Category Archives: Complex Analysis

实分析与复分析

August 15, 2025 zr9558 Leave a comment

先说结论：复分析研究的是性质“最好的”函数，实分析研究的是性质”最差的“函数。

复分析与实分析是数学分析的两个重要分支，它们在研究对象、方法和性质上存在显著差异。

1. 研究对象

复分析：主要研究复变量的函数，即定义在复数域上的函数。复分析的核心对象是解析函数（全纯函数），这类函数在定义域内处处可微，并且满足柯西-黎曼方程（Cauchy-Riemann equations）。

实分析：研究实变量的函数，即定义在实数域上的函数。实分析的对象包括连续函数、可微函数、可积函数等，但更关注的是那些性质较差的函数，如处处连续但处处不可导的函数、狄利克雷函数（Dirichlet function）和康托尔函数（Cantor function）等。

2. 函数的性质

复分析：解析函数具有极强的正则性（regularity），例如：解析函数在其定义域内无限可微。解析函数的局部性质可以完全决定其全局性质（如解析延拓）。解析函数的积分路径可以自由变形（柯西积分定理）。解析函数的实部和虚部是调和函数（满足拉普拉斯方程）。解析函数的零点孤立（除非函数恒为零）。

实分析：实函数的性质更加多样化和复杂，例如：存在处处连续但处处不可导的函数（如魏尔斯特拉斯函数）。存在可积但不可微的函数（如狄利克雷函数）。存在单调连续但导数几乎处处为零的函数（如康托尔函数）。实分析中的函数通常需要更精细的工具（如测度论、勒贝格积分）来研究其性质。可测函数”差不多“是连续函数。

3. 核心定理

复分析：

柯西积分定理（Cauchy’s Integral Theorem）：解析函数在简单闭合路径上的积分为零。
柯西积分公式（Cauchy’s Integral Formula）：解析函数的值可以通过其在边界上的积分表示。
留数定理（Residue Theorem）：计算解析函数在奇点处的积分。
最大模原理（Maximum Modulus Principle）：解析函数的模在边界上取得最大值。

实分析：

勒贝格积分（Lebesgue Integral）：比黎曼积分更一般的积分理论，适用于更广泛的函数类。
单调收敛定理（Monotone Convergence Theorem）：单调递增的非负可测函数序列的积分与极限可交换。
控制收敛定理（Dominated Convergence Theorem）：在控制函数的条件下，极限与积分可交换。
鲁津定理（Luzin’s Theorem）：几乎处处有限的函数在适当条件下可以“几乎连续”。

4. 应用领域

复分析：在物理学中用于研究流体力学、电磁学和量子力学，在工程学中用于信号处理和控制系统，在数论中用于研究素数分布（如黎曼 zeta 函数），在复动力系统中研究分形结构。个人感觉复分析的发展依赖于解析数论的那些大问题。

实分析：在概率论中用于研究随机过程和测度论，在动力系统中研究遍历理论，在经济学中用于优化理论和博弈论。

5. 总结

复分析和实分析虽然在形式上都是研究函数的性质，但复分析更关注那些性质极好的函数（解析函数），而实分析则更关注那些性质复杂甚至“病态”的函数。复分析的许多结果依赖于复数的特殊性质，而实分析则需要更一般的工具（如测度论）来处理实函数的多样性。两者在数学和应用科学中都有广泛的应用，并且相互补充，共同构成了现代分析学的基础。

Complex Analysis

Riemann Zeta 函数（二）

April 27, 2018 zr9558 Leave a comment

在上一篇文章里面，我们已经给出了 Riemann Zeta 函数的定义，

$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}.$

其定义域是 $[1,\infty)\subseteq\mathbb{R}.$ 根据级数与定积分的等价关系可以得到：

当 $s = 1$ 时， $\zeta(1) = \infty;$
当 $s>1$ 时， $\zeta(s)<\infty.$

本文将会重点讲两个内容：

如何把 Riemann Zeta 函数从 $[1,\infty)\subseteq \mathbb{R}$ 上延拓到 $\{s\in \mathbb{C}: \Re(s)>0\}$ 上；
Riemann Zeta 函数在 $\{s\in\mathbb{C}: \Re(s)\geq 1\}$ 上没有零点。

Riemann Zeta 函数定义域的延拓

如果想把 Riemann Zeta 函数的定义域从 $[1,\infty)\subseteq \mathbb{R}$ 延拓到更大的区域 $\{s\in\mathbb{C}:\Re(s)>0\}$ 上，就需要给出 Riemann Zeta 函数在 $\{s\in \mathbb{C}: \Re(s)>0\}$ 上的定义。而且在原始的定义域 $[1,\infty)\subseteq\mathbb{R}$ 上面，新的函数的取值必须与原函数的取值保持一致。

首先，我们将会在 $[1,\infty)\subseteq \mathbb{R}$ 上面证明如下恒等式：

$\zeta(s) = \frac{s}{s-1} - s\int_{1}^{\infty}\frac{\{x\}}{x^{s+1}}dx$ .

