preOrder): [A, B, D, E, C, F, G]
中序遍历 (inOrder): [D, B, E, A, F, C, G]
构建过程：

第一次递归 (构建整棵树):
根节点: preOrder 的第一个元素是 A。所以 A 是根。
划分中序: 在 inOrder [D, B, E, A, F, C, G] 中，A 的位置：
leftInOrder = [D, B, E] (左子树的中序遍历)
rightInOrder = [F, C, G] (右子树的中序遍历)
划分前序:
左子树节点数 leftTreeSize = 3 (来自 leftInOrder 的长度)。
preOrder (除去 A) 是 [B, D, E, C, F, G]。
leftPreOrder = [B, D, E] (接下来的 3 个元素)
rightPreOrder = [C, F, G] (剩下的元素)
现在，根节点 A 的左孩子通过 leftPreOrder: [B, D, E] 和 leftInOrder: [D, B, E] 构建。
根节点 A 的右孩子通过 rightPreOrder: [C, F, G] 和 rightInOrder: [F, C, G] 构建。
第二次递归 (构建 A 的左子树):
输入: preOrder_L = [B, D, E], inOrder_L = [D, B, E]
根节点: preOrder_L 的第一个元素是 B。所以 B 是该子树的根。
划分中序: 在 inOrder_L [D, B, E] 中，B 的位置：
leftInOrder_L_sub = [D]
rightInOrder_L_sub = [E]
划分前序:
左子树节点数 = 1。
preOrder_L (除去 B) 是 [D, E]。
leftPreOrder_L_sub = [D] (接下来的 1 个元素)
rightPreOrder_L_sub = [E] (剩下的元素)
B 的左孩子通过 pre: [D], in: [D] 构建。
B 的右孩子通过 pre: [E], in: [E] 构建。
第三次递归 (构建 B 的左子树):
输入: preOrder = [D], inOrder = [D]
根节点: D。
划分中序: leftInOrder 为空，rightInOrder 为空。
D 是叶子节点。返回 D。
第三次递归 (构建 B 的右子树):
输入: preOrder = [E], inOrder = [E]
根节点: E。
划分中序: leftInOrder 为空，rightInOrder 为空。
E 是叶子节点。返回 E。
此时，B 的左右孩子都已确定：B 的左孩子是 D，右孩子是 E。
第二次递归 (构建 A 的右子树):
输入: preOrder_R = [C, F, G], inOrder_R = [F, C, G]
根节点: preOrder_R 的第一个元素是 C。所以 C 是该子树的根。
划分中序: 在 inOrder_R [F, C, G] 中，C 的位置：
leftInOrder_R_sub = [F]
rightInOrder_R_sub = [G]
划分前序:
左子树节点数 = 1。
preOrder_R (除去 C) 是 [F, G]。
leftPreOrder_R_sub = [F] (接下来的 1 个元素)
rightPreOrder_R_sub = [G] (剩下的元素)
C 的左孩子通过 pre: [F], in: [F] 构建 (得到叶子节点 F)。
C 的右孩子通过 pre: [G], in: [G] 构建 (得到叶子节点 G)。
此时，C 的左右孩子都已确定：C 的左孩子是 F，右孩子是 G。
最终结果：
将所有构建好的子树连接起来，得到的二叉树结构如下：
A / \ B C / \ / \ D E F G
这个过程与使用后序和中序遍历构建树非常相似，主要的区别在于从哪个遍历序列中取根节点（前序是第一个，后序是最后一个），以及如何根据根节点和中序遍历划分对应遍历序列中的子序列。

让我们来看一个例子：
情况 1：树 A:
1 / 2

前序遍历: [1, 2]
后序遍历: [2, 1]
情况 2：树 B:
1 \ 2

前序遍历: [1, 2]
后序遍历: [2, 1]
对于这两棵不同的树，它们的前序遍历和后序遍历结果是完全相同的。因此，仅给出前序 [1, 2] 和后序 [2, 1]，我们无法判断节点 2 究竟是节点 1 的左孩子还是右孩子。

课程云盘

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图论

Fri, 23 May 2025 07:18:13 GMT

本复习资料旨在为学习数据结构与算法课程的本科生提供一份关于图论的全面指导。内容涵盖图的基本概念、常用表示方法、核心遍历算法、最小生成树、最短路径问题、拓扑排序以及环检测等。每个主题都配有生动详尽的说明和C++代码示例，以帮助深入理解并有效备考。
I. 图论基础 (Introduction to Graphs)
图论是计算机科学中一个至关重要的分支，它为建模和解决涉及对象及其关系的问题提供了一个强大的数学框架。
A. 什么是图？ (What is a Graph?)
顶点 (Vertices) 与边 (Edges)：从根本上说，一个图 (Graph) 是由两个集合组成的：一个非空顶点集合 (Vertices 或 Nodes) 和一个边集合 (Edges)。数学上，一个图可以表示为 . 顶点代表图中的基本单元或实体，例如城市、人或网页。边则代表这些实体之间的连接或关系，例如连接城市的道路、人与人之间的友谊或网页之间的超链接。理解这两个基本构成元素——顶点和边——是掌握后续所有图论概念和算法的基石。
图的实际应用举例：图的抽象模型使其能够广泛应用于现实世界的多种场景中。例如：

社交网络：用户是顶点，用户之间的好友或关注关系是边。
交通网络：城市或交叉路口是顶点，连接它们的公路或航线是边，边的权重可以表示距离或时间。
万维网：网页是顶点，网页之间的超链接是边。
任务调度：项目中的各个任务是顶点，任务之间的依赖关系（例如，任务A必须在任务B开始前完成）是边。这类问题常通过拓扑排序解决。
生物信息学：蛋白质或基因是顶点，它们之间的相互作用是边。
这些例子揭示了图作为一种通用建模工具的强大能力，能够将复杂系统简化为可分析的结构。
B. 图的种类 (Types of Graphs)
根据边的特性和图的结构，图可以分为多种类型，理解这些类型对于选择合适的算法至关重要。
无向图 (Undirected Graphs) 与有向图 (Directed Graphs / Digraphs)：

无向图：边没有方向。如果顶点和顶点之间有一条边，那么这条边既可以看作从到，也可以看作从到。这种关系是双向的，例如Facebook上的好友关系。
有向图 (也称有向图)：边具有明确的方向。一条从指向的边表示存在从到的关系，但不一定存在从到的关系。例如Twitter上的关注关系，用户A关注用户B，不代表用户B也关注用户A。边的方向性是图的一个核心属性，它直接影响路径、连通性等概念以及算法的选择。
无权图 (Unweighted Graphs) 与加权图 (Weighted Graphs)：

无权图：图中的所有边都没有关联的权重或成本，每条边都被认为是等价的。
加权图：图中的每条边都关联一个数值，称为权重 (Weight) 或成本 (Cost)。权重可以表示多种含义，如距离、时间、容量、费用等。例如，在地图应用中，连接两个城市的边的权重可以表示它们之间的实际距离。加权图为模拟更复杂的现实场景提供了可能，许多重要的图算法，如Dijkstra最短路径算法和Prim最小生成树算法，都是针对加权图设计的。
简单图 (Simple Graphs), 多重图 (Multigraphs), 自环 (Loops)：

简单图：一个图如果任意两个顶点之间最多只有一条边相连，并且图中不包含自环（即连接顶点到其自身的边），则称其为简单图。大多数基础的图算法，如简单路径查找，通常是在简单图的背景下讨论的。
多重图：允许在同一对顶点之间存在多条平行的边。例如，两个数据中心之间可能有多条光纤连接。
自环 (Loop)：一条连接顶点到其自身的边。例如，在状态图中，一个状态可以通过某个操作转换回自身。在邻接矩阵表示法中，自环对应于主对角线上的非零元素。
其他常见类型 (Brief mention)：

完全图 (Complete Graph)：在一个无向简单图中，如果每对不同的顶点之间都恰好存在一条边，则称该图为完全图。若有个顶点，则完全图有条边。
二分图 (Bipartite Graph)：图的顶点集可以被划分为两个没有交集的子集和，使得图中的每一条边都连接中的一个顶点和中的一个顶点。例如，任务分配问题中，一组顶点代表人，另一组顶点代表任务，边表示某人可以执行某任务。
C. 图的基本术语 (Basic Graph Terminology)
掌握图论的基本术语对于理解和交流图算法至关重要。
顶点的度 (Degree of a Vertex), 入度 (In-degree), 出度 (Out-degree)：

在无向图中，一个顶点的度是指与该顶点相关联（或称入射）的边的数量。在一个不包含自环的简单图中，一个顶点的度最大为，其中是顶点总数。
在有向图中，一个顶点的度分为入度和出度。
入度 (In-degree)：指向该顶点的边的数量。
出度 (Out-degree)：从该顶点指出的边的数量。
顶点的度（或入度/出度）是衡量其连接性的基本指标，并在多种算法（如Kahn拓扑排序算法依赖入度）中扮演关键角色。
邻接顶点 (Adjacent Vertices / Neighbors)：如果两个顶点之间存在一条边，则称这两个顶点是邻接的，它们互为邻居。
路径 (Path)：路径是从一个顶点到另一个顶点的一个顶点序列，其中 , ，并且对于所有 , 都是图中的一条边。路径的长度在无权图中通常指边的数量，在加权图中通常指路径上所有边的权重之和。
闭合路径 (Closed Path)：如果一条路径的起始顶点与终止顶点相同，则称其为闭合路径。
简单路径 (Simple Path)：如果一条路径中所有的顶点都是唯一的（除了可能的起始顶点和终止顶点相同的情况），则称其为简单路径。
环路 (Cycle)：一条简单路径，其第一个和最后一个顶点相同，并且路径中不包含重复的边（除了连接起点和终点的那条边，如果路径长度大于1）。严格来说，一个环路通常要求至少包含3条边（在简单图中）以避免退化情况。环路是图论中的一个核心概念，许多问题（如环检测、拓扑排序）都与之相关。
连通图 (Connected Graphs) 与连通分量 (Connected Components)：

连通图 (特指无向图)：如果一个无向图中任意两个不同的顶点之间都至少存在一条路径，则称该图是连通图。
连通分量 (特指无向图)：无向图的极大连通子图。这意味着一个连通分量本身是连通的，并且如果将原图中不在此分量内的任何其他顶点加入该分量，它将不再连通。一个非连通图可以分解为若干个连通分量。
对于有向图，连通性有更细致的划分，如弱连通（忽略边的方向后是连通的）和强连通（任意两个顶点之间都存在双向路径）。
连通性描述了图的“整体性”。许多图算法的正确执行依赖于图的连通性，或者需要对图的每个连通分量分别应用算法。
图的定义、类型以及相关的基本术语构成了图算法世界的“词汇表”和“语法规则”。对这些基础知识的精确掌握是理解和实现复杂图算法的前提。例如，一个算法的描述可能指定它适用于“加权无向连通图”，如果对这些术语理解不清，就无法正确应用该算法。图的抽象性使其具有极强的建模能力，能够将不同领域的问题（如社交网络分析、物流路径优化、任务依赖管理）统一到同一个数学框架下进行研究和求解。因此，学习图论不仅是掌握一组特定的算法，更是获得一种强大的分析和解决问题的思维模式。
II. 图的表示方法 (Graph Representations)
要在计算机中处理图，首先需要一种方式来存储图的结构，即顶点和它们之间的连接关系。选择合适的表示方法对算法的效率和实现复杂度有显著影响。主要有两种标准的图表示方法：邻接矩阵和邻接表。
A. 邻接矩阵 (Adjacency Matrix)
概念与结构：邻接矩阵是一种通过二维数组来表示图中顶点间连接关系的方法。对于一个包含个顶点的图，邻接矩阵是一个的方阵，记为 adjMatrix。

如果顶点和顶点之间存在一条边：
对于无权图，adjMatrix[i][j] 的值通常设为 1。
对于加权图，adjMatrix[i][j] 的值设为连接顶点和的边的权重。
如果顶点和顶点之间不存在边，则 adjMatrix[i][j] 的值通常设为 0 (对于无权图或表示不存在连接)，或者在某些算法（如最短路径算法）的初始化阶段设为一个特殊值（如无穷大）来表示不直接连通。
对于无向图，由于边和是同一条边，所以邻接矩阵是关于主对角线对称的，即 adjMatrix[i][j] = adjMatrix[j][i]。
如果图中存在自环（即从顶点到其自身的边），则邻接矩阵的主对角线元素 adjMatrix[i][i] 将为非零值（1或权重）。若不允许自环的简单图，主对角线元素通常为0。
邻接矩阵是一种非常直观的表示方式，易于理解和实现。
C++ 代码示例：下面是一个使用 std::vector> 实现邻接矩阵的C++类示例，用于表示一个无权图。它可以很容易地修改以支持加权图（将矩阵元素类型改为权重类型，并将1替换为权重值）。
#include #include #include // For std::out_of_range class GraphAdjMatrix { private: int numVertices; std::vector> adjMatrix; bool isDirected; // 图是否有向 public: // 构造函数：初始化顶点数量和邻接矩阵 GraphAdjMatrix(int V, bool directed = false) : numVertices(V), isDirected(directed) { if (V <= 0) { throw std::invalid_argument("Number of vertices must be positive."); } // 初始化V x V的邻接矩阵，所有元素为0 adjMatrix.resize(numVertices, std::vector(numVertices, 0)); std::cout << "Initialized Adjacency Matrix for " << numVertices << " vertices." << std::endl; } // 添加边 (u, v) // 对于加权图，可以增加一个weight参数，并将矩阵值设为weight void addEdge(int u, int v, int weight = 1) { // 默认为无权图，权重为1 if (u >= 0 && u < numVertices && v >= 0 && v < numVertices) { adjMatrix[u][v] = weight; if (!isDirected) { // 如果是无向图，对称地添加边 adjMatrix[v][u] = weight; } std::cout << "Added edge between " << u << " and " << v << " with weight " << weight << "." << std::endl; } else { throw std::out_of_range("Vertex out of bounds."); } } // 移除边 (u, v) void removeEdge(int u, int v) { if (u >= 0 && u < numVertices && v >= 0 && v < numVertices) { adjMatrix[u][v] = 0; // 或表示无穷大的值 if (!isDirected) { adjMatrix[v][u] = 0; } std::cout << "Removed edge between " << u << " and " << v << "." << std::endl; } else { throw std::out_of_range("Vertex out of bounds."); } } // 检查顶点u和v之间是否存在边 bool hasEdge(int u, int v) { if (u >= 0 && u < numVertices && v >= 0 && v < numVertices) { return adjMatrix[u][v]!= 0; // 对于加权图，只要不是0或无穷大就认为有边 } throw std::out_of_range("Vertex out of bounds."); return false; } // 打印邻接矩阵 void display() { std::cout << "\nAdjacency Matrix:" << std::endl; for (int i = 0; i < numVertices; ++i) { for (int j = 0; j < numVertices; ++j) { std::cout << adjMatrix[i][j] << " "; } std::cout << std::endl; } } // 析构函数 (std::vector 会自动管理内存，如果用裸指针则需要手动释放) ~GraphAdjMatrix() { std::cout << "Adjacency Matrix Graph destroyed." << std::endl; } }; // 主函数示例 // int main() { // try { // GraphAdjMatrix g(4, false); // 创建一个4个顶点的无向图 // g.addEdge(0, 1); // g.addEdge(0, 2); // g.addEdge(1, 2); // g.addEdge(2, 3); // g.display(); // // GraphAdjMatrix dg(3, true); // 创建一个3个顶点的有向图 // dg.addEdge(0, 1, 5); // 加权边 // dg.addEdge(1, 2, 3); // dg.addEdge(0, 2, 10); // dg.display(); // std::cout << "Edge (0,1) exists: " << dg.hasEdge(0,1) << std::endl; // std::cout << "Edge (1,0) exists: " << dg.hasEdge(1,0) << std::endl; // // } catch (const std::exception& e) { // std::cerr << "Exception: " << e.what() << std::endl; // } // return 0; // }
在上述代码中，构造函数初始化一个全零的矩阵。addEdge 方法根据图是否有向来设置矩阵元素。hasEdge 方法可以快速检查边的存在性。
优缺点分析：邻接矩阵的主要优点包括：

快速判断边的存在性：检查顶点和之间是否存在边（即 adjMatrix[u][v] 是否非零）只需要的时间。
添加或删除边操作高效：如果顶点已知，添加或删除一条边也只需要的时间（修改矩阵中的一个元素）。
对于稠密图（图中边的数量接近于），邻接矩阵的空间效率相对可以接受，并且其操作速度较快。
然而，邻接矩阵也有一些显著的缺点：

空间复杂度高：它总是需要的空间，其中是顶点数。对于稀疏图（边的数量远小于），这将导致大量的空间浪费，因为矩阵中的大部分元素都会是0。
遍历顶点的邻接点效率低：要找到一个特定顶点的所有邻接点，需要遍历矩阵中的整行（或整列），这需要的时间。
添加或删除顶点操作成本高：这些操作通常需要重新分配和复制整个矩阵，或者至少是级别的调整，因此非常耗时。
空间与时间复杂度总结：

空间复杂度：
添加顶点： (由于矩阵重建)
添加边：
删除顶点： (由于矩阵重建)
删除边：
查询边 (u,v) 是否存在：
获取顶点的所有邻接点：
B. 邻接表 (Adjacency List)
概念与结构：邻接表是另一种常用的图表示方法，尤其适用于稀疏图。它由一个包含个列表（通常是链表或动态数组，如C++中的 std::vector）的数组（或向量）构成。数组的第个索引（或元素）对应图中的第个顶点，该索引处的列表存储了所有与顶点直接邻接的顶点。

对于无权图，列表中通常只存储邻接顶点的编号。
对于加权图，列表中的每个元素通常是一个序对 (pair)，例如 std::pair，其中第一个元素是邻接顶点的编号，第二个元素是连接当前顶点与该邻接顶点的边的权重。
由于图的结构可能非常复杂，任意两个顶点之间都可能存在关系，这种任意的逻辑关系使得图无法简单地通过固定的存储位置来表示其连接性，而邻接表恰好提供了这种所需的灵活性。
C++ 代码示例：下面是一个使用 std::vector>> 实现邻接表的C++类示例，用于表示一个加权图。对于无权图，可以将 pair 替换为单个 int 来存储邻接顶点。
#include #include #include // 可选，如果想用链表代替vector作为邻接列表 #include // For std::remove_if for removeEdge // 对于加权图，边信息可以包含目标顶点和权重 struct EdgeNode { int to_vertex; int weight; }; class GraphAdjList { private: int numVertices; std::vector> adjList; // 使用list作为邻接列表，方便删除边 bool isDirected; public: // 构造函数 GraphAdjList(int V, bool directed = false) : numVertices(V), isDirected(directed) { if (V <= 0) { throw std::invalid_argument("Number of vertices must be positive."); } adjList.resize(numVertices); std::cout << "Initialized Adjacency List for " << numVertices << " vertices." << std::endl; } // 添加边 (u, v) 权重为 weight void addEdge(int u, int v, int weight) { if (u >= 0 && u < numVertices && v >= 0 && v < numVertices) { adjList[u].push_back({v, weight}); if (!isDirected) { // 如果是无向图，对称地添加边 adjList[v].push_back({u, weight}); } std::cout << "Added edge between " << u << " and " << v << " with weight " << weight << "." << std::endl; } else { throw std::out_of_range("Vertex out of bounds."); } } // 移除边 (u, v) // 注意：对于多重图，这会移除第一个匹配的边。如果需要移除所有匹配边，需修改。 void removeEdge(int u, int v) { if (u >= 0 && u < numVertices && v >= 0 && v < numVertices) { adjList[u].remove_if([v](const EdgeNode& edge){ return edge.to_vertex == v; }); if (!isDirected) { adjList[v].remove_if([u](const EdgeNode& edge){ return edge.to_vertex == u; }); } std::cout << "Attempted to remove edge(s) between " << u << " and " << v << "." << std::endl; } else { throw std::out_of_range("Vertex out of bounds."); } } // 检查顶点u和v之间是否存在边 bool hasEdge(int u, int v) { if (u >= 0 && u < numVertices && v >= 0 && v < numVertices) { for (const auto& edge : adjList[u]) { if (edge.to_vertex == v) { return true; } } return false; } throw std::out_of_range("Vertex out of bounds."); return false; } // 打印邻接表 void display() { std::cout << "\nAdjacency List:" << std::endl; for (int i = 0; i < numVertices; ++i) { std::cout << "Vertex " << i << ":"; for (const auto& edge : adjList[i]) { std::cout << " -> (" << edge.to_vertex << ", w:" << edge.weight << ")"; } std::cout << std::endl; } } // 获取顶点的邻接点列表 (只读访问) const std::list& getNeighbors(int u) const { if (u >= 0 && u < numVertices) { return adjList[u]; } throw std::out_of_range("Vertex out of bounds."); } ~GraphAdjList() { std::cout << "Adjacency List Graph destroyed." << std::endl; } }; // 主函数示例 // int main() { // try { // GraphAdjList g(4, false); // 4个顶点的无向加权图 // g.addEdge(0, 1, 10); // g.addEdge(0, 2, 5); // g.addEdge(1, 2, 2); // g.addEdge(2, 3, 1); // g.display(); // // std::cout << "Neighbors of vertex 0:" << std::endl; // for (const auto& edge : g.getNeighbors(0)) { // std::cout << " To: " << edge.to_vertex << ", Weight: " << edge.weight << std::endl; // } // // std::cout << "Edge (0,1) exists: " << g.hasEdge(0,1) << std::endl; // g.removeEdge(0,1); // std::cout << "Edge (0,1) exists after removal: " << g.hasEdge(0,1) << std::endl; // g.display(); // // } catch (const std::exception& e) { // std::cerr << "Exception: " << e.what() << std::endl; // } // return 0; // }
在此示例中，adjList 是一个 std::vector，其每个元素是一个 std::list。EdgeNode 结构体存储了目标顶点和边的权重。addEdge 方法将新的邻接信息添加到相应顶点的列表中。
优缺点分析：邻接表的主要优点包括：

空间效率高：特别是对于稀疏图，邻接表的空间复杂度为，其中是顶点数，是边数。这是因为它只存储实际存在的边。
高效遍历顶点的邻接点：获取并遍历一个特定顶点的所有邻接点非常高效，时间复杂度与其度数成正比。这对于像深度优先搜索 (DFS) 和广度优先搜索 (BFS) 这样的图遍历算法非常有利。
添加顶点操作相对高效：向图中添加一个新顶点（不包括其边）通常是操作（即在表示邻接表的数组/向量末尾添加一个新列表）。
添加边操作高效：向现有顶点的邻接列表中添加一条边通常是（摊销时间，如果使用 std::vector::push_back 或链表头插）。
邻接表的缺点主要有：

判断边的存在性较慢：要检查顶点和之间是否存在边，最坏情况下需要遍历顶点 (或 ) 的整个邻接列表，时间复杂度为，其中是顶点的度。在稠密图中，这可能接近。
删除边操作可能较慢：与查询类似，删除一条边需要先在顶点的邻接列表中找到并移除（对于无向图，还需要在的列表中找到并移除），这同样可能需要的时间。
根据 61 的观点，邻接链表（一种邻接表的实现方式）在边的读写上可能比矩阵表示稍慢，因为链表中的指针在内存中可能不是连续存储的，这可能影响缓存性能。
空间与时间复杂度总结：

空间复杂度：
添加顶点： (仅指为新顶点创建空邻接列表)
添加边： (摊销)
删除顶点： (最坏情况，因为需要遍历所有其他顶点的邻接列表以移除与被删除顶点相关的边)
删除边 (u,v)： (对于有向图，或无向图中从到的部分) 或 (对于无向图的完整删除)
查询边 (u,v) 是否存在：或
获取顶点的所有邻接点：
C. 如何选择合适的表示方法 (Choosing the Right Representation)
选择邻接矩阵还是邻接表，主要取决于图的特性（特别是密度）和应用中所需执行的主要操作类型。
图的密度：

稠密图 (Dense Graphs)：当边的数量接近于的上限时，图被认为是稠密的。在这种情况下，邻接矩阵的空间开销与相差不大，其的边查询速度可能成为优势。
稀疏图 (Sparse Graphs)：当边的数量远小于（例如与成线性关系）时，图被认为是稀疏的。此时，邻接表的空间复杂度远优于邻接矩阵，是首选的表示方法。21 甚至探讨了对稀疏有向图的邻接矩阵进行压缩存储的方法，以缓解其空间浪费问题，这反过来也强调了邻接表在稀疏图上的自然优势。
主要操作需求：

