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Math Help - Uniqueness in a contraction

  1. #1
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    Uniqueness in a contraction

    Given a complete metric space X and a function f: X->X with a Lipschitz constant of r on (0,1) (i.e contracting function) show that there is a unique point f(x) = x
    Use a sequence with arbitrary x0 and xn=f(xn-1)
    to get x as limit of the sequence.

    How do I get the limit and how do i use this in proving that f(x)=x?
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  2. #2
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    Quote Originally Posted by twerdster View Post
    Given a complete metric space X and a function f: X->X with a Lipschitz constant of r on (0,1) (i.e contracting function) show that there is a unique point f(x) = x
    Use a sequence with arbitrary x0 and xn=f(xn-1)
    to get x as limit of the sequence.

    How do I get the limit and how do i use this in proving that f(x)=x?
    If you define x_n= f(x_{n-1}), starting with chosen x_0, then d(x_n, x_{n+1})= d(f(x_{n-1}, df(x_n))\le rd(x_{n-1}, x_n). Use induction to prove that d(x_{n+1},x_n)\le r^nd(x_1, x_0).

    That isn't quite enough to show that this is a Cauchy sequence. For a Cauchy sequence, you need to show that d(x_n, x_m) goes to 0 as n and m both go to infinity independently. Use the fact that, for n> m, d(x_n, x_m)\le d(x_n, x_{n-1})+ d(x_{n-1}, x_m)\le d(x_n, x_{n-1})+ d(x_{n-1},x_{n-2})+ d(x_{n-2},x_m), etc.

    To prove uniqueness, suppose that f(x)= x and f(y)= y. Then d(x, y)= d(f(x), f(y))\le rd(x, y)
    Last edited by Plato; May 18th 2010 at 06:57 AM. Reason: LaTex
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    If you define x_n= f(x_{n-1}), starting with chosen x_0, then d(x_n, x_{n+1})= d(f(x_{n-1}, df(x_n))\le rd(x_{n-1}, x_n). Use induction to prove that d(x_{n+1},x_n)\le r^nd(x_1, x_0).

    That isn't quite enough to show that this is a Cauchy sequence. For a Cauchy sequence, you need to show that d(x_n, x_m) goes to 0 as n and m both go to infinity independently. Use the fact that, for n> m, d(x_n, x_m)\le d(x_n, x_{n-1})+ d(x_{n-1}, x_m)\le d(x_n, x_{n-1})+ d(x_{n-1},x_{n-2})+ d(x_{n-2},x_m), etc.

    To prove uniqueness, suppose that f(x)= x and f(y)= y. Then d(x, y)= d(f(x), f(y))\le rd(x, y)
    Adding on, to prove that if \xi=\lim_{n\to\infty}x_n then f(\xi)=\xi we note that since the sequence coverges so does every subsequence to the same value and f(\xi)=f\left(\lim_{n\to\infty}x_n\right)\overset{  {\color{red}(*)}}{=}\lim_{n\to\infty}f(x_n)=\lim_{  n\to\infty}x_{n+1}=\xi the red starrred part gotten by continuity (every contraction is continuous)
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  4. #4
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    Why is every contraction necessarily continuous?
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by twerdster View Post
    Why is every contraction necessarily continuous?
    Because every Lipschitz function is uniformly continuous and thus continuous.

    For a fixed x_0\in X you can make d(f(x_0),f(y))<\varepsilon by just taking \delta=\varepsilon. For then, d(x_0,y)<\delta=\varepsilon\implies d(f(x_0),f(y))=rd(x,y)<d(x,y)<\delta=\varepsilon
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