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Proving existence of unique fixed point on a compact space

  1. Nov 25, 2013 #1
    The problem statement, all variables and given/known data.

    Let ##(M,d)## be a metric space and let ##f:M \to M## be a continuous function such that ##d(f(x),f(y))>d(x,y)## for every ##x, y \in M## with ##x≠y##. Prove that ##f## has a unique fixed point

    The attempt at a solution.

    The easy part is always to prove unicity. Suppose there exist ##x,y##, ##x≠y## : ##x## and ##y## are both fixed points. Then ##d(x,y)=d(f(x),f(y)>d(x,y)##, which is absurd. Then, ##x## must be unique.

    Now I have to prove existence and here I had problems:

    First I've tried to prove it trying to repeat the standard proof of the fixed point theorem on a complete m.e. with ##f## being a contraction, I mean:

    Let ##x## be an arbitrary element in M, I've defined a sequence ##\{x_n\}_{n \in \mathbb N} : x_1=x, x_2=f(x_1)=f(x), x_3=f(x_2)=f^2(x_1)## and, in general, ##x_n=f(x_{n-1})=f^{n-1}(x)##.

    As ##M## is compact, there exists ##\{x_{n_k}\}_{k \in \mathbb N}## a convergent subsequence of ##\{x_n\}_{n \in \mathbb N}##. Then, ##x_{n_k} \to z##. I would love to apply ##f## and say ##z=lim_{k \to \infty} x_{n_{k+1}}=f(lim_{k \to \infty} x_{k_n})=f(z)## (using that ##f## is continuous), but then I've realized this is completely wrong as ##f(x_{n_k})=x_{n_k+1}≠x_{n_{k+1}}##.
     
    Last edited: Nov 25, 2013
  2. jcsd
  3. Nov 25, 2013 #2
    I found you a theorem. If you can prove it, I think you have your result.

    Theorem (Intermediate value theorem for metric spaces): given ##
    (S_{1}, d_{1}), (S_{2}, d_{2}), f:E_{1} \rightarrow E_{2}, f ##
    is continuous,## E \subseteq S_{1} ## is connected, then## f(E)
    \subseteq S_{2} ##is connected.

    On further research I find there are a lot of "fixed point" theorems around. Check out the contraction mapping theorem here: http://www.math.tifr.res.in/~publ/ln/tifr26.pdf
     
    Last edited: Nov 25, 2013
  4. Nov 26, 2013 #3
    I'll try to solve it using that theorem, if I can arrive to a solution, I'll post it. Thanks!
     
  5. Nov 26, 2013 #4
    That's good. I have a feeling this is a standard result, and probably quite useful.
     
  6. Nov 26, 2013 #5

    pasmith

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    Homework Helper

    My instinct is to prove that [itex]f[/itex] is invertible, and apply the contraction mapping theorem to [itex]f^{-1}[/itex]. However it's not that straightforward.
    Certainly [itex]f[/itex] is injective: if [itex]x \neq y[/itex] then we have [itex]d(f(x),f(y)) > d(x,y) > 0[/itex] so that necessarily [itex]f(x) \neq f(y)[/itex].

    Unfortunately we are not given that [itex]f[/itex] is surjective (although maybe compactness combined with the condition on [itex]f[/itex] requires that; I haven't been able to show that myself). It is however sufficient to find a subset [itex]U \subset M[/itex] such that [itex]f(U) = U[/itex]. The function [itex]f_{U}: U \to U : x \mapsto f(x)[/itex] will then be invertible.

    How to find such a [itex]U[/itex]? We have a sequence of nested non-empty closed subsets
    [tex]
    M = f^0(M) \supseteq f(M) \supseteq \dots \supseteq f^n(M) \supseteq \dots
    [/tex]
    and one thing to try is to consider the intersection
    [tex]
    I = \bigcap_{n=0}^{\infty} f^n(M)
    [/tex]
    which is non-empty by Cantor's intersection theorem. Certainly any fixed point of [itex]f[/itex] must be in this intersection. If we can show that [itex]f(I) = I[/itex] then we will be done.
     
