sufficient condition for contraction mapping I have a vector valued mapping $F:\mathbb{R}^2\rightarrow\mathbb{R^2}$, I'm wondering whether there's a sufficient condition for it to be a contraction mapping. 
For example, if $F$ is $:\mathbb{R}\rightarrow\mathbb{R}$, and $F\in C^1$, then a sufficient condition is $F'(\cdot)<1$ in all its domain. So for the $\mathbb{R}^2\rightarrow\mathbb{R^2}$ mapping, is there a similar condition, say, the spectrum of its Jacobian matrix is less than 1?
Thanks!
 A: A sufficient condition is that
$$
\|F'(x)\|\le r<1,\tag{1}
$$ 
for all $x=(x_1,_2)\in\mathbb R^2$, for some $r\in(0,1)$.
Indeed, if $(1)$ holds, then 
$$
F(b)-F(a)=\int_0^1 \frac{d}{dt}F(tb+(1-t)a)\,dt=
\int_0^1 F'(tb+(1-t)a)\cdot(b-a)\,dt.
$$
Hence
$$
\|F(b)-F(a)\|\le
\int_0^1 \|F'(tb+(1-t)a)\|\|b-a\|\,dt\le r\|b-a\|.
$$
A: Suppose
$\Vert Df(\mathbf x) \Vert \le K <1 \tag{0}$
for all $\mathbf x \in \Bbb R^2$, where $Df(\mathbf x)$ denotes the Jacobean matrix of $f(\mathbf x)$.  For $\mathbf y, \mathbf z \in \Bbb R^2$, let $\gamma(t):[0, 1] \to \Bbb R^2$ be the path
$\gamma(t) = (1 - t) \mathbf y + t\mathbf z; \tag{1}$
$\gamma(t)$ is clearly a line segment joining $\mathbf y$ and $\mathbf z$.  We have
$\dot \gamma(t) = \mathbf z - \mathbf y, \tag{2}$
a constant.  Then
$f(\mathbf z) - f(\mathbf y) = \int_0^1 \dfrac{f(\gamma(t))}{dt}dt = \int_0^1 Df(\gamma(t)) \dot \gamma(t) dt, \tag{3}$
where the chain rule has been used to establish the rightmost equality, and thus
$\Vert f(\mathbf z) - f(\mathbf y) \Vert = \Vert \int_0^1 Df(\gamma(t)) \dot \gamma(t) dt \Vert \le \int_0^1 \Vert Df(\gamma(t)) \Vert \Vert \dot \gamma(t) \Vert dt$
$\le \int_0^1 K \Vert \mathbf z - \mathbf y \Vert dt = K \Vert \mathbf z - \mathbf y \Vert. \tag{4}$
Since (4) shows that
$\Vert f(\mathbf z) - f(\mathbf y) \Vert \le K \Vert \mathbf z - \mathbf y \Vert \tag{5}$
and $K < 1$, we see that $f(\mathbf x)$ is a contraction under the stated condition (0).
The spectral radius $\rho(Df(\mathbf x))$ of $Df(\mathbf x)$ satisfies $\rho(Df(\mathbf x)) \le \Vert Df(\mathbf x) \Vert$; see here.  Since it is apparently possible that $\rho(Df(\mathbf x)) < 1$ while $\Vert Df(\mathbf x) \Vert \ge 1$, this implies that we can't use the spectral radius alone to establish contractivity in dimensions greater than $1$, at least by the preceding argument.  I don't have an example of such $f(\mathbf x)$ at hand, but I suspect the same could easily be found by some googling around.
Hope this helps.  Good Boxing Day to One and All,
and as ever,
Fiat Lux!!!
