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$\bf命题1：$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$f'\left( x \right)$

证明：由$f\left( x \right) \in {C^1}\left( { - \infty , + \infty } \right)$知，$f'\left( x \right) \in C\left[ {a,b}
\right]$，则

由$\bf{Cantor定理}$知，$f'\left( x \right)$在$\left[ {a,b} \right]$上一致连续，即对任意$\varepsilon > 0$，存在$\delta > 0$，使得对任意的$x,y \in \left[ {a,b} \right]$满足$\left| {x - y} \right| < \delta $时，有\[\left| {f'\left( x \right) - f'\left( y \right)} \right| < \varepsilon \]

由微分中值定理知，存在${\xi _n} \in \left( {x,x + \frac{1}{n}} \right)$，使得
\[{f_n}\left( x \right) = nf'\left( {{\xi _n}} \right)\frac{1}{n} = f'\left( {{\xi _n}} \right)\]
取$N = \frac{1}{\delta }$，则当$n > N$时，对任意$x \in \left[ {a,b} \right]$，有\[\left| {x - {\xi _n}} \right| < \delta \]

从而有\[\left| {f'\left( x \right) - f'\left( {{\xi _n}} \right)} \right| < \varepsilon \]
所以对任意$\varepsilon > 0$，存在$N = \frac{1}{\delta } > 0$，使得当$n > N$时，对任意$x \in \left[ {a,b} \right]$，有\[\left| {f'\left( x \right) - {f_n}\left( x \right)} \right|{\rm{ = }}\left| {f'\left( x \right) - f'\left( {{\xi _n}} \right)} \right| < \varepsilon \]
从而由函数列一致收敛的定义即证

$\bf注1：$$N$的取值由不等式${\left| {x - {\xi _n}} \right| < \delta }$放缩得到

$\bf注2：$由于$f'\left( x \right) \in C\left( { - \infty , + \infty } \right)$，所以\[ \lim \limits_{n \to \infty } {f_n}\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{f\left( {x + \frac{1}{n}} \right) - f\left( x \right)}}{{\frac{1}{n}}} = f'\left( x \right)\]
即${f_n}\left( x \right)$在$\left( { - \infty , + \infty } \right)$上处处收敛于$f'\left( x \right)$