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Cauchy sequence & Fixed point

  1. Feb 24, 2010 #1
    Cauchy sequence & "Fixed" point

    1. The problem statement, all variables and given/known data
    Suppose that f: Rd->Rd and there is a constant c E (0,1) such that
    ||f(x)-f(y)|| ≤ c||x-y|| for all x, y E Rd. Let xo E Rd be an arbitrary point in Rd, let xn+1=f(xn). Prove that
    a) f is continuous everywhere.
    b) (xn) is Cauchy.
    c) (xn) converges to a limit y.
    d) Show that y is a fixed point of f ,that is f(y)=y, and that f has exactly one fixed point.

    2. Relevant equations
    3. The attempt at a solution
    I proved part a & part b, but I have no idea how to prove parts c & d.
    Any help is appreciated!
  2. jcsd
  3. Feb 24, 2010 #2
    Re: Cauchy sequence & "Fixed" point

    Isn't (c) obvious? Since xn is Cauchy and Rd is complete,then xn will converge at some point

    The first part of (d):Just limit both sides and you will get y=f(y)
    Second part:
    Suppose that there is another z such that z=f(z)
    Then [tex]\[
    \left. {\left\| {f(y) - f(z)} \right.} \right\| = \left\| {\left. {y - z} \right\|} \right. \le c\left\| {\left. {y - z} \right\|} \right.
    But since 0<c<1,we conclude that [tex]\[
    \left\| {\left. {y - z} \right\|} \right. = 0
    Thus y=z
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