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Maxima And Integration Proof

  1. Feb 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Let f be continuous and positive on [a,b] and let M denote its maximum value. Prove:

    M = [tex] \lim_{n \rightarrow \infty} \sqrt[n]{\int_a^b (f(x))^n dx}[/tex]


    2. Relevant equations



    3. The attempt at a solution

    Writing out the integral as a limit of Riemann sums, we have [tex] \lim_{n \rightarrow \infty} \sqrt[n]{\int_a^b (f(x))^n dx} = \lim_{n \rightarrow \infty} \sqrt[n]{\frac{b-a}{n}[f^n(x_1) + f^n(x_2) + ... + f^n(\alpha) + ... + f^n(n)]}
    [/tex] where [tex] f(\alpha) = M [/tex].

    This expression can be rewritten as [tex]\lim_{n \rightarrow \infty} \sqrt[n]{\frac{b-a}{n}} \sqrt[n]{f^n(x_1) + f^n(x_2) + ... + f^n(\alpha) + ... + f^n(n)}[/tex]. Now, letting n tend to infinity, the nth root of (b-a)/n converges to 1 and so we need only evaluate the limit of the infinite series. Let all f values raised to the power of n, except for f(alpha) = M, be equal to [tex] \xi^n[/tex]. Then we have [tex]\lim_{n \rightarrow \infty} \sqrt[n]{\int_a^b (f(x))^n dx} =\lim_{n \rightarrow \infty} \sqrt[n]{M^n + \xi^n}[/tex].

    Now, since f is positive everywhere in the interval and remembering there are n terms under the nth root sign, then [tex] \sqrt[n]{M^n} = M < \sqrt[n]{M^n + \xi^n} \le \sqrt[n]{nM^n} = M\sqrt[n]{n} [/tex]. Since the right hand side of the inequality converges to M (since the nth root of n converges so 1 as n goes to infinity), then the limit of the Riemann sums is also M and so we have [tex] M = \lim_{n \rightarrow \infty} \sqrt[n]{\int_a^b (f(x))^n dx}[/tex].


    How does this proof look?
     
  2. jcsd
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