题目： \(\begin{aligned} &\text { (1) 设 } f(x) \in C^{(2)}[0,1], f(0)=f(1)=0, \text { 且当 } x \in(0,1) \text { 时 }\left|f^{\prime \prime}(x)\right| \leq A \\ &\text { 证明: 当 } 0 \leq x \leq 1 \text { 时，有 }\left|f^{\prime}(x)\right| \leq \frac{A}{2} \\ &\text { (2)设 } f(x)(-\infty<x<+\infty) \text { 为二阶可微函数, 且 } M_{k}=\sup _{-\infty<x<+\infty}\left|f^{(k)}(x)\right|<+\infty(k=0,1,2) . \\ &\text { 证明不等式: } M^{2} \leq 2 M_{0} M_{2} \end{aligned}\)
解答：
\[\begin{aligned} &(1) f(0)=f(x)+f^{\prime}(x)(0-x)+\frac{1}{2} f^{\prime \prime}(\xi)(0-x)^{2}, f(1)=f(x)+f^{\prime}(x)(1-x)+\frac{1}{2} f^{\prime \prime}(\eta)(1-x)^{2} \\ &\text { 两式相椷得 } f^{\prime}(x)=\frac{1}{2} f^{\prime \prime}(\xi) x^{2}-\frac{1}{2} f^{\prime \prime}(\eta)(1-x)^{2} \Rightarrow\left|f^{\prime}(x)\right|=\frac{1}{2}\left|f^{\prime \prime}(\xi) x^{2}-f^{\prime \prime}(\eta)(1-x)^{2}\right| \leq \frac{1}{2} A \\ &(2) f(x+h)=f(x)+f^{\prime}(x) h+\frac{1}{2} f^{\prime \prime}(\xi) h^{2}, f(x-h)=f(x)-f^{\prime}(x) h+\frac{1}{2} f^{\prime \prime}(\eta) h^{2} \\ &\text { 两式相减得 } f(x+h)-f(x-h)=2 f^{\prime}(x) h+\frac{1}{2} h^{2}\left[f^{\prime \prime}(\xi)-f^{\prime \prime}(\eta)\right] \\ &\Rightarrow f^{\prime}(x)=\frac{f(x+h)-f(x-h)}{2 h}+\frac{1}{4} h\left[f^{\prime \prime}(\xi)-f^{\prime \prime}(\eta)\right] \leq \frac{M_{0}}{h}+\frac{1}{2} M_{2} h \\ &\Rightarrow f^{\prime}(x) \leq \frac{M_{0}}{h}+\frac{1}{2} M_{2} h \text { 对 } \forall x, h \text { 恒成立. } \Rightarrow M_{1} \leq \sqrt{2 M_{0} M_{2}} \Rightarrow M_{1}^{2} \leq 2 M_{0} M_{2} \end{aligned}\]