Circle on taxicab metric on $\mathbb{R}^2$

Question

The taxicab metric of $\mathbb{R}^n$ is defined by d($\vec{v}$,$\vec{u}$)= $\sum\limits_{i=1}^n$$|v_i−u_i|$. Show that the taxicab metric does satisfy all of the properties of a distance function. What is a circle in the taxicab metric on $\mathbb{R}^2$? Why is it called the taxicab metric? Note that the taxicab metric is actually the distance function defined by the p- norm defined above but with p=1.

Answer 1

For a vector $x\in\mathbb R^n$, define $$\|x\|_1:=\sum_{i=1}^n|x_i|.$$ We want to show that this is a norm. Positive-definiteness is clear from the observation that the the sum of nonnegative numbers is nonnegative, and equal to zero iff each number is zero. If $a\in\mathbb R$ then $$\|ax\|_1 = \sum_{i=1}^n |ax_i| =|a|\sum_{i=1}^n|x_i|, $$ establishing homogeneity. Finally, if $x,y\in\mathbb R^n$ then $$\|x+y\|_1 = \sum_{n=1}^\infty |x_i+y_i|\leqslant \sum_{i=1}^n |x_i| + \sum_{i=1}^n |y_i|, $$ which follows from the triangle inequality on real numbers.

The norm $\|\cdot\|_1$ then defines a metric on $\mathbb R^n$ by $$d_1(x,y) = \|x-y\|_1. $$ It is clear from inspection that $d_1(x,y)=0$ iff $x=y$ and from symmetry that $d_1(x,y)=d_1(y,x)$. It remains to verify the triangle inequality; if $x,y,z\in\mathbb R^n$ then \begin{align} d_1(x,z) &= \sum_{i=1}^n |x_i-z_i| \\ &= \sum_{i=1}^n |x_i-y_i+y_i-z_i|\\ &\leqslant \sum_{i=1}^n |x_i-y_i| + \sum_{i=1}^n |y_i-z_i|\\ &= d_1(x,y) + d_1(y,z). \end{align}

A "circle" in the $l^1$ metric with center $p$ and radius $r$ is the set $$\{x\in\mathbb R^n : d_1(x,p) = r\}.$$

I will leave it to you to figure out what that looks like (in two dimensions, that is!).