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 $U = \mathbb{D}$.



Bounds on the derivative of a Riemann map
I wrote a [short paper](https://arxiv.org/abs/1601.02711) 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 $|z|\le 1-d(z)$. 


