How to check if a transformation $F:\mathbb{R}^2 \to \mathbb{R}^2$ is a contraction mapping? Let
$$(\forall x_1,x_2 \in \mathbb R) \qquad F \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}
= \begin{bmatrix}
   \frac{1}{4}x_1 + \frac{1}{2}x_2 - 1 \\
   \frac{1}{2}x_1 + \frac{1}{4}x_2 + 2 \end{bmatrix}.$$
How do I check/prove that $F:\mathbb{R}^2 \to \mathbb{R}^2$ is a contraction mapping? How to find the constant point?
I know how to proceed in similar cases when dealing with $\mathbb{R} \to \mathbb{R}$. I check supremum of first derivative. But I don't know what to do here.
If anybody could show me on the above example, that would be great.
 A: You have to compute derivative of $F$, lets denote it by $dF$ and show that its norm, considered as a linear operator from $R^{2} to R^{2}$ is less than $1$.
This is a consequence of, sort of, mean value theorem which states that, locally i.e. for small $||h||$
$$F(x+h)-F(x) = (dF)h + o(||h||)$$
$dF$ is a matrix build out of first derivatives of $F$.
In your case it has this form $$\begin{bmatrix}
   \frac{1}{4},\frac{1}{2} \\
   \frac{1}{2}, \frac{1}{4} 
   \end{bmatrix}$$
So, its norm, as a max of its values on vectors $x=(x_{1},x_{2})$  with $x_{1}^{2} + x_{2}^{2} <=1 $ should be less than 1.
In your case, it is easy to see that the norm is $\frac{3}{4}$.
A: $F$ is a very simple map (its linear+constant) so  you can just do the computation directly. For the first component, a simple application of Cauchy Schwarz gives
\begin{align}|F_1(x)-F_1(y)|= \left|\frac14(x_1-y_1) + \frac12(x_2-y_2)\right| 
&\le \frac14|x_1-y_1| + \frac12|x_2-y_2| 
\\&=\binom{\frac14}{\frac12}\cdot \binom{|x_1-y_1|}{|x_2-y_2|} 
\\&\le \left|\binom{\frac14}{\frac12} \right|\left|\binom{|x_1-y_1|}{|x_2-y_2|}  \right| \\
&= \frac{\sqrt 5}{4 }|x-y|
\end{align}
basically the same computation also gives $|F_2(x)-F_2(y)|\le \frac{\sqrt 5}{4 }|x-y|$. Then
$$ |F(x)-F(y)|=\sqrt{|F_2(x)-F_2(y)|^2+|F_2(x)-F_2(y)|^2} \le \frac{\sqrt{10}}4|x-y|$$
and $\sqrt{10}/4 = \sqrt{10/16}<1$ so its a contraction (using the Euclidean norm).
