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Contraction Mapping Theorem Question

  1. May 3, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider the function g : [0,∞) → R defined by g(x) = x + e−2x.

    Given |g(x2) − g(x1)| < |x2 − x1| for all x1, x2 ∈ [0,∞) with x1 ≠ x2.

    Is g a contraction on [0,∞)? Why?

    2. Relevant equations

    I think we are intended to use the given equation and the CMT

    CMT states that |f(x)-f(y)|=c|x-y|, where x,y∈R.

    Which is similar to |g(x2) − g(x1)|<|x2 − x1|

    3. The attempt at a solution

    I really don't know where to begin with this; I'm not simply after the answer though, just a prod in the right direction would be great
  2. jcsd
  3. May 3, 2010 #2
    On my computer, the formula for g(x) displays an unreadable symbol before the e-2x. Also, I confess that I have only seen CMT from browsing my friend's analysis book, and it doesn't exactly say
    Are they equivalent?

    EDIT: The actual statement I read is at http://en.wikipedia.org/wiki/Banach_fixed_point_theorem
    Last edited: May 3, 2010
  4. May 3, 2010 #3
    g is a contraction precisely when there is some positive real number c < 1 with

    [tex]|g(y) - g(x)| < c |y-x|[/tex]

    for all x,y in the domain.

    You have to either find a positive c < 1 that satisfies the inequality above for all [tex]x,y \in [0,\infty)[/tex] or show no positive c < 1 exists which satisfies the inequality. Note that

    [tex]|g(y) - g(x)| < |y-x|[/tex]

    does not show g is a contraction. The constant 0 < c < 1 really is needed (check the proof of the contraction mapping theorem to see why if you're interested; it's a really cool piece of mathematics).
    Last edited: May 3, 2010
  5. May 3, 2010 #4


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    No, it doesn't. That is simply the definition of "contraction map". The CMT states that if f is a contraction map that take set S into itself, then it has exactly one fixed point in S.

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