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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(ξ )=maxx[a,b]f(x)=M
<script id="MathJax-Element-1" type="math/tex; mode=display">f\left( \xi \right){\rm{ = }}\mathop {max}\limits_{x \in \left[ {a,b} \right]} f\left( x \right){\rm{ = }}M</script>

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

|f(x)?f(ξ )|<ε
<script id="MathJax-Element-2" type="math/tex; mode=display">\left| {f\left( x \right) - f\left( \xi \right)} \right| < \varepsilon </script>

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

an(ξ +δξ ?δ(M?ε)ndx)1n=(M?ε)(2δ)1nM?ε(n)
<script id="MathJax-Element-3" type="math/tex; mode=display">{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)</script>而
an(baMndx)1n=M(b?a)1nM(n)
<script id="MathJax-Element-4" type="math/tex; mode=display">{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)</script>所以有
M?ε lim???nanlimˉˉˉˉˉnanM
<script id="MathJax-Element-5" type="math/tex; mode=display">M - \varepsilon \le \mathop {\underline {\lim } }\limits_{n \to \infty } {a_n} \le \mathop {\overline {\lim } }\limits_{n \to \infty } {a_n} \le M</script>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${a_n} = \sqrt[n]{{\int_a^b {{f^n}\left( x \right)} dx}}$是单调递增的

$\bf注2:$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

limnbafn(x)g(x)dx?????????????n=maxx[a,b]f(x)
<script id="MathJax-Element-6" type="math/tex; mode=display">\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)</script>