证明：当 $s=1$ 时，上述等式显然成立，两侧都是 $\infty.$

$\frac{s}{s-1}-s\int_{1}^{\infty}\frac{\{x\}}{x^{s+1}}dx$

$= \frac{s}{s-1} - s\sum_{n=1}^{\infty}\int_{n}^{n+1}\frac{\{x\}}{x^{s+1}}dx$

$= \frac{s}{s-1} - s\sum_{n=1}^{\infty}\int_{n}^{n+1}\frac{x-n}{x^{s+1}}dx$

$= \frac{s}{s-1} - s\sum_{n=1}^{\infty}\bigg(\int_{n}^{n+1}\frac{1}{x^{s}}dx - \int_{n}^{n+1}\frac{n}{x^{s+1}}dx\bigg)$

$= \frac{s}{s-1} - s\int_{1}^{\infty}\frac{1}{x^{s}}dx + \sum_{n=1}^{\infty}n\cdot\int_{n}^{n+1}\frac{s}{x^{s+1}}dx$

$= \sum_{n=1}^{\infty}n\cdot\bigg(\frac{1}{n^{s}}-\frac{1}{(n+1)^{s}}\bigg)$

$= \sum_{n=1}^{\infty}\bigg(\frac{1}{n^{s-1}}-\frac{1}{(n+1)^{s-1}} + \frac{1}{(n+1)^{s}}\bigg)$

$= \sum_{n=1}^{\infty}\frac{1}{n^{s}}.$

从右式的表达式

$\frac{s}{s-1} - s \int_{1}^{\infty}\frac{\{x\}}{x^{s+1}}dx$

可以看出 $\zeta(s)$ 可以延拓到 $\{s \in\mathbb{C}:\Re(s)>0\}$ 上。而且右侧的函数在 $\{s\in\mathbb{C}:\Re(s)>0,s\neq 1\}$ 是解析的，并且 $s=1$ 是该函数的一个极点。进一步的分析可以得到，我们得到一个关于 $(s-1)\zeta(s)$ 的解析函数，而且 $\lim_{s\rightarrow 1}(s-1)\zeta(s)=1.$ 综上所述：

Riemann Zeta 函数可以延拓到 $\{s\in\mathbb{C}:\Re(s)>0\}$ 上；
Riemann Zeta 函数在 $\{s\in\mathbb{C}:\Re(s)>0, s\neq 1\}$ 上是解析的； $s=1$ 是 Riemann Zeta 函数的极点。

Riemann Zeta 函数的非零区域

著名的 Riemann 猜想说的是 $\zeta(s)$ 函数的所有非平凡零点都在直线 $\{s\in\mathbb{C}:\Re(s)=1/2\}$ 上。因此，数学家首先要找出的就是 Riemann Zeta 函数的非零区域。而本篇文章将会证明 Riemann Zeta 函数在 $\{s\in\mathbb{C}:\Re(s)\geq 1\}$ 上面没有零点。

$\Re(s)>1$ 区域

首先，我们要证明当 $\Re(s)>1$ 时， $\zeta(s)\neq 0.$

在这里，就需要使用一个重要的恒等式：当 $\Re(s)>1$ 时，

$\zeta(s) =\sum_{n=1}^{\infty}\frac{1}{n^{s}}$

$= \prod_{p}\bigg(1+\frac{1}{p^{s}}+\frac{1}{p^{2s}}+\cdots\bigg)$

$= \prod_{n=1}^{\infty}\bigg(1-\frac{1}{p_{n}^{s}}\bigg)^{-1},$

其中这里的 $p$ 表示所有的素数相乘，而 $p_{n}$ 表示第 $n$ 个素数。

下面我们证明：

$\bigg|1-\frac{1}{p_{n}^{s}}\bigg|^{-1}\geq 1-\frac{1}{p_{n}^{\sigma}-1} .$

事实上，令 $s = \sigma + i t,$ ，当 $\sigma=\Re(s)>1$ 时，我们有

$\bigg|1-\frac{1}{p_{n}^{s}}\bigg|^{-1} = \bigg(1+\frac{1}{p_{n}^{s}}+\frac{1}{p_{n}^{2s}}+\cdots\bigg)$

$\geq 1-\frac{1}{|p_{n}^{s}|}- \frac{1}{|p_{n}^{2s}|} -\cdots$

$= 1- \frac{1}{p_{n}^{\sigma}} - \frac{1}{p_{n}^{2\sigma}} -\cdots$

$= 1- \frac{1}{p_{n}^{\sigma}-1}.$

因此，

$|\zeta(s)| \geq \prod_{n=1}^{\infty}\bigg|1-\frac{1}{p_{n}^{s}}\bigg|^{-1} \geq\prod_{n=1}^{\infty}\bigg(1-\frac{1}{p_{n}^{\sigma}-1}\bigg).$

