Convergence of a sequence of integrals

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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 g_n(x)= \left( \int_a^b (f(x))^n \, dx \right)^\frac{1}{n} 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 x_o such that |f(x_o)-M| &lt; \frac{ε}{2} , and by continuity if |x-x_o| &lt; δ then |f(x)-f(x_o)| &lt; \frac{ε}{2}<br /> and then triangle inequality it up to show |f(x)-M| &lt; ε

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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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