周民强一道题

陈洪葛 posted @ Jun 09, 2014 11:09:13 AM in 数学分析 , 668 阅读

设$f$是定义在$R$上的实函数,对任意绝对收敛的级数$\displaystyle\sum_{n=1}^{\infty}a_{n}$,若$\displaystyle\sum_{n=1}^{+\infty}f(a_{n})$收敛,则
\[ f(x)=O(x) (x\to 0) \]
证明:若$f(x)$并不是$O(x)$,(当$x\to 0$)时,必要时我们能找到一个单调的序列$x_{n}\to 0$,且对任意的$n$,都有$x_{n}\neq 0$,同时$f(x_{n})$保持着一致的符号,且
\[ \lim_{k\to\infty}\left|\frac{f(x_{k})}{x_{k}}\right|=+\infty \]
对任意的$n>0$,存在这样的$k_{n}\in \mathbf{N}^{+}$,当$k\geq k_{n}$时,有$|x_{k}|\leq \frac{1}{2n^2}$且
\[ \left|\frac{f(x_{k})}{x_{k}}\right|\geq n \]
令$ j_{n}=\left[\frac{1}{n^2|x_{k_{n}}|}\right]$,则
\[ \frac{1}{2n^2}\leq \overbrace{|x_{k_{n}}+x_{k_{n}}+\cdots+x_{k_{n}}|}^{j_{n}\text{个}}\leq \frac{1}{n^2} \]
对$n$求和,就有
\[ \left|\sum_{n=1}^{\infty}\overbrace{x_{k_{n}}+x_{k_{n}}+\cdots+x_{k_{n}}}^{j_{n}\text{个}} \right|\leq \sum_{n=1}^{\infty}\frac{1}{n^2} \]
是收敛的,但这时
\[ \left|\sum_{n=1}^{\infty}\overbrace{f(x_{k_{n}})+f(x_{k_{n}})+\cdots+f(x_{k_{n}})}^{j_{n}\text{个}}\right|\geq \sum_{n=1}^{\infty}\frac{1}{2n}\]
发散。这一矛盾说明了
\[ f(x)=O(x) (x\to 0) \]


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