同时，

$\lim_{n\rightarrow \infty} \bigg(1- \frac{1}{p_{n}^{\sigma}-1}\bigg) = 1 ,$

$1-\frac{1}{p_{n+1}^{\sigma}-1} \geq 1- \frac{1}{p_{n}^{\sigma}-1} ,$

$\sum_{n=1}^{\infty}\frac{1}{p_{n}^{\sigma}}\leq \sum_{n=1}^{\infty}\frac{1}{n^{\sigma}}<\infty$ when $\sigma>1.$

所以，当 $\Re(s)>1$ 时， $\zeta(s) \neq 0.$

$\Re(s) =1$ 直线

Claim 1. 下面我们将会证明恒等式：对于 $\sigma >1, \text{ } t\in\mathbb{R},$

$\Re(\ln\zeta(\sigma + it)) = \sum_{n=2}^{\infty}\frac{\Lambda(n)}{n^{\sigma}\ln(n)}\cos(t\ln(n)) ,$

其中当 $n$ 形如 $p^{\alpha},$ $p$ 是素数， $\alpha \geq 1.$ $\Lambda(n) = \ln(p).$ 而对于其余的 $n,$ $\Lambda(n)=0.$

事实上，根据 Euler 公式，

$\zeta(s) = \prod_{p}\bigg(1-\frac{1}{p^{s}}\bigg)^{-1}.$

令 $s = \sigma + it,$ 可以得到

$\ln\zeta(s) = -\sum_{p}\ln\bigg(1-\frac{1}{p^{s}}\bigg)$

$= \sum_{p}\sum_{\alpha=1}^{\infty}\frac{1}{\alpha p^{\alpha s}}$

$= \sum_{p}\sum_{\alpha=1}^{\infty}\frac{1}{\alpha p^{\alpha\sigma}}\cdot p^{-i\alpha t}$

$= \sum_{p}\sum_{\alpha = 1}^{\infty}\frac{1}{\alpha p^{\alpha\sigma}}\cdot e^{-i\alpha t \ln p}$

进一步，

$\Re(\ln\zeta(s)) = \sum_{p}\sum_{\alpha =1}^{\infty}\frac{1}{\alpha p^{\alpha\sigma}}\cos(\alpha t \ln p)$

并且右侧等于

$RHS = \sum_{n=2}^{\infty}\frac{\Lambda(n)}{n^{\sigma}\ln(n)}\cos(t\ln(n))$

$= \sum_{p}\sum_{\alpha = 1}^{\infty} \frac{\ln(p)}{p^{\alpha\sigma}\ln(p^{\alpha})}\cos(t\ln(p^{\alpha}))$

$= \sum_{p}\sum_{\alpha = 1}^{\infty}\frac{1}{\alpha p^{\alpha\sigma}}\cos(\alpha t\ln p).$

所以，恒等式成立，Claim 1 证明完毕。

Claim 2.

$\Re(3\ln\zeta(\sigma) + 4\ln\zeta(\sigma+it) + \ln\zeta(\sigma+2it))\geq 0,$

其中 $\sigma>1,$ $t\in\mathbb{R}.$ 换句话说

$|\zeta(\sigma)^{3}\zeta(\sigma+it)^{4}\zeta(\sigma+2it)|\geq 1.$

事实上，

从三角函数的性质可以得到：

$3+4\cos(\theta)+\cos(2\theta) = 3 + 4\cos(\theta)+2\cos^{2}(\theta)-1$

$= 2(\cos(\theta)-1)^{2}\geq 0,$

所以，从 Claim 1 可以得到

$\Re(3\ln\zeta(\sigma) + 4\ln\zeta(\sigma+it) + \ln\zeta(\sigma+2it))$

$= \sum_{n=2}^{\infty} \frac{\Lambda(n)}{n^{\sigma}\ln(n)} \cdot ( 3 + 4\cos(t\ln(n)) + \cos(2t\ln(n))) \geq 0.$

进一步地，使用 $\Re(\ln(z)) = \ln(|z|)$ 可以得到

$0\leq 3\ln|\zeta(\sigma)| + 4\ln|\zeta(\sigma+it)| + \ln|\zeta(\sigma+2it)|$

$= \ln|\zeta(\sigma)^{3}\zeta(\sigma+it)^{4}\zeta(\sigma+2it)|,$

可以推导出 $|\zeta(\sigma)^{3}\zeta(\sigma+it)^{4}\zeta(\sigma+2it)|\geq 1.$ 因此 Claim 2 证明完毕。

