📝 题目
1. 5.9 设 $f\left( x\right) \in C\left( {-\infty , + \infty }\right)$ ,存在 $\mathop{\lim }\limits_{{x \rightarrow \pm \infty }}f\left( x\right) = + \infty$ ,且 $f\left( x\right)$ 的最小值 $f\left( a\right) < a$ . 求证: $f\left( {f\left( x\right) }\right)$ 至少在两个点处取到它的最小值.
💡 答案与解析
1. 5.9 令 $F\left( x\right) = f\left( x\right) - a$ ,由 $F\left( a\right) < 0,\mathop{\lim }\limits_{{x \rightarrow \pm \infty }}F\left( x\right) = + \infty$ ,显然 $\exists {a}_{1} < a$ $< {a}_{2}$ ,使得
$$ f\left( {a}_{1}\right) = a = f\left( {a}_{2}\right) \Rightarrow f\left( {f\left( {a}_{1}\right) }\right) = f\left( a\right) = f\left( {f\left( {a}_{2}\right) }\right) . $$
1.5.10 用反证法. 假设 $f\left( x\right)$ 在 $\left\lbrack {a,b}\right\rbrack$ 上无上界,则 $\exists {x}_{n} \in \left\lbrack {a,b}\right\rbrack$ 使得
$$ f\left( {x}_{n}\right) > n\left( {n = 1,2,\cdots }\right) \overset{\text{ 波尔察诺定理 }}{ = }\exists \left\{ {x}_{{n}_{k}}\right\} , $$
及 $c \in \left\lbrack {a,b}\right\rbrack$ 使得 $\displaystyle{\mathop{\lim }\limits_{{k \rightarrow \infty }}{x}_{{n}_{k}} = c}$ . 这样,一方面由 $f\left( x\right)$ 在点 $x = c$ 处上半连续,对 ${\varepsilon }_{0} =$ 1,3 $\delta > 0$ ,使得
$$ f\left( x\right) < f\left( c\right) + 1\;\left( {\forall x \in \left( {c - \delta ,c + \delta }\right) \cap \left\lbrack {a,b}\right\rbrack }\right) . $$
另一方面, $\displaystyle{\mathop{\lim }\limits_{{k \rightarrow \infty }}{x}_{{n}_{k}} = c \Rightarrow \exists K \in N}$ ,使得
$$ \left| {{x}_{{n}_{k}} - c}\right| < \delta \Rightarrow {n}_{k} < f\left( {x}_{{n}_{k}}\right) < f\left( c\right) + 1\;\left( {\forall k > K}\right) , $$
矛盾.
1.5.13 因为 $f\left( x\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上一致连续,所以对 $\forall \varepsilon > 0,\exists \delta > 0$ ,使得 $\forall \xi$ , $\eta \geq 0,\left| {\xi - \eta }\right| < \delta$ 有
$$ \left| {f\left( \xi \right) - f\left( \eta \right) }\right| < \varepsilon /