方企勤 第三章 一元函数积分学 第15题
📝 题目
例 15 设 $f\left( x\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上二阶连续可导, $f\left( 0\right) = f\left( 1\right) = 0$ ,并且 $f\left( x\right) \neq 0\left( {\forall x \in \left( {0,1}\right) }\right)$ . 求证:
$$ {\int }_{0}^{1}\left| \frac{{f}^{\prime \prime }\left( x\right) }{f\left( x\right) }\right| \mathrm{d}x > 4. \tag{3.18} $$
💡 答案解析
证法 1 因为 $f\left( x\right) \neq 0\left( {\forall \in \left( {0,1}\right) }\right)$ ,所以 $f\left( x\right)$ 在(0,1)内同号,不妨设 $f\left( x\right) > 0$ . 又 $f\left( x\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上连续,故存在 ${x}_{0} \in \left( {0,1}\right)$ ,使得 $f\left( {x}_{0}\right) = \mathop{\max }\limits_{{x \in \left\lbrack {0,1}\right\rbrack }}f\left( x\right) > 0$ ,从而
$$ {\int }_{0}^{1}\left| \frac{{f}^{\prime \prime }\left( x\right) }{f\left( x\right) }\right| \mathrm{d}x > \frac{1}{f\left( {x}_{0}\right) }{\int }_{0}^{1}\left| {{f}^{\prime \prime }\left( x\right) }\right| \mathrm{d}x. \tag{3.19} $$
又由微分中值定理,分别存在 $\alpha \in \left( {0,{x}_{0}}\right) ,\beta \in \left( {{x}_{0},1}\right)$ ,使得
$$ {f}^{\prime }\left( \alpha \right) = \frac{f\left( {x}_{0}\right) }{{x}_{0}},\;{f}^{\prime }\left( \beta \right) = - \frac{f\left( {x}_{0}\right) }{1 - {x}_{0}}. \tag{3.20} $$
因此, 根据微积分基本定理, 有
$$ {\int }_{0}^{1}\left| {{f}^{\prime \prime }\left( x\right) }\right| \mathrm{d}x \leq {\int }_{\alpha }^{\beta }\left| {{f}^{\prime \prime }\left( x\right) }\right| \mathrm{d}x \geq \left| {{\int }_{\alpha }^{\beta }{f}^{\prime \prime }\left( x\right) \mathrm{d}x}\right| $$
$$ = \left| {{f}^{\prime }\left( \beta \right) - {f}^{\prime }\left( \alpha \right) }\right| = \frac{f\left( {x}_{0}\right) }{{x}_{0}\left( {1 - {x}_{0}}\right) } $$
$$ \geq {4f}\left( {x}_{0}\right) , \tag{3.21} $$
其中最后一个不等式是由于
$$ \sqrt{{x}_{0}\left( {1 - {x}_{0}}\right) } \leq \frac{{x}_{0} + \left( {1 - {x}_{0}}\right) }{2} = \frac{1}{2} \Rightarrow {x}_{0}\left( {1 - {x}_{0}}\right) \leq \frac{1}{4}. $$
联合 (3.20), (3.21) 与 (3.18)式, 即得结论.
