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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\[\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{\int_a^b {{f^n}\left( x \right)g\left( x \right)} dx}} = \mathop {max}\limits_{x \in \left[ {a,b} \right]} f\left( x \right)\]