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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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    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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    @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