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a) Assume that $\gamma$ is a symmetric convex curve w.r.t. two orthogonal lines (such a curve is the ellipse). Is the following statement true. 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 $\gamma$.

b) Does the same question but only Lipschitz part has affirmative answer, in the class of homeomorphisms.

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Tell us which maps did you try and what are the bound which you can prove. –  Anton Petrunin Jun 19 '13 at 21:33
    
@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. –  djoke Jun 20 '13 at 16:37
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2 Answers

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.

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@Anton, I have made some revision of the previous question. –  djoke Jun 19 '13 at 8:10
    
@Anton: I didn't understand this argument indeed. How do you relate angles with bi-Lipschitz. Maybe it is not quasiconformal at the corrners. –  djoke Jun 20 '13 at 18:31
    
@djoke, the answer is updated, you will see little more than you ask. –  Anton Petrunin Jun 21 '13 at 12:24
    
@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). –  djoke Jun 22 '13 at 5:54
    
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... –  Anton Petrunin Jun 22 '13 at 10:30
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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.

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@Sergei I meant that the map should be a homeomorphism, otherwise the question is trivial. Your projection maybe is not onto? –  djoke Jun 19 '13 at 10:40
    
@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. –  Anton Petrunin Jun 19 '13 at 14:50
    
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 $Lip(f)=\sqrt{2}{4}\pi$. –  djoke Jun 19 '13 at 16:18
    
$Lip(f)=1/4 \sqrt{2} \pi$ –  djoke Jun 19 '13 at 16:19
    
@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. –  Sergei Ivanov Jun 19 '13 at 19:54
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