Claim 3. $\zeta(1+it)\neq 0$ 对于所有的 $\{t\in\mathbb{R}: t\neq 0\}$ 成立。

反证法：假设 $\zeta(s)$ 在 $s=\sigma + it (t\neq 0)$ 存在阶数为 $m$ 的零点。也就是说：

$\lim_{\sigma\rightarrow 1^{+}} \frac{\zeta(\sigma+it)}{(\sigma+it-1)^{m}}=c\neq 0,$ 其中 $m\geq 1.$

从 Riemann Zeta 函数的延拓可以知道， $\lim_{\sigma\rightarrow 1^{+}}(\sigma -1)\zeta(\sigma) = 1.$ 并且 $\zeta(s)$ 在 $\{s\in\mathbb{C}:\Re(s)>0, s\neq 1\}$ 上是解析函数。

从 Claim 2 可以得到：

$|(\sigma-1)^{3}\zeta(\sigma)^{3}(\sigma+it-1)^{-4m}\zeta(\sigma+it)^{4}\zeta(\sigma+2it)|$

$\geq |\sigma-1|^{3}|\sigma-1+it|^{-4m}$

$\geq |\sigma-1|^{3}\cdot |\sigma-1|^{-4m}$

$= \frac{1}{|\sigma-1|^{4m-3}}.$

令 $\sigma\rightarrow 1^{+},$ 可以得到左侧趋近于一个有限的值，但是右侧趋近于无穷，所以得到矛盾。也就是说当 $t\neq 0$ 时， $\zeta(1+it)\neq 0$ 成立。

根据之前的知识， $s= 1$ 是 $\zeta(s)$ 的极点，所以我们得到了本篇文章的主要结论： $\zeta(s)$ 在 $\{s\in\mathbb{C}:\Re(s)\geq 1\}$ 上面没有零点。

总结

本篇文章从 Riemann Zeta 函数的延拓开始，证明了 Riemann Zeta 函数在 $\{s\in\mathbb{C}:\Re(s)\geq 1\}$ 上没有零点。在下一篇文章中，笔者将会证明在 $\Re(s)=1$ 附近一个“狭长”的区域上，Riemann Zeta 函数没有零点。

Complex Analysis

January 25, 2016 zr9558 Leave a comment

学习复分析也已经很多年了，七七八八的也读了不少的书籍和论文。略作总结工作，方便后来人学习参考。复分析是一门历史悠久的学科，主要是研究解析函数，亚纯函数在复球面的性质。下面一一介绍这些基本内容。

300px-Mandel_zoom_00_mandelbrot_set

（1）提到复变函数，首先需要了解复数 (Complex Numbers) 的基本性质和四则运算规则。怎么样计算复数的平方根，极坐标与xy坐标的转换，复数的模之类的。这些在高中的时候基本上都会学过。

（2）复变函数自然是在复平面上来研究问题，此时数学分析里面的求导数之类的运算就会很自然的引入到复平面里面，从而引出解析函数 (Holomorphic Functions / Analytic Functions) 的定义。那么研究解析函数的性质就是关键所在。最关键的地方就是所谓的Cauchy—Riemann公式，这个是判断一个函数是否是解析函数的关键所在。

（3）明白解析函数的定义以及性质之后，就会把数学分析里面的曲线积分 (Line Integrals) 的概念引入复分析中，定义几乎是一致的。在引入了闭曲线和曲线积分之后，就会有出现复分析中的重要的定理：Cauchy积分公式 (Cauchy’s Integral Formula)。这个是复分析的第一个重要定理。

（4）既然是解析函数，那么函数的定义域 (Domain) 就是一个关键的问题。可以从整个定义域去考虑这个函数，也可以从局部来研究这个函数。这个时候研究解析函数的奇点 (Singularity) 就是关键所在，奇点根据性质分成可去奇点 (Removable Singularity)，极点 (Pole)，本性奇点 (Essential Singularity) 三类，围绕这三类奇点，会有各自奇妙的定理。

（5）复变函数中，留数定理 (Residue Theorem) 是一个重要的定理，反映了曲线积分和零点极点的性质。与之类似的幅角定理也展示了类似的关系。

（6）除了积分，导数也是解析函数的一个研究方向。导数加上收敛 (Convergence) 的概念就可以引出 Taylor 级数 (Taylor Series) 和 Laurent 级数 (Laurent Series) 的概念。除此之外，正规族 (Normal Families) 里面有一个非常重要的定理，那就是Arzela定理。

（7）以上都是从分析的角度来研究复分析，如果从几何的角度来说，最重要的定理莫过于 Riemann 映照定理 (Riemann Mapping Theorem)。这个时候一般会介绍线性变换，就是 Mobius 变换 (Mobius Transforms)，把各种各样的单连通区域映射成单位圆。研究 Mobius 变换的保角和交比之类的性质。

