Is contraction mapping equivalent to having spectral radius of Jacobian matrix less than unity?

Question

Contraction mapping theorem (https://en.wikipedia.org/wiki/Contraction_mapping) enables a fixed point iteration. Let $f$ be a contraction mapping defined on a metric space. If the spectral radius of the Jacobian matrix of $f$ at a fixed point is larger than or equal to 1, the fixed point iteration may not converge. So $f$ must have the spactral radius less than 1. Conversely, if the spectral radius is less than 1, the fixed point iteration converges. So it should be a contraction. Thus, I think that a function is a contraction in a region if and only if there exists a fixed point in the region and the spectral radius of the Jacobian matrix at the fixed point is less than 1. My argument is based on whether or not the fixed point iteration works. I think the argument is not so rigorous and I'm not entirely sure it is actually true. Could you give a proof or disproof? Thank you.

Answer 1

A contraction map doesn't have to be differentiable hence doesn't have to have a well-defined Jacobian matrix. Consider $f:\mathbb{R}\to \mathbb{R}$ via $x\mapsto \frac{|x|}{2}$.