Unique Fixed Point: Proving F^3 is a Contraction

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Homework Help Overview

The problem involves a mapping F of a nonempty complete metric space into itself, specifically examining the implications of F^3 being a contraction. The goal is to demonstrate that F has a unique fixed point.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of assuming F does not have a unique fixed point and explore the resulting contradictions. Questions arise regarding the conditions under which f(f(x)) equals x and the necessity of x being a fixed point of f.

Discussion Status

The discussion is ongoing, with participants examining the validity of their arguments and questioning the assumptions made in their reasoning. Some guidance has been offered regarding the logical structure of the arguments presented, but no consensus has been reached.

Contextual Notes

Participants express uncertainty about the implications of their assumptions and the definitions involved, indicating a need for clarification on the nature of fixed points in the context of contraction mappings.

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).
 


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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