方企勤 第三章 一元函数积分学 第3.3题
📝 题目
3. 3.19 设 $f\left( x\right) \in C\left\lbrack {a,b}\right\rbrack$ ,且 $f\left( x\right) \geq 0\left( {\forall x \in \left\lbrack {a,b}\right\rbrack }\right)$ . 求证:
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}{\left\{ {\int }_{a}^{b}{\left\lbrack f\left( x\right) \right\rbrack }^{n}\mathrm{\;d}x\right\} }^{\frac{1}{n}} = \mathop{\max }\limits_{{x \in \left\lbrack {a,b}\right\rbrack }}f\left( x\right) . $$
💡 答案解析
3. 3.19 设
$$ M = \mathop{\max }\limits_{{x \in \left\lbrack {a,b}\right\rbrack }}f\left( x\right) > 0,\;f\left( {x}_{0}\right) = M\;\left( {a < {x}_{0} < b}\right) , $$
则对 $\forall \varepsilon \in \left( {0,M}\right) ,\exists \delta > 0$ ,使得
$$ M - \frac{\varepsilon }{2} < f\left( x\right) \leq M\;\left( {\forall x \in \left( {{x}_{0} - \delta ,{x}_{0} + \delta }\right) }\right) . $$
令 ${I}_{n} = {\left\{ {\int }_{a}^{b}{\left\lbrack f\left( x\right) \right\rbrack }^{n}\mathrm{\;d}x\right\} }^{\frac{1}{n}}$ ,则有
$$ {\left( 2\delta \right) }^{\frac{1}{n}}\left( {M - \frac{\varepsilon }{2}}\right) \leq {\left\{ {\int }_{{x}_{0} - \delta }^{{x}_{0} + \delta }{\left\lbrack f\left( x\right) \right\rbrack }^{n}\mathrm{\;d}x\right\} }^{\frac{1}{n}} \leq {I}_{n} \leq M{\left( b - a\right) }^{\frac{1}{n}}. $$
又因为 $\mathop{\lim }\limits_{{n \rightarrow \infty }}{\left( 2\delta \right) }^{\frac{1}{n}} = 1,\mathop{\lim }\limits_{{n \rightarrow \infty }}{\left( b - a\right) }^{\frac{1}{n}} = 1$ ,所以对上述的 $\varepsilon > 0$ ,存在 $N \in \mathbf{N}$ 使得
$$ {\left( 2\delta \right) }^{\frac{1}{n}}\left( {M - \frac{\varepsilon }{2}}\right) > M - \varepsilon ,\;M{\left( b - a\right) }^{\frac{1}{n}} < M + \varepsilon \;\left( {\forall n > N}\right) $$
$$ \Rightarrow M - \varepsilon < {I}_{n} < M + \varepsilon \;\left( {\forall n > N}\right) . $$