（8）椭圆函数 (Elliptic Functions)，经典的双周期函数 (Double Periodic Functions)。这里有 Weierstrass 理论，是研究 Weierstrass 函数的，有经典的微分方程，以及该函数的性质。以上就是复分析或者复变函数的一些课程介绍，如果有遗漏或者疏忽的地方请大家指教。

推荐书籍：

（1）Complex Analysis，3rd Edition，Lars V.Ahlfors

（2）Complex Analysis，Elias M. Stein

Complex Analysis

Normal Families

October 20, 2014 zr9558 Leave a comment

Reference Book: Joel L.Schiff- Normal Families

Some Classical Theorems

Weierstrass Theorem Let $\{ f_{n}\}$ be a sequence of analytic functions on a domain $\Omega$ which converges uniformly on compact subsets of $\Omega$ to a function $f$ . Then $f$ is analytic in $\Omega$ , and the sequence of derivatives $\{ f_{n}^{(k)}\}$ converges uniformly on compact subsets to $f^{(k)}, k=1,2,3...$ .

Hurwitz Theorem Let $\{ f_{n}\}$ be a sequence of analytic functions on a domain $\Omega$ which converges uniformly on compact subsets of $\Omega$ to a non-constant analytic function $f(z)$ . If $f(z_{0})=0$ for some $z_{0}\in\Omega$ , then for each $r>0$ sufficiently small, there exists an $N=N(r)$ , such that for all $n>N$ , $f_{n}(z)$ has the same number of zeros in $D(z_{0},r)$ as does $f(z)$ . (The zeros are counted according to multiplicity).

The Maximum Principle If $f(z)$ is analytic and non-constant in a region $\Omega$ , then its absolute value $|f(z)|$ has no maximum in $\Omega$ .

The Maximum Principle’ If $f(z)$ is defined and continuous on a closed bounded set $E$ and analytic on the interior of $E$ , then the maximum of $|f(z)|$ on $E$ is assumed on the boundary of $E$ .

Corollary 1.4.1 If $\{ f_{n}\}$ is a sequence of univalent analytic functions in a domain $\Omega$ which converge uniformly on compact subsets of $\Omega$ to a non-constant analytic function $f$ , then $f$ is univalent in $\Omega$ .

Definition 1.5.1 A family of functions $\mathcal{F}$ is locally bounded on a domain $\Omega$ if, for each $z_{0}\in \Omega$ , there is a positive number $M=M(z_{0})$ and a neighbourhood $D(z_{0},r)\subset \Omega$ such that $|f(z)|\leq M$ for all $z\in D(z_{0}, r)$ and all $f\in \mathcal{F}$ .

Theorem 1.5.2 If $\mathcal{F}$ is a family of locally bounded analytic functions on a domain $\Omega$ , then the family of derivatives $\mathcal{F}^{'}=\{ f^{'}: f\in \mathcal{F}\}$ form a locally bounded family in $\Omega$ .

The converse of Theorem 1.5.2 is false, since $\mathcal{F}=\{n: n=1,2,3...\}$ . However, the following partial converse does hold.

Theorem 1.5.3 Let $\mathcal{F}$ be a family of analytic functions on $\Omega$ such that the family of derivatives $\mathcal{F}^{'}$ is locally bounded and suppose that there is some $z_{0}\in \Omega$ with $|f(z_{0})|\leq M<\infty$ for all $f\in \mathcal{F}$ . Then $\mathcal{F}$ is locally bounded. (Hint: find a path connecting $z_{0}$ and $z$ .)

Definition 1.6.1 A family $\mathcal{F}$ of functions defined on a domain $\Omega$ is said to be equicontinuous (spherically continuous) at a point $z^{'}\in \Omega$ if, for each $\epsilon>0$ , there is a $\delta=\delta(\epsilon,z^{'})>0$ such that $|f(z)-f(z^{'})|<\epsilon$ , $(\chi(f(z),f(z^{'}))<\epsilon)$ whenever $|z-z^{'}|<\delta$ , for every $f\in \mathcal{F}$ . Moreover, $\mathcal{F}$ is equicontinuous (spherical continuous) on a subset $E\subset \Omega$ if it is continuous (spherically continuous) at each point of $E$ .

Normal Families of Analytic Functions

Definition 2.1.1 A familiy $\mathcal{F}$ of analytic functions on a domain $\Omega\subset \mathbb{C}$ is normal in $\Omega$ if every sequence of functions $\{f_{n}\}\subset \mathcal{F}$ contains either a subsequence which converges to a limit function $f\not\equiv \infty$ uniformly on each compact subset of $\Omega$ , or a subsequence which converges uniformly to $\infty$ on each compact subset.

The family $\mathcal{F}$ is said to be normal at a point $z_{0}\in\Omega$ if it is normal in some neighbourhood of $z_{0}$ .

