4
$\begingroup$

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:

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

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

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

  • 0
    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
  • 0
    $\int_a^b f(x)^n dx = \sum_{k=1}^m M_k ^n mis(I_k)$ 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
  • 0
    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