如果算法需要频繁地检查任意两个顶点之间是否存在边，邻接矩阵的查询时间非常有利。
如果算法需要频繁地遍历一个顶点的所有邻接点（例如在DFS、BFS等图遍历算法中），邻接表则更为高效，因为它直接提供了邻接点列表。
如果图中顶点的添加和删除操作非常频繁，需要仔细评估两种方法。邻接矩阵的顶点增删成本很高 ()，而邻接表在添加顶点本身时较快，但删除顶点时仍需处理相关的边，可能达到。

下表总结了邻接矩阵和邻接表在各种操作上的性能特点：

数据来源：基于 14 及其他相关片段的分析。
此表格清晰地展示了两种表示方法的权衡。在实际应用和算法竞赛中，由于现实世界中的许多大型图（如社交网络、网页图）通常是稀疏的，邻接表因其空间效率和对遍历操作的良好支持而更为常用。
图的表示方法是图算法得以高效执行的基石。一种算法的理论时间复杂度往往是基于某种特定的图表示（通常是邻接表，因为它对稀疏图更友好）推导出来的。例如，标准的BFS和DFS算法的时间复杂度就是假设图是用邻接表表示的。如果用邻接矩阵，由于查找一个顶点的所有邻居需要时间，这些遍历算法的复杂度可能会变成。这凸显了数据结构选择对算法性能的深远影响。此外，C++标准模板库 (STL) 中的 std::vector 和 std::list（以及用于加权图的 std::pair）为实现邻接表提供了极大的便利，使得开发者可以专注于算法逻辑本身，而不必过多地纠缠于底层数据结构的细节实现。然而，理解这些STL容器大致的性能特征（例如 std::vector::push_back 的摊销复杂度）对于准确分析算法性能仍然是重要的。最终，选择哪种表示方法是一个需要在空间效率、时间效率以及特定操作需求之间进行权衡的决策。
III. 图的遍历算法 (Graph Traversal Algorithms)
图的遍历是指按照某种特定的顺序访问图中的每一个顶点，并且每个顶点仅访问一次。这是许多更复杂图算法的基础。最核心的两种遍历算法是广度优先搜索 (BFS) 和深度优先搜索 (DFS)。
A. 广度优先搜索 (BFS - Breadth-First Search)
算法思想与直观理解：广度优先搜索 (BFS) 是一种系统性地探索图中顶点的方法。它从一个指定的源顶点开始，首先访问所有与源顶点直接相邻的顶点（可以看作是第一层或距离为1的层）。然后，对于第一层中所有已被访问的顶点，BFS会继续访问它们所有未被访问过的邻接点（构成第二层或距离为2的层）。这个过程逐层向外扩展，直到所有从源顶点可达的顶点都被访问过为止。可以将BFS的过程想象成在平静的水面上投下一颗石子，产生的波纹会以石子落点为中心，一圈一圈地向外均匀扩散。BFS的这种“逐层”或“按距离远近”的访问特性，使其天然适用于寻找无权图中的最短路径（以边的数量衡量）。
为了实现这种逐层访问的顺序，BFS通常使用一个队列 (Queue) 数据结构。队列的先进先出 (FIFO) 特性保证了先发现的顶点（离源点更近的顶点）其邻接点会被优先探索。同时，为了避免重复访问同一个顶点（尤其是在包含环的图中可能导致无限循环），BFS会使用一个布尔数组 visited 来标记每个顶点是否已经被访问过。
C++ 代码示例：以下是一个基于邻接表实现的BFS算法的C++代码示例。假设图用 std::vector> adjList 表示（对于无权图），其中 adjList[u] 包含了顶点的所有邻接顶点。
#include #include #include #include // GraphAdjList 使用了 list // (假设 GraphAdjList 类如前文定义，这里仅展示BFS成员函数) // 或者可以是一个独立的BFS函数，接收邻接表和起始顶点作为参数 // 独立BFS函数版本 void bfs_standalone(int startVertex, int numVertices, const std::vector>& adjList) { if (startVertex < 0 || startVertex >= numVertices) { std::cerr << "Error: Start vertex is out of bounds." << std::endl; return; } std::vector visited(numVertices, false); // 标记顶点是否被访问过 std::queue q; // 辅助队列 // 1. 将起始顶点标记为已访问，并将其入队 visited[startVertex] = true; q.push(startVertex); std::cout << "BFS traversal starting from vertex " << startVertex << ": "; // 2. 当队列不为空时，循环继续 while (!q.empty()) { // 3. 从队列中取出一个顶点 int u = q.front(); q.pop(); // 4. 访问该顶点 (例如，打印) std::cout << u << " "; // 5. 遍历该顶点的所有邻

栈

Wed, 14 May 2025 01:10:38 GMT

栈是一种特殊的线性数据结构，遵循后进先出 (Last-In, First-Out, LIFO) 的原则。可以将其想象成一叠盘子：你最后放上去的盘子，会是第一个被取走的。

栈顶 (Top): 允许进行插入和删除操作的一端。
栈底 (Bottom): 栈的另一端，通常固定不动。
压栈/入栈 (Push): 向栈顶添加元素的过程。
弹栈/出栈 (Pop): 从栈顶移除元素的过程。
核心特性总结：

LIFO原则：最后进入栈的元素最先离开。
受限的线性表：只能在栈顶进行操作。

栈主要支持以下几种基本操作：

push(element): 将元素 element 压入栈顶。
前置条件 (对于有界实现)：栈未满。
后置条件： element 成为新的栈顶元素，栈的大小增加1。
pop(): 移除栈顶元素。注意：在 C++ 的 std::stack 中，pop() 只移除元素，不返回值。获取栈顶元素需要使用 top()。
前置条件：栈不能为空。
后置条件：原栈顶元素被移除，其下的元素（如果存在）成为新的栈顶，栈的大小减少1。
top() (或 peek()): 返回对栈顶元素的引用（或值），但不移除它。
前置条件：栈不能为空。
后置条件：栈的状态不改变。
empty(): 检查栈是否为空。
后置条件：如果栈中没有元素，返回 true；否则返回 false。
size(): 返回栈中元素的数量。

栈在计算机科学中有广泛的应用，例如：

函数调用栈 (Call Stack): 管理函数调用的顺序，存储局部变量、参数和返回地址。
表达式求值：
中缀表达式转后缀表达式 (逆波兰表达式)。
后缀表达式求值。
括号匹配：检查代码或表达式中的括号 ((), {}, []) 是否成对出现且正确嵌套。
深度优先搜索 (DFS - Depth First Search): 在图或树的遍历中，栈可以用来存储待访问的节点。
浏览器的前进/后退功能：后退历史可以看作一个栈。
撤销/重做 (Undo/Redo) 操作：用户操作可以被存储在一个栈中。
递归的非递归实现：许多递归算法可以用栈来模拟实现。

在 C++ 中，栈可以通过多种方式实现：

基于数组的实现 (静态或动态)
基于链表的实现
使用 C++ 标准库的 std::stack (适配器类)

使用一个固定大小的数组来存储栈中的元素。需要一个额外的变量（通常称为 topIndex 或 count）来指示栈顶元素在数组中的下一个可用位置或当前栈顶元素的索引。
优点：

实现相对简单。
内存分配在编译时或对象创建时完成，访问速度快（缓存友好）。
缺点：

大小固定：栈的容量在创建时就确定了。如果元素数量超过数组容量，会导致栈溢出 (Stack Overflow)。
空间浪费：如果分配的数组空间远大于实际存储的元素数量，会造成空间浪费。
C++ 代码示例 (静态数组)：
#include #include // 用于 std::out_of_range 和 std::runtime_error const int MAX_SIZE = 100; // 预设栈的最大容量 template class ArrayStack { private: T arr[MAX_SIZE]; int topIndex; // 指向栈顶元素的索引，-1 表示空栈 public: ArrayStack() : topIndex(-1) {} // 压栈 void push(const T& element) { if (isFull()) { throw std::overflow_error("Stack overflow: Cannot push to a full stack."); } arr[++topIndex] = element; } // 弹栈 (只移除，不返回) void pop() { if (isEmpty()) { throw std::underflow_error("Stack underflow: Cannot pop from an empty stack."); } topIndex--; } // 查看栈顶元素 T& top() { if (isEmpty()) { throw std::underflow_error("Stack is empty: Cannot get top element."); } return arr[topIndex]; } // 查看栈顶元素 (const 版本) const T& top() const { if (isEmpty()) { throw std::underflow_error("Stack is empty: Cannot get top element."); } return arr[topIndex]; } // 检查栈是否为空 bool isEmpty() const { return topIndex == -1; } // 检查栈是否已满 bool isFull() const { return topIndex == MAX_SIZE - 1; } // 获取栈中元素数量 int size() const { return topIndex + 1; } // 打印栈内容 (用于调试) void print() const { if (isEmpty()) { std::cout << "Stack is empty." << std::endl; return; } std::cout << "Stack (top to bottom): "; for (int i = topIndex; i >= 0; --i) { std::cout << arr[i] << " "; } std::cout << std::endl; } }; // --- ArrayStack 示例用法 --- void testArrayStack() { std::cout << "--- ArrayStack (Static) 示例 ---" << std::endl; ArrayStack s; std::cout << "Is empty: " << s.isEmpty() << std::endl; // 1 (true) s.push(10); s.push(20); s.push(30); s.print(); // Stack (top to bottom): 30 20 10 std::cout << "Top element: " << s.top() << std::endl; // 30 std::cout << "Size: " << s.size() << std::endl; // 3 std::cout << "Is empty: " << s.isEmpty() << std::endl; // 0 (false) s.pop(); // 移除 30 s.print(); // Stack (top to bottom): 20 10 std::cout << "Top element after pop: " << s.top() << std::endl; // 20 s.pop(); // 移除 20 s.pop(); // 移除 10 s.print(); // Stack is empty. std::cout << "Is empty: " << s.isEmpty() << std::endl; // 1 (true) try { s.pop(); // 尝试从空栈 pop } catch (const std::underflow_error& e) { std::cerr << "Caught exception: " << e.what() << std::endl; } // 填满栈并尝试溢出 ArrayStack charStack; for (int i = 0; i < MAX_SIZE; ++i) { charStack.push('A' + (i % 26)); } std::cout << "Char stack is full: " << charStack.isFull() << std::endl; try { charStack.push('Z'); } catch (const std::overflow_error& e) { std::cerr << "Caught exception: " << e.what() << std::endl; } std::cout << std::endl; }
时间复杂度 (静态数组实现)：

push(element):
pop():
top():
isEmpty():
isFull():
size():
空间复杂度 (静态数组实现)：，其中是数组的最大容量 (MAX_SIZE)。

注意: 对于动态数组实现 (使用 new T[capacity] 和 delete[])，push 操作在需要扩容时的时间复杂度会是 (其中 N 为当前元素数量)，但均摊时间复杂度仍为。

使用链表（通常是单向链表）来存储栈中的元素。链表的头节点通常作为栈顶。
优点：

动态大小：栈的大小可以根据需要动态增长或缩小，没有预设的容量限制（受限于系统可用内存）。
空间高效：只为实际存储的元素分配内存。
缺点：

额外的内存开销：每个节点都需要额外的空间来存储指向下一个节点的指针。
内存不连续：相比数组，可能在缓存利用率上稍差。
C++ 代码示例 (链表)：
#include #include // 用于 std::out_of_range template class LinkedStack { private: struct Node { T data; Node* next; Node(const T& val) : data(val), next(nullptr) {} }; Node* topNode; // 指向栈顶节点 int count; // 栈中元素数量 public: LinkedStack() : topNode(nullptr), count(0) {} ~LinkedStack() { // 析构函数，释放所有节点内存 while (!isEmpty()) { pop(); } } // 压栈 void push(const T& element) { Node* newNode = new Node(element); newNode->next = topNode; topNode = newNode; count++; } // 弹栈 (只移除，不返回) void pop() { if (isEmpty()) { throw std::underflow_error("Stack underflow: Cannot pop from an empty stack."); } Node* temp = topNode; topNode = topNode->next; delete temp; count--; } // 查看栈顶元素 T& top() { if (isEmpty()) { throw std::underflow_error("Stack is empty: Cannot get top element."); } return topNode->data; } // 查看栈顶元素 (const 版本) const T& top() const { if (isEmpty()) { throw std::underflow_error("Stack is empty: Cannot get top element."); } return topNode->data; } // 检查栈是否为空 bool isEmpty() const { return topNode == nullptr; // 或者 count == 0 } // 获取栈中元素数量 int size() const { return count; } // 打印栈内容 (用于调试) void print() const { if (isEmpty()) { std::cout << "LinkedStack is empty." << std::endl; return; } std::cout << "LinkedStack (top to bottom): "; Node* current = topNode; while (current != nullptr) { std::cout << current->data << " "; current = current->next; } std::cout << std::endl; } }; // --- LinkedStack 示例用法 --- void testLinkedStack() { std::cout << "--- LinkedStack 示例 ---" << std::endl; LinkedStack s; std::cout << "Is empty: " << s.isEmpty() << std::endl; // 1 (true) s.push("Hello"); s.push("World"); s.push("C++"); s.print(); // LinkedStack (top to bottom): C++ World Hello std::cout << "Top element: " << s.top() << std::endl; // C++ std::cout << "Size: " << s.size() << std::endl; // 3 s.pop(); // 移除 "C++" s.print(); // LinkedStack (top to bottom): World Hello std::cout << "Top element after pop: " << s.top() << std::endl; // World s.pop(); // 移除 "World" s.pop(); // 移除 "Hello" s.print(); // LinkedStack is empty. std::cout << "Is empty: " << s.isEmpty() << std::endl; // 1 (true) try { std::cout << s.top() << std::endl; // 尝试从空栈获取 top } catch (const std::underflow_error& e) { std::cerr << "Caught exception: " << e.what() << std::endl; } std::cout << std::endl; }
时间复杂度 (链表实现)：

push(element): (在链表头部插入)
pop(): (删除链表头部节点)
top(): (访问链表头部节点)
isEmpty():
size(): (如果维护了一个 count 变量)
空间复杂度 (链表实现)：，其中是栈中元素的数量。每个元素都需要一个节点的空间。

C++ 标准库提供了一个容器适配器 std::stack，它默认使用 std::deque (双端队列) 作为其底层容器，但也可以配置为使用 std::vector 或 std::list。std::stack 封装了底层容器，使其表现出栈的行为。
优点：

标准、可靠、高效：由标准库提供，经过良好测试和优化。
易于使用：无需自己实现，直接包含头文件即可。
灵活性：可以选择底层容器。
缺点：

功能受限：只提供标准的栈操作，不能直接访问底层容器的非栈操作（比如直接访问非栈顶元素）。
C++ 代码示例 (std::stack)：
#include #include // 包含 stack 头文件 #include // 可以用作 std::stack 的底层容器 #include // 也可以用作 std::stack 的底层容器 #include // std::stack 默认的底层容器 // --- std::stack 示例用法 --- void testStdStack() { std::cout << "--- std::stack (default: deque) 示例 ---" << std::endl; std::stack s_deque; // 默认底层容器是 std::deque std::cout << "Is empty: " << s_deque.empty() << std::endl; // 1 (true) s_deque.push(100); s_deque.push(200); s_deque.push(300); // std::stack 没有直接的 print 方法，需要逐个取出打印 (会改变栈) // 或者访问底层容器（但不推荐，破坏封装） std::cout << "Top element: " << s_deque.top() << std::endl; // 300 std::cout << "Size: " << s_deque.size() << std::endl; // 3 s_deque.pop(); // 移除 300 std::cout << "Top element after pop: " << s_deque.top() << std::endl; // 200 std::cout << "Size after pop: " << s_deque.size() << std::endl; // 2 std::cout << "Popping elements: "; while (!s_deque.empty()) { std::cout << s_deque.top() << " "; s_deque.pop(); } std::cout << std::endl; std::cout << "Is empty after all pops: " << s_deque.empty() << std::endl; // 1 (true) // 使用 std::vector 作为底层容器 std::cout << "\n--- std::stack (with std::vector) 示例 ---" << std::endl; std::stack> s_vector; s_vector.push(1.1); s_vector.push(2.2); std::cout << "Top element (vector-based): " << s_vector.top() << std::endl; // 2.2 s_vector.pop(); std::cout << "Top element after pop (vector-based): " << s_vector.top() << std::endl; // 1.1 // 使用 std::list 作为底层容器 std::cout << "\n--- std::stack (with std::list) 示例 ---" << std::endl; std::stack> s_list; s_list.push('a'); s_list.push('b'); std::cout << "Top element (list-based): " << s_list.top() << std::endl; // 'b' s_list.pop(); std::cout << "Top element after pop (list-based): " << s_list.top() << std::endl; // 'a' std::cout << std::endl; }
std::stack 的操作及其复杂度 (通常)：

push(element): 通常 (均摊，取决于底层容器的 push_back 或 push_front)
pop(): 通常 (取决于底层容器的 pop_back 或 pop_front)
top(): (取决于底层容器的 back 或 front)
empty():
size():

std::stack: 强烈推荐在大多数情况下使用。它简单、安全，并且是标准的一部分。
基于数组的实现：
优点：对于固定且较小的最大容量，内存是连续的，可能具有较好的缓存性能。
缺点：容量固定（静态数组）或动态扩容有开销。
适用场景：教学目的，或者在嵌入式系统等对内存控制有严格要求的环境中，且最大容量已知。
基于链表的实现：
优点：真正的动态大小，插入删除操作稳定。
缺点：每个元素有额外的指针开销，内存不连续。
适用场景：教学目的，或者当栈的大小非常动态且难以预估时。

理解 LIFO 原则。
掌握基本操作： push, pop, top, empty, size 的含义和 C++ 中的对应。
熟悉 C++ 中的实现方式：数组（静态/动态）、链表、std::stack。
时间与空间复杂度：能够分析各种操作在不同实现下的复杂度。
应用场景：能够举例说明栈在实际问题中的应用，并理解其工作原理。
边界条件处理：
对空栈进行 pop 或 top 操作（如何处理，如抛出异常）。
对满栈（数组实现）进行 push 操作。
std::stack 的用法：知道它是容器适配器，默认底层容器，以及如何指定其他底层容器。
资源管理：在链表或动态数组实现中，注意内存的分配 (new) 和释放 (delete/delete[])，防止内存泄漏（特别是在析构函数和 pop 中）。

手动实现栈：用静态数组、动态数组（带扩容）和链表分别实现一个栈类。
栈的应用编程：
用栈实现有效的括号匹配检查器。
用栈实现后缀表达式（逆波兰表达式）求值。
用栈实现中缀表达式到后缀表达式的转换。
使用两个栈模拟一个队列。
设计一个栈，除了基本操作外，还支持一个 getMin() 方法，该方法可以在时间内返回栈中的最小元素。
用栈实现递归函数的非递归版本（例如斐波那契数列或树的遍历）。
分析问题：判断某个问题是否适合用栈来解决，并说明理由。
// Main function to run test cases int main() { testArrayStack(); testLinkedStack(); testStdStack(); return 0; }
(请将上述 main 函数与之前的类定义和测试函数放在同一个 .C++ 文件中编译运行，或者分别编译链接。)
祝你复习顺利，考试取得好成绩！

中序线索链表

Wed, 14 May 2025 07:47:19 GMT

好的，我们来调整一下关于“标签”的约定，并更新解释和 C++ 代码。
新的约定：

Tag = 0: 表示对应的指针是一个线索 (Thread)。
lTag = 0: leftChild 指向中序前驱。
rTag = 0: rightChild 指向中序后继。
Tag = 1: 表示对应的指针指向一个子节点 (Child)。
lTag = 1: leftChild 指向左子节点。
rTag = 1: rightChild 指向右子节点。
我们将使用 bool类型的标签，其中 false 代表 0 (线索)，true 代表 1 (子节点)。

（这部分与之前相同，保持不变）

二叉树 (Binary Tree): 一种树形数据结构，每个节点最多有两个子节点，分别称为左子节点 (left child) 和右子节点 (right child)。
中序遍历 (Inorder Traversal): 遍历二叉树的一种方式，顺序是：左子树 -> 根节点 -> 右子树。

（这部分与之前相同，保持不变）
利用空指针域，存放指向该节点在中序遍历序列中的前驱或后继节点的指针，从而优化遍历。

中序线索链表是一种特殊的二叉树，它通过修改节点的结构并利用空指针域来存储中序遍历的前驱和后继信息。

如果一个节点的左子节点指针为空 (nullptr)，则让它指向该节点的中序前驱，并且其左标志位 lTag 设为 0 (表示线索)。
如果一个节点的右子节点指针为空 (nullptr)，则让它指向该节点的中序后继，并且其右标志位 rTag 设为 0 (表示线索)。
如果指针指向的是实际的子节点：

左指针指向左子节点，则 lTag 设为 1。
右指针指向右子节点，则 rTag 设为 1。

template struct ThreadNode { T data; ThreadNode* leftChild; ThreadNode* rightChild; bool lTag; // false (0) if leftChild is a thread, true (1) if it's a child bool rTag; // false (0) if rightChild is a thread, true (1) if it's a child // Constructor: When a node is created, its links are initially not threads pointing to // specific predecessors/successors. They are either null or will point to children. // So, we initialize tags to 'true' (1), indicating they are (or would be) child links. // The threading process will change these to 'false' (0) if they become threads. ThreadNode(T val) : data(val), leftChild(nullptr), rightChild(nullptr), lTag(true), rTag(true) {} };

lTag 为 false (0) 表示 leftChild 是指向中序前驱的线索。
lTag 为 true (1) 表示 leftChild 是指向左子节点的指针。
rTag 为 false (0) 表示 rightChild 是指向中序后继的线索。
rTag 为 true (1) 表示 rightChild 是指向右子节点的指针。

在对普通二叉树进行中序遍历的过程中，设置线索和相应的标志位。指针 pre 指向当前访问节点在中序遍历中的前一个节点。
线索化过程 (inorderThreading):

中序遍历左子树（如果左指针是子节点链接，即 lTag == 1）。
处理当前节点 current:
处理左线索: 如果 current 没有左子节点 (current->leftChild == nullptr)，则将其左指针指向 pre，并设置 current->lTag = false (0, 表示线索)。如果它有左子节点，则 current->lTag = true (1, 表示子节点)。
处理右线索: 如果 pre 不为空且 pre 没有右子节点 (pre->rightChild == nullptr)，则将 pre 的右指针指向 current，并设置 pre->rTag = false (0, 表示线索)。如果 pre 有右子节点，则 pre->rTag = true (1, 表示子节点)。
更新 pre = current。
中序遍历右子树（如果右指针是子节点链接，即 rTag == 1）。
C++ 实现线索化 (按新约定):
template ThreadNode* pre_threading_ptr = nullptr; //全局或成员变量 template void buildInorderThreadsRecursive(ThreadNode* current) { if (current == nullptr) { return; } // 1. Traverse left subtree if it's a child link // For initial threading of an unthreaded tree, we can assume lTag is true if leftChild is not null. // The check current->lTag is more for robustness or re-threading. // A simpler initial threading just calls: buildInorderThreadsRecursive(current->leftChild); // Here, we'll assume tags might have been set by constructor or previous operations. if (current->lTag == true) { // If true (1), it's a child link buildInorderThreadsRecursive(current->leftChild); } // 2. Process current node // Set left thread for 'current' if it has no left child if (current->leftChild == nullptr) { current->lTag = false; // false (0) means THREAD current->leftChild = pre_threading_ptr; } else { current->lTag = true; // true (1) means CHILD (already was or should be) } // Set right thread for 'pre_threading_ptr' if it has no right child if (pre_threading_ptr != nullptr) { if (pre_threading_ptr->rightChild == nullptr) { pre_threading_ptr->rTag = false; // false (0) means THREAD pre_threading_ptr->rightChild = current; } else { // If pre_threading_ptr->rightChild is not null, it's a child. // Its rTag should be true (1). This would typically be set when that child was processed // or by the constructor. pre_threading_ptr->rTag = true; // Ensure it's marked as CHILD } } pre_threading_ptr = current; // Update predecessor // 3. Traverse right subtree if it's a child link if (current->rTag == true) { // If true (1), it's a child link buildInorderThreadsRecursive(current->rightChild); } } template void createInorderThreadedTree(ThreadNode* root) { pre_threading_ptr = nullptr; if (root == nullptr) return; // Call the recursive helper buildInorderThreadsRecursive(root); // The last node visited (pre_threading_ptr) in inorder traversal might have its // rightChild as nullptr. This should be a thread to "nullptr" (no successor). if (pre_threading_ptr != nullptr && pre_threading_ptr->rightChild == nullptr) { pre_threading_ptr->rTag = false; // false (0) means THREAD (to nullptr) // pre_threading_ptr->rightChild remains nullptr } }

