Hartogs's extension theorem

Remarks

A short proof uses the $\bar\partial$ operator. Choose a smooth function $\chi$ with compact support in $D$ that equals $1$ near $K$, and let $g = (1 - \chi) f$, extended by $0$ on $K$. The form $\bar\partial g = -f \, \bar\partial \chi$ is smooth with compact support in $\C^n$, and for $n \ge 2$ the equation $\bar\partial u = \bar\partial g$ has a solution $u$ with compact support. Then $F = g - u$ is holomorphic on $D$. The function $u$ vanishes on the unbounded component of the complement of the support of $\bar\partial g$. So $F = f$ on a nonempty open subset of $D \setminus K$, hence on all of $D \setminus K$ by the identity theorem.

The hypothesis $n \ge 2$ is necessary: $1/z$ is holomorphic on $\C \setminus \set{0}$ and does not extend to $\C$. Connectedness of $D \setminus K$ is also necessary. In the unit ball of $\C^n$ let $K$ be the sphere $\abs{z} = 1/2$. The function equal to $0$ inside $K$ and to $1$ outside is holomorphic on the ball minus $K$, but it has no holomorphic extension.

Example: every holomorphic function on the punctured ball $\set{0 < \abs{z} < 1}$ in $\C^2$ extends holomorphically over the origin.

Going deeper

For a student

The theorem is the first sign that domains in $\C^n$, $n \ge 2$, carry hidden constraints for holomorphic functions. It implies that the zeros and the singularities of holomorphic functions are never isolated when $n \ge 2$. It leads to the notions of domain of holomorphy and envelope of holomorphy, which describe how far all holomorphic functions on a domain extend. The proof sketched above shows the role of the compactly supported solution of $\bar\partial u = v$, which exists only when $n \ge 2$.