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Math Help - contraction mapping theorem

  1. #1
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    contraction mapping theorem

    I need to show f:[0,1]->[0,1] is a contraction mapping. where g is continous map form [0,1] to itself
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  2. #2
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    Re: contraction mapping theorem

    The definition of "contraction mapping" is that |f(a)- f(b)|< |a- b|. A "strict contraction" would have |f(a)- f(b)|<= c|a- b| where c< 1. Can you show either of those? Use the fact that |sin(x)|<= 1 of course.
    Last edited by HallsofIvy; December 21st 2012 at 06:46 AM.
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  3. #3
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    Re: contraction mapping theorem

    Yes I'm looking for the latter strict contraction as you say. ok so |F(x)-F(y)|=|sinx|||x^{0.5}-g(x)|-|y^{0.5}-g(y)| which is less than or equal to [x^{0.5}-g(x)|+|y^0.5-g(y)| by triangle inequality and |sinx|<=1. Hmm tex doesn't seem to be working. Hope you can read it
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    Re: contraction mapping theorem

    It is not enough to use the fact that |sin(x)| <= 1. Suppose g(x) = 0. Then sqrt(x) is not contracting.
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    Re: contraction mapping theorem

    can you help? Maybe use domain and co-domain of the functions?
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  6. #6
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    Re: contraction mapping theorem

    Let g(x) = 0, i.e., f(x) = sin(x) * sqrt(x). Then f'(x) exceeds 1 between 0.7 and 0.8. By the mean value theorem, f(0.8) - f(0.7) > 0.1, so f(x) is not a contraction.

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  7. #7
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    Re: contraction mapping theorem

    look at post 2.
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  8. #8
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    Re: contraction mapping theorem

    Quote Originally Posted by Plato13 View Post
    look at post 2.
    I have looked. What are you saying?
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