Results 1 to 6 of 6

Math Help - Unique fixed point

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    6

    Unique fixed point

    I'm having a problem understanding this:

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

    I'm not sure I understand what they mean by U^n.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by BenH View Post
    I'm having a problem understanding this:

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

    I'm not sure I understand what they mean by U^n.
    They mean the n-fold composition with itself. What is M? Hopefully (for my solution anyways) it's a compact metric space.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2011
    Posts
    6
    Quote Originally Posted by Drexel28 View Post
    They mean the n-fold composition with itself. What is M? Hopefully (for my solution anyways) it's a compact metric space.
    Does the n-fold composition with itself mean 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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Tinyboss's Avatar
    Joined
    Jul 2008
    Posts
    433
    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 \{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; March 12th 2011 at 04:04 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    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 \{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 \text{dist}_U:M\to\mathbb{R}:d(x,U(x)). It's easy to show that this mapping is continuous since d:M\times M\to\mathbb{R} and 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 \text{dist}_U=d\circ j. Thus, since M is compact one has that \text{dist}_U has a minimum point. Use the n-fold composition contraction fact then to show (by contradiction) that this point must be a fixed point--for uniqueness see Tinyboss's response.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Mar 2011
    Posts
    6
    Thanks for your replies! Compactness can not be assumed. I'm not sure, but I think completeness can be assumed. Anyway, lets assume M is complete. Then since U^n is a strict contraction, by Banach fix point theorem there exists a unique fixed point a\in M s.t U^n(a)=a. Performing U on both sides we get
    U(U^n(a))=U(a)\Leftrightarrow U^n(U(a))=U(a).
    This means that U(a) is a fixed point to U^n as well.
    Uniqueness implies that U(a)=a. Thus a is a fixed point to U as well. As for uniqueness, assume that a and b both are fixed points to U
    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.
    0<L<1 because of that U^n is a strict contraction.
    Is this correct?
    Last edited by BenH; March 13th 2011 at 09:28 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. The fixed point
    Posted in the Geometry Forum
    Replies: 0
    Last Post: April 2nd 2011, 10:57 AM
  2. Unique Fixed Point for B-Contraction
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: April 28th 2010, 04:11 PM
  3. [SOLVED] unique point
    Posted in the Differential Geometry Forum
    Replies: 8
    Last Post: January 14th 2010, 02:56 PM
  4. Unique Cluster Point => Convergent?
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 13th 2008, 02:21 PM
  5. fixed point?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 1st 2006, 01:46 AM

Search Tags


/mathhelpforum @mathhelpforum