# 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.

(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?

• 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. Oct 13, 2021 at 5:58

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 $$f(x)=x$$.
• 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 Oct 13, 2021 at 14:50