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Math Help - Real Analysis - Sequences

  1. #1
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    Real Analysis - Sequences

    Hey, I'm pretty sure I already solved part a. of this question, but decided to put it up just in case. Any help would be greatly appreciated. Thanks.

    Let fn(x) = nx / (1 + nx^2)

    a. Find the pointwise limit of (fn) for all x Є (0, infinity).
    b. Is the convergence uniform on (0, infinity)?
    c. Is the convergence uniform on (0,1)?
    d. Is the convergence uniform on (1, infinity)?
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  2. #2
    Member TheMasterMind's Avatar
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    Quote Originally Posted by ajj86 View Post
    Hey, I'm pretty sure I already solved part a. of this question, but decided to put it up just in case. Any help would be greatly appreciated. Thanks.

    Let fn(x) = nx / (1 + nx^2)

    a. Find the pointwise limit of (fn) for all x Є (0, infinity).
    b. Is the convergence uniform on (0, infinity)?
    c. Is the convergence uniform on (0,1)?
    d. Is the convergence uniform on (1, infinity)?
    Note that for x≥1, f_n(x) = nx/(1+nx^2) < nx/(nx^2) = 1/x ≤ 1 so each of the f_n is bounded above by 1 on [1, infinity) so since f_n is continuous on [0, 1] and [0, 1] is compact we have by the extreme value theorem that f_n(x) is bounded above on [0, 1] taking the greater of 1 and the upper bound for f_n(x) on [0, 1], upper bound for f_n(x) on [0, infinity]. Therefore each one of the f_n is bounded above on [0, infinity].

    to find the pointwise limit of the f_n, we see that for x>0

    f_n(x)

    = \frac{nx}{1+nx^2}

    = \frac{1}{(1/(nx) + x)}

    this approaches 1/x as n\forwardarrow\infinity now for every n, f_n(0) = 0, we have

    n---->infinity lim f_n(x) = {0 if x=0, 1/x if x≠0}

    this shows it is not bounded above on [0, infinity]. It follows that the f_n cannot converge uniformly
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