Control on Conformal map

Let $\Omega$ be smooth simply connected open set of $\mathbb{R}^2$ such that $\overline{\Omega}$ is compact. We know that there exists a conformal diffeomorphism $\psi$ from $\mathbb{D}$ to $\Omega$. Can we have a control of the derivative of the " best one" with respect to geometric quantities, that is to say do we have $$\inf_{\phi \in M} \lvert \nabla (\psi \circ \phi)\rvert \leq C$$ where $M$ is the group of conformal diffeo of $\mathbb{D}$ and $C$ depends only on quantities such as the area of $\Omega$, the length and the total curvature of $\partial \Omega$ (which is smooth here).

In fact, it will be more geometrical to try to get

$$\inf_{\phi \in M}\left( \lvert \nabla (\psi \circ \phi)\rvert+ \lvert \nabla (\psi \circ \phi)^{-1}\rvert\right) \leq C$$

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In the earlier question there was also some confusion about whether you want to control just $\sup |\nabla \psi|$, or $\max(\sup |\nabla \psi|, \sup |\nabla \psi^{-1}|)$. The second is more geometrically natural: extreme squeezing is just as bad as extreme stretching. What are your thoughts on this? // I ask to be sure that the question you ask is actually the question you want answered. – user53153 Feb 8 '13 at 19:20
I think, there are equivalent, but i have had some precision. – Paul Feb 8 '13 at 19:37

1 Answer

Note: the answer consists of three loosely related parts. Too much to say.

Before specializing to conformal maps, it's worth discussing general diffeomorphisms with $C^1$ control. There is a nice necessary and sufficient condition for their existence.

Definition 1. A Jordan domain $\Omega$ is inner chordarc if there exists a constant $C$ such that for any $a, b\in \partial \Omega$ and any open Jordan arc $\gamma \subset \Omega$ with endpoints at $a$ and $b$, the shortest path from $a$ to $b$ along $\partial \Omega$ has length at most $C\operatorname{length}(\gamma)$.

Thinking of $\Omega$ as a lake, we can reformulate the condition: if you can walk $C$ times faster than you can swim, then it is always faster to walk along the shore than to swim. All convex domains are obviously inner chordarc, though the constant deteriorates if the domain is long and thin. About the only thing that the inner chordarc condition prohibits is owtward-pointing cusps. Inward cusps and nonzero angles are okay. We have the following very nice result:

Theorem 1. A Jordan domain $\Omega$ is inner chordarc if and only if there exists a diffeomorphism $\psi:\mathbb D\to \Omega$ such that both $|\nabla \psi|$ and $|\nabla (\psi^{-1})|$ are uniformly bounded.

I'm pretty sure that the bounds are quantitative: i.e., $C$ in Definition 1 controls the derivatives in Theorem 1. But you should check with the source:

J. Väisälä, Homeomorphisms of bounded length distortion, Ann. Acad. Sci. Fenn. Ser. A I Math. 12 (1987), 303-312.

Now restrict the consideration to conformal maps. The inner chordarc condition is still necessary, of course. But it is now far from sufficient: for example, the conformal map of the disk onto the square behaves rather badly near the vertices (even though the square is a perfectly nice inner chordarc domain). This is where the Cauchy-Riemann equation shows its rigidity: the angular distortion $\pi$ to $\pi/2$ must be exactly matched by radial distortion, which results in blow up of derivatives at the boundary. By slightly smoothing the corners we can make the derivative very large for a smooth domain. Granted, the smoothened square has a large integral of square curvature.

An example with moderate curvature is a dumbbell-shaped domain:

You can imagine making the middle part arbitrarily thin while keeping curvature uniformly bounded everywhere. This will make the constant $C$ in Definition 1 rather large, and with it the $C^1$ norm of any diffeomorphism onto disk, in particular conformal ones. Examples like this one is why I'm skeptical of the attempt to control diffeomorphisms by curvature of the boundary.

In the positive direction, let's begin with an easy observation:

Suppose $f:\mathbb D\to\Omega$ is a conformal map. Then $$\operatorname{dist}(f(0),\partial \Omega)\le |f'(0)|\le 4\operatorname{dist}(f(0),\partial \Omega)\tag{Koebe}$$

This is a basic Koebe-1/4 estimate, e.g., Corollary 1.4 in Boundary behaviour of conformal maps by Pommerenke. More serious stuff is found in Chapter 3 of the same book, in particular Theorem 3.5. The assumption here is that the boundary $\partial \Omega$ admits a parametrization with Dini-continuous tangent vector (this is somewhat geometric, or can be obtained from geometric conditions). Then it turns out that for the conformal map $f:\mathbb D\to \Omega$ the argument $\arg f'$ is Dini-continuous as well. This implies that its harmonic conjugate $\log |f'|$ is also Dini-continuous (and a bound for it can be obtained). This bound on $\log |f'|$ gives the two-sided bound on $|f'|$, as desired.

Technical point: the harmonic conjugate is determined only up to an additive constant. This is why we need some control of $\log|f'(0)|$, stated above as (Koebe). In terms of your question, this means using a Möbuis map to make sure that the center of disk is mapped into a "thick" part of $\Omega$, preferably into a point that realizes its inradius.

Conclusion: if the domain has a point with inradius comparable to $1$, and if its boundary has a parametrization with decent modulus of continuity for the tangent vector, then there is a conformal diffeomorphism with controlled derivative.

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Thanks you for your nice answer! Indeed i have also realize that it is false for conformal maps. I have almost the proof of the following result: there exists a conformal diffeo from $\Omega$ to $\mathbb{D}$ whose gradient is uniformly bounded by the total curvature and the diameter iff the total curvature is strictly smaller than $\4 pi$ and the diameter smaller than $4$. Which is confirm by your conter example. – Paul Feb 17 '13 at 9:57