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Pointwise Limits and Uniform Convergence Help

  1. Apr 23, 2010 #1
    Hi all
    Came across this question which i have attempted to answer but am sure i have gone wrong somewhere, any help would be appreciated...

    Suppose fn(x):R-->R, where fn(x)=x+(1/n) (n belongs to N). Find the pointwise limits of the sequences (fn), (1/n fn) and (fn2). In each case determine whether the convergence is uniform.

    Now i managed to get all three as being uniformly convergent and am sure i went wrong somewhere. I am still struggling to understand uniform convergence completely. The pointwise limits i computed were x, 0, x2 respectively...sorry i havnt put my working up struggling to use latex.
     
  2. jcsd
  3. Apr 23, 2010 #2
    Your limits are right. (Let [tex]f[/tex] be the limit of [tex]f_n[/tex]). The first sequence is uniformly convergent: given [tex]\epsilon >0[/tex], choose N to be the lowest integer that is greater than:[tex]\frac{1}{\epsilon}[/tex].

    Then, for all [tex]n>N[/tex]:
    [tex]|f-f_n|=|x-(x+\frac{1}{n})|=\frac{1}{n}<\frac{1}{N}<\epsilon[/tex].

    For the second one, note that [tex]sup_x|f_n|[/tex] does not go to 0. This tells you it is not uniformly convergent. To prove this, suppose we are given [tex]\epsilon >0[/tex]. For any choice of N, let [tex]x>\epsilon .N-\frac{1}{N}[/tex].
    Then:

    [tex]|f-f_N|=|0-\frac{x}{N}-\frac{1}{N^2}|=|\frac{x}{N}+\frac{1}{N^2}|>\frac{\epsilon .N-\frac{1}{N}}{N}+\frac{1}{N^2}=\epsilon[/tex].

    The third one is very similar. I'm afraid all this might be a bit bewildering if you don't yet understand uniform convergence though! Keep thinking about it and eventually it will click.

    PS - I may have made some mistakes here as I did it very quickly, so be sure to check everything!
     
    Last edited: Apr 23, 2010
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