If $f$ is nonnegative and continuous on $[a,b]$, then $\left(\int_a^b f(x)^n \ dx\right)^{1/n}\to\max\limits_{[a,b]} f$ [duplicate]

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:

$(\int_a^b f(x)^n \ dx)^{1/n} \ge M$

$(\int_a^b f(x)^n \ dx)^{1/n} \le M(b-a)^{1/n}$

$\Rightarrow \exists \ \lim_n \ (\int_a^b f(x)^n \ dx)^{1/n}=M=\max_I \ f$

marked as duplicate by Guy Fsone, Rolf Hoyer, Peter Taylor, Trevor Gunn, heropupNov 11 '17 at 17:22

• The identity $(\int_a^b f(x)^n \ dx)^{1/n}=(\sum_{k=1}^m M_h^n mis(I_k))^{1/n}$ is unclear to me. – Siminore Aug 12 '12 at 9:47
• $\int_a^b f(x)^n dx = \sum_{k=1}^m M_k ^n mis(I_k)$ because: $\sum_{k=1}^m M_k mis(I_k)-\sum_{k=1}^m m_k mis(I_k)=\sum_{k=1}^m (M_k-m_k) mis(I_k)<\epsilon (b-a)$. Where $m_k$ is the minimum on $\bar{I_k}$, then $\int_a^b f(x)^n dx =\sum_{k=1}^m M_k ^n mis(I_k)=\sum_{k=1}^m m_k ^n mis(I_k)$ – Itachi Aug 12 '12 at 10:09
• In general, an integral is a limit of a Riemann sum like the one you wrote. It is a finite sum only under very special assumptions on $f$. Actually, where did $\epsilon$ go, in your explanation? – Siminore Aug 12 '12 at 12:39

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 $\lim\limits_{n\to\infty} J_n=M$.

• Thanks for the solution!! :) – Itachi Aug 12 '12 at 10:49
• @Itachi, it's o.k. to wait a little for other possible answers and then choose from them the one you like the best, but in the meantime, and since you liked did's answer, perhaps it's a good idea to upvote his answer. – DonAntonio Aug 12 '12 at 12:39
• @DonAntonio: +1. :-)) – Did Aug 12 '12 at 13:12
• @Don, OP is not registered, so s/he cannot upvote did's fine answer. – J. M. is a poor mathematician Aug 12 '12 at 23:10
• I see, @J.M, thanks. But isn't anyone capable to post a question already registered in the forum and at least capable of upvote answers to his own question? I thought this was the situation... – DonAntonio Aug 13 '12 at 2:11