📝 题目
例 13 设 $f\left( x\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上有连续的一阶导数,且 $f\left( x\right) \geq 0$ , ${f}^{\prime }\left( x\right) \leq 0$ . 若 $F\left( x\right) \overset{\text{ 定义 }}{ = }{\int }_{0}^{x}f\left( t\right) \mathrm{d}t$ ,求证: 对任意的 $x \in \left( {0,1}\right)$ ,都有
$$ {xF}\left( 1\right) \leq F\left( x\right) \leq 2{\int }_{0}^{1}F\left( t\right) \mathrm{d}t. $$
第一个不等式的证明
💡 答案与解析
证法 1 注意到 ${F}^{\prime \prime }\left( x\right) = {f}^{\prime }\left( x\right) \leq 0$ ,可知 $F\left( x\right)$ 是单调增加的凸函数. 根据凸函数曲线在所对弦的上方 (参考图 3.2),对任意的 $x \in$ (0,1),因为弦上的高度 ${AB} \leq$ 曲线上的高度 ${AC}$ ,又
$$ {AC} = F\left( x\right) ,\;\frac{AB}{F\left( 1\right) } = \frac{x}{1} \Rightarrow {AB} = {xF}\left( 1\right) , $$
所以 ${xF}\left( 1\right) \leq F\left( x\right)$ .
\begin{center} \includegraphics[max width=0.2\textwidth]{images/023.jpg} \end{center} \hspace*{3em}
图 3.2
证法 2 令 $g\left( x\right) = F\left( x\right) - {xF}\left( 1\right)$ ,则 $g\left( 0\right) = g\left( 1\right) = 0$ . 由积分中值定理, $\exists \xi \in \left( {0,1}\right)$ ,使得 ${g}^{\prime }\left( x\right) = f\left( x\right) - F\left( 1\right) = f\left( x\right) - f\left( \xi \right)$ ,于是
$$ {g}^{\prime }\left( x\right) = f\left( x\right) - f\left( \xi \right) \begin{cases} \geq 0\left( {0 < x \leq \xi }\right) & \Rightarrow g\left( x\right) \uparrow \left( {0 < x \leq \xi }\right) \\ & \Rightarrow g\left( x\right) \geq g\left( 0\right) = 0, \\ \leq 0\left( {\xi \leq x < 1}\right) & \Rightarrow g\left( x\right) \downarrow \left( {\xi \leq x < 1}\right) \\ & \Rightarrow g\left( x\right) \geq g\left( 1\right) = 0, \end{cases} $$
从而 ${xF}\left( 1\right) \leq F\left( x\right)$ .
证法 3 用积分中值定理. 对 $\forall x \in \left( {0,1}\right)$ ,
$$ F\left( x\right) - {xF}\left( 1\right) = F\left( x\right) - x{\int }_{0}^{x}f\left( t\right) \mathrm{d}t - x{\int }_{x}^{1}f\left( t\right) \mathrm{d}t $$
$$ = \left( {1 - x}\right) {\int }_{0}^{x}f\left( t\right) \mathrm{d}t - x{\int }_{x}^{1}f\left( t\right) \mathrm{d}t $$
$$ \text{ (3) }{\xi }_{1} \in \left( {0,x}\right) ,\exists {\xi }_{2} \in \left( {x,1}\right) \text{ ) } $$
$$ = x\left( {1 - x}\right) \left( {f\left( {\xi }_{1}\right) - f\left( {\xi }_{2}\right) }\right) \geq 0 $$
$$ \text{ (因为 }{f}^{\prime }\left( x\right) \leq 0 \Rightarrow f\left( x\right) \downarrow \text{ ), } $$
即得 $F\left( x\right) \geq {xF}\left( 1\right)$ .
证法 4 作辅助函数 $G\left( x\right) = {\int }_{0}^{x}f\left( t\right) \mathrm{d}t - x{\int }_{0}^{1}f\left( t\right) \mathrm{d}t$ ,则 $G\left( 0\right) =$ $G\left( 1\right) = 0$ . 又 ${G}^{\prime \prime }\left( x\right) = {f}^{\prime }\left( x\right) \leq 0$ . 这意味着, $G\left( x\right)$ 是凸函数. 根据凸函数曲线在所对弦的上方,现在曲线 $G\left( x\right)$ 所对的弦就是 $x$ 轴上的线段 $\left\lbrack {0,1}\right\rbrack$ ,即得 $G\left( x\right) \geq 0\left( {\forall x \in \left\lbrack {0,1}\right\rbrack }\right)$ .
证法 5 令 $G\left( x\right) = \left\{ \begin{array}{ll} \frac{F\left( x\right) }{x} & \left( {x > 0}\right) , \\ {f}^{\prime }\left( 0\right) & \left( {x = 0}\right) , \end{array}\right.$ 则 $G\left( x\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上连续, 在(0,1)内可导. 且注意到 $f\left( x\right) \downarrow$ ,由积分中值定理,对 $\forall x \in$ $\left( {0,1}\right) ,\exists \xi \in \left( {0,x}\right)$ ,使得
$$ {G}^{\prime }\left( x\right) = \frac{{xf}\left( x\right) - {xf}\left( \xi \right) }{{x}^{2}} = \frac{x\left( {f\left( x\right) - f\left( \xi \right) }\right) }{{x}^{2}} \leq 0 \Rightarrow G\left( x\right) \downarrow , $$
即得
$$ G\left( 1\right) \leq G\left( x\right) \Rightarrow \frac{F\left( 1\right) }{1} \leq \frac{F\left( x\right) }{x} \Rightarrow {xF}\left( 1\right) \leq F\left( x\right) . $$
证法 6 作变量替换 $u = \frac{t}{x}$ ,则 $F\left( x\right) = x{\int }_{0}^{1}f\left( {ux}\right) \mathrm{d}u$ . 因此,只要
证
$$ x{\int }_{0}^{1}f\left( {ux}\right) \mathrm{d}u \geq x{\int }_{0}^{1}f\left( u\right) \mathrm{d}u\;\left( {\forall x \in \left( {0,1}\right) }\right) . $$
事实上, 因为
$$ {f}^{\prime }\left( x\right) \leq 0 \Rightarrow f\left( x\right) \downarrow \Rightarrow f\left( {ux}\right) \geq f\left( u\right) \;\left( {\forall x \in \left( {0,1}\right) }\right) , $$
所以
$$ x{\int }_{0}^{1}f\left( {ux}\right) \mathrm{d}u - x{\int }_{0}^{1}f\left( u\right) \mathrm{d}u = x{\int }_{0}^{1}\left\lbrack {f\left( {ux}\right) - f\left( u\right) }\right\rbrack \geq 0. $$
即得结论.
