TITLE: Show the sequence converges to M
QUESTION [10 upvotes]: 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?

REPLY [10 votes]: 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 $$
\int_a^b f(x)^n dx \ge \int_u^v f(x)^n dx\ge (M - \epsilon
)^n(v-u)
$$