设函数 $f(x)$ 在 $[-a, a]$ 上具有二阶连续导数, 证明:
(I) 若 $f(x)=0$, 则存在 $\xi \in(-a, a)$, 使得 $f^{\prime \prime}(\xi)=\frac{1}{a^2}[f(a)+f(-a)]$;
(II) 若 $f(x)$ 在 $(-a, a)$ 内取得极值, 则存在 $\eta \in(-a, a)$ 使得
$$
\left|f^{\prime \prime}(\eta)\right| \geq \frac{1}{2 a^2}|f(a)-f(-a)| .
$$
【答案】 (I) 证明: $f(x)=f(0)+f^{\prime}(0) x+\frac{f^{\prime \prime}(\eta)}{2 !} x^2=f^{\prime}(0) x+\frac{f^{\prime \prime}(\eta)}{2 !} x^2, \eta$ 介于 0 与 $x$ 之间, 则 $f(a)=f^{\prime}(0) a+\frac{f^{\prime \prime}\left(\eta_1\right)}{2 !} a^2, 0 < \eta_1 < a$.(1)
$$
f(-a)=f^{\prime}(0)(-a)+\frac{f^{\prime \prime}\left(\eta_2\right)}{2 !} a^2,-a < \eta_2 < 0 \text {, (2) }
$$
(1)+2)得: $f(a)+f(-a)=\frac{a^2}{2}\left[f^{\prime \prime}\left(\eta_1\right)+f^{\prime \prime}\left(\eta_2\right)\right]$.(3)
又 $f^{\prime \prime}(x)$ 在 $\left[\eta_2, \eta_1\right]$ 上连续, 则必有最大值 $M$ 与最小值 $m$, 即
$$
, m \leq f^{\prime \prime}\left(\eta_1\right) \leq M, m \leq f^{\prime \prime}\left(\eta_2\right) \leq M \text {, 从而 } m \leq \frac{f^{\prime \prime}\left(\eta_1\right)+f^{\prime \prime}\left(\eta_2\right)}{2} \leq M \text {. }
$$
由介值定理得: 存在 $\xi \in\left[\eta_2, \eta_1\right] \subset(-a, a)$, 有 $\frac{f^{\prime \prime}\left(\eta_1\right)+f^{\prime \prime}\left(\eta_2\right)}{2}=f^{\prime \prime}(\xi)$,
代入(3)得: $f(a)+f(-a)=a^2 f^{\prime \prime}(\xi)$, 即 $f^{\prime \prime}(\xi)=\frac{f(a)+f(-a)}{a^2}$.

(II) 设 $f(x)$ 在 $x=x_0 \in(-a, a)$ 取极值, 且 $f(x)$ 在 $x=x_0$ 可导, 则 $f^{\prime}\left(x_0\right)=0$.
又 $f(x)=f\left(x_0\right)+f^{\prime}\left(x_0\right)\left(x-x_0\right)+\frac{f^{\prime \prime}(\gamma)}{2 !}\left(x-x_0\right)^2=f\left(x_0\right)+\frac{f^{\prime \prime}(\gamma)}{2 !}\left(x-x_0\right)^2, \gamma$ 介于 0 与 $x$ 之间, 则 $f(-a)=f\left(x_0\right)+\frac{f^{\prime \prime}\left(\gamma_1\right)}{2 !}\left(-a-x_0\right)^2,-a < \gamma_1 < 0$.
$$
f(a)=f\left(x_0\right)+\frac{f^{\prime \prime}\left(\gamma_2\right)}{2 !}\left(a-x_0\right)^2, 0 < \gamma_2 < a .
$$
从而 $|f(a)-f(-a)|=\left|\frac{1}{2}\left(a-x_0\right)^2 f^{\prime \prime}\left(\gamma_2\right)-\frac{1}{2}\left(a+x_0\right)^2 f^{\prime \prime}\left(\gamma_1\right)\right|$

$$
\leq \frac{1}{2}\left|\left(a-x_0\right)^2 f^{\prime \prime}\left(\gamma_2\right)\right|+\frac{1}{2}\left|\left(a+x_0\right)^2 f^{\prime \prime}\left(\gamma_1\right)\right| .
$$
又 $\left|f^{\prime \prime}(x)\right|$ 连续, 设 $M=\max \left\{\left|f^{\prime \prime}\left(\gamma_1\right)\right|\right\}\left|f^{\prime \prime}\left(\gamma_2\right)\right|$, 则
$$
|f(a)-f(-a)| \leq \frac{1}{2} M\left(a+x_0\right)^2+\frac{1}{2} M\left(a-x_0\right)^2=M\left(a^2+x_0{ }^2\right)
$$
又 $x_0 \in(-a, a)$, 则 $|f(a)-f(-a)| \leq M\left(a^2+x_0^2\right) \leq 2 M a^2$,
则 $M \geq \frac{1}{2 a^2}|f(a)-f(-a)|$, 即存在 $\eta=\gamma_1$ 或 $\eta=\gamma_2 \in(-a, a)$,
有 $\left|f^{\prime \prime}(\eta)\right| \geq \frac{1}{2 a^2}|f(a)-f(-a)|$.


系统推荐