Construct on $\mathbb{R}^2$ a norm $\lVert \cdot \rVert$ such that the metric space ($\mathbb{R}^2 , \lVert \cdot \rVert$) is neither isometric to the

Question

Construct on $\mathbb{R}^2$ a norm $\lVert \cdot \rVert$ such that the metric space ($\mathbb{R}^2 , \lVert \cdot \rVert$) is neither isometric to the space ($\mathbb{R}^2 , \lVert \cdot \rVert_2$), nor to the space ($\mathbb{R}^2 , \lVert \cdot \rVert_∞$).

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Any hint ? I know that Isometry is a distance-preserving transformation between metric spaces, and I know what a metric space and norm are, but I don't really see where to start. Any help would be appreciated

Answer 1

The following is for $\mathbb{R}^2$.

Let $\|x\|_*$ $ = \|x\|_1 + \|x\|_\infty$. A little work shows that the unit ball in the $\|\cdot\|_*$ norm has $8$ extreme points. If $L$ is invertible, then the unit ball of the norm $x \mapsto \|Lx\|_*$ also has $8$ extreme points.

In a similar fashion, the closed unit balls of the norms $\|\cdot\|_\infty$ and $x \mapsto \| Lx \|_\infty$ have $4$ extreme points and similarly for the norms $\|\cdot\|_2$ and $x \mapsto \| Lx \|_2$ have an infinite number of extreme points.

Hence $(\mathbb{R}^2, \| \cdot \|_*)$ cannot be isometric to $(\mathbb{R}^2, \| \cdot \|_2)$ or $(\mathbb{R}^2, \| \cdot \|_\infty)$.

Addendum:

If $C$ is a convex set, then $z \in C$ is an extreme point of $C$ iff $z$ is not in the relative interior of any line segment contained in $C$.

If $\alpha$ is an invertible affine map, then $z$ is an extreme point of $C$ iff $\alpha(z)$ is an extreme point of $\alpha(C)$.

Note that the last statement follows because $\alpha, \alpha^{-1}$ map line segments to line segments.

In particular, since an isometry is an invertible affine map, the number of extreme points of the closed unit ball is the same as the number of mapped closed unit ball.