Proving Limit of Integral for Nonnegative Continuous Function on [0,1]

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Homework Statement


f nonnegative continuous function on the interval [0,1]. Let M be the supremum of f on the interval. Prove:
[tex]\lim_{n \rightarrow \infty} \left[ \int^1_0 f(t)^n dt \right] ^{1/n} = M[/tex]


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The Attempt at a Solution


I was trying using Upper Sums but I don't know how to compute the limit.
 
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It seems like you need following intermediate result:

Let [tex]x_1,...x_n[/tex] be real numbers and [tex]w_1,...,w_n[/tex] be positive such that [tex]w_1+...+w_n=1[/tex]. Assume without loss of generality that [tex]x_1[/tex] is the greatest among the [tex]x_i[/tex]. Then

[tex]\lim_{n\rightarrow+\infty}{\sqrt[n]{\sum_{i=1}^k{w_ix_i^n}}}=x_1[/tex].

HINT: Show, using the squeeze theorem, that

[tex]\lim_{n\rightarrow +\infty}{\left(\frac{1}{n} ln\left(\frac{\sum_{i=1}^k{w_ix_i^n}}{x_1^n}\right)\right)}=0[/tex]
 
Just think about this simple case.

Suppose you have a function that is M for a little interval of width [itex]\delta[/itex], and zero otherwise. Never mind that it isn't continuous and all that. Now what's the integral there? And what's the limit? That should put you on the right track.