魏尔施特拉斯分解定理

英文为primary factors或是elementary factors。也有译为“主要因子”的版本。^[1]

对于任意的${\displaystyle n\in \mathbb {N} _{0}}$ ，基本因子${\displaystyle E_{n}(z)}$ 的定义如下：^[2]

${\displaystyle {\begin{aligned}&E_{n}(z)=(1-z)\exp \left(h_{n}(z)\right)\\&h_{n}(z)={\begin{cases}0&{\text{ if }}n=0,\\{\frac {z^{1}}{1}}+{\frac {z^{2}}{2}}+\cdots +{\frac {z^{n}}{n}}&{\text{ otherwise. }}\end{cases}}\end{aligned}}}$ 

其中，级数${\displaystyle h_{n}(z)={\frac {z^{1}}{1}}+{\frac {z^{2}}{2}}+{\frac {z^{3}}{3}}+\cdots +{\frac {z^{n}}{n}}}$ 。

对于级数${\displaystyle h_{n}(z)}$ ，有如下性质。以下性质在后续引理的证明中会用到（主要是3.、4.与5.）。

1. ${\displaystyle |z|<1}$ 的情况下，${\displaystyle h_{n}(z)}$ 可被展开为${\displaystyle {\frac {1}{1-z}}=1+z^{1}+z^{2}+z^{3}+\cdots }$ 。接着两边同时积分，可得${\displaystyle {\begin{aligned}\int {\frac {1}{1-z}}dz=-\log(1-z)={\frac {z^{1}}{1}}+{\frac {z^{2}}{2}}+{\frac {z^{3}}{3}}+\cdots \end{aligned}}}$ 。所以${\displaystyle h_{n}(z)}$ 的极限可以表示为${\displaystyle h_{\infty }(z)=\lim _{n\to \infty }h_{n}(z)=-\log(1-z)}$ 。
2. 因为${\displaystyle (1-z)=\exp(\log(1-z))=\exp \left(-h_{\infty }(z)\right)}$ ，所以${\displaystyle {\frac {1}{1-z}}=\exp \left(h_{\infty }(z)\right)}$ 。
3. 如果将${\displaystyle h_{\infty }(z)}$ 与${\displaystyle h_{n}(z)}$ 之间的差额定义为新的级数${\displaystyle r_{n}(z)=h_{\infty }(z)-h_{n}(z)}$ 。
4. 利用2.与3.改写${\displaystyle E_{n}(z)}$ 的定义式：${\displaystyle {\begin{aligned}&E_{n}(z)=(1-z)\exp \left(h_{n}(z)\right)=(1-z)\exp \left(h_{\infty }(z)-r_{n}(z)\right)\\&\quad =(1-z){\frac {1}{1-z}}\exp \left(-r_{n}(z)\right)=\exp \left(-r_{n}(z)\right)\end{aligned}}}$ 。改写后的基本因子定义式${\displaystyle E_{n}(z)=\exp \left(-r_{n}(z)\right)}$ 将会在后续引理的证明中用到。
5. 将3.的关系写成级数形式：${\displaystyle r_{n}(z)={\frac {z^{n+1}}{n+1}}+{\frac {z^{n+2}}{n+2}}+{\frac {z^{n+3}}{n+3}}+\cdots =\sum _{k=n+1}^{\infty }{\frac {z^{n+1}}{n+1}}={\frac {z^{n+1}}{n+1}}\sum _{k=0}^{\infty }{\frac {n+1}{n+1+k}}z^{k}}$ 。

利用以上性质，可以证明下面的重要引理。该引理在后续证明魏尔施特拉斯分解定理时有关键性作用。^[2]

引理 (15.8, Rudin): 对于${\displaystyle |z|\leq 1,n\in \mathbb {N} _{0}}$ ，

${\displaystyle {\begin{aligned}\left|1-E_{n}(z)\right|\leq |z|^{n+1}\end{aligned}}}$ 成立。

证明：

${\displaystyle n=0}$ 时，${\displaystyle |1-z|\leq |z|}$ 显而易见。所以只讨论${\displaystyle n\geq 1}$ 的情况。

i) 将引理左边的部分（不带绝对值）定义为一个新函数${\displaystyle u_{n}(z)=1-E_{n}(z)=1-(1-z)\exp h_{n}(z)}$ 。后续称此式为式${\displaystyle (1)}$ 。

