Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Convergence of a sequence of integrals

  1. Feb 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Let I=[a,b], f : I to R be continuous and suppose that f(x) >= 0 . If M = sup{f(x):x ε I} show that the sequence $$\left( \int_a^b (f(x))^n \, dx \right)^\frac{1}{n}$$
    converges to M

    3. The attempt at a solution

    Where do I start? I'm thinking of having [tex] g_n(x)= \left( \int_a^b (f(x))^n \, dx \right)^\frac{1}{n}[/tex] and showing that converges to a function g(x) (uniformly?) but that just feels like restating the problem.

    If I can show that there exists [itex]x_o[/itex] such that [itex]|f(x_o)-M| < \frac{ε}{2}[/itex] , and by continuity if [itex]|x-x_o| < δ[/itex] then [itex]|f(x)-f(x_o)| < \frac{ε}{2}
    [/itex] and then triangle inequality it up to show [itex]|f(x)-M| < ε[/itex]

    I still feel this gets me nowhere. Any ideas?
  2. jcsd
  3. Feb 24, 2012 #2


    User Avatar
    Gold Member

    Well you can trivially bound it from above, so just work on bounding it from below. Can you find an interval over which f(x)>M(1-ε), if so ∫f(x)ndx>δ Mn(1-ε)n. What happens to this bound when you raise it to 1/n?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook