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Thread: Unique fixed point

  1. #1
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    Unique fixed point

    I'm having a problem understanding this:

    Show that if $\displaystyle U$ is a continuous map such that $\displaystyle U^n:M \rightarrow M$ is a strict contraction for some fixed integer $\displaystyle n$, then $\displaystyle U$ has a unique fixed point.

    I'm not sure I understand what they mean by $\displaystyle U^n$.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by BenH View Post
    I'm having a problem understanding this:

    Show that if $\displaystyle U$ is a continuous map such that $\displaystyle U^n:M \rightarrow M$ is a strict contraction for some fixed integer $\displaystyle n$, then $\displaystyle U$ has a unique fixed point.

    I'm not sure I understand what they mean by $\displaystyle U^n$.
    They mean the $\displaystyle n$-fold composition with itself. What is $\displaystyle M$? Hopefully (for my solution anyways) it's a compact metric space.
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    They mean the $\displaystyle n$-fold composition with itself. What is $\displaystyle M$? Hopefully (for my solution anyways) it's a compact metric space.
    Does the $\displaystyle n$-fold composition with itself mean $\displaystyle U^n=U\circ U\circ U \circ \cdots \circ U$? It's a metric space, but I don't think you can assume that it is compact.
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  4. #4
    Senior Member Tinyboss's Avatar
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    That's what n-fold composition is, yes. And I believe it's completeness, not compactness, that you need for this result. To get a fixed point, make a sequence $\displaystyle \{x,U(x),U^{(2)}(x),\dots\}$ and show that it's Cauchy (this is where we need the space to be complete). For uniqueness, suppose there are two distinct fixed points and look at their images under U--do they get closer together?.
    Last edited by Tinyboss; Mar 12th 2011 at 04:04 PM.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Tinyboss View Post
    That's what n-fold composition is, yes. And I believe it's completeness, not compactness, that you need for this result. To get a fixed point, make a sequence $\displaystyle \{x,U(x),U^{(2)}(x),\dots\}$ and show that it's Cauchy (this is where we need the space to be complete). For uniqueness, suppose there are two distinct fixed points and look at their images under U--do they get closer together?.
    That is the classic method--but if it's compact there is an eaiser way by considering the map $\displaystyle \text{dist}_U:M\to\mathbb{R}:d(x,U(x))$. It's easy to show that this mapping is continuous since $\displaystyle d:M\times M\to\mathbb{R}$ and $\displaystyle j:M\to M\times M:x\mapsto (x,U(x))$ are both continuous (the first obviously and the second since each coordinate map is continuous) and $\displaystyle \text{dist}_U=d\circ j$. Thus, since $\displaystyle M$ is compact one has that $\displaystyle \text{dist}_U$ has a minimum point. Use the $\displaystyle n$-fold composition contraction fact then to show (by contradiction) that this point must be a fixed point--for uniqueness see Tinyboss's response.
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  6. #6
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    Thanks for your replies! Compactness can not be assumed. I'm not sure, but I think completeness can be assumed. Anyway, lets assume $\displaystyle M$ is complete. Then since $\displaystyle U^n$ is a strict contraction, by Banach fix point theorem there exists a unique fixed point $\displaystyle a\in M$ s.t $\displaystyle U^n(a)=a$. Performing $\displaystyle U$ on both sides we get
    $\displaystyle U(U^n(a))=U(a)\Leftrightarrow U^n(U(a))=U(a)$.
    This means that $\displaystyle U(a)$ is a fixed point to $\displaystyle U^n$ as well.
    Uniqueness implies that $\displaystyle U(a)=a$. Thus $\displaystyle a$ is a fixed point to $\displaystyle U$ as well. As for uniqueness, assume that $\displaystyle a$ and $\displaystyle b$ both are fixed points to $\displaystyle U$
    $\displaystyle d_M (a,b)=d_M(U(a),U(b))=d_M(U(U(a)),U(U(b)))=\cdots =d_M(U^n(a),U^n(b))\leq Ld_M(a,b)\Rightarrow d_M(a,b)=0 \Leftrightarrow a=b$.
    $\displaystyle 0<L<1$ because of that $\displaystyle U^n$ is a strict contraction.
    Is this correct?
    Last edited by BenH; Mar 13th 2011 at 09:28 AM.
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