What do balls looks like in the metric $\max\{\lvert x_1-y_1\rvert,\lvert x_2^2-y_2^2 \rvert\}$

Question

I am working in the metric space $(X,d)$, where $X=\mathbb{R}\times[0,\infty)$ and $d\colon X\times X\to \mathbb{R}$ defined by $$ d\big[(x_1,x_2),(y_1,y_2)\big]=\max\big\{\lvert x_1-y_1\rvert,\lvert x_2^2-y_2^2 \rvert\big\}, $$ and I am trying to get a handle on what balls look like. As an example, I was trying to establish the boundary of the ball $B_1\big((0,0)\big)$ centred at the origin and radius 1. I believe the ball intersects the coordinate axes at $(\pm1,0)$ and $(0,1)$, but I am struggling to determine whether, for example, the boundary is curved (parabolic?) or whether is assumes the shape of the square, sides parallel to the coordinate axes, as I have seen in similar examples. Would anyone be able to explain what this ball looks like, and perhaps explain an intuitive way of apply this to similar examples (i.e. other metrics defined in terms of the max function)?

Thanks.

Answer 1

I think the first term should be $|x_1-y_1|$, otherwise it is independent of $y_1$.

The ball $B((x_1,x_2),1)$ is a rectangle $$x_1-1<y_1<x_1+1,\\\sqrt{x_2^2-1}<y_2<\sqrt{x_2^2+1}$$

Of course, it is trimmed at $y_2=0$ if $x_2<1$

Answer 2

Consider a map $f : X\rightarrow D\subset (\mathbb{R}^2,\|\ \|_\infty) ,\ f(x_1,x_2) = (x_1,x_2^2) $ where $D=\mathbb{R}\times [0,\infty)$. Then it is an isometry.

$f$ sends a line $ \{(t,y)|t\in \mathbb{R}\}$ onto line and a half line $\{ (x,t )| t\geq 0\}$ onto $\{ (x,t )| t\geq 0\}$. And so does $f^{-1}$.

So any square in $D$, which is $R$-ball, is mapped onto a rectangle.