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visits | member for | 2 years, 4 months |
seen | Jun 24 '13 at 20:36 | |
stats | profile views | 557 |
Jun 25 |
awarded | Tumbleweed |
Jun 24 |
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Lipschitz parametrization of a symmetric convex curve
@ Anton, I came back to the original question, since there is not a counterexample yet. |
Jun 24 |
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Lipschitz parametrization of a symmetric convex curve
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Jun 22 |
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Lipschitz parametrization of a symmetric convex curve
@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). |
Jun 20 |
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Lipschitz parametrization of a symmetric convex curve
@Anton: I didn't understand this argument indeed. How do you relate angles with bi-Lipschitz. Maybe it is not quasiconformal at the corrners. |
Jun 20 |
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Lipschitz parametrization of a symmetric convex curve
@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. |
Jun 19 |
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Lipschitz parametrization of a symmetric convex curve
$Lip(f)=1/4 \sqrt{2} \pi$ |
Jun 19 |
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Lipschitz parametrization of a symmetric convex curve
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 $Lip(f)=\sqrt{2}{4}\pi$. |
Jun 19 |
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Lipschitz parametrization of a symmetric convex curve
@Sergei I meant that the map should be a homeomorphism, otherwise the question is trivial. Your projection maybe is not onto? |
Jun 19 |
revised |
Lipschitz parametrization of a symmetric convex curve
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Jun 19 |
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Lipschitz parametrization of a symmetric convex curve
@Anton, I have made some revision of the previous question. |
Jun 19 |
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Lipschitz parametrization of a symmetric convex curve
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Jun 19 |
revised |
Lipschitz parametrization of a symmetric convex curve
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Jun 18 |
revised |
Lipschitz parametrization of a symmetric convex curve
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Jun 18 |
asked | Lipschitz parametrization of a symmetric convex curve |
Jun 7 |
accepted | Riemann isometry vs Euclidean bi-Lipschitz mapping |
Jun 7 |
asked | Riemann isometry vs Euclidean bi-Lipschitz mapping |
May 21 |
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Lipschitz map of the circle onto a triangle
Then you obtain $L^9$ growth of bi-Lipschitz constant. |
May 20 |
asked | Lipschitz map of the circle onto a triangle |
May 19 |
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Lipschitz map of the ellipse
The metric from the Euclidean plane is assumed and Mixon has the answer. |