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Let $(M,d)$ be a metric space, $M$ compact. If $f:M \to M$ is continuous and weakly contractive (i.e. $d(f(x), f(y)) < d(x,y) , \forall x,y \in M$), then $\exists x_0 \in M $ unique s.t $f(x_0)=x_0$ . Suggestion: $g: M \to R$ with $g(x)=d(x, f(x)) $ is continuous relative to $d$.

I was able to prove the continuity of $g$, but no the existence of the fixed point.

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MariaU
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  • Hint: Let $x_1\in M$ be arbitrary and define a sequence $(x_n)$ recursively by $x_{n+1}=f(x_n)$. Since $M$ is compact, this sequence has a convergent subsequence. Try to prove that the limit of this convergent subsequence is a fixed point of $f$. – Mark McClure Dec 01 '15 at 01:36
  • a continuous function on a compact set reaches (i.e. attains) its maximum and minimum. Show the minimum of $g$ is $0$. @MarkMcClure Your suggestions works for complete (not necessarily compact) metric spaces provided there is $0<a<1$ with $d(f(x), f(y)) \le a d(x,y) , \forall x,y \in M$ – Mirko Dec 01 '15 at 03:14
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    Thank you @Mirko your suggestion was really helpful – MariaU Dec 01 '15 at 04:03
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    Possible duplicate of Contractive Operators on Compact Spaces – Sahiba Arora Jan 03 '18 at 18:40
  • duplicate of https://math.stackexchange.com/questions/2779756/edelstein-theorem – am70 Mar 29 '23 at 16:31

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