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Let $X,Y$ be metric space and suppose that $f:X\rightarrow Y$ is a uniform embedding; i.e.: $$ \omega(d_X(x,z))\leq d_Y(f(x),f(z)) \leq \Omega(d_X(x,z)), $$ where $\omega\leq \Omega$ are both strictly increasing continuous functions mapping $[0,\infty)$ to itself and which fix $0$.

Is $f$ a quasisymmetry? I.e.: does there exist a monotone function $\eta:[0,\infty)\rightarrow [0,\infty)$ satisfying $$ \frac{d_Y(f(x),f(y))}{d_{Y}(f(x),f(z))} \leq \eta\left(\frac{d_X(x,y)}{d_X(x,z)}\right) . $$

Note on Edit: I have reduced my previous question down to this more general one; since quasisymmetries preserve the doubling property.

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  • $\begingroup$ If it were clear, the proof would also most likely provide a clear estimate. But it's just false. Just playing zigzag, you have a surjective 1-Lipschitz map from the reals (or half-reals) onto the 1-skeleton of a regular trivalent tree, which is not doubling. $\endgroup$
    – YCor
    Commented Mar 28, 2022 at 9:12
  • $\begingroup$ For a Lipschitz map the inverse simply doesn't exist. $\endgroup$
    – YCor
    Commented Mar 28, 2022 at 15:54
  • $\begingroup$ But the map need not be injective, and injectivity is not a necessary condition. And even for bijective maps between doubling spaces, uniform continuity can fail... $\endgroup$
    – YCor
    Commented Mar 28, 2022 at 16:53
  • $\begingroup$ this question has Lipschitz in the title but not in the question itself. It is also strangely stated. are you assuming that $f$ is a bijection? Please restate the question clearly. $\endgroup$
    – Vitali Kapovitch
    Commented Mar 28, 2022 at 18:01
  • $\begingroup$ @YCor and VitaliKpovitch I have reduced the earlier incarnation of my question down to the current formulation: when are uniform embeddings quasi-symmetries (if their moduli are well-behaved). $\endgroup$
    – Carlos_Petterson
    Commented Mar 29, 2022 at 6:58

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This is false. For example take $f: [0,\infty)\to [0,\infty)$ given by $$ f(x)=\begin{cases}0& \text{ if } x=0\\ e^{-1/x}& \text{ if } x>0 \end{cases}$$

Then $\lim_{x\to 0}\frac{f(2x)}{f(x)}=\infty$ so $f$ is not quasisymmetric. But $f$ is continuous and monotone hence gives a homeomorphism onto the image when restricted to $[0,1]$. Since $[0,1]$ is compact $f$ is uniformly continuous on it. Same for $f^{-1}$.

Let me now answer the original question (which the OP erased) which asked for a weaker conclusion which is still false. The question was the following.

Let $f: X\to Y$ be a bijection such that both $f$ and $f^{-1}$ are uniformly continuous. Suppose $X$ is doubling i.e. there exists a constant $C>$ such any ball of radius $r$ can be covered by at most $C$ balls of radius $r/2$.

Question: Does this imply that $Y$ is also doubling?

The answer is NO.

The answer is yes if $f$ is bi-Lipschitz in which case the statement is trivial but any weaker conditions on the modulus of continuity should not be sufficient.

Here is an example of a metric of the closed $2$-disk which is not doubling.

Take a 2-disk $\bar D=\{x^2+y^2\le 1, z=0\}$ in the $xy$ plane in $\mathbb R^3$ and attach 10 vertical intervals of length 2 at different points of the interior of the disk. The doubling constant of the resulting space $\tilde X_1$ on scale 1 will be about 10.

The space $\tilde X_1$ is no longer a disk but we can make it one by replacing the vertical intervals by very thin "fingers" (graphs of bump functions on tiny disks) of the same height 2 so that the the new space $X_1$ is Hausdorff close to $\tilde X_1$ and is a graph of a function from $\bar D$ to $\mathbb R$.

Now take a very small $\epsilon$, take small disk $D_\epsilon$ in $D$ if radius $\epsilon$ away from the fingers and repeat the procedure on $D_\epsilon$ on scale $\epsilon$ by attaching 100 vertical intervals of length $2\epsilon$. Replace the smaller intervals by fingers. then the doubling constant on scale $\epsilon$ will be 100. Iterate. The limit space $X_\infty$ will be a graph of a function on $\bar D$ but will not be doubling. The projection map $f: X_\infty\to \bar D$ is a homeomorphism. It's 1-Lipschitz. The inverse is uniformly continuous because the spaces involved are compact.

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  • $\begingroup$ This is an extremely nice example. So in your opinion what would you expect that we need of the inverse $f^{-1}$ since your example shows that uniform continuity of $f^{-1}$ isn't enough? My intuition was based on the snowflaking $f:(X,d)\mapsto(X,d^{\alpha})$ ($\alpha\in (0,1)$) in which case $f$ has moduli $\alpha$ and $\cdot^{1/\alpha}$ . $\endgroup$
    – Carlos_Petterson
    Commented Mar 29, 2022 at 12:54
  • $\begingroup$ @Carl_Petterson As I said I think you need $f$ to be bi-Lipschitz. I see no reason why any weaker modulus of continuity would suffice. For example should be possible to construct counterexamples when $f$ is bi-Hölder but not bi-Lipschitz. $\endgroup$
    – Vitali Kapovitch
    Commented Mar 29, 2022 at 13:00
  • $\begingroup$ Ah okay, so then I guess the "correct condition" is simply $f$ is quasi-symmetric and we cannot hope for much more? $\endgroup$
    – Carlos_Petterson
    Commented Mar 29, 2022 at 13:01
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    $\begingroup$ ok, yes, that would work too. $\endgroup$
    – Vitali Kapovitch
    Commented Mar 29, 2022 at 13:01
  • $\begingroup$ Cool or weakly quasi-symmetric (assuming $X$ is connected); ok cool, then that also covers both examples (bi-Lipschitz and the snowflake). Thanks Vitali really nice construction btw, it really helped :) $\endgroup$
    – Carlos_Petterson
    Commented Mar 29, 2022 at 13:02

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