📝 题目
例 13 设 $f\left( x\right)$ 在 $\left( {-\infty , + \infty }\right)$ 可微, $f\left( 0\right) = 0$ ,且处处有 $\left| {{f}^{\prime }\left( x\right) }\right|$ $\leq \left| {f\left( x\right) }\right|$ . 求证: $f\left( x\right) \equiv 0\left( {\forall x \in \mathbf{R}}\right)$ .
💡 答案与解析
证 (1) 证 $f\left( x\right) \equiv 0\left( {\forall x \in \left\lbrack {0,1}\right\rbrack }\right)$ . 对 $\forall x > 0$ ,利用微分中值定理, 推出
$$ \left| {f\left( x\right) }\right| = \left| {f\left( x\right) - f\left( 0\right) }\right| = \left| {{f}^{\prime }\left( {x}_{1}\right) }\right| x \leq \left| {f\left( {x}_{1}\right) }\right| x $$
$$ \left( {0 < {x}_{1} < x}\right) \text{ . } $$
用 ${x}_{1}$ 代替上式中的 $x$ ,有 $\left| {f\left( {x}_{1}\right) }\right| \leq \left| {f\left( {x}_{2}\right) }\right| {x}_{1}\left( {0 < {x}_{2} < {x}_{1}}\right)$ . 同理
$$ \left| {f\left( {x}_{2}\right) }\right| \leq \left| {f\left( {x}_{3}\right) }\right| {x}_{2}\;\left( {0 < {x}_{3} < {x}_{2}}\right) ,\;\cdots $$
$$ \left| {f\left( {x}_{n - 1}\right) }\right| \leq \left| {f\left( {x}_{n}\right) }\right| {x}_{n - 1}\;\left( {0 < {x}_{n} < {x}_{n - 1}}\right) . $$
将以上几个不等式的左边和右边分别相乘, 得
$$ \left| {f\left( x\right) }\right| \leq x \cdot {x}_{1} \cdot \cdots \cdot {x}_{n - 1}\left| {f\left( {x}_{n}\right) }\right| \leq {x}^{n}\left| {f\left( {x}_{n}\right) }\right| . $$
因为 $f\left( x\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上连续,故有 ${M}_{1} > 0$ ,使得
$$ \left| {f\left( x\right) }\right| \leq {M}_{1}\;\left( {\forall x \in \left\lbrack {0,1}\right\rbrack }\right) . $$
于是,当 $0 \leq x < 1$ 时,有 $\left| {f\left( x\right) }\right| \leq {M}_{1}{x}^{n} \rightarrow 0\left( {\text{ 当 }n \rightarrow \infty }\right)$ . 由此得出 $f\left( x\right) \equiv 0\left( {0 \leq x < 1}\right)$ . 又由连续性,有 $f\left( 1\right) = 0$ . 从而
$$ f\left( x\right) \equiv 0\;\left( {\forall x \in \left\lbrack {0,1}\right\rbrack }\right) . $$
(2)证 $f\left( x\right) \equiv 0\left( {x \geq 0}\right)$ . 用数学归纳法. (1) 已证 $f\left( x\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 中恒等于 0,今设 $f\left( x\right)$ 在 $\left\lbrack {0,n}\right\rbrack$ 上恒等于 0 . 令 ${f}_{n}\left( x\right) = f\left( {x + n}\right)$ , 则有
$$ {f}_{n}\left( 0\right) = 0,\;\left| {{f}_{n}^{\prime }\left( x\right) }\right| = \left| {{f}^{\prime }\left( {x + n}\right) }\right| \leq \left| {f\left( {x + n}\right) }\right| = \left| {{f}_{n}\left( x\right) }\right| . $$
用 (1) 的结果,便可推出 ${f}_{n}\left( x\right) \equiv 0\left( {\forall x \in \left\lbrack {0,1}\right\rbrack }\right)$ ,即
$$ f\left( x\right) \equiv 0\;\left( {\forall x \in \left\lbrack {n,n + 1}\right\rbrack }\right) . $$
从而 $f\left( x\right) \equiv 0\left( {\forall x \in \left\lbrack {0,n + 1}\right\rbrack }\right)$ . 这样利用数学归纳法原理便可推得 $\forall n \in \mathbf{N}$ ,有 $f\left( x\right) \equiv 0\left( {\forall x \in \left\lbrack {0,n}\right\rbrack }\right)$ ,这显然蕴含着 $f\left( x\right)$ 在 $\lbrack 0,\infty )$ 上恒等于 0 .
(3) 证 $f\left( x\right) \equiv 0\left( {\forall x \leq 0}\right)$ . 令 ${f}_{ - }\left( x\right) = f\left( {-x}\right)$ ,则有 ${f}_{ - }\left( 0\right) = 0$ , 且
$$ \left| {{f}_{ - }^{\prime }\left( x\right) }\right| = \left| {{f}^{\prime }\left( {-x}\right) }\right| \leq \left| {f\left( {-x}\right) }\right| = \left| {{f}_{ - }\left( x\right) }\right| . $$
利用 (2) 的结果,便可推出 ${f}_{ - }\left( x\right) \equiv 0\left( {\forall x \geq 0}\right)$ ,即
$$ f\left( x\right) \equiv 0\;\left( {\forall x \leq 0}\right) . $$