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Math Help - Contraction mapping theorem

  1. #1
    Senior Member Danneedshelp's Avatar
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    Contraction mapping theorem

    Let f be a function defined on all of \mathbb{R}, and assume there is a constant c such that 0<c<1 and |f(x)-f(y)|\leq\\k|x-y| for all x,y\in\mathbb{R}.

    a) Show f is continuous on \mathbb{R}.

    Let c be a real number and \epsilon>0. Then |f(x)-f(c)|\leq\\k|x-c| by definition. So, let \delta=\frac{\epsilon}{k}. It follows that |x-c|<\delta implies |f(x)-f(c)|<k\frac{\epsilon}{k}=\epsilon.

    b) Pick some point y_{1}\in\mathbb{R} and construct the sequence

    (y_{1},f(y_{1}),f(f(y_{1})),...).

    In general, if y_{n+1}=f(y_{n}), show that the resulting sequence (y_{n}) is a Cauchy sequence. Hence, we may let y=limy_{n}.

    I am not sure how to approach this one. Some help getting started would be appreciated. It seems almost trivial, but I am not sure how formalize it.
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  2. #2
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    just write |f(f(x))-f(x)| < k|f(x)-x|, |f(f(f(x)))-f(f(x))| < k^2 |f(x)-x| and so on...
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  3. #3
    Senior Member Danneedshelp's Avatar
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    So in general |f(x)-f^{n}(x)|\leq\\k^{n}|f(x)-x|. I am not sure how this proves the sequence is Cauchy. Why did you choose f(x)-x in the expression k^{i}|f(x)-x|?

    Do I do something like this?

    |f^{n}(x)-f^{m}(x)|=|f^{n}(x)-f(x)+f(x)-f^{m}(x)|<br />
\leq\\|f^{n}(x)-f(x)|+|f(x)-f^{m}(x)|<br />
\leq\\(k^{n}+k^{m})|f(x)-x|

    How do I control the size of (k^{n}+k^{m})|f(x)-x|?
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  4. #4
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    Quote Originally Posted by Danneedshelp View Post
    So in general |f(x)-f^{n}(x)|\leq\\k^{n}|f(x)-x|. I am not sure how this proves the sequence is Cauchy. Why did you choose f(x)-x in the expression k^{i}|f(x)-x|?

    Do I do something like this?

    |f^{n}(x)-f^{m}(x)|=|f^{n}(x)-f(x)+f(x)-f^{m}(x)|<br />
\leq\\|f^{n}(x)-f(x)|+|f(x)-f^{m}(x)|<br />
\leq\\(k^{n}+k^{m})|f(x)-x|

    How do I control the size of (k^{n}+k^{m})|f(x)-x|?
    first off, you get |f^{n-1}(x)-f^{n}(x)|\leq\\k^{n-1}|f(x)-x|, not |f(x)-f^{n}(x)|\leq\\k^{n}|f(x)-x|. then let m>n and use triangle inequality: |f^m(x)-f^{n}(x)|\leq\\ \sum_{k=n}^{m-1} |f^{k+1}(x)-f^k(x)|<\ldots and u get geometric series. u can see this bound goes to zero if m and n go to infinity.
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