算法：

找到中序遍历的第一个节点：从根节点开始，只要当前节点的 lTag 为 1 (表示是子节点链接) 且左子节点存在，就一直向左下走。
循环访问节点并找到下一个节点：
访问当前节点 current。
找到中序后继：
如果 current->rTag == false (0, 右指针是线索)，那么 current->rightChild 就是中序后继。
如果 current->rTag == true (1, 右指针指向右子树)，那么中序后继是其右子树中“最左下”的节点（即从 current->rightChild 开始，只要 lTag == 1 就一直向左下走）。
C++ 实现中序遍历 (按新约定):
template void inorderTraversalThreaded(ThreadNode* root) { if (root == nullptr) { std::cout << "Tree is empty." << std::endl; return; } // 1. Find the first node in inorder traversal ThreadNode* current = root; // Go to the leftmost node, following child links while (current != nullptr && current->lTag == true) { // true (1) means it's a child link current = current->leftChild; } // Now 'current' is the leftmost node (first in inorder) // 2. Traverse the tree using threads std::cout << "Inorder traversal (threaded): "; while (current != nullptr) { std::cout << current->data << " "; // Move to the inorder successor if (current->rTag == false) { // false (0) means right child is a THREAD current = current->rightChild; // Successor is directly pointed by the thread } else { // rTag is true (1), meaning right child is a normal CHILD pointer // Successor is the leftmost node in the right subtree current = current->rightChild; // Go to the leftmost node of this right subtree while (current != nullptr && current->lTag == true) { // true (1) means child link current = current->leftChild; } } } std::cout << std::endl; }

#include // Node structure for Inorder Threaded Binary Tree template struct ThreadNode { T data; ThreadNode* leftChild; ThreadNode* rightChild; bool lTag; // false (0) if leftChild is a thread, true (1) if it's a child bool rTag; // false (0) if rightChild is a thread, true (1) if it's a child // Constructor: Initialize tags to true (1), indicating child links by default. // If a pointer (e.g. leftChild) remains nullptr, the threading process // will set its corresponding lTag to false (0). ThreadNode(T val) : data(val), leftChild(nullptr), rightChild(nullptr), lTag(true), rTag(true) {} }; template ThreadNode* pre_node_for_threading_g = nullptr; // Global for simplicity template void buildInorderThreadsRecursive(ThreadNode* current) { if (current == nullptr) { return; } // 1. Traverse left subtree if it's a child link. // Assumes that if current->leftChild is not NULL, current->lTag is initially true (1). if (current->lTag == true) { buildInorderThreadsRecursive(current->leftChild); } // 2. Process current node // If current node has no left child, its left pointer becomes a thread. if (current->leftChild == nullptr) { current->lTag = false; // false (0) indicates THREAD current->leftChild = pre_node_for_threading_g; } else { // If it has a left child, ensure lTag is true (1) for CHILD. // This is typically already true from constructor or previous steps if tree was built with tags. current->lTag = true; } // If the predecessor has no right child, its right pointer becomes a thread to current. if (pre_node_for_threading_g != nullptr) { if (pre_node_for_threading_g->rightChild == nullptr) { pre_node_for_threading_g->rTag = false; // false (0) indicates THREAD pre_node_for_threading_g->rightChild = current; } else { // If predecessor has a right child, ensure its rTag is true (1) for CHILD. pre_node_for_threading_g->rTag = true; } } pre_node_for_threading_g = current; // Update predecessor // 3. Traverse right subtree if it's a child link. if (current->rTag == true) { buildInorderThreadsRecursive(current->rightChild); } } template void createInorderThreadedTree(ThreadNode* root) { pre_node_for_threading_g = nullptr; if (root == nullptr) return; buildInorderThreadsRecursive(root); // The last node visited (pre_node_for_threading_g) should have its right thread // point to nullptr (or a header node in more complex setups). // Its rTag should be false (0) if rightChild is indeed a thread (even to nullptr). if (pre_node_for_threading_g != nullptr && pre_node_for_threading_g->rightChild == nullptr) { pre_node_for_threading_g->rTag = false; // false (0) means THREAD to nullptr } } template void inorderTraversalThreaded(ThreadNode* root) { if (root == nullptr) { std::cout << "Tree is empty." << std::endl; return; } ThreadNode* current = root; // Find the first node (leftmost) while (current != nullptr && current->lTag == true) { // true (1) means CHILD link current = current->leftChild; } std::cout << "Inorder Traversal (Threaded): "; while (current != nullptr) { std::cout << current->data << " "; if (current->rTag == false) { // false (0) means rChild is THREAD current = current->rightChild; } else { // rTag is true (1), rChild is a CHILD current = current->rightChild; // Go to right child // Then find the leftmost node in this right subtree while (current != nullptr && current->lTag == true) { // true (1) means CHILD link current = current->leftChild; } } } std::cout << std::endl; } int main() { // Construct a sample binary tree. Tags are initially true (1) by constructor. // 4 // / \ // 2 5 // / \ \ // 1 3 6 // Inorder: 1, 2, 3, 4, 5, 6 ThreadNode* root = new ThreadNode(4); root->leftChild = new ThreadNode(2); // root->lTag remains true (1) root->rightChild = new ThreadNode(5); // root->rTag remains true (1) root->leftChild->leftChild = new ThreadNode(1); root->leftChild->rightChild = new ThreadNode(3); root->rightChild->rightChild = new ThreadNode(6); std::cout << "Original tree structure created. Tags default to 1 (child link potential)." << std::endl; createInorderThreadedTree(root); std::cout << "Tree has been threaded." << std::endl; inorderTraversalThreaded(root); // Expected output: 1 2 3 4 5 6 // Manually check some threads for node '3' (inorder: 1, 2, *3*, 4, 5, 6) // Predecessor of 3 is 2. Successor of 3 is 4. // Node 3's leftChild will be a thread to 2 (lTag=0). // Node 3's rightChild will be a thread to 4 (rTag=0). ThreadNode* node3 = root->leftChild->rightChild; // This is node 3 if (node3->lTag == false) { // lTag is 0 (false) for thread std::cout << "Node 3's left thread (lTag=0) points to: " << (node3->leftChild ? std::to_string(node3->leftChild->data) : "nullptr") << " (Expected: 2)" << std::endl; } else { std::cout << "Node 3's left is a child (lTag=1)." << std::endl; } if (node3->rTag == false) { // rTag is 0 (false) for thread std::cout << "Node 3's right thread (rTag=0) points to: " << (node3->rightChild ? std::to_string(node3->rightChild->data) : "nullptr") << " (Expected: 4)" << std::endl; } else { std::cout << "Node 3's right is a child (rTag=1)." << std::endl; } // Manually check threads for node '1' (inorder: *1*, 2, 3, 4, 5, 6) // Predecessor of 1 is null. Successor of 1 is 2. // Node 1's leftChild will be a thread to nullptr (lTag=0). // Node 1's rightChild will be a thread to 2 (rTag=0). ThreadNode* node1 = root->leftChild->leftChild; // This is node 1 if (node1->lTag == false) { std::cout << "Node 1's left thread (lTag=0) points to: " << (node1->leftChild ? std::to_string(node1->leftChild->data) : "nullptr") << " (Expected: nullptr for this simple setup)" << std::endl; } if (node1->rTag == false) { std::cout << "Node 1's right thread (rTag=0) points to: " << (node1->rightChild ? std::to_string(node1->rightChild->data) : "nullptr") << " (Expected: 2)" << std::endl; } // Memory cleanup would be needed here for a complete program // delete ... (complex for threaded trees, omitted for brevity) return 0; }

（这部分与之前相同，保持不变）
优点：

快速遍历。
查找前驱/后继容易。
空间利用。
缺点：

结构复杂。
插入和删除复杂。

核心目的：利用空指针，实现不需要栈/递归的快速中序遍历，并方便查找中序前驱/后继。
节点变化：增加 lTag 和 rTag (布尔型)。
leftChild 指针：
lTag = false (0): 指向中序前驱 (是线索)。
lTag = true (1): 指向左子节点。
rightChild 指针：
rTag = false (0): 指向中序后继 (是线索)。
rTag = true (1): 指向右子节点。
线索化：在中序遍历过程中，用一个 pre 指针记录前驱，建立当前节点与 pre 之间的线索，并正确设置 Tag 为 false (0)。如果指针是指向子节点的，则 Tag 为 true (1)。
遍历：找到第一个节点（一直沿 lTag=true (1) 的左子链走到底），然后利用 rTag=false (0) 的右线索或右子树 (rTag=true (1)) 的最左节点不断找后继。
这个版本应该符合你对标签值 0 和 1 的具体要求了。祝你期末备考顺利！

Bootstrap

Fri, 02 May 2025 16:55:31 GMT

The bootstrap method estimates standard errors and computes confidence intervals for a statistic . It aims to estimate , the variance of , which depends on the unknown distribution function . The method involves two steps:

Step 1: Estimate with .
Step 2: Approximate using simulation.
For , Step 1 gives . In more complex cases, Step 2 is needed to approximate because a simple formula cannot be derived.

Theorem (Law of Large Numbers). Let be i.i.d. with and . Let . Then, for every , as . We say that converges in probability to and write .
Let i.i.d. then by law of large numbers, . Let , then

Hence, we can use the sample variance of the simulated values to approximate .

We can approximate by simulation. To simulate from the distribution of when the data are assumed to have distribution , we simulate from and then compute . Drawing an observation from is equivalent to drawing one point at random from the original data set. Thus, to simulate it suffices to draw observations with replacement from .
Bootstrap Variance Estimation

Draw .
Compute .
Repeat steps 1 and times, to get .
Let

There are three methods to construct bootstrap confidence intervals.
See All of Statistic p 111
主要通过程序计算.

chap1-2

Fri, 02 May 2025 16:55:31 GMT

Section 3 introduces the fundamental definitions of probability functions on sets endowed with a -algebra structure. Key concepts discussed include the interchange of and , properties of disjoint unions (), and DeMorgan's laws.
Section 4 elaborates on conditional probability and independence, presenting them as generalizations of probability theory with broad applicability. The conditional probability is defined computationally as , while independence is defined as , implying . The section concludes with a discussion of simulation methods for approximating probabilities in complex real-world scenarios, along with an introduction to error estimation, which is further explored in Chapter 4 in the context of confidence intervals.
Section 5 introduces the concept of random variables as functions mapping from a sample space (equipped with a sigma algebra) to . It distinguishes between the probability mass function (pmf) for discrete distributions and the probability density function (pdf) for continuous distributions, denoted by . The cumulative distribution function (cdf) is defined as , where represents the sample space. The notion of equality in distribution is defined as if and only if . However, it is noted that does not necessarily imply (e.g., ). The section also outlines the properties of cdfs, including being nondecreasing, having a lower limit of 0, an upper limit of 1, and being right continuous. Additionally, it establishes that .
Section 6 focuses on discrete random variables, whose sample space is at most countable, and explores transformations between random variables, i.e., . In the case where is one-to-one, the pmf of is given by ( and are discrete). When is piecewise one-to-one, the definition of requires a case-by-case analysis.
Section 7 introduces continuous random variables, where the cdf is given by . The quantile of order of the distribution of is defined as a value such that and , also known as the () th percentile of . It is noted that the quantile may not be unique and that . For transformations, the pdf of is given by for in , the support of . For continuous random vectors, refers to the Jacobian of the transformation.
Section 8 discusses the expectation , and .
Section 9 covers specific expectations, including the mean , the variance , and the moment generating function (mgf) (if it exists for ). The derivatives of are explored, noting that , , and is called the th moment of .
Section 10 presents important inequalities: the existence of implies the existence of for . Markov's Inequality is stated as for nonnegative and positive . Chebyshev's Inequality is given as: for with , for we have . Jensen's Inequality states that for a convex function (), we have .
Chapter 2 delves into multivariate distributions. Section 1 introduces the basic definitions and calculation techniques for two random variables. Section 2 presents transformations and the moment generating function (mgf) technique, which is effective for linear functions of random variables. If then .
Section 3 introduces conditional distributions and expectations. The conditional pmf is given by , and the conditional pdf is . The probability and . The conditional expectation of ( is a function of ) is . The conditional variance is . Conditional expectation satisfies , which can serve as an alternative definition of conditional expectation.
Section 4 introduces the concept of independent random variables. If have the joint pdf , then are independent if and only if can be written as , where and are nonnegative functions. For independent , we have . If is the joint mgf, then are independent if and only if .
Section 5 introduces the correlation coefficient , where the covariance . Independence implies that random variables are uncorrelated.
Section 6 extends the discussion to several random variables. are said to be mutually independent if and only if . Consequently, . Let be a -dimensional random vector, . If is an matrix of random variables then the mean of is , the variance-covariance matrix is (), and and . All variance-covariance matrices are positive semi-definite, because where is a random variable.
Section 7 discusses transformations for several random variables. Section 8 addresses the linear combinations of random variables. If are iid, then these random variables constitute a random sample of size from that common distribution. The sample variance is defined by .

chap3

Sun, 18 May 2025 15:50:28 GMT

, has the Bernouli distribution, with pmf:

.
For , the pmf is

The mgf of a binomial distribution is

Then .

A distribution with a pmf of the form is called a negative binomial distribution, which means the probability that exactly successes in the first trials and a success on the th trial. Its mgf is , for .
If , then and . We say that has a geometric distribution.

, , we say has a multinomial distribution with parameters and if

The conditional pmf is interesting.

We say that has a hypergeometric distribution with parameters . The mean of is and .

if then . (check by mgf)

The definition of Gamma function is

has the pmf

is also called the exponential distribution with parameter . for .
The characteristic function of is

If then .

When , then the exponential distribution has mean and variance .
Remark. Some texts define the PDF of as , while others use .

Consider the special case for given . has the pdf

and the mgf

is said to have a chi-square distribution ( -distribution) and any of this form is called a chi-square pdf. Because -distribution is a subfamily of -distribution, we have .
Theorem. Let have a distribution. If , then exists and it is given by
Proof.
Note that

Make the change of variable in the above integral. This results in

This simplifies to the desired result provided that .
□

Let , , then has the beta distribution, with pdf:

And

Let have the standard normal distribution then . is the CDF of .
Definition (Normal Distribution). We say a random variable has a normal distribution if its pdf is
The parameters and are the mean and variance of , respectively.
Theorem. If then . Particularly,

Calculate the CDF of .
Proof. Because , where is , the cdf for is, for ,

That is,

And

If we change the variable of integration by writing , then

Hence the pdf of the continuous-type random variable is

Since is a pdf

hence, it must be that and thus is .
□
Theorem. are iid, then .
Prove by the characteristic function.

We say that follows a bivariate normal distribution if its pdf is given by

Where

and , for , and satisfies . Clearly, this function is positive everywhere in . As we show in Section 3.5.2, it is a pdf with the mgf given by:

Then

Consider the random vector where are iid random variables. Then the density of is

We abbreviate this by saying that .
For the general case, suppose is an , symmetric and positive semi-definite matrix. The spectral decomposition is , where , are the eigenvalues of and the columns of , are the corresponding eigenvectors. The matrix is orthogonal, i.e., . .
Definition (Multivariate Normal). Say an -dimensional random vector has a multivariate normal distribution if its mgf is
where is a symmetric, positive semi-definite matrix and . We abbreviate this by saying that has a distribution.
If then . And we have

Further, the mgf of is given by

Theorem. If and is positive definite then .
Proof.
It's a consequence of Theorem 4.
□
Theorem (Linearity). Suppose . Let where is and . Then .
Theorem. Suppose . Let where is and . Then .
Proof.
Calculate the mgf of .
□
Particularly, suppose

Then .
Theorem (decomposite to independent subvector). , , then and are independent iff , i.e.
Proof.
The joint mgf is

The product of their marginal mgfs is

Then iff .
□
Theorem (conditional distribution). are defined as above, then the conditional distribution of is

Proof.
See proof from Hogg p.204.
□

indepedent; ; has the pdf

By Theorem 2, for ,

Thus

We will show the technique to compute the pdf of .
Let indepedent. Then the joint pdf of , say is

Define a new random variable by

The transformation technique used to obtain the pdf of . The equations

define a transformation from to . Calculate the Jacobian of the transformation

Accordingly, the joint pdf of and is given by

where .
The marginal pdf of is then

In this integral let and it is seen that

The distribution of the random variable is usually called a -distribution. , with max at . As , the -distribution converges to the .

Let , . The joint pdf of and is then

We define the new random variable

and we propose finding the pdf of . The equations

By calculations, we have

The distribution of this random variable is usually called an -distribution, .
By Theorem 2, for , ,

In particular,

Our final note concerns an important result for the later chapters on inference for normal random variables. It is a corollary to the -distribution derived bove and is often referred to as Student's Theorem.
Theorem (Student's theorem). Let be iid random variables each having a normal distribution with mean and variance . Define the random variables Then
has a distribution.
and are independent.
has a distribution.
The random variable has a Student -distribution with degrees of freedom.
Proof.
Note that we have proved part (a) in Corollary 3.4.1. Let . Because are iid random variables, has a multivariate normal distribution , where denotes a vector whose components are all 1 . Let . Note that . Define the random vector by . Consider the following transformation:

Because is a linear transformation of multivariate normal random vector, by Theorem 3.5.2 it has a multivariate normal distribution with mean

where denotes a vector whose components are all 0 , and covariance matrix

Because is the first component of , we can also obtain part (a) by Theorem 3.5.1. Next, because the covariances are is independent of . But . Hence, is independent of , also. Thus part (b) is true.
Consider the random variable

Each term in this sum is the square of a random variable and, hence, has a distribution (Theorem 3.4.1). Because the summands are independent, it follows from Corollary 3.3 .1 that is a random variable. Note the following identity:

By part (b), the two terms on the right side of the last equation are independent. Further, the second term is the square of a standard normal random variable and, hence, has a distribution. Taking mgfs of both sides, we have

Solving for the mgf of on the right side we obtain part (c). Finally, part (d) follows immediately from parts (a)-(c) upon writing , (3.6.9), as

□

For distributions with pdfs , supports , means and variances , with positive mixing probabilities where . Let and consider the function

chap4

Fri, 02 May 2025 16:55:31 GMT

We continue the concepts of samples and statistics in chapter 2 while introducing the main tools of inference: confidence intervals and tests of hypotheses.
In a typical statistic problem, we have but don't know its pdf or pmf . There are two cases: completely unkown and known down to a parameter which may be a vector. Now we consider the second case, e.g. with unkown , with unknown .
idd is called a random sample of size , and a function of the sample is called a statistic. Once the sample is drawn, then is called the realization of , where and is the realization of the sample.

In chapter 6 and 7 we discuss several theories of estimation in general. We briefly discuss the maximum likelihood estimator (mle) and then use it to obtain point estimators. Our discussion is for the continuous case.
The information is involved in the likelihood function . An often-used estimate is the value of that provides a maximum of . If unique, this is called the maximum likelihood estimator (mle) and we denote it as . In practice, it's easier to work with , and solves .

Suppose with density , . The log of the likelihood function is given by
The first partial of the log-likelihood with respect to is
Setting this partial to 0 then we obtain the solution , thus the statistic is the mle of . Because we have hence is an unbiased estimator of .

Let be one of zero. Let , , denote the probability of success. Then the pmf of is
If is a random sample on , then
Taking logs, we have
The partial derivative of is
Thus , then is an unbiased estimator of .

Let then .
The two partial derivatives simplify to

Setting these to 0 and solving simultaneously, we see that the mles are
We kown that is unbiased estimator for and is an unbiased estimator of . Thus for the mle of , , which is a biased estimator of . Note that the bias of is which converges to 0 as . However is the preferred estimator of .

Let be iid with the uniform density; i.e.

Because is in the support, differentiation is not helpful here. The likelihood function can be written as

The function is a decreasing function of for all and is otherwise. So the maximum occurs at the smallest value that can assume; i.e. the mle is .

In this section we briefly discuss a histogram of the sample, which is an estimate of the pmf, or the pdf, , of .

Assume discrete with pmf . Let be a random sample on . Suppose the space of is finite, i.e. . An intuitive estimate of is the relative frequency of in the sample.

These estimators constitute the nonparametric estimate of the pmf . Because

is an unbiased estimator of .
Next suppose infinite, i.e.. In practice, we select a value and make the groupings

Let be the proportion of sample items that are greater than or equal to . Then the estimates form our estimate of .
A histogram is a barplot of versus . There are two cases to consider. For the first case, supppose the values represent qualitative categories, e.g. hair colors of a population of people. Such histograms are usually called bar charts. An example is helpful here.

For the second case, assume that the values in the space are ordinal in nature; i.e. the natural ordering of the is numerically meaningful. In this case, the usual histogram is an abutting bar chart with heights that are plotted in the natural order of the , as in the following example.

For this section, assume that the random sample is from a continuous random variable with continuous pdf . For an arbitrary but fixed point and a given , consider the interval . By the mean value theorem for integrals, we have for some , , that

Let the sample items fall in , which suggests the following nonparmetric extimate of at a given :

More formally, a nonparametric estimator of is

Then

Hence is approximately an unbiased estimator of the density .

Recall that the random variable of interest has density , where is unknown. In Section 4.1, we discussed estimating by a statistic , where is a sample from the distribution of . But how much did miss ? In this section, we embody this estimate of error in terms of a confidence interval, which we now formally define:

For , define for .
Motivated by the CLT, we have

A practical problem of interest is the comparison of two distributions, and . In this section, we compare the means of and , denoted by and . In particular, we obtain confidence intervals for the difference . Let be a random sample from the distribution of and from , and all independent of one another. Let and and , which is an unbiased estimator of . This difference is the numerator of the pivot random variable. By independence of the samples,

Let and be the sample variances. Then estimating the variances by the sample variances, consider the random variable

By CLT, this pivot variable has an approximate distribution. This leads to the approximate confidence interval for given by

where is the standard error of . This is a large sample confidence interval for .

Omitted...

Omitted...

Definition (order statistic).

Note that the support of an be partitioned into mutually disjoint sets that map onto the support of namely, . One of these sets is , and the others can be found by permuting the s in all possible way, thus the Jacobian equal to 1.

Omitted...

continuous rv with cdf . For , define the th distribution percentile to be , where . Let be the order statistics. Then

Then serves as a confidence interval for , the quantile of order .

See All of statistics Chapter 10.

We partition the parameter space into two disjoint sets and and we wish to test

We call the null hypothesis and the alternative hypothesis. Let be a r.v. and be the range of . To test a hypothesis, we aim to find the rejection region(critical region) . If we reject , otherwise retain .
Definition (critical region (rejection region)). A test of versus is based on a subset of . This set is called the critical region (rejection region) and its corresponding decision rule (test) is

Usually, the rejection region (critical region ) is of the form

where is a test statistic and is called a critical value.
Definition (power function, size (significance level), level). The power function of a test with rejection region is defined by
The size of a test is defined to be
A test is said to have level[1] if its size is less than or equal to .
Definition (type I error, type II error). Rejecting when is true is called a type I error. Retaining when is true is called a type II error.
Definition (two-side test, one-side test). A test of the form
is called a two-side test. A test of the form
is called a one-side test. The most common tests are two-sided.

Next we consider four widely used tests: the Wald test, the test, the permutation test, and the likelihood ratio test.

Let be a scalar parameter, let be an estimate of and let be the estimated standard error of .
Definition (The Wald Test). Consider testing . Assume that the estimator is asymptotically Normal, i.e.
The size Wald test is: reject when .
means that for ,

Thus

where .
Remark. An alternative version of the Wald test statistic is where is the standard error computed at . Both versions of the test is valid.
Let us consider the power of the Wald test when the null hypothesis is false.

Theorem (The rejection region of size Wald test). The size Wald test rejects versus iff where
Thus, testing the typothesis is equivalent to checking whether the null value is in the confidence interval.
Remark. When we reject we often say that the result is statistically significant.

