$\bf命题1：$$（\bf{Bendixon判别法}）$设$\sum\limits_{n = 1}^\infty {{u_n}\left( x \right)} $为$\left[ {a,b} \right]$上的可微函数项级数，且$\sum\limits_{n = 1}^\infty {{u_n}^\prime \left( x \right)} $的部分和函数列在$\left[ {a,b} \right]$上一致有界
证明：如果$\sum\limits_{n = 1}^\infty {{u_n}\left( x \right)} $在$\left[ {a,b}
\right]$上收敛，则必在$\left[ {a,b} \right]$上一致收敛
证明：由一致有界的定义知，存在$C>0$，使得对每个正整数$n$和每个$x \in \left[ {a,b}
\right]$，有
\[\left| {\sum\limits_{k = 1}^n {{u_k}^\prime \left( x \right)} }
\right| \le C\]
对任给$\varepsilon > 0$，取区间$\left[ {a,b}
\right]$的等距分划$\left\{ {{x_0},{x_1}, \cdots ,{x_m}}
\right\}$，使得当$m$充分大时，分划的细度\[\Delta {x_i} = \frac{{b - a}}{m} <
\frac{\varepsilon }{{4C}}\]
由于$\sum\limits_{n = 1}^\infty {{u_n}\left( x
\right)} $在$\left[ {a,b} \right]$上处处收敛，则由$\bf{Cauchy收敛准则}$知，对任给$\varepsilon
>0$，存在$N>0$，使得当$n>N$时，对任意正整数$p$和分划的每个分点${x_i}$，同时成立
\[\left| {\sum\limits_{k = n + 1}^{n + p} {{u_k}\left( {{x_i}} \right)} }
\right| < \frac{\varepsilon }{2},i = 0,1, \cdots ,m\]
于是对任意$x \in \left[
{a,b} \right]$，不妨设$x \in \left[ {{x_{i - 1}},{x_i}} \right]$，由微分中值定理知，存在${\xi
_i} \in \left( {x,{x_i}} \right)$，使得
\begin{align*}
\left| {\sum\limits_{k
= n + 1}^{n + p} {{u_k}\left( x \right)} } \right| &= \left| {\sum\limits_{k
= n + 1}^{n + p} {{u_k}\left( {{x_i}} \right)} + \sum\limits_{k = n + 1}^{n + p}
{\left( {{u_k}\left( x \right) - {u_k}\left( {{x_i}} \right)} \right)} }
\right|\\&
\le \left| {\sum\limits_{k = n + 1}^{n + p} {{u_k}\left(
{{x_i}} \right)} } \right| + \left| {\sum\limits_{k = 1}^n {{u_k}^\prime \left(
{{\xi _i}} \right)} } \right|\left| {x - {x_i}} \right|\\&
<
\frac{\varepsilon }{2} + 2C\left| {x - {x_i}} \right| \le \frac{\varepsilon }{2}
+ \frac{\varepsilon }{2} =
\varepsilon
\end{align*}
从而由函数项级数一致收敛的$\bf{Cauchy准则}$即证
原文地址：http://www.cnblogs.com/ly758241/p/3706442.html
