# If $T^m$ is a contraction, has $T$ got a fixed point? [duplicate]

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.

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

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 $$x_0$$.