I need to use a version of Hartog's extension theorem, that I did not find by googling around. However I think I found a solution myself, and wanted to ask if this is correct. Any feedback would be appreciated 🙂
Most versions of Hartog's extension theorem are similar to this one from Wikipedia:
Let $f$ be a holomorphic function on a set $G \setminus K$, where $G$ is an open subset of $\mathbb C^n (n ≥ 2)$ and $K$ is a compact subset of $G$. If the complement $G \setminus K$ is connected, then $f$ can be extended to a unique holomorphic function on $G$.
However, I've got a holomorphic function defined on an open set $U \setminus Z$, where $U \subset \mathbb C^n$ is open, and $Z$ is an analytic set of complex codimension $\geq 2$. So in general $Z$ will not be compact.
My idea is to proceed by induction on $n$. If $n = 2$, $Z$ is a discrete set of points, so this case is covered by the version of Hartog's theorem cited above.
For $n > 2$, I would like to choose some hyperplane
$H \subset \mathbb C^n$, such that for some fixed $v \in \mathbb C^n \setminus H$,
\begin{equation} \tag{$*$}
\dim (H + z \cdot v) \cap Z = \dim Z – 1, \quad \forall z \in \mathbb C.\end{equation} This happens if and only if $H + z \cdot v$ does not contain any irreducible component of $Z$. Choosing such an $H$ is always possible, since each irreducible component is contained in a Zariski closed subset of hyperplanes.
So by induction we can extend $f$ to any of the slices $U \cap (H + z \cdot v)$, and this gives a well-defined function
$$ f : U \to \mathbb C. $$
To see that this extension is holomorphic, suppose $p \in H+z \cdot v$, and $z_1, \dotsc, z_{n-1}$ are coordinates on $H + z \cdot v$. Then by Cauchy's formula on $H + z \cdot v$ one has
$$f(p) = \int_{|w_1| = r} \frac{f(w_1, z_2(p), \dotsc, z_{n-1}(p), z(p))}{w_1 – z_1} dw_1$$
And since $\frac{\partial f}{\partial \overline z} = 0$ outside of $Z$, the same is true for the integral on the right hand side. So $f$ is holomorphic everywhere.
Best Answer
I am not sure about all the details, but since you assume to deal with complex analytic sets whose codimension is at least 2, your result is similar to Hassler Whitney's version ([1] §2.5, pp. 50-51, theorem 5B) of Hartogs' theorem on separate analyticity:
Theorem. Let $H\subset \Bbb C^n$ be a open set and let $V$ be an analytic set of codimension $\ge 2$ in $H$. Then any holomorphic function $f$ in $H\setminus V$ has a unique holomorphic extension to the whole $H$.
His proof does not uses directly Cauchy's formula, but he uses other results he proved earlier by using it. This perhaps is not a direct answer to your question, but it may help yourself checking it.
References
[1] Hassler Whitney, Complex analytic varieties, (English) Addison-Wesley Series in Mathematics. Reading, Massachussets-Menlo Park, California-London-Don Mills, Ontario: Addison-Wesley Publishing Company. XII, 399 p. (1972), MR0387634, Zbl 0265.32008.