3
$\begingroup$

Assume that $\gamma$ is a convex curve which is symmetric with respect to two orthogonal lines (the ellipse is such a curve).

  1. I want to know if there exists a $(l,L)$-bi-Lipschitz mapping of the unit circle onto $\gamma$ such that $L=\mathbf{diam}(\gamma)/2$ and $l=\mathbf{dist}(\gamma,0)$, where $0$ is the center of $\gamma$.

  2. Here is a weaker version of the above question: Does there exist a $L$-Lipschitz homeomorphism of the unit circle onto $\gamma$ such that $L=\mathbf{diam}(\gamma)/2$?

$\endgroup$
2
  • $\begingroup$ Tell us which maps did you try and what are the bound which you can prove. $\endgroup$ Jun 19, 2013 at 21:33
  • $\begingroup$ @Anton. I guess that arc-length parametrization g is the best mapping for part. You can prove that $Lip(g)\le \pi/2 diam(γ)/2$, but this is not that I expect. $\endgroup$
    – djoke
    Jun 20, 2013 at 16:37

2 Answers 2

2
$\begingroup$

Consider the convex hull $L_\varepsilon$ of the points $(\pm 1,0)$ and the $\varepsilon$-disc centered at the origin for small $\varepsilon>0$.

If there is a $(\varepsilon,1)$-bi-Lipschitz map from the unit disc onto $L_\varepsilon$ then all the angles of curve bounding $L_\varepsilon$ has to be at least $\pi{\cdot}\varepsilon$ (see the sketch below).

On the other hand, $L_\varepsilon$ has two corners with angles about $2{\cdot} \varepsilon$.


The sketch. Assume $f$ maps the half-plane $H$ to the angle $A$ and has Lipschitz constants $\varepsilon$ and $1$, yet assume that $f(0)=0$ and $0$ is the tip of $A$.

Essentially we need to show that angle measure of $A$ is at least $\varepsilon{\cdot}\pi$. Note that if $B_r(x)$ lies in $H$ then $B_{\varepsilon\cdot r}(f(x))$ lies in $A$. It follows that the image of an $\alpha$-angle in $H$ with tip at $0$ intersects unit circle along an arc of length $\varepsilon{\cdot}\alpha+o(\alpha)$. Hence the result follows.

$\endgroup$
6
  • $\begingroup$ @Anton, I have made some revision of the previous question. $\endgroup$
    – djoke
    Jun 19, 2013 at 8:10
  • $\begingroup$ @Anton: I didn't understand this argument indeed. How do you relate angles with bi-Lipschitz. Maybe it is not quasiconformal at the corrners. $\endgroup$
    – djoke
    Jun 20, 2013 at 18:31
  • $\begingroup$ @djoke, the answer is updated, you will see little more than you ask. $\endgroup$ Jun 21, 2013 at 12:24
  • $\begingroup$ @Anton, Did you forget to put $\epsilon \alpha+o(\alpha)$. Remember that I am asking of bi-Lipschitz mappings between curves (not between open domains). $\endgroup$
    – djoke
    Jun 22, 2013 at 5:54
  • $\begingroup$ sure, I will correct it. No I did not realized that you need bi-Lipschitz mappings between curves; for the curves the question look much simpler... $\endgroup$ Jun 22, 2013 at 10:30
1
$\begingroup$

For the Lipschitz-only part, the answer is yes. More generally, if $\alpha$ and $\beta$ are any two convex curves and $\beta$ fits inside an $L$-homothetic copy of $\alpha$, then there exist an $L$-Lipschitz map $f$ from $\alpha$ onto $\beta$.

Indeed, by rescaling we may assume that $\beta$ is inside $\alpha$, and $L=1$. In this case, let $f$ be the nearest-point projection to $\beta$, i.e. $f(x)$ is the point on $\beta$ nearest to $x$. It is easy to see that this map is well-defined in the outer region of $\beta$ and does not increase distances.

$\endgroup$
5
  • $\begingroup$ @Sergei I meant that the map should be a homeomorphism, otherwise the question is trivial. Your projection maybe is not onto? $\endgroup$
    – djoke
    Jun 19, 2013 at 10:40
  • $\begingroup$ @djoke, it is not hard to perturb the map into a homeomorphism by making the Lipschitz constant little worse. On the other hand if you want to keep the constant then you can not do this --- square is an example. $\endgroup$ Jun 19, 2013 at 14:50
  • $\begingroup$ The square in not a counterexample. Namely, assume as we may that the square $|\gamma|=2\pi$. Then by using arc-length parametrization $f: S^1\to \gamma$ we obtain that $Lip(f)=\sqrt{2}{4}\pi$. $\endgroup$
    – djoke
    Jun 19, 2013 at 16:18
  • $\begingroup$ $Lip(f)=1/4 \sqrt{2} \pi$ $\endgroup$
    – djoke
    Jun 19, 2013 at 16:19
  • $\begingroup$ @djoke: the map is onto, but indeed it fails to be injective if the target has corners. I overlooked that. It seems that it can be approximated by a homeomorphism with the same Lipschitz constant if the target is peicewise smooth. I am not sure about the case of infinitely many corners. $\endgroup$ Jun 19, 2013 at 19:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.