Let $f:\mathbb C\to\mathbb C$ be a complex function. As usual, we say that $f$ is differentiable at $z_0$ if the following limit exists $$ \lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0} $$ Can $f$ be differentiable only in the closed unit disc? I cannot find any simple example.


1 Answer 1


Let $$f(z)=(|z|-1)^2\chi_{\mathbb{C}\backslash\overline{\mathbb{D}}}(z)$$ where $\chi_E(z)$ is one if $z\in E$ and otherwise is $0$

It is clear that $f$ is complex differentiable on the open unit disc and it is not outside the closed unit disc (since $(|z|-1)^2$ is not complex differentiable). It remains to prove that it is differentiable on the boundary. It suffices to prove that

$$\lim_{z\to z_0\in\partial\mathbb{D}\\z\in \mathbb{C}\backslash\overline{\mathbb{D}}}\frac1{z-z_0}(|z|-1)^2=0\\ \lim \left|\frac1{z-z_0}(|z|-1)^2\right|\le \lim \frac{1}{|z|-1}(|z|-1)^2=0$$

  • 1
    $\begingroup$ Thanks @Caffeine. It is a nice example. $\endgroup$
    – boaz
    Nov 22, 2020 at 10:23

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