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Let U be a simply connected domain with smooth boundary in the complex plane, and let $\mathbb D$ be the unit disc. Is there a nice sufficient condition for the existence of a biholomorphic map $f:\mathbb{D} \xrightarrow{\simeq} U$ with $$\sup_{z \in \mathbb{D}}\,\,\,\, \left| f^\prime(z)\right| \le 1$$

For comparison, $\left\| f^\prime \right\|_{L^1(S^1)}$ is independent of the choice of f (determined by the length of the boundary of U), while $\left\|f^\prime\right\|_{L^\infty(S^1)}$ can be as bad as you like even when $U = \mathbb{D}$.

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  • $\begingroup$ James: You might want to also indicate your guess for what a sufficient condition might be. Namely, the guess is that if $U$ is convex and contained in $\mathbb D$, then there there should exist a Riemann mapping $f:\mathbb D\to U$ with derivative $\le 1$. [James and I discussed that problem in person earlier today] $\endgroup$ – André Henriques May 3 '14 at 4:32
  • $\begingroup$ Andre: I left off that guess as it turned out to be incorrect. If you take a convex region with corners (e.g. a semidisk), the derivatives of the Riemann map won't lie in $H^\infty$. You can approximate this by regions with smooth boundary, and the derivatives of these Riemann maps will be large. $\endgroup$ – James Tener May 3 '14 at 18:49
  • $\begingroup$ James: It seems like a slightly odd (by which I mean not very natural-looking) question. For example, if your domain has Dini-smooth boundary, then the derivative will extend continuously to the boundary. Hence a sufficiently small rescaling of your domain will satisfy your condition. Potentially you could get something out of this by looking at the constants, but somehow I doubt it will be very nice. What applications do you have in mind? $\endgroup$ – Lasse Rempe-Gillen May 4 '14 at 18:15
  • $\begingroup$ Lasse: I admit that it's a slightly odd question, but maybe this will put it in slightly more context. Assume U is inside the disk and contains 0. Then the natural restriction map $H^2(\mathbb{D}) \to L^2(\partial U)$ is a contraction when U is given harmonic measure with pole at 0. I was curious for what domains this remains true if instead you use arclength measure on $\partial U$, which immediately holds whenever there exists an f with $\left\|f^\prime\right\|_{H^\infty} \le 1$. $\endgroup$ – James Tener May 5 '14 at 19:32
  • $\begingroup$ (continued) Your point about rescaling domains is a good one. I'd imagine there is a trade-off between "how much U differs from a disk" and "how small you have to scale U to get the property I want". My hope was that there was a way of quantifying this relationship, but perhaps this is unlikely. My original motivation was to show that a certain infinite tensor product of these operators was bounded, for which it would be great if they were contractions (although I have a different way of getting the end result now). $\endgroup$ – James Tener May 5 '14 at 19:35
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I wrote a short paper to answer this question. Here's a summary. Let's shift $U$ so it contains $0$ and normalize the biholomorphic map $f : \mathbb D\to U$ by $f(0)=0$.

By the maximum principle, it suffices to have $|f'|\le 1$ near the boundary of $\mathbb D$, which can be expressed in terms of the harmonic measure of $U$ about $0$: its density must be at least $\frac{1}{2\pi}$ everywhere. Thinking of harmonic measure as the exit distribution of Brownian motion starting at $0$, we can see three ways in which it can have low density:

  1. Some part of the boundary is far away, hard to reach
  2. Some part of the boundary is too close to $0$, so it intercepts too many particles.
  3. Some part of the boundary is too curved, hard to squeeze in (Brownian motion has a hard time traveling through narrow corridors).

Let's assume $U$ is convex. With such a domain we associate three radii:

  • Outer radius $R_O$ is the smallest radius of a disk centered at $0$ and containing $U$
  • Inner radius $R_I$ is the largest radius of a disk centered at $0$ and contained in $U$
  • Curvature radius $R_C$ is the minimal radius of curvature of $\partial U$. (Equivalently, $R_C$ is the largest radius $R$ such that $U$ can be written as a union of open disks of radius $R$.)

Note that $R_O\ge R_I$ and $R_O\ge R_C$, while there is no general relation between $R_I$ and $R_C$.

