# Prove that $f$ has a unique fixed point.

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

$$c_{n} := \sup_{x,y \in X, x \neq y} \frac{\rho(f^{n}(x),f^{n}(y))}{\rho(x,y)}.$$

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?

• Since we don't know $c_1<1$, it might not be possible to show such $\lambda$ exists. The easier way to deal with this problem is to define a sequence by letting $x_0$ to be arbitrary and $x_{n+1}=f(x_n)$. Then show the sequence is Cauchy using the hypothesis. – Marco Aug 31 '18 at 0:10
• Got it, thanks! – Mike D Aug 31 '18 at 0:16

So if $\sum_nc_n<+\infty$, $c_n\underset{n\infty}{\rightarrow}0$ so there exists a $n\in\mathbf{N}$ such that $c_n<1$. So there exists $n\in\mathbf{N}$ such that $f^n$ is a contraction. So by the fixed-point theorem, $f^n$ has an unique fixed-point $x$.

Because $f^{n+1}(x)=f^n\big(f(x)\big)=f\big(f^n(x)\big)=f(x)$, 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 $y=x$.