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This question already has an answer here:

Let $(X,d)$ be a metric space and complete and $T: X\to X$ a mapping such that $T^m$ is a contraction. Show that $T$ has a unique fixed point.

It is clear that $T^m$ has a unique fixed point (Banach theorem) but, how can you use it to prove the existence of a fixed point of $T$?

Thank you.

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marked as duplicate by grand_chat, Community Feb 19 at 18:06

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  • $\begingroup$ Indeed, I looked up before writing and I did not find it. Thank you very much. $\endgroup$ – Rubén Fernández Fuertes Feb 19 at 18:06
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If $T^mx_0=x_0$, $T^m(Tx_0)=T(T^mx_0)=Tx_0$, so $Tx_0$ is fixed by $T^m$. By unicity, it must be $x_0$.

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  • $\begingroup$ I thought it must be easy but not as easy. Now I feel a little bit upset haha Many thanks! $\endgroup$ – Rubén Fernández Fuertes Feb 19 at 17:46

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