📝 题目
例 11 设 $f\left( x\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上连续可导,且 $f\left( 0\right) = 0,f\left( 1\right) = 1$ ,求证:
$$ {\int }_{0}^{1}\left| {f\left( x\right) - {f}^{\prime }\left( x\right) }\right| \geq {\mathrm{e}}^{-1}. \tag{3.14} $$
💡 答案与解析
证 设 $v\left( x\right)$ 满足
$$ {v}^{\prime }\left( x\right) = - v\left( x\right) ,\;v\left( x\right) > 0\;\left( {x \in \left( {0,1}\right) }\right) . \tag{3. 15} $$
显然 $v\left( x\right) \overset{\text{ 定义 }}{ = }{\mathrm{e}}^{-x}$ 满足 (3.15) 式. 于是
$$ \left| {f\left( x\right) - {f}^{\prime }\left( x\right) }\right| = \frac{\left| v\left( x\right) f\left( x\right) - v\left( x\right) {f}^{\prime }\left( x\right) \right| }{v\left( x\right) } $$
$$ = \frac{\left| {\left( v\left( x\right) f\left( x\right) \right) }^{\prime }\right| }{v\left( x\right) } = {\mathrm{e}}^{x}\left| {\left( {\mathrm{e}}^{-x}f\left( x\right) \right) }^{\prime }\right| $$
$$ \geq \left| {\left( {\mathrm{e}}^{-x}f\left( x\right) \right) }^{\prime }\right| \geq {\left( {\mathrm{e}}^{-x}f\left( x\right) \right) }^{\prime } $$
$$ \left( {\forall x \in \left( {0,1}\right) }\right) \text{ . } $$
所以
$$ {\int }_{0}^{1}\left| {f\left( x\right) - {f}^{\prime }\left( x\right) }\right| \mathrm{d}x \geq {\int }_{0}^{1}{\left( {\mathrm{e}}^{-x}f\left( x\right) \right) }^{\prime }\mathrm{d}x = {\left. {\mathrm{e}}^{-x}f\left( x\right) \right| }_{0}^{1} = {\mathrm{e}}^{-1}. $$