📝 题目
例 17 设 $a > 0,{f}^{\prime }\left( x\right)$ 在 $\left\lbrack {0,a}\right\rbrack$ 上连续,求证:
$$ \left| {f\left( 0\right) }\right| \leq \frac{1}{a}{\int }_{0}^{a}\left| {f\left( x\right) }\right| \mathrm{d}x + {\int }_{0}^{a}\left| {{f}^{\prime }\left( x\right) }\right| \mathrm{d}x. $$
💡 答案与解析
证 根据积分中值定理,存在 $\xi \in \left\lbrack {0,a}\right\rbrack$ ,使得
$$ \left| {f\left( \xi \right) }\right| = \frac{1}{a}{\int }_{0}^{a}\left| {f\left( x\right) }\right| \mathrm{d}x. \tag{3.27} $$
又由牛顿-莱布尼茨公式, 有
$$ f\left( \xi \right) - f\left( 0\right) = {\int }_{0}^{\xi }{f}^{\prime }\left( x\right) \mathrm{d}x $$
$$ \Rightarrow \left| {f\left( 0\right) }\right| \leq \left| {f\left( \xi \right) }\right| + \left| {{\int }_{0}^{\xi }{f}^{\prime }\left( x\right) \mathrm{d}x}\right| . \tag{3.28} $$
联立 (3.27) 与 (3.28) 式, 解得
$$ \left| {f\left( 0\right) }\right| \leq \left| {f\left( \xi \right) }\right| + \left| {{\int }_{0}^{\xi }{f}^{\prime }\left( x\right) \mathrm{d}x}\right| $$
$$ \leq \frac{1}{a}{\int }_{0}^{a}\left| {f\left( x\right) }\right| \mathrm{d}x + {\int }_{0}^{a}\left| {{f}^{\prime }\left( x\right) }\right| \mathrm{d}x. $$