$\bf命题1:$设$f\left( x \right) \in {C^1}\left( { - \infty , + \infty }
\right)$,令
fn(x)=n[f(x+1n)?f(x)]
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 $时,有
∣∣f′(x)?f′(y)∣∣<ε
由微分中值定理知,存在${\xi _n} \in \left( {x,x + \frac{1}{n}} \right)$,使得
fn(x)=nf′(ξn)1n=f′(ξn)
取$N = \frac{1}{\delta }$,则当$n > N$时,对任意$x \in \left[ {a,b}
\right]$,有
|x?ξn|<δ
从而有
∣∣f′(x)?f′(ξn)∣∣<ε
所以对任意$\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
∣∣f′(x)?fn(x)∣∣=∣∣f′(x)?f′(ξn)∣∣<ε
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
limn→∞fn(x)=limn→∞f(x+1n)?f(x)1n=f′(x)
即${f_n}\left( x \right)$在$\left( { - \infty , + \infty }
\right)$上处处收敛于$f‘\left( x \right)$