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(X,d) is a complete metric space. If $f:X\rightarrow X$ is such that there exists $n\in \mathbb{N}$ for which $f^{n}$ is contracting, show that $f$ admits a unique fixed point.

I think the method is to prove that $f^{n}$ contracting $\Rightarrow $ $f$ is contracting, hence we can apply the Banach fixed point theorem. But I have no idea how to prove this. Maybe by a recursion?

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    $\begingroup$ Welcome to MSE. A question should be written in such a way that it can be understood even by someone who did not read the title. $\endgroup$ Oct 13, 2021 at 5:58

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By the Banach fixed point theorem, $f^n$ has a unique fixed point $x$. Then $f(x)$ is also a fixed point of $f^n$, and so by uniqueness $f(x)=x$.

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    $\begingroup$ I dont get how you come to the conclusion that $f(x)$ is also a fixed point of $f^{ n}$. EDIT: GOT IT IM SO DUMB $\endgroup$
    – SerIddi
    Oct 13, 2021 at 14:50

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