📝 题目
例 27 设 $f\left( x\right)$ 在 $\lbrack 0, + \infty )$ 上连续, $\displaystyle{\int }_{0}^{+\infty }\left| {f\left( x\right) }\right| \mathrm{d}x$ 收敛,并且
$$ \left| {f\left( x\right) }\right| \leq {\int }_{0}^{x}\left| {f\left( t\right) }\right| \mathrm{d}t\;\left( {x \geq 0}\right) , \tag{3.37} $$
求证 $f\left( x\right) \equiv 0$ .
分析 注意到 (3.37) 式右端的导数恰是 (3.37) 式的左端, 因此想到用 (3.37) 式右端去除 (3.37) 式的两端, 使得左端凑成对数导数, 但是又遇到 (3.37) 式右端可能等于零而不能做除数的麻烦. 于是想到用添加 $\varepsilon$ 的技巧.
💡 答案与解析
证 对 $\forall \varepsilon > 0$ ,
$$ \left| {f\left( x\right) }\right| \leq {\int }_{0}^{x}\left| {f\left( t\right) }\right| \mathrm{d}t \Rightarrow \left| {f\left( x\right) }\right| < \varepsilon + {\int }_{0}^{x}\left| {f\left( t\right) }\right| \mathrm{d}t\;\left( {x \geq 0}\right) $$
$$ \Rightarrow \frac{\left| f\left( x\right) \right| }{\varepsilon + {\int }_{0}^{x}\left| {f\left( t\right) }\right| \mathrm{d}t} < 1\;\left( {x \geq 0}\right) $$
$$ \Rightarrow \frac{\mathrm{d}}{\mathrm{d}x}\ln \left\{ {\varepsilon + {\int }_{0}^{x}\left| {f\left( t\right) }\right| \mathrm{d}t}\right\} < 1\;\left( {x \geq 0}\right) . $$
再对上式两边从 0 到 $x$ 积分,得
$$ {\left. \ln \left\{ \varepsilon + {\int }_{0}^{x}\left| f\left( t\right) \right| \mathrm{d}t\right\} \right| }_{0}^{x} \leq x \Rightarrow \frac{\varepsilon + {\int }_{0}^{x}\left| {f\left( t\right) }\right| \mathrm{d}t}{\varepsilon } \leq {\mathrm{e}}^{x} $$
$$ \Rightarrow \varepsilon + {\int }_{0}^{x}\left| {f\left( t\right) }\right| \mathrm{d}t \leq \varepsilon {\mathrm{e}}^{x}. $$
令 $\varepsilon \rightarrow 0$ ,即得
$$ 0 \leq {\int }_{0}^{x}\left| {f\left( t\right) }\right| \mathrm{d}t \leq 0 \Rightarrow {\int }_{0}^{x}\left| {f\left( t\right) }\right| \mathrm{d}t \equiv 0\left( {\forall x \geq 0}\right) $$
$$ \Rightarrow f\left( x\right) \equiv 0\left( {\forall x \geq 0}\right) . $$
\subsubsection{六、定积分的极限}