For every we have a size test with rejection region . Then

-value is the smallest level at which we can reject .
Remark (Warning!). Do not confuse the -value with . The -value is not the probability that the null hypothesis is true.
Suppose that the size test is of the form

Then

where is the observed value of .
The -value is the probability (under ) of observing a value of the test statistic the same as or more extreme than what was actually observed.

The Pearson's test is used for multinomial data. If has a multinomial distribution, then the MLE of is
need not to be independent!
Definition (Pearson's statistic). Pearson's statistic is
where is the expected value of under .
Under , . Hence the test: reject if has asymptotic level . The p-value is where is the observed value of the test statistic.

The definition is useless.↩︎

chap5

Fri, 02 May 2025 16:55:31 GMT

In this section, we formalize the notion of a sequence of random variables getting "close" to another random variable as . We say converges in probability to , denoted as , if for all .
One way to demonstrate convergence in probability is to use Chebyshev's Theorem: for a random variable with variance , and for any , we have .
Let be a sequence of independent and identically distributed (i.i.d.) random variables with mean and variance . Let . Then .
Proof.
The mean and variance of are and , respectively. By Chebyshev's Theorem, for all , we have

□
The Weak Law of Large Numbers (WLLN) implies that all the mass of the distribution of converges to . In a sense, for large , is close to . But how close is it? For example, if we were to estimate by , what can we say about the error of estimation? Actually, a Strong Law of Large Numbers (SLLN) can be proved. Moreover, we can weaken the hypothesis of the WLLN to require only that the are i.i.d. with finite mean . Thus, the SLLN is a first moment theorem, while the WLLN requires the existence of the second moment.
Next, we list several theorems concerning convergence in probability, which is closed under linearity.
Theorem. Suppose and . Then .
Proof.
Using the triangle inequality, we have

□
Theorem. For any constant , if , then .
Theorem. If and the real function is continuous at , then . (Note: the constant is necessary.)
Proof.
Since is continuous at , for all , there exists such that for . Thus, implies . Substituting for , we obtain

□
This theorem gives us many useful results, e.g., if , then , (for ), and (for ).
Theorem. Suppose and . Then .
Proof.
We can write

Since and , we have and . Therefore,

Alternatively, .
□

In many situations we can show statistic convergence without the distribution function of the statistic. But how close is the statistic to the estimator?

Theorem (Stirling's formula).

Convergence in distribution is weaker than convergence in probability. Thus convergence in distribution is often called weak convergence.
Theorem. If converges to in probability, then converges to in distribution.
^4b6469
Proof.
Let be a point of continuity of . For every ,

Basd on this inequality and the fact that we see that

To get a lower bound, we proceed similarly with the complement to show that

Hence

Using a relationship between and , it follows that

Letting gives us the desired result.
□

Theorem. If constant, then .
Let be given. Then

The converse is not true.
Theorem. then .
The proof is similar to the above theorem.
We often use this result as follows. Suppose it is difficult to show that converges to in distributino, but it is easy to show that converges in distribution to and that converges to 0 in probability. Hence by this last theorem. as desired.
The next two theorems state general results.
Theorem. and continuous on the support of . Then .
An often-used application of this theorem occurs when we have a sequence of random variables which converges in distribution to a standard normal random variable . Because the distribution of is , it follows by the above theorem that converges in distribution to a distribution.
Theorem (Slutsky's theorem). If and then .
The proof is similar to ^4b6469

Theorem.
One way of thinking of a sequence that is bounded in probability (or one that is converging to a random variable in distribution) is that the probability mass of is not escaping to . At times we can use boundedness in probability instead of convergence in distribution. A property we will need later is given in the following theorem:
Theorem. Let be a sequence of random variables bounded in probability and let be a sequence of random variables that converges to 0 in probability. Then

Let is given. Choose and an integer such that

Then

Remark. It's similar to .

The -method is employed to determine the asymptotic distribution of a function of a random variable, given the distribution of the random variable itself. This is analogous to problems discussed in previous chapters, such as Theorem 9 and Theorem 10 (Slutsky's theorem).
Little-o Notation
The notation signifies that converges to 0 in probability relative to , formally:

Big-O Notation
The notation indicates that is bounded in probability as .
Theorem (Theorem 5.2.8. ). If is a sequence of random variables that is bounded in probability and , then as .
Proof of Theorem 5.2.8
Let . Then, there exist and such that for , . Since , we have:

Theorem (Theorem 5.2.9 (-Method)). Let be a sequence of random variables such that If is differentiable at and , then
Proof of Theorem 5.2.9
Since , it follows that

Because , it implies that is bounded in probability. Therefore, by Theorem 14 (Theorem 5.2.8.) , in probability. Hence, the result follows.

It's difficult to obtain , but quite easier from the mgf that corresponds to the cdf .

Prove by characteristic function .
The Central Limit Theorem is saying that when is large, fixed positive integer, the random variable has an approximate normal distribution with mean and variance . We can equivalently state the conclusion of the Central Limit Theorem as

This is often a convenient formulation to use.
Remark. We know that and have approximately normal distributions, provided that is large enough. Later, we find that other statistics also have approximate normal distributions, and this is the reason that the normal distribution is so improtant to statisticians. That is, while not many underlying distributions are normal, the distributions of statistics calculated from random samples arising from these distributions are often close to being normal.
We can combine -method with Central Limit Theorem. Assume that is a random sample on which has finite mean and variance . Then by the Central Limit Theorem, we have

Hence by the -method, we have

for a continuous transformation such that .

This section discusses asymptotic concepts for sequences of random vectors.
Definition (convergence in probability). Let be a sequence of -dimensional random vectors, and let be a random vector, all defined on the same sample space. We say that converges in probability to if for all . As in the univariate case, we write .
Theorem (Theorem 5.4.1). Let be a sequence of -dimensional random vectors, and let be a random vector, all defined on the same sample space. Then if and only if for all .
Based on Theorem 18 (Theorem 5.4.1), many theorems involving convergence in probability can be extended to the multivariate setting.
Let be a sequence of i.i.d. random vectors with common mean vector and variance-covariance matrix . Denote the vector of means by . By the Weak Law of Large Numbers, in probability for each . Hence, by Theorem 18 (Theorem 5.4.1), in probability.
Now consider the analog of the sample variances. Let . Define the sample variances and covariances by

If we define the matrix , then in probability.
Definition (convergence in distribution). Let be a sequence of random vectors with having distribution function , and let be a random vector with distribution function . Then converges in distribution to if for all points at which is continuous. We write .
Theorem. Let be a sequence of random vectors that converges in distribution to a random vector , and let be a function that is continuous on the support of . Then converges in distribution to .
Theorem. Let be a sequence of random vectors with having distribution function and moment generating function . Let be a random vector with distribution function and moment generating function . Then converges in distribution to if and only if, for some , for all such that .
Theorem (Multivariate Central Limit Theorem). Let be a sequence of i.i.d. random vectors with common mean vector and variance-covariance matrix which is positive definite. Assume that the common moment generating function exists in an open neighborhood of . Let Then converges in distribution to a distribution.
\begin{proof}
Let be a vector in the stipulated neighborhood of . The moment generating function of is

where . Note that are i.i.d. with mean 0 and variance . Hence, by the standard Central Limit Theorem,

Then is the MGF of evaluated at 1. Therefore, we must have

Because the last quantity is the moment generating function of a distribution, the result follows.
\end{proof}
Theorem (Theorem 5.4.5). Let be a sequence of -dimensional random vectors. Suppose . Let be an matrix of constants, and let be an -dimensional vector of constants. Then .
Theorem (Theorem 5.4.6). Let be a sequence of -dimensional random vectors. Suppose Let be a transformation such that and the matrix of partial derivatives, are continuous and do not vanish in a neighborhood of . Let at . Then

chap6

Mon, 05 May 2025 13:34:55 GMT

We need some regularity conditions.

Definition (Assumptions 6.1.1 (Regularity Conditions)). Regularity conditions (R0)-(R5) are
(R0) The cdfs are distinct; i.e., .
(R1) The pdfs have common support for all .
(R2) The point is an interior point in .
(R3) The pdf is twice differentiable as a function of .
(R4) The integral can be differentiated twice under the integral sign as a function of .
(R5) The pdf is three times differentiable as a function of . Further, for all , there exist a constant and a function such that with , for all and all in the support of .

Since

thus

Apply again,

We define the Fisher information:

The function is called the score function. Recall the calculation of MLE, the mle solves

Theorem (Rao-Cramér Lower Bound). Let be iid with common pdf for . Assume that the regularity conditions (R0)-(R4) hold. Let be a statistic with mean . Then
Corollary. Under the assumptions of Theorem 2 (Rao-Cramér Lower Bound) , if is an unbiased estimator of , so that , then the Rao-Cramér inequality becomes
Definition (Efficiency). The efficiency of that estimator is

Definition (Efficient Estimator). Let be an unbiased estimator of a parameter . The statistic is called an efficient estimator of iff

Theorem. Assume are iid with pdf for such that the regularity conditions (R0)-(R5) are satisfied. Suppose further that the Fisher information satisfies . Then any consistent sequence of solutions of the mle equations satisfies
Definition. Let be independent and identically distributed with probability density function . Suppose is an estimator of such that . Then
The asymptotic efficiency of is defined to be
The estimator is said to be asymptotically efficient if the ratio in part (a) is 1 .
Let be another estimator such that . Then the asymptotic relative efficiency of to is the reciprocal of the ratio of their respective asymptotic variances; i.e.,

See All of statistic.
Definition (likelihood ratio test). Consider testing The likelihood ratio statistic is where is the MLE and is the MLE when is restricted to lie in .
When , .[1] We have

The -value for the test is , where is the observed value. We reject at level if -value.
We reject at level when , where .
Consider another statistic

Under , , then we construct a test[2]

Note that under , .
The third test is called Rao's score test. The scores are

Define the statistic

Recall that

Assume that . Then the Fisher information is given by

Recall that

We also have

The diagonal entries of are

Similar to Corollary 3, if is an unbiased estimate of , then

We call it efficient if attaining the lower bound.
if is diagonal.

Theorem. Let be iid with pdf for . Assume the regularity conditions hold. Then
The likelihood equation, has a solution such that .
For any sequence that satisfies (1),
For , we have

where

Suppose are i.i.d. . In this case and . Then

Then

Let , then

Note that is unbiased, is biased.

Defined in Hogg↩︎
This test is often referred to as a Wald-type test.↩︎

chap7

Sun, 11 May 2025 16:31:45 GMT

This chapter focuses on finding "optimal" point estimators, particularly Minimum Variance Unbiased Estimators (MVUEs), by leveraging the concept of sufficiency.

Goal: Find estimators that are close to the true parameter value.
Properties Reviewed:
Consistency: Estimator converges in probability to the parameter as sample size increases.
Unbiasedness: Expected value of the estimator equals the parameter.
Minimum Variance Unbiased Estimator (MVUE):
Definition. An estimator is an MVUE of if:
It is unbiased: .
Its variance is less than or equal to the variance of every other unbiased estimator of .

Loss Functions and Risk:
Decision function[1] : is the function of the observed value of the statistic which is the point estimate of . One value of the decision function, say , is called a decision.
Loss Function : Measures the "cost" of estimating with . Common example: Squared Error Loss .
Risk Function : Expected value of the loss function, .
Minimizing risk uniformly is usually impossible.
Minimax Principle: Choose the decision rule that minimizes the maximum risk: for all other rules .
Likelihood Principle: If two experiments yield likelihood functions and that are proportional (), they contain the same information about , and the inference should be the same.

Idea: A statistic is sufficient if it captures all the information about contained in the sample. Knowing the value of is equivalent to knowing the original sample for the purpose of inference about .
Formal Definition:
Definition. A statistic with pdf/pmf is sufficient for if and only if the ratio does not depend on . (This means the conditional distribution of the sample given does not depend on ).

Neyman Factorization Theorem: A much easier way to find sufficient statistics.
Theorem (Neyman). is sufficient for if and only if the joint pdf/pmf can be factored as: where does not depend on . The function depends on the data only through the statistic .

Example (Normal Mean, Known Variance): Let with known. The joint pdf is: Here, and is the second term, which does not involve . Thus, is a sufficient statistic for .
Example (Poisson): Let . Joint pmf: Here and . Thus, is sufficient for .

Rao-Blackwell Theorem: Provides a method to improve any unbiased estimator by conditioning on a sufficient statistic.
Theorem (Rao-Blackwell). Let be a sufficient statistic for , and let be any unbiased estimator of . Define . Then:
is a statistic (does not depend on ).
(it's unbiased).
. Equality holds iff is already a function of .

Implication: To find an MVUE, we only need to consider functions of a sufficient statistic.
Sufficiency and MLEs:
Theorem. If a sufficient statistic exists and a unique Maximum Likelihood Estimator (MLE) exists, then must be a function of .

Often, the MLE is a good starting point to find an MVUE. It might be biased, but we can adjust it (if necessary) to make it unbiased, and it will still be a function of the sufficient statistic.
Example (Exponential): (mean ). Pdf is .
is sufficient.
MLE is .
(biased).
MVUE is .

Completeness: A property of the family of distributions of a statistic.
Definition. A family of pdfs/pmfs for a statistic is complete if for all implies (almost everywhere w.r.t. the distributions in the family).

Intuition: If a family is complete, the only unbiased estimator of 0 that is a function of is the function .
Lehmann-Scheffé Theorem: Connects sufficiency, completeness, and MVUEs.
Theorem (Lehmann-Scheffé). If is a complete sufficient statistic for , and is an unbiased estimator of , then is the unique MVUE of . (Uniqueness means any other MVUE must be equal to almost everywhere).

Complete Sufficient Statistic: A statistic that is both sufficient for and whose family of distributions is complete.
Example (Uniform): .
is sufficient.
The family of pdfs for is for . This family can be shown to be complete.
.
Therefore, is the unique MVUE of .

A large class of common distributions where finding complete sufficient statistics is straightforward.
Definition (Regular Exponential Class): A family is a regular exponential class if:
for .
The support does not depend on .
is a nontrivial continuous function of .
and are continuous (continuous case) or is nontrivial (discrete case).
Key Result: For a random sample from a regular exponential class distribution:
Theorem. The statistic is a complete sufficient statistic for .

Implication: If you can write the pdf/pmf in the exponential form, you can immediately identify the complete sufficient statistic. Then, if you find any function of that statistic which is unbiased for , it is automatically the unique MVUE by Lehmann-Scheffé.
Example (Normal Mean, Known Variance): , known.
.
Here . So is complete sufficient.
Since , is the unique MVUE.

Often interested in estimating , not just .
If is a complete sufficient statistic for , we seek a function such that . If found, is the unique MVUE of .
Techniques:
Find an unbiased estimator (perhaps not a function of ) and compute . (Rao-Blackwell).
Guess a function (often related to the MLE of ) and check if . Adjust if necessary.
Example (Bernoulli Variance): . Estimate .
is complete sufficient. is MVUE of .
MLE of is .
.
MVUE of is .
Bootstrap Standard Errors: Can compute standard errors for MVUEs using bootstrap resampling (nonparametric or parametric).

Concepts extend to vector parameters .
Joint Sufficiency: A vector of statistics is jointly sufficient if the definition or factorization theorem holds.
Completeness: for all implies .
Rao-Blackwell & Lehmann-Scheffé: Apply similarly. If is jointly complete sufficient for , and is unbiased for , then is the unique MVUE of .
Exponential Class (Multiparameter): If regular, then for are jointly complete sufficient statistics for .
Example (Normal): .
.
, .
, are jointly complete sufficient.
Equivalently, and are jointly complete sufficient.
Since and , and are the unique MVUEs of and .

Minimal Sufficient Statistic: A sufficient statistic that is a function of every other sufficient statistic. It achieves the maximum possible data reduction while retaining sufficiency.
Often, complete sufficient statistics are minimal.
If an MLE is itself sufficient, it is often minimal.
Ancillary Statistic: A statistic whose distribution does not depend on the parameter .
Location-Invariant: If , a statistic is location-invariant if . Such statistics are ancillary for . Ex: , Range.
Scale-Invariant: If , a statistic is scale-invariant if for . Such statistics are ancillary for . Ex: .
Location- and Scale-Invariant: Combination of the above. Ex: .

Key Relationship (Basu's Theorem):
Theorem (Basu). If is a complete sufficient statistic for , then is independent of any ancillary statistic .

The converse also holds: If is sufficient and is independent of an ancillary statistic , then must be complete.
This provides a powerful tool for proving independence.
Example (Normal and ):
Let the model be .
If is known, is complete sufficient for . is location-invariant, hence ancillary for . By Basu's Theorem, and are independent.
If is known, is complete sufficient for . is not ancillary for (its distribution depends on ). We cannot use Basu's theorem here (though they are still independent).
If Sufficiency is Not Complete: An ancillary statistic might still provide information about the precision of an estimator based on the sufficient statistic (e.g., Example 7.9.5).

or decision rule↩︎

Convergence of Random Variables

Fri, 02 May 2025 16:55:31 GMT

See All of Statistic Chapter 5.

We say that converges almost surely to , written if

We say that converges in to , written if

hw1

Fri, 02 May 2025 16:55:31 GMT

(a)

Then , is the median. Or say that .
(b)

Then .
(c)

Then .

Then .

Characteristic function , by the inverse formula

Then is discrete random variable with .
The mgf of is

hw2

Fri, 02 May 2025 16:55:31 GMT

Let then .

The pdf of is

Then

The pdf of is

The pdf of is

Hence has an -distribution with parameters and .

Let

Then we have the transformation from to . The Jacobian is

Then the joint distribution of is

The distribution of is

Hence has an -distribution with parameter .

Exercise. is a random sample of (normal is not required) with , compute the moments of sample mean and sample variance and .

Exercise. is a random sample from , compute , where is the sample deviation.

By student's theorem, . Then .

Exercise. Suppose , and two samples are independent. Then
(1) ;
(2) Furthermore, if , then . where is called mixed samples Variance.
(1) Let , , then . Let

The equation

defines a transformation from to . Calculate the Jacobian of this transformation

The pdf of and are

The joint pdf of is

Then

Therefore .
(2) , . . Then

By student's theorem, and are independent. Thus are independent. Thus are independent.

Therefore

where and .

hw3

Fri, 02 May 2025 16:55:31 GMT

5.1 7 9 5.2 7 18 5.3 11 12

For

If then thus .
If then

Then for any , we have

Therefore , i.e. in probability.

Since , is not an unbiased estimator of . But has the mean thus is an unbiased estimator of .

To assure the existence, we use the characteristic functions.

Denote .
By the inverse formula

Then

Denote , then

Thus , which means

Let , using the -method, we have

Therefore the limiting distribution of is .

We know that

Let then . Using the -method, we have

Therefore

We know that

Let then . Using the -method, we have

Therefore

which means the variance of is essentially free of .

hw4

Fri, 02 May 2025 16:55:31 GMT

• Section 4.1: 2, 6, 8 • Section 4.2: 3, 9, 10

(a)

The normal probability model is credible.
(b)
For normal distributions, the mles are

Locate on the plot:

Overlay the normal pdf with these estimates:

(c)

where is the number of successes (pitchers weighing over 215 pounds) and is the probability of success (a pitcher weighing over 215 pounds).

Let then

(d)

where is the CDF of the standard normal distribution . Then

The estimator

where

Its variance is

Its mgf is

For Poisson distribution with mean , we have

The likelihood function is

The first parital of the log-likelihood with respect to is

Setting this partial to 0 and solving for , we obtain the solution , thus the mle of is

The mle of the pmf is

Proof.
(a) The characteristic function of is

Then the characteristic function of is

Then .
(b) define to be the upper critical point of a -distribution with degrees of freedom, i.e. . Using a simple algebraic derivation, we obtain

Then an approximate confidence interval for is given by

(c)

□

(a)

, and means , where . We obtain

Then a confidence interval for is given by

(b)
By student's theorem, the rv has a -distribution with degrees of freedom. Define

We obtain

Then a confidence interval for is given by

(c)
Omitted

Then

Then

By the defintion of -distribution, the rv

has a -distribution with . Thus

Denote to be . Then

Then the confidence interval for is given by

Then

hw5

Fri, 02 May 2025 16:55:31 GMT

• Section 4.2: 17, 18, 19, 23, 25, 27(a)(c) • Section 4.4: 6(a), 7, 9, 26, 30, 31
Exercise.
The statistic model is

Let are random sample of distribution . To calculate the mle of ,

Then

Thus the mle of is

Thus . . When is large,

Let where is the standard normal distribution. Then the confidence interval of is (we replace by its estimator )

Put then

Exercise.
(a) Trivial.
(b) .

(c)
A confidence interval for is

Exercise.

Denote , then

Thus .

Let , then the confidence interval for is

Exercise.
Let be a random sample of , be of . Then . Therefore . Let and denote the realized values of the statistics and . Let then the confidence interval of is

Exercise.
Denote , then

Thus

Let

which is an unbiased estimator of . By Student's theorem,

Then

And

is an unbiased estimator of . Since is independent of and is independent of by Student's theorem, is independent of and . Thus

Denote , , then the confidence interval of is

We know that thus

i.e.

Exercise ((a)(c)).
(a)

Then

(c)
Trivial.
Exercise.
Let denote the 3 order statistics based on the random sample , then the joint pdf of is given by

Then

Exercise.

That is

Exercise.

Then

Exercise.

Exercise.

For binomial distribution , thus

is a confidence interval for .
The interval means that the median number has the probability lying the interval .
Exercise.
(a)

Then

(b)
and , then

Then the confidence interval for can be

hw6

Fri, 02 May 2025 16:55:31 GMT

• Section 4.5: 3, 5, 8, 11 • Section 4.6: 7, 8, 9, 10
Exercise.

The power function is

Exercise.
The rejection set is

The power of the test is

The significance level of the test is

When is false, , then

Exercise.

The power function is

Let and then

We have

Exercise.
(1) Clearly, , then

Let then .
(2) the power function is

Exercise.
(a) Let's test the null hypothesis that . Write this as , where . The nonparametric plug-in estimate of is , with estimated standard error

By Student's theorem, the random variable

Let

The critical region of is

(b) Let , then

，因此在显著性水平下有理由拒绝零假设，即休斯顿市中心悬浮颗粒密度大于墨尔本.
Exercise.
(a) .
(b) , let be random sample of , . Then the estimator of is . Define

Then , and

(c) then

所以认为广告战役成功.
dist = NormalDistribution[0, 1] CDF[dist, 2.53907] 1 - %
Exercise.

Then .
Exercise.

When ,

fDist = FRatioDistribution[12, 10] Quantile[fDist, 0.95]

hw7

Fri, 02 May 2025 16:55:31 GMT

Section 4.7：2， 5， 8， 9 （for ed7：1，4, 8, 9） Section 6.1: 2, 4, 6, 9, 10 (for ed7: 2, 4, 5, 8, 9)
Exercise.

Let denote the frequency of ,

In trials we observe . We will test versus . Since , , , , the test statistic is

The -value is

which is the smallest level at which we reject . Since , is rejected at level of significance.
1 - CDF[ChiSquareDistribution[3], 8.38095]
Exercise.
The number of each type is multinomial with probability . His theory of inheritance predicts that is equal to

In trials he observed . We will test versus . Since, , and 34.75 , the test statistic is

The value for a is . Since is not larger than we do not reject the null. The p-value is

which is not evidence against .
Exercise.
Let denote the frequncy of , , . Let

We will test versus . Denote

Fix , , then the test statistic is

We reject at significant level , iff -value , iff , where . Thus . The smallest value of is .
alpha = 0.05; Quantile[ChiSquareDistribution[11], 1 - alpha]
Reduce[{8/3 k^2 >= 19.6}, k, Reals]
Exercise.
Let's analyze the data and perform the chi-squared goodness-of-fit test.
The observed data is:

:
:
:
:
:
The total number of observations is .
(a) Compute the corresponding chi-square goodness-of-fit statistic.

Estimate the Poisson parameter : As hinted, we treat the category as for computing the sample mean (), which is the estimate for .
Calculate Expected Probabilities under :
Using with ():

Calculate Expected Frequencies ():
With :

Calculate the Chi-Square Statistic ():
The formula is: Plugging in the values: So, the chi-square goodness-of-fit statistic is approximately 7.228.
(b) How many degrees of freedom are associated with this chi-square?
The degrees of freedom () for a chi-squared goodness-of-fit test are calculated as:

Where:

= number of categories = 5 (categories are 0, 1, 2, 3, >3)
= number of parameters estimated from the data = 1 (we estimated )
Therefore, .
There are 3 degrees of freedom associated with this chi-square test.
(c) Do these data result in the rejection of the Poisson model at the significance level?

