📝 题目
例 14 设 $f\left( x\right) \in C\left\lbrack {0,1}\right\rbrack$ ,且 $\displaystyle{\int }_{0}^{1}f\left( x\right) \mathrm{d}x = 0,{\int }_{0}^{1}{xf}\left( x\right) \mathrm{d}x = 1$ . 求证:
$$ \mathop{\max }\limits_{{0 \leq x \leq 1}}\left| {f\left( x\right) }\right| > 4 $$
💡 答案与解析
证 用反证法. 如果 $\left| {f\left( x\right) }\right| \leq 4$ ,那么
$$ 1 = \left| {{\int }_{0}^{1}\left( {x - \frac{1}{2}}\right) f\left( x\right) \mathrm{d}x}\right| \leq {\int }_{0}^{1}\left| {x - \frac{1}{2}}\right| \left| {f\left( x\right) }\right| \mathrm{d}x $$
$$ \leq 4{\int }_{0}^{1}\left| {x - \frac{1}{2}}\right| \mathrm{d}x = 1, $$
由此推出 $\displaystyle{\int }_{0}^{1}\left| {x - \frac{1}{2}}\right| \left| {f\left( x\right) }\right| \mathrm{d}x = 1$ ,从而
$$ {\int }_{0}^{1}\left( {4 - \left| {f\left( x\right) }\right| }\right) \left| {x - \frac{1}{2}}\right| \mathrm{d}x = 0, $$
即得 $\left| {f\left( x\right) }\right| \equiv 4$ . 又 $f\left( x\right)$ 连续,可见 $f\left( x\right) \equiv 4$ 或 $f\left( x\right) \equiv - 4$ . 这些都与 $\displaystyle{\int }_{0}^{1}f\left( x\right) \mathrm{d}x = 0$ 矛盾.