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So I've been trying to find a solution for this all afternoon, but haven't found a good place to start:

Prove that if $f:[a,b]tomathbf{R}^+$ is a continuous function with maximum value $M$, then
$$
lim_{ntoinfty}left(int_a^b f(x)^n,dxright)^{1/n} = M
$$

Here are some of the paths I've considered, though none have been very successful:

(1) Considering the sequence of functions for all increasing integer $n$ and trying to show that the sequence converges. We've had plenty of work on converging sequences, but with the integral expression, I am not sure how to simplify.

(2) Showing that that sequence is increasing (again, how?) and then showing there to be a supremum at $M$. I'm not sure how the maximum of the function arrives in this problem.

(3) Mean value theorems for integrals

If anyone could give me a solid place to start or perhaps point me to a place where this question has been asked before (I can't seem to find it), I would be very grateful.

If you have done the following question, then you probably know how to do the one in the post:

Given $m$ positive real numbers $a_1, dots, a_n$, prove that
$$
lim_{nto +infty} (a_1^n + dots + a_m^n)^{1/n} = max_{j} a_j.
$$

Proof

The trick is to use the squeeze theorem, and note that $$ max_j a_j leqslant (a_1^n + dots + a_m^n)^{1/n} leqslant m^{1/n} a_j. $$ Now let $n to +infty$, and note that $lim_n m^{1/n} = 1$.

Using the similar trick, we could do the original one. The problem is $f$ is "continuous". Even $f$ could reach $M$ at some point $x_0$, we cannot directly use it. So we only requires that for each $varepsilon >0$, find those $x$ s.t. $M-varepsilon leqslant f(x) leqslant M$. So we have the following proof.

Proof

Let $x_0 in [a, b]$ be the point where $f(c) =M$ [this is possible, since $fin mathcal C[a,b]$]. For each $epsilon > 0$, by the definition of continuity, there exists an interval $[c,d]$ that contains $x_0$ s.t. $$ xin [c,d] implies M-varepsilon leqslant f(x) leqslant M. $$ Thus $$ (d-c)^{1/n} (M-varepsilon) leqslant left(int_c^d f^nright)^{1/n}leqslant left(int_a^b f^nright)^{1/n} leqslant M, $$ and then $$M-varepsilon leqslant varliminf_n left(int_a^b f^nright)^{1/n} leqslant varlimsup_n left(int_a^b f^nright)^{1/n} leqslant M. $$ Now let $varepsilon to 0^+$, then the upper limit and the lower limit are equal to $M$, hence the limit equation.

Remark

Technically, this is actually not an application of the squeezing theorem [the end of the inequality chains are not the same], but the idea is similar. If no tools of upper/lower limits are allowed, then the whole proof could be completed in $varepsilon$-$N$ format.