  7. Nov 26, 2013 #6
    Hmm, I don't know how to prove that ##I=\bigcap_{n=0}^{\infty} f^n(M)## satisfies ##I=f(I)##. Suppose I could prove ##I=f(I)##, is it true that ##I## is compact?, because if not, all I am going to write is absolutely wrong:
    Lets prove that ##I## consists of only one point. Suppose there is more than one element. Define ##f:I \times I \to \mathbb R## as ##f(x,y)=d(x,y)##,##f## is a continuous function on a compact metric space (cartesian product of compact sets is compact), then attains a maximum. Then, there exist ##x_1,x_2## such that ##diam(I)=d(x_1,x_2)##. By hypothesis, there are ##y_1, y_2 \in I: f(x_1)=y_1, f(x_2)=y_2##, but ##d(x_1,x_2)<d(f(x_1),f(x_2))##, which is absurd by definition of diameter.

    As ##I## consists of only one point, then we have ##\{x\}=I=f(I)=\{f(x)\}##.

    So, we've proved existence by defining that nested sequence of sets. In my first post I've proved unicity. Now there are two things left to prove: how can I show that ##I=f(I)##? and, is it true that ##I## is compact?

    Your previous answer was very helpful, pasmith. Thanks!
     
  8. Nov 26, 2013 #7

    Dick

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    You just put your finger on the problem. If M is compact then d:MxM->R attains a maximum. So M must consist of a single point for the reasons you just indicated. So it obviously has a trivial fixed point. I notice 'compact' is only in the title of the thread and your proof. The statement you quoted doesn't mention 'compact'.
     
    Last edited: Nov 26, 2013
  9. Nov 26, 2013 #8
    Oh, that's simply because I'm an idiot and I didn't notice that I haven't wrote ##M## a COMPACT metric space. That is one of the hypothesis of the problem, and now I cannot edit my original post :(

    Wait, now that it's been clarified ##M## is compact, the function I've defined is from ##I \times I \to \mathbb R##, I need compactness of ##I## to say the function attains a maximum
     
  10. Nov 26, 2013 #9
    One of the hypothesis of the exercise is that ##M## is a compact metric space, I've forgot to write that.
     
  11. Nov 26, 2013 #10

    Dick

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    Sure, sure. But if M is compact, doesn't your argument show M consists of a single point? Forget I.
     
    Last edited: Nov 26, 2013
  12. Nov 26, 2013 #11
    Yes, but my argument is based on the fact that ##I=f(I)## and that ##I## is compact, these are two facts I've assumed to be true without proving them because I didn't know how to.

    I had another idea to prove the statement without using Cantor's intersection theorem but there is something that doesn't convince me, if you want/can check it, maybe you realize where my mistake is:

    In a previous post I've proved that if a fixed point exists, then it must be unique. Now I want to prove existence:
    Consider ##f:M \to \mathbb R## defined as ##f(x)=d(x,f(x))##. Since ##f## is continuous, ##f## attains a maximum. Let ##x## be te point where ##f## attains it maximum and let ##D## be the maximum, ##D=d(x,f(x))##, if ##D=0 \implies d(x,f(x))=0 \implies x=f(x)##, so we are done. If ##D>0##, then by hypothesis, ##D=d(x,f(x))<d(f(x),f(f(x)))##, which is absurd since ##D## is the maximum. Then it must be ##D=0## and ##x=f(x)##.

    If I read this proof, I think it's correct, but one conclusion of this proof is that for any ##z \in M##, ##0≤d(z,f(z)≤d(x,f(x))=D=0##, then ##d(z,f(z))=0## for every ##z## so all the points are fixed points!!!. There must be something wrong here and I can't see where.
     
  13. Nov 26, 2013 #12

    Dick

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    All points are fixed points, and there aren't many of them. d:MxM->R is a continuous function from the compact space MxM into R. So it attains a maximum. So there is a D (diam(M)) such that d(x,y)<=D for all x and y in M. There is also an x' and y' such that d(x',y')=D. That d(f(x'),f(y'))>d(x',y')=D is obvious rubbish. So x'=y' and D=0. M consists of a single point. It's the same argument you applied to I.
     
  14. Nov 26, 2013 #13
    Ooooh, so the conclusion of my proof would be ##|M|=1##, I thought about it but I still kept suspecting there was something wrong with the proof. Thanks!!!
     
  15. Nov 26, 2013 #14

    Dick

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    Welcome! The whole thing does seem like kind of a cheat. Didn't need to use f is continuous. Etc, etc. I still wonder if they didn't mean to ask a somewhat different question. But it is true that compact metric spaces don't have mappings like that. Unless the metric space is has a single point.
     
    Last edited: Nov 26, 2013
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