Hartogs figure

Example

Let $n \ge 2$ and write a point of $\C^n$ as $z = (z', z_n)$ with $z' \in \C^{n-1}$, and $\abs{z'} = \max_{j < n} \abs{z_j}$. Let $\Delta^n = \set{z : \abs{z'} < 1, \abs{z_n} < 1}$ be the unit polydisc, and fix $0 < r < 1$ and $0 < s < 1$. The Hartogs figure is the connected open set $$H = \set{z \in \Delta^n : \abs{z'} < r \text{ or } \abs{z_n} > s}.$$ It is the union of a thin column $\set{\abs{z'} < r}$ and a shell $\set{s < \abs{z_n} < 1}$ inside $\Delta^n$.

Every holomorphic function $f$ on $H$ extends to a holomorphic function on all of $\Delta^n$. Fix $t$ with $s < t < 1$ and put $$F(z', z_n) = \frac{1}{2\pi i} \int_{\abs{\zeta} = t} \frac{f(z', \zeta)}{\zeta - z_n} \, d\zeta \quad \text{for } \abs{z'} < 1, \ \abs{z_n} < t.$$ The circle $\abs{\zeta} = t$ lies in the shell, so the integrand is defined, and $F$ is holomorphic by differentiation under the integral sign. For $\abs{z'} < r$, the one-variable function $\zeta \mapsto f(z', \zeta)$ is holomorphic on the whole unit disc, so the Cauchy integral formula gives $F = f$ there. The sets where $F$ and $f$ are both defined form a connected open set, so the identity theorem gives $F = f$ on all of it. Gluing $F$ and $f$ gives the extension to $\Delta^n$.