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If $f$ is a mapping of a complete metric space $(X, d)$ into itself and $f^N$(composite $f$ for $N$ times) is a contraction mapping for some positive integer $N$, then $f$ has precisely one fixed point. (Banach fixed point theorem is applicable)

I tried to show that $f$ is also a contraction function. I considered the sequence $x,f(x),f^2 (x),... $, but then fail pathetically in showing that it's cauchy. Please tell me if my direction is correct, any new ideas are appreciated. Thanks.


marked as duplicate by Martin Sleziak, Ali Caglayan, Davide Giraudo, Claude Leibovici, Namaste Feb 13 '15 at 12:25

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$f^N$ has one fixed point $\alpha$, then $f^N(\alpha)=\alpha$ and $f^{N+1}(\alpha)=f(\alpha)$ so $f^N(f(\alpha))=f(\alpha)$, hence $f(\alpha)$ is also a fixed point. By uniqueness we have $f(\alpha)=\alpha$.

  • $\begingroup$ It's so simple!!!! $\endgroup$ – A. Chu Aug 12 '14 at 15:11

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