1. The Bergman metric on unit disk
\newcommand{\abs}[1]{\left| #1 \right|}
1.1 Schwarz’s lemma
Lemma 1.1. If f(z)∈Hol(D)f(z) \in Hol(\mathbb{D})f(z)∈Hol(D), and f(0)=0f(0) = 0f(0)=0, then
∣f(z)∣≤∣z∣,and∣f′(0)∣≤1.\left| f(z) \right| \le |z|, and |f'(0)| \le 1.
∣f(z)∣≤∣z∣,and∣f′(0)∣≤1.
Equality holds at some point zzz if and only if f(z)=eiγzf(z) = e^{i \gamma} zf(z)=eiγz, where γ\gammaγ is a real constant.
We shall use the invariant form of Schwarz’s lemma. A Mo¨bius transformationM\ddot{o}bius\ ...
格式转换 mdx --> dictionary
未读这是一个非常详细的指南，介绍了如何将 Mdict 格式的字典转换为 macOS 原生字典，以下是整理和翻译的步骤，便于参考和使用：
环境准备
创建工作目录：
12mkdir -p ~/Developer/ConvertDict/cd ~/Developer/ConvertDict/
安装 Python 3 和 Homebrew：
请参考网上教程安装 Homebrew 和 Python 3（例如 Python 3.9.7）。Homebrew 安装过程中会自动安装 Command Line Tools for Xcode。
安装 Python 依赖包：
1pip3 install lxml beautifulsoup4 html5lib
安装 lzo：
1234brew install lzoexport C_INCLUDE_PATH=/opt/homebrew/Cellar/lzo/2.10/include/lzo:/opt/homebrew/Cellar/lzo/2.10/include/export LIBRARY_PATH=/opt/homebrew/libpi ...
Topology
未读Supplementary Exercises: Nets
We have already seen that sequences are “adequate” to detect limit points, continuous functions, and compact sets in metrizable spaces. There is a generalization of the notion of sequence, called a net, that will do the same thing for an arbitrary topological space. We give the relevant definitions here, and leave the proofs as exercises. Recall that a relation ⪯\preceq⪯ on a set AAA is called a partial order relation if the following conditions hold:
α⪯α\alpha \p ...
Functional Analysis
未读Definition. If M\mathcal{M}M is a subset of the Hilbert space H\mathscr{H}H, then the orthogonal complement of M\mathcal{M}M, denoted M⊥\mathcal{M}^{\perp}M⊥,is the set of vectors in H\mathscr{H}H orthogonal every vector in M\mathcal{M}M.
Remark. M⊥\mathcal{M}^{\perp}M⊥ is a closed subspaces of H\mathscr{H}H.
Proof. Let xxx,yyy ∈M⊥\in \mathscr{M}^{\perp}∈M⊥, ∀z∈M\forall z \in \mathcal{M}∀z∈M, <z,x>=<z,y>=0<{z},{x}> = <{z},{y}> = 0<z,x>=<z,y>=0. Then,
<z,x+y ...
Number Theory
未读 定义： 我们用 σ(n)\sigma(n)σ(n) 表示自然数 nnn 的所有因子之和，即
σ(n)=∑m∣nm\sigma(n) = \sum_{m|n}{m}
σ(n)=m∣n∑m
(1). 若一个自然数 nnn , 做素因数分解：n=p1a1⋅⋅⋅pkakn = p_{1}^{a_1}\cdot\cdot\cdot p_{k}^{a_k}n=p1a1⋅⋅⋅pkak, 则有
σ(n)=p1a1+1−1p1−1⋅⋅⋅pkak+1−1pk−1\sigma(n) = \dfrac{p_{1}^{a_1+1} - 1}{p_1 - 1} \cdot\cdot\cdot \dfrac{p_{k}^{a_k+1} - 1}{p_k - 1}
σ(n)=p1−1p1a1+1−1⋅⋅⋅pk−1pkak+1−1
证明：
注意到
∏i=1k∑j=0αipij=∑(j1,⋅⋅⋅,jk)∈∏i=1k([0,αi]⋂N)p1j1⋅⋅⋅pkjk\prod_{i=1}^{k} {\sum_{j=0}^{ { {\alpha } _ {i} } } {p_{i}^{j} } ...
