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Uniform convergence

  1. Mar 26, 2013 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    Suppose that ##s_n(x)## converges uniformly to ##s(x)## on ##[b, ∞)##.

    If ##lim_{x→∞} s_n(x) = a_n## for each n and ##lim_{n→∞} a_n = a## prove that :

    ##lim_{x→∞} s(x) = a##

    2. Relevant equations

    ##\space ε/N##

    3. The attempt at a solution

    I see a quick way to do this one using first principle definitions.

    ##\forall ε>0, \exists N(ε) \space | \space n > N(ε) \Rightarrow |s_n(x) - s(x)| < ε/3, \space \forall x \in [b, ∞)##

    ##\forall ε>0, \exists N_1 \space | \space x > N_1 \Rightarrow |s_n(x) - a_n| < ε/3##

    ##\forall ε>0, \exists N_2 \space | \space n > N_2 \Rightarrow |a_n - a| < ε/3##

    We want to prove :

    ##\forall ε>0, \exists N \space | \space x > N \Rightarrow |s(x) - a| < ε##

    So :

    ##|s(x) - a| = |s(x) - s_n(x) + s_n(x) - a_n + a_n - a| ≤ |s_n(x) - s(x)| + |s_n(x) - a_n| + |a_n - a| < ε/3 + ε/3 + ε/3 = ε##

    Does this look okay?
     
  2. jcsd
  3. Mar 26, 2013 #2

    Dick

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    Almost. You should have an N1, N2, N3 in the first three statements. Then you just want to say how your final N is related to those three N's.
     
  4. Mar 26, 2013 #3

    Zondrina

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    Ah, I see. I thought being explicit about ##N(ε)## would make the context clear.

    So I should also say choosing ##N = max\{N_1, N_2, N_3\}##?
     
  5. Mar 26, 2013 #4

    Dick

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    Yes, that's it. But now that I've stared at it for a bit there is a second problem. The ##N_2## is also going to depend on n. There's a bit more work to do. You have to choose an ##N_2## independent of n.
     
    Last edited: Mar 26, 2013
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