证明：由连续函数的最值定理知，存在$xi in left[ {a,b} ight]$，使得$$fleft( xi ight){ m{ = }}mathop {max}limits_{x in left[ {a,b} ight]} fleft( x ight){ m{ = }}M$$
从而由$fleft( x ight) in Cleft[ {a,b} ight]$知，$fleft( x ight)$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 } ight) cap left[ {a,b} ight]$，有[left| {fleft( x ight) - fleft( xi ight)} ight| < varepsilon ]
即有$fleft( x ight) > M - varepsilon $，令${a_n} = sqrt[n]{{int_a^b {{f^n}left( x ight)} dx}}$，从而可知[{a_n} ge {left( {int_{xi - delta }^{xi + delta } {{{left( {M - varepsilon } ight)}^n}} dx} ight)^{frac{1}{n}}} = left( {M - varepsilon } ight){left( {2delta } ight)^{frac{1}{n}}} o M - varepsilon left( {n o infty } ight)]而[{a_n} le {left( {int_a^b {{M^n}} dx} ight)^{frac{1}{n}}} = M{left( {b - a} ight)^{frac{1}{n}}} o Mleft( {n o infty } ight)]所以有[M - varepsilon le mathop {underline {lim } }limits_{n o infty } {a_n} le mathop {overline {lim } }limits_{n o infty } {a_n} le M]令$varepsilon o 0$，则$mathop {underline {lim } }limits_{n o infty } {a_n} = mathop {overline {lim } }limits_{n o infty } {a_n} = M$，从而命题得证
$f注1：$我们由$f{Hddot older不等式}$可知${a_n} = sqrt[n]{{int_a^{a + 1} {{f^n}left( x ight)} dx}}$是单调递增的(证明)
$f注2：$我们同理可证
$f命题：$设正值函数$fleft( x ight),gleft( x ight) in Cleft[ {a,b} ight]$，则[mathop {lim }limits_{n o infty } sqrt[n]{{int_a^b {{f^n}left( x ight)gleft( x ight)} dx}} = mathop {max}limits_{x in left[ {a,b} ight]} fleft( x ight)]