What will be a circle look like considering this distance function?

I am working on some exercises in the book Geometry: A Metric Approach with Models by R.S. Millman. He defines the following map: $$d_S(P,Q):\mathbb R^2\times\mathbb R^2\to\mathbb R\\\ d_S(P,Q)=\max\{~|x_2-x_1|,~|y_2-y_1|~\}$$ where in $P(x_1,y_1),~Q(x_2,y_2)$ are points in $\mathbb R^2$ and then wants the reader to show that $d_S$ is a distance function on $\mathbb R^2$.

I was thinking to myself of what would be a circle looked like regarding this function in Cartesian plane? I fixed a point, for example the origin, as a centre and then probed the case with a given positive constant $r$ as a radius. I just got two horizon $|y|=r$ or two vertical $|x|=r$ lines. the results didn't have good taste. Any help? Was I right? Thanks

Solution 1:

Yes. What you obtained is right. In general, the $p$-metric, where $p \geq 1$, is defined as $$d_S(P,Q) = \left(\vert x_2 - x_1 \vert^p + \vert y_2 - y_1 \vert^p\right)^{1/p}$$ Now it is easy to prove that as $p \to \infty$, we get the metric $\max\{|x_2-x_1|,|y_2-y_1|\}$. Now you can use a plotting software and see how unit ball looks like for different $p$'s. You will find the for $p=1$, it is a diamond and as you increase $p$ you encompass "more" and "more" region in the plane. For $p=2$, you get the "usual" circle and for $p \to \infty$, you get the "unit square".

The diagram below shows the unit ball for different $p$'s. The inner most diamond is the case for $p=1$, which is also called as the taxi-cab metric. As $p$ increases, the "area" enclosed by the unit ball slowly increases. The outer unit square is the metric you get when you let $p \to \infty$, i.e., you get the metric $\max\{|x_2-x_1|,|y_2-y_1|\}$.

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Solution 2:

We set

$$\max\{|x_2 - x_1|, |y_2 - y_1|\} = r$$

with $r > 0$ being a "radius".

Take a center for the "circle" -- say $(0, 0)$, to make this

$$\max\{|x|, |y|\} = r$$.

Now, this means at least one of $|x|$ and $|y|$ is $r$, and the other less. So there is a set of all $x$ for which $y = r$ and $|x| < r$, which is a horizontal line segment. For $y = -r$, we have another such segment, on the other side of the x-axis. Repeating for $x = r$ and $|y| < r$ and $x = -r$ and $|y| < r$, we get two vertical line segments. Altogether, we get a square with side lengths $2r$.