# Banach theorem example

By Banach fixed point theorem, if a metric on a metric space $X$ is such that $d(f(x),f(y))\leq K d(x,y)$ for $K\in (0,1)$ then $f$ has one unique fixed point.

Is there an example where $d(f(x),f(y))\leq K d(x,y)$ does not have a fixed point if $K=1$?

What if $X$ is a compact space?

-
$X$ must be complete –  Marco Castronovo Dec 10 '11 at 23:22
To see that the conclusion isn't always true when the space is complete, let the space be the real line minus the singleton of $0$, and let $f(x)=x/2$. –  Michael Hardy Dec 11 '11 at 0:44
I don't think that space is complete Michael: the sequence $( \frac{1}{n})$ is Cauchy without a limit in the given space. –  Geoff Robinson Dec 29 '11 at 9:31

For $X=\mathbb{R}$, $d$ the usual metric, and $f(x)=x+1$, we have $d(f(x),f(y)) = |(x+1)-(y+1)| = |x-y| = d(x,y)$, so it satisfies the desired inequality with $K=1$, but $f$ clearly has no fixed point.

For a compact example, take $X=S^1$ with the distance inherited by embedding it in $\mathbb{R}^2$ as the unit circle, and $f$ a rotation that is not by an angle that is a multiple of $2\pi$. Again, $d(f(x),f(y))=d(x,y)$ for all $x$ and $y$ , but $f$ has no fixed points.

-
If $K=1$, you can even have the stronger condition $d(f(x),f(y))<Kd(x,y)$ without having a fixed point (provided the space is not compact). For example, let the space be the real line with the usual metric, and let $f(x)=\sqrt{x^2+1}$.
If the space is compact, then $d(f(x),f(y))/d(x,y)$ must attain its maximum value $C$, which must be less than $1$. Then we have $d(f(x),f(y))\leq Cd(x,y)$ with $0<C<1$ so you'd have the usual conclusion about contraction mappings having a unique fixed point, which his attractive, if it's a complete metric space.
The explanation here is false because $d(f(x),f(y))/d(x,y)$ is not defined on the diagonal (where $x = y$) and the complement of the diagonal in $C \times C$ isn't compact, hence the justification for the maximum being achieved is wrong. In fact on a compact space one can have $d(f(x),f(y)) < d(x,y)$ without $f$ being a contraction on a nbd. of the fixed point. For ex, let $f(x) = x/(1+x)$ with $C = [0,1]$. The fixed point is $x = 0$ and $|f(x)-f(y)|/|x-y|$ is arb. close to 1 when $x$ and $y$ are suff. close to 0. Neither $f$ nor any iterate of $f$ is a contraction on any neighborhood of 0. –  KCd Dec 29 '11 at 13:13