方企勤 第三章 一元函数积分学 第14题
📝 题目
例 14 设 ${f}^{\prime }\left( x\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上连续,求证:
$$ {\int }_{0}^{1}\left| {f\left( x\right) }\right| \mathrm{d}x \leq \max \left\{ {{\int }_{0}^{1}\left| {{f}^{\prime }\left( x\right) }\right| \mathrm{d}x,\left| {{\int }_{0}^{1}f\left( x\right) \mathrm{d}x}\right| }\right\} . $$
💡 答案解析
证 分两种情况讨论. (1) 如果 $f\left( x\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上不变号,则
$$ {\int }_{0}^{1}\left| {f\left( x\right) }\right| \mathrm{d}x = \left| {{\int }_{0}^{1}f\left( x\right) \mathrm{d}x}\right| $$
$$ \leq \max \left\{ {{\int }_{0}^{1}\left| {{f}^{\prime }\left( x\right) }\right| \mathrm{d}x,\left| {{\int }_{0}^{1}f\left( x\right) \mathrm{d}x}\right| }\right\} , $$
即要证的不等式成立.
(2)如果 $f\left( x\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上变号,则存在 ${x}_{0} \in \left( {0,1}\right)$ ,使得
$$ f\left( {x}_{0}\right) = 0. $$
又因为 $f\left( x\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上连续,存在 ${x}_{M} \in \left( {0,1}\right)$ ,使得
$$ f\left( {x}_{M}\right) = \mathop{\max }\limits_{{x \in \left( {0,1}\right) }}\left| {f\left( x\right) }\right| > 0, $$
故有
$$ {\int }_{0}^{1}\left| {f\left( x\right) }\right| \mathrm{d}x \leq \mathop{\max }\limits_{{x \in \left( {0,1}\right) }}\left| {f\left( x\right) }\right| = \left| {f\left( {x}_{M}\right) - f\left( {x}_{0}\right) }\right| $$
(用微积分基本定理)
$$ = \left| {{\int }_{{x}_{0}}^{{x}_{M}}{f}^{\prime }\left( x\right) \mathrm{d}x}\right| \leq {\int }_{0}^{1}\left| {{f}^{\prime }\left( x\right) }\right| \mathrm{d}x $$
$$ \leq \max \left\{ {{\int }_{0}^{1}\left| {{f}^{\prime }\left( x\right) }\right| \mathrm{d}x,\left| {{\int }_{0}^{1}f\left( x\right) \mathrm{d}x}\right| }\right\} , $$
即要证的不等式成立.