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Homework Help: Integral Convergence Theorem (limit of integrals)

  1. Apr 13, 2010 #1
    1. The problem statement, all variables and given/known data
    ra27.JPG

    2. Relevant equations
    3. The attempt at a solution
    I think it is related to the integral convergence theorem.
    ra24.JPG
    But how can we deal with the function g(x)? Is this problem supposed to be a direct use of the theorem or do we have to start the proof from scratch? How should we begin the proof?

    Any help is appreciated!
     
  2. jcsd
  3. Apr 13, 2010 #2

    Gib Z

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    You can use the Integral convergence theorem right after you show t_n(x) = f_n(x)g(x) converges uniformly to f(x)g(x).
     
  4. Apr 13, 2010 #3
    OK, so we need to show that sup|f_n(x)g(x)-f(x)g(x)|->0.

    sup|f_n(x)g(x)-f(x)g(x)|‚ȧsup|f_n(x)-f(x)|+sup|g(x)|

    How can we deal with sup|g(x)|?
     
  5. Apr 13, 2010 #4

    Gib Z

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    Sorry, how did you get the second line? Shouldn't the Right hand side be the product of those two terms, not the sum.
     
  6. Apr 13, 2010 #5
    oops...you're right.
    |f_n(x)g(x)-f(x)g(x)|=|f_n(x)-f(x)||g(x)|

    How can we show sup|f_n(x)g(x)-f(x)g(x)|->0 then?
     
    Last edited: Apr 14, 2010
  7. Apr 13, 2010 #6

    Gib Z

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    Look at both the terms on the right hand side. The first you know something about f_n and f to help you, and the extreme value theorem says that since g is an element of C[0,1], it attains a maximum value in that interval as well.
     
  8. Apr 13, 2010 #7
    OK, that makes sense. But how can we handle the supremum? Will our bounds be preserved after taking the supremum? Why or why not?
     
  9. Apr 13, 2010 #8

    Gib Z

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    The extreme value theorem states g(x) attains both a maximum and a minimum in [0,1]. Do you understand why sup |g(x)| = max{ |min g(x)| , |max g(x)| } ?

    Which ever one of them it is, let it be a constant A.

    Since f_n converges uniformly to f, you know that for a chosen positive chosen, let it be [itex]\epsilon / A[/itex], then [itex]|f_n (x) - f(x) | < \epsilon / A[/itex] for sufficiently large values of n.

    Put this together, theres not much left to finish the question.
     
  10. Apr 13, 2010 #9
    But is it necessarily true that
    sup{|f_n(x)f(x)||g(x)|} = sup{|f_n(x)f(x)|} sup{|g(x)|} ??
     
  11. Apr 13, 2010 #10

    Gib Z

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    No, but you have learned that for two sets A and B, and their product set AB

    sup AB is less or equal to sup A sup B. You can use that to finish this question.
     
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