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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:

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

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

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

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    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.2012-08-12
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    $\int_a^b f(x)^n dx = \sum_{k=1}^m M_k ^n mis(I_k)$ 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)$2012-08-12
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    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?2012-08-12

1 Answers 1

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

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    Thanks for the solution!! :)2012-08-12
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    @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.2012-08-12
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    @DonAntonio: +1. :-))2012-08-12
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    @Don, OP is not registered, so s/he cannot upvote did's fine answer.2012-08-12
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    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...2012-08-13
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    @Don, unregistered users can ask and answer questions, as well as comment on their questions and answers to them, and accept answers to those questions. Voting, however, is only for registered users with a certain amount of rep. Unless OP registers *and* [acquires a certain amount of rep](http://math.stackexchange.com/privileges/vote-up), no dice.2012-08-13