549565
2024-07-02 20:42:04 227人阅读
$\bf命题2:$设$f\left( x \right) \in C\left( { - \infty , + \infty } \right)$,令
<script id="MathJax-Element-1" type="math/tex; mode=display">{f_n}\left( x \right) = \sum\limits_{k = 0}^{n - 1} {\frac{1}{n}} f\left( {x + \frac{k}{n}} \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$\int_0^1 {f\left( {x + t} \right)dt}$
证明:由$f\left( x \right) \in C\left( { - \infty , + \infty } \right)$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 $时,有
|f(x)?f(y)|<ε <script id="MathJax-Element-2" type="math/tex; mode=display">\left| {f\left( x \right) - f\left( y \right)} \right| < \varepsilon </script>
取$N = \frac{1}{\delta }$,则当$n > N$时,对任意$x \in \left[ {a,b} \right]$,$t \in \left[ {\frac{k}{n},\frac{{k + 1}}{n}} \right]$,有
<script id="MathJax-Element-3" type="math/tex; mode=display">\left| {\left( {x + \frac{k}{n}} \right) - \left( {x + t} \right)} \right| < \delta </script>从而有
<script id="MathJax-Element-4" type="math/tex; mode=display">\left| {f\left( {x + \frac{k}{n}} \right) - f\left( {x + t} \right)} \right| < \varepsilon </script>所以对任意$\varepsilon > 0$,存在$N = \frac{1}{\delta } > 0$,使得当$n > N$时,对任意$x \in \left[ {a,b} \right]$,有
∣∣∣fn(x)?∫10f(x+t)dt∣∣∣=∣∣∣∣∑k=0n?1∫k+1nkn[f(x+kn)?f(x+t)]dt∣∣∣∣<ε <script id="MathJax-Element-5" type="math/tex; mode=display">\left| {{f_n}\left( x \right) - \int_0^1 {f\left( {x + t} \right)dt} } \right| = \left| {\sum\limits_{k = 0}^{n - 1} {\int_{\frac{k}{n}}^{\frac{{k + 1}}{n}} {\left[ {f\left( {x + \frac{k}{n}} \right) - f\left( {x + t} \right)} \right]dt} } } \right| < \varepsilon </script>
从而由函数列一致收敛的定义即证$\bf注1:$$\int_0^1 {f\left( {x + t} \right)dt} {\rm{ = }}\sum\limits_{k = 0}^{n - 1} {\int_{\frac{k}{n}}^{\frac{{k + 1}}{n}} {f\left( {x + t} \right)dt} } $
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