68285

证明：由连续函数的最值定理知，存在$\xi  \in \left[ {a,b} \right]$，使得$$f\left( \xi  \right){\rm{ = }}\mathop {max}\limits_{x \in \left[ {a,b} \right]} f\left( x \right){\rm{ = }}M$$

从而由$f\left( x \right) \in C\left[ {a,b} \right]$知，$f\left( x \right)$在点$\xi $连续，则对任给$\varepsilon  > 0$，存在$\delta  > 0$，使得对任意$x \in \left( {\xi  - \delta ,\xi  + \delta } \right) \cap \left[ {a,b} \right]$，有\[\left| {f\left( x \right) - f\left( \xi  \right)} \right| < \varepsilon \]

即有$f\left( x \right) > M - \varepsilon $，令${a_n} = \sqrt[n]{{\int_a^b {{f^n}\left( x \right)} dx}}$，从而可知\[{a_n} \ge {\left( {\int_{\xi  - \delta }^{\xi  + \delta } {{{\left( {M - \varepsilon } \right)}^n}} dx} \right)^{\frac{1}{n}}} = \left( {M - \varepsilon } \right){\left( {2\delta } \right)^{\frac{1}{n}}} \to M - \varepsilon \left( {n \to \infty } \right)\]而\[{a_n} \le {\left( {\int_a^b {{M^n}} dx} \right)^{\frac{1}{n}}} = M{\left( {b - a} \right)^{\frac{1}{n}}} \to M\left( {n \to \infty } \right)\]所以有\[M - \varepsilon  \le \mathop {\underline {\lim } }\limits_{n \to \infty } {a_n} \le \mathop {\overline {\lim } }\limits_{n \to \infty } {a_n} \le M\]令$\varepsilon  \to 0$，则$\mathop {\underline {\lim } }\limits_{n \to \infty } {a_n} = \mathop {\overline {\lim } }\limits_{n \to \infty } {a_n} = M$，从而命题得证

$\bf注1：$我们由$\bf{H\ddot older不等式}$可知${a_n} = \sqrt[n]{{\int_a^{a + 1} {{f^n}\left( x \right)} dx}}$是单调递增的(证明)

$\bf注2：$我们同理可证

$\bf命题：$设正值函数$f\left( x \right),g\left( x \right) \in C\left[ {a,b} \right]$，则\[\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{\int_a^b {{f^n}\left( x \right)g\left( x \right)} dx}} = \mathop {max}\limits_{x \in \left[ {a,b} \right]} f\left( x \right)\]