If $f^k$ is a strict contraction for some integer $k>1$, show that $f$ has a unique fixed point.

Question

Let M be complete and let $f:M\to M$ be continuous. Assume that $f^k$ is a strict contraction for some integer $k>1$ (to avoid any misinterpretation, the notation $f^k$ means the composition of $f$ with itself $k$ times).

Question: How do I show that $f$ has a unique fixed point?

I know that a map $f: M \to M$ is a strict contraction if there exists $0\leq \alpha < 1$ such that $d(\,f(x), \,f(y)) \leq \alpha\, d(x,y)$ is satisfied for all $x,y\in M$. Furthermore, I know that if $M$ is complete and $f$ is a strict contraction, then $f$ has a unique fixed point.

Since $f^k$ is a strict contraction, we know that there exists $0\leq\alpha < 1$ such that $d(\,f^k(x), \,f^k(y)) \leq \alpha \,d(x,y)$ for all $x,y \in M$. What I've tried thus far is to show that $d(\,f^k(x),\,f^k(y)) \leq \alpha \, d(x,y)$ implies $d(\,f(x), \,f(y)) \leq \alpha\, d(x,y)$ as well, but I didn't get very far..

Thanks in advance!

Answer 1

$f^k$ has a unique fixed point $x$. $f^k(f(x))=f^{k+1}(x)=f(f^k(x))=f(x)$ so $f(x)$ is a fixed point for $f^k$ as well, and so $f(x)=x$. If $y$ is a fixed point for $f$ then $f^k(y)=y$ and so $x=y$.