If $f$ is nonnegative and continuous on $[a,b]$, then $\left(\int_a^b f(x)^n \ dx\right)^{1/n}\to\max\limits_{[a,b]} f$ I've been working on the following problem: 

Show that if $f\in C[a, b]$ , $f\ge 0$ on $[a, b]$, then $\left(\int_a^b f(x)^n \,dx\right)^{1/n}$ converges when $n\to\infty$ and the limit is $\max_If$ with $I=[a, b]$.

This is my solution:
For  Weierstrass $f$ has maximum, $\exists \  \xi : f(\xi)=M$; and as $f$ is defined on $[a,b]$, $f$ is U.C., then:
$\forall \epsilon >0  \ \exists \delta >0: \forall x,y \in [a,b]: |x-y|< \delta \Rightarrow |f(x)-f(y)|< \epsilon$
Let $[a,b]=\bigcup_{k=1}^m I_k$, with $mis(I_k)<\delta$, and $M_k=max_{\bar(I_k)}  f(x)$
$(\int_a^b f(x)^n \  dx)^{1/n}=(\sum_{k=1}^m M_h^n mis(I_k))^{1/n}=(M_1^n mis(I_1)+...+M^n mis(I_j)+...M_m^n mis(I_m))^{1/n}=$
=$M ((M_1/M)^n mis(I_1)+...+mis(I_j)+...+(M_m/M)^n mis(I_m))^{1/n}$
Then:
$(\int_a^b f(x)^n \  dx)^{1/n} \ge M$
$(\int_a^b f(x)^n \  dx)^{1/n} \le M(b-a)^{1/n}$
$\Rightarrow \exists \  \lim_n \ (\int_a^b f(x)^n \  dx)^{1/n}=M=\max_I \ f$
 A: We know the end of the story, don't we, so let us try to use only estimates that will prove in the end that the limit is what it is. 
Let $M=\max\{f(x)\,;\,x\in[a,b]\}$, let $u\gt0$ with $u\lt M$ and, for every $n$, let $J_n=\left(\int_a^bf^n\right)^{1/n}$.


*

*On the one hand, $0\leqslant f\leqslant M$ on $[a,b]$ hence $J_n\leqslant\left(\int_a^bM^n\right)^{1/n}=(b-a)^{1/n}\cdot M$. 

*On the other hand, there exists some $\xi$ in $[a,b]$ such that $f(\xi)=M$. Furthermore, $f$ is continuous at $\xi$ hence there exists an interval $K$ of length $v\gt0$ which contains $\xi$ and such that $f\geqslant M-u$ on $K$. Since $f\geqslant0$ on $[a,b]$, this yields $J_n\geqslant\left(\int_K(M-u)^n\right)^{1/n}=v^{1/n}\cdot (M-u)$.
Now is the time to collect our estimates... 
Namely, for every $n$, $v^{1/n}\cdot (M-u)\leqslant J_n\leqslant (b-a)^{1/n}\cdot M$. When $n\to\infty$, $(b-a)^{1/n}\to1$ and $v^{1/n}\to1$ hence $M-u\leqslant\liminf\limits_{n\to\infty} J_n\leqslant\limsup\limits_{n\to\infty} J_n\leqslant M$. 
Finally, this holds for every $u\gt0$ with $u\lt M$ hence $\lim\limits_{n\to\infty} J_n=M$.
