Does a holomorphic function on the unit disk with continuous radial limits have a continuous extension to the closed disk? Question: does there exist a holomorphic function $f$ defined on the unit disk $D$ such that 


*

*$\forall t \in \mathbb{R}, \exists \lim _{r\rightarrow 1^-} f(re^{it})\in\mathbb{C}$;

*the periodic function $\tilde{f} :\mathbb{R} \rightarrow \mathbb{C}, t\mapsto\lim _{r\rightarrow 1^-} f(re^{it})\in\mathbb{C}$ is continuous;

*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. 
 A: 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: 
$$f(1-(1+i) t) = \exp(4/t^4)\to\infty \quad \text{as }t\to 0+$$
