Results 1 to 4 of 4

Math Help - Uniform convergence

  1. #1
    Senior Member Twig's Avatar
    Joined
    Mar 2008
    From
    Gothenburg
    Posts
    396

    Uniform convergence

    Hi!

    Problem: Show that  f_{n}(x)=nx^{n}(1-x) converges "pointwise", donīt know if this is the correct english word for it. And show that the convergence is not uniform on [0,1] .

    Solution:

    Itīs easy to find that f_{n}(x) \to f(x)=0 \; \; , \forall x \in [0,1]

    How do I show that the convergence is not uniform?

    We have that \lim_{n\to \infty} f_{n}(x) = f(x) is continous on [0,1] , so we have to proceed right?

    By letting d(x)=|f_{n}(x)-f(x)| , we have to show that d(x) \to 0 as n\to \infty

    I tried taking the derivative but didnīt come up with anything good.

    Thx!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jun 2009
    Posts
    113
    I believe all the functions f_n are positive and attain their maximum at x=1/2 (using the derivative). Hence the supremum norm of each f_n is n(1/2)^{n+1} and this implies uniform convergence!

    Perhaps I have commited a mistake with elemental calculus, but the way of attacking is clearly this, to observe what's going on with the supremum of the f_n.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2006
    From
    Florida
    Posts
    228
    Quote Originally Posted by Enrique2 View Post
    I believe all the functions f_n are positive and attain their maximum at x=1/2 (using the derivative). Hence the supremum norm of each f_n is n(1/2)^{n+1} and this implies uniform convergence!

    Perhaps I have commited a mistake with elemental calculus, but the way of attacking is clearly this, to observe what's going on with the supremum of the f_n.
    Actually taking the derivative shows that the max of f_n occurs when x=\frac{n}{n+1}. So f_n\left(\frac{n}{n+1}\right)=n\left(1-\frac{1}{n+1}\right)^n\frac{1}{n}=\left(1-\frac{1}{n+1}\right)^n\to e^{-1} (not sure about the final limit being e^{-1} but i know it's not 0).
    Thus \sup|f-f_n|=e^{-1}\neq{0}. So no uniform convergence for you.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Twig's Avatar
    Joined
    Mar 2008
    From
    Gothenburg
    Posts
    396
    I was supposed to show that the convergence was not uniform, so that is good =)

    I donīt know where I got it wrong, I got that x=\frac{-n}{n+1}

    thx
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Uniform convergence vs pointwise convergence
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 16th 2012, 12:03 AM
  2. Replies: 1
    Last Post: October 31st 2010, 08:09 PM
  3. Pointwise convergence to uniform convergence
    Posted in the Calculus Forum
    Replies: 13
    Last Post: November 29th 2009, 09:25 AM
  4. Pointwise Convergence vs. Uniform Convergence
    Posted in the Calculus Forum
    Replies: 8
    Last Post: October 31st 2007, 06:47 PM
  5. Uniform Continuous and Uniform Convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 28th 2007, 03:51 PM

Search Tags


/mathhelpforum @mathhelpforum