📝 题目
3. 3.20 设 $f\left( x\right)$ 在 $\lbrack 0, + \infty )$ 上单调上升,函数
$$ F\left( x\right) \frac{\text{ 定义 }\left\{ \begin{array}{ll} \frac{1}{x}{\int }_{0}^{x}f\left( t\right) \mathrm{d}t, & x > 0, \\ f\left( {0 + 0}\right) , & x =
💡 答案与解析
3. 3.20 对 $\forall {x}_{2} > {x}_{1} > 0$ ,
$$ F\left( {x}_{1}\right) = \frac{1}{{x}_{1}}{\int }_{0}^{{x}_{1}}f\left( t\right) \mathrm{d}t\overset{u = {x}_{1}t}{ = }{\int }_{0}^{1}f\left( {{x}_{1}u}\right) \mathrm{d}u, $$
$$ F\left( {x}_{2}\right) = \frac{1}{{x}_{2}}{\int }_{0}^{{x}_{2}}f\left( t\right) \mathrm{d}t\overset{u = {x}_{2}t}{ = }{\int }_{0}^{1}f\left( {{x}_{2}u}\right) \mathrm{d}u. $$
因为 $f\left( x\right) \uparrow$ ,所以
$$ F\left( {x}_{1}\right) = {\int }_{0}^{1}f\left( {{x}_{1}u}\right) \mathrm{d}u \leq {\int }_{0}^{1}f\left( {{x}_{2}u}\right) \mathrm{d}u = F\left( {x}_{2}\right) \Rightarrow F\left( x\right) \uparrow . $$
再由 $f\left( x\right)$ 单调上升,有
$$ f\left( {0 + 0}\right) \leq f\left( {xu}\right) \leq f\left( x\right) \;\left( {\forall x > 0,0 \leq u \leq 1}\right) $$
推出
$$ f\left( {0 + 0}\right) = F\left( 0\right) \leq F\left( x\right) = {\int }_{0}^{1}f\left( {xu}\right) \mathrm{d}u \leq f\left( x\right) , $$
由此根据夹挤准则, 有
$$ \mathop{\lim }\limits_{{x \rightarrow 0 + 0}}F\left( x\right) = f\left( {0 + 0}\right) = F\left( 0\right) . $$