Results 1 to 8 of 8

Math Help - Any ideas on how to show this?

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    14

    Any ideas on how to show this?

    f is continuous and
    <br />
f:[a,b] \rightarrow ]0,\infty[<br />

    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}<br />
    Last edited by KZA459; February 28th 2009 at 10:26 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Abu-Khalil's Avatar
    Joined
    Oct 2008
    From
    Santiago
    Posts
    148
    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?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2008
    Posts
    14
    oops and yeah the hypothesis might help eh?

    f is cts and
    <br />
f:[a,b] \rightarrow ]0,\infty[<br />

    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}}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member Abu-Khalil's Avatar
    Joined
    Oct 2008
    From
    Santiago
    Posts
    148
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Dec 2008
    Posts
    14
    Well I got to that step, but I can't see how that f(\xi) relates to the supremum.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member Abu-Khalil's Avatar
    Joined
    Oct 2008
    From
    Santiago
    Posts
    148
    Quote Originally Posted by KZA459 View Post
    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 -><-
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by KZA459 View Post
    f is continuous and
    <br />
f:[a,b] \rightarrow ]0,\infty[<br />

    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}<br />
    Look here.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Dec 2008
    Posts
    14
    ah thanks a lot!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Any ideas here please? :-)
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 5th 2010, 08:56 AM
  2. Any ideas for tanh(x/2)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 20th 2010, 10:22 AM
  3. Any ideas...
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: February 28th 2010, 10:19 PM
  4. AS coursework, Ideas?
    Posted in the Math Topics Forum
    Replies: 0
    Last Post: November 4th 2009, 10:34 AM
  5. i need some ideas
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 6th 2008, 08:22 AM

Search Tags


/mathhelpforum @mathhelpforum