📝 题目
例 10 设 $f\left( x\right) ,g\left( x\right) \in R\left\lbrack {a,b}\right\rbrack$ ,求证:
$$ \left| {{\int }_{a}^{b}f\left( x\right) g\left( x\right) \mathrm{d}x}\right| \leq \sqrt{{\int }_{a}^{b}{f}^{2}\left( x\right) \mathrm{d}x} \cdot \sqrt{{\int }_{a}^{b}{g}^{2}\left( x\right) \mathrm{d}x}. $$
💡 答案与解析
证 对 $\forall t \in \mathbf{R}$ ,积分
$$ {\int }_{a}^{b}{\left\lbrack tf\left( x\right) + g\left( x\right) \right\rbrack }^{2}\mathrm{\;d}x = {t}^{2}{\int }_{a}^{b}{f}^{2}\left( x\right) \mathrm{d}x + {2t}{\int }_{a}^{b}f\left( x\right) \mathrm{d}x{\int }_{a}^{b}g\left( x\right) \mathrm{d}x $$
$$ + {\int }_{a}^{b}{g}^{2}\left( x\right) \mathrm{d}x \geq 0. $$
根据二次三项式的恒正条件是判别式 ${b}^{2} - {ac} \leq 0$ ,故有
$$ {\left( {\int }_{a}^{b}f\left( x\right) \mathrm{d}x{\int }_{a}^{b}g\left( x\right) \mathrm{d}x\right) }^{2} \leq {\int }_{a}^{b}{f}^{2}\left( x\right) \mathrm{d}x \cdot {\int }_{a}^{b}{g}^{2}\left( x\right) \mathrm{d}x, $$
即得
$$ \left| {{\int }_{a}^{b}f\left( x\right) g\left( x\right) \mathrm{d}x}\right| \leq \sqrt{{\int }_{a}^{b}{f}^{2}\left( x\right) \mathrm{d}x} \cdot \sqrt{{\int }_{a}^{b}{g}^{2}\left( x\right) \mathrm{d}x}. $$