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Math Help - Contraction Functions

  1. #1
    Super Member Aryth's Avatar
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    Contraction Functions

    I just learned about these yesterday, so I'm having a little trouble with this problem.

    If T: [0,1] to [0,1] and if there is a real number alpha with 0<= alpha < 1 such that |T'(x)| <= alpha (0 <= x <= 1) where T' is the derivative of T, prove that T is a contraction on [0,1].

    Any help would be appreciated!
    Last edited by Aryth; February 22nd 2014 at 02:04 PM.
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  2. #2
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    Re: Contraction Functions

    Quote Originally Posted by Aryth View Post
    If T: [0,1] to [0,1] and if there is a real number alpha with 0<= alpha < 1 such that |T'(x)| <= alpha (0 <= x <= 1) where T' is the derivative of T, prove that T is a contraction on [0,1].
    Please review what you have posted. The reason being that is the usual definition of a Contraction Function.
    We don't prove definitions.
    If your text material give a different definition please post it.

    What is the exact wording of the question?
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  3. #3
    Super Member Aryth's Avatar
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    Re: Contraction Functions

    Quote Originally Posted by Plato View Post
    Please review what you have posted. The reason being that is the usual definition of a Contraction Function.
    We don't prove definitions.
    If your text material give a different definition please post it.

    What is the exact wording of the question?
    That is the exact wording of the question. O.O

    The definition I've been given is:

    Let <M,p> be a metric space with a function T: M to M. We say that T is a contraction on M if there exists an alpha in R with 0 <= alpha < 1 such that p(Tx,Ty) <= alpha*p(x,y) (x,y in M) with alpha independent of x and y.
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  4. #4
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    Re: Contraction Functions

    Quote Originally Posted by Aryth View Post
    That is the exact wording of the question.
    The definition I've been given is:
    Let <M,p> be a metric space with a function T: M to M. We say that T is a contraction on M if there exists an alpha in R with 0 <= alpha < 1 such that p(Tx,Ty) <= alpha*p(x,y) (x,y in M) with alpha independent of x and y.
    Then you are given that:
    Quote Originally Posted by Aryth View Post
    If T: [0,1] to [0,1] and if there is a real number alpha with 0<= alpha < 1 such that |T'(x)| <= alpha (0 <= x <= 1) where T' is the derivative of T, prove that T is a contraction on [0,1].
    So what is there to prove?
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  5. #5
    Super Member Aryth's Avatar
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    Re: Contraction Functions

    That's the problem, I suppose... I don't see how making the derivative bounded gives me that T is a contraction by definition...
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