📝 题目
例 12 设 $f\left( x\right)$ 无穷次可微,且
$$ \left| {{f}^{\left( n\right) }\left( x\right) }\right| \leq M,\;f\left( {1/n}\right) = 0\;\left( {n = 1,2,\cdots }\right) . $$
求证: $f\left( x\right) \equiv 0$ .
💡 答案与解析
证法 1 记 ${R}_{N}\left( x\right) \overset{\text{ 定义 }}{ = }f\left( x\right) - \mathop{\sum }\limits_{{n = 0}}^{N}\frac{{f}^{\left( n\right) }\left( 0\right) }{n!}{x}^{n}$ . 则 $\exists 0 < \theta < 1$ ,使得
$$ \left| {{R}_{N}\left( x\right) }\right| = \left| {\frac{{f}^{\left( N + 1\right) }\left( {\theta x}\right) }{\left( {N + 1}\right) !}{x}^{N + 1}}\right| \leq \frac{M{\left| x\right| }^{N + 1}}{\left( {N + 1}\right) !} \rightarrow 0\;\left( {N \rightarrow + \infty }\right) , $$
从而
$$ f\left( x\right) = \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}\frac{{f}^{\left( n\right) }\left( 0\right) }{n!}{x}^{n},\;x \in \left( {-\infty , + \infty }\right) . $$
下面用反证法. 假设 $f\left( x\right) ≢ 0$ ,那么 $\exists m$ ,使得 ${f}^{\left( m\right) }\left( 0\right) \neq 0$ ,而
$f\left( 0\right) = {f}^{\prime }\left( 0\right) = \cdots = {f}^{\left( m - 1\right) }\left( 0\right) = 0$ . 于是
$$ f\left( x\right) = \mathop{\sum }\limits_{{n = m}}^{{+\infty }}\frac{{f}^{\left( n\right) }\left( 0\right) }{n!}{x}^{n} = {x}^{m}\varphi \left( x\right) , $$
其中 $\varphi \left( x\right) \frac{\text{ 定义 }}{}\mathop{\sum }\limits_{{n = m}}^{{+\infty }}\frac{{f}^{\left( n\right) }\left( 0\right) }{n!}{x}^{n - m}$ . 显然 $\varphi \left( x\right)$ 连续,并且
$$ \varphi \left( 0\right) = \frac{{f}^{\left( m\right) }\left( 0\right) }{m!} \neq 0. $$
另一方面,由 $f\left( {1/n}\right) = 0\left( {n = 1,2,\cdots }\right)$ 得出 $\varphi \left( {1/n}\right) = 0\left( {n = 1,2,\cdots }\right)$ , 从而 $\varphi \left( 0\right) = 0$ . 这样引出矛盾. 所以 $f\left( x\right) \equiv 0$ .
证法 2 (1) 用数学归纳法证明. 对任意整数 $k \geq 0$ ,存在对 $n$ 严格下降的序列 ${x}_{n}^{\left( k\right) } \rightarrow 0\left( {n \rightarrow + \infty }\right)$ ,使得 ${f}^{\left( k\right) }\left( {x}_{n}^{\left( k\right) }\right) = 0\left( {n = 1,2,\cdots }\right)$ . 当 $k = 0$ 时,由题设可知取 ${x}_{n}^{\left( 0\right) } = \frac{1}{n}\left( {n = 1,2,\cdots }\right)$ 便符合要求. 今设 ${x}_{n}^{\left( k\right) } \rightarrow 0\left( {n \rightarrow \infty }\right) ,{x}_{n}^{\left( k\right) } > {x}_{n + 1}^{\left( k\right) }$ 且 ${f}^{\left( k\right) }\left( {x}_{n}^{\left( k\right) }\right) = 0\left( {n = 1,2,\cdots }\right)$ . 由于 ${f}^{\left( k\right) }\left( x\right)$ 可微,根据罗尔定理,对 $\forall n \in \mathbf{N},\exists {x}_{n}^{\left( k + 1\right) } \in \left( {{x}_{n + 1}^{\left( k\right) },{x}_{n}^{\left( k\right) }}\right)$ ,使得 ${f}^{\left( k + 1\right) }\left( {x}_{n}^{\left( k + 1\right) }\right) = 0\left( {n = 1,2,\cdots }\right)$ . 显然 ${x}_{n}^{\left( k + 1\right) } \rightarrow 0$ 且
$$ {x}_{n}^{\left( k + 1\right) } > {x}_{n + 1}^{\left( k + 1\right) }\;\left( {n = 1,2,\cdots }\right) . $$
(2)因为 ${f}^{\left( k\right) }\left( x\right)$ 连续及 ${f}^{\left( k\right) }\left( {x}_{n}^{\left( k\right) }\right) = 0$ ,令 $\displaystyle{n \rightarrow + \infty}$ ,即得
$$ {f}^{\left( k\right) }\left( 0\right) = 0\;\left( {k = 0,1,2,\cdots }\right) . $$
(3)由(2),对 $\forall x \in \mathbf{R}$ ,按泰勒公式,有
$$ \left| {f\left( x\right) }\right| = \left| {\frac{{f}^{\left( k\right) }\left( {\theta x}\right) }{k!}{x}^{k}}\right| \leq \frac{M{\left| x\right| }^{k}}{k!}. $$
令 $\displaystyle{k \rightarrow \infty}$ ,即得 $f\left( x\right) \equiv 0$ .