Proving Contraction Mapping: The Linear Map $T$

[ozkan12]

The linear map

$T:{R}^{2}\to {R}^{2}$, $T(x,y)=\left(\frac{8x+8y}{10},\frac{x+y}{10}\right)$ is not a contraction with respect to the Euclidean metric, but is a contraction with respect to

$d(x,y)=\sum_{i=1}^{n}\left| {x}_{i}-{y}_{i} \right| $, $x,y\in {R}^{n}$

contraction constant= 9/10

My questions:

Please show that T is not contraction with respect to Euclidian Metric.

Please show that T is contraction with respect to $d(x,y)=\sum_{i=1}^{n}\left| {x}_{i}-{y}_{i} \right |$ $x,y\in {R}^{n}$

 

[jvicens]

Hello,

Thank you for sharing this interesting problem. I will be happy to show why $T$ is not a contraction with respect to the Euclidean metric and why it is a contraction with respect to the metric $d(x,y)$.

First, let's define the Euclidean metric as $d_E(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}$.

To show that $T$ is not a contraction with respect to $d_E$, we need to find two points $x,y\in{R}^2$ such that $d_E(T(x),T(y))>d_E(x,y)$. Let's choose $x=(1,0)$ and $y=(0,1)$. Then, we have $T(x)=\left(\frac{8}{10},\frac{1}{10}\right)$ and $T(y)=\left(\frac{8}{10},\frac{1}{10}\right)$. Therefore, $d_E(T(x),T(y))=\sqrt{(\frac{8}{10}-\frac{1}{10})^2+(\frac{1}{10}-\frac{1}{10})^2}=\sqrt{\frac{49}{100}}=\frac{7}{10}$. However, $d_E(x,y)=\sqrt{(1-0)^2+(0-1)^2}=\sqrt{2}$. Since $\frac{7}{10}>\sqrt{2}$, we can conclude that $T$ is not a contraction with respect to $d_E$.

Next, let's show that $T$ is a contraction with respect to $d(x,y)$. We can rewrite the metric $d(x,y)$ as $d(x,y)=|x_1-y_1|+|x_2-y_2|$. Now, for any two points $x,y\in{R}^2$, we have $d(T(x),T(y))=|T(x)_1-T(y)_1|+|T(x)_2-T(y)_2|=\left|\frac{8x_1+8x_2}{10}-\frac{8y_1+8y_2}{10}\right|+\left|\frac{x_1+x_2}{10}-\frac{y_1+y_2}{10}\right|=\