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Assume $f : [a,b] \to R$ is continuous and $f(x) \ge 0$ for all $x \in [a,b]$, and $M = \sup\{f(x) : x \in [a,b]\}$. Show that $$\lim_{n\to\infty}\left[\int_a^bf(x)^ndx\right]^{1/n}$$ converges to $M$.

I know you have to get $|[\int_a^b[f(x)^ndx]^{1/n}-M| < \epsilon$ by the definition of a convergent sequence, but I'm not sure how to simplify the left side of this inequality. Are you suppose to look at the function inside the sequence as a composition of two functions maybe $f(x)$ and $g(x) = x^n$ or use a change of variable to better integrate this function?

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Hint: We want to prove that the limit is $\max f$. So the part of $[a,b]$ where $f$ is not close of its max should not contribute to the integral. You already know that $f\le M \implies \displaystyle\left(\int_a^b f^n(x)dx\right)^{1/n}\le (b-a)^{1/n}M$, whose limit is $M$, so it remains to prove the other inequality.

When $f$ is continuous, there is at least one (non-empty) interval $[u,v]$ such as $$ x\in [u,v] \implies f(x) \ge M - \epsilon $$ Then $$ \int_a^b f(x)^n dx \ge \int_u^v f(x)^n dx\ge (M - \epsilon )^n(v-u) $$

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – user642796 Nov 20 '14 at 20:08

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