Let $(X,d)$ be a compact doubling metric space with doubling constant $C>0$. Let $\{\mathbb{X}_n\}_{n=0}^{\infty}$ be a sequences of finite subsets of $X$ with $$ \left\{B\left(x_k,\frac1{n}\right)\right\}_{k=0}^{\mathbb{X}_n} \mbox{covers } X \mbox{ and } \#\mathbb{X}_n \mbox{is the $\frac1{k}$-covering number of $X$}. $$ Fix some $x^{\star}\in X$ and consider the associated sequences of $1$-Lipschitz maps $$ K_n:\,x\mapsto \left(d(x,x_n)-d(x_n,x^{\star})\right)_{x_n\in \mathbb{X}_n}. $$

In the case where $(X,d)$ is a compact Riemannian manifold, then this paper of Katz and Katz (with un unpublished quantitative version found here) shows the Kuratowksi embedding by first showing (in their proof) that the $K_n$ approximate the Fréchet embedding $$ K_{\infty}:x\mapsto \left(d(x,\cdot)-d(\cdot,x^{\star})\right) \in \ell^{\infty}. $$

What I mean is, if $(X,d)$ is compact (and possibly doubling as above) then can we always find sets $\{\mathbb{X}_n\}_{n=0}^{\infty}$ such that $$ \lim\limits_{n \uparrow \infty}\,\max_{x\in X}\,|K_n(x)-K_{\infty}(x)|=0? $$

I'm assuming so, but I really can't see what the obstruction would be/why I can't find this result in the literature...

For instance, if $\emptyset\neq X'\subsetneq X$ and $X$ is a compact Riemannian manifold without boundary, then the construction should work by only considering points in $X'$. Unless I'm missing something, this should work and I guess it should work for any doubling space by applying Assouad and ``replacing'' Lipschitz with Hölder?

  • $\begingroup$ It seems that your formulation is wrong --- the functions under max have different domains. It can be made right, and then it is easy to prove. $\endgroup$ Mar 1, 2022 at 22:54
  • $\begingroup$ @AntonPetrunin It is fixed, the target embedding is the Fréchet one. $\endgroup$ Mar 2, 2022 at 10:01


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