For $a,b>0$ define $$ \phi(a,b) = \begin{cases} \frac{\log a-\log b}{a-b} ,\quad & a\ne b \\ \frac{1}{a} ,\quad &a=b \end{cases} $$

Theorem. If $$ (R_O-R_C) \phi( R_I, R_C) + \frac12 \log R_C\le 0 \tag1$$ then a conformal map $f:\mathbb D\to U$ with $f(0)=0$ satisfies $|f'|\le 1$ in $\mathbb D$.

Remark: when $U$ is the disk of radius $R$, the left side of (1) is equal to $\frac12 \log R$, i.e., the requirement is $R\le 1$. So the condition (1) is sharp in some sense.

Idea of proof: for $z\in \mathbb D$, let $d=\operatorname{dist}(f(z), \partial U)$. It suffices to show $d(z)\le 1-|z|$ when $z$ is close to $\partial\mathbb D$. To do this, estimate the hyperbolic distance from $0$ to $f(z)$ from above, using the convexity of $U$ together with the existence of large disks in $U$. By the conformal invariance of the hyperbolic metric, a lower bound on it gives an upper bound on $|z|$, which leads to $|z|\le 1-d(z)$.

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  • $\begingroup$ Thank you for posting this, I missed your paper when it first came out (although I just now rediscovered an earlier form of it in my inbox)! I'm still quite interested in studying the spectrum of the absolute value of the restriction operator I mentioned. It seems like your approach would give an upper bound on its norm. Do you have a sense of how sharp this is? $\endgroup$ – James Tener Jan 21 at 2:45
  • $\begingroup$ It's sharp when $U$ is a concentric disk (the estimate just needs rescaling as in Prop. 4.9 here). Very much not sharp when $R_C$ is small, because for the restriction operator it does not really matter whether $U$ has a sharp corner, while for the conformal map onto $U$ it does. The restriction is more naturally studied in terms of Carleson condition. Indeed, the arclength measure on $\partial U$, when $U$ is a convex subdomain of $\mathbb D$, seems to satisfy Carleson's condition with a uniform $C$. $\endgroup$ – user86028 Jan 21 at 15:51
  • $\begingroup$ The Carleson measure approach yields an explicit bound (Duren's $H^p$ spaces, remark on page 163) but it's rough. OTOH we have the Fejér-Riesz inequality: the integral of $|f|^2$ over any diameter of $\mathbb D$ is at most half of the integral over $\partial \mathbb D$. So, if $U$ is a needle-like domain converging to a diameter of $\mathbb D$, the restriction operator has norm $\to 1$. It would be nice to bound the norm of restriction by something like $2$ for general convex $U\subset \mathbb D$. The bound can't be $1$, as is shown by $f(z)=z+i$ with $U=$ upper halfdisk. $\endgroup$ – user86028 Jan 22 at 15:24
  • $\begingroup$ A general bound like that would be very interesting. I wrote a short article on this topic with Mihai Putinar (arxiv.org/abs/1612.03970), but I don't think we have much to say on this topic. If you are interested, we could discuss this problem in more detail $\endgroup$ – James Tener Jan 22 at 23:25
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Just to start, there are some restrictions given by the Koebe $\frac{1}{4}$-theorem, which says that given an injective conformal map, \begin{align} f:\mathbb{D}\rightarrow \mathbb{C}, \end{align} the image of $f$ contains the ball $B\left(f(0),\frac{\lvert f'(0) \rvert}{4}\right)\subset \mathbb{C}.$

It is also worth mentioning the nice generalization of Koebe's theorem to quasi-conformal maps by Gehring & Astala here: http://projecteuclid.org/download/pdf_1/euclid.mmj/1029003136

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  • $\begingroup$ I appreciate the information on the Koebe 1/4-theorem, but your example with the disk of radius 1/5 confuses me. Multiplication by 1/5 is exactly the kind of map I'm looking for. Looking at subsets of the unit disk, the question seems like it should be about the geometry of the region, and disks are as nice as it gets. $\endgroup$ – James Tener May 3 '14 at 13:51
  • $\begingroup$ This is what I get for answering a question extremely late at night. I answered the opposite of what you asked. I'm sorry James. The Koebe theorem will give you tell you something about how BIG a domain you can map the disk to (by applying it to the inverse map). Let me think more about the question. $\endgroup$ – Andy Sanders May 3 '14 at 20:01

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