1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    HallsofIvy

    User Avatar
    Science Advisor

    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.

     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook