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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.

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    $\begingroup$ Your function is not even a well defined function. It certainly not a complex differentiable function inside the disk. $\endgroup$ Jun 7, 2020 at 5:05
  • $\begingroup$ Why not? Maybe I should clarify: it's defined on the disc centered at z=1 and with a radius of 1, so it can be made single valued, and it is easy to check that it's differentiable at z=0. $\endgroup$
    – EthanK
    Jun 7, 2020 at 14:09
  • $\begingroup$ @EthanK: it is not possible to define as an analytic function $z^{\frac32}$ on a neighbourhood of $0$. To define it, you need a branch of the logarithm, which is defined only on simply connected sets in $\Bbb{C}-\{0\}$. $\endgroup$
    – user515010
    Jun 10, 2020 at 12:40
  • $\begingroup$ A counterexample to the differentiability is still easy: consider $f(z)=\ln(1-z)(1-z)^2$ on $\bar{\mathbb{D}}-\{1\}$ extended on the whole disc with $f(1)=0$. It is easy to see that $f(e^{ix})$ is not twice differentiable, and so $f'$ is not differentiable in your sense. $\endgroup$
    – user515010
    Jun 10, 2020 at 12:44

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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 )

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  • $\begingroup$ How easy is it to see that complex differentiability at a single point implies complex differentiability on some open containing that point...? $\endgroup$ Jun 7, 2020 at 19:44
  • $\begingroup$ @paul you can have complex differentiability at an isolated point only eg $z\bar z$ at zero, but the point here is that complex differentiability requires the function to be defined in a neighborhood and the limit defining it exists when approaching from all directions and that's where the $z^{3/2}$ argument fails as there is no way to define that to be continuos in a neighborhood of zero (and as in the real case differentiability implies continuity at a given point) - $\endgroup$
    – Conrad
    Jun 7, 2020 at 20:47
  • $\begingroup$ thanks. I guess all my intuition is for holomorphic/complex-diff on an open (non-empty)... $\endgroup$ Jun 7, 2020 at 22:14
  • $\begingroup$ @paul I agree - that is the usual meaning (the rest are generally curiosities), while boundary behaviour is a very interesting and complicated topic as I tried to convey above $\endgroup$
    – Conrad
    Jun 7, 2020 at 22:24
  • $\begingroup$ Yes, I edited my question to make what I was asking more clear hopefully. Your example is interesting, and the continuity of the derivatives does imply that it's differentiable in my sense, but I really want to know if there exists a function that's differentiable in my sense but has a discontinuous derivative. $\endgroup$
    – EthanK
    Jun 8, 2020 at 3:45

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