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时间:2014-05-04 19:12:16      阅读:351      评论:0      收藏:0      [点我收藏+]

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$\bf命题1:$设$f\left( x \right) \in {C^1}\left( { - \infty , + \infty } \right)$,令

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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

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由微分中值定理知,存在${\xi _n} \in \left( {x,x + \frac{1}{n}} \right)$,使得

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取$N = \frac{1}{\delta }$,则当$n > N$时,对任意$x \in \left[ {a,b} \right]$,有
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从而有

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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
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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

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即${f_n}\left( x \right)$在$\left( { - \infty , + \infty } \right)$上处处收敛于$f‘\left( x \right)$

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26566

标签:style   class   ext   color   width   set   

原文地址:http://www.cnblogs.com/ly758241/p/3706461.html

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