Theorem 2.1.2 A family of analytic functions $\mathcal{F}$ is normal in a domain $\Omega$ if and only if $\mathcal{F}$ is normal at each point in $\Omega$ .

Theorem 2.2.1 Arzela-Ascoli Theorem. If a sequence $\{f_{n}\}$ of continuous functions converges uniformly on a compact set $K$ to a limit function $f\not\equiv \infty$ , then $\{f_{n}\}$ is equicontinuous on $K$ , and $f$ is continuous. Conversely, if $\{f_{n}\}$ is equicontinuous and locally bounded on $\Omega$ , then a subsequence can be extracted from $\{f_{n}\}$ which converges locally uniformly in $\Omega$ to a (continuous) limit function $f$ .

Montel’s Theorem If $\mathcal{F}$ is a locally bounded family of analytic functions on a domain $\Omega$ , then $\mathcal{F}$ is a normal family in $\Omega$ .

Koebe Distortion Theorem Let $f(z)$ be analytic univalent in a domain $\Omega$ and $K$ a compact subset of $\Omega$ . Then there exists a constant $c=c(\Omega, K)$ such that for any $z,w\in K$ , $c^{-1}\leq |f^{'}(z)| / |f^{'}(w)| \leq c$ .

Vitali-Porter Theorem Let $\{f_{n}\}$ be a locally bounded sequence of analytic functions in a domain $\Omega$ such that $\lim_{n\rightarrow \infty}f_{n}(z)$ exists for each $z$ belonging to a set $E\subset \Omega$ which has an accumulation point in $\Omega$ . Then $\{ f_{n}\}$ converges uniformly on compact subsets of $\Omega$ to an analytic function.

Proof. From Montel’s Theorem, $\{ f_{n}\}$ is normal, extract a subsequence $\{ f_{n_{k}}\}$ which converges normally to an analytic function $f$ . Then $\lim_{k\rightarrow \infty} f_{n_{k}}(z)=f(z)$ for each $z\in E$ . Suppose, however, that $\{ f_{n}\}$ does not converge uniformly on compact subsets of $\Omega$ to $f$ . Then there exists some $\epsilon>0$ , a compact subset $K\subset \Omega$ , as well as a subsequence $\{f_{m_{j}}\}$ and points $z_{j}\in K$ satisfying $|f_{m_{j}}(z_{j})- f(z_{j})| \geq \epsilon,$ $j=1,2,3,...$ . Now $\{ f_{m_{j}}\}$ itself has a subsequence which converges uniformly on compact subsets to an analytic function $g$ , and $g\not\equiv f$ from above. However, since $f$ and $g$ must agree at all points of $E$ , the Identity Theorem for analytic functions implies $f\equiv g$ on $\Omega$ , a contradiction which establishes the theorem.

Fundamental Normality Test Let $\mathcal{F}$ be the family of analytic functions on a domain $\Omega$ which omit two fixed values $a$ and $b$ in $\mathbb{C}$ . Then $\mathcal{F}$ is normal in $\Omega$ .

Generalized Normality Test Suppose that $\mathcal{F}$ is a family of analytic functions in a domain $\Omega$ which omit a value $a\in \mathbb{C}$ and such that no function of $\mathcal{F}$ assumes the value $b\in \mathbb{C}$ at more that $p$ points. Then $\mathcal{F}$ is normal in $\Omega$ .

2.3 Examples:

Assume $U$ is the unit disk in the complex plane, $\Omega$ is a region (connected open set) in $\mathbb{C}$ .

1. $\mathcal{F}=\{ f_{n}(z)=z^{n}: n=1,2,3...\}$ in $U$ . Then $\mathcal{F}$ is normal in $U$ , but not compact since $0 \notin \mathcal{F}$ . In the domain $U^{'}: |z|>1$ , $\mathcal{F}$ is normal.

2. $\mathcal{F}=\{ f_{n}(z)=\frac{z}{n}: n=1,2,3...\}$ is a normal family in $\mathcal{C}$ but not compact.

3. $\mathcal{F}=\{ f: f$ analytic in $\Omega$ and $|f|\leq M \}$ . Then $\mathcal{F}$ is normal in $\Omega$ and compact.

4. $\mathcal{F}=\{ f: f$ analytic in $\Omega$ and $\Re f>0\}$ . Then $\mathcal{F}$ is normal but not compact. Hint: $\mathcal{G}=\{g=e^{-f}:f\in \mathcal{F}\}$ is a uniformly bounded family.

5. $\mathcal{S}=\{ f: f$ analytic, univalent in $U$ , $f(0)=0, f^{'}(0)=1 \}$ . These are the normalised “Schlicht” functions in $U$ . $\mathcal{S}$ is normal and compact.

