方企勤 第一章 分析基础 第13题
📝 题目
例 13 设 $f\left( x\right)$ 是在 $\lbrack 0, + \infty )$ 上的非负连续函数,且满足对 $\forall {x}_{1},{x}_{2} \geq 0$ 有 $f\left( {{x}_{1} + {x}_{2}}\right) \leq f\left( {x}_{1}\right) + f\left( {x}_{2}\right)$ . 求证:
$$ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{f\left( x\right) }{x} = \mathop{\inf }\limits_{{x > 0}}\frac{f\left( x\right) }{x}. $$
💡 答案解析
证 记 $m = \mathop{\inf }\limits_{{x > 0}}\frac{f\left( x\right) }{x}$ ,则 $\displaystyle{0 \leq m < + \infty}$ . 根据下确界定义,对 $\forall \varepsilon$ $> 0$ ,存在 ${x}_{0} > 0$ ,使得 $m \leq \frac{f\left( {x}_{0}\right) }{{x}_{0}} < m + \frac{\varepsilon }{2}$ . 对 $\forall x > 0$ ,将 $\frac{x}{{x}_{0}}$ 进行整数部分和小数部分的分解:
$$ \frac{x}{{x}_{0}} = \left\lbrack \frac{x}{{x}_{0}}\right\rbrack + \left\{ \frac{x}{{x}_{0}}\right\} \Rightarrow x = \left\lbrack \frac{x}{{x}_{0}}\right\rbrack {x}_{0} + \left\{ \frac{x}{{x}_{0}}\right\} {x}_{0} $$
$$ \Rightarrow f\left( x\right) \leq \left\lbrack \frac{x}{{x}_{0}}\right\rbrack f\left( {x}_{0}\right) + f\left( {\left\{ \frac{x}{{x}_{0}}\right\} {x}_{0}}\right) . $$
故有
$$ \frac{f\left( x\right) }{x} \leq \frac{\left\lbrack \frac{x}{{x}_{0}}\right\rbrack }{\frac{x}{{x}_{0}}}\frac{f\left( {x}_{0}\right) }{{x}_{0}} + \frac{f\left( {\left\{ \frac{x}{{x}_{0}}\right\} {x}_{0}}\right) }{x} $$
$$ \leq \frac{f\left( {x}_{0}\right) }{{x}_{0}} + \frac{1}{x} \cdot f\left( {\left\{ \frac{x}{{x}_{0}}\right\} {x}_{0}}\right) , \tag{5.10} $$
其中 $\left\lbrack *\right\rbrack ,\left( *\right)$ 分别表示 $*$ 的整数部分和分数部分,注意到 $0 \leq$ $\left\{ \frac{x}{{x}_{0}}\right\} {x}_{0} \leq {x}_{0}$ ,便知 $f\left( {\left\{ \frac{x}{{x}_{0}}\right\} {x}_{0}}\right)$ 有界,所以对上述的 $\varepsilon > 0$ , $\exists X > 0$ ,使得对 $\forall x > X$ ,有
$$ 0 \leq \frac{1}{x} \cdot f\left( {\left\{ \frac{x}{{x}_{0}}\right\} {x}_{0}}\right) < \frac{\varepsilon }{2}. $$
于是由 (5.10) 式有
$$ m \leq \frac{f\left( x\right) }{x} \leq m + \varepsilon ,\;\forall x > X. $$