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Homework Help: Analysis Comp question: function that's a contraction mapping when composed w/itself

  1. Oct 10, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose F is mapping of a nonempty complete metric space into itself, and that
    F^3 = F o F o F is a contraction (o's denote composition). Show that f has a unique fixed point.


    3. The attempt at a solution
    Isn't this kind of a trick question? Suppose f does not have a unique fixed point. Then there does not exist a unique x such that f(x)=x. Hence there doesn't exist a unique x s.t. f(f(x))=f(x)=x. Hence there doesn't exist a unique x s.t. f(f(f(x)))=f(f(x))=f(x)=x.

    But this contradicts the assumption that F^3 is a contraction, since any contraction on a complete metric space has a unique fixed point.

    What am I missing here, this is too easy!
     
  2. jcsd
  3. Oct 10, 2011 #2
    Re: Analysis Comp question: function that's a contraction mapping when composed w/its

    Why can't there exists an x such that f(f(x))=x??
    And why must f(f(x))=f(x)? You don't know that f(x) is a fixed point of f...
     
  4. Oct 10, 2011 #3
    Re: Analysis Comp question: function that's a contraction mapping when composed w/its

    If I move towards a contradiction by supposing the contrary, that there doesn't exist a unique fixed point for f, then there is no unique x st f(x)=x (they may exist but not unique). Let y=f(x). Then there is no unique y st f(y)=y ----> no unique f(x) st f(f(x))=f(x), and further, there is no unique x st f(x)=x, so no unique x st f(f(x))=f(x)=x.

    Continuing in this fashion leads to a contradiction that there is no unique fixed point for f^3, since by hypothesis f^3 is a contraction mapping (which implies f^3 has a unique fixed point). Hence the assumption that f has no unique fixed point is false.

    Am I missing the forest for the trees? What's wrong with this argument? (Sorry for mixing up the F's and f's earlier, and also while this is a graduate comp question i'm actually an undergrad. Supposedly the graduate analysis course at my university is comparable to an honors undergrad analysis elsewhere).
     
  5. Oct 11, 2011 #4
    Re: Analysis Comp question: function that's a contraction mapping when composed w/its

    You have now given an argument why there can't exist an x such that f(f(x))=f(x)=x. This is ok.
    But why can't there exist an x such that f(f(x))=x without x=f(x)????
     
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