Continuity of Derivative of Analytic Function on Disc Boundary Suppose we have an analytic function $f(z)$ defined on a closed disc $D$ in the complex plane, so that it is complex-differentiable inside and on the boundary of $D$ (see edit below). I know that this implies that $f$ is infinitely differentiable on the interior of $D$, which can be proven using Cauchy's integral.
However, this approach fails when $z$ is on the boundary of $D$, so my question is does $f'$ necessarily have to be continuous or even
bounded on the boundary of $D$? Are there counterexamples?
I've already found a simple counterexample showing that $f'$ doesn't need to be differentiable:
\begin{equation}
f(z) = z^\frac{3}{2}, \; z \in \bar B_1(1)
\end{equation}
However, its derivative is still continuous.
EDIT: Sorry for the confusion, when I said that it it complex differentiable on the boundary of the disc, I meant that we use the usual definition of differentiability, but when we take the limit of the difference quotient we only consider open neighborhoods in the relative topology of the disc (so it might not be able to analytically extend to an open neighborhood in $\mathbb{C}$). This question is just about whether or not that derivative is necessarily continuous.
 A: Your question is phrased poorly as complex differentiability, means differentiability when $z$ approaches from all directions towards a point and in your example, $z^{3/2}$ is not continuously defined at $0$ on a full disc around zero - sure, one can define it continuously on a slit disc (so in particular from the inside of your disc $B_1(1)$) and then $f'(z) \to 0, z \to 0, z \in B_1(1)$ but that doesn't make $f$ differentiable (in the complex sense) at $0$ only $f$ differentiable in the real sense on $\partial B_1(1)$ which is a real manifold of dimension $1$ (a circle, or locally an arc)
There are easy examples of functions for which $f^{(k)}, k \le N$ converges from inside the disc at any point on the boundary but the boundary value $f$ (which is then $N$ real differentiable) doesn't extend analytically at any point (just take $\sum \frac{z^{2^n}}{n^2}$ on the unit disc and integrate it $N$ times, so the result $f_N$ has the unit disc as a natural boundary but it's derivatives up to order $N$ converge to the boundary and they are continuos) and with more care one can arrange so $f$ is infinitely differentiable ($C^{\infty}$) on the boundary, or even stronger all the derivatives converge from inside the circle at any boundary point, but the circle is still a natural boundary.
Summarizing - the convergence of the derivative to a boundary point from inside the circle doesn't tell you anything about the existence of the derivative there in the complex sense (which again generally means that $f$ extends analytically on a small ball etc )
