# When are uniform embeddings quasisymetric

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.

• 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.
– YCor
Mar 28 at 9:12
• For a Lipschitz map the inverse simply doesn't exist.
– YCor
Mar 28 at 15:54
• 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...
– YCor
Mar 28 at 16:53
• 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. Mar 28 at 18:01
• @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). Mar 29 at 6:58

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 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.

• 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 $\cdot^{1/\alpha}$ . Mar 29 at 12:54
• @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. Mar 29 at 13:00
• Ah okay, so then I guess the "correct condition" is simply $f$ is quasi-symmetric and we cannot hope for much more? Mar 29 at 13:01
• ok, yes, that would work too. Mar 29 at 13:01
• 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 :) Mar 29 at 13:02