Hypotheses:
: The data follows a Poisson distribution.
: The data does not follow a Poisson distribution.
Significance Level: .
Test Statistic: .
Degrees of Freedom: .
Critical Value: We need to find the critical value from the chi-squared distribution with 3 degrees of freedom for . This is the value such that . Looking up this value in a chi-squared table or using statistical software:
Decision Rule: Reject if the test statistic is greater than the critical value .
Comparison: . Our test statistic is less than the critical value.
Conclusion: Since the test statistic () is not greater than the critical value (), we fail to reject the null hypothesis () at the significance level.
Therefore, based on this test, these data do not result in the rejection of the Poisson model at the significance level. There is not sufficient evidence to conclude that the data does not follow a Poisson distribution.
Exercise.
(a)

Let

i..

(b)

reaches its maximum iff reaches its minimum. Thus .
Exercise.
(a)

reaches its maximum at .
(b)
Find the pdf of .

Thus

Therefore

(c)
Definition (Consistency). Let be a random variable with cdf , . Let be a sample from the distribution of and let denote a statistic. We say is a consistent estimator of if

By the invariance of MLE, we have

Need to show that for any given ,

We have

Exercise.
First to find the mle of . Let

Let

Then

Denote

Then the mle of is

To show is consistent, it suffices to show is consistent, which is consequence of the Law of Large number.
Exercise.
,

Then

The realization is

Exercise.

The pmf of is

Thus

If then

reaches its maximum at , thus .
If then . Therefore

hw8

Sun, 04 May 2025 15:57:10 GMT

Section 6.2: 9, 10, 11, 12, 16(ed7: 14)
Exercise.
Proof.

So is an unbiased estimator of .

The fisher information is

The Rao–Cramér lower bound in this case is

The variance of is

The efficiency is

□
Exercise.

Let then .

The efficiency is

Exercise.

Thus is unbiased estimator of .

The efficiency is

Exercise.
(a)
We say if the pdf of

The characteristic function of is

Then

Thus .
(b)

Let , , where for .
(c)
When , then . The confidence interval is

Exercise.
(a)

The efficiency is

(b)
The log-likelihood function is

Let

i.e.

The solution of Equation (1), , is the mle of .

(c)

By Student's theorem,

Then the characteristic function of is

The characteristic function of is

By the inverse formula, the asymptotic pdf of is

Thus .

hw9

Wed, 07 May 2025 15:56:14 GMT

Section 6.3: 6, 7,11,16,19 Section 6.4: 2, 3, 5, 6
Exercise.

Then

We reject at level when

Remark. Here we use instead of , just for fun.
Exercise.
Under the general alternatives,

Thus

Therefore

(* 设置参数 *) alpha = 0.05; n = 10; (* 定义您的方程表达式 *) equation = -n Log[n] - n - n Log[W] + W; (* 计算所需的分布分位数 *) lowerQuantile = Quantile[ChiSquareDistribution[1], alpha/2]; upperQuantile = Quantile[ChiSquareDistribution[1], 1 - alpha/2]; (* 联立两个不等式并求解 W *) Reduce[ (equation <= lowerQuantile) && (equation >= upperQuantile), W ]
Solve Equation (1) , we get

i.e.

Exercise.
(a)

To obtain the null distribution of , we first have . Then

(b)
\[Theta]0 = 3; n = 5; alpha = 0.05; Reduce[2/\[Theta]0 W - 8 n Log[W] - 8 n + 8 n Log[\[Theta]0] + 8 n Log[4 n] >= Quantile[ChiSquareDistribution[1], 1 - alpha], W]

Then .
Exercise.
(a)

Let then we have the solution . Since , denote

(b)

The Wald-test is

(c)

The Rao's score statistic is

Exercise.
(a)

where and . Assuming all , then

Then

(b)
When is true, we have

Then

Thus

Hence .
Exercise.
(a)
Let , then

Let , then the mles of , and are

(b)

Let , then the mles of and are

Exercise.

The mles of and are

Exercise.

We have , then . The equality holds iff , .
Thus , .

Then

Therefore, is unbiased, is not unbiased.
-(1/2)*Integrate[(1/(2*\[Rho])^n)*n*x*(\[Theta] + \[Rho] - x)^(n - 1), {x, \[Theta] - \[Rho], \[Theta] + \[Rho]}] + (1/2)*Integrate[(1/(2*\[Rho])^n)*n*x*(x - \[Theta] + \[Rho])^(n - 1), {x, \[Theta] - \[Rho], \[Theta] + \[Rho]}]
Exercise.
(a)
The mle of and are

Then

Thus

where , thus the mle of is

(b)

Thus the mle of is

hw10

Wed, 14 May 2025 15:33:39 GMT

Section 6.5: 4, 5(b), 6, 7; 6(b): This Λ is a function of what F-statistic that would actually be used in this test? Section 7.1: 2, 4, 5, 8
Exercise.
Consider testing

The likelihood ratio statistic is

where

then

We reject at level provided

We know that

Then decreases when and increases when ; . Thus the solution of Equation (1) gives

for some .
Exercise.

The likelihood ratio statistic is

Let then

When ,

Thus

So the test can be based on .
Exercise.
(a)

(b)
Under ,

Exercise.

Let , then

Under ,

Then

Exercise.

is an unbiased estimator of .

Exercise.

The equality holds iff

Exercise.

When , is smaller maximum risk; otherwise, is.

Exercise.

Let , then

Thus, does not minimize . (If so, , which is a contradiction.)

hw11

Tue, 20 May 2025 09:11:44 GMT

Exercise.

Let then

Then

does not depend on . Thus is sufficient.
Exercise.

Then

does not depend on . Thus is sufficient.
Exercise.
For ,

does not depend on . Thus is sufficient.
In the general case,

does not depend on . Thus is sufficient.
Exercise.

Then the Likelihood function is

By Neyman Factorization Theorem,

is the sufficient statistic for .
Exercise.

Exercise.
7.2.1

The unbiased sufficient estimator is

7.2.2

The unbiased sufficient estimator is

7.2.3

The unbiased sufficient estimator is

Exercise.

Denote

Then

We have the identity:

Thus

Therefore

We have the indentity

Then

Hence

Markov Chain

Thu, 15 May 2025 09:30:08 GMT

好的，马尔可夫链具有一些非常重要的基本性质，这些性质是我们理解和应用它的基础。以下是一些核心的性质：

马尔可夫性 (Markov Property) - 无记忆性
定义：这是马尔可夫链最根本的性质。它指系统的下一个状态只依赖于其当前状态，而与它过去的状态序列无关。
数学表达：。
意义：大大简化了对随机过程的分析，因为我们不需要追踪系统的全部历史。
状态空间 (State Space)
定义：系统所有可能状态的集合。
分类：
离散状态空间 (Discrete State Space)：状态是可数的（有限个或无限可数个），例如天气（晴、阴、雨）、骰子点数（1, 2, 3, 4, 5, 6）。我们通常讨论的马尔可夫链都是离散状态的。
连续状态空间 (Continuous State Space)：状态是连续的，例如某个物理量的精确值。这通常涉及到更复杂的马尔可夫过程。
有限状态空间 (Finite State Space)：状态的个数是有限的。
无限状态空间 (Infinite State Space)：状态的个数是无限的。
转移概率 (Transition Probabilities)
定义：从一个状态转移到另一个状态的概率，记为或。
性质：
（概率非负且不大于1）
（对于任意状态，从出发到所有可能状态的转移概率之和为1，其中是状态空间）
时齐性 (Time Homogeneity) / 稳态转移概率 (Stationary Transition Probabilities)
定义：转移概率不随时间的变化而变化。也就是说，在任何时刻从状态转移到状态的概率都是相同的。
数学表达：对所有成立。
意义：大多数我们初学的马尔可夫链都假设是时齐的，这使得分析更加简单。如果转移概率随时间变化，则称为非时齐马尔可夫链。
n步转移概率 (n-step Transition Probabilities)
定义：从状态经过步转移到状态的概率，记为。
计算：可以通过 Chapman-Kolmogorov 方程计算：。特别地，状态转移矩阵的 n 次幂的元就是。
以下是一些更深入的、描述马尔可夫链长期行为的性质：

可达性 (Accessibility)
定义：如果从状态出发，经过有限步（可能为0步）能够到达状态 (即对某个成立)，则称状态从状态可达，记为。
互通性 (Communication)
定义：如果状态和状态相互可达 (即且 )，则称状态和状态是互通的，记为。
性质：互通是一种等价关系，可以将状态空间划分为若干个互通类。
不可约性 (Irreducibility)
定义：如果一个马尔可夫链中所有状态都属于同一个互通类（即从任意状态都可以到达任意其他状态），则称该马尔可夫链是不可约的。
意义：不可约马尔可夫链的行为更为整体和一致。
常返性 (Recurrence) 与瞬逝性 (Transience)
常返状态：如果从状态出发，最终能以概率1返回到状态，则称状态是常返的。
瞬逝状态：如果从状态出发，最终返回到状态的概率小于1（意味着有可能永远不再返回），则称状态是瞬逝的。
性质：
在不可约马尔可夫链中，所有状态要么都是常返的，要么都是瞬逝的。
对于有限状态空间的不可约马尔可夫链，所有状态都是常返的。
周期性 (Periodicity)
定义：状态的周期是使得的所有步数的最大公约数。如果对所有成立，则定义或 (不同教材定义可能略有差异，但核心是不能返回自身)。
非周期性：如果一个状态的周期，则称该状态是非周期的。
性质：同一个互通类中的所有状态具有相同的周期。如果一个不可约马尔可夫链的所有状态都是非周期的，则称该链是非周期的。
平稳分布 (Stationary Distribution) / 极限分布 (Limiting Distribution)
平稳分布 ()：一个概率分布向量，其中是系统处于状态的概率，并且满足以及（即）。如果系统当前的概率分布是，那么经过一步转移后，其概率分布仍然是。
极限分布：当时间时，的极限（如果存在）。
重要定理：对于一个不可约、非周期、正常返（对于有限状态空间，不可约即正常返）的马尔可夫链，存在唯一的平稳分布，并且该平稳分布也是极限分布，即 (与初始状态无关)。
意义：描述了马尔可夫链在长时间运行后的稳定状态。
这些是马尔可夫链的一些最基本的性质，理解它们对于进一步学习和应用马尔可夫链至关重要。根据研究的深入程度，还会有更多细致的性质和分类。

要证明一个随机过程是一个马尔可夫链，核心在于验证它是否满足马尔可夫性质 (Markov Property)，并且明确其状态空间 (State Space) 和时间参数 (Time Parameter)。
Definition (Markov Property). A stochastic process has the Markov property if, for any in , for all and . In other words, the future is independent of the past given the present.
以下是证明一个过程是马尔可夫链的一般步骤和关键点：

定义状态空间 (Define the State Space ):
首先，你需要清晰地定义系统所有可能的状态。状态空间可以是有限的，也可以是无限可数的。
确保每个状态都是互斥的，即系统在任何时刻只能处于一个状态。
例子：
天气模型：
硬币投掷：
排队系统中的顾客数量：
定义时间参数 (Define the Time Parameter ):
确定过程是离散时间的还是连续时间的。对于初学者，我们通常关注离散时间马尔可夫链，时间参数或 {}.
或表示在时间步 (或时刻 ) 系统所处的状态。
验证马尔可夫性质 (Verify the Markov Property): 这是最关键的一步。马尔可夫性质指出，给定现在，未来与过去独立。更正式地说，对于任意时间步和任意状态序列：要证明这一点，你需要：
明确条件概率的含义：理解是在事件发生的条件下事件发生的概率。
利用过程的定义和规则：根据你所研究的随机过程的具体规则和描述，推导其一步转移的条件概率。
展示“无记忆性”：
考虑在已知当前状态以及过去所有状态的情况下，系统在下一个时刻转移到状态的概率。
然后，证明这个概率仅仅依赖于当前状态，而与过去的状态无关。也就是说，那些过去的状态信息对于预测下一个状态是多余的。如何操作？
直接从定义出发：有些过程的定义本身就隐含了这种无记忆性。例如，如果规则明确说明“下一步只看当前”，那么马尔可夫性质就比较容易验证。
通过计算条件概率：
写出的表达式。
利用过程的内在机制，看看这个表达式是否可以简化为只依赖于和的形式，即。
如果可以，那么马尔可夫性质就得到了证明。
（可选，但常伴随）确定转移概率 (Determine Transition Probabilities):
一旦证明了马尔可夫性质，通常下一步是确定（或者说明如何确定）一步转移概率。
如果这些转移概率不依赖于时间，那么这个马尔可夫链是时齐的 (Time-Homogeneous)。大多数基础讨论中的马尔可夫链都是时齐的。如果依赖于时间，即，则它是非时齐的 (Non-Time-Homogeneous)。证明过程本身是马尔可夫链时，并不强制要求时齐性，但通常会关注这一点。
一个简单的例子来说明思路：简化的天气模型
假设我们有一个天气模型，规则如下：

如果今天下雨，明天有的概率下雨，的概率晴天。
如果今天晴天，明天有的概率下雨，的概率晴天。
明天的天气只取决于今天的天气。
证明步骤：

状态空间：
时间参数：离散时间， (天数)
验证马尔可夫性质：我们要证明：

考虑 : 根据模型规则，“明天的天气只取决于今天的天气”。所以，即使我们知道的具体天气情况，只要，下一天是雨天的概率就是 0.7。因此，。同时，。两者相等。
类似地，可以对所有可能的当前状态和下一个状态进行验证：

由于对于所有情况，未来状态的概率只依赖于当前状态，所以该过程满足马尔可夫性质。

转移概率（时齐）：

这些概率不随天数变化，所以这是一个时齐马尔可夫链。
总结关键：
证明的核心是严格地、形式化地展示出“给定现在，未来与过去无关”。这通常需要仔细分析随机过程的规则和定义，并运用条件概率的知识。对于一些复杂的过程，这可能涉及到更多的数学推导。

Models-Statistical-Inference-and-Learning

Fri, 02 May 2025 16:55:31 GMT

This is the Chapter 6 of All of Statistics.

A typical statistical inference question: Given a sample how do we infer ?

In general, a parametric model takes the form

where is an unknown parameter (or vector) taking values in the parameter space .

A nonparametric model is a set that cannot be parmeterized by a finite number of parameters. For example, is nonparametric.

Remark. We first discuss nonparametric inference. Two dominant approaches to statistical inference are frequentist inferenceand Bayesian inference.

If is a parametric model, we write

The subscript indicates that the probability or expectation is w.r.t. ; it does not mean we are averaging over . Similarly, write for the variance.

Many inferential problems can be identified as one of three types:

Estimation
Confidence sets
Hypothesis testing

Point estimation refers to provide a single "best guess" of some quantity of interest, such as cdf, pdf, regression function....
Denote a point estimate of by or .
Remark. Note that is fixed, unknown quantity, depens on the data so is a r.v.

Say unbiased if .
Unbiasedness is considered less important than before.
Say consistent if .
The distribution of is called the sampling distribution. The standard deviation of is called the standard error, denoted by :

Often, depends on the unknown . The estimated se is denoted by .

The mean squared error (MSE) is defined by

We have

Which means the estimator is not always asymptotically Normal. But point estimators often have this property.

Note that is random and is fixed.
Remark (A interpret of confidence in tervals).

Point estimators often have a limite Normal distribution, meaning that . In this case we can sonstruct (approximate) confidence intervals as follows.

Parametric Inference

Fri, 02 May 2025 16:55:31 GMT

See All of Statistic Chap 9

There are two methods for estimating , the method of moments and the method of maximum likelihood.

When estimating a parameter , we may only be interested in a function , called the parameter of interest. Other components of are nuisance parameters. For instance, if and we want to estimate , then is the parameter of interest, with being the nuisance parameter. As an example, if and we want to know the fraction of the population with test scores larger than 1, then .

Suppose that the parameter has components. For , define the moment

and the sample moment

Definition (method of moments estimator). The method of moments estimator is defined to be the value of such that

Let . Then , and . By Definition 1 (method of moments estimator),

Let . Then . We need to solve the quations

The solution is

Theorem (Theorem 9.6). Let denote the method of moments estimator. Under appropriate conditions on the model, the following statements hold:
The estimate exists with probability tending to 1.
The estimate is consistent: .
The estimate is asymptotically Normal: where and

Let i.i.d. with pdf .
Definition (Likelihood function). The likelihood function is defined by The log-likelihood function is defined by .
The likelihood function is just the joint density of the data , except that we treat it is a function of the parameter . Thus, .
Definition (maximum likelihood estimator). The maximum likelihood estimator MLE, denoted by , is the value of that maximizes .
Often, it is easier to work with the log-likelihood.
Remark. If we multiply by any positive constant (not depending on ) then this will not change the mle. Hence, we shall often drop constants in the likelihood function.

Let . The parameter is , and the likelihood function (ignoring some constants) is:

where is the sample mean and . The last equality above follows from the fact that , which can be verified by writing and then expanding the square.
The log-likelihood is

Solving the equations

we conclude that and . It can be verified that these are indeed global maxima of the likelihood.

Here is an example that many people find confusing. Let . Recall that

Consider a fixed value of . Suppose for some . Then, and hence . It follows that if any . Therefore, if where .
Now consider any . For every we then have that so that . In conclusion,

Now is strictly decreasing over the interval . Hence, .

The MLE possesses many properties that make it an appealing choice choice of estimator

The MLE is consistent: where denotes the true value of the parameter ;
The MLE is equivariant: if is the MLE of then is the MLE of ;[1]
The MLE is asymptotically Normal: ; also, the estimated standard error se can often be computed analytically;
The MLE is asymptotically optimal or efficient: roughly, this means that among all well-behaved estimators, the MLE has the smallest variance, at least for large samples;
The MLE is approximately the Bayes estimator. (This point will be explained later.)
In sufficiently complicated problems, the MLE will no longer be a good estimator.
We focus on the simpler situations where the MLE works well, assuming that the model satisfies certain regularity conditions, which are essentially smoothness conditions on .

Consistency means that the mle converges in probability to the true value.
Definition (Kullback-Leibler distance). If and are pdf's, define the Kullback-Leibler distance[2] between and to be It can be shown that and . For any write to mean .
We will say that the model is identifiable if implies that 0. This means that different values of the parameter correspond to different distributions.
Theorem. Let denote the true value of . Define and . Suppose that and that, for every , Let denote the MLE. Then .

when admits an inverse.↩︎
This is not a distance in the formal sense because is not symmetric.↩︎

simple-approximation

Fri, 02 May 2025 16:55:31 GMT

If we want to verify a property that holds for general Borel-measurable function , we can follow four steps as below.

Verify the property when is indicator function.
Verify the property when is nonnegative simple function.
Verify the property when is Borel-measurable function
Verify the property when is General Borel-measurable function
An example is as follow.

Theorem. Let be a random variable on a probability space and let be a Borel-measurable function on . Then
and if this quantity is finite, then

Proof.
Step 1. Indicator functions. (Omitted)
Step 2. Nonnegative simple functions. (Trivial because of the linearity)
Step 3. Nonnegative Borel-measurable functions. Let be an arbitraty nonnegative Borel-measurable function defined on . For each positive integer , define the sets

Remark. 这样的定义是为了保证在时，每个划分越来越细，同时 . 而且对于任意，是的加细（的划分点都包含在的划分点集内）
For eac fixed , the sets correspond to the partition

At the next stage , the partition points include all those at stage and new partition points at the midpoints between te old ones. Because of this fact, the simple functions

satisfy . Furthermore, these functions become more and more accurate approximations of as becomes larger; indeed for every . From Step 2, we knoe that

for every . Letting and using the Monotone Convergence Theorem, on both sides of the equation, we obtain

This proves when is a nonnegative Borel-measurable function.
Step 4. General Borel-measurable function. (consider where and are both nonnegative Borel-measurable functions)
□

Yau contest

Sat, 17 May 2025 14:18:42 GMT

D

Okay, this problem (Problem 6 from your image) is from the area of statistical decision theory, specifically focusing on estimation theory and minimax estimators. Since you're a beginner in statistics, understanding this problem and its solution will require familiarity with several key concepts.
Here's a breakdown of the topics and materials you'll need to study:
I. Foundational Probability and Statistics Concepts:

Random Variables:
Definition: What a random variable is (discrete and continuous).
Probability Distributions: Understanding probability mass functions (PMF) for discrete variables and probability density functions (PDF) for continuous variables.
Expected Value (Mean): .
Variance: .
Specific Probability Distributions:
Normal (Gaussian) Distribution: .
Its PDF: .
Properties: Mean , variance .
The problem states . This means each is normally distributed with mean and variance .
Independent Random Variables:
Understanding what it means for random variables to be independent. This is crucial as the problem states are independent.
II. Statistical Estimation Theory:

Parameter Estimation:
The goal is to estimate an unknown parameter (here, the mean vector ) based on observed data (here, ).
Estimator: A function of the data used to estimate the parameter. In this problem, the estimator being considered is itself, meaning is used to estimate .
Loss Function :
A function that quantifies the "cost" or "loss" incurred when the true parameter is and our estimate is .
In this problem, the loss function is a weighted squared error loss: This means errors in estimating are weighted by . Errors for components with smaller (and thus smaller variance for ) are penalized more heavily relative to their scale if the in the denominator were not there. The in the denominator effectively normalizes the squared error by the variance of the observation if .
Risk Function :
The expected loss of an estimator when the true parameter is .
. The expectation is taken over the distribution of , given .
For the estimator (meaning ): Since , is the variance of , which is .
So, .
This shows that the estimator has a constant risk (it doesn't depend on the true value of ). An estimator with constant risk is called an equalizer rule.
III. Bayesian Statistics (Used in the Proof Strategy):

Prior Distribution :
In Bayesian statistics, the unknown parameter is treated as a random variable having a prior distribution, which reflects our beliefs about before observing data.
The solution considers independent priors . This is a common choice for mathematical convenience.
Posterior Distribution :
After observing data , the prior belief is updated to a posterior distribution using Bayes' theorem: , where is the likelihood of the data given the parameter.
Bayes Estimator :
An estimator that minimizes the Bayes risk (defined below) for a given prior .
For squared error loss $ (di - \theta_i)^2 -th component of the Bayes estimator is the mean of the posterior distribution of : $\delta{\pi,i}(X) = \mathbb{E}[\theta_i | X]$.
The problem uses a slightly modified loss. For the loss , the Bayes estimator for will still be (since the and are independent across ).
The "Recall" part of the problem gives a formula for the posterior mean when the likelihood is normal and the prior is normal (this is a standard result for conjugate priors):
If and , then the posterior distribution of is also normal: The posterior mean is .
In the problem's context, for estimating :
(so is , is , variance is ).
Prior (so , prior variance is ).
The Bayes estimator for (posterior mean ) is: The solution in the image has . This is slightly different from what I derived for the posterior mean.
Let's re-check the posterior mean given in the problem: .
Here: , , . And for the prior: , .
So, .
The solution's notation is indeed the posterior mean, which simplifies to .
Bayes Risk (or in the solution):
The expected risk with respect to the prior distribution of : .
The Bayes estimator minimizes this Bayes risk. The minimum Bayes risk is .
The solution calculates the Bayes risk for the Bayes estimator (where parameterizes the prior variance).
. This is not quite right.
The Bayes risk for is .
It's also .
Alternatively, for squared error type losses, the Bayes risk of the Bayes estimator is .
The solution states .
This term is or related to posterior variance.
The posterior variance is .
The Bayes risk for component under loss is .
So for loss , it would be .
The solution has . This is slightly different.
Let's check: The Bayes risk for component under squared error loss.
For the weighted loss , the Bayes risk for is .
The solution states the associated Bayes risk is .
This formula for the Bayes risk of a Bayes estimator under normal-normal model with weighted squared error loss is a standard result that you'd find in textbooks on Bayesian decision theory.
IV. Minimax Estimation:

Minimax Estimator :
An estimator is minimax if it minimizes the maximum possible risk:
Finding minimax estimators can be hard.
Proving Minimaxity (Key Theorem Used in the Solution):
A common way to prove an estimator is minimax is to:
Show that is an equalizer rule, meaning its risk is constant (doesn't depend on ). Let this constant risk be .
Find a sequence of prior distributions such that the Bayes risk of the Bayes estimator for , denoted , converges to as .
If both conditions hold, then is minimax.
(This relies on the fact that and ).
The theorem states: If is an equalizer rule with risk , and if there's a sequence of priors such that the corresponding Bayes risks converge to , then is minimax.
Applying the Theorem to the Problem:

