Is a holomorphic $f\colon U\to\mathbb{C}$ with continuous extension to $\overline{U}$ Lipschitz continuous on $\partial U$?

Question

Let $U\subset\mathbb{C}$ be a bounded connected open subset with smooth boundary $\partial U$. Suppose that we have a holomorphic function $f\colon U\to\mathbb{C}$ that can be continuously extended to the closure $\overline{U}$. Is $f$ then Lipschitz continuous on the boundary $\partial U$?

In other words, does there exist a constant $L\geq0$ such that $$|f(z_{1})-f(z_{2})|\leq L|z_{1}-z_{2}|$$ for all $z_{1},z_{2}\in\partial U$?

According to this MSE post one has $|f'(z)|\leq 1/\rho(z)$ for all $z\in U$, where $\rho(z)$ denotes the distance from $z$ to $\partial U$. Because of this bound, I have the feeling that it may happen that $|f'(z)|$ blows up near $\partial U$. So I suspect that my claim is false in general.

Any suggestions or examples would be greatly appreciated.

Answer 1

An example: $U = B(1, 1)$ (the disk with center $1$ and radius $1$) and $f(z) = \sqrt z$ (the principal branch of the square root with $-\pi/2 < \arg \sqrt z < \pi/2$). $f$ can be continuously extended to the closure of the disk (with $f(0) = 0$). But $$ f(z) - f(w) = \frac{1}{f(z)+f(w)} (z-w) $$ shows that $f$ is not Lipschitz continuous near the origin.

Answer 2

I think this already fails for the unit disc $U=\mathbb{D}$. Let $A(\mathbb{D})$ be the disc algebra (the Banach algebra of all $f \in C(\overline{\mathbb{D}},\mathbb{C})$ which are holomorphic in $\mathbb{D}$, endowed with the maximum norm $\|\cdot\|$). Let $$ M_n:=\{f \in A(\mathbb{D}): |f(z)-f(w)| \le n |z-w| ~~ (z,w \in \partial \mathbb{D}) \}, ~ n \in \mathbb{N}. $$
Then each $M_n$ is closed, and each $M_n$ has empty interior: If $f \in M_n$ and $\varepsilon > 0$, then the function $g(z):= f(z)+ \varepsilon z^m$ has distance $\|f-g\| \le \varepsilon$ but $g \notin M_n$ for $m$ sufficiently big. Therefore $\bigcup_n M_n$ is of first category. By Baire's Theorem most functions in $A(\mathbb{D})$ are not Lipschitz continuous on $\partial \mathbb{D}$.