方企勤 第一章 分析基础 第2题
📝 题目
例 2 . 设 $f\left( x\right) \in C\left( {a,b}\right) ,{x}_{1},{x}_{2},\cdots ,{x}_{n} \in \left( {a,b}\right)$ . 求证: $\exists \xi \in$ (a, b),使得
$$ f\left( \xi \right) = \frac{1}{n}\mathop{\sum }\limits_{{k = 1}}^{n}f\left( {x}_{k}\right) . \tag{5.1} $$
💡 答案解析
证 设
$$ f\left( {x}_{{k}_{m}}\right) = \min \left\{ {f\left( {x}_{1}\right) ,f\left( {x}_{2}\right) ,\cdots ,f\left( {x}_{n}\right) }\right\} , $$
$$ f\left( {x}_{{k}_{M}}\right) = \max \left\{ {f\left( {x}_{1}\right) ,f\left( {x}_{2}\right) ,\cdots ,f\left( {x}_{n}\right) }\right\} . $$
如果 ${x}_{{k}_{m}} = {x}_{{k}_{M}}$ ,则有
$$ f\left( {x}_{{k}_{m}}\right) = f\left( {x}_{{k}_{M}}\right) \Rightarrow f\left( {x}_{1}\right) = f\left( {x}_{2}\right) = \cdots = f\left( {x}_{n}\right) . $$
这时任取 ${x}_{k}\left( {k = 1,2,\cdots }\right)$ 为 $\xi$ ,都符合要求.
当 ${x}_{{k}_{m}} \neq {x}_{{k}_{M}}$ 时,令
$$ \alpha = \min \left\{ {{x}_{{k}_{m}},{x}_{{k}_{M}}}\right\} ,\;\beta = \max \left\{ {{x}_{{k}_{m}},{x}_{{k}_{M}}}\right\} , $$
则有
$$ f\left( {x}_{{k}_{m}}\right) \leq f\left( {x}_{k}\right) \leq f\left( {x}_{{k}_{M}}\right) \;\left( {k = 1,2,\cdots }\right) . $$
由此推出 $f\left( {x}_{{k}_{m}}\right) \leq \frac{1}{n}\mathop{\sum }\limits_{{k = 1}}^{n}f\left( {x}_{k}\right) \leq f\left( {x}_{{k}_{M}}\right)$ ,在区间 $\left\lbrack {\alpha ,\beta }\right\rbrack$ 上应用介值定理,则 $\exists \xi \in \left\lbrack {\alpha ,\beta }\right\rbrack \subset \left( {a,b}\right)$ ,使得 (5.1) 成立.