This is an example from a text to show that this mapping does not have a fixed point because it is contractive but not a contraction:
I am not sure what the difference is between contractive and contraction. Doesn't the function satisfy all the requirements of the fixed point theorem/
On
another example, the hyperbola (upper branch) $$ 3y^2 - 4yx + x^2 = 5 $$ or $$ y=\frac{2x+\sqrt{15+x^2}}{3}$$ When $x\geq 0,$ we see $\sqrt{15+x^2} > x$ so that $2x+\sqrt{15+x^2} > 3x$
The trouble occurs because the derivative gets arbitrarily close to $1$
$$ y'=\frac{x+2\sqrt{15+x^2}}{3\sqrt{15+x^2}}$$
as $x$ increases without bound
A contraction map is a map $f$ such that there exists a $0 \le k < 1$ such that $$|f(x) - f(y)| \le k|x - y|$$ for all $x$ and $y$ in the domain.
A contractive map, also called a shrinking map, is a map $f$ for which $$|f(x) - f(y)| < |x - y|$$ for all $x$ and $y$ in the domain. Not all contractive maps are contraction maps, as the example points out. Although contraction maps must have a (unique) fixed point, contractive maps may not have any fixed point.
On compact metric spaces, contractive and contraction maps are the same.