Unique Fixed Point: Proving F^3 is a Contraction

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diligence
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Homework Statement


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.

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!
 
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diligence said:

Homework Statement


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.


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.

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...
 


micromass said:
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...

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).