# Proof of Hartogs's theorem

I'd be very grateful if someone could help me understand the proof of Hartogs's theorem appearing in Huybrechts' "Complex Geometry." The statement is:

Let $\mathbb{P}^n \subset \mathbb{C}^n$ be the unit polydisc. Let $\mathbb{P}_c:= \{ \boldsymbol{z} : 0\leq |z_i|<c\}$ for some $0<c<1$. Then if $f: \mathbb{P}^n -\bar{\mathbb{P}}_c \rightarrow \mathbb{C}$ is holomorphic, then $f$ extends to a holomorphic function on $\mathbb{P}$.

The proof goes as follows: For fixed $\boldsymbol{w} \in \mathbb{P}^{n-1}$, the function $f_w: z \mapsto f(z,\boldsymbol{w})$ defines a holomorphic function on the annular region $A= \{z: c<|z|<1 \}$ in the complex plain. Let $f_w= \Sigma_{k=-\infty}^{\infty} a_k(\boldsymbol{w})z^k$ be the Laurent expansion of $f_w$ in this region. Then the $a_k$ define holomorphic functions on $\mathbb{P}^{n-1}$ (by a previous lemma) and if $k<0$ then $a_k$ vanishes whenever some $w_i>c$ (since $f_w$ then extends to the whole disc) and so vanishes on all of $\mathbb{P}^{n-1}$. Therefore we can write $f|_{A \times \mathbb{P}^{n-1}}= \Sigma_{k=0}^{\infty} a_k(\boldsymbol{w})z^k$.

I understand everything up to this point. What I can't understand is why this sum of holomorphic functions defines a holomorphic function on all of the unit polydisc. Presumably it is supposed to converge uniformly on compact subsets or something? Huybrechts says something like "the $a_k$ attain their suprema on the boundary and so uniform convergence is implied by uniform convergence on the annular region" and I have no idea what boundary or annular region he's talking about. A priori I don't know why I should really know anything about the uniformity of the convergence of the sum outside of a single copy of $A$ i.e. when $\boldsymbol{w}$ is fixed.

Thanks for your time and sorry if this is mostly nonsense.

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This may not be the answer you're looking for but it is how I would approach this theorem. Convergence matters rely on Cauchy's theorem anyway so why don't use that directly in a proof?

Let $c < r < R < 1$ and define $f_r: \{z \mid r < |z| < 1\} \times \mathbb{P}^{n-1} \to \mathbb{C}$ by

$$f_r(z, w) = \frac{1}{2 \pi i}\int_{|v|=r} \frac{f(v, w)}{v - z}dv.$$

Then $f_r$ is holomorphic and vanishes if $|w_k| > c$ for some index $k$ since $v \mapsto f(v, w)$ is then holomorphic within the contour $|v|=r$. This implies that $f_r$ is identically zero. In particular $f$ has the following representation on $\{z \mid r < |z| < R\} \times \mathbb{P}^{n-1}$

$$f(z, w) = \frac{1}{2 \pi i} \int_{|v| = R} \frac{f(v, w)}{v - z}dv.$$

But the right hand side defines a holomorphic function on $\{z \mid |z| < R\} \times \mathbb{P}^{n-1}$.

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Thanks WimC. That seems much easier. –  Heather Huskinson Nov 24 '12 at 19:22