📝 题目
例 19 设 $f\left( x\right) \in {C}^{1}\lbrack 0,\infty ),x{f}^{\prime }\left( x\right)$ 在 $\lbrack 0,\infty )$ 上有界,并且
$$ \frac{1}{x}{\int }_{x}^{2x}\left| {f\left( t\right) }\right| \mathrm{d}t \rightarrow 0\;\left( {x \rightarrow + \infty }\right) . $$
求证: $f\left( x\right) \rightarrow 0\left( {x \rightarrow + \infty }\right)$ .
💡 答案与解析
证 令 $M\overset{\text{ 定义 }}{ \approx }\mathop{\sup }\limits_{{x \geq 0}}\left| {x{f}^{\prime }\left( x\right) }\right|$ ,则有 $\left| {{f}^{\prime }\left( x\right) }\right| \leq \frac{M}{x}\left( {\forall x > 0}\right)$ . 对 $\forall {x}_{0} > 0$ 及 $x \geq {x}_{0}$ ,由微积分基本定理,
$$ \left| {f\left( x\right) - f\left( {x}_{0}\right) }\right| = \left| {{\int }_{{x}_{0}}^{x}{f}^{\prime }\left( t\right) \mathrm{d}t}\right| \leq {\int }_{{x}_{0}}^{x}\left| {{f}^{\prime }\left( t\right) }\right| \mathrm{d}t $$
$$ \leq {\int }_{{x}_{0}}^{x}\frac{M}{t}\mathrm{\;d}t \leq \frac{M}{{x}_{0}}\left( {x - {x}_{0}}\right) , $$
即得
$$ \left| {f\left( x\right) }\right| \geq \left| {f\left( {x}_{0}\right) }\right| - \frac{M}{{x}_{0}}\left( {x - {x}_{0}}\right) \;\left( {\forall x \geq {x}_{0} > 0}\right) . $$
(7.34)
我们先证 $\exists X > 0$ ,使得
$$ \left| {f\left( x\right) }\right| < M\;\left( {\forall x > X}\right) . \tag{7. 35} $$
用反证法. 假定使 (7.35) 式成立的 $X$ 不存在,那么 $\forall n \in \mathbf{N},\exists {x}_{n} >$ $n$ ,使得 $\left| {f\left( {x}_{n}\right) }\right| \geq M$ . 由 (7.34) 式有
$$ \left| {f\left( x\right) }\right| \geq \left| {f\left( {x}_{n}\right) }\right| - \frac{M}{{x}_{n}}\left( {x - {x}_{n}}\right) $$
$$ = \left| {f\left( {x}_{n}\right) }\right| \left\{ {1 - \frac{1}{{T}_{n}}\left( {x - {x}_{n}}\right) }\right\} \;\left( {x \geq {x}_{n}}\right) , $$
其中 ${T}_{n}\frac{\text{ 定义 }{x}_{n}\left| {f\left( {x}_{n}\right) }\right| }{M} \geq {x}_{n}$ ,因此
$$ \frac{1}{{x}_{n}}{\int }_{{x}_{n}}^{2{x}_{n}}\left| {f\left( x\right) }\right| \mathrm{d}x \geq \frac{1}{{x}_{n}}{\int }_{{x}_{n}}^{2{x}_{n}}\left| {f\left( {x}_{n}\right) }\right| \left\{ {1 - \frac{1}{{x}_{n}}\left( {x - {x}_{n}}\right) }\right\} \mathrm{d}x $$
$$ = \frac{\left| f\left( {x}_{n}\right) \right| }{{x}_{n}} \cdot \frac{1}{2}{x}_{n} = \frac{\left| f\left( {x}_{n}\right) \right| }{2} \geq \frac{M}{2} > 0. $$
这与 ${x}_{n} \rightarrow + \infty \left( {n \rightarrow \infty }\right)$ 时
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{{x}_{n}}{\int }_{{x}_{n}}^{2{x}_{n}}\left| {f\left( x\right) }\right| \mathrm{d}x = 0 $$
矛盾. 于是使 (7.35) 式成立的 $X$ 的确存在.
现在 $\forall {x}_{0} > X$ ,由 (7.35) 式有 $\left| {f\left( {x}_{0}\right) }\right| < M$ ,从而 $T$ 定义 $\frac{{x}_{0}\left| {f\left( {x}_{0}\right) }\right| }{M} < {x}_{0}$ ,因此,根据 (7.34) 式有
$$ {\int }_{{x}_{0}}^{2{x}_{0}}\left| {f\left( x\right) }\right| \mathrm{d}x \geq {\int }_{{x}_{0}}^{{x}_{0} + T}\left| {f\left( x\right) }\right| \mathrm{d}x \geq T\left| {f\left( {x}_{0}\right) }\right| - \frac{M}{{x}_{0}} \cdot \frac{1}{2}{T}^{2} $$
$$ = T\left( {\left| {f\left( {x}_{0}\right) }\right| - \frac{MT}{2{x}_{0}}}\right) = \frac{T}{2}\left| {f\left( {x}_{0}\right) }\right| = \frac{{x}_{0}{\left| f\left( {x}_{0}\right) \right| }^{2}}{2M}, $$
即得
$$ {\left| f\left( {x}_{0}\right) \right| }^{2} \leq \frac{2M}{{x}_{0}}{\int }_{{x}_{0}}^{2{x}_{0}}\left| {f\left( x\right) }\right| \mathrm{d}x\;\left( {\forall {x}_{0} > X}\right) . $$
改写成
$$ 0 \leq \left| {f\left( x\right) }\right| \leq {\left( \frac{2M}{x}{\int }_{x}^{2x}\left| f\left( t\right) \right| \mathrm{d}t\right) }^{\frac{1}{2}}\;\left( {\forall x > X}\right) . $$
由极限两边夹挤准则即得结论.