Assume that $D\_n\subset D$, where $D$ is the unit disk is an increasing sequence of Jordan domains with smooth boundaries such that $\cup\_{n=1}^\infty D\_n=D$ and let $f\_n: D \to D\_n$ be conformal mapping such that $f(0)=0$ and $\arg(f\_n'(0))=0$. Is $f\_n$ uniformly convergent to $Id$.



Uniform convergence of conformal mappings
The answer is no. Consider the following sequence $D\_n$. Take the disc $|z|<1-1/n=r\_n$,
and remove from it the arc $\{ z:|z|=r\_{n-1},|\arg z|<\pi-1/n\}$ and the interval
$(r\_{n-1},r\_n)$. The result is a simply connected region $D\_n$.
This $D\_n$ contains $|z|<r\_{n-1}$, so the union of $D\_n$ is the whole unit disc,
and they are nested. Small deformation will make these regions Jordan, smooth, etc.


Evidently the convergence of the conformal maps $f:D\to D\_n$ is uniform on compact subsets
of $D$ (this follows from Caratheodory thm), but not uniform in $D$, because near the point $z=-1$ the function $f\_n(z)$ takes values close to every point of the unit circle.
(A small neighborhood of $-1$ blows up under $f\_n$ to almost
the whole ring $r\_{n-1}<|z|<r\_n$.)


