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Uniform convergence of series of functions

  1. Sep 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Is the series [tex]\sum \frac{x^n}{1 + x^n}[/tex] uniformly convergent on [tex][0,1)[/tex] ?


    2. Relevant equations



    3. The attempt at a solution
    I'm guessing that it is NOT uniformly convergent but I don't know how to show it. But I would note that it is quite easy to show that the series is uniformly convergent on [tex][0,a][/tex] for [tex]\forall a, 0 < a < 1[/tex] (i.e. straightforward application of the Weierstrass M-Test).

    I'm aware of the following fact is useful to check whether a series of functions is "not" uniformly convergent. We know that if [tex]\sum g_k[/tex] is uniformly convergent on [tex]S \subseteq \mathbb{R}[/tex], then [tex]\lim_{n \to \infty} \sup \{|g_k(x)| : x \in S\} = 0[/tex].

    Hence motivated by this fact, a good trick would be to use derivatives to find the maximum. But in this case, it doesn't work so well. For convienence, set [tex]g_n(x) = \frac{x^n}{1 + x^n}[/tex] and in this case we can drop the absolute value since [tex]x \in [0,1)[/tex]. And if my differentation is correct, we should have [tex]g_n'(x) = \frac{nx^{n-1}}{(1 + x^n)^2}[/tex] and set this to zero to find the maximum / minimum and we see that [tex]g_n'(x) = 0[/tex] implies [tex]x = 0[/tex] --- and I'm very certain that [tex]x = 0[/tex], so that [tex]g_n(x) = 0[/tex], is not the maximum of [tex]g_n[/tex], and in fact it is clear that it is the minimum. Thus, the above attempt to find the supremum doesn't work so well.

    I also thought about using the Cauchy criterion for uniform convergence, but there does not seem to be any "convenient" indicies for me to choose to show that it fails the criterion.


    Any help is greatly appreciated!
     
  2. jcsd
  3. Sep 11, 2009 #2

    AKG

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    Science Advisor
    Homework Helper

    Given k, what is the set:

    [tex]\sup \{|g_k(x)| : x \in [0,\, 1)\}[/tex]

    Find out what that set is, then it's easy to compute its sup, and then computing the lim of those sups is going to be easy, and your proof will be done.
     
  4. Sep 11, 2009 #3
    Thanks for the reply. I thought a little bit more on the problem yesterday and I think I made it way too complicated in my previous attempt.

    It's simply as follows. The number is [tex]\sup \{|\frac{x^n}{1 + x^n}| : x \in [0,1)\} = \frac{1}{2}[/tex] and this is seen by setting [tex]x = 1[/tex]. And of course, if that is equal to 1/2, then its limit is still 1/2 and thus, this series cannot be uniformly convergent.
     
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