This first theorem you have pointed out following from the continuity of f, what is it?

this is the continuity of $f$ in each point $x$ for which $f(x) = \max_{[a,b]} f$.

@jlang The preimage of the open set $(M-\epsilon,M+1)$ is open. (And you rather need that $f(x)\color{red}>M-\epsilon$ on a nonempty open interval)

@jlang I made it more explicit.

@mookid So basically on this sub interval [u,v] of [a,b] f(x) achieves it's least upper bound or at least gets very close to it?

@mookid how do you select N in order to insure that $|(M-\epsilon)^n(u-v)-M| \le \epsilon$

@jlang first message: this is it! second message: do not forget the nth root

@jlang: Not so fast: at least from that inequality you have $\left[\int_a^bf(x)^ndx\right]^{1/n} \geq (M-\epsilon)(v-u)^{1/n}$.

@jlang I added details concerning the intuition leading to my Hint, hoping to make this clearer.

@John oops I forgot about that... You should need to select N though so that $(u-v)^{1/n}$ becomes one though so the inequality becomes $|(M-\epsilon)^n-M|=\epsilon$ right?

@mookid thank you! In your new edit, when you say "the other inequality" which one are you referring to?

I am referring to $\lim(...) \ge M$

@mookid: In your latest edit I guess you only have LHS $\leq M(b-a)^{1/n}$ (Not a big problem though).

@John I let something to be done for OP ;) be careful, you wrote the inequality the wrong way.

@mookid: I mean this: $f\le M \implies \displaystyle\left(\int_a^b f^n(x)dx\right)^{1/n}\le \displaystyle\left(\int_a^b M^ndx\right)^{1/n} \le M(b-a)^{1/n}$.

I think I am more confused now? Why do you want to prove that the limit is $\ge M$?

@John oh yes, thanks!

@jlang to prove $\lim(...) = M, $ you prove $\lim(...) \le M, $ and then $\lim(...) \ge M $

@mookid oh yeah of course! But how do you prove that other way? As limit goes to infinity of $(M-\epsilon)(u-v)^{1/n}$ gives you $(M-\epsilon)$. Which on that small subinterval is less than f(x)?

@jlang stop here. This is true for every $\epsilon$. So...

I'm not sure how to relate that back to the necessary lim of the sequence?

@mookid or can you just choose an epsilon to fore fill the inequality?

@mookid oh wait nevermind, you know $(M-\epsilon) \le f(x) \le M$ to this limit is also less than or equal to M. Thus, the limit must be M? But clearly, $(M-\epsilon) \le M$. I was just over thinking it.

@mookid oops in regards to my previous comment, you don't want that limit to be less than or equal to M.

Just in case if it is still not clear, first of all, you do not know if limit exists as $n\to \infty$. But the first inequality tell you that $\limsup (\int_a^b f^n(x) dx)^{1/n} \leq M$, while the second one tells you that $\liminf (\int_a^b f^n(x) dx)^{1/n} \geq M-\epsilon$. Thus $M \geq \limsup (\int_a^b f^n(x) dx)^{1/n} \geq \liminf (\int_a^b f^n(x) dx)^{1/n} \geq M - \epsilon$. But $\epsilon$ is arbitrary, which forces that limit exist and is equal to $M$. @jlang

@jling : Also remark that this fact is extremely useful in PDE, because it is always easier to find integral bound than pointwise bound. The result we prove here give us a link from integral bound to pointwise bound, which is pretty amazing.