Theorem: Suppose $$U\in\mathbb{R}^n$$ is a convex set. A differentiable function $$f:U\in\mathbb{R}^n \to U$$ is a contraction mapping if $$sup( \| f' \| )<1$$ .
Proof:

Suppose $$x$$ and $$y$$ are two arbitrary points in $$U$$.   And $$L$$  is the line segment connecting $$x$$ and $$y$$.  Then by the multivariable mean value theorem, $$\| f(x) - f(y) \|_\infty \leq \| x - y \| _\infty \underset{c \in L}{\text{max}} \| f'(c) \|_\infty$$.  Then by setting $$k=sup( \| f' \|_\infty )$$, thus $$0 \leq k<1$$.  Then

$$\| f(x) - f(y) \|_\infty \leq k \| x - y \| _\infty .$$

The last inequality states $$f$$ as a contraction mapping with the norm $$\| \; \|_\infty$$.

Define $$x_0=x ,$$ $$x_{n+1} = f(x_n)$$

Thus $$\| x_{n+1} - x_n \|_\infty = \| f(x_n) - f(x_{n-1}) \|_\infty \leq k \| x_n - x_{n-1} \|_\infty,$$ so it follows by induction that $$\| x_{n+1} - x_n \|_\infty \leq k \| x_n - x_{n-1} \|_\infty.$$  Then if $$0 \leq n \leq m$$, then

$$\| x_n - x_m \|_\infty$$

$$\leq \| x_n - x_{n+1} \|_\infty +\dotsb+\| x_{m-1} - x_m \|_\infty$$

$$\leq ( k^n +\dotsb+ k^{m-1} )\| x_1 - x_0 \|_\infty$$

$$\leq k^n \| x_1 - x_0 \|_\infty ( 1 + k + k^2 + \dots )$$

$$\leq \frac{ k^n \| x_1 - x_0 \|_\infty }{ 1 - k }$$

By equivalence of norms, $$m \| \; \| \leq \| \; \|_\infty \leq M \| \; \|$$.

$$m \| x_n - x_m \| \leq \| x_n - x_m \|_\infty \leq \frac{ k^n \| x_1 - x_0 \|_\infty }{ 1 - k } \leq \frac{ k^n M \| x_1 - x_0 \| }{ 1 - k }$$

Therefore

$$\| x_n - x_m \| \leq \| x_n - x_m \|_\infty \leq \frac{ k^n \| x_1 - x_0 \|_\infty }{ 1 - k } \leq \frac{M}{m} \frac{ k^n \| x_1 - x_0 \| }{ 1 - k }$$.

As $$n \to \infty$$, $$\| x_n - x_m \| \to 0$$.  Hence for an arbitrary norm, $$\{x_n\}_0^\infty$$ does converge to one point if $$sup( \| f' \|_\infty )<1$$.  By the same reasoning, it can be shown that if $$sup( \| f' \| )<1$$, then the sequence $$\{ x_n \}_0^\infty$$ converges in terms of $$\| \; \|_\infty$$.  That is, regardless of the norm actually used, $$f$$ is a contraction mapping.