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How to show a series of functions is not uniformly convergent

  1. Sep 13, 2009 #1
    Ok, I chose to ask about ways to determine if a series of functions is NOT uniformly convergent because I think that would best answer the overall difficulties I have with uniform convergence. I have a good idea of what uniform convergence is, I can give the definition, and if the problem is uniformly convergent there is a good chance I can show it. But if the problem is not then I don't think that "None of my tests work" is a rigorous mathematical argument to deny uniform convergence and none of the tests I have give explicit conditions for non-uniformity.

    1. The problem statement, all variables and given/known data
    Here is just one example. Show whether [tex]\sum \sin( \frac{x}{n^2})[/tex] converges uniformly for the interval "All x".


    2. Relevant equations
    The first time around I showed that this series was convergent by comparing it to [tex]\sum\frac{x}{n^2}[/tex] and was thus able to show that it was absolutely convergent for all x.
    [tex]\sum \left|\sin( \frac{x}{n^2})\right|\leq\sum\frac{x}{n^2}[/tex]

    3. The attempt at a solution
    Now that I want to prove uniform convergence I'm wanting to do the same thing except I need to make the comparing series not dependent on x, make it a series of constants (which is essentialy the Weierstrass test, correct?). Unfortunately the way I see it, no matter what x I choose to make [tex]\sum\frac{x}{n^2}[/tex] a constant, there will always be a greater x that will ruin the M-Test inequality. It is the connection from here to non-uniform convergence which I can't seem to make.

    I appreciate any helpful comments. I also have a few other confusions which seem to result from the interval to be checked. Particularly when it is an open interval where the series of functions is undefined at either endpoint.
     
  2. jcsd
  3. Nov 8, 2011 #2
    Oh I wanna ask the same thing. How can I prove something is *not* uniformly convergent?

    Too bad there's no answer here. :confused:
     
  4. Nov 8, 2011 #3
    Last edited by a moderator: Apr 26, 2017
  5. Nov 8, 2011 #4
    Limn→∞sin(x/n2) = 0 for any fixed x.

    Given any n can you find x s.t. |sin(x/n2) - 0| > 1/2? I'm letting epsilon = 1/2.
     
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