方企勤 第一章 分析基础 第28题
📝 题目
例 28 设函数 $f\left( x\right)$ 在 $\left\lbrack {a,b}\right\rbrack$ 上单调上升, $f\left( a\right) > a,f\left( b\right) < b$ . 求证: $\exists c \in \left\lbrack {a,b}\right\rbrack$ ,使得 $f\left( c\right) = c$ .
💡 答案解析
证 将 $\left\lbrack {a,b}\right\rbrack$ 二等分,分点记为 ${c}_{0}$ ,
$$ \left\{ \begin{array}{l} \text{ 若 }f\left( {c}_{0}\right) = {c}_{0},\text{ 取 }c = {c}_{0}\text{ 即符合要求, } \\ \text{ 若 }f\left( {c}_{0}\right) > {c}_{0},\text{ 取 }\left\lbrack {{a}_{1},{b}_{1}}\right\rbrack = \left\lbrack {{c}_{0},b}\right\rbrack , \\ \text{ 若 }f\left( {c}_{0}\right) < {c}_{0},\text{ 取 }\left\lbrack {{a}_{1},{b}_{1}}\right\rbrack = \left\lbrack {a,{c}_{0}}\right\rbrack , \end{array}\right. $$
$$ \Rightarrow f\left( {a}_{1}\right) > {a}_{1},f\left( {b}_{1}\right) < {b}_{1}\text{ . } $$
将 $\left\lbrack {{a}_{1},{b}_{1}}\right\rbrack$ 二等分,分点记为 ${c}_{1}$ ,
$$ \left\{ \begin{array}{l} \text{ 若 }f\left( {c}_{1}\right) = {c}_{1},\text{ 取 }c = {c}_{1}\text{ 即符合要求, } \\ \text{ 若 }f\left( {c}_{1}\right) > {c}_{1},\text{ 取 }\left\lbrack {{a}_{2},{b}_{2}}\right\rbrack = \left\lbrack {{c}_{1},{b}_{1}}\right\rbrack , \\ \text{ 若 }f\left( {c}_{1}\right) < {c}_{1},\text{ 取 }\left\lbrack {{a}_{2},{b}_{2}}\right\rbrack = \left\lbrack {{a}_{1},{c}_{1}}\right\rbrack , \end{array}\right. $$
$$ \Rightarrow f\left( {a}_{2}\right) > {a}_{2},f\left( {b}_{2}\right) < {b}_{2}. $$
如此继续下去,要么到某一步时,取到一个分点 ${c}_{n}$ ,使得 $f\left( {c}_{n}\right) = {c}_{n}$ $\Rightarrow c = {c}_{n}$ ; 要么这些步骤可无限进行下去,产生一串闭区间 $\left\lbrack {{a}_{n},{b}_{n}}\right\rbrack$ , 并具有如下三条性质:
$$ \left\lbrack {{a}_{n + 1},{b}_{n + 1}}\right\rbrack \subset \left\lbrack {{a}_{n},{b}_{n}}\right\rbrack \;\left( {n = 1,2,\cdots }\right) ; $$
$$ {b}_{n} - {a}_{n} = \frac{1}{{2}^{n}}\left( {b - a}\right) \rightarrow 0\;\left( {n \rightarrow \infty }\right) ; $$
$$ f\left( {a}_{n}\right) > {a}_{n},\;f\left( {b}_{n}\right) < {b}_{n}. $$
因此,根据区间套定理, $\exists c$ 使得 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = c = \mathop{\lim }\limits_{{n \rightarrow \infty }}{b}_{n}}$ . 并由 $f\left( x\right)$ 的单调性, 有
$$ f\left( {c - 0}\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {a}_{n}\right) \geq \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = c, $$
$$ f\left( {c + 0}\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {b}_{n}\right) \leq \mathop{\lim }\limits_{{n \rightarrow \infty }}{b}_{n} = c. $$
由此,利用 $f\left( x\right)$ 的单调性,得到
$$ c \leq f\left( {c - 0}\right) \leq f\left( c\right) \leq f\left( {c + 0}\right) \leq c $$
$$ \Rightarrow f\left( c\right) = c\text{ . } $$