Normal Families of Meromorphic Functions

Assume a function $f(z)$ is analytic in a neighbourhood of $a$ , except perhaps at $a$ itself. In other words, $f(z)$ shall be analytic in a region $0<|z-a|<\delta$ . The point $a$ is called an isolated singularity of $f(z)$ . There are three cases about an isolated singularity. The first one is a removable singularity, the second one is a pole, the third one is an essential singularity. A function $f(z)$ which is analytic in a region $\Omega$ , except for poles, is said to be meromorphic in $\Omega$ .

The chordal distance $\chi(z_{1}, z_{2})$ between $z_{1}$ and $z_{2}$ is

$\chi(z_{1}, z_{2}) = \frac{|z_{1}-z_{2}|}{\sqrt{1+|z_{1}|^{2}}\sqrt{1+|z_{2}|^{2}}}$ if $z_{1}$ and $z_{2}$ are in the finite plane, and

$\chi(z_{1}, \infty) = \frac{1}{\sqrt{1+|z_{1}|^{2}}},$ if $z_{2}=\infty$ . Clearly, $\chi(z_{1}, z_{2})\leq 1$ , and $\chi(z_{1}^{-1}, z_{2}^{-1}) = \chi(z_{1}, z_{2})$ . The chordal metric and spherical metric are uniformly equivalent and generate the same open sets on the Riemann sphere.

Definition 1.2.1 A sequence of functions $\{ f_{n}\}$ converges spherically uniformly to $f$ on a set $E\subset \mathbb{C}$ if, for any $\epsilon>0$ , there is a number $n_{0}$ such that $n\geq n_{0}$ implies $\chi(f(z), f_{n}(z))<\epsilon$ , for all $z\in E$ .

Definition 3.1.1 A family $\mathcal{F}$ of meromorphic functions in a domain $\Omega$ is normal in $\Omega$ if every sequence $\{ f_{n} \} \subset \mathcal{F}$ contains a subsequence which converges spherically uniformly on compact subsets of $\Omega$ .

Theorem 3.1.3 Let $\{ f_{n}\}$ be a sequence of meromorphic functions on a domain $\Omega$ . Then $\{ f_{n}\}$ converges spherically uniformly on compact subsets of $\Omega$ to $f$ if and only if about each point $z_{0}\in \Omega$ there is a closed disk $K(z_{0},r)$ in which $|f_{n}-f|\rightarrow 0$ or $|1/f_{n} - 1/f| \rightarrow 0$ uniformly as $n\rightarrow \infty$ .

Corollary 3.1.4 Let $\{ f_{n}\}$ be a sequence of meromorphic functions on $\Omega$ which converges spherically uniformly on compact subsets to $f$ . Then $f$ is either a meromorphic function on $\Omega$ or identically equal to $\infty$ .

Corollary 3.1.5 Let $\{ f_{n}\}$ be a sequence of analytic functions on a domain $\Omega$ which converge spherically uniformly on compact subsets of $\Omega$ to $f$ . Then $f$ is either analytic on $\Omega$ or identically equal to $\infty$ .

Theorem 3.2.1 A family $\mathcal{F}$ of meromorphic functions in a domain $\Omega$ is normal if and only if $\mathcal{F}$ is spherically equicontinuous in $\Omega$ .

Fundamental Normality Test Let $\mathcal{F}$ be a family of meromorphic functions on a domain $\Omega$ which omit three distinct values $a, b, c \in \mathbb{C}$ . Then $\mathcal{F}$ is normal in $\Omega$ .

Vitali-Porter Theorem Let $\{f_{n}\}$ be a sequence belonging to a spherically equicontinuous family of meromorphic functions such that $\{ f_{n}(z)\}$ converges spherically on a point set $E$ having an accumulation point in $\Omega$ . Then $\{ f_{n}\}$ converges spherically uniformly on compact subsets of $\Omega$ .

Let $f(z)$ be meromorphic on a domain $\Omega$ . If $z\in \Omega$ is not a pole, the derivative in the spherical metric, called the spherical derivative, is given by $f^{\#}(z) =\lim_{z^{'}\rightarrow z}\frac{\chi(f(z),f(z^{'}))}{|z-z^{'}|} =\frac{|f^{'}(z) |}{1+|f(z)|^{2}}$ . If $\zeta$ is a pole of $f(z)$ , define $f^{\#}(\zeta) = \lim_{z\rightarrow \zeta} \frac{|f^{'}(z)|}{1+|f(z)|^{2}}$ .

Marty’s Theorem A family $\mathcal{F}$ of meromorphic functions on a domain $\Omega$ is normal if and only if for each compact subset $K\subset \Omega$ , there exists a constant $C=C(K)$ such that spherical derivative $f^{\#}(z) =\frac{|f^{'}(z) |}{1+|f(z)|^{2}}\leq C, z\in K, f\in \mathcal{F},$ that is, $f^{\#}$ is locally bounded.

