Show that the following mapping is a contraction. I have the following problem from a past paper:
"Show that the mapping,
$$T(x_1,x_2)=\left(\frac{x_1+2x_2}5-1,\frac{x_1-2x_2}7+1\right)$$
is a contraction on $(\mathbb R^2,d_\infty)$."
I understand that the way to approach this problem is to consider two points in $\mathbb R^2$, say $x=(x_1,x_2)$ and $y=(y_1,y_2)$, and then to consider $d_\infty(T(x),T(y))$, however, I am tripping myself up with the fact that we have both components $x_1$ and $x_2$ included in both arguments of the mapping, and so, I am not too sure how to continue.
On the other hand, if I had the mapping, $T(x_1, x_2) = (\frac{x_1}2,\frac{x_2}3)$, it is easy to see that in this case,
$$d_\infty(T(x),T(y))=d_\infty\left(\frac12|x_1-x_2|,\frac13|y_1-y_2|\right)$$ 
because each argument of the mapping includes only one of the components of the input vectors.
How should I proceed with the former example?
 A: Let's denote for convenience $\Delta x=y-x$, so that $y=x+\Delta x$, and consider
$$
\Delta T=T(x+\Delta x)-T(x)=(\Delta T_1, \Delta T_2)=\Bigl(\frac{\Delta x_1+2\Delta x_2}{5},\frac{\Delta x_1-2\Delta x_2}{7}\Bigr).
$$
We need to estimate
$$
d_\infty(T(y),T(x))=\max\{|\Delta T_1|,|\Delta T_2|\}.
$$
Let's do one component at at time
$$
|\Delta T_1|=\Bigl|\frac{\Delta x_1+2\Delta x_2}{5}\Bigr|\le\frac15\underbrace{|\Delta x_1|}_{\le d_\infty(y,x)}+
\frac25\underbrace{|\Delta x_2|}_{\le d_\infty(y,x)}\le \Bigl(\frac15+\frac25\Big)d_\infty(y,x)=\frac35 d_\infty(y,x).
$$
Similarly,
$$
|\Delta T_2|=\Bigl|\frac{\Delta x_1-2\Delta x_2}{7}\Bigr|\le
\frac17\underbrace{|\Delta x_1|}_{\le d_\infty(y,x)}+
\frac27\underbrace{|\Delta x_2|}_{\le d_\infty(y,x)}\le \Bigl(\frac17+\frac27\Big)d_\infty(y,x)=\frac37 d_\infty(y,x).
$$
Therefore
$$
d_\infty(T(y),T(x))=\max\{|\Delta T_1|,|\Delta T_2|\}\le\max\left\{\frac35 d_\infty(y,x),\frac37 d_\infty(y,x) \right\}\le\frac35 d_\infty(y,x).
$$
A: $$
\begin{align}
&d_\infty[T(x_1,x_2)-T(y_1,y_2)]\\
&=d_\infty\left[\left(\frac{x_1+2x_2}5-1,\frac{x_1-2x_2}7+1\right)-\left(\frac{y_1+2y_2}5-1,\frac{y_1-2y_2}7+1\right)\right]\\
&=d_\infty\left[\left(\frac{(x_1-y_1)+2(x_2-y_2)}5,\frac{(x_1-y_1)-2(x_2-y_2)}7\right)\right]\\
&=\max\left[\left|\frac{(x_1-y_1)+2(x_2-y_2)}5\right|,\left|\frac{(x_1-y_1)-2(x_2-y_2)}7\right|\right]\\
&\le\max\left[\,\tfrac15\left|x_1-y_1\right|+\tfrac25\left|x_2-y_2\right|,\tfrac17\left|x_1-y_1\right|+\tfrac27\left|x_2-y_2\right|\,\right]\\[3pt]
&=\tfrac15\left|x_1-y_1\right|+\tfrac25\left|x_2-y_2\right|\\[3pt]
&\le\tfrac15\max\left[\,\left|x_1-y_1\right|,\left|x_2-y_2\right|\,\right]+\tfrac25\max\left[\,\left|x_1-y_1\right|,\left|x_2-y_2\right|\,\right]\\[3pt]
&=\tfrac35d_\infty\left[x_1-y_1,x_2-y_2\right]
\end{align}
$$
