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Contraction mapping theorem

  1. Apr 21, 2007 #1
    Here is the question from our book:
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    Let [itex](X,d)[/itex] be a complete metric space, and let [itex]f:X\to X[/itex] and [itex]g:X\to X[/itex] be two strict contractions on [itex]X[/itex] with contraction coefficients [itex]c[/itex] and [itex]c'[/itex] respectively. From the Contraction Mapping Theorem we know that [itex]f[/itex] has some fixed point [itex]x_0[/itex], and [itex]g[/itex] has some fixed point [itex]y_0[/itex]. Suppose we know that there is an [itex]\epsilon > 0[/itex] such that [itex]d(f(x),g(x)) \leq \epsilon[/itex] for all [itex]x \in X[/itex]. Show that [itex]d(x_0,y_0) \leq \epsilon/(1-\max(c,c'))[/itex]. Thus nearby contractions have nearby fixed points.
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    Below is my proof. I believe everything I have is correct. My question is why do we not conclude that [itex]d(x_0,y_0) < \epsilon/(1-\min(c,c'))[/itex] as I believe this is a stronger result that follows (with very slight modification) from my proof below.

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    Proof.

    Let [itex]m = \max(c,c')[/itex].

    So we have [itex]d(f(x),f(y)) < md(x,y)[/itex] and [itex]d(g(x),g(y)) < md(x,y)[/itex] for all [itex]x,y \in X[/itex].

    Since [itex]x_0[/itex] is a fixed point of [itex]f[/itex], and [itex]y_0[/itex] is a fixed point of [itex]g[/itex], we know that [itex]f(x_0) = x_0[/itex], and [itex]g(y_0) = y_0[/itex]. Hence we get the following

    [tex]
    \begin{align*}
    d(x_0,y_0) & = d(f(x_0),g(y_0)) \\
    & \leq d(f(x_0),f(y_0)) + d(f(y_0),g(y_0)) \\
    & \leq md(x_0,y_0) + \epsilon
    \end{align*}
    [/tex]

    So [itex]d(x_0,y_0) \leq md(x_0,y_0) + \epsilon[/itex].

    Hence, [itex]d(x_0,y_0)(1 - m) \leq \epsilon[/itex].

    Therefore [tex]d(x_0,y_0) \leq \dfrac{\epsilon}{1 - \max(c,c')}[/tex].

    ------------

    I can't see anything wrong, but there is probably some subtlety that I am missing. Any ideas? Thanks!


    Definitions if needed.

    Strict contraction: Let [itex]f:X\to X[/itex] be a map, then [itex]f[/itex] is a strict contraction if [itex]d(f(x),f(y) < d(x,y)[/itex].

    Fixed point: x is a fixed point of a function f if f(x) = x.

    Contraction mapping theorem. Let [itex](X,d)[/itex] be a metric space, and let [itex]f:X\to X[/itex] be a strict contraction. Then f can have at most one fixed point. Moreover, if we assume X is complete and non-empty then f has exactly one fixed point.
     
    Last edited: Apr 21, 2007
  2. jcsd
  3. Apr 21, 2007 #2

    Hurkyl

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    I don't see a problem.
     
  4. Apr 21, 2007 #3

    AKG

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    The proof looks fine, and yes, with a slight modification of your proof you can derive the stronger result that replaces 'max' with 'min'.
     
  5. Apr 21, 2007 #4
    Thanks! (10 char)
     
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