Any ideas on how to show this?

**KZA459** · February 28th 2009, 09:14 AM

f is continuous and
$ f:[a,b] \rightarrow ]0,\infty[ $

Show that the supremum of f(x) on [a,b] equals

sup f(x) = lim{n->infinity} (1/(b-a) * int[(f(x))^n])^1/n

The integral is from a to b

Latex Attempt:
$sup_{a{\leq}x{\leq}b}f(x) = lim_{n\rightarrow\infty} [\frac{1}{b-a}\int_a^b (f(x))^n]^\frac{1}{n} $

**Abu-Khalil** · February 28th 2009, 09:24 AM

This? $\sup_{a\leq x \leq b}f(x)=\lim_{n\to\infty}\left\{\frac{1}{b-a}\int_a^bf(x)^ndx\right\}^{\frac{1}{n}}$

It's $f$ continuous or not explicit?

**KZA459** · February 28th 2009, 09:25 AM

oops and yeah the hypothesis might help eh?

f is cts and
$ f:[a,b] \rightarrow ]0,\infty[ $

sorry!

yep, this

$\sup_{a\leq x \leq b}f(x) = \lim_{n\to\infty}\left\{\frac{1}{b-a}\int_a^b(f(x))^ndx\right\}^{\frac{1}{n}}$

**Abu-Khalil** · February 28th 2009, 09:34 AM

It's curious 'cause if $f$ is continuous then using MVT for integrals you know there exists $\xi\in(a,b)$ such as $f(\xi)^n=\frac{1}{b-a}\int_a^bf(x)^ndx$ .

So
$\lim_{n\to\infty}\left\{\frac{1}{b-a}\int_a^b(f(x))^ndx\right\}^{\frac{1}{n}}=\lim_{n \to\infty}\left\{f(\xi)^n\right\}^\frac{1}{n}=f(\x i)$ or not? But if $f$ is an increasing function the premise fails :S

**KZA459** · February 28th 2009, 09:40 AM

Well I got to that step, but I can't see how that $f(\xi)$ relates to the supremum.

**Abu-Khalil** · February 28th 2009, 09:42 AM

Originally Posted by KZA459

Well I got to that step, but I can't see how that $f(\xi)$ relates to the supremum.

Also, all continuous function is enclosed in a closed interval -><-

**ThePerfectHacker** · February 28th 2009, 11:03 AM

Originally Posted by KZA459

f is continuous and
$ f:[a,b] \rightarrow ]0,\infty[ $

Show that the supremum of f(x) on [a,b] equals

sup f(x) = lim{n->infinity} (1/(b-a) * int[(f(x))^n])^1/n

The integral is from a to b

Latex Attempt:
$sup_{a{\leq}x{\leq}b}f(x) = lim_{n\rightarrow\infty} [\frac{1}{b-a}\int_a^b (f(x))^n]^\frac{1}{n} $

Look here.

**KZA459** · February 28th 2009, 11:04 AM

ah thanks a lot!

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