Complex Analysis

Plane Hyperbolic Geometry

October 15, 2013 zr9558 Leave a comment

Assume

$\mathbb{D}=\{ z: |z|<1\}$ is the unit disc on the complex plane $\mathbb{C},$

$\mathbb{H}=\{z: \Im{z}>0\}$ is the upper half plane on the complex plane,

$\mathbb{B}=\{ z: |\Im{z}|<\pi/2\}$ is the band between $y=-\pi/2$ and $y=\pi/2$ .

Definition 1. Hyperbolic metric on the unit disc.

The hyperbolic metric on the unit disc $\mathbb{D}$ is defined as

$\rho_{\mathbb{D}}(z)=\frac{2}{1-|z|^{2}} |dz|$ for all $z \in \mathbb{D} .$

If $\phi : U \rightarrow \mathbb{D}$ is a conformal mapping, where $U \subseteq \mathbb{C}$ , then we can also define the hyperbolic metric on the domain U,

$\rho_{U}(z)=\frac{2 |\phi^{'}(z)|}{1-|\phi(z)|^{2}} |dz|$ for all $z\in U.$

From above and $\phi(z)=(z-i)/(z+i)$ is a conformal mapping which maps the upper half plane $\mathbb{H}$ onto the unit disc $\mathbb{D}.$ From above formula, we can calculate the hyperbolic metric on $\mathbb{H}$ is

$\rho_{\mathbb{H}}(z)=\frac{1}{\Im{z}} |dz|$ for all $z\in \mathbb{H}.$

The hyperbolic metric on the band $\mathbb{B}$ is

$\rho_{\mathbb{B}}(z)=\frac{1}{\cos \Im{z} } |dz|$ for all $z\in \mathbb{B}.$

Similarly, we can define the one dimensional hyperbolic metric. On the real line $\mathbb{R}$ , if the interval $I=(-1,1)$ , then the restriction of the hyperbolic metric on the unit disc $\mathbb{D}$ is

$\rho_{I}(x)= \frac{2}{1-x^{2}} dx$ for all $x \in (-1,1).$

This is called the hyperbolic metric of the interval I.

Using the same idea, we can extend the definition of hyperbolic metric on any real interval $I=(a,b)$ . Since there exists a linear map $\phi$ which maps a to -1 and b to 1, i.e. $\phi(x)=(2x-b-a)/(b-a)$ . Its derivative is $\phi^{'}(x)= 2/(b-a)$ . Therefore, the hyperbolic metric on the interval I is

$\rho_{(a,b)}(x)=\frac{2|\phi^{'}(x)|}{1-|\phi(x)|^{2}} dx= \frac{b-a}{(x-a)(b-x)} dx= (\frac{1}{x-a}+ \frac{1}{b-x}) dx$ for all $x\in (a,b).$

Moreover, assume $(c,d) \subseteq (a,b)$ , then the hyperbolic distance between c and d is

$\int_{c}^{d} \rho_{(a,b)}(x) dx = \int_{c}^{d} (\frac{1}{x-a} + \frac{1}{b-x}) dx = (\ln\frac{x-a}{b-x}) |_{x=c}^{x=d} = \ln \frac{(d-a)(b-c)}{(b-d)(c-a)}.$

If we use the notation of cross ratio, then assume $l=(a,c), j=(c,d), r=(d,b),$ $t=(a,b)$ . Therefore, the hyperbolic distance between c and d in the interval (a,b) equals to

$\ln \frac{(|l|+|j|)\cdot (|j|+|r|)}{|l| \cdot |r|} = \ln (1+ \frac{|t|\cdot |j|}{|l| \cdot |r|}) = \ln (1+ Cr(t,j)),$

where $Cr(t,j)= (|t|\cdot |j|) / (|l| \cdot |r|).$

Definition 2. (Curvature of conformal metric)

Let $\rho$ be a $C^{2}$ positive function on an open subset $U \subseteq \mathbb{C}$ . Then the curvature of the metric $\rho(z)|dz|$ is given by

$K(z)=-\frac{(\Delta \ln \rho)(z)}{\rho^{2}(z)},$

where $\Delta$ is the Laplacian operator $\Delta= \frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}}.$

Remark. Use the identities

$\frac{\partial}{\partial \overline{z}} =\frac{1}{2} (\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}),$

$\frac{\partial}{\partial z} =\frac{1}{2} (\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}),$

we get

$\Delta=\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} = 4 \frac{\partial^{2}}{\partial z \partial \overline{z}}.$

Theorem 1.

The curvature of hyperbolic metric of the unit disc $\mathbb{D}$ , the upper half plane $\mathbb{H}$ and the band $\mathbb{B}$ is -1.

Theorem 2.

If $\phi: U\rightarrow \mathbb{D}$ is a conformal mapping, where $U \subseteq \mathbb{C}$ is an open subset of the complex plane $\mathbb{C}$ . From above, the hyperbolic metric on U is