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This question already has an answer here:

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.

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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 $f(\alpha)=\alpha$.

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

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