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If $K$ be a compact metric space and $\Phi:K \to K$ be such that $d(\Phi(x),\Phi(y))<d(x,y)$, Show that $\Phi$ has a unique fixed point.

Here is my approach but not sure

To show that $\Phi$ has a unique fixed point, it is suffice to show $K$ is complete and $\Phi$ is a contraction.

Notice that $K$ is compact, hence it is totally bounded and complete, hence $K$ is complete. Furthermore, pick $c:=1/2$, then for all $x,y \in K$, we have $$d(\Phi(x),\Phi(y)) < \frac{1}{2} d(x,y)) < d(x,y)$$ and then we know such $\Phi$ by definition is a contraction; Hence, by contraction principle we know $\Phi$ has a unique fixed point.

Is the above proof works?

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    $\begingroup$ Where would you get $d(\Phi(x),\Phi(y)) < \frac{1}{2} d(x,y)$ from? $\endgroup$ Nov 19, 2014 at 18:47
  • $\begingroup$ Xiao, @Daniel Fischer, Thanks, I think I'm thinking another way around, (originally, I just think since $d(\Phi(x),\Phi(y)) < d(x,y)$ hence $d(\Phi(x),\Phi(y)) < c d(x,y)$ for some $0\le c<1$ also true, but I'm wrong. ); So could you please say more words about how to fix/ or rewrite the proof ? Thanks $\endgroup$
    – Fianra
    Nov 19, 2014 at 19:34
  • $\begingroup$ @Xiao, Thanks, but I don't quite understand how to use Brouwer fixed-point theorem, do you mean I need to exploit compactness and create a ball or something like this? (actually I just know Banach contraction principle..), could you say more words here? $\endgroup$
    – Fianra
    Nov 19, 2014 at 19:53
  • $\begingroup$ @Fianra Actually Banach contraction principle is fine, you need to use the fact that $K$ is compact. Let me write out more argument in a bit. I am on my cell phone atm, but you can look here math.stackexchange.com/questions/118536/… $\endgroup$
    – Xiao
    Nov 19, 2014 at 20:09

1 Answer 1

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Your assumptions for $\Phi$ are slightly weaker than what's needed for the contraction mapping principle, but the assumption that K is compact is stronger than is needed for the contraction mapping principle, so the idea is to repeat the proof of the contraction mapping principle using the compactness of K to fill in for the missing properties of $\Phi$.

First note that $\Phi$ is continuous (for any $\epsilon$ pick $\delta = \epsilon$, etc.). Suppose $\Phi$ has no fixed point. Then $\forall (x) \; d(x, \Phi (x)) > 0$. Now since $d(\cdot , \Phi (\cdot))$ and $d(\Phi (\cdot), \Phi^2 (\cdot))$ are continuous (as compositions of continuous functions) and both are always non-zero, their ratio $r(\cdot):= d(\cdot, \Phi (\cdot))/d(\Phi (\cdot), \Phi^2 (\cdot))$ is also continuous. Since K is compact, there is an $x \in K$ where $r(x)$ is maximal. Let this maximal value be denoted by q. By hypotheses about $\Phi$, we have $q<1$.

Now for any point $x \in K$, consider the sequence $\left(\Phi^k (x)\right)_{k=0}^{\infty}$. We have $d(\Phi (x),\Phi^2 (x))<q d(x,\Phi (x))$. Inductively, $d(\Phi (x)^k,\Phi^{k+1} (x))<q^k d(x,\Phi (x))$. This is now the situation faced in the proof of the contraction mapping principle. The sequence has a cluster point y, which must be a limit point because for any $\epsilon$ we can find a large value k where both $d(\Phi^k (x),y)<\epsilon$ and $\forall (j>k)$: $$ d(\Phi (x)^k,\Phi^j (x)) \\ \leq d(\Phi (x)^k,\Phi^{k+1} (x))+d(\Phi (x)^{k+1},\Phi^{k+2} (x))+...+d(\Phi (x)^{j-1},\Phi^j (x)) \\ \leq q^k + q^{k+1}+...+q^{j-1} \leq \frac{q^k-q^j}{1-q}\leq \epsilon $$

y must be a fixed point, because $d(y, \Phi (y))\leq d(y,\Phi^k (x)) + d(\Phi (y),\Phi^k (x))\leq d(y,\Phi^k (x)) + q d(y,\Phi^{k-1} (x))$, which can be made arbitrarily small. This clearly contradicts the assumption that $\Phi$ had no fixed points.

Now showing that the fixed point is unique is easy: If there were two fixed points y and z, then $d(y,z)=d(\Phi (y), \Phi (z)) < d(y,z)$, contradiction.

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