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Uniform Convergence

  1. Oct 9, 2012 #1
    I'm wondering about uniform convergence. We're looking at it in my complex analysis class. We are using uniform convergence of a series of functions, to say that we can interchange integration of the sum, that is: [itex]\int\sum b_{j}z^{j}dz[/itex]=[itex]\sum\int b_{j}z^{j}dz[/itex]=[itex]\int f(z)dz[/itex]

    On an intuitive level I don't understand why uniform convergence is necessary. I figured that since the integral is linear this is trivial. I was wondering if someone could explain this to me. Maybe elaborate on what can break down, so that they aren't equal if [itex]\sum b_{j}z^{j}[/itex] doesn't uniformly converge to [itex]f(z)[/itex]
  2. jcsd
  3. Oct 10, 2012 #2


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    Think about this simple example of what non-uniform convergence can do- fj(x) is 0 if x< 0, x/n if [itex]0\le x\le 1/n[/itex], and 1 if x> n. fj is continuous for all j but the lim of the sequence, f(x)= 0 if [itex]x\le 0[/itex], 1 if x> 0 is not continuous at x= 0.

    As for your remark about the integral being linear: that would tell you that
    [tex]\int\sum f_j(x) dx= \sum \int f_j(x)dx[/tex]
    for any finite sum, not for infinite sums. Both the integral and infinite sum are defined in terms of limits and it is "uniformity" that allows us to swap limits.
  4. Oct 10, 2012 #3


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    Dearly Missed

    To follow up on Halls' comment:
    The limit of an infinite sum might, for example, be a non-integrable function, if the convergence is not uniform of the partial sums.
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