Show that $\lim((\int_{a}^{b}f^{n})^\frac{1}{n})=\sup\{f(x):x\in[a,b]\}$ [duplicate]

Another exercise (this one is 7.2.18) from "Introduction to Real Analysis" by Bartle and Sherbert that I'm struggling with:

Let $f$ be continuous on $[a,b]$, let $f(x)>=0$ for $x\in[a,b]$, and let $M_{n}:=(\int_{a}^{b}f^{n})^\frac{1}{n}$. Show that $\lim(M_{n})=\sup\{f(x):x\in[a,b]\}$.

I think that as $f$ is continuous, thus bounded, on given interval, one should start with representing $M_{n}$ as $(M^{n}\int_{a}^{b}(\frac{f}{M})^{n})^\frac{1}{n}$, where $M$ is essential supremum of $f$ on given interval, and then proceed from there; however, I'm not sure how to formalize further steps...

Note that this particular exercise is following exercises on MVT for integrals. So another approach I tried was to follow from the fact that there exists $c_{1},\ldots,c_{n}\in[a,b]$ such that $M_{n}=(f(c_{1})\int_{a}^{b}f^{n-1})^\frac{1}{n}=\ldots=(f(c_{1})\ldots f(c_{n})(b-a))^\frac{1}{n}$, but to no avail.

marked as duplicate by sdcvvc, Davide Giraudo, Sasha, GEdgar, Dennis GulkoMar 2 '13 at 14:52

• Indeed it's a duplicate of this post - sorry for this, I tried searching other posts before posting my question, but haven't encountered this one. Still I hope answers provided below by Andreas and Davide will further clarify it - at least I find them very valuable. – Crni Mar 2 '13 at 14:26

Fix $\varepsilon>0$, and let $x_0\in [a,b]$ such that $f(x_0)=M$ like in the OP. Take $\delta>0$ such that if $|x-x_0|\leqslant \delta$ and $x\in [a,b]$ then $$M-\varepsilon< f(x)<M+\varepsilon .$$ Call $I:=[a,b]\cap (x_0-\delta,x_0+\delta)$; then the length of $I$ is at least $\delta$ and smaller than $2\delta$. Hence $$M^n(b-a)\geqslant \int_a^bf(x)^ndx\geqslant \int_I f(x)^ndx\geqslant \delta (M-\varepsilon)^n.$$ Taking the power $\frac 1n$, this gives that $$M(b-a)^{\frac 1n}\geqslant \left(\int_a^bf(x)^ndx\right)^{\frac 1n}\geqslant \delta^{\frac 1n}(M-\varepsilon).$$ The first inequality gives that $$\limsup_{n\to +\infty}\left(\int_a^bf(x)^ndx\right)^{\frac 1n}\leqslant M,$$ and the second one that for each $\varepsilon>0$, $$\liminf_{n\to +\infty}\left(\int_a^bf(x)^ndx\right)^{\frac 1n}\geqslant M-\varepsilon.$$ The conclusion follows.
By Jensen's inequality (see for instance this post) you have that $$m_n := \frac{M_n}{(b-a)^{\frac1n}} = \frac{1}{(b-a)^{\frac 1n}} \left( \int_a^bf^n \right)^{\frac1n}$$ is increasing, therefore it either converges or diverges. Note that $\lim_{n\to\infty} m_n=\lim_{n\to\infty}M_n$.
Since $0\leq f\leq\max f$ you have $$M_n \leq \max f$$ so we conclude that $M_n$ converges and $\lim_{n\to\infty}M_n\leq\max f$. Now it suffices to show that $\lim_{n\to\infty}M_n\geq\max f$.
Since $f$ is continuous, for all $\varepsilon>0$ the set $$E_\varepsilon := \{x\in(a,b):~f(x)\geq\max f-\varepsilon\}$$ has positive measure... can you go on from here? (Hint: give an estimate of $\int_{E_\varepsilon} f$)