# Any ideas on how to show this?

• Feb 28th 2009, 09:14 AM
KZA459
Any ideas on how to show this?
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}
$
• Feb 28th 2009, 09:24 AM
Abu-Khalil
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?
• Feb 28th 2009, 09:25 AM
KZA459
oops and yeah the hypothesis might help eh?

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

sorry!

yep, this

$\sup_{a\leq x \leq b}f(x) = \lim_{n\to\infty}\left\{\frac{1}{b-a}\int_a^b(f(x))^ndx\right\}^{\frac{1}{n}}$
• Feb 28th 2009, 09:34 AM
Abu-Khalil
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
• Feb 28th 2009, 09:40 AM
KZA459
Well I got to that step, but I can't see how that $f(\xi)$ relates to the supremum.
• Feb 28th 2009, 09:42 AM
Abu-Khalil
Quote:

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 -><-
• Feb 28th 2009, 11:03 AM
ThePerfectHacker
Quote:

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.
• Feb 28th 2009, 11:04 AM
KZA459
ah thanks a lot!