The estimator (where ) has a constant risk . So, is an equalizer rule.
The solution considers a sequence of priors , so is the prior.
The Bayes risk for this sequence of priors is (as per the solution's formula).
The solution then takes the limit as (this corresponds to a sequence of priors that become increasingly "non-informative" or "flat"). As , each term .
So, . There's a discrepancy here. The constant risk of is . The limit of Bayes risks is .
For to be minimax, the limit of Bayes risks should converge to the constant risk of , which is . Let's re-read the solution carefully:
"The associated Bayes risk is ."
This simplifies to . This part is consistent.
"Clearly, as , , which is identical to the Bayes risk of ."
This limit calculation in the solution image is incorrect.
, not . The solution seems to have a typo in this limit evaluation or in the expression for . Let's look at the Bayes risk of the estimator itself, under the prior .
. Since (constant),
.
The theorem for minimaxity is: an equalizer rule with risk is minimax if , or if there's a sequence of priors such that . The solution strategy is sound: show is an equalizer, and its risk is the limit of some Bayes risks.
The risk of is . This is the .
We need to find a sequence of priors (here, parameterized by ) such that the minimal Bayes risk converges to .
The Bayes estimator component is .
The Bayes risk associated with this estimator is (as given in the solution, which I re-derived as ):
.
As , . This does not equal . Possible misunderstanding or error in the provided solution text:
The statement in the solution "Clearly, as , " is incorrect for the given .
Perhaps the intended Bayes risk calculation or the prior setup was different. Let's assume the theorem application is standard. For to be minimax, its constant risk (2022) must be the limit of the Bayes risks of the Bayes estimators for a sequence of priors. What if the estimator itself is a Bayes estimator or a limit of Bayes estimators?
The Bayes estimator is .
As , .
So, .
This means the estimator is a limit of Bayes estimators.
If an estimator is Bayes for some prior (or a limit of Bayes estimators) and has constant risk, then it is minimax. This is another standard result.
has constant risk .
.
This approach (showing an equalizer rule is a limit of Bayes rules) is a common technique. Let's review the general conditions for minimaxity often used:
If an estimator has constant risk (is an equalizer rule) and is Bayes with respect to some prior, it is minimax.
If an estimator is an equalizer rule with risk , and if for any there exists a prior such that the Bayes risk (i.e., the supremum of Bayes risks is ), then is minimax. This is equivalent to finding a sequence such that . The solution's text "This implies that with gives a least favorable sequence of priors and is minimax" suggests they are trying to use the latter condition.
The error must be in the expression for or its limit. Let's re-examine the calculation of the Bayes risk of .
The risk of an estimator for under loss is .
The Bayes risk for component is .
For the Bayes estimator , the Bayes risk is .
This is known to be . (The expectation is over the marginal distribution of ).
.
So, the -th component of Bayes risk for the Bayes estimator is .
Then the total Bayes risk is .
Now, let's take the limit as : So, This matches the constant risk of .
So, with this corrected formula for the Bayes risk of the Bayes estimator, the solution's logic holds. The discrepancy was that the solution in the image has .
My derived Bayes risk for the Bayes estimator is . Let's check the solution's expression for .
.
This is the formula written in the solution. If this is the Bayes risk, then its limit as is indeed . There is a confusion about which specific formula for Bayes risk is being used or if there's a standard result that leads to the solution's expression for .
Commonly, .
If , then for squared error loss , the Bayes risk is .
For weighted squared error , Bayes risk is .
Here . So it should be . Why does the solution use ?
Let (precision of prior) and (precision of data).
Posterior mean . Here , so .
Posterior variance .
The Bayes risk of the Bayes estimator for is .
Since the data itself is random, is actually constant (not dependent on value for normal-normal case). So .
So the Bayes risk for under squared error is .
Then for the weighted loss , the Bayes risk is .
. It seems my derivation of the Bayes risk is , whose limit is .
The solution states the Bayes risk is .
Let's break down the solution's term :
This is .
If this is the correct Bayes risk for the Bayes estimator , then its limit as is indeed . So, there is a conflict between the standard formula for the Bayes risk and the expression the solution provides for .
The Bayes risk is .
.
.
The quantity is the mean squared error of the Bayes estimator . For Bayesian estimation under a specific prior, this MSE averaged over the prior of and data is the component of the Bayes risk.
The Bayes risk for estimating with under squared error loss is . No, this is not it.
It's .
Or .
The Bayes risk for squared error for is (if is used) or simpler, .
So the weighted Bayes risk is .
If this is correct, then the limit is . The solution's claim that "Clearly, as , " seems to imply that the terms of their sum should go to 1.
If , this limit is .
If , this limit is .

傅里叶分析初步

Fri, 09 May 2025 02:22:06 GMT

傅里叶级数的三种主要收敛类型——逐点收敛 (Pointwise Convergence)、一致收敛 (Uniform Convergence) 和绝对收敛 (Absolute Convergence)——之间存在明确的层级关系和重要的区别。
首先，我们来定义这三种收敛性，并针对傅里叶级数 (或其实数形式 ) 进行说明：

逐点收敛 (Pointwise Convergence)
定义: 对于级数，如果对于定义域中的每一个固定的，数列部分和收敛于一个值，则称该级数在处逐点收敛于。如果对定义域中所有都如此，则称级数逐点收敛于函数。
傅里叶级数: (或对应实数形式的部分和) 当时，对于每个固定的，其极限等于 (或在不连续点等于 )。
特点: 不同点的收敛速度可能不同。极限函数不一定是连续的，即使级数的每一项都是连续的。
一致收敛 (Uniform Convergence)
定义: 对于级数，如果部分和序列一致收敛于函数，即对于任意，存在一个与无关的，使得当时，对定义域中的所有，都有。这等价于。
傅里叶级数: 。
特点: 收敛速度不依赖于。如果级数的每一项都是连续的，并且级数一致收敛于，则极限函数也必定是连续的。这是非常重要的性质。
绝对收敛 (Absolute Convergence)
定义 (对于函数项级数): 如果级数对于每个都收敛，则称级数在该点绝对收敛。
傅里叶级数: 对于傅里叶级数，我们通常关心其系数的绝对收敛性，即 (或等价地 ) 是否收敛。这是因为。如果系数级数收敛，那么根据Weierstrass M-判别法，傅里叶级数必然一致收敛且绝对收敛（作为函数项级数，即对每个都收敛，因为它等于）。
特点: 绝对收敛是一个很强的条件。
它们之间的关系：

绝对收敛一致收敛 (对于傅里叶级数，指系数绝对收敛)
如果傅里叶级数的系数满足，那么根据 Weierstrass M-判别法 (取，因为对所有成立)，傅里叶级数必定一致收敛。
同时，由于收敛，所以傅里叶级数本身也作为函数项级数绝对收敛于每个点。
一致收敛逐点收敛
这是由定义直接得出的。如果一个级数在整个定义域上一致地逼近一个函数，那么它在定义域内的每一个特定点也必然逼近该函数在该点的值。
总结层级关系：
对于傅里叶级数：
系数绝对收敛 () 一致收敛逐点收敛
反向关系（通常不成立）：

逐点收敛一致收敛
一个傅里叶级数可能在每个点都收敛到函数值（或跳跃点的中点值），但不是一致收敛。例如，对于有跳跃间断点的函数，即使其傅里叶级数在跳跃点收敛到中点值，在跳跃点附近也会出现吉布斯现象 (Gibbs Phenomenon)，这妨碍了一致收敛。
即使对于连续函数，其傅里叶级数也可能逐点收敛（例如，根据Carleson定理，函数的傅里叶级数几乎处处逐点收敛），但不一定一致收敛。Lusin曾构造出连续函数，其傅里叶级数在某些点发散。
一致收敛绝对收敛 (指系数的绝对收敛)
存在一致收敛的傅里叶级数，但其系数的绝对和是发散的。例如，某些Hölder连续函数（如且）的傅里叶级数是一致收敛的，但其系数不一定绝对收敛。
逐点收敛绝对收敛
这从前面两条就可以推断出来。例如，条件收敛的数值级数（如）就是逐点收敛但不绝对收敛的例子。函数项级数也有类似情况。
影响收敛性的函数性质：

函数的连续性与光滑性起着关键作用：
逐点收敛: 函数有界变差 (Dirichlet-Jordan 条件)；满足Dini条件；Hölder连续 () 等都能保证逐点收敛。
一致收敛: 函数连续且有界变差；函数Hölder连续 ()；函数是 (一阶连续可微) 等都能保证一致收敛。一般来说，函数越光滑，其傅里叶系数衰减越快，越容易一致收敛。
绝对收敛 (): 需要更强的光滑性。例如，根据Bernstein定理，若且，则系数绝对收敛。若函数是且 (对于周期的区间，二阶导数连续)，则系数通常绝对收敛。如果函数可以解析延拓到包含实轴的一个带状区域，其傅里叶系数也会指数衰减，从而绝对收敛。
简而言之：

绝对收敛是最强的，它保证了一致收敛和逐点收敛。
一致收敛比逐点收敛强，它保证了逐点收敛，并且如果级数项连续，则极限函数也连续。
逐点收敛是最弱的，不保证另外两者。
理解这些区别对于分析傅里叶级数的行为至关重要。
傅里叶级数收敛性分析：逐点收敛、一致收敛与绝对收敛的比较
引言：傅里叶级数收敛性问题
傅里叶级数是一种强大的数学工具，它允许我们将一个周期函数 f(x) 表示为一系列正弦和余弦函数（或复指数函数）的无穷和 [1]。对于周期为 2L 的函数 f(x)，其傅里叶级数通常写作：

其中傅里叶系数和通过积分计算得出 [3]：

或者使用复指数形式：

其中

傅里叶级数的概念起源于19世纪初傅里叶对热传导问题的研究 [2]。然而，一个基本且深刻的问题随之产生：这个无穷级数在何种意义下收敛？并且，它是否收敛到原始函数 f(x)？ [6]。
傅里叶级数的收敛性并非理所当然，它取决于函数 f(x) 的性质以及我们所考虑的收敛类型 [5]。历史上，关于傅里叶级数收敛性的探讨充满了挑战，傅里叶本人的断言曾受到拉格朗日等数学家的质疑，而后来的研究（如du Bois-Reymond）也揭示了连续函数的傅里叶级数也可能在某点发散 [2]。这表明，理解傅里叶级数的收敛性需要严谨的数学框架，区分不同的收敛模式至关重要。级数表示中的 ~ 符号提醒我们这仅仅是一个形式上的展开，并不自动保证等式成立 [5]。
核心挑战在于，傅里叶级数的构建依赖于函数的全局积分性质（计算系数），而收敛性则可能涉及函数在某一点的局部性质或整个区间上的整体性质。这种全局与局部的联系是傅里叶分析的核心，也是其复杂性的来源。
本报告旨在深入探讨傅里叶级数的三种主要收敛类型：逐点收敛（pointwise convergence）、一致收敛（uniform convergence）和绝对收敛（absolute convergence）[8]。我们将详细阐述它们的定义、收敛条件（包括狄利克雷条件、若尔当判别法、狄尼判别法、伯恩斯坦定理等）、蕴含关系、局限性以及它们在函数性质保持（如连续性、可微性）和收敛速度方面的差异。理解这些不同类型的收敛对于理论分析和实际应用（如信号处理、偏微分方程求解）都具有重要意义 [1]。不同的收敛类型对应着傅里叶级数逼近原函数的不同“质量”等级，某些应用（如需要保证近似误差在整个区间一致小）可能需要比逐点收敛更强的收敛模式，如一致收敛 [11]。
收敛类型的定义
为了精确比较不同的收敛模式，我们首先需要明确它们的定义。令表示傅里叶级数的前 N 项部分和：
对于实数形式：

对于复数形式：

Definition (Pointwise Convergence). 逐点收敛 (Pointwise Convergence) 定义: 傅里叶级数 S(x) 在集合 I 上逐点收敛于函数 f(x)，如果对于 I 中的每一个 x，都有 [11]。
解释: 这种收敛性是在每个点 x 上独立考虑的。对于任意给定的点 x 和任意小的正数，我们总能找到一个足够大的（这个可能依赖于 x 和），使得当时，。收敛的速度在不同点上可能存在显著差异。
Definition (Uniform Convergence). 一致收敛 (Uniform Convergence) 定义: 傅里叶级数在集合 I 上一致收敛于函数 f(x)，如果 [11]。等价地，对于任意给定的，存在一个（不依赖于 x），使得对于所有和 I 中的所有 x，都有 [20]。
解释: 这是一种更强的收敛形式，要求级数在整个区间 I 上以“相同的速度”逼近函数 f(x)。部分和的图像最终会完全落在 f(x) 图像周围一个宽度为的“带子”内。一致收敛关注的是在整个区间上的最大逼近误差。
Definition (Absolute Convergence). 绝对收敛 (Absolute Convergence) 定义: 傅里叶级数称为绝对收敛，如果其系数的绝对值构成的级数收敛：对于实数形式：。对于复数形式： [3]。
解释: 绝对收敛是关于傅里叶系数本身大小的一个条件。它不直接描述如何逼近 f(x)，但它蕴含了非常好的收敛性质。一个重要的特性是，绝对收敛的级数，其各项可以任意重排而不改变级数的和 [24]。这与逐点收敛和一致收敛形成了对比，后两者关注的是部分和函数序列与目标函数 f(x) 之间的逼近关系。
这三种定义揭示了它们关注点的不同：逐点收敛关注每个独立点上的极限行为；一致收敛关注整个区间上逼近误差的最大值；绝对收敛则关注系数的衰减速度。正是这种关注点的差异导致了它们各自成立的条件以及所蕴含的性质有所不同。特别是，一致收敛要求存在一个对所有 x 都有效的，这比逐点收敛的要求（可以依赖于 x）严格得多 [22]，也解释了为何在某些情况下（如吉布斯现象）一致收敛会失效。
逐点收敛
逐点收敛是傅里叶级数收敛性中最基本的一种形式，它描述了级数在单个点上是否趋向于函数（或其某个特定值）。
收敛条件:判断傅里叶级数是否逐点收敛有多种充分条件，这些条件通常与函数的局部光滑性或有界变差性质有关。

狄利克雷条件 (Dirichlet Conditions): 这是最经典的判据之一。若周期为 2L 的函数 f(x) 在一个周期内满足：(1) 绝对可积 ()；(2) 只有有限个极大值和极小值；(3) 只有有限个第一类间断点（左右极限均存在且有限），则其傅里叶级数在每一点 x 都收敛 [1]。
若尔当判别法 (Jordan's Test - Bounded Variation): 如果 f(x) 是一个周期为 2 且在上具有有界变差 (bounded variation) 的函数，则其傅里叶级数在每一点都逐点收敛 [8]。有界变差函数意味着函数图像的总“上下波动”是有限的，这类函数允许有跳跃间断点，但不能有无限次的振荡。这个条件比经典的狄利克雷条件更广泛。
狄尼判别法 (Dini's Test): 这是一个更精细的局部判据。如果 f 局部可积，并且对于某点和某个值 S，积分对于某个收敛，则 f 的傅里叶级数在点收敛于 S [6]。这个条件考察了函数在点附近关于值 S 的对称部分的平均收敛速度。
局部光滑性条件: 如果函数 f 在点处满足 Hölder 条件（即 , ），或者在点可微（即左右导数存在且相等），则其傅里叶级数在点收敛于 [8]。即使在跳跃间断点，如果左右导数存在，级数也会收敛 [8]。
L2 条件 (Carleson's Theorem): 对于平方可积函数 (), 其傅里叶级数几乎处处 (almost everywhere) 逐点收敛于 f(x) [6]。这意味着不收敛的点集测度为零。
收敛值:当傅里叶级数在点 x 逐点收敛时，其收敛值取决于 f(x) 在该点的性质：

连续点: 如果 f(x) 在 x 点连续且满足上述任一收敛条件，则傅里叶级数收敛于 f(x) [1]。
间断点: 如果 f(x) 在 x 点存在第一类间断点（跳跃间断点）且满足收敛条件，则傅里叶级数收敛于该点左右极限的算术平均值，即 [1]。
局限性与现象:逐点收敛虽然基本，但有其局限性。
吉布斯现象 (Gibbs Phenomenon): 在函数 f(x) 的跳跃间断点附近，傅里叶级数的部分和会表现出“过冲”和“下冲”的振荡行为。当时，这些振荡的峰值并不会趋于零，而是趋向于一个超过函数实际跳跃高度约 9% 的值。虽然振荡区域会向间断点压缩，但峰值高度不变，这表明在包含间断点的任何邻域内，收敛都不是一致的 [1]。
可能发散: 存在这样的连续函数，其傅里叶级数在某（些）点发散（du Bois-Reymond 构造）[2]。甚至存在 L1 可积函数（比连续函数更弱的条件），其傅里叶级数处处发散（Kolmogorov 构造）[6]。
逐点收敛的条件，如狄尼判别法或可微性，都集中在考察函数在收敛点邻域的行为 [6]。这体现了逐点收敛的“局部性”特征，即级数在一点的收敛性主要由函数在该点附近的性质决定，这与需要全局信息的傅里叶系数计算形成对比。黎曼的局部化原理 (localization principle) 进一步强调了这一点：如果两个函数在某点的一个小邻域内相等，那么它们的傅里叶级数在该点的收敛性态（收敛或发散，以及收敛值）必然相同 [29]。然而，du Bois-Reymond 和 Kolmogorov 的反例揭示了逐点收敛的深刻局限性 [2]。它们表明，仅仅是函数的基本性质，如连续性或 L1 可积性，并不足以保证傅里叶级数处处收敛。这促使数学家们寻找更强的函数条件（如有界变差、满足狄尼条件等）来确保逐点收敛，或者转向研究其他类型的收敛（如 L2 收敛、一致收敛、可和性）。卡尔松定理 (L2 函数傅里叶级数几乎处处收敛) 则表明，对于在许多应用中至关重要的 L2 函数类，逐点发散是“罕见”的 [6]。
一致收敛
一致收敛是比逐点收敛更强的收敛模式，它要求傅里叶级数在整个区间上以统一的速率逼近函数。
收敛条件:确保傅里叶级数一致收敛的条件通常要求函数具有更好的整体光滑性和连续性。

连续性 + 有界变差 (Jordan's Test): 如果函数 f(x) 在上连续且具有有界变差，并且满足周期性条件，则其傅里叶级数在上一致收敛于 f(x) [8]。
Hölder 连续性: 如果周期函数 f(x) 在整个区间上满足 Hölder 条件（阶数），则其傅里叶级数一致收敛 [8]。这包括了 Lipschitz 连续函数（）的情况。
狄尼-利普希茨判别法 (Dini-Lipschitz Test): 如果 f(x) 是连续的周期函数，且其连续模满足当时，则其傅里叶级数一致收敛 [27]。这个条件比 Hölder 连续性稍弱，但仍保证了一致收敛。
连续可微性 (C1): 如果函数 f(x) 连续，其导数 f′(x) 分段连续（piecewise continuous），且满足周期性边界条件，则 f(x) 的傅里叶级数一致收敛于 f(x) [12]。这是实际应用中一个常用且相对容易验证的条件。
更高阶光滑性 (Ck,k，即 k 阶连续可微的周期函数），其傅里叶级数同样一致收敛。函数的阶数越高，通常收敛速度越快 [29]。
系数绝对收敛: 如果 f(x) 连续，且其傅里叶系数绝对可和 ( 或 )，则傅里叶级数一致收敛 [6]。
蕴含的性质:一致收敛具有重要的理论意义。

保持连续性: 如果构成级数的各项函数（cos(nx),sin(nx)）是连续的，且级数一致收敛，那么其和函数 f(x) 也必定是连续的 [16]。这意味着，如果一个函数的傅里叶级数一致收敛，那么该函数必须是连续的（并且满足周期性边界条件才能在整个周期上连续）。
逐项积分: 一致收敛的函数项级数可以逐项积分 [16]。这意味着对于一致收敛的傅里叶级数，我们可以通过对其各项积分来得到原函数积分的傅里叶级数（常数项需要调整）。
逐项微分: 逐项微分需要更强的条件，即微分后得到的级数需要一致收敛 [16]。（详见第7节）。
一致逼近: 一致收敛保证了对于任意给定的精度，总能找到一个有限项的部分和，使得在整个区间 I 上的每一点 x，逼近误差都小于。
与逐点收敛的关系:一致收敛必然蕴含逐点收敛 [10]。然而，反之不成立。最典型的例子就是吉布斯现象：在跳跃间断点附近，傅里叶级数虽然逐点收敛（收敛到跳跃中点），但由于持续存在的过冲，收敛在包含该点的任何邻域内，收敛都不是一致的 [1]。此外，也存在连续函数，其傅里叶级数逐点收敛但非一致收敛 [8]。例如，某些精心构造的级数或在特定点收敛速度特别慢的级数 [12]。
一致收敛与函数的全局光滑性和连续性紧密相关。跳跃间断点是其主要障碍，因为吉布斯现象导致在间断点附近的最大误差无法任意小 [33]。因此，函数的连续性（特别是周期延拓后的连续性，即）是一致收敛的必要条件 [30]。确保一致收敛的充分条件，如若尔当判别法（连续+有界变差）或 Hölder 连续性，都对函数的整体行为施加了限制，排除了剧烈振荡或不连续性 [8]。值得注意的是，光滑性（f 连续，f′ 分段连续，f(−L)=f(L)）足以保证 f 的傅里叶级数 S(f) 一致收敛 [12]。其证明通常依赖于将 f 的系数与 f′ 的系数通过分部积分联系起来（），然后利用 f′ 的系数满足的贝塞尔不等式或帕塞瓦尔等式（这仅要求 f′ 是 L2 或分段连续）以及柯西-施瓦茨不等式来证明和收敛得足够快（至少像），从而保证 S(f) 的绝对收敛，进而保证一致收敛 [20]。这个过程并不要求 f′ 的傅里叶级数 S(f′) 本身一致收敛。
绝对收敛
绝对收敛是傅里叶级数最强的收敛形式之一，它关注的是系数本身的大小。
收敛条件:绝对收敛要求傅里叶系数衰减得足够快。

伯恩斯坦定理 (Bernstein's Theorem): 如果周期函数 f(x) 属于 Hölder 类 (或 Lip()), 且阶数 , 则其傅里叶级数绝对收敛 [8]。这个的条件是临界的；存在的 Hölder 函数，其傅里叶级数不绝对收敛 [8]。
Zygmund 定理: 如果 f(x) 具有有界变差且属于 Hölder 类 (对于某个 ), 则其傅里叶级数绝对收敛 [8]。这表明有界变差本身不足以保证绝对收敛，但结合任意阶的 Hölder 连续性即可。
足够的光滑性 (Ck): 如果函数 f(x) 足够光滑，例如且 , 则其傅里叶系数衰减速度至少为 , 这保证了或收敛，即级数绝对收敛 [29]。然而，如果仅仅是（f 和 f′ 连续），系数衰减速度通常为，不足以保证绝对收敛，尽管此时级数是一致收敛的 [20]。函数 f 及其导数 f′ 均属于 L2 空间也不能保证绝对收敛 [8]。
维纳代数 (Wiener's Algebra): 具有绝对收敛傅里叶级数的连续函数构成的集合被称为维纳代数，记作 [8]。伯恩斯坦定理和 Zygmund定理给出的就是函数属于维纳代数的充分条件。
蕴含的性质:绝对收敛是非常好的性质。

