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How to show that if $f^N$ is a contraction mapping, then $f$ has a unique fixed point.

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

The Banach fixed point theorem is already proven so we can show the following:

$f^N$ has a 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 $f(\alpha)=\alpha$

However, this only shows that $f^N$ has a unique fixed point because the Banach theorem states that the contraction mapping has a fixed point. $f^N$ is a contraction mapping but f is assumed to be not a contraction mapping. How can we show that $f$ also has a unique fixed point?


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