📝 题目
例 11 设 $f\left( x\right)$ 在 $\left( {-\infty , + \infty }\right)$ 上二次连续可微,
$$ \left| {f\left( x\right) }\right| \leq 1,\;{\left| f\left( 0\right) \right| }^{2} + {\left| {f}^{\prime }\left( 0\right) \right| }^{2} = 4. $$
求证: $\exists \xi \in \mathbf{R}$ ,使得 $f\left( \xi \right) + {f}^{\prime \prime }\left( \xi \right) = 0$ .
💡 答案与解析
证 考虑函数 $F\left( x\right) \overset{\text{ 定义 }}{ = }{\left| f\left( x\right) \right| }^{2} + {\left| {f}^{\prime }\left( x\right) \right| }^{2}$ ,则 $F\left( 0\right) = 4$ ,
$$ {F}^{\prime }\left( x\right) = 2{f}^{\prime }\left( x\right) \left\lbrack {f\left( x\right) + {f}^{\prime \prime }\left( x\right) }\right\rbrack . $$
为了证明结论,只要找到一个点 $\xi$ ,它既是 ${F}^{\prime }\left( x\right)$ 的零点,又是 ${f}^{\prime }\left( x\right)$ 的非零点就行了.
$\forall X > 0$ ,由题设 $\left| {f\left( x\right) }\right| \leq 1$ ,利用微分中值定理, $\exists {\xi }_{1} \in \left( {-X,0}\right)$ 及 ${\xi }_{2} \in \left( {0,X}\right)$ ,使得
$$ \left| {{f}^{\prime }\left( {\xi }_{1}\right) }\right| = \left| \frac{f\left( 0\right) - f\left( {-X}\right) }{X}\right| \leq \frac{2}{X}, $$
$$ \left| {{f}^{\prime }\left( {\xi }_{2}\right) }\right| = \left| \frac{f\left( X\right) - f\left( 0\right) }{X}\right| \leq \frac{2}{X}. $$
由此可见,如果取 $X = 2$ ,那么
$$ \max \left( {F\left( {\xi }_{1}\right) ,F\left( {\xi }_{2}\right) }\right) \leq 1 + \frac{4}{{X}^{2}} = 2 < F\left( 0\right) . $$
因此 $\exists \xi \in \left( {{\xi }_{1},{\xi }_{2}}\right)$ ,使得 $F\left( \xi \right) = \mathop{\max }\limits_{{{\xi }_{1} \leq x \leq {\xi }_{2}}}F\left( x\right)$ . 因为 $\xi$ 是极值点,所以 ${F}^{\prime }\left( \xi \right) = 0$ . 另外
$$ {\left| {f}^{\prime }\left( \xi \right) \right| }^{2} = F\left( \xi \right) - {\left| f\left( \xi \right) \right| }^{2} \geq F\left( 0\right) - 1 = 3, $$
所以 ${f}^{\prime }\left( \xi \right) \neq 0$ . 于是有 $f\left( \xi \right) + {f}^{\prime \prime }\left( \xi \right) = \frac{{F}^{\prime }\left( \xi \right) }{2{f}^{\prime }\left( \xi \right) } = 0$ .