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Let $(X,d)$ be a complete metric space, and let $f:X\to X$ be eventually contractive (there exists $p\in\mathbb{N}$ such that $g:=f^{\circ p}$ is contractive). Then $f$ has a unique fixed point.

Is there a way to prove this without using the Banach contraction theorem? Would appreciate a hint.

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  • $\begingroup$ You have to explain what "eventually contracttive" means here $\endgroup$ – Orest Bucicovschi Oct 13 '17 at 3:12
  • $\begingroup$ Please see my updated question. $\endgroup$ – sequence Oct 13 '17 at 3:31
  • $\begingroup$ So you know how to prove it using Banach f p theorem, and want a different proof? $\endgroup$ – Orest Bucicovschi Oct 13 '17 at 3:34
  • $\begingroup$ Using the Banach theorem is actually just applying it. So I think this problem likely asks for more. $\endgroup$ – sequence Oct 13 '17 at 3:35
  • $\begingroup$ I don't think it is immediate. The contraction mapping theorem would imply that there is a unique fixed point for the power $f^{\circ p}$ only, not for $f$ itself. $\endgroup$ – timur Oct 13 '17 at 3:42
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I suppose $f$ is continuous...

If $f^p$ is a contraction, then $(f^{np}(x))_n$ converges to the fixed point $x_0$ of $f^p$ for every $x\in X$. But then $f^{np+1}(x)$ converges to $f(x_0)$. But it also has to converge to $x_0$ (take $x'= f(x)$ as initial point). We conclude that $f(x_0) = x_0$.

Note that the fixed point will be the limit of $f^n(x)$ for every initial point $x$. The convergence rate is of the order $q^{\frac{n}{p}}$.

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