📝 题目
例 25 设 $f\left( x\right)$ 在 $\mathbf{R}$ 上有二阶连续导数,且有
$$ {M}_{0} = {\int }_{-\infty }^{+\infty }\left| {f\left( x\right) }\right| \mathrm{d}x < + \infty ,\;{M}_{2} = {\int }_{-\infty }^{+\infty }\left| {{f}^{\prime \prime }\left( x\right) }\right| \mathrm{d}x < + \infty . $$
求证: (1) ${M}_{1} = {\int }_{-\infty }^{\infty }\left| {{f}^{\prime }\left( x\right) }\right| \mathrm{d}x < + \infty$ ; (2) ${M}_{1}^{2} \leq 4{M}_{0}{M}_{2}$ .
💡 答案与解析
证(1) $\forall h > 0$ ,由积分的恒等变换式(由分部积分可证)
$$ f\left( {x + h}\right) = f\left( x\right) + {f}^{\prime }\left( x\right) h + {\int }_{x}^{x + h}\left( {x + h - t}\right) {f}^{\prime \prime }\left( t\right) \mathrm{d}t, $$
可得导函数及其积分的估计式:
$$ \left| {{f}^{\prime }\left( x\right) }\right| \leq \frac{1}{h}\left\lbrack {\left| {f\left( {x + h}\right) }\right| + \left| {f\left( x\right) }\right| + {\int }_{x}^{x + h}\left( {x + h - t}\right) \left| {{f}^{\prime \prime }\left( t\right) }\right| \mathrm{d}t}\right\rbrack , $$
$$ {\int }_{a}^{b}\left| {{f}^{\prime }\left( x\right) }\right| \mathrm{d}x \leq \frac{1}{h}\left\lbrack {{\int }_{a}^{b}\left| {f\left( {x + h}\right) }\right| \mathrm{d}x + {\int }_{a}^{b}\left| {f\left( x\right) }\right| \mathrm{d}x}\right. $$
$$ \left. {+{\int }_{a}^{b}\mathrm{\;d}x{\int }_{x}^{x + h}\left( {x + h - t}\right) \left| {{f}^{\prime \prime }\left( t\right) }\right| \mathrm{d}t}\right\rbrack $$
$$ \leq \frac{1}{h}\left\lbrack {2{\int }_{-\infty }^{+\infty }\left| {f\left( x\right) }\right| \mathrm{d}x + {\iint }_{D}\left( {x + h - t}\right) \left| {{f}^{\prime \prime }\left( t\right) }\right| \mathrm{d}t\mathrm{\;d}x}\right\rbrack , $$
(7.43)
其中区域 $D$ 由直线 $t = x,t = x + h,x = a,x = b$ 围成. 根据重积分化累次积分公式得
$$ {\iint }_{D}\left( {x + h - t}\right) \left| {{f}^{\prime \prime }\left( t\right) }\right| \mathrm{d}t\mathrm{\;d}x $$
$$ = {\int }_{a}^{a + h}\mathrm{\;d}t{\int }_{a}^{t}\left( {x + h - t}\right) \left| {{f}^{\prime \prime }\left( t\right) }\right| \mathrm{d}x $$
$$ + {\int }_{a + h}^{b}\mathrm{\;d}t{\int }_{t - h}^{t}\left( {x + h - t}\right) \left| {{f}^{\prime \prime }\left( t\right) }\right| \mathrm{d}x $$
$$ + {\int }_{b}^{b + h}\mathrm{\;d}t{\int }_{t - h}^{b}\left( {x + h - t}\right) \left| {{f}^{\prime \prime }\left( t\right) }\right| \mathrm{d}x $$
$$ \leq \frac{{h}^{2}}{2}{\int }_{a}^{a + h}\left| {{f}^{\prime \prime }\left( t\right) }\right| \mathrm{d}t + \frac{{h}^{2}}{2}{\int }_{a + h}^{b}\left| {{f}^{\prime \prime }\left( t\right) }\right| \mathrm{d}t + \frac{{h}^{2}}{2}{\int }_{b}^{b + h}\left| {{f}^{\prime \prime }\left( t\right) }\right| \mathrm{d}t $$
$$ = \frac{{h}^{2}}{2}{\int }_{a}^{b + h}\left| {{f}^{\prime \prime }\left( t\right) }\right| \mathrm{d}t. \tag{7.44} $$
联立 (7.43) 和 (7.44) 式, 有
$$ {\int }_{a}^{b}\left| {{f}^{\prime }\left( x\right) }\right| \mathrm{d}x \leq \frac{1}{h}\left\lbrack {2{\int }_{-\infty }^{\infty }\left| {f\left( x\right) }\right| \mathrm{d}x + \frac{{h}^{2}}{2}{\int }_{a}^{b + h}\left| {{f}^{\prime \prime }\left( t\right) }\right| \mathrm{d}t}\right\rbrack $$
$$ \leq \frac{2}{h}{M}_{0} + \frac{h}{2}{M}_{2}. $$
令 $\displaystyle{a \rightarrow - \infty ,b \rightarrow + \infty}$ ,即得 ${M}_{1} = {\int }_{-\infty }^{+\infty }\left| {{f}^{\prime }\left( x\right) }\right| \mathrm{d}x \leq \frac{2}{h}{M}_{0} + \frac{h}{2}{M}_{2}$ .
(2)取 $h = 2\sqrt{{M}_{0}/{M}_{2}}$ ,代入(1)即得
$$ {M}_{1} \leq 2\sqrt{{M}_{0}{M}_{2}}\text{ 或 }{M}_{1}^{2} \leq 4{M}_{0}{M}_{2}. $$