Quasi-projections and isometries


Discontinuous nearest-point projection



Continuous quasi-projection



An isometry must incur sqrt(2) distance cost






The shortest circle is a hexagon

Let {\|\cdot\|} be some norm on {{\mathbb R}^2}. The norm induces a metric, and the metric yields a notion of curve length: the supremum of sums of distances over partitions. The unit circle {C=\{x\in \mathbb R^2\colon \|x\|=1\}} is a closed curve; how small can its length be under the norm?

For the Euclidean norm, the length of unit circle is {2\pi\approx 6.28}. But it can be less than that: if {C} is a regular hexagon, its length is exactly {6}. Indeed, each of the sides of {C} is a unit vector with respect to the norm defined by {C}, being a parallel translate of a vector connecting the center to a vertex.

Hexagon as unit disk
Hexagon as unit disk

To show that {6} cannot be beaten, suppose that {C} is the unit circle for some norm. Fix a point {p\in C}. Draw the circle {\{x\colon \|x-p\|=1\}}; it will cross {C} at some point {q}. The points {p,q,q-p, -p, -q, p-q} are vertices of a hexagon inscribed in {C}. Since every side of the hexagon has length {1}, the length of {C} is at least {6}.

It takes more effort to prove that the regular hexagon and its affine images, are the only unit circles of length {6}; a proof can be found in Geometry of Spheres in Normed Spaces by Juan Jorge Schäffer.

Diameter vs radius, part II

A set A in a metric space X has diameter \mathrm{diam}\, A=\sup_{a,b\in A} |a-b| and radius \mathrm{rad}\, A = \inf_{x\in X}\sup_{a\in A} |a-x|. It’s easier to find the radius by expressing it in a different form: the smallest (or infimal) value of r such that the \bigcap_{a\in A} \overline{B}(a,r)\ne\varnothing, where \overline{B}(a,r) is the closed ball of radius r with center a.

Suppose X is a normed linear space and f\colon A\to X is a map that does not increase distances (hence does not increase the diameter of any set). I already said that the radius may increase under f, but my example was incorrect. Here is a correct one: the set A consists of 3 points in red.

Radius increases under a nonexpanding map

The blue hexagon is the unit sphere in this space. The three points in red have distance 2 from one another. So do their images under f, but the radius increases from 1 to 2/\sqrt{3}. The set f(A) consists of three vertices of the regular hexagon (in green) circumscribed about the blue one.

I think this example is as bad as it gets in two dimensions: that is, we should have \mathrm{rad}\, f(A) \le \frac{2}{\sqrt{3}} \mathrm{rad}\, A in any 2-dimensional normed space. Informally, the worst case is when the unit ball is triangular, which it can’t be because of the symmetry requirement. The hexagon is the next worst thing.

In higher dimensions the constant cannot be smaller than \frac{2}{\sqrt{3}}, since the above construction can be implemented in a subspace. I don’t know whether the constant grows with dimension or not (either way it can’t exceed 2, as remarked before).