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26566

2024-07-02 20:41:38   215人阅读

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

fn(x)=n[f(x+1n)?f(x)]
<script id="MathJax-Element-1" type="math/tex; mode=display">{f_n}\left( x \right) = n\left[ {f\left( {x + \frac{1}{n}} \right) - f\left( x \right)} \right]</script>

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
<script id="MathJax-Element-2" type="math/tex; mode=display">\left| {f‘\left( x \right) - f‘\left( y \right)} \right| < \varepsilon </script>

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)
<script id="MathJax-Element-3" type="math/tex; mode=display">{f_n}\left( x \right) = nf‘\left( {{\xi _n}} \right)\frac{1}{n} = f‘\left( {{\xi _n}} \right)</script>
取$N = \frac{1}{\delta }$，则当$n > N$时，对任意$x \in \left[ {a,b} \right]$，有
|x?ξn|<δ
<script id="MathJax-Element-4" type="math/tex; mode=display">\left| {x - {\xi _n}} \right| < \delta </script>

从而有

∣∣f′(x)?f′(ξn)∣∣<ε
<script id="MathJax-Element-5" type="math/tex; mode=display">\left| {f‘\left( x \right) - f‘\left( {{\xi _n}} \right)} \right| < \varepsilon </script>
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
<script id="MathJax-Element-6" type="math/tex; mode=display">\left| {f‘\left( x \right) - {f_n}\left( x \right)} \right|{\rm{ = }}\left| {f‘\left( x \right) - f‘\left( {{\xi _n}} \right)} \right| < \varepsilon </script>
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)
<script id="MathJax-Element-7" type="math/tex; mode=display"> \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)</script>
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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