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If $p(t)=P(e^{it})=r(t)e^{it}$, where $r(t)>0$ is a mapping between $[0,2π]$ i.e. between the unit circle $T$ and a curve, then it seems that $\mathrm{Lip}(p)=\mathrm{Lip}(P)$ but I have't the proof.

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I doubt it. Try this example: $r(t)=1+\epsilon \cos 2t$.. it may or may not work. –  user31373 Jul 12 '12 at 19:40
    
@LeonidKovalev It is not a good example. Maybe some complicated example should be given. –  Marijan Jul 12 '12 at 20:06
    
It is possible to prove that $\text{Lip}(p)\le \text{Lip}(P)$. Let $f:[0,2\pi) \to S^1, f(t)=e^{it}$ (which is clearly invertible, with inverse $g:S^1\to [0,2\pi), g(z)=\arg z\mod 2\pi$.) For every $t,s \in [0,2\pi)$ with $t\ne s$ we have $$ \frac{|p(t)-p(s)|}{|t-s|}=\frac{|P(f(t))-P(f(s))|}{|f(t)-f(s)|}\cdot\frac{|f(t)-‌​f(s)|}{|t-s|}\le \text{Lip}(P)\text{Lip}(f). $$ Since $\text{Lip}(f)=1$, it follows that $\text{Lip}(p) \le \text{Lip}(P)$. As for the other inequality I can't tell whether it's true! –  Mercy Jul 12 '12 at 23:52
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@Mercy Of course this direction is trivial, because $|e^{it}-e^{is}|\le |t-s|$. –  Marijan Jul 13 '12 at 6:15

1 Answer 1

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Turns out this is true. We may assume that $\mathrm{Lip}(p)=1$, that is, $\dot r^2+r^2\le 1$ where the dot indicates derivative with respect to $t$. Note that $r\le 1$. Our goal is to show that $|r(t)e^{it}-r(0)|\le |e^{it}-1|$ for all $t$. After squaring this simplifies to $$(1)\qquad \qquad 2\cos t\,(1-r(0)r(t))\le 2-r(0)^2-r(t)^2$$ which is trivially true when $\cos t\le 0$. So we only have to deal with $0<t<\pi/2$.

We lose no generality in assuming that $r(t)\le r(0)$. It is convenient to write $r(0)=\cos \alpha$ where $0\le \alpha\le \pi/2$. Integrating differential inequality $\dot r\ge -\sqrt{1-r^2}$, we find that $r(t)\ge \cos(\alpha+t)$. The inequality (1) can be written as $$(2)\qquad\qquad r(t)^2-2r(t)\cos\alpha\cos t+\cos^2\alpha-2(1-\cos t)\le 0$$ Since the left hand side is a convex function of $r(t)$, it suffices to verify (2) at the endpoints $r(t)=\cos(\alpha+t)$ and $r(t)=\cos\alpha$.

Plugging $r(t)=\cos(\alpha+t)$ into (2) we get $-(1-\cos t)^2\le 0$. Plugging $r(t)=\cos \alpha$ into (2) we get $-2\sin^2 \alpha \, (1-\cos t)\le 0$. QED

There are somewhat similar lemmas in Radial extension of a bi-Lipschitz parametrization of a starlike Jordan curve by Kalaj and in On quasiconformal self-mappings of the unit disk satisfying Poisson's equation by Kalaj and Pavlović, although in both cases they are about modulus-preserving maps rather than argument-preserving.

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