TITLE: Banach theorem example
QUESTION [6 upvotes]: 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?

REPLY [5 votes]: 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 $f(x)=\sqrt{x^2+1}$.
Later note: This next paragraph has problems, but I suspect it can be fixed by adding something.....
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.