Absolutely continuous and rectifiable boundary

Question

Assume that $\gamma$ is a Rectifiable curve in $\mathbf{C}$ and ssume that $f$ is a bounded holomorphic function on the unit disk $U$ such that if $z_n$ converges to a boundary point of $\mathbf{U}$, then $f(z_n)$ up to a sub-sequence converges to a point $w\in\gamma$. The question arises is this mapping $f$ absolutely continuous on $\partial U$.

Answer 1

The answer is "yes". (I assume that your curve is meant to be Jordan, i.e. a simple closed curve? Otherwise, some more thought would need to be put into the argument.)

If $\gamma$ is the unit circle, then your function $f$ is a proper self-map of the unit disc, and hence a finite Blaschke product.

In general, we can write $f=\phi\circ\tilde{f}$, where $\tilde{f}$ is as above and $\phi\colon U\to V$ is a conformal map, where $\partial V$ is a rectifiable curve.

The boundary behaviour of conformal maps is well-understood. Since the boundary is locally connected by assumption, the map $\phi$ and hence $f$ extends continuously. If $\gamma$ is $C^{\infty}$, then $\phi$ is $C^{\infty}$ on $\partial U$, if $\gamma$ is Dini-smooth, then $\phi$ is $C^1$. (See e.g. Pommerenke, Boundary Behaviour of Conformal Maps.)

The case of rectifiable boundary is covered by a theorem of F. and M. Riesz, which states indeed that the map $\phi$ is absolutely continuous on $\partial U$ (see e.g. Collingwood and Lohwater, The Theory of Cluster Sets, Theorem 3.3).