Bounds on the derivative of a Riemann map 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 $U = \mathbb{D}$.
 A: 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
A: 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:


*

*Some part of the boundary is far away, hard to reach 

*Some part of the boundary is too close to $0$, so it intercepts too many particles.   

*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 $|z|\le 1-d(z)$.  
