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Math Help - Show That a Function is Contractive

  1. #1
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    Question Show That a Function is Contractive

    Question:
    Show that the following function is contractive on the indicated intervals. Determine the best values of [lamda] in Equation (2).

    abs(x)^(2/3) on abs(x) < or = 1/3

    Equation (2):
    abs(F(x)-F(y)) < or = [lamda]*abs(x-y)

    Any help with this problem would be greatly appreciated! Thanks!
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  2. #2
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    Quote Originally Posted by theamazingjenny View Post
    Question:
    Show that the following function is contractive on the indicated intervals. Determine the best values of [lamda] in Equation (2).

    abs(x)^(2/3) on abs(x) < or = 1/3

    Equation (2):
    abs(F(x)-F(y)) < or = [lamda]*abs(x-y)

    Any help with this problem would be greatly appreciated! Thanks!
    I do not think this function is contractive.

    Assume that it is,
    |F(x)-F(y)| <= K*|x-y| for -1/3<=x,y<=1/3
    Take y=0 to get,
    |F(x)|<= K*|x| for -1/3<=x<=1/3
    Then certainly,
    |F(x)|<= K*|x| for 0<x<=1/3
    Thus,
    x^{2/3} <= K*x for 0<x<=1/3
    But then,
    x^{-1/3} <= K
    Then certainly
    x^{-1/3} <= K+3
    But this is false if you choose,
    x>1/(K+3)^3
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  3. #3
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    Quote Originally Posted by theamazingjenny View Post
    Question:
    Show that the following function is contractive on the indicated intervals. Determine the best values of [lamda] in Equation (2).

    abs(x)^(2/3) on abs(x) < or = 1/3

    Equation (2):
    abs(F(x)-F(y)) < or = [lamda]*abs(x-y)

    Any help with this problem would be greatly appreciated! Thanks!
    A simpler demonstration that this is not a contraction is obtained by putting
    x=1/8, and y=0, then:

    abs( abs(x)^2/3 - abs(0)^2/3) = 1/4 > x - y = 1/8

    RonL
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