Is the holomorphic extension of some $f(z) : D(0,1) \rightarrow \mathbb C$, continuous on the closure, still continuous at $z_0 \in \partial D(0,1)$? Assume that $f : D(0,1) \rightarrow \mathbb C$ is holomorphic, continuous on $\overline{D(0,1)}$, and admits a holomorphic extension on some open subset $D(0,1) \subset U$.
Assume that there exists some $z_0 \in \partial D(0,1) \cap U^c$. Can $f : U \rightarrow \mathbb C$ be extended continuously at $z_0$ ?
In other words, we know that the limit $$\lim \limits_{z \to z_0, z \in D(0,1)}f(z)$$ exists but does
$$\lim \limits_{z \to z_0, z \in U}f(z)$$
necessarily exist ?
I'm particularly interested in the case where, say, $z_0 = 1$ and $U$ is shaped like $$U = D(0,1) \cup \{z \in \mathbb C : |z| < R, z \neq 1, |\arg(z-1)| > \phi\}$$
for some $R > 1$ and $0 < \phi < \frac{\pi}{2}$ (See Figure VI.6 at p.390 of Analytic Combinatorics).
This seems somewhat related to what they call the Lindelöf Principle in Chapter 6.2 of Geometric Function Theory.
 A: A comment that got too long and gives a sketch to show that the question here has a negative answer at least in the $\Delta$ analytic case above where $$U = D(0,1) \cup \{z \in \mathbb C : |z| < R, z \neq 1, |\arg(z-1)| > \phi\}$$ for some $R>0, 0<\phi<\pi/2$, or similar cases where the boundary of $U$ is a Jordan curve that touches the unit circle at $1$ (or any $|z_0|=1$), and has one-sided tangents there that make an acute angle with the $x$ axis (the radial line through $z_0$)
Take $\phi$ a Riemann map from $U$ onto $\mathbb D$ and notice that it extends continuously and injectively to a map between $J=\partial U$ and $C$ (the unit circle), while by a rotation we can assume $\phi(1)=1$
Then $\phi(C)=J_1$ is a Jordan curve that stays within a non-tangential domain at $1$ in the unit disc by the properties of $U, J$
Take a function $g$ that is analytic in the unit disc and has a non-tangential limit at $1$ but st the unrestricted limit at $1$ doesn't exist
(one can use the infinite Blaschke product $g(z)=\frac{1-e^{\frac{-2z}{1-z}}}{e-e^{-\frac{1+z}{1-z}}}$ which has non tangential limit $1/e$ at $1$ but has absolute value $1$ on the unit circle minus $1$ for example).
Then if $f=g \circ \phi^{-1}$ we clearly have that $f$ is analytic on $U$, continuous on $U-1$ but also continuous on the closed unit disc since by construction $\phi^{-1}(\bar {\mathbb D})=J_1 \cup D_1$ where $D_1$ is the inner domain of $J_1$ and is contained in a non-tangential domain of the unit disc at $1$ where $g$ is continuous.
However if $f$ were continuous on $U$ so at $1$, it would follow that $g=f \circ \phi$ is continuous on the closed unit disc which contradicts our construction of $g$
Note that $f$ above is unimodular on $J-1$ ($|f(z)|=1$ there), $f(1)=1/e$ if we approach $1$ "nontangentially" but also has infinitely many zeroes in $U$ that approach tangentially $1$ (the images under $\phi^{-1}$ of the zeroes of $g$ which are $z_n=\frac{in\pi}{in\pi-1}, n \in \mathbb Z$)
