Results 1 to 5 of 5

Math Help - proving pointwise convergence for piecewise function

  1. #1
    Senior Member sfspitfire23's Avatar
    Joined
    Oct 2009
    Posts
    273

    proving pointwise convergence for piecewise function

    Show that the sequence {f_n} defined by

    f_n(x) = \left\{<br />
\begin{array}{c l}<br />
  n & x\geqslant n \\<br />
  0 & x<n<br />
\end{array}<br />
\right.<br />



    converges pointwise on \mathbb{R} to the zero function f(x)=0

    I know that I can pick any N such that when x < n and n > N i can get |f_n(x) - f(x)| = |f_n(x)| < epsilon. But I'm not sure how to show the convergence for x >orequalto n...
    Last edited by sfspitfire23; May 7th 2010 at 11:21 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by sfspitfire23 View Post
    Show that the sequence {f_n} defined by
    f_n(x) = n if x >orequalto n
    = 0 if x < n

    converges pointwise on R to the zero function f(x)=0

    I know that I can pick any N such that when x < n and n > N i can get |f_n(x) - f(x)| = |f_n(x)| < epsilon. But I'm not sure how to show the convergence for x >orequalto n...
    Pointwise convergence means that you can pick the x before the N. So given a fixed x all you have to do is to take N > x. Then, for that x, f_n(x) will be 0 for all n>N.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member sfspitfire23's Avatar
    Joined
    Oct 2009
    Posts
    273
    So, Let \epsilon >0 and fix x\in \mathbb{R} . Then, take N>x. Then we have f_n(x)=0, f(x)=0. So, for n>N we have |0-0|=0<\epsilon .

    How is this? Also, this convergence is NOT uniform, right?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Quote Originally Posted by sfspitfire23 View Post
    So, Let \epsilon >0 and fix x\in \mathbb{R} . Then, take N>x. Then we have f_n(x)=0, f(x)=0. So, for n>N we have |0-0|=0<\epsilon .

    How is this? Also, this convergence is NOT uniform, right?

    Thanks!
    Correct and correct!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    May 2010
    Posts
    1

    wrong?

    Doesn't this violate the fact that n>N if you say N>x?

    Remember x≥n so
    N>x≥n which contradicts n>N.

    Right?
    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 15th 2012, 11:03 PM
  2. Replies: 14
    Last Post: October 7th 2011, 09:45 PM
  3. Function series pointwise convergence question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 28th 2009, 03:23 PM
  4. Pointwise convergence to uniform convergence
    Posted in the Calculus Forum
    Replies: 13
    Last Post: November 29th 2009, 08:25 AM
  5. Pointwise Convergence vs. Uniform Convergence
    Posted in the Calculus Forum
    Replies: 8
    Last Post: October 31st 2007, 05:47 PM

Search Tags


/mathhelpforum @mathhelpforum