Real Analysis
未读 首先我们给出一个总的思路：(或者说前提小结，也可以说是摘要)
我们从集 XXX 的某些子集所成的环 R\mathscr{R}R, 以及环 R\mathscr{R}R 上的测度 μ\muμ 出发. 根据环 R\mathscr{R}R 做集类 H(R)\mathscr{H(R)}H(R), 它是一个 σ−\sigma-σ−环. 然后由测度 μ\muμ, 在 H(R)\mathscr{H(R)}H(R) 上作出由 μ\muμ 引出的外测度 μ∗\mu ^{*}μ∗, μ∗\mu^*μ∗ 是 μ\muμ 的“延拓”, 即对于 ∀E∈R\forall E \in \mathscr{R}∀E∈R, 有 μ∗(E)=μ(E)\mu^*(E)=\mu(E)μ∗(E)=μ(E). 外测度 μ∗\mu^*μ∗ 具有测度的一部分性质, 但不一定有可加性, 一般来说只有次可加性. 但是, 我们可以利用 CaratheodoryCaratheodoryCaratheodory 条件分出了一类集, 即 μ∗−\mu^*-μ∗− 可测集的全体 (R∗⊂H(R))\left(\mathscr{R}^* \ ...
Real Analysis
未读Definition 1:
Let XXX is a fixed set(or a Basic Space), and denote E:={E:E⊂X}\mathcal{E} := \left\{ E:E\subset X \right\}E:={E:E⊂X}. Call E\mathcal{E}E is a set class on XXX.
Definition 2:
Let E\mathcal{E}E is a set class on XXX, and if its elements satisfy:
∀E1,E2∈E\forall E_1,E_2 \in \mathcal{E}∀E1,E2∈E, have E1∪E2∈EE_1 \cup E_2 \in \mathcal{E}E1∪E2∈E and E1−E2∈EE_1-E_2 \in \mathcal{E}E1−E2∈E, then we called E\mathcal{E}E is a Ring on XXX and denote: R\mathcal{R}R.
In particular, if we h ...
Fourier Analysis
未读Fourier Analysis
Problem 1
One can construct Riemann integrable functions on [0,1][0,1][0,1] that have a dense set of discontinuities as follows.
(a) Let f(x)=0f(x)=0f(x)=0, when x<0x<0x<0, and f(x)=1f(x)=1f(x)=1 if x≥0x\geq0x≥0. Choose a countable dense sequence {rn}\{r_n\}{rn} in [0,1][0,1][0,1]. Then, show that function
F(x)=∑n=1∞1n2f(x−rn)F(x)=\sum_{n=1}^{\infty} \dfrac{1}{n^2} f(x-r_n)
F(x)=n=1∑∞n21f(x−rn)
is integrable and has discontinuities at all points of the sequence { ...
analysis
未读皮卡存在唯一性定理
1. 定义：
LipschitzLipschitzLipschitz 条件：
设f(x)f(x)f(x)是定义在R{\mathbb{R}}R上的连续函数，若∀x1,x2∈R\forall x_1 , x_2 \in {\mathbb{R}}∀x1,x2∈R，都存在LLL,使得：
∣f(x1)−f(x2)∣≤L∣x1−x2∣|f(x_1)-f(x_2)| \leq L|x_1 - x_2|
∣f(x1)−f(x2)∣≤L∣x1−x2∣
则称f(x)f(x)f(x)在R{\mathbb{R}}R上满足LipschitzLipschitzLipschitz条件
2. 定理内容
设初值问题：
(E):dydx=f(x,y),y(x0)=y0(E):\quad\dfrac{\mathrm{d} y}{\mathrm{d} x}=f(x,y),\quad y(x_0)=y_0
(E):dxdy=f(x,y),y(x0)=y0
其中f(x,y)f(x,y)f(x,y)在矩形区域：
∣x−x0∣≤a,∣y−y0∣≤b|x-x_0| \leq a ,\quad ...
