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Let X be a metric space. A self-map $\Phi$ on X is said to ve a pseudocontraction if $d(\Phi(x),\Phi(y))<d(x,y) $ holds for all distinct $x,y \in X$.

(a) If $\Phi \in X^X$ is a pseudocontraction, then ($d(\Phi^{m+1}(x)),\Phi^m(x)$) is a decreasing sequence, and hence converges. (Here $\Phi^1:=\Phi$ and $\Phi^{m+1}:=\Phi \circ \Phi^m, m=1,2,...$).Use this to show that if ($\Phi^m(x)$) has a convergent subsequence, then $d(x^*,\Phi(x^*))=d(\Phi(x^*),\Phi^2(x^*) )$

(b) Prove Edelstein's Fixed point Theorem: If $X$ is a compact metric space and $\Phi \in X^X$ a pseudocontraction, then there exists a unique $x^*$ such that $\Phi(x^*)=x^*$. Moreover, we have $lim \Phi^m(x)=x^*$ for any $x \in X$.

The problem is taken from Ok(2007) Chapter C6 Exercise 50. The main difficulty here, for me, is part (a). I am sure if (a) can be proved, then we can use (a) to prove b. But how to prove (a)? The question seems a little bit abstract.

peter xue
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  • What have you tried? – ACB Jan 17 '22 at 06:38
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    I think there is a typo in (a). Let $X=[-10,10]$ and $\Phi(x)=-\frac 9{10}x$. Then $\Phi(10)=-9$, $\Phi^2(10)=81/10$, $\Phi^3(10)=-729/100$ and $\Phi^4(10)=6561/1000$ and clearly $|\Phi(10)-\Phi^3(10)|<|\Phi(10)-\Phi^4(10)|$. – MaoWao Jan 17 '22 at 07:50
  • Sorry, my bad, I have revised the typo. – peter xue Jan 17 '22 at 08:33

1 Answers1

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Welcome to MSE!
(a) We only need to prove $d(\Phi^{m+1}(x),\Phi^{m}(x))$ is decreasing. However, it is directily from $d(\Phi(x),\Phi(y))<d(x,y)$, where we can choose $y=\Phi(x)$. If $(\Phi^{m}(x))$ has a convergent subsequence, which can be denote as $(\Phi^{m_k}(x))$, then $\lim_{k\to \infty}d(\Phi^{m_k}(x),x^{\ast})=0$. Therefore, $$ d(x^{\ast},\Phi(x^{\ast}))=\lim_{k\to \infty}d(\Phi^{m_{k+1}}(x),\Phi(\Phi^{m_{k+1}}(x)))\le \lim_{k\to \infty}d(\Phi(\Phi^{m_{k}}(x)),\Phi^2(\Phi^{m_{k}}(x)))=d(\Phi(x^{\ast}),\Phi^2(x^{\ast})) $$ Since $m_{k+1}\ge m_k+1$. In addition, $d(x^{\ast},\Phi(x^{\ast}))\ge d(\Phi(x^{\ast}),\Phi^{2}(x^{\ast}))$ is obvious. And $(b)$ can be obtained directly from $(a)$.

Mr.xue
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  • Thank you. But I find I still can't fully prove (b). If $X$ is compact, it is sequentially compact, then every subsequence of $X$ converges in $X$, so we can use the conclusion of part(a). But how to use the equality of (a) to get $d(x,\Phi(x^))=0$,i.e. $\Phi(x^)=x^$? – peter xue Jan 17 '22 at 11:00
  • @peterxue If $x^{\ast}\ne \Phi(x^{\ast})$, then we must have $d(\Phi(x^{\ast}),\Phi^2(x^{\ast}))< d(x^{\ast},\Phi(x^{\ast}))$. – Mr.xue Jan 17 '22 at 11:33
  • The last stupid question: can you be more specific on showing $lim\Phi^m(x)=x^∗$?I try to use the conclusion that $\Phi(x^)=x^$, but again, failed.Thanks. – peter xue Jan 17 '22 at 12:39
  • @peterxue I just denote $x^{\ast}$ as the limit, – Mr.xue Jan 18 '22 at 11:41
  • I know $lim \Phi^{m_k}(x)=\Phi(x^)=x^$, but how to show $lim \Phi^{m_k}(x)=lim \Phi^{m}(x)$ . If we can show the latter, then $lim \Phi^{m}(x)=x^*$, right? @Mr.xue – peter xue Jan 18 '22 at 14:14
  • @peterxue You are right. However, we don't need the existence of $\lim_{m\to \infty} \Phi^{m}(x)$. We only need a Fixed point that is the limit of the subsequence $\Phi^{m_k}(x)$. – Mr.xue Jan 19 '22 at 05:16