📝 题目
例 17 设 $f\left( x\right)$ 在 $\left\lbrack {a,b}\right\rbrack$ 上可微,且 $\left| {{f}^{\prime }\left( x\right) }\right| \leq M$ . 求证:
$$ \left| {\frac{1}{b - a}{\int }_{a}^{b}f\left( x\right) \mathrm{d}x - \frac{f\left( a\right) + f\left( b\right) }{2}}\right| \leq \frac{M\left( {b - a}\right) }{4}\left( {1 - {\theta }^{2}}\right) , $$
其中 $\theta \overset{\text{ 定义 }}{ = }\frac{f\left( b\right) - f\left( a\right) }{M\left( {b - a}\right) }$ .
💡 答案与解析
证法 1 根据微分中值定理,
$$ f\left( x\right) \leq \left\{ \begin{array}{ll} f\left( a\right) + M\left( {x - a}\right) & \left( {a \leq x \leq \alpha }\right) , \\ f\left( b\right) + M\left( {b - x}\right) & \left( {\alpha \leq x \leq b}\right) , \end{array}\right. $$
其中 $\alpha = \frac{1}{2}\left( {a + b + \theta \left( {b - a}\right) }\right)$ . 因此,
$$ \frac{1}{b - a}{\int }_{a}^{b}f\left( x\right) \mathrm{d}x \leq \frac{1}{b - a}{\int }_{a}^{a}\left\lbrack {f\left( a\right) + M\left( {x - a}\right) }\right\rbrack \mathrm{d}x $$
$$ + \frac{1}{b - a}{\int }_{a}^{b}\left\lbrack {f\left( b\right) + M\left( {b - x}\right) }\right\rbrack \mathrm{d}x $$
$$ = \frac{a - a}{b - a}f\left( a\right) + \frac{b - a}{b - a}f\left( b\right) $$
$$ + \frac{M}{2\left( {b - a}\right) }\left\lbrack {{\left( \alpha - a\right) }^{2} + {\left( b - \alpha \right) }^{2}}\right\rbrack $$
$$ = \frac{1}{2}\left( {1 + \theta }\right) f\left( a\right) + \frac{1}{2}\left( {1 - \theta }\right) f\left( b\right) $$
$$ + \frac{M\left( {b - a}\right) }{8}\left\lbrack {{\left( 1 + \theta \right) }^{2} + {\left( 1 - \theta \right) }^{2}}\right\rbrack $$
$$ = \frac{1}{2}\left( {f\left( a\right) + f\left( b\right) }\right) + \frac{M\left( {b - a}\right) }{4}\left( {1 - {\theta }^{2}}\right) . $$
(7.29)
同理
$$ f\left( x\right) \geq \left\{ \begin{array}{ll} f\left( a\right) - M\left( {x - a}\right) & \left( {a \leq x \leq \beta }\right) , \\ f\left( b\right) - M\left( {b - x}\right) & \left( {\beta \leq x \leq b}\right) , \end{array}\right. $$
其中 $\beta = \frac{1}{2}\left( {a + b - \theta \left( {b - a}\right) }\right)$ . 因此
$$ \frac{1}{b - a}{\int }_{a}^{b}f\left( x\right) \mathrm{d}x \geq \frac{1}{b - a}{\int }_{a}^{\beta }\left\lbrack {f\left( a\right) - M\left( {x - a}\right) }\right\rbrack \mathrm{d}x $$
$$ + \frac{1}{b - a}{\int }_{\beta }^{b}\left\lbrack {f\left( b\right) - M\left( {b - x}\right) }\right\rbrack \mathrm{d}x $$
$$ = \frac{\beta - a}{b - a}f\left( a\right) + \frac{b - \beta }{b - a}f\left( b\right) $$
$$ - \frac{M}{2\left( {b - a}\right) }\left\lbrack {{\left( \beta - a\right) }^{2} + {\left( b - \beta \right) }^{2}}\right\rbrack $$
$$ = \frac{1}{2}\left( {1 - \theta }\right) f\left( a\right) + \frac{1}{2}\left( {1 + \theta }\right) f\left( b\right) $$
$$ - \frac{M\left( {b - a}\right) }{8}\left\lbrack {{\left( 1 - \theta \right) }^{2} + {\left( 1 + \theta \right) }^{2}}\right\rbrack $$
$$ = \frac{1}{2}\left\lbrack {f\left( a\right) + f\left( b\right) }\right\rbrack - \frac{M\left( {b - a}\right) }{4}\left( {1 - {\theta }^{2}}\right) . \tag{7.30} $$
联立 (7.29) 和 (7.30) 式即得结论.
