# Thread: Any ideas on how to show this?

1. ## Any ideas on how to show this?

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

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

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

It's $\displaystyle f$ continuous or not explicit?

3. oops and yeah the hypothesis might help eh?

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

sorry!

yep, this

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

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

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

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

6. Originally Posted by KZA459
Well I got to that step, but I can't see how that $\displaystyle f(\xi)$ relates to the supremum.
Also, all continuous function is enclosed in a closed interval -><-

7. Originally Posted by KZA459
f is continuous and
$\displaystyle f:[a,b] \rightarrow ]0,\infty[$

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

8. ah thanks a lot!