关于函数列一致收敛的专题讨论

$\bf命题：$设$f\left( x \right) \in {C^1}\left( { - \infty , + \infty } \right)$，令\[{f_n}\left( x \right) = n\left[ {f\left( {x + \frac{1}{n}} \right) - f\left( x \right)} \right]\]

证明：对任意$x \in \left[ {a,b} \right] \subset \left( { - \infty , + \infty } \right)$，有${f_n}\left( x \right)$一致收敛于$f'\left( x \right)$

参考答案

$\bf命题：$设$f\left( x \right) \in C\left( { - \infty , + \infty } \right)$，令\[{f_n}\left( x \right) = \sum\limits_{k = 0}^{n - 1} {\frac{1}{n}} f\left( {x + \frac{k}{n}} \right)\]

证明：对任意$x \in \left[ {a,b} \right] \subset \left( { - \infty , + \infty } \right)$，有${f_n}\left( x \right)$一致收敛于$\int_0^1 {f\left( {x + t} \right)dt}$

参考答案

$\bf命题：$$\left( \bf{Dini定理} \right)$设函数列$\left\{ {{f_n}\left( x \right)} \right\}$的每一项及其极限函数$f\left( x \right)$

均在$x \in \left[ {a,b} \right]$上连续，且对每个$ x \in \left[ {a,b} \right]$有$\left\{ {{f_n}\left( x \right)} \right\}$为单调数列，则函数列$\left\{ {{f_n}\left( x \right)} \right\}$在$\left[ {a,b} \right]$上一致收敛于${f\left( x \right)}$

方法一

$\bf命题：$