蕴含一致收敛: 绝对收敛的傅里叶级数必然一致收敛 [6]。这是因为，根据 Weierstrass M 判别法，若，则一致收敛。
和函数连续: 由于绝对收敛蕴含一致收敛，而一致收敛保持连续性，因此绝对收敛的傅里叶级数的和函数必然是连续的。
允许重排: 绝对收敛级数的一个重要特性是，级数的各项可以任意重新排列，其和保持不变 [24]。这在进行某些代数运算或变换时非常有用。
与一致收敛的关系:绝对收敛严格强于一致收敛。存在一致收敛但非绝对收敛的傅里叶级数 [21]。一个经典的例子是级数 [57]。这个级数可以使用狄利克雷判别法证明其一致收敛，但其系数的绝对值构成的级数发散（因为发散）[53]。
绝对收敛本质上是对傅里叶系数衰减速度的要求。它要求系数的绝对值之和有限，这意味着系数衰减速度必须快于 [39]。像或 Hölder 这样的光滑性条件恰好能保证这种快速衰减 [8]。而一致收敛则依赖于级数项之间的相互抵消（特别是在三角级数中，如狄利克雷判别法所示 [53]），以及部分和与函数之间的最大误差。的例子清晰地展示了这种区别：尽管系数的衰减相对较慢，不足以保证绝对收敛，但项的振荡和的单调递减性相结合，通过狄利克雷判别法可以保证级数一致收敛 [57]。这说明，即使系数大小衰减不够快，精巧的项间抵消也可能导致一致收敛。
收敛类型比较分析
傅里叶级数的三种主要收敛类型——逐点收敛、一致收敛和绝对收敛——构成了理解级数逼近性质的层级结构。它们之间存在明确的蕴含关系，对函数性质的要求也逐级递增。
层级关系与蕴含:这三种收敛类型构成了一个严格的强弱层级：绝对收敛一致收敛逐点收敛 [6]。

绝对收敛蕴含一致收敛是因为，如果，根据 Weierstrass M-判别法，由于，级数必然一致收敛 [6]。
一致收敛蕴含逐点收敛是根据它们各自的定义：如果级数在整个区间上一致地逼近函数，那么在区间内的每一点上自然也逼近函数 [11]。
反向不成立:
逐点收敛一致收敛：吉布斯现象是典型的反例，方波的傅里叶级数逐点收敛（在间断点收敛到均值），但在包含间断点的区间上不一致收敛 [1]。
一致收敛绝对收敛：级数是一致收敛的，但其系数绝对值之和发散 [57]。
函数性质要求:实现不同类型的收敛对函数的光滑性/正则性有不同的要求：

逐点收敛: 要求相对较弱，可以是局部条件（如在某点可微、满足 Dini 条件）或全局条件（如有界变差、属于 L2 空间（几乎处处收敛））。允许函数存在某些类型的不连续点（如跳跃间断点）[1]。
一致收敛: 要求更强的全局性质。函数通常需要是连续的（且满足周期性边界条件），并且具有一定的整体光滑度，如有界变差（Jordan 判别法）、满足 Hölder 条件或 C1 光滑 [8]。
绝对收敛: 要求最高的全局光滑性。通常需要函数属于 C2 或更高阶的光滑类，或者满足 Hölder 条件且阶数（Bernstein 定理）[8]。
收敛速度与系数衰减:函数的光滑程度直接影响其傅里叶系数的衰减速度，进而决定了收敛的速度和类型 [38]。
光滑性系数衰减:

f 有跳跃间断点 (分段连续): [38]。
f 连续, f′ 分段连续 (C1 分段光滑): [38]。
(k 阶连续可微): 或 (如果绝对连续或分段连续) [28]。
f 满足 Hölder 条件 : [8]。
(m阶导数 Hölder 连续) + 有限振荡: [62]。
衰减收敛类型/速度:

O(1/n) 衰减：通常只保证 L2 收敛和某些条件下的逐点收敛（可能伴随 Gibbs 现象）。收敛速度相对较慢。L2 误差 [39]。
且：
p=2 (C1 分段光滑): 保证一致收敛和绝对收敛。L2 误差 [39]。
p>1.5 (对应 Hölder ): 保证绝对收敛（从而一致收敛）[8]。
p 越大（函数越光滑），收敛速度越快。一致收敛误差通常与系数衰减速度相关，例如对于函数，误差可能为 [11] 或更快。
函数性质的保持:

连续性: 一致收敛可以保持连续性，即如果级数各项连续且一致收敛，则和函数也连续 [16]。逐点收敛则不保证保持连续性 [11]。绝对收敛因蕴含一致收敛，其和函数也是连续的。
可微性: 傅里叶级数的逐项微分需要更强的条件，通常要求微分后得到的级数一致收敛（见第7节）[16]。仅仅是原级数的一致收敛或绝对收敛通常不足以保证逐项微分的有效性。
表1：傅里叶级数收敛类型比较

核心关系在于函数的光滑性、系数衰减速度和收敛性之间的紧密联系。更光滑的函数通常对应更快的傅里叶系数衰减速度，而更快的系数衰减则能保证更强的收敛模式（从逐点到一致再到绝对）和更快的收敛速度 [39]。例如，从分段连续 () 到 C1 分段光滑 () 再到 C2 ( 或更快)，系数衰减速度加快，收敛性也从可能仅 L2 收敛或伴有 Gibbs 现象的逐点收敛，提升到一致收敛，再到绝对收敛。伯恩斯坦定理关于 Hölder 的阈值 [8] 明确指出了保证绝对收敛所需的最低光滑度界限。然而，各种收敛类型所需的条件差异显著，反映了它们衡量逼近程度的不同侧重。逐点收敛依赖局部性质 [8]，一致收敛需要全局误差控制 [27]，而绝对收敛完全取决于系数的大小 [23]。这解释了为何函数可能满足逐点收敛的局部条件却因全局问题（如间断点）而无法一致收敛 [33]，或者通过项间抵消实现一致收敛却因系数衰减不够快而无法绝对收敛 [57]。
逐项微分
傅里叶级数的一个重要应用是在求解微分方程等问题中，这自然引出了一个问题：能否对傅里叶级数进行逐项微分，并且得到的级数是否收敛到原函数的导数？答案是肯定的，但需要满足比级数本身收敛更强的条件。
Theorem (Statement). 定理陈述:若函数 f(x) 在上满足以下条件 [31]：
f(x) 处处连续。
f′(x) 在上分段光滑 (piecewise smooth)。这意味着 f′(x) 本身及其导数 f′′(x) 都是分段连续的。
f(x) 满足周期性边界条件。则 f(x) 的傅里叶级数可以逐项微分，得到 f′(x) 的傅里叶级数：并且这个微分得到的级数在 f′(x) 的所有连续点处收敛于 f′(x)。
一致收敛的角色:逐项微分的合理性依赖于级数收敛的性质。根据函数项级数的基本理论，若要保证和函数的导数等于级数逐项微分后的和，通常要求微分后得到的级数是一致收敛的 [16]。
原级数 S(f) 的一致收敛: 定理中的条件（f 连续，f′ 分段光滑，f(−L)=f(L)）实际上保证了原级数 S(f) 是一致收敛的 [31]。这是因为 f′ 分段光滑意味着 f′ 有界变差，从而 f 满足 Hölder 条件（），结合连续性和周期性边界条件，足以保证 S(f) 一致收敛。
微分级数 S(f′) 的收敛: 定理的关键在于确保微分后的级数 S(f′) 收敛。
条件 f′(x) 分段光滑保证了 f′(x) 本身满足傅里叶级数逐点收敛的条件（如狄利克雷条件或若尔当判别法）[17]。
因此，微分得到的级数（即 f′ 的傅里叶级数）至少在 f′ 的连续点处逐点收敛于 f′(x)。
如果 f′ 本身是连续的（即），那么 S(f′) 也会一致收敛。
系数关系:通过分部积分可以建立 f′(x) 的傅里叶系数 () 与 f(x) 的傅里叶系数 () 之间的关系 [20]：

由于且，边界项为零。
因此：

同理可得：

(注意：)
这与逐项微分得到的级数形式一致。
复数形式下，关系更简洁：。
光滑性的影响:函数 f 的光滑性越好，其导数 f′ 的光滑性也越好。
如果且，那么且。
这意味着 f′ 至少是 C1 的，根据一致收敛的条件，f′ 的傅里叶级数 S(f′) 必然一致收敛。
因此，对于足够光滑的函数（至少 C2），逐项微分是完全合理的，并且得到的级数一致收敛于导函数 [39]。
逐项微分并非无条件成立。
关键在于微分运算会给系数带来一个因子 n () [20]。
如果原级数的系数衰减速度不够快（例如仅为 O(1/n)），那么乘以 n 后的系数可能不再趋于零，导致微分后的级数发散 [35]。
这就是为什么定理要求 f 连续且 f′ 分段光滑：f 连续保证了至少以的速度衰减（因为 f′ 分段光滑意味着 f′ 是 L2 的），这样乘以 n 后的系数仍然以 O(1/n) 的速度衰减，足以保证 f′ 的傅里叶级数 S(f′) 至少逐点收敛（根据有界变差函数的收敛性）[31]。
结论
核心关系与层级: 这三种收敛类型构成了一个清晰的强弱层级：绝对收敛是最强的，它蕴含一致收敛；一致收敛强于逐点收敛。
然而，这些蕴含关系的反向通常不成立，表明它们各自描述了级数逼近函数的不同侧面。
光滑性是关键: 函数的光滑程度是决定其傅里叶级数收敛行为的核心因素。
更光滑的函数（具有更多阶连续导数，或满足更高阶的 Hölder 条件）通常对应着更快的傅里叶系数衰减速度。
这种更快的衰减是实现更强收敛模式（从逐点到一致，再到绝对）和更快收敛速度的基础。
例如，从分段连续到 C1 再到 C2，系数衰减率从 O(1/n) 提升到 O(1/n2) 或更快，收敛性也相应增强。
条件与局限:

逐点收敛依赖于函数的局部性质（如 Dini 条件、可微性）或全局的有界变差性质。它允许函数存在跳跃间断点（级数收敛到均值），但面临 Gibbs 现象和可能发散的问题（即使对连续函数）。
一致收敛要求函数具有全局连续性和一定的光滑度（如连续且有界变差、Hölder 连续、C1）。它保证了逼近误差在整个区间上一致趋于零，并能保持函数的连续性。
绝对收敛对函数光滑性要求最高（如 C2、Hölder ），它保证了系数绝对值级数的收敛，从而得到最强的收敛保证，包括一致收敛和允许项重排。
应用意义: 理解这些收敛类型的差异对于理论研究和实际应用至关重要 [1]。

在信号处理中，L2 收敛（均方收敛）与信号能量相关。
在数值计算和近似理论中，一致收敛提供了全局误差界限，保证了近似的可靠性。
在偏微分方程的级数解中，逐项积分和微分的合理性通常依赖于（微分后级数的）一致收敛。
绝对收敛则为级数的代数运算提供了便利。
尽管存在收敛性方面的复杂性和挑战（如发散的反例），傅里叶级数仍然是一个极其强大的工具。
对于工程和物理中遇到的大多数“行为良好”的函数（通常足够光滑），其傅里叶级数都具有良好的收敛性质 [1]。
即使对于连续但不够光滑的函数，可和性理论（如 Cesàro 平均）也提供了替代的收敛方式，保证了结果的稳健性 [2]。
深入理解傅里叶级数的不同收敛模式及其条件，是有效运用这一工具解决科学和工程问题的基础。

实数的完备性

Fri, 02 May 2025 16:55:30 GMT

44．证明下列命题是等价的：
（I）Dedekind 分割原理；
（II）确界存在原理；
（III）单调有界原理；
（IV）区间套定理；
（V）有限覆盖定理；
（VI）聚点原理；
（VII）有界数列必有收敛子列；
（VIII）Cauchy 收敛准则．
证明：
：只要证明，若数集有上界，则必有唯一的上确界。
若有最大值，则就是的上确界。
设为无限集且无最大值。做实数集的分割，其中为的一切上界所组成之集，为的补集。于是
（i），
（ii），
（iii）任取，若存在使，则成为的一个上界，从而，这与（ii）矛盾。故对任意，均有。
因此，是一个 Dedekind 分割。由分割原理，存在唯一，使对任意有

因不可能是的最大值（若是的最大值，则，从而，矛盾！）故对一切，均有，即不能成立。这样，只有成立，即是中的最小者，也就是的上界中的最小者，故有唯一的上确界。
：不妨假设单调增加且有上界。据确界存在原理，必有上确界，即

兹证就是当时的极限。
事实上，由上确界定义知对任意，存在，使

再据单调增加，可知对任何均有

因此，当时有，即

：设闭区间列满足条件：
（i），
（ii）。
由（i）知数列单调增加且有上界，故收敛。令

又因对每一 , 是的上界，且是的最小上界，故

由（1），（2）可知，即

若还存在实数使

则，故。
：设是闭区间的一个开覆盖。假如不能被中任何有限个开集覆盖，将等分为两个区间，则其中至少有一个区间不能被中任何有限个开集覆盖，记此区间为。再等分，同样至少有一个不能被中任何有限个开集覆盖，记此区间为。如此继续下去，得到一列闭区间

适合
（i）任何一个都不能被中任何有限个开集覆盖。
（ii）。
（iii）。
据区间套定理，存在唯一实数，且

因覆盖了，故中至少有一个开集从而至少有一个开区间，使得

由极限性质知存在，当时有

即。因此，覆盖了。这与（i）发生矛盾。
：设为有界的无限集。令

则。假如没有聚点，那么对任意，存在的邻域，使中至多含有的有限个点，即是有限集。显然，当走遍时，这些邻域就覆盖了，即

据有限覆盖定理，存在有限个邻域，它们足以覆盖，即

从而。由此得到为有限集，此为矛盾。因此，必有聚点。
：设是有界无穷数列。若是由有限个实数重复出现而构成的数列，则至少有一个数要重复出现无穷多次。设重复出现的项为，则

即是的一个收敛子列。
现设确由无穷多个不同的实数组成，则此数列为一有界无穷集。据聚点原理，至少有一聚点。于是，对任何，中必含有的无穷多项，从而在中可以选出的一个项使。因是任意正整数，故得的一个子列，使得，因而

：必要性是显然的。
兹证充分性。据条件，对，存在，当时有

于是，

令

则，故有界。因此存在收敛子列，设

于是由不等式

可知，。
：设是全体实数的任意一个分割。因，故可任取，则。将等分为二，若分点就取右半区间并记为；若，则取左半区间并记为。总之，。如此继续下去，得到闭区间列，满足
（i），
（ii），
（iii）。
由（i），（ii）可知数列

是 Cauchy 数列，因而收敛。设其极限为。若，可证必为的最大值。事实上，假如存在而有，取正数，则

由（iii），每个，这与收敛于发生矛盾。因此，为的最大值。此时显然无最小值。类似地可证，若在中，则必是的最小值，此时无最大值。
注：本题中八个等价的命题从不同的角度刻画了实数的连续性（又称完备性）。

chap1

Fri, 02 May 2025 16:55:29 GMT

第一节介绍正整数（自然数）集合具有的两个重要的性质，这两个性质还是等价的。它们分别是良序原理（任意非空正整数集有最小数）和数学归纳法原理（若正整数集含有 1 且若则那么 .）
Proposition. 1 是最小的正整数
Proposition. 数学归纳法原理和良序原理是等价的
这个证明比较繁琐
接下来是一个应用
Proposition. 大于 1 的整数要么是素数，要么在不计次序的情况下可以唯一分解为素数的乘积
若合数不能写成素数乘积，不妨设是最小的这样的合数（由良序原理），由于是合数，就可以写作其中 . 于是要么是素数，要么是素数的乘积（因为是最小的不能写成素数乘积的合数），故也是素数乘积，故矛盾！
其次证明唯一性：若存在合数，它的因子分解不是唯一的，，不妨设是最小的这样的数（由良序原理）显然所有的都和互异，否则可以消去，这与的最小性矛盾。不妨设，不妨设，考虑，显然，将写成素数乘积，于是，同时，这是不同的分解，因为前者中没有，于是取代成为最小的因子分解不唯一的合数，矛盾！
Proposition. 没有最大的素数
若是前面所有的素数，那么又是一个素数。
Proposition. 若是素数，那么是无理数
若则，即，那么左边有奇数个素因子，右边有偶数个，矛盾！
Proposition. 对于非平方自然数，，，那么没有最小数，没有最大数

Definition (分割). 基本定义有理数的一个集合叫做一个分割，倘若
（i）至少含有一个有理数，但不含有所有有理数；
（ii）若及，其中为有理数，则；
（iii）不含最大的有理数．
Proposition. 若则
中的元素称为的下数，不在的有理数称为的上数，上数集记作 .
Proposition. 对于，，那么是一个分割. 且是的最小上数.
上述分割叫做一个有理分割，若是由此从做出的有理分割，则记作 .
Definition (分割之间的序关系). 若且，那么 . 类似定义 . 简单验证可知这是序关系.
Proposition (分割的加法). ，则是分割，记作 .
Proposition (有理数的阿基米德性质). 若，则存在，使得 .
定义是一个分割. 的真上数指的是中的非最小数.
Proposition (分割的减法). 若是分割，则存在唯一使得 . 记作 .
置，则，若使得，那么 .
Proposition (分割的乘法). ，是一个分割，记作 .
Definition (分割的绝对值). 若则；若则 .
若或那么，否则 .
Proposition (分割的除法). 对于，是一个分割，且 . 记作或 .
若，那么
Proposition. 对于任意分割，存在唯一分割满足 .

hw1

Fri, 02 May 2025 16:55:29 GMT

Exercise.
证明：是代数。
证明是代数。
证明是代数。
Definition (代数). 代数是一个非空集合和一个二元运算，满足：
封闭性：，
我们通常将代数记为。

简单验证：

简单验证：

简单验证

Exercise. 证明关于矩阵乘法是非 Abel 代数
首先，我们需要明确非 Abel 代数的定义：
Definition (非 Abel 代数). 一个代数称为非 Abel 代数（或非交换代数），如果存在，使得。
现在证明关于矩阵乘法是非 Abel 代数：
首先验证关于矩阵乘法是一个代数：
对于任意

我们有矩阵乘法下

其中每个元素都，于是，故关于矩阵乘法是一个代数。
接下来验证关于矩阵乘法是非 Abel 代数：
取

于是

于是关于矩阵乘法是非 Abel 代数。
Exercise. 证明是 Abel 环
即证为 Abel 群，为 Abel 代数，且有乘法单位元，满足乘法结合律
Definition (Abel 环). Abel 环是指满足以下条件的代数结构：
是 Abel 群
是半群（满足结合律）且有单位元
乘法对加法满足分配律
乘法满足交换律（即，）
首先验证是 Abel 群：

封闭性：对于任意，设，，则，其中若则，若则。由于整数对加法封闭，所以，因此。
结合律：对于任意，有，这是因为多项式系数的加法满足结合律。
单位元：零多项式是加法单位元，对任意，有。
逆元：对于任意，其加法逆元为，满足。
交换律：对于任意，有，这是因为多项式系数的加法满足交换律。
接下来验证是 Abel 代数：

封闭性：对于任意，，它们的乘积为，其中。由于整数对乘法和加法都封闭，所以，因此。
结合律：对于任意，有，这是多项式乘法的基本性质。
单位元：多项式是乘法单位元，对任意，有。
交换律：对于任意，有，这是因为多项式乘法满足交换律。
最后验证乘法对加法的分配律：
对于任意，有：

综上所述，是 Abel 环。
Exercise. 是素数，，其中表示模余的剩余类，证明：是 Abel 域
即证为 Abel 环，且任一非零元有乘法逆元

Exercise.
定义。证明：是偏序集，而非全序集
设为非空集合，为的子集构成的集合，为集合的包含关系，证明为偏序集，而非全序集

Exercise. 将整数集良序化

hw3

Fri, 02 May 2025 16:55:29 GMT

Exercise. 证明：Cauchy 列改变前任意有限项仍为 Cauchy 列．

假设改变的有限项中指标最大的为，那么对于任意的，存在正整数是的当时，

故得证 !
Exercise. 用几何方法证明命题 1．1．10．（提示：有理数的稠密性）
Proposition (命题 1.1.10). 是中的 Cauchy 列的小数表示．
显然！
Exercise. 证明：若是有理 Cauchy 列，则也是有理 Cauchy 列．

显然 . 由于是有理 Cauchy 列，故对于任意，存在，使得对于任意 , 有

于是

因此也是有理 Cauchy 列.
Exercise. 证明：
（1）戴德金分割有理分割；
（2）有理分割有理 Cauchy 列的等价类．
Definition (戴德金分割).
Definition (有理分割).
戴德金分割有理分割：根据戴德金原理，对于实数域的戴德金分割，要么下类中存在最大数，要么上类中存在最小数。无论如何，记这个数字为 , 那么

或者

这是个有理分割。显然有理分割戴德金分割。故戴德金分割有理分割

若是有理分割，分点为，它对应等价类。若
Exercise. 设数列有界．证明：数列极限存在只有唯一的聚点．
若是数列的极限，这等价于对于任意的，存在，使得

显然是的聚点，假设也是聚点，那么存在，使得

于是对于，有

由任意性可知

只有唯一聚点。
另一个方向是显然的！
Exercise. 证明确界存在定理．
根据戴德金定理：只要证明，若数集有上界，则必有唯一的上确界。
1．若有最大值，则就是的上确界。
2．设为无限集且无最大值。做实数集的分割，其中为的一切上界所组成之集，为的补集。于是
（a）．
（b），
（c）任取，若存在使，则成为的一个上界，从而，这与（b）矛盾。故对任意，均有。
因此，是一个 Dedekind 分割。由分割原理，存在唯一，使对任意有

因不可能是的最大值（若是的最大值，则，从而．矛盾！）故对一切，均有，即不能成立。这样，只有成立，即是中的最小者，也就是的上界中的最小者，故有唯一的上确界。
Exercise. 设是正整数，给定个实数，构造数列使得是其聚点．
我们可以构造数列，. 接下来令

其中 . 于是是的聚点.

hw4

Fri, 02 May 2025 16:55:29 GMT

Exercise.
Proof.

因此

□
Exercise.
(1)
Proof.

将写作，于是在内稠密，由于连续，所以在内稠密.
□
(2)
Proof.
为了验证等分布，只需验证对于任意，，我们有

这等价于证明：

我们先证明如下引理：
Lemma. 若是周期 1 的连续函数，是无理数，那么

证明思路是是采用三角级数逼近.
Proof.
首先我们验证上式对于都成立，若，那么显然成立. 若，那么

显然，如果都满足 Lemma 3，那么也满足，其中 . 故 Lemma 3 对于三角多项式都成立.
接着我们对于任意的连续，存在三角多项式，使得

同时

因此

□

□
(3)
Proof.
由前两问，这是显然的
□
Exercise.

其中

与无关，可以任意小.
故

收敛
Exercise.
时发散，时收敛.

hw5

Fri, 02 May 2025 16:55:29 GMT

Proof.
For any , . Thus . For any , then . Then , thus . Hence .
□

Theorem (theorem 2.14 in baby rudin). Let be the set of all sequences whose elements are the digits and . This set is uncountable.
The elements of are sequences like
Let be a countable subset of , and let consist of the sequences . We construct a sequence as follows. If the th digit in is , we let the th digit of be , and vice versa. Then the sequence differs from every member of in at least one place; hence . But clearly , so that is a proper subset of .
We have shown that every countable subset of is a proper subset of . It follows that is uncountable (for otherwise would be a proper subset of , which is absurd).
Therefore, .

Proof.
采用蔓延的思想，对于任意，考虑，记

显然 . 这里可能为，可能为 . 显然区间 . 并且为开集，这是因为对于任意，存在，若，这与定义矛盾！故 .
存在如下映射

显然是单射，因此 .
□

不存在

(1) 若 , 则，不是的内点，也就是说的“附近”不包含于中. 对于任意充分小的，. 由疏集的定义，对于中的任意非空开集，若，那么存在使得，存在，选取的子集，这个集合与不相交，因此是疏集.
若是疏集，但 . 那么存在，使得存在 . 选取区间，那么它有非空开子集与不相交，这不可能发生在的拓扑上！
因此，是疏集等价于 .
(2) 这是显然的.

is a Julia set.
JuliaSetPlot[2 z^2 + 3 z + 1, z, ColorFunction -> None]

hw6

Fri, 02 May 2025 16:55:29 GMT

Exercise.

(1) 显然且

(2) 显然
(3) 不妨设，若那么显然有

若，则

因此是一个距离.
Exercise.

(1) 显然，若，那么 . 因此 .
(2) 显然 .
(3)

因此是上的一个距离.
Exercise.
是开集，则是闭集，但是

故是开集，故既开又闭，但不是全集也不是空集. 这样的定义不合适

hw7

Fri, 02 May 2025 16:55:29 GMT

Exercise.
Definition (inner product space).
Proof.
(1) Let

Then for any , a scalar, we have

Let then

(2)
Let

Then for any , scalar , we have

□
Exercise.

means for any , there exists s.t. for any satisfying , we have

not existing means that there exists , for any , there exists satisfying and

Exercise.
Proof.
(1) Pick any sequence , . Denote . For any , there exists , s.t.

