# If $f^N$ is contraction function, show that $f$ has precisely one fixed point. [duplicate]

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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, NamasteFeb 13 '15 at 12:25

• Can you show that $f^N$ has precisely one fixed point? – Hamou Aug 12 '14 at 15:06
$f^N$ 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 $f(\alpha)=\alpha$.