proof that metric on $\mathbb{R}^2$ cannot have counterintuitive property

Question

Suppose $d$ is a metric on $\mathbb{R}^2$ whose induced topology is the same as the euclidean topology and such that $$\left\{w \in \mathbb{R}^2 : d (w,0)=1\right\}$$is nonempty and compact. Assume that for arbitrary $w=\begin{pmatrix}u\\v\end{pmatrix}\in \mathbb{R}^2$ with $d(w,0)=1$ there is $1>t_0^w > 0$ such that $$d\left( t\cdot \begin{pmatrix}u\\v\end{pmatrix} , 0\right)>d\left( s\cdot \begin{pmatrix}u\\v\end{pmatrix} , 0\right)\quad\forall t_0^w>t>s>0.$$Is there a global $t_0>0$, such that $t_0^w > t_0$ for all $w \in \mathbb{R}^2$ with $d (w,0)=1$?

Answer 1

No, there is no positive $t_0$ in general, and there exists a distance function $d$ such that $t_0$ cannot be positive, no matter how the $t_0^w$ are chosen.

Consider the continuous function $$f(x_1, x_2) := \begin{cases}2 x_2\cdot \max(0, (1-|x_2^2/x_1-1|)) & \text{ if }x_1 \neq 0\\ 0 & \text{otherwise}\end{cases}$$ on $\mathbb R^2$.

Now use this $f$ to define the metric $d(x,y) := |x_1-y_1| + |x_2-y_2| + |f(x) - f(y)|$. This metric induces the usual euclidean topology on $\mathbb R^2$. This can be seen by the fact that $d(x,y) = d_1((x,f(x)), (y,f(y)))$, where $d_1$ is the taxicab metric on $\mathbb R^3$ and the map $x \mapsto (x,f(x))$ is an embedding of $\mathbb R^2$ into $\mathbb R^3$.

Now consider any $w = (u, v)$ with $d(w,0) = 1$. Then for sufficiently small $t$, $$d(tw, 0) = t|u| + t|v| + 2t\left|v \cdot \max(0, (1-|tv^2/u-1|))\right|$$ $$= t\cdot(|u|+|v|) + \begin{cases}2t^2|v^3/u| & \text{ if } u > 0 \\ 0 & \text{ otherwise}\end{cases}$$

which is strictly ascending in $t$. So there does exist a $t^w_0$ as desired.

Fix a choice of all $t^w_0$, and let $t_0$ be the infimum of all $t^w_0$.

However, for every small $\varepsilon > 0$, consider the vector $\tilde w := (\varepsilon, 1)$, and let $w := \frac{1}{1+\varepsilon}\tilde w$.

For sufficently small $\varepsilon$, we have $0 < \varepsilon < 2\varepsilon < 1 < \frac{1}{1+\varepsilon}$.

Then it is true that $d(0, \varepsilon \tilde w) = \varepsilon + \varepsilon^2 + 2\varepsilon$, but $d(0, 2\varepsilon \tilde w) = 2\varepsilon + 2\varepsilon^2 + 0 < d(0, \varepsilon \tilde w)$ and $d(0, w) = 1$.

Choosing $t = 2\varepsilon(1+\varepsilon), s = \varepsilon(1+\varepsilon)$, we just found out that $t > s$ but $d(0, sw) > d(0,tw)$.

Therefore $t_0^w$ must be smaller than $2\varepsilon(1+\varepsilon)$, and $t_0$ must be smaller than that. Unfortunately, when $\varepsilon$ goes to $0$, so does $\varepsilon(1+\varepsilon)$, so $t_0$ cannot be positive. $~~~~\square$

Here's a plot of $f$ for reference: