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.

Distance function is a submetry

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:

These aren't the quotients I'm looking for

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.

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