设 $f(x)$ 二阶连续可导, $f(0)=f(1)=0$, $\dps{\max_{0\leq x\leq 1}f(x)=2}$. 证明: $$\bex \min_{0\leq x\leq 1}f''(x)\leq -16. \eex$$
证明: 设 $$\bex \xi\in (0,1),\st f(\xi)=\max_{0\leq x\leq 1}f(x)=2\ra f'(\xi)=0. \eex$$ 在 $\xi$ 处由 Taylor 展式, $$\beex \bea 0=f(0)=f(\xi)+f'(\xi)(-\xi)+\cfrac{f''(\eta)}{2}(-\xi)^2,&0<\eta<\xi,\\ 0=f(1)=f(\xi)+f'(\xi)(1-\xi)+\cfrac{f''(\zeta)}{2}(1-\xi)^2,&\xi<\zeta<1. \eea \eeex$$ 于是 $$\bex f''(\eta)=-\cfrac{4}{\eta^2},\quad f''(\zeta)=-\cfrac{4}{(1-\xi)^2}. \eex$$ 若 $0<\xi\leq \cfrac{1}{2}$, 则 $$\bex \min_{0\leq x\leq 1}f''(x)\leq f''(\eta)\leq -16; \eex$$ 若 $\cfrac{1}{2}<\xi<1$, 则 $$\bex \min_{0\leq x\leq 1}f''(x)\leq f''(\zeta)=-16. \eex$$
