# Three weddings and a funeral

This is a continuation of the series on submetries. First, an example: the distance function to a closed convex set $K\subset \mathbb R^n$ is a submetry from $\mathbb R^n\setminus K$ onto $(0,\infty)$. Note that the best regularity we can expect from such distance function is $C^{1,1}$, the Lipschitzness of first derivatives. The second derivative does not exist at the points where level curves make the transition from straight to circular.

Now back to an attempt to introduce duality between immetries (=isometric embeddings) and submetries. Recall that the Lipschitz dual $X^\sharp$ of a pointed metric space $X\ni 0$ is the set of all Lipschitz functions $\varphi\colon X\to\mathbb R$ such that $f(0)=0$. Given the Lipschitz norm, $X^\sharp$ becomes a Banach space. Any closed ball $B_r(a)\subset X$ can be written as
(*) $B_r(a)=\lbrace x\in X\colon |\varphi(x)-\varphi(a)|\le r \text{ for all } \varphi\in X^\sharp \text{ such that }\|\varphi\|\le 1\rbrace$.
Indeed, $\subset$ is obvious, and the reverse inclusion follows by considering $\varphi(x)=d(x,a)-d(0,a)$.

It’s natural to use (*) to relate immetries to submetries, because the former are defined by $f^{-1}(B_r(f(x)))=B_r(x)$ and the latter by $f(B_r(x))=B_r(f(x))$.

In the previous post I considered four implications:

1. If $f$ is an immetry, then $f^\sharp$ is a submetry
2. If $f$ is a submetry, then $f^\sharp$ is an immetry
3. If $f^\sharp$ is an immetry, then $f$ is a submetry
4. If $f^\sharp$ is a submetry, then $f$ is an immetry

I proved (1) already. Implication (2) is also easy to prove: if $f\colon Y\to X$ is a submetry, then $\|\varphi\circ f\|=\|\varphi\|$ for every $\varphi\in X^\sharp$, because pairs of points that (almost) realize $\|\varphi\|$ can be lifted through $f$.

Here is a proof of (4). If $f^{\sharp}\colon X^\sharp\to Y^\sharp$ is a submetry, then $\lbrace \varphi\in X^\sharp \colon \|\varphi\|\le 1\rbrace = \lbrace \psi\circ f\colon \psi\in Y^\sharp, \|\psi\|\le 1\rbrace$. We can use this in (*) to obtain, for any $a\in X$ and $r\ge 0$,
$B_r(a)=\lbrace x\in X\colon |\psi(f(x))-\psi(f(a))|\le r \text{ for all } \psi\in Y^\sharp \text{ such that }\|\psi\|\le 1\rbrace$.
The latter set is nothing but $\lbrace x\in X\colon f(x)\in B_r(f(a))\rbrace$, which means $f$ is an immetry.

I already noted that (3) fails for general metric spaces, the inclusion $f\colon \mathbb Q\to \mathbb R$ being a counterexample. However, this counterexample is not impressive, because $f(B_r(x))$ is dense in $B_r(f(x))$. One could still hope that (3) holds for proper metric spaces. By definition, a metric space is proper if every closed ball is compact. (Or, equivalently, every bounded sequence has a convergent subsequence.) In metric geometry properness is a very common assumption which excludes incomplete spaces and infinite-dimensional spaces. Its relevance to submetries can be seen from their definition $f(B_r(x))=B_r(f(x))$. It requires the image $f(B_r(x))$ to be closed, which is not very likely to happen unless $B_r(x)$ is compact. (No such issues arise on the immetry side, because $f^{-1}(B_r(f(x)))$ is always closed.)

However, it turns out (3) is false even for compact spaces. The counterexample was already present on this blog:

Indeed, the Lipschitz norm of a function $\varphi$ defined on an interval is equal to the essential supremum of $|\varphi'|$. The composition with the metric quotient map given above preserves $\mathrm{ess\,sup}\, |\varphi'|$. Hence, the induced map $[0,2]^\sharp\to [0,3]^\sharp$ is an immetry.

This site uses Akismet to reduce spam. Learn how your comment data is processed.