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Prove the Contraction Mapping Theorem.

  1. Feb 8, 2014 #1
    Prove the Contraction Mapping Theorem.

    Let ##(X,d)## be a complete metric space and ##g : X \rightarrow X## be a map such that ##\forall x,y \in X, d(g(x), g(y)) \le \lambda d(x,y)## for some ##0<\lambda < 1##.Then ##g## has a unique fixed point ##x^* \in X ##, and it attracts everything, i.e. for any ##x_0 \in X## , the sequence of iterates ##x_0, g(x_0), g(g(x_0))##, ... converges to the fixed point ##x^* \in X##.

    The hint I am given are for existence and convergence - prove that the sequence is Cauchy. For uniqueness, choose two fixed points of ##g## and apply the map to both.

    My approach is After we got to ##d(x_{n+1}, x_n) \le \lambda^n d(x_1, x_0)##,

    then assuming that ##m > n##, ##d(x_m, x_n) \le d(x_m, x_{m-1}) + d(x_{m-1}, x_{m-2}) + ... + d(x_{n+1}, x_{n})## since each term of the right hand size is 0 ...so if we add up all the 0 terms, we get 0 on the right hand size. Therefore, ##d(x_m, x_n)## is 0. If this is right, then the question I have here is how do I guarantee that ##x_{m-1} > x_n##?
     
  2. jcsd
  3. Feb 9, 2014 #2

    Dick

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    Assuming you are trying to prove the sequence is Cauchy and that ##x_n=g^n(x_0)##, why do you think the terms on the right side are zero??
     
  4. Feb 9, 2014 #3
    I realize my reasoning a little bit off since the difference between m and n could be infinity , is any difference way to argue that the sum of the right hand side are zeros?
     
  5. Feb 9, 2014 #4

    Dick

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    They aren't zeros. Try to bound it with a geometric series.
     
  6. Feb 9, 2014 #5
    Use geometric in the right hand side and could you show a step or two for that?
     
  7. Feb 9, 2014 #6

    Dick

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    You've got ##d(x_{n+1}, x_n) \le \lambda^n d(x_1, x_0)##. Use it!
     
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