1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Uniform convergence

  1. Feb 14, 2010 #1
    1. The problem statement, all variables and given/known data

    fn(x) = 0 if x [tex]\leq[/tex] n
    = x-n if x[tex]\geq[/tex] n

    Is fn(x) uniformly convergent on [a,b]?
    Is it uniformly convergent on R?

    2. Relevant equations



    3. The attempt at a solution

    I think limn -> infinity fn(x)= 0
    However I do not know what supx in [a,b] |fn(x)| would be, I think it could be 0 but I don't know how to show this.

    I'm unsure about over R as well?
     
  2. jcsd
  3. Feb 14, 2010 #2
    Have you looked at a couple of specific functions?

    On an interval [a,b], the sequence f_n is eventually constant, isn't it?
     
  4. Feb 14, 2010 #3
    I have maybe shown that it is uniformly convergent on the interval [a,b] but I still don't know about over the real numbers. My instinct is that it isn't but I'm not sure how to show this.
     
  5. Feb 14, 2010 #4
    It is not.

    What is your definition of uniform convergence? Here's the one I learned: A sequence of functions f_n converges to f uniformly if for any eps>0, there exists an N such that |f_n(x) - f(x)|< eps for all n>N and x in the domain of f.

    To show that a sequence does not converge uniformly (after showing it does converge pointwise), you could give yourself an arbitarily large n and show that |f_n(x) - f(x)| > 1 for some x depending on n.
     
  6. Feb 14, 2010 #5
    I've been using:

    f_n converges to f uniformly if there exists an N, for all n>N, m_n := sup|f_n(x) - f(x)| exists and tends to 0 as n tends to infinity.

    I thought that on the interval [a,b] f_n(x) and f(x) would equal 0 as n tends to infinity?
     
  7. Feb 14, 2010 #6
    Correct. Explicitly, if n > b, then f_n(x) = 0 for all x in [a,b].

    Let's start with "sup|f_n(x) - f(x)| exists". You can show that f_n -> 0 pointwise, correct? Pick your favorite number n. Does sup_x |f_n(x) - 0| exist?
     
  8. Feb 14, 2010 #7
    So, say n=10, then sup | f_10(x) - 0 | = sup | x - 10 | = b-10 ?
     
  9. Feb 14, 2010 #8
    the issue here is the set over which we are looking at the function. at some point n=N will be bigger than b so f will be zero for all x in this interval for all values of n>=N. hence uniform convergence. for the second one have you heard of the cauchy criterion for uniform convergence? look it up. notice that when you subtract different functions from the sequence they leave an integer remainder. this does not get arbitrarily small...
     
  10. Feb 14, 2010 #9
    Yes, but only if your function is restricted to an interval [a,b] with b > 10. What happens when you let your function vary over the entire real line? Is there an upper bound on the difference |f_10(x)-0| = x-10 if you let x get arbitrarily large?
     
  11. Feb 14, 2010 #10
    Thanks, I think I understand now :)
     
  12. Feb 14, 2010 #11
    also the sequence x-n is simply a shifting down the y axis of the function y=x. also it intersects the x-axis at x=n. this sequences is essentially going move out at a constant velocity in the x direction. these lines continue out to infinity without getting arbitrarily close to one another nor close to any concievable function.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook