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Uniform convergence

  1. Nov 4, 2013 #1
    1. The problem statement, all variables and given/known data

    Is the sequence of function ##f_1, f_2,f_3,\ldots## on ##[0,1]## uniformly convergent if ##f_n(x) = \frac{x}{1+nx^2}##?

    2. The attempt at a solution

    I got the following but I think I did it wrong.

    For ##f_n(x) = \frac{x}{1+nx^2}##, I got if ##f_n \to0## then we must find ##\epsilon>0## an ##N## such that for ##n>N## implies ##|f_n-0|<\epsilon.## So ##f_n(x) = \frac{x}{1+nx^2}##; ##\lim_{n\to\infty}f_n(x) =0##. Then for ##\epsilon>0## we have ##|f_n(x)-f(x)| = |\frac{x}{1+nx^2}|\le |\frac{1}{1+n}|<|\frac{1}{n}|<\epsilon## thus ##N = \frac{1}{\epsilon}##. But I think this is wrong since ##|1+nx^2|<|1+n|##? How can I show it is uniformly convergent?
     
  2. jcsd
  3. Nov 4, 2013 #2

    jbunniii

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    ##f_n \rightarrow 0## uniformly if and only if ##\sup f_n \rightarrow 0##. So I would suggest that you start by finding the maximum value of ##f_n##.
     
  4. Nov 4, 2013 #3
    jbunniii -

    We haven't been taught that way yet. We still haven't defined "derivative" or proved ##f_n \to 0## uniformly iff ##\sup f_n \to 0## so I can't use it to prove my problem. What I was thinking is how can I divide into two regions?
     
  5. Nov 4, 2013 #4

    jbunniii

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    Consider the denominator: ##1 + nx^2##. This is not quite a perfect square, but it would be if we added the missing term (actually, subtracting is more useful here): ##1 - 2\sqrt{n} x + nx^2 = (1 - \sqrt{n}x)^2##. Now this is a square, so it is nonnegative. What can you conclude?
     
  6. Nov 4, 2013 #5
    Thanks, jbunniii for the hint!
     
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