Vtyjkyuk

This is a practice problem for an exam I am taking.

Let $(X,rho )$ be a complete metric space and $f: X rightarrow X$ a function. Writing $f^n$ for the $n$-th iterate of $f$, denote

$$c_n := sup_x,y in X, x neq y fracrho(f^n(x),f^n(y))rho(x,y).$$

Assuming that $Sigma _n=1^infty c_n < infty $, prove that $f$ has a unique fixed point in $X$.

Since we are dealing with a complete metric space, I only need to show that $f$ is a contraction mapping. I know that given $x$ and $y$ in $X$,

$$ rho(f^n(x),f^n(y)) leq c_n rho(x,y).$$

Ultimately , I want to find some $lambda < 1$ such that $rho(f(x),f(y)) leq lambda rho(x,y).$ Any suggestions on how to proceed?

So if $sum_nc_n<+infty$, $c_nundersetninftyrightarrow0$ so there exists a $ninmathbfN$ such that $c_n<1$. So there exists $ninmathbfN$ such that $f^n$ is a contraction. So by the fixed-point theorem, $f^n$ has an unique fixed-point $x$.

Because $f^n+1(x)=f^nbig(f(x)big)=fbig(f^n(x)big)=f(x)$, we conclude that $f(x)$ is also a fixed-point of $f^n$ so by unicity $f(x)=x$ and hence $f$ has a fixed-point. Unicity comes from the fact if $y$ is another fixed point of $f$, it is also a fixed-point of $f^n$ so $y=x$.