📝 题目
例 9 设 ${f}^{\prime \prime }\left( x\right) > 0\left( {\forall x \in \left\lbrack {0,1}\right\rbrack }\right)$ ,求证 $\displaystyle{\int }_{0}^{1}f\left( {x}^{\lambda }\right) \mathrm{d}x \geq f\left( \frac{1}{\lambda + 1}\right)$ , 其中 $\lambda$ 为任意正实数.
💡 答案与解析
证法 1 由题设条件, $f\left( u\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上是凹函数,从而曲线 $y = f\left( u\right)$ 在 $u = \frac{1}{\lambda + 1}$ 处的切线上方,即有
$$ f\left( u\right) \geq f\left( \frac{1}{\lambda + 1}\right) + {f}^{\prime }\left( \frac{1}{\lambda + 1}\right) \left( {u - \frac{1}{\lambda + 1}}\right) \;\left( {\forall u \in \left( {0,1}\right) }\right) . $$
(3.9)
注意到 $x \in \left\lbrack {0,1}\right\rbrack \Rightarrow {x}^{\lambda } \in \left\lbrack {0,1}\right\rbrack$ ,令 $u = {x}^{\lambda }$ 并代入 (3.9) 式,我们有
$$ f\left( {x}^{\lambda }\right) \geq f\left( \frac{1}{\lambda + 1}\right) + {f}^{\prime }\left( \frac{1}{\lambda + 1}\right) \left( {{x}^{\lambda } - \frac{1}{\lambda + 1}}\right) \;\left( {\forall x \in \left\lbrack {0,1}\right\rbrack }\right) , $$
从而
$$ {\int }_{0}^{1}f\left( {x}^{\lambda }\right) \mathrm{d}x \geq f\left( \frac{1}{\lambda + 1}\right) + {f}^{\prime }\left( \frac{1}{\lambda + 1}\right) {\int }_{0}^{1}\left( {{x}^{\lambda } - \frac{1}{\lambda + 1}}\right) \mathrm{d}x $$
$$ = f\left( \frac{1}{\lambda + 1}\right) \text{ . } $$
证法 ${2f}\left( u\right)$ 在 $u = \frac{1}{\lambda + 1}$ 处的二阶泰勒展开式为
$$ f\left( u\right) = f\left( \frac{1}{\lambda + 1}\right) + {f}^{\prime }\left( \frac{1}{\lambda + 1}\right) \left( {u - \frac{1}{\lambda + 1}}\right) $$
$$ + \frac{{f}^{\prime \prime }\left( \xi \right) }{2!}{\left( u - \frac{1}{\lambda + 1}\right) }^{2}, \tag{3.10} $$
其中 $\xi$ 在 $u$ 与 $\frac{1}{\lambda + 1}$ 之间. 注意到 $x \in \left\lbrack {0,1}\right\rbrack \Rightarrow {x}^{\lambda } \in \left\lbrack {0,1}\right\rbrack$ ,令 $u = {x}^{\lambda }$ 并代入 (3.10) 式, 我们有
$$ f\left( {x}^{\lambda }\right) = f\left( \frac{1}{\lambda + 1}\right) + {f}^{\prime }\left( \frac{1}{\lambda + 1}\right) \left( {{x}^{\lambda } - \frac{1}{\lambda + 1}}\right) $$
$$ + \frac{{f}^{\prime \prime }\left( \xi \right) }{2!}{\left( {x}^{\lambda } - \frac{1}{\lambda + 1}\right) }^{2}, $$
从而
$$ {\int }_{0}^{1}f\left( {x}^{\lambda }\right) \mathrm{d}x = f\left( \frac{1}{\lambda + 1}\right) + {f}^{\prime }\left( \frac{1}{\lambda + 1}\right) {\int }_{0}^{1}\left( {{x}^{\lambda } - \frac{1}{\lambda + 1}}\right) \mathrm{d}x $$
$$ + {\int }_{0}^{1}\frac{{f}^{\prime \prime }\left( \xi \right) }{2!}{\left( {x}^{\lambda } - \frac{1}{\lambda + 1}\right) }^{2}\mathrm{\;d}x $$
$$ \geq f\left( \frac{1}{\lambda + 1}\right) + {f}^{\prime }\left( \frac{1}{\lambda + 1}\right) {\int }_{0}^{1}\left( {{x}^{\lambda } - \frac{1}{\lambda + 1}}\right) \mathrm{d}x $$
$$ = f\left( \frac{1}{\lambda + 1}\right) \text{ . } $$