If $T^m$ is a contraction, has $T$ got a fixed point?

Question

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.

Answer 1

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 $x_0$.