Lipschitz does not imply fixed points

Question

I have the following problem in mind:

Let us say we have a function $f:X\rightarrow X$ (X is a complete metric space) and it respects that if $x\neq y$ then :

$d(f(x),f(y))<d(x,y)$.

My trouble is I want to find such a map which does not have a fixed point (unilike that case where we have $d(f(x),f(y))\leq d(x,y)$ and we can show that it has a unique fixed point).

Could someone give me a hint? Thanks

Answer 1

Wikipedia's article on the Banach fixed-point theorem gives the counterexample $$ T:[1,\infty)\to [1,\infty)\qquad T(x) = x+\frac1x $$

For $X=\mathbb R$ we could take $f(x)=\sqrt{x^2+1}$.

Answer 2

Another counterxample is $\dfrac 12 (x+\sqrt{x^2+1})$.