# proving fixed point

Let $(X,d)$ be a complete metric space and let $f: X \to X$ be a mapping such that for each $n \geq 1$, there exists a constant $c_n$ such that $d(f^n(x),f^n(y)) \leq c_n d(x,y)$ for all $x,y \in X$, where $\sum_{n=1}^{\infty} c_n < \infty$.

I want to prove that $f$ has a unique fixed point.

• Hint: Look up a proof of "Banach's fixed point theorem" and try to adapt it to your situation. – m.s Jul 23 '17 at 13:11

By the convergence of the series, there is some $N$ such that $c_n<1$ for For all $n>N$. Then $f^n$ is a contraction and has a unique fixed point $x_n$. This also shows that $f$ itself can have at most one fixed point. But $x_n$ is also a (hence the) fixed point of $f^{kn}$, i.e., $x_n=x_{kn}$. Then also $x_n=x_{n^2+n}=x_{n+1}$ and finally $$f(x_n)=f^{n+1}(x_n)=f^{n+1}(x_{n+1})=x_{n+1}=x_n.$$