证法 2 证法 1 中的 ${x}_{0}$ 是 $f\left( x\right)$ 的极大值点,故有 ${f}^{\prime }\left( {x}_{0}\right) = 0$ . 根据牛顿-莱布尼茨公式与积分中值定理,对 $\forall x \in \left( {0,{x}_{0}}\right\rbrack$ ,有
$$ \left| {f\left( x\right) }\right| = \left| {f\left( x\right) - f\left( 0\right) }\right| \overset{\text{ 由牛-莱公式 }}{ = }\left| {{\int }_{0}^{x}{f}^{\prime }\left( t\right) \mathrm{d}t}\right| $$
$$ \leq {\int }_{0}^{x}\left| {{f}^{\prime }\left( t\right) }\right| \mathrm{d}t \leq {\int }_{0}^{{x}_{0}}\left| {{f}^{\prime }\left( t\right) }\right| \mathrm{d}t $$
$$ \overset{\text{ 由积分中值定理 }}{ = }{x}_{0}\left| {{f}^{\prime }\left( \xi \right) }\right| \;\left( {\xi \in \left\lbrack {0,{x}_{0}}\right\rbrack }\right) \tag{3.22} $$
及
$$ \left| {{f}^{\prime }\left( \xi \right) }\right| = \left| {{f}^{\prime }\left( \xi \right) - {f}^{\prime }\left( {x}_{0}\right) }\right| $$
$$ \overset{\text{ 由牛-莱公式 }}{ = }\left| {{\int }_{\xi }^{{x}_{0}}{f}^{\prime \prime }\left( t\right) \mathrm{d}t}\right| $$
$$ \leq {\int }_{0}^{{x}_{0}}\left| {{f}^{\prime \prime }\left( t\right) }\right| \mathrm{d}t. \tag{3.23} $$
联合 (3.22) 与 (3.23) 式,特别当 $x = {x}_{0}$ 时,有
$$ \left| {f\left( {x}_{0}\right) }\right| \leq {x}_{0}{\int }_{0}^{{x}_{0}}\left| {{f}^{\prime \prime }\left( x\right) }\right| \mathrm{d}x. \tag{3.24} $$
同理,对 $\forall x \in \left\lbrack {{x}_{0},1}\right\rbrack$ ,有
$$ \left| {f\left( x\right) }\right| \leq {\int }_{x}^{1}\left| {{f}^{\prime }\left( t\right) }\right| \mathrm{d}t \leq {\int }_{{x}_{0}}^{1}\left| {{f}^{\prime }\left( t\right) }\right| \mathrm{d}t $$
$$ = \left( {1 - {x}_{0}}\right) \left| {{f}^{\prime }\left( \eta \right) }\right| \;\left( {\eta \in \left( {{x}_{0},1}\right) }\right) , $$
$$ \left| {{f}^{\prime }\left( \eta \right) }\right| = \left| {{f}^{\prime }\left( \eta \right) - {f}^{\prime }\left( {x}_{0}\right) }\right| \leq {\int }_{{x}_{0}}^{\eta }{f}^{\prime \prime }\left( t\right) \mathrm{d}t \leq {\int }_{{x}_{0}}^{1}\left| {{f}^{\prime \prime }\left( t\right) }\right| \mathrm{d}t. $$
联合上面两式,特别当 $x = {x}_{0}$ 时,有
$$ \left| {f\left( {x}_{0}\right) }\right| \leq \left( {1 - {x}_{0}}\right) {\int }_{{x}_{0}}^{1}\left| {{f}^{\prime \prime }\left( x\right) }\right| \mathrm{d}x. \tag{3.25} $$
联合 (3.19),(3.24),(3.25)式,有
$$ {\int }_{0}^{1}\left| \frac{{f}^{\prime \prime }\left( x\right) }{f\left( x\right) }\right| \mathrm{d}x > \frac{1}{\left| f\left( {x}_{0}\right) \right| }{\int }_{0}^{1}\left| {{f}^{\prime \prime }\left( x\right) }\right| \mathrm{d}x $$
$$ = \frac{1}{\left| f\left( {x}_{0}\right) \right| }\left\{ {{\int }_{0}^{{x}_{0}}\left| {{f}^{\prime \prime }\left( x\right) }\right| \mathrm{d}x + {\int }_{{x}_{0}}^{1}\left| {{f}^{\prime \prime }\left( x\right) }\right| \mathrm{d}x}\right\} $$
$$ = \frac{1}{\left| f\left( {x}_{0}\right) \right| }\left\{ {\frac{\left| f\left( {x}_{0}\right) \right| }{{x}_{0}} + \frac{\left| f\left( {x}_{0}\right) \right| }{1 - {x}_{0}}}\right\} = \frac{1}{{x}_{0}} + \frac{1}{1 - {x}_{0}} $$
$$ = \frac{1}{{x}_{0}\left( {1 - {x}_{0}}\right) } \geq 4. $$