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Math Help - Pointwise and uniform convergence.

  1. #1
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    Pointwise and uniform convergence.

    Hi I've got some questions with answers I'm working through but can't understand some of the points and was hoping someone could explain them to me!
    The question was
    Let fn(x) = nx/(1+n^2x^2)

    (i) Prove that (fn) converges pointwise on R and find the limit function.
    (ii) Is the convergence uniform on [0, 1]?
    (iii) Is the convergence uniform on [1,1)?

    The first part I'm fine with and get the limit function being fn \to 0 as I think fn \leqslant 1/n|x|

    For the second part I said no as I thought Sup fn was just a half. In my answers it says Sup |fn(x)| \geqslant  fn(1/n) \to1/2
    which is kind of what I got but don't see where they get the first part from.

    For iii) the answers have sup |fn(x)| \leqslant1/n \to 0 and again while I felt it would tend to 0 I don't understand the reasoning!

    Sorry if this seems stupid but I just can't see the jump in the answers to where they got what they did! If anyone could explain I'd be very grateful!
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  2. #2
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    Quote Originally Posted by shmounal View Post
    Hi I've got some questions with answers I'm working through but can't understand some of the points and was hoping someone could explain them to me!
    The question was
    Let fn(x) = nx/(1+n^2x^2)

    (i) Prove that (fn) converges pointwise on R and find the limit function.
    (ii) Is the convergence uniform on [0, 1]?
    (iii) Is the convergence uniform on [1,1)? Shouldn't that be [1,∞)?

    The first part I'm fine with and get the limit function being fn \to 0 as I think fn \leqslant 1/n|x| Correct, except that the case when x=0 should be dealt with separately, because 1/|x| is not defined then.

    For the second part I said no as I thought Sup fn was just a half. In my answers it says Sup |fn(x)| \geqslant  fn(1/n) \to1/2
    which is kind of what I got but don't see where they get the first part from. Your answer is correct, but it's probably best to include the fact that the sup is attained when x=1/n.

    For iii) the answers have sup |fn(x)| \leqslant1/n \to 0 and again while I felt it would tend to 0 I don't understand the reasoning! [Assuming that the 1 should be ∞, as noted above.] The function \color{blue}f_n(x) is decreasing on the interval [1,∞) (because its derivative is negative), so its max value occurs at the start of the interval, when x=1, and \color{blue}f_n(1) = n/(1+n^2) < 1/n.
    ..
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  3. #3
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    Reaching for f(x) = 0.

    Pick an arbitrary value, a.

    Prove that everywhere on the interval of interest that we can get closer to f(x) = 0 than 'a'.

    For example, using a = 0.01, if I choose |x| > 0.01, I see that if I now select n > 9999, everything on |x| > 0.01 is closer to f(x) = 0 than 'a'. Now, notice that 'a' is arbitrary. Because you can choose what you want, this leads to the completed proof. Of course, you should do it with 'a' and not 0.01.
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  4. #4
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    That's awesome, thank you both for your help. Hopefully a question like that will come up tomorrow!
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