证法 2 对 $\forall \left| \lambda \right| \leq 1$ ,令 $\gamma \frac{\text{ 定义 }a + b}{2} + \frac{\lambda }{2}\left( {b - a}\right)$ . 用分部积分法推得
$$ {\int }_{a}^{b}f\left( x\right) \mathrm{d}x = {\int }_{a}^{b}f\left( x\right) \mathrm{d}\left( {x - \gamma }\right) $$
$$ = \left( {b - \gamma }\right) f\left( b\right) + \left( {\gamma - a}\right) f\left( a\right) $$
$$ + {\int }_{a}^{\gamma }{f}^{\prime }\left( x\right) \left( {\gamma - x}\right) \mathrm{d}x - {\int }_{\gamma }^{b}{f}^{\prime }\left( x\right) \left( {x - \gamma }\right) \mathrm{d}x. $$
两边除以(b - a)并根据积分中值定理,容易推出
$$ \frac{1}{b - a}{\int }_{a}^{b}f\left( x\right) \mathrm{d}x - \frac{1}{2}\left\lbrack {f\left( a\right) + f\left( b\right) }\right\rbrack $$
$$ = \frac{1}{2}\lambda \left\lbrack {f\left( a\right) - f\left( b\right) }\right\rbrack + \frac{b - a}{8}\left\lbrack {{\left( 1 + \lambda \right) }^{2}{f}^{\prime }\left( \xi \right) }\right. $$
$$ \left. {-{\left( 1 - \lambda \right) }^{2}{f}^{\prime }\left( \eta \right) }\right\rbrack \text{ . } \tag{7.31} $$
在 (7.31) 式中取 $\lambda = \theta$ ,有
$$ \frac{1}{b - a}{\int }_{a}^{b}f\left( x\right) \mathrm{d}x - \frac{1}{2}\left\lbrack {f\left( a\right) + f\left( b\right) }\right\rbrack $$
$$ \leq - \frac{1}{2}M\left( {b - a}\right) {\theta }^{2} + \frac{M}{8}\left( {b - a}\right) \left\lbrack {{\left( 1 + \theta \right) }^{2} + {\left( 1 - \theta \right) }^{2}}\right\rbrack $$
$$ = \frac{M}{4}\left( {b - a}\right) \left( {1 - {\theta }^{2}}\right) . \tag{7.32} $$
在 (7.31) 式中取 $\lambda = - \theta$ ,有
$$ \frac{1}{b - a}{\int }_{a}^{b}f\left( x\right) \mathrm{d}x - \frac{1}{2}\left\lbrack {f\left( a\right) + f\left( b\right) }\right\rbrack $$
$$ \geq \frac{1}{2}M\left( {b - a}\right) {\theta }^{2} - \frac{M}{8}\left( {b - a}\right) \left\lbrack {{\left( 1 + \theta \right) }^{2} + {\left( 1 - \theta \right) }^{2}}\right\rbrack $$
$$ = - \frac{M}{4}\left( {b - a}\right) \left( {1 - {\theta }^{2}}\right) . \tag{7.33} $$
联立 (7.32) 和 (7.33) 式即得结论.