Convergence of a sequence of integrals

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SUMMARY

The sequence $$\left( \int_a^b (f(x))^n \, dx \right)^\frac{1}{n}$$ converges to the supremum M of a continuous, non-negative function f defined on the interval I=[a,b]. The approach involves defining g_n(x) and demonstrating its convergence to a function g(x) uniformly. By establishing a point x_o where |f(x_o)-M| < ε/2 and applying continuity, one can derive bounds that lead to the conclusion that the sequence converges to M as n approaches infinity.

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Homework Statement




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


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?
 
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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?
 

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