Then

For any , we have

Thus converges. By Heine's theorem,

exists.
(2) When , we have

Let , we have

Let then , as . But

Thus does not exists, while exists, exists. Hence

does not exists.
□
Exercise.

Proof.
(1) for any , check that

因为关于充分大的单调有界趋于，且级数收敛. 由 Abel test，级数收敛.
(2) 任意给定，则

令，那么下面两个级数收敛

考虑数列，于是

故不存在.
□
Exercise.

Proof.
For any given , , the limit

exists. And

Then

i.e. the limit exists, i.e. converges on .
□
Exercise.
Proof.
(1)
For any , for given , exists. And

By dominate convergence theorem, converge on .
(2)
Let , then

Thus diverges on .
□
Exercise.
Proof.
Pick any then for any , we have

Similar for , then is uniformly continuous on .
For any , pick then for any , we have

Thus is uniformly continuous on . Hence is uniformly continuous on .
□
Exercise.
Proof.
For any , for any given we have

exists.

By dominate convergence theorem, converge on .
□

hw8

Fri, 02 May 2025 16:55:30 GMT

Exercise.
Prove the important limit .
Proof.

When ,

Thus for ,

Let , then

i.e.

Since is continuous,

□
Exercise.
If is strictly monotonically increasing on , , then .
Proof.
Since and is strictly monotonically increasing on , for any , there exists , s.t.

Otherwise for some , which is a contradiction. Since is strictly monotonically increasing, we have

Thus . Since is arbitrary, we have

□
Exercise.
Find .
似乎题出错了.

Exercise.
is defined in . Prove that if , then but the converse is not necessarily true.
Proof.
If then for any , there exsits , s.t.

Then for any , , then

Thus .
Conversely, a conterexample is

For , the limit but .
□
Exercise.
Let , denote , and . Then for ,
Proof.
By the definition of and , there are two sequences , s.t.

For any , there are infinitely many elements of lying in , and infinitely many elements of lying in . WLOG, assume that and . By the continuity of , applying the IVT, there exists s.t. . Thus we have a sequence with .
□
Exercise.
(1) Does there exist a function that is continuous at irrational points and discontinuous at rational points? (2) Does there exist a function that is continuous at rational points and discontinuous at irrational points?
(1) Yes.
Example (Raindrop function). Define Thomae's function as follows: Then is continuous at every irrational point and discontinuous at every rational point.
(2) No, there does not exist a function that is continuous at rational points and discontinuous at irrational points. The set of points at which a function is continuous must be a set (a countable intersection of open sets). The set of rational numbers is not a set.
Theorem. The set of points at which a function is continuous must be a set.
Proof.
Let be a function. Let be the set of points at which is continuous. We want to show that is a set.
For each and each , define

is open for each , since it is the union of open intervals.
We claim that if and only if for all , . In other words, we claim that if and only if for every there exists an open interval such that and .
Then if and only if for every there exists such that if , then . Therefore, if and only if for all .
Thus, . Since each is open, is a set.
Now we prove the if and only if statement.
If is continuous at , then for every , there exists such that if , then . Let and , then and . So for all .
If for all , , then for every there exists an open interval such that and . Let be given. Choose large enough such that . Then there exists an open interval such that and . Choose such that . Then if then . So is continuous at .
□
Exercise.
Prove that the Cantor function on is continuous.
Definition (Cantor function). The Cantor function is defined as follows:
Write in base 3 : where .
If has a 1 in its ternary expansion, truncate the expansion at the first occurrence of a 1. Change every digit after that first 1 to 0.
Replace all 2's with 1's.
Interpret the result as a binary number. This gives the value of the Cantor function .
Proof.
Let . We want to show that for every , there exists a such that if , then .
Let . Choose such that .
Choose . Suppose .
Write and .
Since , we must have for all .
Let and be the binary representations of and respectively.
Since for all , the first digits of the ternary expansions of and are the same.
When we apply the Cantor function, we truncate the ternary expansion at the first occurrence of 1, and then change all 2's to 1's. Therefore, the first digits of the binary expansions of and are either the same, or differ by the truncation and replacement rules.
In any case, .
We can also say .
Thus, is continuous.
□
Alternative proof:
Proof.
Let .
Let where .
The Cantor function is continuous everywhere except at points where has a 1 in its ternary expansion.
If has a 1 in its ternary expansion, truncate the expansion at the first occurrence of a 1. Change every digit after that first 1 to 0.
Replace all 2's with 1's.
Interpret the result as a binary number.
Let in base 3.
Then in base 2, where if and if .
Let in base 3.
Then in base 2, where if and if .
Since , we have .
Therefore, is continuous at .
□

hw9

Fri, 02 May 2025 16:55:30 GMT

Exercise.
Proof.
(1)

(2)

正确的复数极化恒等式为

□
Exercise.
Proof.
显然

距离函数

就不满足正齐次性.
□
Exercise.
Proof.
For any satisfying , by Holder's inequality,

Thus

Similarly,

Pick s.t.

Then

□
Exercise.
在数学领域中，术语“NLP”通常指的是非线性规划 (Nonlinear Programming)。
非线性规划是优化理论的一个分支，研究的是在非线性等式或不等式约束下，如何最小化或最大化一个非线性目标函数的问题。
一个一般的非线性规划问题可以表述为：
最小化（或最大化）
约束条件为：

其中：

是决策变量向量。
是目标函数，它是一个非线性函数。
是不等式约束函数，它们可以是线性的或非线性的。
是等式约束函数，它们可以是线性的或非线性的。
是变量的定义域或简单的边界约束，通常是一个凸集。
如果目标函数或任何一个约束函数或是非线性的，那么这个问题就被称为非线性规划问题。作为对比，如果目标函数和所有约束函数都是线性的，那么问题就是线性规划（LP）问题。
非线性规划问题的求解通常比线性规划复杂得多，一般没有统一的，能在多项式时间内找到全局最优解的方法（除非问题具有特定的结构，如凸规划）。常用的求解方法包括梯度下降法，牛顿法，序列二次规划（SQP）等。
请注意，您可能更熟悉在人工智能和计算机科学领域中的“NLP”，它代表的是自然语言处理（Natural Language Processing）。但在数学的优化或运筹学等领域中，“NLP”几乎总是指代非线性规划。
Exercise.
Proof.
Check:

is normed linear space.
is complete.
To check that is a linear space, we need to verify that it satisfies the axioms of a vector space.

Closure under addition: If , then is also continuous on . Thus, .
Closure under scalar multiplication: If and is a scalar, then is also continuous on . Thus, .
Commutativity of addition: For all , for all , so .
Associativity of addition: For all , for all , so .
Existence of additive identity: The zero function for all is continuous on , so . For any , for all , so .
Existence of additive inverse: If , then is also continuous on , so . For any , for all , so .
Compatibility of scalar multiplication with field multiplication: For all scalars and , for all , so .
Identity element of scalar multiplication: For all , for all , so .
Distributivity of scalar multiplication with respect to vector addition: For all scalars and , for all , so .
Distributivity of scalar multiplication with respect to field addition: For all scalars and , for all , so .
Since satisfies all the axioms of a vector space, it is a linear space.
Definition (Normed linear space). A normed linear space is a linear space on which there is defined a real-valued function that satisfies (i) for all , and if and only if .
(ii) for all .
(iii) for all and all scalars .
Clearly, .

Definition (Complete Space). A metric space is called complete if every Cauchy sequence in converges to a limit that is also in .
Suppose there is a Cauchy sequence of functions in . Check that converges in . Let , then for any , there exists s.t. ,

Let then

Check that . For any , there exists , s.t.

Then

, then we are done.
□

hw10

Thu, 08 May 2025 15:50:01 GMT

Exercise.
For any , put , then

Thus is uniformly continuous on .
Exercise.
这个习题很有挑战性！
Remark. 参见 Stein 习题 3.16 (d)

时一致收敛，时存在反例.
反例参见《实分析中的反例》汪林

证明太长了

A simpler version in Harmonic Analysis, Katznelson, p.33 :

Exercise.
du Bois-Reymond showed an counterexample.
参见于品《数学分析讲义 123》p.660

Exercise.
题目可能出错了，在上无定义.
Since and are continuous differentiable out of the neighborhood of 0, is continuous differentiable.
For ,

For ,
Exercise.
在的阶-Holder 常数定义为

这是上导数定义的推广（推广到分数阶）.
(1)
对于，当时，

于是

类似的，对于，

当时，

于是

因此

(2)
For , small enough

Thus

Similarly, for ,

When ,

Then

Hence

Exercise.
若则称；若还有，，则称 .
由此，(1) 是显然的.
(2)
Too hard.
选择合适的希尔伯特空间和带有合适边界条件的定义域（例如，Dirichlet条件下的，或上的)，算子就是一个无界正算子。

hw11

Mon, 19 May 2025 17:09:57 GMT

Exercise.
For , , then , thus ,

Let , then , i.e.

We are done!
Exercise.
(1)
There exists , such that

Then for any

where .
(2)
If

for some . Let , then . Thus is continuous at .
Definition (变分).
Definition (Holder 条件).
Exercise.
Proof.
由于将个分点一次插入与将这个分点逐个插入，所得到的都是分割，因此我们可以考虑将这个分点逐个插入的情况。设步长为 .
设

先考虑在中插入一个分点，记对应的新分割为，不妨设被插入小区间中，于是

这样，对应于分割的小区间有个，它们包含分割的个完全相同的小区间以及被分成的两个子区间 , . 记

则

由于

所以 , 即 . 从而有

对于在中插入两个分点对应的分割，可以看成在中插入一个分点，于是有 , 且 . 从而

分割是由分割增加个分点所得的新分割，则

对于不等式的第二部分：当从添加一个分点得到时，这个新分点必然落入的某个子区间，设其长度为。根据单点插入的结果，我们有： . 由于是的一个加细分割（或者是本身），其任何子区间的长度都不会超过的最大子区间长度，即。所以，对于。现在我们可以进行迭代：
这就证明了不等式的第二部分：。综上所述，对于下和，我们已经完整证明了：命题证毕。
□
Exercise.
只需要证明

对于任意给定的的划分 , 由定义可知

令就有

另一方面，对于任意给定的的划分 , 由 Exercise 2,

令就有

再对取就有

故

类似可证：

Definition (上积分的定义). 设是一个有界函数。对于的一个划分，其中，令。上达布和定义为。上积分定义为所有上达布和的下确界：
Definition (下积分的定义). 设是一个有界函数。对于的一个划分，其中，令。下达布和定义为。下积分定义为所有下达布和的上确界：
Definition (Darboux 上和与下和). 设是定义在上的有界函数, 是的一个分割. 记和 . 定义关于分割的 Darboux 上和为定义关于分割的 Darboux 下和为
Exercise.
由于闭区间单调函数有界，单调函数间断点至多可数，由 Lebesgue 定理可知，闭区间单调函数 Riemann 可积.
Theorem (Lebesgue 定理). 是有界的. 则黎曼可积当且仅当的不连续点集是勒贝格零测集.
下面我们给出一个初等的证明：

测地曲率和测地线(计算版本)

Fri, 02 May 2025 16:55:29 GMT

通常我们把欧式平面看作是二维平直空间 (Gauss 曲率为零)，而把给定第一基本形式的抽象曲面称为二维弯曲空间. 本章的目标就是研究二维弯曲空间中的几何学.

对于正则参数曲面满足方程 . 是曲面上的一条曲线，它的方程是，其中是曲线的弧长参数. 那么的参数方程为

我们的目的是沿着曲线建立一个新的正交标架场，兼顾和，定义如下

直观上，是将曲线的切向量绕着的单位法向量正向旋转得到的. 改标架场沿着曲线的运动公式为

其中是曲线的曲率向量在曲面的法向量上的正交投影，故恰好是曲面上的曲线的法曲率；是曲线的曲率向量在曲面的切平面上的正交投影。这里的的计算公式是

把最后的式子展开得到

曲线作为曲面内的曲线的测地曲率和它作为空间曲线的曲率的关系式是

这里是曲线的次法向量和曲面的单位法向量之间的夹角．
沿曲线的上述正交标架场的运动公式中的不是属于曲面的内蕴几何学的量，它的计算公式是

称为曲线的测地挠率. 实际上，测地挠率和法曲率的性质相同，都是曲面在任意一点的切方向的函数，与曲线本身的弯曲性无关.

Definition (测地线). 在曲面上测地曲率恒等于零的曲线称为曲面上的测地线. 曲面上的测地线是属于曲面的内蕴几何学的概念.
因为平面曲线的测地曲率就是它的相对曲率，因此平面上的测地线就是该平面上的直线．由此可见，曲面上的测地线的概念是平面上的直线概念的推广．

曲面上的测地线作为的外围空间中的曲线的特征是：或者曲线本身是直线，或者它的主法向量处处是曲面的法向量．从运动学观点来看，测地线的特征是：如果在曲面上运动的质点只受到将它约束在曲面上的力的作用（即作用力的方向垂直于曲面），则点的轨迹是曲面上的测地线．

曲面上的测地线所满足的内在特征是：对于它在曲面内的任意一个有固定端点的变分而言，的弧长是变分曲线的弧长的临界值．

极小曲面

Fri, 02 May 2025 16:55:29 GMT

参见彭家贵《微分几何教材》
Definition (极小曲面). 中平均曲率处处恒为零的曲面称为极小曲面.

第一基本形式系数

中的曲率函数

即曲面为极小曲面．

空间曲线(不严谨)

Fri, 02 May 2025 16:55:29 GMT

对于中的正则曲线 () ，引进参数使得

被称为曲线的 (弧长参数. 一个显然的事实是，正则曲线的参数为正则参数的特征是 . 一般的计算中，未必是弧长参数.
关于空间曲线的理论，最重要的是沿曲线定义的 Frenet 标架和 Frenet 公式.
假定空间曲线的参数方程为，其中是弧长参数，那么它的
单位切向量：

曲率：

曲率非零时，主法向量 (这样的定义是自然的)：

次法向量：

挠率[1]

曲线的 Frenet 公式是

上式都是关于弧长参数的导数.

首先根据定义知道 . 其次根据的定义知道

根据最原本的定义可知

接下来待定系数求解，分别与作内积得到

从而

我们完成了 Frenet 公式的推导证明！

在曲线的一般参数下，设曲线的参数方程是，则它的单位切向量是

假定曲线的弧长参数是，则，所以

因此

故

由此得到曲线的曲率是

次法向量是

这样，曲线的主法向量是

再利用 Frenet 公式得到

此时，单位切向量，主法向量和次法向量的导数是

Theorem (空间曲线基本定理). 给定两个连续可微函数其中，则在三维欧式空间中存在一条空间曲线，以为弧长参数，以为曲率，以为挠率，并且这样的曲线的形状是完全确定的.
Remark. 在给定的情况下，Frenet 公式构成了向量的微分方程组，可以求解.

两条相交曲线在交点附近的接近程度是用所谓的切触阶来刻画的.

平面曲线可以看作空间曲线的特例，即的空间曲线. 空间曲线求曲率的公式照用. 特别的是，平面本身有定向，将其单位切向量正向 (逆时针) 旋转便得到法向量 (唯一确定). 于是

正向旋转得到[2]

相对曲率

相对曲率与曲率的关系是 . 正号表示曲线朝着的方向弯曲，负号表示曲线的主法向量为 .

已知参数方程，直接套公式爆算：

已知一般方程：

直接通过对方程求导，求解出 .

根据曲线论基本定理具有这样常数曲率和挠率的曲线必定是圆螺旋线

Remark. 本题也可以根据 Frenet 标架直接求解微分方程组.

挠率一开始的定义就是使得的函数，下面这个只是计算式.↩︎
不一定是主法向量↩︎

曲面的整体性质

Fri, 02 May 2025 16:55:29 GMT

参见《微分几何学习指导》徐森林

曲面上点为脐点，当且仅当存在使得

曲面论(不严谨)

Fri, 02 May 2025 16:55:29 GMT

为了说明构成曲面的完全的不变量系统，需要曲面论的基本公式，也就是曲面的自然标架场的求导公式．它们在曲面的理论中扮演基本的角色，相当于曲线论中的 Frenet 公式．
Einstein 和式约定：

规定希腊字母作为指标的取值范围为 . 规定拉丁字母作为指标的取值范围为 .

Theorem (Gauss 绝妙定理). 曲面的 Gauss 曲率是曲面在保长变换下的不变量.
事实上，由 Gauss 方程可知

因此曲面的 Gauss 曲率是由它的第一基本形式完全确定的. 在的情形，Gauss 曲率用曲面的第一基本形式的表达式是

在曲面的等温参数系下，，则 Gauss 曲率的表达式是

Remark. Gauss 绝妙定理是微分几何学发展过程中的里程碑，开创了内蕴几何学的新时代，进而引发了 Riemann 几何学.

曲面形式(不严谨)

Fri, 02 May 2025 16:55:29 GMT

参考微分几何例题详解和习题汇编·陈维恒.
正则参数曲面[1] 是满足的连续可微映射

在每一点处有确定的标架，其中

称一个变换为正则参数变换，若

而且有足够的可微性.

其中 .

直纹面可以表示为

其中是直纹面的准现，是直母线的方向向量.

可展曲面是一种特殊的直纹面. 它的切平面沿每一条直母线是不变的. 曲面是可展曲面的充要条件是

Exercise. 例题 3.2 写出单叶双曲面作为直纹面的参数方程．
直接强行因式分解即可.

Exercise. 例题 3.5 证明：一个正则参数曲面是球面的一部分的充分必要条件是，它的所有法线都经过一个固定点．
必要性显然，下面考虑充分性：假定曲面的所有法线都经过一个固定点 , 即存在函数使得

下面证明是常值函数. 将 Equation (1) 分别对求导数得到

因为是切向量，又因为单位向量函数的偏导数必定与其它自身正交，因此用与 Equation (2) 做内积得到 . 这说明，于是

因此 .

Exercise. 例题 3.6 证明：旋转面的法线必定与旋转轴平行或相交；反过来，如果一个正则参数曲面的所有法线都与一条固定的直线相交，则它必定是旋转面．
必要性显然. 因为旋转面的参数方程为

其中 .
充分性的证明需要发挥一点几何想象力．不妨假定曲面的所有法线都经过－轴，用一个通过－轴，并且与平面的夹角为的平面截曲面，其截线的参数方程可以假设为，这就是说该截线上的点到－轴的距离是，到平面的距离是，这里是该截线上的参数，是任意的固定值．现在要证明：函数与无关，因而该曲面就是一个旋转面.[2]

Exercise. 例题 3.10 设球面的参数方程是（参看例题 3．3（1））求它的第一基本形式．
直接计算 .

先写成第一基本形式，然后再配方换元.

Exercise. 求单参数平面族的包络：

命

则

将方程组中的参数消去得到

这是一张柱面，属于可展曲面的一种．写成参数方程的形式是

曲面第一基本形式描写曲面上与度量有关的性质，曲面第二基本形式描写曲面的形状.
设曲面的参数方程是，其单位法向量为，则曲面的第二基本形式为

其中

在计算系数时，必须用单位法向量. 初学者容易犯用代替的错误.
法曲率：

正则参数曲面在任意一个固定点，其法曲率必定在两个彼此正交的切方向上分别取最大值和最小值．曲面在一个固定点处沿各个切方向的法曲率的最大值和最小值称为曲面在该点的主曲率，记为，达到这最大值和最小值的切方向称为曲面在该点的主方向．这个事实可以通过直接计算来证实（参看例题 4．5）．若曲面在点的两个彼此正交的主方向单位向量是，对应的主曲率是，则曲面在点沿着与主方向的夹角为的切方向的法曲率是

这就是著名的 Euler 公式．
Gauss 曲率:

平均曲率:

法曲率为零的切方向称为渐近方向．曲面只在双曲点和抛物点有渐近方向．曲面上其切方向处处是曲面的渐近方向的曲线称为曲面上的渐近曲线．渐近曲线的微分方程是

曲面上点为脐点，当且仅当存在使得

这题计算量太大

正则曲面是二维流形的例子，曲面的每一个正则参数表示给出了一个局部坐标系↩︎
很明显，当变化时，上面的截线就扫出曲面 .↩︎

数学类解析几何

Fri, 02 May 2025 16:55:29 GMT

先自己画图看看能不能用几何直观秒了.

见《解析几何》·吕林根·第四版
二次曲面的一般方程为

适当选取坐标系，二次曲面的方程总可以化为下列五个方程之一：

.
.
.
.
.
通过适当选取坐标系，二次曲面的方程总可以化成以下十七种形式之一：

椭球面

单叶双曲面

双叶双曲面

二次锥面

椭圆抛物面

双曲抛物面

椭圆柱面

双曲柱面

抛物柱面

虚椭球面

点（虚母线二次锥面）

虚椭圆柱面

交于一条实直线的一对共轭虚平面

一对相交平面

一对平行平面

一对平行的共轭虚平面

一对重合平面

对进行正交变换化为标准型即可. 或者直接配方.

Differential-Geometry

Fri, 02 May 2025 16:55:29 GMT

Exercise. 八，（本题 12 分）在空间直角坐标系中，设悬链面的方程为．（1）证明：悬链面是极小曲面；（2）求悬链面上 Gauss 曲率取到最大值的所有点．
（1）计算偏导数，得

于是曲面的单位法向量为

又它们的二阶偏导数为

故得第一基本形式系数

中的曲率函数

即曲面为极小曲面．
（2）Gauss 曲率．故取到最大值 -1 当且仅当，对应悬链面上的点为单位圆周

Exercise. 八，（12 分）（i）设庞加莱上半平面带有黎曼度量证明的测地线为圆心在轴上的上半圆或者为平行于轴的直线；
（ii）对于半径为的球面称经过球心的平面与球面的交线为大圆．求球面上的测地线方程，并说明球面上的大圆是测地线；
（iii）阐明（i），（ii）的几何意义

数学分析

Fri, 02 May 2025 16:55:04 GMT

证明考虑泰勒展开

于是

令那么

Appendix

Tue, 06 May 2025 07:27:10 GMT

cmc决赛

Fri, 09 May 2025 02:43:00 GMT

优先采用教材上提供的基本方法和自然的思路.
换序问题在时间比较富裕的情况下一定要严格证明.
中值定理一定是自然的构造, 不要想的奇奇怪怪.
填空题是基本盘, 一定不能错, 多检查, 可以猜答案.
遇事不绝, 分部积分.
遇事不绝, 不妨设标准型.
数学类高年级组误判率较高! 重代数轻分析!
第一天考完下午和晚上是最好玩的, 抓紧时机面基各路大神, 笔者当初考完带着一堆小伙伴去撸串, 干了十瓶可乐.
考完当天晚上就会出成绩, 误判几率不低, 可以尽力配合老师查分.

Jordan 分解
同时（正交）相似上三角/对角化[1]
两个半正定矩阵可以同时合同对角化.
实对称矩阵和正定矩阵显然可以同时合同对角化
矩阵打洞
摄动法，可以不妨设可逆
和有完全一样的非零 Jordan 块
与所有可逆矩阵可交换的矩阵是数量矩阵 .
对，有 .
.
是半正定矩阵，等价于所有主子式[2]非负，等价于的所有特征值为非负实数，等价于存在使得 .
若可逆，则是的多项式. 考虑特征多项式和哈密顿凯莱定理 (结合 )
是的多项式. 注意到 .
瑞丽商
反对称实矩阵的特征值实部为 0.
循环行列式
求解 . 有唯一解的充要条件是无相同特征值. 证明考虑分块 Jordan 爆算. 这也是对于交结数的证明.
特征多项式等于极小多项式的等价条件.
覆盖定理
实正规矩阵[3]的正交相似标准型.
相似的正规矩阵必然酉相似. 若为实正规，则正交相似.
酉相似实矩阵必然实正交相似.
是复正规矩阵的充要条件是的共轭转置是的多项式
三对角矩阵的行列式递推
Perron 判别法

正交性要求特征值全为实数，或者实对称...↩︎
主子式是指取任意行和对应的列交叉出来的子式的行列式，代数主子式还要考虑逆序数.↩︎
↩︎

vim

Fri, 02 May 2025 16:55:04 GMT

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参考资料
• Vim 官方文档：Vim documentation
• 菜鸟教程 Vim 教程：菜鸟教程 - Vim 教程