第二个不等式的证明
证法 1 曲边梯形面积 $\displaystyle{\int }_{0}^{1}F\left( t\right) \mathrm{d}t \geq$ 弦下方的梯形面积,即
$$ {\int }_{0}^{1}F\left( t\right) \mathrm{d}t \geq \frac{1}{2}\left( {F\left( 0\right) + F\left( 1\right) }\right) \overset{\text{ 因为 }F\left( 0\right) = 0}{ = }\frac{1}{2}F\left( 1\right) \geq \frac{1}{2}F\left( x\right) . $$
证法 2 利用第一个不等式,两边从 0 到 1 积分,得
$$ 2{\int }_{0}^{1}F\left( x\right) \mathrm{d}x \geq {2F}\left( 1\right) {\int }_{0}^{1}x\mathrm{\;d}x = {2F}\left( 1\right) \left( {\left. \frac{1}{2}{x}^{2}\right| }_{0}^{1}\right) = F\left( 1\right) . $$
又
$$ {F}^{\prime }\left( x\right) = f\left( x\right) > 0 \Rightarrow F\left( x\right) \uparrow \Rightarrow F\left( x\right) \leq F\left( 1\right) \leq 2{\int }_{0}^{1}F\left( t\right) \mathrm{d}t. $$
证法 3 用分部积分法.
$$ {\int }_{0}^{1}F\left( t\right) \mathrm{d}t = {\left. tF\left( t\right) \right| }_{0}^{1} - {\int }_{0}^{1}{tf}\left( t\right) \mathrm{d}t = F\left( 1\right) - {\int }_{0}^{1}{tf}\left( t\right) \mathrm{d}t, $$
$$ {\int }_{0}^{1}F\left( t\right) \mathrm{d}t = {\int }_{0}^{1}F\left( t\right) \mathrm{d}\left( {t - 1}\right) $$
$$ = {\left. \left( t - 1\right) F\left( t\right) \right| }_{0}^{1} - {\int }_{0}^{1}\left( {t - 1}\right) f\left( t\right) \mathrm{d}t $$
$$ = - {\int }_{0}^{1}\left( {t - 1}\right) f\left( t\right) \mathrm{d}t. $$
将以上两式相加, 得到
$$ 2{\int }_{0}^{1}F\left( t\right) \mathrm{d}t = F\left( 1\right) - {\int }_{0}^{1}\left( {{2t} - 1}\right) f\left( t\right) \mathrm{d}t $$
$$ = F\left( 1\right) - {\int }_{0}^{1}f\left( t\right) \mathrm{d}\left\lbrack {t\left( {t - 1}\right) }\right\rbrack $$
$$ = F\left( 1\right) - {\int }_{0}^{1}t\left( {t - 1}\right) {f}^{\prime }\left( t\right) \mathrm{d}t $$
$$ = F\left( 1\right) + {\int }_{0}^{1}t\left( {1 - t}\right) {f}^{\prime }\left( t\right) \mathrm{d}t $$
$$ \geq F\left( 1\right) \geq F\left( x\right) \text{ (因为 }F\left( x\right) \uparrow \text{ ). } $$
证法 4 将第二个不等式改写为 $F\left( x\right) - {\int }_{0}^{1}F\left( t\right) \mathrm{d}t \leq {\int }_{0}^{1}F\left( t\right) \mathrm{d}t$ . 用分部积分法.
$$ \text{ 不等式左端 } = F\left( x\right) - {\int }_{0}^{1}F\left( t\right) \mathrm{d}t $$
$$ = {\left. F\left( x\right) - tF\left( t\right) \right| }_{0}^{1} + {\int }_{0}^{1}{tf}\left( t\right) \mathrm{d}t $$
$$ = F\left( x\right) - F\left( 1\right) + {\int }_{0}^{1}{tf}\left( t\right) \mathrm{d}t $$
$$ \leq {\int }_{0}^{1}{tf}\left( t\right) \mathrm{d}t $$
(因为 ${F}^{\prime }\left( x\right) = f\left( x\right) \geq 0 \Rightarrow F\left( x\right) \uparrow \Rightarrow F\left( x\right) \leq F\left( 1\right)$ ).
现在为了证明不等式左端 $\leq$ 右端,只要证明 ${tf}\left( t\right) \leq F\left( t\right)$ . 事实上,令 $g\left( t\right) = {tf}\left( t\right) - F\left( t\right)$ ,则
$$ {g}^{\prime }\left( t\right) = f\left( t\right) + t{f}^{\prime }\left( t\right) - f\left( t\right) = t{f}^{\prime }\left( t\right) \leq 0 \Rightarrow g\left( t\right) \downarrow , $$
从而 $g\left( t\right) \leq g\left( 0\right) = 0 \Rightarrow {tf}\left( t\right) \leq F\left( t\right)$ ,即得结论.