📝 题目
例 18 设 $f\left( x\right)$ 在 $\left( {-\infty , + \infty }\right)$ 上二次连续可微,并满足:
(1) $f\left( x\right) \leq {f}^{\prime \prime }\left( x\right)$ ;
(2) $\mathop{\lim }\limits_{{x \rightarrow \pm \infty }}{\mathrm{e}}^{-\left| x\right| }f\left( x\right) = 0$ .
求证: $f\left( x\right) \leq 0\left( {\forall x \in \left( {-\infty , + \infty }\right) }\right)$ .
💡 答案与解析
证法 $1\forall {x}_{0} \in \mathbf{R}$ 及 $\forall x \geq {x}_{0}$ ,由条件 (1) 有
$$ 0 \leq {\int }_{{x}_{0}}^{x}{\mathrm{e}}^{t}\left( {{f}^{\prime \prime }\left( t\right) - f\left( t\right) }\right) \mathrm{d}t $$
$$ = {\int }_{{x}_{0}}^{x}{\mathrm{e}}^{t}\left( {{f}^{\prime \prime }\left( t\right) + {f}^{\prime }\left( t\right) }\right) \mathrm{d}t - {\int }_{{x}_{0}}^{x}{\mathrm{e}}^{t}\left( {{f}^{\prime }\left( t\right) + f\left( t\right) }\right) \mathrm{d}t $$
$$ = {\left. {\mathrm{e}}^{t}{f}^{\prime }\left( t\right) \right| }_{{x}_{0}}^{x} - {\left. {\mathrm{e}}^{t}f\left( t\right) \right| }_{{x}_{0}}^{x} $$
$$ = {\mathrm{e}}^{x}\left( {{f}^{\prime }\left( x\right) - f\left( x\right) }\right) - {\mathrm{e}}^{{x}_{0}}\left\lbrack {{f}^{\prime }\left( {x}_{0}\right) - f\left( {x}_{0}\right) }\right\rbrack . $$
两边同时乘以 ${\mathrm{e}}^{-{2x}}$ ,得
$$ {\mathrm{e}}^{-x}\left\lbrack {{f}^{\prime }\left( x\right) - f\left( x\right) }\right\rbrack \geq {\mathrm{e}}^{{x}_{0}}\left\lbrack {{f}^{\prime }\left( {x}_{0}\right) - f\left( {x}_{0}\right) }\right\rbrack {\mathrm{e}}^{-{2x}}. $$
两边从 ${x}_{0}$ 到 $x$ 积分,推出
$$ {\mathrm{e}}^{-x}f\left( x\right) - {\mathrm{e}}^{-{x}_{0}}f\left( {x}_{0}\right) \geq \frac{1}{2}\left( {{\mathrm{e}}^{-2{x}_{0}} - {\mathrm{e}}^{-{2x}}}\right) {\mathrm{e}}^{{x}_{0}}\left\lbrack {{f}^{\prime }\left( {x}_{0}\right) - f\left( {x}_{0}\right) }\right\rbrack . $$
令 $\displaystyle{x \rightarrow + \infty}$ ,由条件 (2) 得
$$ - {\mathrm{e}}^{-{x}_{0}}f\left( {x}_{0}\right) \geq \frac{1}{2}{\mathrm{e}}^{-{x}_{0}}\left\lbrack {{f}^{\prime }\left( {x}_{0}\right) - f\left( {x}_{0}\right) }\right\rbrack , $$
即
$$ {f}^{\prime }\left( {x}_{0}\right) + f\left( {x}_{0}\right) \leq 0 : \left( {\forall {x}_{0} \in \mathbf{R}}\right) . $$
因此 $\frac{\mathrm{d}}{\mathrm{d}x}\left( {{\mathrm{e}}^{x}f\left( x\right) }\right) = {\mathrm{e}}^{x}\left( {{f}^{\prime }\left( x\right) + f\left( x\right) }\right) \leq 0\left( {\forall x \in \mathbf{R}}\right)$ . 从而 ${\mathrm{e}}^{x}f\left( x\right)$ 单调下降,即 ${\mathrm{e}}^{x}f\left( x\right) \leq {\mathrm{e}}^{{x}_{0}}f\left( {x}_{0}\right) \left( {\forall x \geq {x}_{0}}\right)$ . 令 $\displaystyle{x}_{0} \rightarrow - \infty}$ ,由条件 (2) 推出 ${\mathrm{e}}^{x}f\left( x\right) \leq 0$ ,即得 $f\left( x\right) \leq 0\left( {\forall x \in \mathbf{R}}\right)$ .
证法 2 考虑函数 $F\left( x\right) = {c}_{1}{\mathrm{e}}^{x} + {c}_{2}{\mathrm{e}}^{-x} - f\left( x\right) \left( {{c}_{1},{c}_{2} > 0}\right)$ ,则有
$$ {F}^{\prime \prime }\left( x\right) = {c}_{1}{\mathrm{e}}^{x} + {c}_{2}{\mathrm{e}}^{-x} - {f}^{\prime \prime }\left( x\right) $$
$$ \leq {c}_{1}{\mathrm{e}}^{x} + {c}_{2}{\mathrm{e}}^{-x} - f\left( x\right) = F\left( x\right) \;\left( {\forall x \in \mathbf{R}}\right) , $$
且
$$ \mathop{\lim }\limits_{{x \rightarrow + \infty }}{\mathrm{e}}^{-x}F\left( x\right) = {c}_{1}\text{ 或 }F\left( x\right) \sim {c}_{1}{\mathrm{e}}^{x}\;\left( {x \rightarrow + \infty }\right) , $$
$$ \mathop{\lim }\limits_{{x \rightarrow - \infty }}{\mathrm{e}}^{x}F\left( x\right) = {c}_{2}\text{ 或 }F\left( x\right) \sim {c}_{2}{\mathrm{e}}^{-x}\;\left( {x \rightarrow - \infty }\right) , $$
从而 $\mathop{\lim }\limits_{{x \rightarrow \pm \infty }}F\left( x\right) = + \infty$ . 由此可见, $F\left( x\right)$ 在 $\left( {-\infty , + \infty }\right)$ 上达到最小值. 记 $F\left( x\right)$ 达到最小值的点为 ${x}^{ * }$ ,则有 $F\left( {x}^{ * }\right) \leq F\left( x\right) \left( {\forall x \in \mathbf{R}}\right)$ ,且 ${F}^{\prime \prime }\left( {x}^{ * }\right) \geq 0$ (否则 ${x}^{ * }$ 为极大值点). 这样
$$ F\left( x\right) \geq F\left( {x}^{ * }\right) \geq {F}^{\prime \prime }\left( {x}^{ * }\right) \geq 0\;\left( {\forall x \in \mathbf{R}}\right) , $$
即 ${c}_{1}{\mathrm{e}}^{x} + {c}_{2}{\mathrm{e}}^{-x} \geq f\left( x\right) \left( {\forall x \in \mathbf{R}}\right)$ . 再令 ${c}_{1} \rightarrow 0,{c}_{2} \rightarrow 0$ ,即得
$$ f\left( x\right) \leq 0\;\left( {\forall x \in \mathbf{R}}\right) . $$