Boundary behaviour of holomorphic function on unit disk

Question

Let $\mathbb{D}=\{z \in \mathbb{C} \ | \ |z|<1 \} $ be the open unit disk in the complex plane. I would like to see explicit examples of the following phenomena:

• a holomorphic function $f$ on $\mathbb{D}$ which extends continuously to the boundary but has no holomorphic extension beyond any boundary point (i.e. on sets of the form $\mathbb{D} \cup B(z_0,r)$ for some $r>0$ and $z_0 \in \partial \mathbb{D}$)
• a holomorphic function $f$ on $\mathbb{D}$ which is bounded on $\mathbb{D}$ but has no holomorphic extension beyond any boundary point (i.e. on sets of the form $\mathbb{D} \cup B(z_0,r)$ for some $r>0$ and $z_0 \in \partial \mathbb{D}$)

Such functions should exist according to this answer to a previous post. Thanks for any reference/advice.

Answer 1

For the first take $$f(z) =\sum_{k=1}^{\infty} \frac{z^n}{n^2} $$

Answer 2

For the first, take a function $f \in C^3(\partial \mathbb{D})$ such that $f$ is nowhere $C^4$ on $\partial \mathbb{D}$. Let $P(r,\theta)$ be the Poisson kernel. Let $u\left(r e^{i \theta}\right)=\frac{1}{2 \pi} \int_{-\pi}^{\pi} P_{r}(\theta-t) f\left(e^{i t}\right) \mathrm{d} t, \quad 0 \leq r<1$. Then $u \in C(\bar{\mathbb{D}})$ and is harmonic in $\mathbb{D}$. Let $v=\int_{0}^{y} u_{x}(x, t) d t -\int_{0}^{x} u_{y}(t, 0) d t$, i.e., $v$ is the harmonic conjugate of $u$. Then $u+iv$ is holomorphic in $\mathbb{D}$. Since $u$ is nowhere $C^4$ on $\partial \mathbb{D}$, $u+iv$ has no holomorphic extension beyond any boundary point, for otherwise, $u$ would be smooth near this point.