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A sequence of functions evaluated at a sequence

  1. Feb 24, 2015 #1
    What are the rules if you have a sequence [itex]f_n[/itex] of real-valued functions on [itex]\mathbb R[/itex] and consider the sequence [itex]f_n(x_n)[/itex], where [itex]x_n[/itex] is some sequence of real numbers that converges: [itex]x_n \to x[/itex]. All I have found is an exercise in Baby Rudin that says that if [itex]f_n \to f[/itex] uniformly on [itex]E[/itex], then [itex]f_n(x_n) \to f(x)[/itex] if [itex]x_n \to x[/itex] is in [itex]E[/itex]. But the exercise seems to indicate that it is possible to have [itex]f_n(x_n) \to f(x)[/itex] for every sequence [itex]x_n\to x[/itex] without having [itex]f_n \to f[/itex] uniformly. (I believe the canonical example [itex]f_n(x) = x^n[/itex] on [itex]E = [0,1][/itex] works here.)

    I ask because I recently had a colleague who claimed that if [itex]x_n \to x[/itex], then [itex] \left( 1 + \frac{x}{n} \right)^n \to e^x[/itex]. She asked what the rule that made this possible was, and I replied that I wasn't sure if it was, in fact, true, since [itex]f_n(x) = \left( 1 + \frac xn \right)^n[/itex] obviously does not converge uniformly to [itex]e^x[/itex] on [itex]\mathbb R[/itex] (even though, obviously, [itex]f_n \to e^x[/itex] pointwise).
     
  2. jcsd
  3. Feb 24, 2015 #2

    PeroK

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    For ##n## large enough:

    ##f_n(x_n)## is close to ##e^{x_n}##; and ##x_n## is close to ##x## so ##e^{x_n}## is close to ##e^x##
     
  4. Feb 24, 2015 #3
    So is it always true that if [itex]f_n \to f[/itex] pointwise and [itex]x_n \to x[/itex], then [itex]f_n(x_n) \to f(x)[/itex]?
     
  5. Feb 24, 2015 #4

    PeroK

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    Why don't you try to prove it?

    Hint: Does ##f_n(x)## converge uniformly to ##e^x## on any bounded interval?
     
    Last edited: Feb 24, 2015
  6. Feb 24, 2015 #5
    Ok, I think I see the strategy you are suggesting. Since [itex]\{ x_n \}[/itex] is bounded (since it converges), we have [itex]-M \leq x_n \leq M[/itex] for all [itex]n[/itex] and for some [itex]M[/itex]. I can restrict my attention to [itex][-M,M][/itex]. On that interval, [itex]\left( 1 + \frac xn \right)^n[/itex] converges uniformly to [itex]e^x[/itex], and I can apply the result of the Baby Rudin exercise.
     
  7. Feb 25, 2015 #6
    Here is what I have so far. I have decided to confine my attention to ##[0,\infty)##. Suppose ##f_n \to f## uniformly on a set ##E##, where each ##f_n## is continuous. (The latter is a hypothesis I inadvertently omitted from my previous posts.) Let ##\{ x_n \}## be a sequence of points in ##E## with ##x_n \to x##. Then, given ##\epsilon > 0##, there exists ##N \in \mathbb N## such that ##n > N## implies ##|f_n(x) - f(x)| < \epsilon/2## for all ##x\in E## and ##|f(x_n) - f(x)| < \epsilon/2##, since the limit function ##f## is continuous. Then for ##n > N##,

    ##|f_n(x_n) - f(x)| < |f_n(x_n) - f(x_n)| + |f(x_n) - f(x)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon##.

    The terms of ##\{ x_n \}## are bounded; suppose they are contained in an interval ##[-M,M]##. Then

    ##M_n = \sup_{x\in [-M,M]} \left| \left(1 + \frac{x}{n} \right)^n - e^x \right|##

    is a decreasing sequence bounded from below by ##0##; hence it converges to ##0##; hence ##\left( 1 + \frac{x}{n} \right)^n \to e^x## uniformly on ##[-M,M]##, which is what we wanted.
     
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