Spectral radius of Jacobian and global convergence of function iterations

Question

Suppose we have a continuously differentiable mapping $f:\mathbb{R}^n\rightarrow \mathbb{R^n} $ with a unique fixed point $x^*$. If the spectral radius of the Jacobian $\rho(\nabla f(x))$ is less than unity at $x^*$, then function iterations $x_t=f(x_{t-1})$ converge to $x^*$ locally around $x^*$.

Suppose $\rho\left( \nabla f(x) \right)\leq \delta< 1$ for all $x\in \mathbb{R}^n$. Do function iterations converge to $x^*$ from any starting point $x_0\in \mathbb{R}^n$?

A related question is this: Derivative-based sufficient conditions for contraction mappings for multivariate continuous functions

Answer 1

Your first sentence is not true. Consider the mapping $f: \mathbb R^2 \to \mathbb R^2$ given by $$ f(x,y) = (x/2 + 10 y,\; y/2)$$ The only fixed point is the origin, and the Jacobian matrix is $$ \pmatrix{1/2 & 10\cr 0 & 1/2\cr}$$ which has spectral radius $1/2$, but this is not locally a contraction.

What is true is that if $f$ is locally Lipschitz with constant $k$ at all points (i.e. each $x$ has a neighbourhood $U(x)$ such that $\|f(x)-f(y)\| \le k \|x - y\|$ for $y \in U$), then $f$ is globally Lipschitz with constant $k$. Namely, for any $x$ and $y$, we use compactness to show that the line segment from $x$ to $y$ is contained in the union of finitely many neighbourhoods $U(z_i)$, and thus there is an increasing sequence $t_0 = 0,\; t_1,\; \ldots, \; t_N = 1$ such that $x_i = x + t_i (y-x)$ and $x_{i+1} = x + t_{i+1}(y-x)$ are both in some $U(z_j)$. Then $$\|f(y) - f(x)\| \le \sum_{i=0}^{N-1} \|f(x_{i+1}) - f(x_i)\| \le k \sum_{i=0}^{N-1} \|x_{i+1} - x_i\| = k \|x - y\|$$