Does a holomorphic function on the unit disk with continuous radial limits have a continuous extension to the closed disk?

Question

Question: does there exist a holomorphic function $f$ defined on the unit disk $D$ such that

1. $\forall t \in \mathbb{R}, \exists \lim _{r\rightarrow 1^-} f(re^{it})\in\mathbb{C}$;
2. the periodic function $\tilde{f} :\mathbb{R} \rightarrow \mathbb{C}, t\mapsto\lim _{r\rightarrow 1^-} f(re^{it})\in\mathbb{C}$ is continuous;
3. the function $\bar f : \bar{D}\rightarrow\mathbb{C}, z\mapsto \begin{cases} f(z)\ \textrm{if} \ \ z\in D\\ \tilde{f}(t)\ \textrm{if} \ \ z=e^{it} \ \textrm{for some} \ t\in\mathbb{R} \end{cases}$ is discontinuous

?

I found an example that satisfies 1 and 3 but when it comes to 2, $\tilde{f}$ is just everywhere continuous except for a (periodic) point, i.e. the function $$f : D \rightarrow \mathbb{C}, z\mapsto \exp\left(-\frac{1}{1-z}\right) $$

and I start guessing that this is the best that could be done.

Edit: notice that if we were dealing with harmonic functions instead of holomorphic functions, actually such an $f$ exists, i.e. the derivative with respect to the angle of the Poisson's kernel: $$f :D\rightarrow \mathbb{R}, re^{i\vartheta}\mapsto\frac{\partial}{\partial\vartheta} \frac{1-r^2}{|1-re^{i\vartheta}|^2}.$$ However, the conjugate of $f$, say $g$, has a boundary that is discontinuous, so $f+ig$ fails to give us the example we were looking for.

Answer 1

The function $f$ you have can be modified into the following counterexample: $$f(z) = \exp\left(-\frac{1}{(z-1)^4}\right)$$ For any $\epsilon>0$ the restriction of this function to the set $$ \Omega = \left\{z : \arg(z-1) \in \bigcup_{k=0}^3\left(\frac{\pi k}{2} -\frac{\pi}{8}+\epsilon, \frac{\pi k}{2} +\frac{\pi}{8}-\epsilon\right)\right\} $$ tends to $0$ as $z\to 1$ while $z\in\Omega$. Indeed, for $z\in\Omega$ we have $$ \arg((z-1)^4) \in \left(-\frac{\pi}{2}+4\epsilon, \frac{\pi}{2}-4\epsilon\right) $$ hence the same holds for $\arg(1/(z-1)^4)$, which implies $\operatorname{Re}(-1/(z-1)^4)\to -\infty$ as $z\to 1$.

The radial limits of $f$ exist everywhere: at $1$ it is $0$ by the above, and at other points $e^{it}$ it is simply $f(e^{it})$. These boundary values are continuous, because as $t\to 0 $, $e^{it}\to 1$ and $e^{it}\in\Omega$ for sufficiently small $t$ (indeed, $\arg(e^{it}-1)\to \pm \pi/2$).

Yet $f$ is not continuous on the closed disk as it's not even bounded: $$f(1-(1+i) t) = \exp(4/t^4)\to\infty \quad \text{as }t\to 0+$$