运用性质4.与5.改写式${\displaystyle (1)}$ ：

${\displaystyle {\begin{aligned}u_{n}(z)=1-E_{n}(z)=1-\exp \left(-r_{n}(z)\right)=1-\exp \left(-{\frac {z^{n+1}}{n+1}}\sum _{k=0}^{\infty }{\frac {n+1}{n+1+k}}z^{k}\right)=1-\exp(-{\frac {z^{n+1}}{n+1}})\exp {(-{\frac {z^{n+2}}{n+1}}{\frac {n+1}{n+2}})}\exp {(-{\frac {z^{n+3}}{n+1}}{\frac {n+1}{n+3}})}...\end{aligned}}}$ 

将指数部分展开后可得（为了简洁，系数用字母${\displaystyle b,c,d}$ 表示）：${\displaystyle u_{n}(z)=1-(1-b_{n+1}z^{n+1}+b_{2n+2}z^{2n+2}+...)(1-c_{n+2}z^{n+2}+c_{2n+4}z^{2n+4}+...)(1-d_{n+3}z^{n+3}+d_{2n+4}z^{2n+4}+...)...}$ 

整理后可得，${\displaystyle u_{n}(z)}$ 可以用一个新的级数来表示：${\displaystyle u_{n}(z)=1-(1-b_{n+1}z^{n+1}-c_{n+2}z^{n+2}-d_{n+3}z^{n+3}+...)}$ 。将系数统一用${\displaystyle a}$ （如${\displaystyle a_{0}=b_{n+1},a_{1}=c_{n+2}}$ ）来标注的话，${\displaystyle u_{n}(z)=z^{n+1}\sum _{k=0}^{\infty }a_{k}z^{k}}$ 。

将该结果微分，可得：

${\displaystyle u'_{n}(z)=a_{0}(n+1)z^{n}+a_{1}(n+2)z^{n+1}+a_{2}(n+3)z^{n+2}+...}$ 

ii) 将式${\displaystyle (1)}$ 直接微分，可得

${\displaystyle {\begin{aligned}&u_{n}^{\prime }(z)=-E_{n}^{\prime }(z)=\exp h_{n}(z)-(1-z)h_{n}^{\prime }(z)\exp h_{n}(z)\\&\quad =\exp h_{n}(z)-(1-z){\frac {1-zn}{1-z}}\exp h_{n}(z)=z^{n}\exp h_{n}(z)\end{aligned}}}$ 

将指数部分展开可得。

${\displaystyle u_{n}^{\prime }(z)=z^{n}\exp h_{n}(z)=z^{n}\exp({\frac {z^{1}}{1}}+{\frac {z^{2}}{2}}+\cdots +{\frac {z^{n}}{n}})=z^{n}\exp(z)\exp({\frac {z^{2}}{2}})\exp({\frac {z^{3}}{3}})...\exp({\frac {z^{n}}{n}})=z^{n}(\sum _{k=0}^{\infty }{\frac {x^{k}}{k!}})(\sum _{l=0}^{\infty }{\frac {x^{2l}}{2^{l}l!}})...(\sum _{j=0}^{\infty }{\frac {x^{nj}}{n^{j}j!}})}$ 

结论1：比较i)与ii)的结果。比较${\displaystyle z^{n}}$ 项可知，${\displaystyle a_{0}(n+1)=1\to a_{0}={\frac {1}{n+1}}}$ 。同样的方法比较后续项可知，${\displaystyle a_{k}}$ 皆为正的实数。

iii) 基于${\displaystyle u_{n}}$ 新设一个级数${\displaystyle v_{n}(z)={\frac {u_{n}(z)}{z^{n+1}}}=\sum _{k=0}^{\infty }a_{k}z^{k}}$ 。因为极点是一个可消极点，所以这也是一个整函数。计算${\displaystyle v_{n}(1)={\frac {u_{n}(1)}{1^{n+1}}}=1-E_{n}(1)=1-[(1-1)\exp \left(h_{n}(1)\right)]=1}$ 

所以在给定的条件${\displaystyle |z|\leq 1}$ 下，运用绝对值不等式的基本性质和结论1：

${\displaystyle \left|v_{n}(z)\right|\leq \sum _{k=0}^{\infty }\left|a_{k}\|z\right|^{k}\leq \sum _{k=0}^{\infty }a_{k}=v_{n}(1)=1{\text{ if }}|z|\leq 1}$ 

即，${\displaystyle \left|v_{n}(z)\right|=\left|{\frac {u_{n}(z)}{z^{n+1}}}\right|\leq 1\to \left|1-E_{n}(z)\right|\leq |z|^{n+1}}$ 成立。引理(15.8)证明完毕。

• 复分析
• 卡尔·魏尔施特拉斯

• Alford的《复分析》